groups-rings-fields

[white grotto]

so the only polynome is 0

okay

what does a nilpotent function do to rank space exactly?

shrink it by 1 dimension minimum each time

?

[agile burrow]

I think the easiest way to see this is to note that $H^i(\mathsf{C}(\mathscr{F})(A^{\bullet})) = \mathrm{ker}(\mathrm{coker} d^{i}_{\mathscr{F}(A)} \to \mathscr{F}(A^{i+1}))$. Then apply $\mathscr{F}$ to $A^{i-1} \to A^i \to K^{i+1} \to 0$, where $K^{i+1} = \mathrm{ker}(d^{i+1})$, and use right-exactness to consider $\mathscr{F}(K^{i+1})$ as a cokernel.

[cloud walrus]

walter

[agile burrow]

oh that looks terrible

[chilly ocean]

can you translate that from wtf to english

[south patrol]

lol walter i thought you were answering this question on lin alg and was worried why you were typing for so long

xd

[agile burrow]

I don't have much to say about nilpotence

also sorry if I interrupted lol

[south patrol]

Me neither yeah because lin agl is weird

nah you didn't aha

[south patrol]

Yeah, it shrinks the dimension by >= 1 each time

In such a way that eventually get 0, of course

[chilly ocean]

@south patrol can I ask what you used to study AA

[south patrol]

lectures basically lol

i am still studying anyway owo

[chilly ocean]

[south patrol]

lol sorry

is there any particular topic you wanna learn about

[chilly ocean]

i don't have access to lectures because i'm still in high school but i am studying group theory right now

from a introductory book

also I don't understand this question :o

[agile burrow]

I learned from both Aluffi's book and Dummit and Foote

[chilly ocean]

i heard aluffi takes a category theory approach

[agile burrow]

yeah, he introduces some basics of the language for a sort of "unified" exposition

some people say it's done well, others disagree. to each their own i guess

[agile burrow]

What part is confusing?

[chilly ocean]

I don't know how to show closure, existence of inverse and identity

I mean, a^n is the identity given that a^n = 1

[agile burrow]

Ok, so let's start with showing that the identity is in H

[chilly ocean]

as for inverses actually, if we have $a^m$ the inverse is just $a^{n-m}$

[cloud walrus]

Ayanokoji

[agile burrow]

What is n in this context?

[chilly ocean]

the power of the element

[chilly ocean]

in H the identity is just a^n no?

[agile burrow]

Oh, I think I see the confusion

H is the set of elements of finite order, meaning it contains the elements g of the group such that g^n = 1 for some positive integer n. That doesn't mean the same n holds for all elements in the group. Does that make sense?

[chilly ocean]

oh

that makes sense

so it's the set of all elements with finite order

[agile burrow]

Yes

[chilly ocean]

how would I show that the identity is in H though

[agile burrow]

Well, can you think of a positive integer such that 1^n = 1?

I guess maybe in this context it's less confusing to use e for the identity element

[chilly ocean]

e^anything = e though

[agile burrow]

Right, but we just need to show that there exists some positive integer n such that e^n = e

In particular, we can let n = 1 -- we know e^1 = e so e has finite order

[chilly ocean]

i think i got it from here

thank you

[agile burrow]

Happy I could help 🙂

Also, maybe this is obvious to you but it tripped me up a little when I was thinking about it so I might as well point it out. You can also view the cohomology as the cokernel of the inclusion of the kernel into the image (which I think makes the whole ker / im thing easier to see). In particular, there's a corresponding result for left exact functors that is maybe worth writing out explicitly.

[coral shale]

i dont agree with u on those roots

bruh

roots can be found by inspection

not wolfram

the hell?

1+i has modulus root 2

no way its 4th power has modulus 0

im so confused

ohjhh ok my bad i f ked up

im still kinda confused

right nvm ur good

yes the question is simple

[south patrol]

Ye

[lethal dune]

The part I don’t understand is why homology is the kernel of that map

[agile burrow]

Do you see why there's a well-defined map from the cokernel of the differential to the next group?

[lethal dune]

Yes

[agile burrow]

Right, lifting representatives is fine because the differential will send anything in the image of the previous map to zero. Then the kernel of this map is just the collection of classes whose representatives are in the kernel of the differential

[lethal dune]

Ohh ok

Now I see

Thanks

[chilly ocean]

Can anyone help with this?

no clue how to approach this

[slim dragon]

what does it mean to be a subgroup

[formal ermine]

you have to show that all of the subgroup axioms hold

[chilly ocean]

x, y in H implies xy in H
inverse in H
identity in H

[slim dragon]

show these all hold

[chilly ocean]

how do I show that closure holds

[slim dragon]

what does x in H mean?

[chilly ocean]

x in H has the property that for some element g in G gx = xg

[slim dragon]

err

not quite

the element g does nto change

[chilly ocean]

right

my bad

[slim dragon]

ok so we know that gx = xg

what does it mean for y to be in H?

[chilly ocean]

same thing

gy = yg

[slim dragon]

okay

then what do we need to check to show that xy is in H?

[chilly ocean]

oh

g(xy) = (xy)g

[slim dragon]

can u use the facts that x and y are in H to show this?

yes, is this true

?

[chilly ocean]

well

we have to show this is true

no?

[slim dragon]

yes

is it?

[chilly ocean]

yes? I think?

i mean

[slim dragon]

prove it

start with (xy)g

use the fact that y and x are in H

xyg = ...

[chilly ocean]

wait do i use associativity property or something lol

probably not

uhhhhhh

sorry i don't know how to approach this

[slim dragon]

uhm

you only know two things

[chilly ocean]

yes

[slim dragon]

x in H, y in H

[chilly ocean]

gx = xg

[slim dragon]

so what is the only obvious thing you can do to modify xyg

[chilly ocean]

oh

[slim dragon]

xyg = ...

[chilly ocean]

xyg = xgy

[slim dragon]

yes

[chilly ocean]

= gxy

[slim dragon]

then the next only obvious thing

so putting the start and end togethert

xyg = gxy

so (xy) in H

done.

[chilly ocean]

thank you

that actually makes sense

[slim dragon]

good!

this is a very common pattern

[chilly ocean]

when proving that something is a subgroup, we always start with two elements x and y in H

and try to apply whatever set operation it is to them

[slim dragon]

with some property

here the property is xg = gx

but in general youll have some property that both x and y have

and you want to show xy also has this property

and its typically done in a similar fashion

[chilly ocean]

i see

so I have to show that inverses and the identity element exist in H

[slim dragon]

yes

to be more clear

if x is in H then x^{-1} is in H

[chilly ocean]

the identity is x = g^{-1} no? lol

wait

inverse*

or no

the property is that

[slim dragon]

huh

[chilly ocean]

sorry I'm a bit confused

[slim dragon]

same idea..

what does it mean for x to be in H

[chilly ocean]

gx = xg

[slim dragon]

what would it mean for x inverse to be in H..

[chilly ocean]

oh

let x^{-1} = y then gy = yg

[slim dragon]

no need to rewrite it

$x^{-1}g = gx^{-1}$

[cloud walrus]

moonside

[slim dragon]

okay

[chilly ocean]

yeah

[slim dragon]

can you start from $xg = gx$ to get the above?

[cloud walrus]

moonside

[chilly ocean]

hm

i mean

yes I can

multiply by the inverse of x?

