groups-rings-fields

[delicate orchid]

to show how different maps compose

the point of commutative diagrams is that it doesn't matter which "path" you take between two points, the functions given by composing along each eage will be equal to one another

for example, in the second one you posted

[ebon gyro]

im following btw my internet cut for a bit

[delicate orchid]

it just says that $\varphi = \pi \circ \overline{\varphi}$

[cloud walrus]

wew ladz

[coral spindle]

It is called a diagram, specifically a commutative diagram. It contains a lot of useful information in one picture. Category theory uses these diagrams heavily.

[delicate orchid]

(and the dashed arrow means that there is exactly one choice of \overline{\varphi} that works)

[barren sierra]

basically it's just notation

rather than write out all of the compositions and stuff

it's a nice visualization

[chilly ocean]

Yeah I can see it is a nice visualization

Makes me want to learn it

[barren sierra]

algebra is very pretty

I highly recommend learning it

[chilly ocean]

Okay thanks for the clarification. I appreciate it.

[round hull]

also seeing and understanding a diagram is different from feeling it

[ebon gyro]

i cant afford to blackbox here since i dont know shit about algebraic groups

[delicate orchid]

all the more reason to blackbox

I never actually learnt what a group is I just blackboxed it

[bleak abyss]

Did I allow you to lose internet?

And fair enough

It's healthy to try and work this out

So I'll take their definition as given

So parabolic subgroup means quotient is complete, solvable + parabolic = Borel

Or eh progress is nice I'll just think about it for GLn

Borel is upper triangular, parabolic is block upper triangular, and unipotent radical is "the diagonal blocks are identity"

[ebon gyro]

this sounds right from the few definitions i do know

[bleak abyss]

Might be better off watching the lecture video tbh

[bleak abyss]

Oh so, I forgot to say one thing about earlier

So yeah GLn

Borel is upper triangular, a parabolic is gonna be some set of block upper triangular matrices. The associated Levi subgroup is just, take the parabolic subgroup and set off diagonal blocks to 0, unipotent radical is about setting the diagonal blocks to the identity

[frigid lark]

I feel like I'm missing something, my response was along the lines of: for the trace we use the independence of characters to show there exists an s in K such that Tr(s) = b in F, not 0, and then we use the fact that Tr is a F linear map, and F is a group. For the norm we just use the fact that if [K : F] = p^n = q, then for a any in F, N(a) = a^q, which is a power of a Frobenius automorphism.

[glossy crag]

The degree is n, not p^n.

[frigid lark]

ah

[glossy crag]

I think you could argue like this @frigid lark: let $K$ have size $q$, $F/K$ be of degree $n$, then $F$ has size $q^n$. The Frobenius automorphism $(x\mapsto x^q)$ is the generator of the Galois group, so $N(x)=x\cdot x^q\cdot x^{q^2}\cdots x^{q^{n-1}}=x^{\sum_{j=0}^{n-1}q^j}=x^{\frac{q^n-1}{q-1}}$, so the kernel of the norm homomorphism consists of the roots of $x^{\frac{q^n-1}{q-1}}-1$. The polynomial $x^{q^{n}-1}-1$ splits over $F$ and has distinct roots, so the polynomial $x^{\frac{q^n-1}{q-1}}-1$ does too, thus the kernel has size $\frac{q^n-1}{q-1}$ and $N(F^\times)\subset K^\times$ has size $\frac{q^n-1}{\frac{q^n-1}{q-1}}=q-1$, therefore $N(F^\times)=K^\times$.

[cloud walrus]

leave_no_norm

[frigid lark]

oh, nice

[frigid lark]

Is the trace argument fine?

[glossy crag]

yes, F/K is separable if and only if Tr\neq0

and a functional V->K is surjective if and only if it's non-zero

[glossy crag]

extensions of finite fields are separable

[solemn dew]

Nevermind I got it

[delicate orchid]

dude you don't have to delete your posts

we chill

[solemn dew]

ϕ(e) = ϕ(ee) = ϕ(e)ϕ(e) = ϕ(e)e' => e' = ϕ(e)

Was wondering how they concluded that in the proof

I am newbie

https://tenor.com/view/kto-kounotoritoken-kounotori-lbow-storkholders-gif-25676336

[ashen heron]

phi(e)=phi(e)phi(e) use the inverse thingy of phi(e) to get e'= phi(e)

In a group c=cc implies c is the identity

[median pawn]

image:

[cloud walrus]

hausdorff

[glossy crag]

Do you know what "algebraic" means for an element of a field extension?

