groups-rings-fields

[pastel cliff]

oh but it's not necessarily a PID i think

[cloud walrus]

Volkenborn

[thorn delta]

ker phi does not need to be 0

[pastel cliff]

<= might work by induction

[thorn delta]

for example the 0 map squares to itself but has nonzero kernel

any projection operator from linear algebra generally has nonzero kernel as well and squares to itself

[ionic spade]

OK, but if that doesn't work how do I use the idempotent condition of phi in the problem statement? Sort of stuck on that.

Like that should have something to do with the kernel but I am having trouble putting my finger on how I need to bridge those two together.

[thorn delta]

do you know about exact sequences?

[ionic spade]

Kind of? How would I use them?

[thorn delta]

you can construct a split exact sequence to do this

you can also just do it a little more manually if u prefer tho

[ionic spade]

I'd prefer this way, I don't think my textbook has presented exact sequences at all.

The problem is from Advanced Linear Algebra by Roman

[thorn delta]

basically, M = ker(phi) \oplus im(phi) if each element of M is uniquely a sum m + n for m in ker(phi) and n in im(phi)

so you just have to play around with the relation phi^2 = phi to figure it out

[ionic spade]

Ok that helps a lot, I’ll use that and see how it goes.

Thanks @thorn delta

[thorn delta]

npnp

[thorn delta]

as another hint, it may help to think about the geometric interpretation of projections in linear algebra

[pastel cliff]

what is this homomorphism?

[south patrol]

what would be the natural hom from Z to Q for example

[pastel cliff]

inclusion?

nah that's silly lol

im not sure actually, forgive me if it's a common example, but im assuming it'd be fraction related

[south patrol]

Okay maybe that was a bit of a confusing thing ll

But what what is the natural like "Inclusion" R -> S^-1 R

obviously needn't be an inclusion if we aren't working with domains etc as said in the statement

[chilly ocean]

Why is the Krull dimension of a PID that's not a field equal to 1? If it's not a field and there's a strict chain of prime ideals starting in P, then since P is prime it's also maximal and can only be strictly contained in the whole ring so shouldn't it have dimension 0?

[agile burrow]

there is one prime ideal which isn't maximal

[chilly ocean]

oh duh, (0)

[chilly ocean]

Is there any reason to write 'K[X]' with a capital 'x' instead of 'K[x]' ?

[delicate bloom]

maybe to remind new students that polynomials are formal objects and not functions, otherwise no

[south patrol]

sometimes it looks cooler

[pastel cliff]

semi related to previous question but i also dont have a strong understanding of what a noetherian/artinian ring are

is a noetherian ring kinda like a weaker PID?

[thorn delta]

sorta but its quite a bit weaker because noetherian rings don't even need to be integral domains

nor commutative

[pastel cliff]

hmmm i guess there's no point to thining of it like that then

[thorn delta]

there is probably better intuition but i just think of it as a nice ring where nice things happen

[pastel cliff]

it's not being very nice to me i dont think

maybe this is a silly problem to have but ACC feels kinda like a contrived condition ig

but then again i felt like that about group axioms when i first started learning this so

the duality of man

[agile burrow]

I remember asking some people why Noetherian rings are useful to study when I first learned about it so I'll try to recall some of their answers. They may not make sense immediately (they certainly didn't for me!) but hearing that it isn't an arbitrary condition to consider is usually motivation enough for me:

• The category of finitely generated modules over a Noetherian ring is abelian, which can be nice to consider from a module theoretic/representation theoretic perspective.

• Noetherian rings have just enough structure to allow for a well-behaved ideal theory. For example, every ideal contains a product of prime ideals. As a consequence, if you impose some additional structure on your ring then you can ensure every ideal factors (uniquely) into a product of prime ideals. These are called Dedekind domains and are useful in algebraic number theory.

• A lot of constructions preserve when a ring is Noetherian, such as taking polynomial rings, quotients, completions, and localizations. Many of the rings we study just end up being Noetherian, so we might as well use that property to gain a better understanding.

[pastel cliff]

idk what half of that stuff is but it's stuff i've like vaguely heard about and is at least nice to know thank you walter

i'll just trust it hesitantly for now

like i know it's not irrelevant i think i just need time with it to really trust it

[agile burrow]

that's fair

[pastel cliff]

noether and i just aren't there yet

[agile burrow]

But yeah, I mean Proposition 2.1 becomes straightforward in Noetherian rings. I just view it as a bit of structure we can tack onto our ring without losing much generality

[pastel cliff]

these are my prof's lecture notes i just thought it was funny that he jumps from saying its proof is straightforward to that

where does hilbert's basis thm fit into this? is it just a way of creating new noetherian rings to extend those "nice" things

[agile burrow]

Yeah, Hilbert's basis theorem shows that taking polynomial ring still preserves being Noetherian. This is one of the most common ways of constructing new rings so it's very good and useful that we get to keep this property

It's also useful in algebraic geometry. The fact that polynomial rings over, say, fields are Noetherian implies that if you have a set of points which are the solutions to an arbitrary collection of polynomials, then in fact they are the solutions to a finite set of polynomials

[pastel cliff]

ok semi silly question, how is having a maximal ideal different from ACC

[agile burrow]

just because you have a maximal ideal doesn't mean that all the other ideals are contained in that ideal. It's also not the case that there can't be an infinite chain of ideals between a given ideal and the maximal ideal containing it

Oh, just because you mentioned it earlier, I'll add a brief remark about artinian rings

The descending chain condition ends up being quite strong. In fact, it actually implies the ascending chain condition (which is non obvious) so in particular all Artinian rings are also Noetherian. Furthermore, an Artinian integral domain must be a field.

[tender bough]

what is a basis that respects a direct sum decomposition?

[eager willow]

Just indexed such that the block matrix makes sense

[sage lodge]

is any element of a coset a representative element in the quotient group?

i'm pretty sure this follows from the fact that cosets are disjoint

[pastel cliff]

moreso from the equivalence relation

[pastel cliff]

https://kconrad.math.uconn.edu/blurbs/ringtheory/noetherian-ring.pdf reading this and im wondering, how are noetherian rings usually taught or presented?

my prof taught us ascending chain condition and then proved finitely generated ideals as a result but the latter seems both more to-the-point and intuitive

i guess this agrees with tht but finite generation seems much easier/palpable

[rustic crown]

prove the equivalence of the 3 statements and then define noetherian by any of them :3

or equivalently define it using one of them and then prove the equivalence. it's usually not a problem because most people will do this proof right after the definition as all these 3 statements are used very often

[tender bough]

this seems to imply that ad_y is closed on the orthogonal complement? Why?

[sharp peak]

Can you have rings with uncountably many ideals?

[coral shale]

{f : Z -> Z}, pointwise addition and multiplication

this is a ring? think it has uncountable ideals if so

[wooden ember]

Another equivalent condition to being noetherian if you learnt the chain condition is that every ideal is finitely generated. I personally find that characterization of noetherian rings to sound a lot more useful than the other one.

[rotund aurora]

Why not? Let X={x_i : i\in I} where I is uncountable, and consider the polynomial ring in uncountably many variables R[X]. Each (x_i) is a different ideal

[rotund aurora]

I agree. I think the most natural context to introduce noetherian rings is in linear algebra over rings. Modules are just vector spaces except that the set of scalars is a ring. So in general you cannot divide by scalars. Vector spaces are pretty nice, but modules are tricky. Many of the usual properties of vector spaces do not carry over to modules, but we would like to determine when properties of vector spaces kind of carry over to modules with some restrictions. Finitely generated modules are kind of nice, but submodules of finitely generated modules need not be finitely generated, and this sucks (non-noetherian rings are an example). Let M be a finitely generated R-module. Every ideal of R is (isomorphic to) a submodule of M, therefore if we expect that every submodule of M be fintiely generated, then R should be noetherian. Surprisingly, this is also a sufficient condition.

[sharp peak]

Is Spec(k[x,y]) = Spec((k[x])[y])?

[coral shale]

k[x][y] = k[x,y] for k a ring?

[sharp peak]

For k an algebraically closed field

[coral shale]

which is a ring

[sharp peak]

yeah

[long nebula]

Without choice, does [\prod_{i\in \bN}\bR] have any maximal ideals besides (I_k = {(x_i)_{i \in \bN} \mid x_k = 0})?