[slim dragon]

ok

show it

[ruby sundial]

This is crazy to me

because

you can think of every commutative ring as embedded in a field

[eager willow]

well, no, every integral domain

[chilly ocean]

zerodivisor moment

[ruby sundial]

wait

yeah just integral donains

this sucks

every commutative ring is in some integral domain though

[eager willow]

but the right idea is that commutative rings are secretly just rings of functions over a topological space where the value at each point can be taken in a field

[ruby sundial]

wait

[eager willow]

no every subring of an integral domain is integral

[ruby sundial]

The way in thinking of it is this

[eager willow]

every ring has a maximal ideal, hence a prime ideal, hence quotients into an integral domain

[ruby sundial]

im conflating surjection with embedding

when it should be injection

[eager willow]

haha sometimes I've heard the verb 'covers' be used here. Every ring covers a field

[ruby sundial]

yeah this is funny tho lol

[eager willow]

every ring covers a division ring I should say, and the division rings are fields when they are commutative

[ruby sundial]

also rn im trying to understand how intersection of two affine opens is the same as fiber product over the scheme

i was told to think of how localization commutes with tensor product

and then glueing

[eager willow]

that's just using the right universal property to describe the intersections

open embeddings U1 -> X, U2 -> X, the universal property for the fiber product is like the simplest thing that maps into X via factoring through both U1 and U2, and that's the intersection embedding to both U1 and U2

but I think putting all the pieces together correctly to show the actual universal property you'd need those facts about localization

setwise its more intuitive though, you pick out the elements of U1 x U2 that look like (u1, u2) where like, both of the ui are just the names given to elements of the Ui, but identified as fundamentally the same element of X. So you're really looking at one element of X and how its called in the subsets U1 and U2 that it appears in. The pair (u1, u2) is like a pair of nicknames for the exact same element of X. So it's the intersection

[next obsidian]

Prove the following:

Maps Y -> U are the same as maps Y -> X whose set-theoretic image lands in U

This shows that U represents the functor sending Y to maps to X which land in U

From this and the functorial description of fibered products you see that U x_X V represents the same functor as U\cap V

This is my unironic suggestion to see it at a high level

[pastel cliff]

are noetherian rings at all related to the idea of localizations of a ring

[chilly ocean]

What makes you think so?

[pastel cliff]

they're near each other in my prof's lecture notes

[eager willow]

localizations and noetherian rings are both deeply connected with topological notions underlying all commutative rings and the foundation of algebraic geometry

but they are not so directly related by their definition

being noetherian for example is not a local property

[next obsidian]

It is

But not at points

Anyway no, I would not say they’re related

They’re both related to ring theory, but I would not say they’re related unless you’re trying to draw really weak comparisons

[eager willow]

? am I wrong about what local property means?

[next obsidian]

Well like it depends yeah?

The nice thing is in most situations (sometimes needing a finiteness hypothesis) it’s equivalent to say it holds at all points and holding on an open cover

But in this case they disagree so what “local” means is kinda not clear but usually it means at primes I think

But I like to specify what sort of local it means

Like affine-local

Is a term ppl use

And (locally) Noetherian is affine-local

But cannot be checked on stalks

¯_(ツ)_/¯

[eager willow]

ok sure, I was using 'can be checked on stalks' but I guess the topological definition should be wrt open covers oops

[next obsidian]

Yeah

That’s what makes it nasty when they don’t agree

Again usually with some amount of finiteness you can spread out something from a stalk to a nbd so they agree often

But sometimes that means assuming things are Noetherian so it doesn’t really@work when wanting to ask if Noetherian is local

[eager willow]

lol nice. well that was a good way to scare off someone asking an honest algebra question :\

[next obsidian]

Nah I think they just are afk

I have spoken to them a lot

[tribal niche]

I know that the group R/Z is isomorphic to the unit circle by the first isomorphism theorem

[next obsidian]

I thought you were gonna end there

So I was gonna say “congrats”

[tribal niche]

By considering the homomorphism (x,y) ↦ (exp(2pi i x), exp(2pi i y)), we also have R^2/Z^2 ≅ S^1 x S^1, right?

[south patrol]

indeed

in fact this works as topological groups ig

[next obsidian]

If you use the “obvious” copy of Z^2 (R^2/Z^2) ≈ (R/Z)^2

So that’s another way to see that isomorphism

[tribal niche]

ah ok, the question's asking me to show R^2/Z^2 ≅ [0,1) x [0,1), so I was worried I was doing something wrong since it seemed too easy

[south patrol]

Hm is this as topological spaces though

[tribal niche]

no just groups

[south patrol]

cause that is different to just using isomorphism theorems

Wait but [0,1) as a group

what does that mean

I am confused lol

[tribal niche]

i assume modulo 1

[south patrol]

Oh hm okie

[main needle]

Any hints on how to find the order of the normalizer/the generators of the subgroup $H = <(12), (34), (5,6),(7,8)>$ of $S_8$? I thought of using orbit stabilizer thm but I got stuck on finding the size of conjugacy class of H

[cloud walrus]

ru0xffian

[marsh goblet]

Hey, I have a question about this result that ‘we see at once’
For reference, this is from the modules over principal rings section of the modules chapter in Lang’s algebra (pg 150)

I’m pretty sure that given such a direct sum decomposition, the y_i are independent as (y_i) being cyclic implies it is isomorphic to R/(a). Therefore it has a as an exponent and so there exist a_i such that a_iy_i = 0 for all I

However, if the y_i are independent, I’m not really sure how to show that we can construct that direct sum decomposition :/

[thorn delta]

It suffices to show that every element on the module in the LHS is uniquely a sum a1y1 + … + am ym

Take two representations of an element on the LHS, subtract them, and apply independence

[marsh goblet]

Ah right that makes sense, thanks so much

[thorn delta]

npnp

[obtuse bear]

Any conjugate of H will be another subgroup of the same form <(ab), (cd), (ef), (gh)>, (i.e. a subgroup generated by four disjoint 2-cycles) and all such subgroups are conjugate since going from one to another amounts to "relabeling" the 8 elements of the underlying set, which can always be done via conjugation. So there's only one orbit. To compute its size you need to count how many distinct subgroups there are of this form. Probably something involving the multinomial coefficient?

[restive birch]

How would i got about finding the generators/relations for $S_3$?

[cloud walrus]

henryduke

[restive birch]

Im not great with generators/relations, i dont really understand how to derive them

[next obsidian]

No, a monoid is a set with an associative operation with the existence of an identity

It does not mean that elements have inverses

It makes it possible to define what inverse even means, but it doesn’t ask that everything be invertible

No

Consider Z

Just by example you see that this is not true

Or maybe Z\{0} under multiplication

If it’s finite

Then the answer is yes

No

Cancellation is equivalent to saying the map x -> yx is injective for all y

(Multiplying on the left)

But this doesn’t mean 1 will be in the image

Taking Z for example again, multiplying by 2 is injective but doesn’t have 1 in the image

But this gives a proof for finite monoids, because then an injection is also surjevtive by pigeonhole

So 1 is in the image, so yx = 1 for some x

What does divisibility mean here?

Then yes

By taking c = 1

No

Again, take Z as an example

Lol

[lethal lava]

This is on page 860 of Dummit and Foote. Could anyone explain why it's true? CG is a group ring and G is a finite group.

[eager willow]

It says it's a vector space. It's clearly an abelian group. Can you guess how you might multiply by an element of C? Maybe first think how you multiply by elements of CG.

[lethal lava]

I think multiplication by an element of C looks like scaling in each component, so we could just define the action of C to be this scaling?

that seems to give a vector space structure

but I’m not sure why the dimension is at most n

[spice whale]

i got that the lie algebra is made of matrices of the form
x 0
0 y
and of the form
0 z
0 0
but I'm confused because these don't constitute a space closed under addition

[chilly ocean]

How did you go about computing the Lie algebra in the first place?

How did you even get something that isn't a vector space?