[median pawn]

yes

doesn't have to be a finite extension tho

[glossy crag]

a^2\in K\implies the minimal polynomial of a over K has degree 2

If a has a minimal polynomial of degree n, then K(a)={c_0+c_1a+...+c_n-1a^n-1 : c_i\in K}

[glossy crag]

to be very explicit in this special case, multiply the numerator and denominator by c-d\alpha

this is similar to how you invert a complex number: write 1/x+iy=x-iy/(x+iy)(x-iy)=x-iy/x^2+y^2

[glossy crag]

The general statement is this.

[median pawn]

cool cool

[median pawn]

got it, thanks!

[wooden ember]

so we have universal coefficient theorems when working over PID's when you apply Hom or tensor. Can you get analogous theorems when applying any additive right exact covariant / left exact contravariant functor to a chain complex, involving the appropriate derived functors?

I don't see any reason why not but just to be sure

[lethal dune]

using spectral sequences, it can be done but it may not be as elegant as UCT

[wooden ember]

is it at all useful to do this?

[lethal dune]

sometimes

[wooden ember]

I guess the fact that I don't really have any derived functors in mind other than Tor and Ext doesn't help to imagine this haha

[lethal dune]

there's a "longer" version of UCT when you don't have PID

reason is Tor_2 may not vanish

[wooden ember]

yeah i guessed as much

[lethal dune]

if you're familiar with ss( just for a strip is enough) you can try to prove UCT

[wooden ember]

im not familiar with spectral sequences unfortunately

[lethal dune]

never too late to pick up

[wooden ember]

fair enough but not during exam season

thanks for the help

[lethal dune]

yeah lol

Think this is right, Tor_p(H_q(X), R) ⟹ H_{p+q}(X; R)

[wet zodiac]

why is first iso so important

[formal ermine]

because it's so useful

[wet zodiac]

example

[wraith cargo]

take any surjective homomorphism
BAM
You get an isomorphisms if you ignore the kernel

😎

[delicate orchid]

every normal subgroup appears the kernel of some homomorphism

that's a nice one

anyway first iso basically is what a quotient is

[coral spindle]

If you want to show an isomorphism involving the quotient, chances are you use the first iso.

Wanna prove the chinese remainder theorem? Use first iso.
Wanna prove that R/Z is isomorphic to the group of unit norm complex numbers? Use first iso.
Wanna prove the other iso theorems? You use first iso.

[summer path]

Wanna summarize undergrad algebra? First iso.

[formal ermine]

50% of my algebra homework was "the splitting field is normal because it's a splitting field per proposition X and separable because it's over Q which has characteristic per proposition Y"

[glossy crag]

https://math.stackexchange.com/questions/4725795/proof-of-the-double-centraliser-theorem-in-lorenzs-algebra-ii
I've put my question about similarity of algebras on MSE, if anyone wants to take a look. If you don't feel like dealing with it or are not familiar with the topic, consider at least upvoting so that the question gets some exposure.

[delicate orchid]

I just think nobody here can actually answer this soz boss

[glossy crag]

I am surprised, you guys can handle Hodge theory and parabolic induction and whatnot, but not a simple question on noncommutative algebra?

[glossy crag]

In any case, drop a brother an upvote, maybe someone on MSE can answer this.

[delicate orchid]

well some people can, I cannot

[delicate orchid]

lololololol

[glossy crag]

Dang, my appeal to ego bait did not work.

[delicate orchid]

I removed my ego via several controlled psychic blasts

[wooden ember]

it's cause non commutative algebra doesnt exist

[glossy crag]

Rebuttal: 2x2 matrices.

[wooden ember]

not a thing

i only work with diagonal matrices

[glossy crag]

But what about quaternions and omg, did you know you can use them in graphics engines ???