[cloud walrus]

孙山

[sharp peak]

@coral shale wait do the prime ideals correspond (bijectively) spec(k[x,y]) <----> spec(k[x]) x spec(k[y])

[coral shale]

i wouldn't know

i stated what i knew

[wooden ember]

Not in a natural way: the prime ideals of k[x] and k[y] are maximal so speck[x] and speck[y] are just k by weak nullstellensatz. However (x-a,y-b) are not the only prime ideals of spec(k[x,y])

True for mSpec though (would just be a rephrasing of weak nullstellensatz)

[sharp peak]

cool

But if we take a disjoint union say Spec(A) \coprod Spec(B) that would be nautrally in bijection with Spec(A times B)

wait is \coprod_{i=1}^{\infty}\Spec(Z/2Z) in bijection with Spec(\prod_{i=1}^{\infty}Z/2Z)?

[chilly ocean]

If the minimum size of maximal algebraic independent sets in K is 0, why can't there be a larger maximal algebraic independent set?

[gritty sparrow]

Let m be a maximal ideal. For $a \in m$ define $S_{a} = { i \in \mathbb{N} \mid a_i = 0}$. Define $F_{m} = { S_{a} \mid a \in m}$. It can be shown that this is an ultrafilter on $\mathbb{N}$. If m is not of the form that you gave, it will infact be a non principal ultrafilter, but there are models of ZF where there are no non principal ultrafilters on $\mathbb{N}$ so we see that atleast some weak form of choice is needed.

[cloud walrus]

chmonkeynumber1fan

[long nebula]

ahhh okay

ty!

love the nickname

[gritty sparrow]

Thx

[vernal pine]

Is $\langle a, b \mid ba = a^{-1}b\rangle$ virtually abelian?

[cloud walrus]

rob

[vernal pine]

It's isomorphic to $\mathbb{Z} \rtimes \mathbb{Z}$ with operation $(m_1, n_1) \odot (m_2, n_2) = (m_1 + (-1)^{n_1}m_2, n_1 + n_2)$

[cloud walrus]

rob

[delicate bloom]

since it looks like n_1 being even would make it abelian, then Z rtimes 2Z would be a subgroup of finite index I guess

[charred bison]

how do you show that for any g, x in a group G, (x g x^-1)^n = x g^n x^-1

[sharp peak]

?

[agile burrow]

Try explicitly writing it out for say n=2 and n=3

[delicate bloom]

induction, work out what n=2 and n=3 by hand if you don't see it

whoops sniped

[charred bison]

gotcha

[formal ermine]

is there any intuition to how a generated group will look like?

I have an intuition for $\gen{x}$

[cloud walrus]

illuminator3 (I/you)

[formal ermine]

but I'm missing one for all of the other cases

e.g. $[G, G] \coloneqq \gen{aba\inv b\inv : a,b \in G}$

[cloud walrus]

illuminator3 (I/you)

[south patrol]

All "words" in the elements in the generating set and their inverses

[formal ermine]

words?

[south patrol]

Essentially all finite products of elements in the generating set and their inverses

[coral spindle]

There is no good intuition for what this kind of group "looks like" in general that I know of, and I would argue at all.

[formal ermine]

ah

thanks

[wooden ember]

depends what you mean by "look like". You could always look at the associated cayley complex to get some sort of a geometric intuition but idk if it's very useful most of the time

[coral spindle]

The problem is that the cayley graph can be arbitrarily complicated, so I don't think there's really any hope of being able to visualise that in general

[foggy merlin]

I have to find all invertible elements of the ring
Z[sqrt(3)]. I know they correspond to integer solutions of
a^2 - 3b^2 = 1 (I proved this already)
but now I don't know how to explicitly find all of them, I know sqrt(3) + 2 generates an infinite set of solutions, but I don't know if there are solutions that have another form. Does one proceed here with ring theory argument or would you need more number theory for this one?

[next obsidian]

Can’t you like

Write out general forms of elements

They look like a + bsqrt(3)

Then multiply two of these

Or maybe that’s how you got to that equation

[south patrol]

i believe that leads to what stockfish gets

[next obsidian]

I mean

[south patrol]

well

[next obsidian]

Why not then say like it’s things of the form a + bsqrt(3) where a = +-sqrt(1 + 3b^2)

So you need 1 + 3b^2 to be a square

Or I guess not

Wait duh a is an integer

So yeah

[elder wave]

is that not like some sort of norm trick

ah nvm

[next obsidian]

Okay so you want to solve 1 + 3b^2 = c^2 so you need b^2 = (c^2 - 1)/3

[elder wave]

misread

[next obsidian]

Idk

Maybe classify maximal ideals

And say it’s the complement of them

[rustic crown]

if u is a unit, then look at v = +- u * (2 + sqrt(3))^n some integer n such that 1 <= v < (2+sqrt(3)) as real numbers

show this implies v = 1

one way to do this is by noticing that 1 < (v + 1/v) < 3+sqrt(3), so if v = a + sqrt(3)b, then 2 <= 2a < 4.7ish and so a = 1,2

which means b = 0,1

so v = 1 or 2 + sqrt(3)

the second one is obviously bad

as v < 2+sqrt(3)

so every unit is a power of (2+sqrt(3)) (upto +-1)

.<

i had to edit all of the messages >.<

why me make so many mistakes >.<

[foggy merlin]

scheme theorist back in action I see

[foggy merlin]

I don't fully get the line of reasoning here, what do your initial assumptions of the form of v add to the equation. well I see the idea, I'll try to work it out thanks

[latent anvil]

How is this scheme theory?

Isn't this just saying every non unit is contained in a maximal ideal?

[foggy merlin]

it is not, it is just trying to use unnecessary abstractness for something that apparently didn't need an abstract solution

[latent anvil]

ah

I guess so

[rustic crown]

oh so the broad idea is similar to how you show in group theory that if g^n = 1 then order of g divides n

you look at the remainder and argue that it's weird

so here if you think of 2+sqrt(3) as "the fundamental" unit

then given any other unit, we try to get a smoller unit

something between 1 and 2+sqrt(3)

and argue that this is bad

[next obsidian]

Yeah my maximal ideal thing was a pretty crap idea, I only said it because maximal ideals of Z[x] are known so it’s realistic to classify the maximal ideals of this ring

[rustic crown]

if u was negative then we replace u with -u and it's still a unit

now since 2+sqrt(3) > 1

you can multiply by a suitable power to get v in that range

ig i wrote it in a very messy way >.<

[rustic crown]

but this is the idea i'm following

[foggy merlin]

this was on sheet one of my algebra course, I uh... but that's cool that we know this I guess, I would like to know too

[latent anvil]

so Z[sqrt(3)] = Z[x]/(x^2 - 3) right?

Prime ideals of this ring then correspond to primes J of Z[x] containing x^2 - 3

But we can just write J down

Also for reasons that may or may not be obvious to you, J has to be maximal

So we want to look at J = (p, f(x)) where p is a prime number and f is irreducible over Fp

hmm this does seem tricky though

[foggy merlin]

thanks, I see the broader picture, I guess fiddling around with the inequalities you wrote will work out eventually

[rustic crown]

is there an elementary way to see this, or did you have dim Z[x] = 2 in head?

[foggy merlin]

reality check: wait how do you prove again that Z[x] is a ufd, this would follow from that fact right?

[latent anvil]

That's what I had in mind but I'm pretty sure there's easier ways to see it

[rustic crown]

you compare it with Q[x] and use gauss

[latent anvil]

Does finite ring extension preserving dimension count as elementary?

bc Z < Z[sqrt(3)] is finite

[rustic crown]

i read about dimension theory a while ago, and i forgot a lot of it already

[latent anvil]

It's hard stuff!!

This is a pretty easy result though

At least imo

[rustic crown]

right

[latent anvil]

Idk though

I'm not sure how much more elementary you can get

Like if you know stuff about stuff

You could say oh it's a dedekind domain

[next obsidian]

So what you want is for f to be a lift of an irreducible factor of x^2-3 mod p

[rustic crown]

yea lol

[next obsidian]

So for p = 2 it’s obvious and for p = 3

[latent anvil]

Agreed okay

[next obsidian]

But that’s why I asked when 3 is a square mod p

[latent anvil]

Right

[next obsidian]

So this method is just a NT thing

[latent anvil]

yeah I mean I think it's going to be quadratic reciprocity man idk

[next obsidian]

Probably but me no kno NT

☠️

[rustic crown]

this happens when p = 1, 11 mod 12 i think

[foggy merlin]

idk, I found it kind of a cool application, since via basically just the definition of a ring you can prove this equation has infinitely many integer solutions

[chilly ocean]

If R is localized first at S then at the image of T, then this is the same as localizing at TS. So is every element of (TS)^(-1)R of the form r/st for s in S and t in T?