[hot lake]

G looks 2-dimensional so its Lie algebra should be two-dimensional too

[cloud walrus]

rakko

[chilly ocean]

@spice whale

[spice whale]

hm

[spice whale]

yeah i was doing something weird with like saying D(A*A²) = D(A*)A² + A*D(A)A + A*AD(A)

[chilly ocean]

[spice whale]

yeah no clue either

[spice whale]

my main problem is trying to prove the converse

or related

[chilly ocean]

Have you never seen a differential equation before?

[spice whale]

like i got that DA(0) = a b
0 2a

[spice whale]

I'm not sure what to set up

...

[chilly ocean]

a(t) = 1 + at works. So does a(t) = exp(at).

[spice whale]

how did you get these

[chilly ocean]

I stared at it.

A more systematic way to do it is with the matrix exponential.

[spice whale]

I'm confused as to what you're doing

in general

[chilly ocean]

I take a matrix $$M = \begin{pmatrix} a & b \ 0 & 2a \end{pmatrix}$$ and I want to show that it is of the form $A'(0)$ for some curve $$A(t) = \begin{pmatrix} a(t) & b(t) \ 0 & a(t)^2 \end{pmatrix}$$ in $G$.

[cloud walrus]

rakko

[spice whale]

ok

that makes sense thanks

[chilly ocean]

In this case, the differential equations you need to solve are relatively simple, so you can just write down A(t) that works.

[spice whale]

yeah he does this with O_n

i think

[chilly ocean]

Ignore that message. That only works if you already know the Lie algebra is that.

Let me think for a moment on the matrix exponential thing. I am not so sure anymore.

Ok, yeah, I just needed a sanity check. The matrix exponential absolutely works here.

how undelete message?

Generally if G is a Lie group and H is a Lie subgroup, the one-parameter subgroups of H are those of G whose initial velocities lie in the Lie algebra of H.

Taking a(t) = 1 + at works, but then you won't get a one-parameter subgroup for A(t). (This isn't required, but it is nice.)

[chilly ocean]

can't

[tribal furnace]

ask a mod

[glossy crag]

To show that Z has no composition series, is it enough to say "Z has no simple subgroups" or am I missing something?

[void cosmos]

suppose it does

every subgroup of Z is normal anyways

so suppose 0= a_0Z subs a_1Z subs... subs a_nZ = Z such that ai+1Z/aiZ is simple for any i <n

[glossy crag]

Thanks, I know the rest.

I just wanted to double-check (if there were a composition series, it'd have to end with a simple subgroup, but there are none in Z).

[void cosmos]

ocool

what

[glossy crag]

If you have a composition series, the penultimate element G_1 in the chain has to be a simple group, since the quotient G_1/G_0=G_1/0=G_1 is assumed to be simple.

Every non-trivial subgroup aZ of Z has the non-trivial subgroup a^2Z. Isn't that enough?

[void cosmos]

its that ai+1Z/aiZ is of prime order --> a group is cyclic so if d|order of group then g^d is a subgroup of order n/d where g is a generator

[glossy crag]

I don't know what you're talking about, what do subgroups of Z/nZ have to do with this? Yes, Z/pZ is the only simple cyclic group, I'm talking of something else.

If 0=G_0<=G_1<=...<=G_n=G is a composition series, then G_1 has to be simple. Z has no simple non-zero subgroups, the end.

[void cosmos]

it being a composition series means Gs are normal subgroups of Z such that Gi+1/Gi is simple

G1 has to be normal

and it is

are you confusing that the G_0s themselves have to be simple?

Gi+1/Gi are not subgroups of Z

@glossy crag

[glossy crag]

Man, what are you babbling about, I don't understand.

Forget the normality condition for a moment, Z is abelian so everything's normal.

[void cosmos]

define a composition series for me please

[glossy crag]

A finite tower of subgroups G_i, each normal in the next one (something that loses meaning in Z), such that the consecutive factor groups are simple (or such that every G_i is maximal in its successor, whichever).

[void cosmos]

now

how does what your saying prove your statement

[glossy crag]

From this definition follows that if 0<=G_1<=...<=G_{n-1}<=G is a composition series, then G_1 is a simple group.

Explain to me how that's wrong.

[void cosmos]

0 = G1 <=...<=G^n=G

[glossy crag]

Yes. Therefore G_1 is necessarily simple. Anything wrong there?

No, wait

[void cosmos]

G1 is the trivial group

[glossy crag]

0=G_0<=G_1<=...

[void cosmos]

or G_0 yea whatever

[glossy crag]

G_1 is simple, yes?

[void cosmos]

G1/G0 is

[glossy crag]

....

G_0 is 0, so G_1/G_0 is isomorphic to G_1

Hell, you could even say they're equal

[lethal dune]

Having a composition series is equivalent to saying it's Arti and Noeth. Is Z Arti?

[glossy crag]

I know that much, I was talking about an "elementary" proof, not involving chain conditions.

[void cosmos]

it being a composition series also says each subgroup is a maximal proper normal subgroup of the next

[lethal dune]

well I believe it does imply then

[void cosmos]

Xi+1/Xi for 0<=i<n is simple

if your tower is of length n

[glossy crag]

Yes, and? I'm saying G_1 is simple, you're telling me I'm wrong and replying with something irrelevant.

[void cosmos]

yes

mb

how do you intent to show Z has no simple subgroups

[glossy crag]

Because if aZ is a non-zero subgroup then it has the non-trivial subgroup a^2Z.

[void cosmos]

and?

[glossy crag]

And that makes aZ not simple. Every subgroup of Z is of this form, what else is there to say?

[void cosmos]

yea

okay cool

this should work ig

idk i just thought of contradicting it havingf a composition series by orders

[glossy crag]

Man, you wasted like 15 minutes of my time with this...

[void cosmos]

of the factors themself

[south patrol]

Z Is not simple and any nontrivial subgroup is isomorphic to Z

[glossy crag]

Or that, whichever.

[south patrol]

indeed

[void cosmos]

yea brain fart

mb

[glossy crag]

Happens to the best of us.

[lethal lava]

Still haven’t figured it out—is what I said on the right track?

[eager willow]

yeah seemed to be on the right track I think you have the correct action in mind, so now try to think if you can come up with an n-dimensional complex vector space that these vector spaces are sitting inside. That guarantees the dimension is \le n

[long nebula]

isn't this just wrong?

Z and Z/6Z for example

[elder wave]

Yeah

It’s true for a PIR

[long nebula]

okay I'll email my prof

ty!

[hollow parrot]

Is one able to restrict the direct product theorem, if H, K normal to G, H intersect K = {e} then HK isom H x K to...if HK is a subgroup of G, H and K normal to HK and H intersect K = {e} then HK isom H x K?

The proof we use seems valid for both cases but I'm unsure

[jovial kelp]

hi there! Would anyone have any advice on how to prove this surjectivity? Also, does the fact that A is an integral domain have any weight on part (a)?

how I'm reading that is that the mapping takes a polynomial in A and maps it to the constant coefficient

[chilly ocean]

Indeed.

So if you're given a in A, what polynomial can you pick that gets mapped to a?

[jovial kelp]

any of them right?

as long as the const coeffiicent is a

[chilly ocean]

As long as the constant coefficient is a, yes.

The constant polynomial whose only term is a works.

I suggest you work through the exercise and see if you use the fact that A is an integral domain or not.

[jovial kelp]

sounds good!

[chilly ocean]

A hint for part b, if you need it, is ||use the answer to part (a)||.

It's pretty much a one liner if you know the right theorem.

[jovial kelp]

okay so about that

for (a), the ker(h) would just be when our constant coefficients are 0

[chilly ocean]

Right.

[jovial kelp]

so something like x^2 + 2x would be in there

[chilly ocean]

Mmhm.

Can you write down the kernel in any nice way?

[void cosmos]

try the isomorphism phi:HxK -->HK; phi(h,k) = hk

[chilly ocean]

Here's a hint: ||if the constant term is zero, you can factor out x. What ideal does this look like?||

[jovial kelp]

Sure let me try! $ker(h) = {x \in A[x] : h(x) = 0}$?

oh wait

haha that doesn't help

[chilly ocean]

x in A[x]?

x isn't an arbitrary element of A[x], so the set on the right doesn't make a lot of sense.

It is true that x is in the kernel of h, though.

[hollow parrot]

This is what we used to prove the original theorem, which to me, appears unchanged if u restrict to the subgroup HK

[void cosmos]

try to prove it

post it here

i will tell you if there is any mistake ( if i can haha )

[chilly ocean]

One way to write "p(x) is in the kernel of h" is that p(x) can be written as x times something.

[jovial kelp]

so simply x * 0?

oh no wait

p(x)h(0)?

[chilly ocean]

Nope.

I'll write it out in detail.

[jovial kelp]

okay thank you (._.)

[chilly ocean]

If p(x) is in the kernel of h, then p(x) has no constant term. This means that we can factor x out; this, itself, means precisely that there exists a polynomial q(x) such that p(x) = xq(x).

Does this remind you of anything?

Any certain kind of ideal, perhaps?

[jovial kelp]

oh I think so! give me a sec to look over my notes

I know I've seen that before

[chilly ocean]

👀

[jovial kelp]

it's prob pretty obvious but I want to make sure

I want to say <x> here, but that seems wrong immediately

[chilly ocean]

That is exactly it. <x> is the set of all polynomials of the form xq(x) for some q(x).

[jovial kelp]

oh

pfft haha

[chilly ocean]

So we at least know that the kernel of h is contained in the ideal <x>. Is it equal to <x>?

(What's the constant term of an element of <x>?)

[jovial kelp]

oh right

yeah it would be

since the constant term of x here is 0

[chilly ocean]

Works for me.

So the kernel of h is exactly the ideal <x>.

[jovial kelp]

ahhh, so then we know that ker(h) = <x>

so then for (b), we know that from the 1st isomorphism theorem (I think that's what it's called)? that A[x]/ker(h) is congruent to A

and ker(h) = <x>

oh wait that's about groups. But I think there's a matching one for rings too

[chilly ocean]

There is.

The first isomorphism theorem in this case will say that A[x] / ker(h) is (ring) isomorphic to im(h).

What happens when you use your answer to the first part?

[jovial kelp]

hmmmm

just to double check im(h) is the image of h?

[chilly ocean]

Yeah.

[jovial kelp]

Since h is a surjective ring homomorphism, then im(h) = A?

[chilly ocean]

Exactly.

[jovial kelp]

whoop whoop!

[chilly ocean]

So you have the answer to the second part.

[jovial kelp]

yep that makes total sense! And since ker(h) is an ideal, then it has that absorption property, so we use that to "factor out" an x term and show that <x> is our kernel

just one last question if you don't mind

[chilly ocean]

I am not particularly sure what this means, but I feel confident that you know how to prove that ker(h) = <x> so I won't press on it.

(If I asked you to give an airtight proof, you could.)

[jovial kelp]

fair enough. I get the idea, I think I just need to word it better haha

yeah I thin kso

I'm having a little trouble starting off the surjective proof. So normally you would take an element from your co-domain and show that something in the domain maps to it

but like with polynomials I'm not sure how to word that, if you know what I mean

[chilly ocean]

Just do this.

[jovial kelp]

like we could take the constant polynomials, that maps to itself

[chilly ocean]

That's all you need to do.

[jovial kelp]

oh and that's enough to show it's surjective? Oh yeah I guess it would

since all of the codomain would be accounted for then

[chilly ocean]

Right.

[jovial kelp]

awesome, I appreciate your help! I'm going to write this up real quick 🙂

[chilly ocean]

Given a in A, you only need to find one p(x) with h(p(x)) = a.

Just one is good enough for surjectivity.

[jovial kelp]

oh right!

[chilly ocean]

Also, you were worried about whether or not you needed A to be an integral domain here. Did you use this fact at all?

[jovial kelp]

nope not at all haha

[chilly ocean]

Yup.

[jovial kelp]

just a red herring 😛

[south patrol]

Indeed, though I suppose if the question asked for evaluation at another value a besides 0 you would need it to be an integral domain to show that A[x]/(x-a) cong A under this

[chilly ocean]

All rings are integral domains anyways.

[jovial kelp]

then my marriage is an integral domain too, since we used a ring

[south patrol]

Lol nice

Ringed spaces were invented when Grothendieck listened to Beyonce

He liked it so he put a ring on it

[jovial kelp]

https://tenor.com/view/ring-beyonce-singleladies-putaringonit-dance-gif-4724560

[south patrol]

lol

[rotund aurora]

does someone know where the term "ring" comes from? its actually so random

[jovial kelp]

I believe it comes from a German word

[rotund aurora]

https://math.stackexchange.com/questions/61497/why-are-rings-called-rings#:~:text=The name "ring" is derived,certain rings of algebraic integers.

[delicate bloom]

I'm probably wrong but I think the german word for integers is 'zahlenring' so it sort of just came from shortening this

[jovial kelp]

hold on I looked this up the other day

and according to an online translator, "zahlenring" means "number ring", roughly translated

same energy

aaah

[elder wave]

The german word would be „ganze Zahlen“

[toxic zephyr]

in the dihedral group with 6 elts (D3 or D6 whatever), the set {x^0,x^1,...,x^5} would contain the set {x^-1,x^-2,...,x^-5} right?
would {x^0,x^1,...,x^5}={x^-3,x^-2,...,x^2,x^3}?

as in... x is a variable of an element in D6. would those sets be the same/contain each other for all x in the group

[south patrol]

Yes, what you've written will just be the subgroup generated by x

Since the order of x is <= 6

[toxic zephyr]

just want to make sure i'm understanding. and since the order is 6 we have that {x^0,x^1,...,x^5}={x^-3,x^-2,...,x^2,x^3}? and because the group is finite the subgroup generated by x is the same as the subgroup generated by x inverse?

[south patrol]

Ye

Well

Last bit doesn't need finiteness

[toxic zephyr]

okay cool thank you. i'm coding a program for a research project related to word maps on dihedral group, and it would be easier to consider word maps that are products of the form x^kc where k is nonnegative and < the order of the group instead of -order<k<+order. feels wrong to just completely disregard inverse terms but since any power of the inverse is a nonnegative power of the elt i guess it's okay

[maiden heath]

Is this correct?

[cloud walrus]

Volkenborn

[ionic spade]

I'm studying the proof to the above theorem, but the book I'm reading states that the reverse direction (M noetherian + finitely generated ---> R is noetherian) is obvious.

I'm not sure how this is evident. Maybe it's because you can take some ideal $\mathcal{I}\subseteq R$, and since [\mathcal{I} M={\sum_{i=1}^nr_uv_i~|~r_i\in \mathcal{I}, v_i\in M}] is a submodule of $M$, then maybe you can make some kind of ACC with the different ideals $\mathcal{I}_k$?

[cloud walrus]

Volkenborn

[ionic spade]

I'm probably overthinking this, why is this super simple?

[elder wave]

R is a finitely generated R module

[ionic spade]

right but not every finitely generated R module is R

or isomorphic to it

[elder wave]

You’re assuming that every finitely generated R-Module is noetherian or not

[ionic spade]

oh wait

LOL

[elder wave]

R is one of these finitely generated R modules and then you have that submodules correspond to ideals etc

[ionic spade]

I'm a complete idiot, thanks @elder wave

[elder wave]

You’re welcome

It happens

[civic tapir]

I'm told that it is clear that the free abelian group generated by the columns of the first pic mod the free abelian group generated by the columns of the second pic is equal to Z/nZ. Why is this clear?