[latent anvil]

yes, ST is the set of products s*t for s in S and t in T

[next obsidian]

the reason this double localization thing is true is because (r/s)/t = r/st lol

I think I just proved it via univ property and didn’t look back

[latent anvil]

I mean I guess lol

Oh

Chmonkey you just made me realize something

I think a certain functor is strong monoidal

[next obsidian]

Omg did u come to grips with ur sexuality?

[latent anvil]

No, I am still bicurious

(curious about both men and twinks)

[next obsidian]

I meant sapiosexual since you like to jerk off with category theory so much heyooooo got ‘em lollll ownedddd

[latent anvil]

What happened to not be self hating about category brain

We were working on this

[next obsidian]

Internalized cat phobia

[rustic crown]

[foggy merlin]

will fear of category theory ever go away

[rustic crown]

yee

cats are cute

[coral shale]

what in the abstract algebra is going on here

[latent anvil]

Yeah!

It's a very fun field

I have been doing a lot of category theory recently

With triangulated categories and stuff

[foggy merlin]

what about it is fun (genuine question)

[latent anvil]

I mean it would depend on your taste in math, I guess? Why is analysis or algebra fun?

I like how natural everything feels. You define things in the right way and a beautiful theory pops out

I like how it emphasizes the perspective that mathematical things are defined by how they interact with one another

I like how it provides the right tools for generalizing old ideas to new contexts

I can be more specific if you want me to give examples or w/e

[foggy merlin]

I personally find math the most motivating when it allows me to answer a classical question, or lets just say. The best way to motivate a new layer of abstraction to me is to give a problem in the current layer i.e. a question I understand with my current knowledge, whose solution requires something new in order to solve it.

The problem I had when I guess just seeing small pieces of cat theory thus far is that it just takes bits of it that either seem very general for no reason or that are specifically tailored for that field you're using it in. I just haven't felt like I got a new tool that gives me more insight. I know that's not true and that it's supposed to be useful/powerful, I just never came across that in my courses

[latent anvil]

Sure, I don't think that category is universally applicable to math

I think if you think it's general for no reason then you probably haven't seen the math that requires its use

I do homological algebra and this is the field that literally led to the invention of category theory

[foggy merlin]

the pieces of algebraic topology I saw do seem like they could use category theory, as the diagrams just get so huge at some point

[latent anvil]

There are things in algebraic topology and commutative algebra that fundamentally cannot be stated without talking about the "Ext" and "tor" functors

Eg the universal coefficient theorem for cohomology

These are objects which are (1) defined by an abstract categorical process, called deriving a functor, (2) can still actually be calculated in explicit cases, and (3) show up very naturally in algebra as kinds of obstructions

And I want to be clear like these aren't specialist tools within these fields, they're usually taught in a first year graduate algebra course and are on most qualifying exams

Even like the tensor product of two modules is hard to think about without thinking about it categorically (ie by its universal property instead of its elements), and that's one of the most fundamental things in algebra

[chilly ocean]

True

[foggy merlin]

Yes the tensor product was always easier to think about lile this, but what problems are these objects used for. What do they help us prove/understand?

Not the tensor product. But what you talked about

[chilly ocean]

You can use homological algebra to study Banach spaces

[rotund aurora]

I have no idea what I'm talking about, but Wiles's proof did use category theory no? I say this because Stockfish commented about classical problems

Maybe I just made this up completelly xdd

[foggy merlin]

I was hoping for a slightly more specific answer than that. But sure I guess

[foggy merlin]

Maybe. Not like I can read it or sth💀 but I guess that would be cool if the proof did have to partly rely on purely cat theoretical objects

[latent anvil]

Wiles proof extensively used category theory

[rotund aurora]

[latent anvil]

All modern algebraic geometry is built on a very categorical foundation

The stuff wiles was doing especially so

[latent anvil]

Ext and tor? As I said, look at the universal coefficient theorem

If you continue doing algebraic topology, algebraic geometry, or commutative algebra you will run into Ext and tor all of the time

[latent anvil]

The proof used cohomology for grothendieck topoi!

See the comments here https://mathoverflow.net/q/35746/136523

[rotund aurora]

I was just reading this

[latent anvil]

Explicitly, here's how Ext and tor show up

You have a short exact sequence 0 -> L -> M -> N -> 0

So roughly this means M is a module containing L as a submodule and with M/L = N

You then tensor with some new module P

The sequence 0 -> L (×) P -> M (×) P -> N (×) P -> 0 will not generally be exact

It's exact on the right but the first map won't always be injective

[foggy merlin]

That does sound pretty cool. But at the same time that also sounds like it would take years of studying for this thing to even become remotely readable for me

[latent anvil]

But we can exactly measure how much it falls to be injective using the modules Tor_i(L, P), Tor_i(M, P), and Tor_i(N, P)

[latent anvil]

Yeah, this is a very complicated example

I think it's fine to say category theory doesn't seem necessary to you right now

[foggy merlin]

This I remember from commalg

[latent anvil]

Right

So how about this

Let A be a ring

I, J ideals of A

In general we don't have IJ = I cap J, yeah?

It would be useful to have a nice criterion somehow, or be able to measure it in a sort of computable way

[foggy merlin]

And with the mathematical path I'm taking rn unfortunately not in the future either.

[latent anvil]

Well it turns out (I cap J)/IJ is isomorphic to Tor_1(A/I, A/J)

What path are you taking?

Chmonkey took his first AG class just to say he didn't like it

[foggy merlin]

More operations research related math atm. I.e. combinatorics, probability and just linalg mixed with probability stuff.

[latent anvil]

Oh fun

Then yes I'd agree

The stuff you were posting above seemed like algebraic number theory and I was skeptical (that you would not need category theory)

[foggy merlin]

This does seem like a cool fact

[rotund aurora]

In linear algebra, there is the natural isomorphism between a vector space and its double dual

[latent anvil]

That's not true

Nor is it interesting on its own

[foggy merlin]

I know there are people that try to formulate problems from that field in category theoretic ways (there always are these kinds of people) but it hasnt been popular so far

[latent anvil]

There's a natural transformation V -> V^** but it won't be an isomorphism unless V is infinite dimensional

I don't have a very positive view of applied category theory

[rotund aurora]

oh okay, my bad.

[latent anvil]

It seems like people start with the question of "how can we apply category theory to this?" and then just define a bunch of things

Instead of doing anything novel

[foggy merlin]

Yeah. Thats what I felt like too. Its never been about finding patterns or phenomena in the subject to develop a new technique/formalism, which is what research should be about

[latent anvil]

Right. I think it's wrong to assume that's what all uses category theory look like though

It was invented because of naturally occurring problems, but those problems were in algebra, algebraic to topology, and algebraic geometry. You do not need it for everything

[foggy merlin]

Well at least in the field I mentioned

[foggy merlin]

Yeah it does have more natural applications in some parts if language processing and various other fields. Its just that in certain fields they simply try to force it because they can

I guess the field that looks most appealing if I happen to change seems to be topological data analysis. It's probably of great use there

[foggy merlin]

I got ghost pinged

[elder wave]

👻

[celest cairn]

Let $\omega$ be a primitive 10th root of unity, where $\omega \neq 1.$ Find the minimal polynomial of $\omega^{24}+\omega^{34}$.
Is the minimal polynomial $x^{4}-2x^{3}+4x^{2}+8x+16$?

[cloud walrus]

Sapphire Gaming

[south patrol]

That is an interesting way to write 2w^4

[chilly ocean]

It should +2 but other than that you should be correct

[celest cairn]

Yea I meant +2

Sweet, thanks 🙂

[sharp peak]

What are the prime ideals in \prod_{i=1}^{\infty}Z/2Z

Is it just 1, the product of the zero ideals

[chilly ocean]

This is a lot more non-trivial than that.

Prime ideals in this ring are in a bijective order-preserving correspondence with ultrafilters on the index set.