[pastel cliff]

\claim $R$ is isomorphic to the localization of $\Z$ at the set $\Z \sans\ang{2}$

[cloud walrus]

sebbb

[pastel cliff]

i posted this a while ago, it's obvious by defn that they're the same but how do you do this "formally"

[south patrol]

what is R

[pastel cliff]

oop forgor to post that, it's subring of Q where all elements have odd denominator

[south patrol]

Oh okay

[pastel cliff]

like they're literally the same by defn no

[south patrol]

Hm I think they probably are under most constructions of Q lol but uh

You can just manually define a map between the two in the "obvious" way anyway

But yes the proof is not very enlightening lol

[pastel cliff]

this being a map from R -> Z \times Z\ (2)

i wonder if this is stupid but is there any connection between group actions and localization? group actions could be defined as special maps from products into a group and localization is kinda the opposite, being defined as a map from a ring into a product that includess that ring

[chilly ocean]

I think you mean Z/(2)

[pastel cliff]

yeah

odd integers

[chilly ocean]

ah

[pastel cliff]

goober question - how is localizing "adding inverses"? we're only doing doing that for a subset of a ring when we localize it right?

if that's a dumb question then what's a better question to ask

[next obsidian]

You’ve added inverses for a subset

[pastel cliff]

connecting this to local rings now then, if im understanding my notes correctly, localizing a ring R at the complement of a prime ideal gives a local ring (ie one with a unique maximal ideal)?

[lethal lava]

I haven’t been able to think of one, I feel like I’m missing something here

[chilly ocean]

Are there any useful techniques or theorems for proving a ring is integral over another? I'm interested in the case of a quotient and its base ring.

[weak flower]

An inverse limit of graded vector spaces/algebras is formed degree-wise. On the other hand, graded vector spaces are just comodules over certain monoid bigebrae.

So are there any sort of general description of inverse limits of comodules over co-/bigebrae, or conditions under which they exist? (It not obvious even in the case of infinite dimensional cogebrae over a field that they should exist)

[cloud walrus]

koaaa

[royal matrix]

yes

[little root]

how is k[x, x^{-1}] usually defined? it seems like you can define it as k[x, y] / (xy - 1), or as the image of the evaluation map k[x, y] -> k(x) that sends f |-> f(x, x^{-1}), or as the localization k[x] at {1, x, x^2, ...}. these are all equivalent, but is any of these more convenient? also, is there an obvious way to see that these are equivalent?

[rustic crown]

ig this most commonly appears as a localization, so that's most natural in some sense?

you think of k[x] as nice functions on a line

and then k[x, x^-1] are nice functions on the line without the origin

and what you're doing is more or less the "obvious" way to prove they are equivalent

[edgy plume]

I have a question, say K and N are groups. is the intersection of K and N also a group? and why

[rustic crown]

i.e. construct a simple map in one direction and show it's a bijection

[rustic crown]

are K and N arbitrary groups? or subgroups of some fixed group?

(in the former case, talking about intersection is kinda weird, and you shouldn't expect it to behave nicely...)

[edgy plume]

oh sorry, they are both subgroups of another group, say G

thus, same binary operation

[rustic crown]

yep, in this case it will form a subgroup.

try thinking about what all things you need to check

all of them should follow quite easily from the fact that both N and K are subgroups

[edgy plume]

ohhhh that's a nice insight, thank you! i will try to da a subgroup test on it

[jovial flint]

In the definiton of D, are m,n and r arbitrary?

[rustic crown]

yep

(btw you have those relations for each choice of m, n, r and you're going to quotient by all such elements not some specific m, n, r)

[jovial flint]

what do you mean by this?

oh you mean these?

[rustic crown]

i mean the module D is generated by all choices of elements, m, n, m', n', r in M x N x M x N x R

[jovial flint]

ah I see, thanks

[rustic crown]

i felt the word "arbitrary" could mean like you're choosing specific elements

just wanted to say that explicity

[jovial flint]

also, in this case wouldnt a general coset look like r\delta(m,n)+D?

[rustic crown]

nope

a general coset would be a finite sum

so would look like

[cloud walrus]

det

[rustic crown]

under very special cases this finite sum would collapse into a single term

[jovial flint]

ohhh

I get it !

thank you!

[rustic crown]

for example, if one of your module is a quotient of R by some ideal

[cloud walrus]

det

[jovial flint]

I still havent really parsed the idea of tensor products

will try to understand that after i finish with this proof

[rustic crown]

yea, the construction of the tensor product is really weird

and you would never use the construction. only thing it tells you is that it exists

the real use of tensor product is its universal property

[jovial flint]

I see

[rustic crown]

which says that any R-bilinear map M x N --> P can be thought of as an R-linear map M ⊗_R N --> P

so dealing with these harder maps, is same as dealing with simpler maps but from a complicated source

[jovial flint]

thats a nice way of thinking about it

[rustic crown]

[formal ermine]

what's $\Set{\frac{f}{g} | f, g \in \bK[x], g \neq 0} = \bK(x)$ called in english

[cloud walrus]

yes yes yes no
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[rustic crown]

fraction field of K[x], field of rational functions over K (in the variable x), etc

[formal ermine]

ah ok thanks!

is there anything special to it? like why does it have its own "symbol"?

[rustic crown]

the notation is quite useful in field theory ig

say you have two fields K contained in L, and a bunch of elements alpha_1, ..., alpha_n inside L

then K[alpha_1, ..., alpha_n] is the smallest subring of L, containing K and those elements, while K(alpha_1, ..., alpha_n) is the smallest subfield of L containing K and those elements

[formal ermine]

ah

[agile swan]

Hi I have written my problem

[chilly ocean]

this doesn't look like abstract algebra

[agile swan]

I don’t know where to post

😅

[chilly ocean]

looks like some kind of algorithm so maybe in #numerical-analysis

[tender bough]

lecture notes have this basis for special lie algebra sl(2)

why does the E disappear here?

[sweet echo]

It should say (λ+2)Ev

The point is Ev is also an eigenvector of H

[tender bough]

ahh, that makes much sense, given the context below

thanks!

[glossy crag]

A fellow KC chad, I see...

[jovial flint]

yes

i do not have the mental fortitude to read dummit and foote

[glossy crag]

I found them pretty readable tbh, tho nothing beats KC. I pray that he releases a proper textbook in my lifetime, preferably on algebraic NT (or w/e the fuck else, I'd read anything by him).

[jovial flint]

dummit and foote is too verbose in my experiemce

takes some time to sit down and read

[glossy crag]

That's true, but they're also pretty thorough and have lots of examples.

[jovial flint]

yes

[glossy crag]

For more advanced stuff, check out Isaacs' "Algebra", I think if you like KC, you'll like him, he's an incredibly efficient expositor.

[south patrol]

Love KC

Look up a random fact and find a 20 page handout on it

[glossy crag]

The man is a godsend to anyone studying algebra. I wonder, are there equivalent personages in other fields (e.g. analysis or probability theory)?

[south patrol]

Ye idk

[glossy crag]

On that note, reading about Krull-Schmidt, can anyone explain this last step to me, I'm blanking? How does $\prod_{i=1}^nK_i=\prod_{j=1}^mK_j$ imply n=m?

[south patrol]

Tao feels similar for those sometimes w his blog

But not rly

[glossy crag]

Nvm, I think I got it. There's a previous characterisation of internal direct products, wherein directness is equivalent to $K_i\cap\prod_{j=1}^{i-1}K_j=1$ for all i>=2, so that would imply K_j is 1 for j>n.

[restive birch]

A few questions:
How would one prove that Q and R under addition are not isomorphic?
How does multiplication or addition work on R^2?

[glossy crag]

One is countable, the other isn't, there can be no bijection between them.
Coordinate-wise.

[south patrol]

Well, how multiplication works on R^2 depends on context XD

[sweet echo]

Are chains and cochains just the same thing but different notation/relabelling? I don't yet see the point in distinguishing between the two

[glossy crag]

Pretty much, it's strictly cosmetic. There are situations where degree goes up (e.g. differential forms), so it's more convenient to use cochains than flip everything.

[restive birch]

ah, so orders of infinity.
what does coordinate-wise mean? (x,y) * (a,b) = (xa, yb)?

[sweet echo]

For example, if we had a projective resolution P2 --> P1 --> P0 --> M and we applied Hom(-,D), its more convenient to call the result a cochain complex?

Just because of how the indices work out

[glossy crag]

If that's what you mean by R^2 (direct product of R with R), then yes. All depends on your implicit meaning of "R^2".

[glossy crag]

Was Hom(-,D) the contravariant one? If so, then yes.

[sweet echo]

Yeah contra, the point being people decided to give cochains their own name and notation bc that form of chain came up enough to warrant doing so?

[glossy crag]

Exactly.

[sweet echo]

Great thank you

[tender bough]

help, I super don't get this...mostly the remark in the second screenshot. How is the Lie bracket taken? The lie bracket is alternating, so F^2 = [F, F] will have to be zero, and F^3 = [F, [F, F]] = [F, 0] = F0 - 0F is zero again, and every power thereafter is zero?

btw are you greek? I like how you just typed lambda

[sweet echo]

Are you asking what the explicit lie bracket on matrices is?

I just have a Greek keyboard on my phone haha

Or are you asking about F^n

Which yeah I would assume is nested lie brackets

[tender bough]

but how are they not all zero for n >= 2?

[sweet echo]

Hmmm

[rustic crown]

i don't really understand what they're saying in the blue underline. but F^n(v) is definitely application of F, n times on v. so F(F(...F(v)...))

that's indeed what the next lemma says

v^(n+1) is F applied on v^(n)

maybe, when they say squares of F-matrices are 0, they're talking about [F, F] = 0 and not F*F

but it's definitely phrased in a very confusing way

and everything looks good to me in the first image

because E increases the eigenvalue

and you can't have infinitely many distinct eigenvalues

so if you keep applying E many time, you'll eventually find some v_lambda which becomes 0 on application of E

[hot lake]

I think remark 5.3 is saying that F(F(v lambda)) can be nonzero 0, despite F² and [F,F] being both 0 and that can seem surprising. The axioms of lie algebra representation require that A(B(v)) - B(A(v)) = A,B, and you can't deduce F(F(v)) = 0 from them. If you try to apply that to F and F you get 0 = 0 instead

[tender bough]

I may have left out crucial context, this is talking about representation

[tender bough]

yeah I thought that

[hot lake]

if you were to misread it and think it's a group representation then you would get F(F(v)) = 0 (despite the algebra not being a group ?)

[tender bough]

here is with more context

[hot lake]

so it's saying to not jump from [F,F] = 0 or F²=0 to F(F(v)) = 0 because that would be wrong

[tender bough]

where the V refers to, I think this V in definition 2.24

[hot lake]

yeah you want [phi(F),phi(F)] = phi([F,F]) = phi(0) = 0 and not phi(F)² = phi(F²) = 0

as in you take lie brackets and not matrix multiplication

[rustic crown]

wait what does F^2 mean if not F o F?

[tender bough]

so when they wrote down H v_lambda = lambda v_lambda in equation (4), they actually mean phi(H)(v_lambda) = lambda v_lambda???

[hot lake]

it means phi(F)² because it's F applied to the representation

but phi(F)² isn't related to phi(F²)

so it's not 0

despite F² being 0

[hot lake]

yes

[tender bough]

😵‍💫

[rustic crown]

oh i see. i was thinking of F as sitting inside the universal enveloping algebra

that totally makes sense now

because like the matrix of F in the lie algebra is like

[1 0]```

[hot lake]

yes

in general there is no meaning to multiplying together elements of lie algebras

so if you even think F²=0 you are going the wrong way

[rustic crown]

yea, i remember using the notation f for the element in the lie algebra and F for the thing in the universal enveloping algebra

[rustic crown]

oh they do say 2 x 2 matrix, i didn't read that >.<
my bad

[hot lake]

ah yeah that rules out the enveloping algebra

[tender bough]

thanks guys, that helps a ton. Sometimes I wish I could just call my instructor anytime and ask stupid questions.

[restive birch]

just to double check my work/understanding, if $f : A\timesB \rightarrow B\timesA$ is defined by $f(a,b) \rightarrow (b,a)$, then $f$ would be an isomorphism, and therefore $A\timesB$ and $B\timesA$ are isomorphic?

uhh... bot?

[rustic crown]

yeep

[formal ermine]

yeah

[rustic crown]

texit is sad >.<

[formal ermine]

texit is down

[chilly ocean]

This is simple enough that you shouldn't need to write it in LaTeX.

Not everything needs to be TeX'd.

[tender bough]

,tex $\text{hello}$

[pastel cliff]

#changelog

[restive birch]

how would i go about either of these problems? this stuff is breaking my brain

[chilly ocean]

Checking that something is a subgroup is routine.

You should have a list of properties to check it against.

[elder wave]

Second part is iso theorem

[chilly ocean]

Come on.

That's way overkill here.

[restive birch]

the book hasnt even talked about what subgroups are

[chilly ocean]

I genuinely doubt that.

[restive birch]

subgroups are chapter 2

maybe it mentioned it in passing and i missed it

[elder wave]

True

lol

[restive birch]

its probably intuitive tho

[tender bough]

in this case you just need to check that the operation is closed

[hot lake]

what's exercise 26 of section 1

[restive birch]

just a group on the same operation, that is contained in the bigger group?

[rustic crown]

(sorry couldn't resist :p)

[tender bough]

hello, I can't read

[elder wave]

I thought about sending it as well

light mode

[restive birch]

serves you right

[tribal furnace]

that's exactly what a subgroup is

[restive birch]

alright, that makes sense

[chilly ocean]

I think they should learn what a subgroup is from their own book instead of having everyone come out of the woodworks to try and get their second of fame.

[hot lake]

maybe it's explained in exercise 26 of section 1

[elder wave]

You can't take my second of fame from me like that

[formal ermine]

@restive birch a subgroup is a subset of the entire group with the following properties:

• the neutral element is in the subgroup
• the inverse of every element of the subgroup is also in the subgroup
• if a and b are in the subgroup then ab is in the subgroup

[little root]

subgroups are indeed defined in exercise 26 of section 1

[restive birch]

yep, my bad

[chilly ocean]

Why are you reading their book for them?

[restive birch]

that wasnt my main question tho, i was asking about the question as a whole

i can guess what a subgroup is, but i still dont understand how to solve the question

[little root]

do you have an idea of proving how phi(G) is a subgroup of H?

[restive birch]

i can try, hold on

[little root]

if you have a formal definition of both subgroup and homomorphism, then it should just be applying the definitions

[restive birch]

ah, yep, makes sense- and if my understanding is right here, then if phi is an isomorphism, phi(G) would not only be a subgroup of H, but would be equivalent to H? or am i missing something?

[formal ermine]

only if phi is surjective

phi(G) is yet another notation for the image of phi

[restive birch]

but if phi is an isomorphism, it must be surjective

right?