[sharp peak]

I am trying to show that \coprod{i=1}^{\infty}\Spec(Z/2Z) in bijection with Spec(\prod{i=1}^{\infty}Z/2Z)

[next obsidian]

It’s not true

[latent anvil]

This is not as scary as you make it seem! That ring is Hom_Set({index set}, Z/2Z), ie the powerset ring of the index set

If we remember the duality between ideals and filters of a boolean algebra the bijection falls out

Also both orders are discrete, ie everything is maximal, so while it is order preserving that's not very interesting

[sharp peak]

How would you show that it is not true without mentioning ultrafilters

[next obsidian]

You probably don’t

[latent anvil]

Any proof you give will be about ultrafilters

Fundamentally

[next obsidian]

You could maybe compute the global functions or something of the infinite disjoint union if you’re crazy and show it isn’t the product of Z/2Z? Idk if that’s true tho lol

[sharp peak]

The reason why I say that is cause the question comes from earlier questions in the problem sets for my AG class

So I presumed it wasn't too bad

[latent anvil]

It's not true I think

Yeah it's not

[next obsidian]

¯_(ツ)_/¯

Any question involving Spec Prod Z/2Z is about ultrafilters

Is my take

[latent anvil]

Global sections are maps to A^1_Z yeah

And it's a coproduct

[latent anvil]

This is fundamentally correct

Doesn't mean you can't solve the problem adam

[sharp peak]

We have a countably infinite number of prime ideals on the left hand side. And the right hand side is??

[latent anvil]

The left hand side doesn't have prime ideals, it has points

But yes, this will work!

Try to figure out a way to show there are uncountably many primes of the countable direct product

(in fact there are always 2^(2^κ) ultrafilters on a set X of infinite cardinality κ)

[chilly ocean]

Neat.

It only feels scary because I never think about anything like these things.

[latent anvil]

Yeah, I get that. I think people overestimate how scary anything logicy is

You can get cool stuff from logic, eg you can prove the algebraic closure of a field exists very cleanly using ultraproducts

(or basically equivalently by the compactness theorem)

[sharp peak]

Since the right hand side is quasi compact and the lhs is not quasi compact then we get that there cannot exist a bijection?

[tender bough]

how is tr_C different from tr_{C^n}?

context: I think my professor is abbreviating this:

[tender bough]

nvm I think it's a typo

[chilly ocean]

Is every element not contained in some maximal ideal?

[agile burrow]

every non-unit is contained in a maximal ideal

in a commutative ring at least

[chilly ocean]

I want to know if I can always find a maximal ideal that does not contain any given nonzero nonunit element.

[delicate orchid]

ok that makes a bit more sense

[foggy merlin]

that shouldn't be possible in a local ring

[chilly ocean]

If the ring is not a local ring, then might this be true?

[gritty sparrow]

It is true in no ring since any ideal containing 1 is non maximal

[foggy merlin]

I have my doubts, one can probably construct examples from this idea to generalize it

[foggy merlin]

I think username0000 meant it should also not be a unit

[gritty sparrow]

Well then in any field the statement is vacuously true lol

[foggy merlin]

it's also true in Z, but not true in local rings for example

[gritty sparrow]

Yeah so if you have a non zero non unit x (x) is not the whole ring, so it can be enlarged to a maximal ideal

In a local ring anything outside the maximal ideal is a unit so your ctr example doesn’t work

[foggy merlin]

you can take the product of two local rings, then it shouldn't work either

[cloud walrus]

Volkenborn

[ionic spade]

But if M is an R-module, don't these two conditions (finitely generated and free) mean that M just has a finite basis?

Why is the given information sufficient to show that M is a vector space (since the base ring is then a field apparently)? Surely there are free modules with finite basis with a non-field base ring.

Am I missing something obvious?

[gritty sparrow]

There are, the point being that if that is all there is then the base ring is a field

[ionic spade]

Oh so if there is no additional structure imposed on the module (besides finitely generated and free) then the base ring is a field?

I don't immediately see why that is obvious though

[gritty sparrow]

Well notexactly. The point is that if R is not a field it will have some finitely generated module that is not free

It will still also have finitely generated free modules in addition

[ionic spade]

Ohhhh, that makes sense, thanks @gritty sparrow.

[tough raven]

This is possible iff the element is not in the intersection of all maximal ideals.
The intersection of all maximal ideals is called the Jacobson radical, and I don't think it's often {0}.

IDK anything about Jacobson radicals, but hopefully this leads to Googlable results about when it's {0}.

[tough raven]

Are there any epis in (Comm)Ring other than quotienting by ideals and localisation?
More precisely and concretely, given a (commutative) ring S and a subring R such that any two homomorphisms to any other (commutative) ring that agree on R must agree on S (ie coincide), is S necessarily a localisation of R?

[chilly ocean]

Finding the minimal polynomial of c seems pretty hard, is there a way to show that Q(5^(1/7)+7^(1/5))=Q(5^(1/7),7^(1/5))

[hollow blaze]

in a boolean algebra, what is the intuition for the ordering of the induced lattice?

is it just a <= b if we can derive b from a? This seems close to right, but not exactly

[thorn glade]

Hello, i dont understand a single thing i had written down. Could someone help me understand why 1 + 0 = 0 ? (both pics are related)

say * is a law on E. If i have x*e = x and that e = 0, why would 1 + 0 = 0 and not 1 + 0 = 1 ?

[formal ermine]

I have no idea what you wrote down either

the translation and rotation contributes to that

please just translate it manually

[thorn glade]

well, it is about neutral element of a law on left or right side

[formal ermine]

law?

[thorn glade]

E is a set, * is the internal composition law of E and e element of E. If we have $\forall x /in E, x*e=x$ then e is a neutral element of E on the right side

[cloud walrus]

🖤𝓼𝓪𝓫𝓮𝓵𝓵𝓪⛓🌹

[thorn glade]

the star

[formal ermine]

oh that's called the operation

[thorn glade]

right

[formal ermine]

are we working with groups?

[thorn glade]

not yet

it is just said that * is operation on E

[formal ermine]

ah

ok

what's your question?

[thorn glade]

is there such "neutral element on left and right side" ?

[formal ermine]

yes

[thorn glade]

alright

[formal ermine]

you mean double sided neutral element, or?

[thorn glade]

and in my case, we have x*e = x

i dont know

im confused

why did i write 1 + 0 = 0

we supposed e neutral on right side

i dont get it

[formal ermine]

yeah it doesn't really make sense

maybe a typo?

[thorn glade]

probably

all the rest was fine until i saw this thing

thank you either way chief !

[rotund aurora]

You don't want to find that by hand lol

By multiplicativity of the degree, you just need to show that the degree of 5^(1/7)+7^(1/5) is strictly greater than 5 and 7. Like 35 only has two prime factors, so there are not a lot of possibilities. EDIT: This is for showing Q(5^(1/7),7^(1/5))=Q(5^(1/7)+7^(1/5)), to show that your number is algebraic, you just need to see that the second field is contained in the first

then you will be done

I'm not sure how straightforward is that tho

[formal ermine]

I don't understand why $\bR[x^2 + 1] \cong \bC$. like I get that $\bR[x^2 + 1] = \Set{p(x) + (x^2+1)q(x) | p(x),q(x) \in \bR[x]}$, but how is that the complex numbers?

[cloud walrus]

illuminator3 (I/you)

[glossy crag]

Not $\bR[x^2+1]$, rather $\bR[x]/(x^2+1)$.

[formal ermine]

that's the same though

[glossy crag]

The set ${p(x) + (x^2+1)q(x) | p(x),q(x) \in \bR[x]}$ is equal to $\bR[x]$ because 1 and x^2+1 are coprime, and $\bR[x]$ is definitely not isomorphic to $\bC$ (because it's not a field).

[cloud walrus]

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

[formal ermine]

what's the set representation of $\bR[x]/(x^2+1)$

[cloud walrus]

illuminator3 (I/you)

[rotund aurora]

C

You are just quotiening out by x^2+1, which means that x^2=-1 and x is something independent of R

That's just the usual i

[formal ermine]

wdym?

what I thought was

[cloud walrus]

Ocean Man

[formal ermine]

$$\bR[x]/(x^2 + 1) = \bR[x]/\Set{(x^2+1)q(x) | q(x) \in \bR[x]} = \Set{p(x) + (x^2+1)q(x) | p(x), q(x) \in \bR[x]}$$

[cloud walrus]

illuminator3 (I/you)

[glossy crag]

What you wrote down is some subset of $\bR[x]$ (actually the entirety of $\bR[x]$, but that's not important), what you're looking for is the quotient ring $\bR[x]/(x^2+1)$.

[cloud walrus]

Ocean Man

[formal ermine]

but isn't R[x] the ring we're generating an ideal from? not R?

because x^2 + 1 is in R[x]

[glossy crag]

Yes, it is, but once you've generated the ideal you quotient R[x] by it and get something new.

[formal ermine]

are the cosets for rings different than those for groups?

[glossy crag]

No, it's the same thing.

[formal ermine]

I thought $G/I = \Set{g + I | g \in G}$

[cloud walrus]

illuminator3 (I/you)

[glossy crag]

That's right, what you wrote down was something else.

[formal ermine]

how

,, \bR[x]/(x^2 + 1) = \Set{p(x) + \gen{x^2+1} | p(x) \in \bR[x]}

?

[cloud walrus]

illuminator3 (I/you)

[formal ermine]

^ that's just applying this definition, no?

[glossy crag]

Because {p+q(x^2+1)|p,q in R[x]} and {p+(x^2+1)|p in R[x]} are two different things.

One is a subset of R[x], the other is the set of equivalence classes of some relation.

[formal ermine]

that one was before quotening out

[formal ermine]

?

[glossy crag]

Yes, this is correct.

[maiden heath]

Hi sorry for the absence. Yeah I understand that you have the basis of the trivial and standard representation. And it is possible to have a general formula for those basis. But I was wondering if it were possible to have something similar for the sign representation?

[formal ermine]

then $\gen{x^2 + 1} = \Set{q(x)(x^2 + 1) | q(x) \in \bR[x]}$

[cloud walrus]

illuminator3 (I/you)

[formal ermine]

the definition of a generated ideal

?

[rotund aurora]

just think of it algebraically

[glossy crag]

Yes, but what you had written before was some set of polynomials (that happened to be all of R[x], although that's beside the point.

[glossy crag]

This is the set of cosets wrpt to the subgroup (x^2+1)

The notation g+N is meant to be taken symbolically, you don't actually write out what N looks like.

[rotund aurora]

x is sent to "something" that will satisfy x^2=-1 and is linearly independent of 1 over R (meaning, you will have a field of dimension 2 over R). That's just C

[formal ermine]

,, \bC \cong \Set{\Set{p(x) + (x^2+1)q(x) | p(x) \in \bR[x]}, q(x) \in \bR[x]}

?

[cloud walrus]

illuminator3 (I/you)

[rotund aurora]

no

well

you are not specifying an operation

[formal ermine]

doesn't matter for the moment

I wanna know the set representation

[rotund aurora]

it does?

[formal ermine]

x is sent to "something" that will satisfy x^2=-1
why?

[rotund aurora]

I have never heard that word

in algebra sets are not really relevant, so idk why you worry about set representations of C if you are doing algebra

[glossy crag]

The set representation is every complex number a+bi corresponds to the coset $(a+bx)+I$

[cloud walrus]

Ocean Man

[formal ermine]

it's confusing if both of you are talking at the same time

[glossy crag]

i.e. 1 complex number corresponds to a whole multitude of polynomials, namely all the ones that look like (a+bx) plus something times (x^2+1)

You wanted the literal set representation, this is what it is.

[rotund aurora]

This is not something deep

[formal ermine]

?

[rotund aurora]

if you know what a quotient is, it should be obvious

[formal ermine]

I know what a quotient is and it isn't

[rotund aurora]

x^2+1 is "sent" to 0, agreed?

[formal ermine]

no

what, why?

[glossy crag]

Say you have an element g, a normal subgroup N of G and by some happenstance g^2 lies in N. Then the coset g+N will have the property (gN)^2=1.

[rotund aurora]

ok so, just go to your book and look the definitions

[glossy crag]

OK, I'm excusing myself from this in order not to confuse things further, you do it @rotund aurora

[formal ermine]

I'm looking at the lecture script

I'm taking an algebra course

[rotund aurora]

Im not going to explain this to you, because I think any book out there would do a much better job than I would do. You are asking what quotients of rings are

[formal ermine]

quotient is all possible cosets

no?

[delicate bloom]

0 + (x^2+1) is one such coset

[formal ermine]

yes

[rotund aurora]

Do you know modular arithmetic?

(genuine question, so I can make an example)

[formal ermine]

yes

[rotund aurora]

so work modulo n. Then a=b modulo n if and only if a-b is divisible by n. This is to say, a-b is in the ideal (n) of Z. Agreed?

[formal ermine]

yes

[rotund aurora]

Then we do the same thing in general for any ideal I of any ring R. We say a=b modulo I if and only if a-b is in I

[formal ermine]

yes

[rotund aurora]

then x^2+1=0 modulo (x^2+1) because x^2+1 is in (x^2+1). And x^2+1=0 implies x^2=-1 (this should be proven, and you really want be able to make this kind of operations, this is why we quotient out by ideals and not by subrings)

[formal ermine]

why is working in (x^2+1) like mod x^2+1

[rotund aurora]

"mod" is just a way of talking, but it is the same as in modular arithmetic

You have a ring R and an ideal I. Then, you define a relation $\sim$ by $a\sim b$ if and only if $a-b\in I$. This turns out to be not only an equivalence relation, but a congruence

[cloud walrus]

Croqueta

[formal ermine]

ah

[chilly ocean]

I think I get the idea [Q(5^(1/7)+7^(1/5):Q] is clearly not 1, so I need to check 5^(1/7)+7^(1/5) is not a root of of some 5 or 7 degree polynomial in Q[x] right?

[rotund aurora]

Meaning, that if $a\sim b$ and $c\sim d$ then $a+c\sim b+d$ and $ac\sim bd$

[cloud walrus]

Croqueta

[formal ermine]

ye

[rotund aurora]

and well, in $\bR[x]/(x^2+1)$ you have $0\sim x^2+1$, since $x^2+1-0=x^2+1\in (x^2+1)$, and since this relation is a congruence, $x^2\sim -1$

[cloud walrus]

Croqueta

[rotund aurora]

And that's essentially it

[formal ermine]

still don't get how quotening out gives us C

[rotund aurora]

Because this quotient ring is generated by our "new" $x$ and $x^2=-1$, so all elements of $\bR[x]/(x^2+1)$ have the form $a+bx$ where $a,b\in \bR$ and $x$ satisfies the relation $x^2=-1$.

[cloud walrus]

Croqueta

[rotund aurora]

and that's just the definition of complex numbers

[formal ermine]

why is everything in $\bR[x]/(x^2+1)$ of the form $a + bx$

[cloud walrus]

illuminator3 (I/you)

[formal ermine]

ah wait

cuz if we get degree 2 or higher the x^2 = -1 applies

so we have a restriction to degree 0 and 1?

[rotund aurora]

yes

like, expressions like $x^3+x^2+1$ still make sense, but $x^3+x^2+1=-x$, see why?

[cloud walrus]

Croqueta

[formal ermine]

yeah

okay so I get how x^2 ~ -1 but I don't understand why quotening the ideal makes that equivalence relation always true in the resulting ring

[rotund aurora]

mmh I'm not sure what you mean

[formal ermine]

in R[x]/(x^2+1) we have that x^2 = -1

but why do we have that

[rotund aurora]

didn't you say you got it already?

[formal ermine]

I get how x^2 = -1 in (x^2+1)

but why in R[x]/(x^2+1)

[rotund aurora]

no no, I probably explained it wrong

the first expression is not "in" but "modulo" (but you don't need to use that word, is just to make the analogy with the usual mod arithmetic).

The answer to the second question is just what I explained

[formal ermine]

ah I think I get it

just like in $\bZ/2\bZ$ the case $3 + \bZ$ is the same as $1 + \bZ$

[cloud walrus]

illuminator3 (I/you)

[formal ermine]

so it's modulo sort of?

[rotund aurora]

Again, only focus on algebraic relations. The set of the ring R[x]/(x^2+1) can still well be R[x], just as in modular arithmetic, say Z/2Z, it makes sense to talk about 3. But when you quotient out, you are introducing new relations, at first, you make those up, then you want these relations to be well behaved under the operations of your original ring

[rotund aurora]

yes

But like, cosets are a way to cope

well I suppose maybe not exactly

[formal ermine]

wait nvm I think I got it but I don't lol

[rotund aurora]

well actually I think strictly speaking the underlying set should be the cosets lol, but I never think in terms of cosets or anything like that. What you said above is the point I'm trying to make, 1+Z=3+Z

[formal ermine]

what is x^3 in (x^2+1)

[rotund aurora]

-x

x^2=-x+x(x^2+1)

[formal ermine]

ahhh right

[rotund aurora]

but not "in" (x^2+1)

[formal ermine]

ye my terminology in english is pretty messed up lol

[rotund aurora]

The first isomorphism theorem should tell you a lot about this

[formal ermine]

which one

[rotund aurora]

consider the homomorphism $\bR[x]\to \mathbb C$ that fixes $\bR$ and sends $x\mapsto i$. What is the kernel?