[formal ermine]

oh yeah I apparently didn't see that part in your message lol

yeah

an isomorphism is a bijection

[little root]

yep! a neat way of saying this is phi is surjective <==> the image is H. you also have an analogous fact that phi is injective <==> the kernel is trivial.

the first fact holds for all functions (not just in groups) but the second fact requires the algebraic structure. you may recognize it from linear algebra

[restive birch]

the extent of my linear algebra knowledge is exactly 7 3b1b videos

[little root]

haha fair. it is an extremely useful criterion for checking injectivity

if 3b1b talks about rank nullity that's one way to think of it (i can't remember if he does)

[restive birch]

im in multivariate in school right now, so we'll be learning a bit of linear algebra soon, hopefully that'll help

[little root]

i would recommend learning abstract linear algebra before abstract algebra, it's like a warm up to a lot of the stuff you'll see in abstract algebra

in school you'll probably see matrices and stuff, but theres a way to do linear algebra without matrices or coordinates and stuff

[formal ermine]

phi is injective <==> the kernel is trivial
this is a good exercise imo

[restive birch]

remind me, what does trivial mean here?

[formal ermine]

it's { e }

only the neutral element

[little root]

(a good linear algebra book imo is axler's linear algebra done right)

[restive birch]

i would, but im really enjoying reading through this textbook, and taking a 250-page detour in linear algebra first is a hard sell

[formal ermine]

you can totally do abstract algebra without full linear algebra knowledge

watching 3b1b's series is enough imho

[restive birch]

i should finish that

i should probably also understand why the calculations for 2x2 determinants work

[formal ermine]

it's just basic geometry

[restive birch]

i remember he showed a diagram at one point, but i was way too tired at the time to pause and work it out

[little root]

sure, it's up to you. the stuff i'm referring to is less like matrices/determinants/gaussian elimination, but more like vector spaces (a specific example of a group), linear transformations (a specific type of group homomorphism), subspaces (a specific type of subgroup), etc. it's just a suggestion

[restive birch]

i can go back and do that now

[formal ermine]

isn't a vector space more like a module

[little root]

yeah well but it's also an abelian group :)

[restive birch]

i think i have a relatively decent understanding of vector spaces/linear transformations from the 3b1b series

[little root]

but yeah it's a module. i would definitely recommend learning linear algebra before modules

[tender bough]

@hot lake @rustic crown why is V necessarily equal to the span of these vectors? Is it possible that F^k = 0 for some k < dim V?

[rustic crown]

its by irreducibility, the span is stable under the action E, F and H

wait i don't see any irreducibility assumption in that image

[hot lake]

from irreducibility, the span of v lambda by every possible chain of E F H has to be the whole space

[tender bough]

@rustic crown irreducibility is mentioned in the next page, maybe it's a typo here

We call V an irreducible finite-dimensional representation of sl(2) if it is generated by a highest weight vector v_lambda, that is, v = span {v_lambda ^(n) | n >= 0}

[hot lake]

ah they make it into a definition of irreducible ?

[rustic crown]

weird

[hot lake]

irreducible means there is no smaller space that is still a representation (stable by the actions of E F H)

the space generated by v lambda and all possible applications of E F H is one

so irreducible means it is equal to the whole space

and in this case

H can't make anything new

and E only goes backwards

so F is the only one making any progress

[tender bough]

at this point I should just send the lecture notes 😂

[rustic crown]

without the assumption of irreducibility that statement is false

[tender bough]

it's on page 17

[rustic crown]

you can look at the representation V_1 ⊕ V_2 say

this is generated by v_{-1}, v_{1}; v_{-2}, v_{0}, v_{2}

the highest weight vector is v_{2} but the span we're looking at only picks up V_2

[restive birch]

Im having trouble with the first part of question 14 here. any tips?

[chilly ocean]

Is x^2+y^2-1 irreducible in C[x,y] ?

[little root]

yes, you can try Eisenstein's criterion by viewing x^2 + y^2 - 1 as a polynomial in x with coefficients in C[y] (which is a UFD)

[pastel cliff]

mightve asked this before but how are UFD's related to local rings/localization

[restive birch]

is it true that phi(k * g^-1) under G is equal to phi(k) * phi(g)^-1 under H, if phi is a homomorphism from G → H?

[rustic crown]

i don't think they have any serious relation

[chilly ocean]

Yes.

phi(kg^{-1}) = phi(k)phi(g)^{-1}.

I don't know what the "under G" and "under H" parts mean, though.

[restive birch]

as in, the operation * being done to the different elements is the operation of G in the first section and that of H in the second

[chilly ocean]

That was clear from the context.

Ok.

[restive birch]

im sorry i put 16 extra characters

[chilly ocean]

[rustic crown]

[chilly ocean]

I was going to say that you don't need to specify that a group operation is being done if it's the only possible thing it could mean in this context, and that when people do so they usually don't say "under G" but rather "where * means the operation of G".

I literally just didn't know what you meant. I'm sorry if I came off as aggressive.

[restive birch]

sorry, mustve misread the tone

the '.' throws me off

[chilly ocean]

Most people seem to here.

You guys are scared of punctuation?

[restive birch]

its just kind of convention ig

[rustic crown]

me is >.<

[restive birch]

given that its harder to convey tone through text alone, different punctuation has taken on different meaning

[chilly ocean]

Yes, and a simple direct argument is possible. Assuming otherwise, you can easily reduce to the case where x^2 + y^2 - 1 = (ax + by + c)(dx + ey + f). Comparing the coefficients of x^2, y^2, and the constant term, you get d = 1/a, e = 1/b, and f = -1/c. Comparing the coefficients of xy, x, and y, it follows from that that a^2 + b^2 = 0, a^2 = c^2, and b^2 = c^2. So a = b = c = 0, a contradiction.

Eisenstein's criterion does indeed give a faster solution, but here's in case you'd prefer a lower tech solution for whatever reason.

[rustic crown]

ig we can also use the fact that monic polynomials with deg =2, 3 are irred in D[x] if and only if they don't have a root. here D is any integral domain i believe. which is probably abstracting out the work you just did

[next obsidian]

They aren’t super related, but here’s a super surprising fact which does involve both

A_S is what you might have seen as S^-1A before

[chilly ocean]

Hair

[muted epoch]

can anyone explain this to me in a non-formal way?

I don't understand what it's saying

[chilly ocean]

A polynomial acts on V by letting x act by your operator.

Write down a few small examples and it should be clearer.

For example, 1 + 2x is the operator I + 2T.

The map sends a polynomial to the associated operator.

[muted epoch]

hmmm this makes a little more sense now that you explain it like that

[chilly ocean]

So phi_T(1 + 2x) = I + 2T, in symbols.

phi_T(1 + x^3 + 12312312x^4) = I + T^3 + 12312312T^4.

etc.

[muted epoch]

how would this be an endomorphism?

oh wait sorry

[chilly ocean]

How is phi_T(polynomial) an endomorphism?

[muted epoch]

Yes

I thought I understood but I'm struggling to conceptualize this

[chilly ocean]

It's a linear combination of I, T, T^2, and so on.

Each is an endomorphism.

[muted epoch]

ohhhh okay

I don't think I have a very good 'math' brain

I can't read these textbooks :/

group theory is by far the hardest math challenge I've faced so far. Idk how yall do it

[chilly ocean]

So $\varphi_T(a_0 + a_1x + a_2x^2 + \cdots + a_nx^n)$ would be $$a_0I + a_1T + a_2T^2 + \cdots + a_nT^n \in \mathrm{End}_K(V).$$

that looks awful

[cloud walrus]

rakko

[chilly ocean]

It takes time.

[muted epoch]

what's your favorite type of math @chilly ocean ?