[formal ermine]

some of the isomorphism theorems are called homomorphism theorems in my native language

[cloud walrus]

Croqueta

[formal ermine]

ahhhhhh

(x^2+1) lol

[rotund aurora]

so that's it

[formal ermine]

yeah and the function is surjective

[rotund aurora]

If you can show that, you will be done. But I'm not sure how straightforward is that, I have not done a lot of this exercises myself

but if you want to show the number is algebraic, you just need to show that the field Q(5^(1/7)+7^(1/5)) is included in Q(5^(1/7),7^(1/5))

since the latter is algebraic and finite

[chilly ocean]

Showing algebraic is easy as it's a sum algebraic numbers but finding it's degree is pain in the ass showing 5^(1/7)+7^(1/5) is not a root of a 5,7 degree polynomial sounds like some annoying calculations

[delicate bloom]

here's a obvious example, $$\left[\bQ\left(\sqrt[3]{7+\sqrt{50}}+\sqrt[3]{7-\sqrt{50}}\right) : \bQ\right] > 1$$ right?

[cloud walrus]

Merosity

[lavish nexus]

isn't this just 2

[next obsidian]

If it has a cbrt

I would be very surprised if it was 2

,w minimal polynomial of cbrt(7 + sqrt(50)) + cbrt(7 - sqrt(50))

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Chmonkey#9767

[lavish nexus]

lmao

[next obsidian]

Bruh

[delicate bloom]

lol

[next obsidian]

I thought you meant degree 2

Tbf

Which felt very weird

Somehow this literallt being the number 2

Makes me feel better

[delicate bloom]

is there a quick cheap way of proving the degree is 1

like it's fixed by permuting sqrt(2) with -sqrt(2), but that's not really exhaustive

[lavish nexus]

just cube it

[delicate bloom]

cause I don't care that it's 2 personally, just that it's a rational number and so should be able to show it's fixed by all field automorphisms

[delicate bloom]

ew no

you cube it

[next obsidian]

Chmowned

[restive birch]

how would i go about this?

[chilly ocean]

it inherits it from Z.. call the equivalence class in Z/nZ of some a in Z [a]. Then ([a]+[b])+[c] = ([a+b]) + [c] = [(a + b)] + [c] = [(a + b) + c] = [a + (b +c)] = [a] + [(b+c)] = [a] + ([b + c]) = [a] + ([b] + [c])

same for multiplication, just replacing + with *

[rotund aurora]

Are there reasonable geometric tools that will allow you to construct every real number? Or at least, every algebraic real number?

[restive birch]

does this imply that all groups are closed under their binary operation?

[rotund aurora]

a binary operation on a set is closed by definition

[restive birch]

okay, thats what i was missing

thanks

[chilly ocean]

yeah a binary operation * is a map from G x G -> G so necessarily it takes values in G

[formal ermine]

closed means its output values are all in G

[hallow wing]

How do this

[glossy crag]

Why does the first isomorphism hold? Is it always so that if L/K is an extension, then $L\otimes_{K}K[x]/I\cong L[x]/I$?

[cloud walrus]

Ocean Man

[glossy crag]

Oh wait, I is not necessarily an ideal in L[x]. OK, what if we substitute I with K[x]f and L[x]f respectively?

[willow mason]

why is finite needed?

the extension i had in mind was, (m/n)a = sum a m times, minus a (n-1) times

[glossy crag]

Because a Q-vector space must necessarily be infinite.

[willow mason]

and why?

[glossy crag]

Because if M is such a vector space (or over any infinite field, not only Q) and x is non-zero, then Qx has infinitely many elements.

If ax=bx, then (a-b)x=0 and since x is non-zero, a-b=0 and a=b. So for different values of a you get different vectors ax.

So a finite Z-module can never become a Q-module.

[willow mason]

i dont understand your reasoning

[glossy crag]

Is this part not clear to you?

[willow mason]

the part above

[glossy crag]

This part shows Qx is an infinite set. It is contained in M, so M itself must be infinite.

[willow mason]

where does this fail then?

[glossy crag]

I just outlined to you why in a Q-vector space different elements act differently (for general modules this is called being "torsion-free"). If n is the order of M and x is a non-zero element, then nx=2nx (both are zero). But this couldn't happen if M were a Q-vector space.

[glossy crag]

Plus, using your definition you'd get 1/2*x=0 for all x, another thing that can't happen in a Q-vector space.

[willow mason]

hmm

feels weird

[glossy crag]

Not rly imo. There's a very basic explanation (the one I gave you): a Q-vector space cannot be finite as a set. M is, so M can't be a Q-vector space.

If you know what ring homomorphisms are I can give you another explanation (a little more high-brow, though I don't think it's more helpful than the one I gave before).

[willow mason]

so here, why cant we have torsion elements

zero div but in context of modules

a-b != 0 and x!=0 but (a-b)x = 0 in M

[glossy crag]

That can't happen if (a-b) is invertible, which is the case in a field if a-b!=0.

[willow mason]

ahh

kk

[glossy crag]

If an element a is invertible, then from ax=0 follows x=0 (multiply both sides by the inverse of a).

[willow mason]

that makes sense

what was the way with homomorphism

[glossy crag]

R-module structures on M correspond with ring homomorphisms of R into End(M) (the ring of endomorphisms of the abelian group M). Ring homomorphisms out of fields are injective, so if M is a K-module, a copy of K has to sit in End(M). However if M is finite, then so is End(M) (as a set of functions from a finite set to itself), so if K is infinite there can be no ring homomorphism and hence no K-module structure on M.

[willow mason]

makes sense

thank you

[solar glacier]

Question regarding the definition of an Ideal in ring theory

[vivid tiger]

just ask the question

the product is from anything in the ring, the sum is from anything in the ideal

[solar glacier]

Ok

So right ideal

Is when ra \in I for r in R and a in I

Sorry that’s left ideal

Or wait

[vivid tiger]

you gave the right ideal

so a right ideal is where multiplication has the ideal element on the right.

left is where ideal is on the left

[eager willow]

What

[vivid tiger]

so if you have a ring R

[glass grail]

what

[solar glacier]

Then

ar \in I

Means I is left ideal

For r in R

And@if I is both left and right ideal the. It’s an ideal ?

[vivid tiger]

yep

[solar glacier]

Ok so the@general term ideal implies both left and right ideal

[vivid tiger]

"ideal" is usually used when it's either clear which it is or when it's abelian, or when it doesn't matter

[eager willow]

Left ideals have a 'left action' by the ring and right ideals have a right action. Same with left and right modules, ideals are submodules.

[solar glacier]

He’s right

[eager willow]

So no a left ideal is on the right not the left

[solar glacier]

So I had it backwards

So a left ideal is ra \ I for r in R a in I

[vivid tiger]

err wait what? rx in I with r in R and x in I is definitely a right ideal nope

[glass grail]

https://mathworld.wolfram.com/LeftIdeal.html

[eager willow]

Nope

[vivid tiger]

nope, I mixed it up

[solar glacier]

Wikipedia confirms

[vivid tiger]

apologies

[solar glacier]

Left ideal is mult by left by ring element

So if I’m given a Boolean ring

And it says every finitely generated ideal is principle

Is that ideal both left and right

[glass grail]

i think a boolean ring is commutative so theyre the same thing in that case

[solar glacier]

Ok so for commutative rings

And ideal is. Both left and right

Otherwise you need to specify left or right

[glass grail]

yeah cause if ra is in I then ar is in I since ra=ar

and converse

[eager willow]

Noncommutative rings include endomorphism rings of most commutative rings, it's worth knowing some noncommutative stuff but algebraic geometry is done classically in the commutative setting and it's where the best examples for understanding are.

Also principle is not left and right. Being both left and right ideals is just called being an ideal

[solar glacier]

So Boolean rings are commutative

[vivid tiger]

boolean rings indeed are abelian

[solar glacier]

Oh I proved that yesterday

That R is commutative if it’s Boolean

It uses the old (a+b) = a + ab + ba + a trick

Ok now back to my proof of if I is an ideal in Boolean ring that is finitelt@generated then it’s principle

[tender bough]

This is from my lecture notes. So this is using a formula for the Killing form on the general linear lie algebra and applying Cartan's Criterion: a Lie algebra is semi-simple iff its Killing form is non-degenerate.
But it only tests for the unit matrix and an arbitrary B, how does this imply that k(A, B) = 0 for any A and B?

oooOO, that's actually the definition of "degenerate bilinear form"..

[solar glacier]

question is this a good start for this proof

[solar glacier]

this proof

[cloud walrus]

MyMathYourMath

[next obsidian]

It’s principal not principle

[solar glacier]

sorry

[next obsidian]

And induction sounds like a good idea

[solar glacier]

okie doke then

can i just show That

$I=(a_1,...,a_{k-2}, a_{k-1}+a_k+a_{k-1}a_k):=I'$

[cloud walrus]

MyMathYourMath

[solar glacier]

then $I$ has $k-1$ generators

[cloud walrus]

MyMathYourMath

[solar glacier]

then done by indution

[solar glacier]

is this a good idea ^

clearly $I' \subset I=(a_1,...,a_k$)$

[cloud walrus]

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

[next obsidian]

If you can show that they’re the same ideal then yeah

You want to just cut down the number of generators by 1

[solar glacier]

yep

and then done by induction

ok i canshow this

thanks!

[toxic zephyr]

is there some sort of algebraic object that acts as something like a boolean where
0+0=0
0+1=1+0=1
1+1=1
?

[south patrol]

I mean boolean algebras are already algebraic enough right

unless you want another name for the same thing

[toxic zephyr]

don't know what that is, so that's probably the answer to my question lol

[ripe basalt]

you probably do know

you've seen it before anyways

[celest cairn]

I wanted to find the minimal polynomial of $\omega+\omega^{6}$, where $\omega$ is a primitive 7th root of unity, and $\omega \neq 1.$
I don’t know what to do for this last part.

[cloud walrus]

Sapphire Gaming

[next obsidian]

Mf, 1 is not a primitive root of unity

You don’t need to specify it isn’t 1

[fast stratus]

(A) make sense… but what do they mean by order r in (b)

[thorn delta]

like, |H| = r

[fast stratus]

Haha, I know that…. Why should it be normal

It’s just an integer right… and certainly not every subgrp is normal… and they are not saying anything special about r

[celest cairn]

Ok.. so what do I do for the last part?

[thorn delta]

they are saying there is only one subgroup of order r

(ignore the message i deleted)

like, the group may have other subgroups, but any subgroup of order r is exactly H

[fast stratus]

Got it… and (a) says gHg-1 is subgroup

Now we know |gHg^-1| = |H| (proved before) and since only subgp with order r is H, H = gHg-1, thus H is normal

[thorn delta]

nice

[restive birch]

am i missing something, or is this just {1, -1, i, -i}?

[teal void]

for n=4 sure but not anything else

[chilly ocean]

suppose G is a group and H normal in G, i am wondering when G/H is isomorphic to some subgroup K of G

for instance, if V is a vector space and W a subspace of V, then there is always a subspace W' of V which is isomorphic to V/W, but this doesnt seem to hold for groups

[formal ermine]

first isomorphism theorem maybe?

if you have a homomorphism phi : G -> G' then G/ker(phi) is isomorphic to im(phi)

[chilly ocean]

yes..? how does that help

[formal ermine]

G/H is isomorphic to K if you can find a homomorphism whose kernel is H and whose image is K

[chilly ocean]

It's apparently true for finite abelian groups. No longer true for infinite finitely-generated abelian groups, since, for example, Z has no subgroup isomorphic to Z/2Z.

I say "apparently" because I googled it.

lol could you share what you found i couldnt find anything (but didnt look too hard)

https://groupprops.subwiki.org/wiki/Quotient_group_need_not_be_isomorphic_to_any_subgroup

This website carried me through the group theory part of my algebra course.

[chilly ocean]

Finding a homomorphism G -> G with kernel H and image K sounds harder than just writing down H and K.

[formal ermine]

fair

[chilly ocean]

thank you! exactly what I was looking for. very funny that the hypothesis I was curious about is in fact a definition: "endomorphism kernel"

[celest cairn]

I wanted to find the minimal polynomial of $\omega+\omega^{6}$, where $\omega$ is a primitive 7th root of unity.
I don’t know what to do for this last part.

[cloud walrus]

Sapphire Gaming

@solemn hollow

[celest cairn]

I know $\omega^{6}+\omega^{5}+…+1 = 0$ and $w^{7}-1 = 0.$

[cloud walrus]

Sapphire Gaming

@solemn hollow

[celest cairn]

Ah ok, thank you 🙂

[cloud walrus]

@solemn hollow

[cloud walrus]

@solemn hollow

[solemn hollow]

@celest cairn

[celest cairn]

Yes

Sorry I miss typed it.

[celest cairn]

@solemn hollow

[sharp dirge]

can every element of a ring be written as sum of 1's ?

[glossy crag]

No, Q is a ring, try writing 1/2 as a sum of 1's.

[sharp dirge]

oof, thanks

my mind was stuck on integer rings lmao

[glossy crag]

If I want to show R/(p^n) is indecomposable as an R-module (i.e. not a non-trivial direct sum of its submodules), is the following enough: if R/(p^n) were decomposable, its submodules A and B would be homomorphic images of it and hence cyclic themselves and since p^n annihilates the module, they would have to be of the form R/(p^a), R/(p^b) for a,b<n. However such a module is annihilated by p^{max(a,b)}, which properly divides p^n (for a,b<n), a contradiction.

[sharp dirge]

when it says $Hom_{R}(R,R)$ it means the module homomorphisms right

[cloud walrus]

pewdssssssss

[sharp dirge]

when R is a ring

[formal ermine]

why are factorizations unique up to a unit (ik how you prove this but I don't understand the definition)

like, why does it make sense to define factorization with a unit in the front

[sharp dirge]

because 4 = (2)(2) = (-2)(-2)

in the integers

we can have -1 infront of some of the factors

[formal ermine]

I forgot to mention that everything is commutative

I don't really get your point... 2*2 has the unit 1 in the front, but so does (-2)*(-2)

[sharp dirge]

okay ...

so for for any integer

there can be many factorizations

if we do not make it unique upto units

[formal ermine]

1 is the only unit in Z

so I don't really get your analogy

[sharp dirge]

-1 too

[formal ermine]

right

[sharp dirge]

so 4 = 2 * 2 = -1 * 2 * -1 * 2

[formal ermine]

wut

[sharp dirge]

the -1 and 1 are meaningless essentially

in factorizations

hard to explain 😅

[formal ermine]

but the -1 appears twice...?

[sharp dirge]

yes

im saying that having units in the factorization is pointless

[formal ermine]

if $r \in R$ and $r \neq 0$ then $r = \epsilon p_1 \ldots p_n$ with $\epsilon \in R^x$ and $p_i$ prime elements

[sharp dirge]

ye

but you can simplyfy

thats what the definition has a unit in front

[formal ermine]

...what?

[cloud walrus]

illuminator3 (I/you)

[sharp dirge]

units in Z are 1 and -1, dividing and multipling by them are not meaningful essentially

[formal ermine]

why is it not meaningful to multiply by a unit?

[sharp dirge]

because we can just multiply by its inverse

[formal ermine]

that doesn't really explain my question

[sharp dirge]

Youre question is why is the definition require "unique upto units" yes?

[formal ermine]

no

[sharp dirge]

.

[formal ermine]

read the part in the parenthesis

and then my second message

[sharp dirge]

"i dont understand the definition"

[formal ermine]

like, why does it make sense to define factorization with a unit in the front

[sharp dirge]

you can just let the unit in the front be 1 if you like

it doesnt really matter

[formal ermine]

that is not my question

[sharp dirge]

the point is, if we have two factorizations. they will only differ by units

[formal ermine]

that is not my question

[delicate orchid]

R^x for units

[formal ermine]

why does it make sense to do that?

[formal ermine]

like why do we define factorization like this?

when will we ever encounter a situation where that unit is necessary

[sharp dirge]

i told you, in alot of cases, its just a 1

[delicate orchid]

In Z it’s definitely needed, as pewds has explained

[formal ermine]

ahhhh cuz for negative numbers

yeh

[delicate orchid]

Yur

[formal ermine]

is there a generalization of that "for negative numbers"?