[chilly ocean]

Hard question since the runner-ups are so close, but I would say some kind of differential geometry is best for me.

[muted epoch]

I'm thinking of taking differential geometry next semester actually

it seems interesting but I'm worried that since I struggle with abstract alg so much that maybe I shouldn't take it

I'm guessing it's more applied than pure/abstract tho right?

[chilly ocean]

Depends on the type of differential geometry.

It's very much a "pure" subject but has a ton of intersection with "applied" ones.

[muted epoch]

yeah that's why I think it'd be so interesting to take as a class

(or learn about on my own at some point)

[restive birch]

Right there with you. Ive spent nearly 2 weeks reading through just 20 pages of group theory with exercises. shit hurts my brain

[muted epoch]

it's just so abstract like... I don't know where I'm at or what I'm doing lmaooo

[restive birch]

I agree, its kinda hard to orient yourself without examples

[muted epoch]

are you taking a class or self-studying?

[restive birch]

Self

[muted epoch]

impressive

I'm taking a class rn but the professor usually just references the textbook, which has few intuitive examples, so it's a struggle

[restive birch]

Whats the textbook?

[muted epoch]

let me find it

http://homepage.divms.uiowa.edu/~goodman/algebrabook.dir/book.2.6.pdf

Algebra Abstract and Concrete Edition 2.6 by Frederick M. Goodman

[restive birch]

im working through dummit and foote rn

[muted epoch]

I've heard of that one

[restive birch]

Its got some examples, but not many

[muted epoch]

does it have solutions as well?

[restive birch]

To the exercises? Maybe, but i dont think so

[muted epoch]

feel like it'd help me to just see the answer to all the exercises

[restive birch]

I wasnt able to find them anyways

[muted epoch]

oof

[north parrot]

Is Q[x,x^-1] integral over Q[x]? At first i thought so because you can just multiply any element of the former by x^d to eliminate any negative terms, but that would mean your polynomial is not monic....

[eager willow]

there are no nontrivial integral elements in the field of fractions over any integral domain UFD

[proud bear]

i don't think this is true. (-1+sqrt(-3))/2 is integral over Z[sqrt(-3)] for example (root of x^2+x+1)

[eager willow]

oh hmm

[proud bear]

but i think it's still true Q[x,x^-1] is not integral over Q[x]

[eager willow]

well x^{-1} isn't integral I was just trying to think what's the quickest generalization of that kind of thing

ok I think it's that n^-1 isn't integral whenever n is a nonunit in an integral domain?

hmm maybe that's only over UFDs

[pastel cliff]

what book is this

[obsidian sleet]

(DF10.4.25) Let $R$ be a subring of the commutative ring $S$ and let $x$ be an indeterminate over $S$. Prove that $S[x]$ and $S\otimes_R R[x]$ are isomorphic as $S$-algebras.

[cloud walrus]

locally cpt HD Abelian cofe (bad

[obsidian sleet]

i need help on this

so my thoughts so far are i want to show that S[x] satisfies the universal property that the tensor product does so that they're isomorphic

for R-algebras and not just R modules

that is

i want to find a bilinear map from S \times R[x] to S[x]

i think this map should also be surjective

then given any bilinear map from S \times R[x] to some R-algebra L, i should fine a corresponding S-alg map from S[x] to L which that makes thigns commute

[obsidian sleet]

my hunch is that the bilinear map here should be just to send (s,\sum r_ix^i) to \sum sr_ix^i

i dont feel too sure that this will actually span S[x]

its definitely bilinear

[next obsidian]

Elements of the tensor products are sums of these

So spanning is actually easy, a polynomial Sum s_ix^i is mapped to by Sun (s_i (x) x^i)

Anyway, there’s a lot of higher-level ways to view this, and I’m not sure what the most useful way for you is hahaha

Maybe the best way is to just come up with the map like you did

But you can see it functorially too, and prove both of these are the free S-algebra on one generator

Or recognize that tensoring with S is the extension of scalars and see what that univ property is

Etc etc

[obsidian sleet]

oh wait we are supposed to view S\times R[x] as Z module right

in the construction

or no maybe i confused

[next obsidian]

Uhhh

[obsidian sleet]

oog

[next obsidian]

No?

[obsidian sleet]

yeah i think im confusing with something from before pain

[next obsidian]

I mean idk

The map you induce from the tensor product is a priori only linear

You have to see by hand that it’s a ring map

[next obsidian]

Matsumura

[obsidian sleet]

let me drop the universal property im using here for tensor product

[next obsidian]

To induce the map you need to use that it’s R-bilinear

If you’re like doing stuff over Z-you’re only using normal bilinearity which isn’t good enough to induce a map from the tensor product over R

[obsidian sleet]

so here S\times R[x] is an R-module and i have R-bilinear maps out of this into some R-algebra L

[next obsidian]

Sure

But I mean this isn’t like

[obsidian sleet]

and from this i want to find an S-algebra map from S[x] to L

corresponding

to whatever i start with

[next obsidian]

Sorry what are you trying to do?

You already described the map from the tensor product to S[x] right?

[obsidian sleet]

im trying to show that S[x] is isomorphic to S\otimes_R R[x] as S-algebras

by showing that S[x] satisfies the universal property

[next obsidian]

No this won’t be enough

[obsidian sleet]

o

[next obsidian]

this will only ever get an isomorphism as like, modules

[obsidian sleet]

i see

[next obsidian]

I mean you could then examine the map from the tensor product

And see by hand that it’s multiplicative

But that’s painful

You should exhibit a map S[x] -> S (x)_R R[x]

You can do this by viewing the latter as an S-algebra via the S on the left

So you only have to specify where x goes then extend multiplicatively and additively

So say like, sx

Where do you think that would go?

[obsidian sleet]

s (x) 1_Rx maybe

[next obsidian]

Yeah

That’s correct

So in that case where does x go?

[obsidian sleet]

like

"itself" ?

[next obsidian]

Well that doesn’t quite make sense in the tensor product

[obsidian sleet]

oh i see

[next obsidian]

But it like kinda does

[obsidian sleet]

i should write it like 1_S (x) 1_Rx

[next obsidian]

Yeah exactly

So do you see how the tensor product is an S-algebra?

Say we wanted to take an element

Sum s_i (x) f_i(x)

How do we multiply an element s’ into it?

To have it be an S-algebra

Or alternatively: what’s the ring map S -> S (x)_R R[x]

[obsidian sleet]

ok let me see

so i was thinking about this cus i saw how the tensor product of R-algebras is an R-algebra

i was thinking about viewing S as an R-algebra or something

[next obsidian]

Right

That’s how you form this

And get a ring

But to make it an S-algebra

You have to only use the fact that S is here

You extend scalars

Cuz there’s a priori no way to multiply stuf in R by things in S

[obsidian sleet]

so should s' times [Sum s_i (x) f_i(x)] be Sum s's_i (x) f_i(x) ?

[next obsidian]

Yeah!

[obsidian sleet]

in my head s' looks like s' (x) 1_R

[next obsidian]

So can you figure out the ring map S -> S (x)_R R[x]?

That’s the map

[obsidian sleet]

ahhhh

i see

[next obsidian]

s’ -> s’ (x) 1

It’s clearly a ring map

[obsidian sleet]

Yes

[next obsidian]

Cuz it all just happens in the left

Okay so now

Specificfuinf tjat x goes to 1 (x) x

Determines a ring map from S[x]

Because like you send f(x) to f(1 (x) x)

This makes sense because
1: we know how to take powers of 1 (x) x)

[obsidian sleet]

i see

[next obsidian]

2: we know how to interpret the scalars

Now it’ll be important to actually know the map

So write down what Sum s_ix^i goes to

[obsidian sleet]

Sum s_i (x) x^i