[sharp dirge]

unit in front is a -1 then

in Z

[delicate orchid]

“Up to a unit” is that generalisation if you mean what I think you do

Maybe look at like

M_2(C) or something

Actually that’s not a ufd is it, lol

Uhhh R[x] with R being the reals

And then take like 5x^2-10x+15 = 5(x+1)(x-3) or (5x+5)(x-3) or (x+1)(5x-15)

We want all of these factorisations to be “equivalent”

And they are, up to a multiplication of the unit 5

brb

[formal ermine]

(x+1)(x-3) would also be a factorization?

[delicate orchid]

Nah because (x+1)(x-3) isn’t 5x^2-10x+15

Maybe I’m explaining it badly

[formal ermine]

I think this is coming from me not understanding something else

so

what does it really mean for two elements to be associated?

I know in notation it's a = br for a unit r

but like, is there any intuitive meaning to it?

[sharp dirge]

yeah

factoring into units is not meaningful

in field, everything not 0 is unit. and you cant factor in a field

[formal ermine]

let's leave the factorization aside for a minute. I want to understand the association thing first

[sharp dirge]

4 = 4* 1

we factored 4 into 4*1

[formal ermine]

???

[sharp dirge]

1 is unit

[formal ermine]

5 is a unit in R

but 4 isn't 4 * 5

[sharp dirge]

wot

[delicate orchid]

but 4 is 5*4/5

So they’re associated

[formal ermine]

ye but we say a ~ b iff a = br for a unit r

[delicate orchid]

(4/5) is defo a unit in R

[formal ermine]

yes

[delicate orchid]

I don’t see your point then

[formal ermine]

I want to know if there's an intuitive meaning to association

what does it mean for two elements to be associated?

[delicate orchid]

Right ok time to spew some wishy washy “intuition”

If two elements are associated then their prime decompositions are “basically” the same (they’re the same up to multiplication by some unit)

[formal ermine]

ohh so association only makes sense when you take prime decomposition into account?

cuz it was like really confusing, we defined association way before ufds

[delicate orchid]

It makes sense because the definition is well defined

Anyway

Prime decomposition is cool because it tells you a lot about how that element interacts with ideals in the ring

[formal ermine]

nono I mean like makes sense as in it actually has a meaning

[formal ermine]

mh

[delicate orchid]

Ok I see

[sharp dirge]

factorization is trivial in fields anyway

cuz everything is already unit

associates is supposed to generalize the similarity between 1 and -1 in Z, (the units)

[delicate orchid]

This might give more intuition behind why we don’t care about unit factorisation - the ideal generated by any unit is the entire ring, so we don’t really get any new info from it

Again this is entirely non-rigorous and wishy-washy but meh

You get what I mean

[formal ermine]

ah

[formal ermine]

what exactly does it tell us?

[delicate orchid]

It tells you which ideals that element is in

Like if p is a factor of n then clearly n is in (p)

[formal ermine]

how do units break this? like if we have r | x for a unit r then why don't we care if r is in the prime decomposition of x?

[delicate orchid]

Because the info we get is
r | x => x in R

Which is trivial in the truest sense

[formal ermine]

ohhhhhhh

it's finally starting to make sense

[delicate orchid]

(The reason why I say that we care about the structure of prime ideals is related to the zariski topology over a ring and the spectrum of a ring if you want to do some Googling)

(Don’t worry about it though)

[formal ermine]

thank you!

I'll think some more about this and come back maybe later or tomorrow

[smoky cypress]

In $\bQ(\zeta)$ where $\zeta$ is the $n$th root of unity, we have $\Gal(\bQ(\zeta)/\bQ)=(\bZ/n\bZ)^*$. Say $H$ is a subgroup, then is the corresponding field $\bQ(\alpha)$ where $\alpha=\sum_{g\in H}\zeta^g$?

[cloud walrus]

CS person

[zenith trellis]

H is a subgroup of which group ?

[smoky cypress]

Of (Z/nZ)*

[pliant raptor]

Could someone give me a hint? I understand that you can write $0=\bigcap_{n=1}^{\infty}Q\oplus nM$, for some $\mathbb{Z}$-module $M$, but how would I get a contradiction?

[cloud walrus]

ImHackingXD

[lethal dune]

projectives over PID are free...

[pliant raptor]

This is from Dummit chapter 10.5, it references that only in chapter 12 will be proved that a finitely generated Z-module is projective iff free, so I'm not sure if that is the objective of the exercise

[lethal dune]

let me check prop 30(4)

ok have you proved that the intersection is empty@pliant raptor ?

[pliant raptor]

You mean the zero module? Yeah, that's what I wrote above

[lethal dune]

nQ =Q right?

for any nonzero q ∈Q, show that q lies in that intersection

hint: ||(keep taking q/n)||

which contradicts the intersection is empty

[pliant raptor]

ahhhh ofc, 0 is always an element of nM

thx

[lethal dune]

[restive birch]

how would i go about this? is the number 1 relevant to all possible binary options? (G in this case is just any group)

[formal ermine]

hint: write x^2 as xx

[next obsidian]

1 is the identity of G

[restive birch]

that wasnt given, at least not where i can see

[next obsidian]

This is the notation for the identity

[restive birch]

huh

[next obsidian]

Either that or e

Or 0 if it’s being written additivelt

Anyway, there’s nothing deep happening at all, 1 is not the number 1 which magically is in G or anything, it’s just defined to be the identity

[restive birch]

alright

notation seems slightly weird but w/e

[next obsidian]

I mean

1g = g

That’s what it does

This seems normal based on your experience with like, actual numbers

[void cosmos]

nothing says the group is multiplicative tho

[next obsidian]

That’s all

[restive birch]

exactly

[next obsidian]

Bro it wrote x^2

[south patrol]

Lol

[void cosmos]

x^2=x+x

in additive gropus

[restive birch]

exactly

[formal ermine]

let 1 be the additive neutral element

[next obsidian]

No…

[restive birch]

depends on the operation

[south patrol]

then you wouldn't write x^2

[next obsidian]

That’s 2x

[south patrol]

Lmao

[next obsidian]

If you’re writing it additicet then x^2 doesn’t make sense

[void cosmos]

i am sure you can write g^n with additive groups

[next obsidian]

That’s not a thing you can write

[void cosmos]

x^2 means apply the operation in question two times

[south patrol]

Groups are basically assumed multiplicative tbh

[void cosmos]

i am sure you can find examples of this notation with additive groups

[south patrol]

Unless they are abelian

[formal ermine]

my literature defines g^n likes this

[restive birch]

it uses x^n in the definition of order

[void cosmos]

yes exactly

o meaning the operation

[void cosmos]

not necessairly multiplication

[south patrol]

Pain

[chilly ocean]

Brain cells on the floor.

[restive birch]

what did i start

[next obsidian]

If you want to go jerk for about the possibility that it means something that makes no sense and then spend forever unable to solve a simple problem go ahead

Or you can accept that 1 is the identity and move on with your lives

[south patrol]

Lol

[void cosmos]

nope just rephrasing what the original poster meant

and what probably he or she got confsueda bout

confused*

[restive birch]

yea i was just missing that notation, especially since this textbook called the indentity of a group e further up

[void cosmos]

yes

1 is not always if not not most of the time called the identity

for me atleast

[shell brook]

Why are you like this

[void cosmos]

me?

[shell brook]

Yes

[void cosmos]

dk what you mean tbh

[south patrol]

Lol

Idk isn't it like the default here

[void cosmos]

if it is then my bad

[south patrol]

Idk how the question can be read otherwise

[void cosmos]

but most of the textbooks i came through went with e

and g^n almost always means apply operation n times

not multiplication only

[restive birch]

thats definitely what the question intended, i see that now, but it is certainly odd since 1 is a number that isnt always the indentity

[chilly ocean]

It is literally just notation.

[south patrol]

1 is common notation for the multiplicative identity

[chilly ocean]

You are fighting over symbols.

And conventions.

[void cosmos]

im not fighting

who is

[restive birch]

sure, but nothing told us this group was multiplicative

[formal ermine]

nothing told us what 1 is either

[chilly ocean]

oh my god

[south patrol]

I think it is obvious from notation

[void cosmos]

it is obvious from the question itself

[south patrol]

*context

[void cosmos]

i was just saying what henryduke had in mind

thats it

[restive birch]

given that my understanding is shaky, im trying to make very few assumptions based on what it seems like the question is implying, i wasn't aware of that notation and while i might have a pedantic quarrel with it, it probably doesn't get in the way much, so im going to go back to the problem now

thanks for the info tho

[void cosmos]

do you know how to do it