[wooden ember]

damn okay you changed my mind

[glossy crag]

e^{i\pi}+1=0 is the most beautiful thing in the world, subscribe to my math blog.

[wooden ember]

i think you mean e^{i tau/2} + 1 = 0

[glossy crag]

this is everything!!!

[white oxide]

im so confused i have no idea where to start for number 1 or any for that matter

like

would i just map sqrt(2) to itself since it's already in the field

idk where to start

oh nvm

i just have to permute the generating set in a way that preserves the elements of F

[chilly ocean]

How I can show that "for every d | n, there exists exactly one subgroup with order d" implies that G is cyclic where G is a group of order n?

I am new to abstract algebra

[glossy crag]

If you know the Euler totient function $\varphi$ and its property $\sum_{d\mid n}\varphi(d)=n$, you can show it that way.

[cloud walrus]

leave_no_norm

[barren sierra]

^^^ that's the only proof I know

[lusty marlin]

Consider an arbitrary element g ∈ G. g has order d, where d is some divisor of n. Therefore g generates a cyclic subgroup of order d, which has ϕ(d) generators. Since there is a unique subgroup of order d, there can be at most ϕ(d) elements of order d in G, for any divisor d of n. Since as written above, the sum of ϕ(d) over all divisors of an integer n is equal to n, it is trivial to proceed from there and prove that the group must be cyclic.

[south patrol]

Also uh if you want more of a link like

The identity on the sum of phi(d) can be computed just by counting the number of elements of each order in Z/nZ

So actually you can rephrase the proof as saying that the number of elements of given order in G is no greater than the number of elements in the cyclic group of the same size

Hence all the numbers must agree

And so G is cyclic

[chilly ocean]

I understand. Thanks!

I have a question

Can I use this as a theorem?

[lusty marlin]

It is a common enough result to be used as a theorem.

[chilly ocean]

Oh ok. Thanks!

[chilly radish]

Funny, the only proof I know of the aforementioned identity is using the fact every cyclic group has exactly one subgroup for each number dividing its order

Oh wait

My bad

I misread the question

I thought it was asking for G cyclic not to provr that G is cyclic

Btw, there's a nice proof of this using the sylow theorems

[chilly ocean]

Why there are at most phi(d) elements of order d in G? Isn't the number of elements with order d exactly phi(d), if d | n?

[chilly radish]

||It follows that G is a direct product of its sylow subgroups and that those subgroups are cyclic, then you use CRT||

[lusty marlin]

No, d | n does not, in general, imply that there are exactly ϕ(d) elements of order d in G.

[barren sierra]

nice

sylow theorems
not possible

[lusty marlin]

In fact, there need not be elements of order d in a finite group for all divisors d of the group order.

[barren sierra]

y r u booing me I'm right

[chilly radish]

I think they mean that if g is of order d, then the subgroup it generates should have exactly phi(d) elements of order d

Which is true

[lusty marlin]

They asked 'in G', which was the label given to the whole group in their question.

[chilly radish]

I guess, but I do think that's what they were getting at

[white oxide]

why is an extension mapping sqrt(3) to -sqrt(3), alpha 1 to alpha 3 and i to i valid? idk if my algebra's wrong or not but under that correspondence alpha 2 was not sent to a conjugate of x^3 - 2

[rocky cloak]

Under that correspondence alpha2 is fixed (mapped to itself)

[white oxide]

hm I must’ve fucked up on calculations then

I’ll check again

Thanks!

[gilded frost]

can any canellative monoid be extended to a group

[formal ermine]

what do you mean with "extended to"?

[gilded frost]

is there always an invertible supermonoid

im thinking by adding inverses to elements that dont have them

and adding elements as necessary to make composition closed

in the way that the naturals are extended to the integers

turning the monoid into a group under addition

is that generally possible?

[coral spindle]

Consider the submonoid {0, 1} of Z under multiplication. Imagine it embeds into a monoid M with inverses. Let v be the inverse of zero in this monoid M. Then 1 = v * 0 = v * 0 * 0 = 1 * 0 = 0.

Oh, ok. You want cancellative monoids.

I believe so, yes. I think you can construct something similar to the field of fractions for a ring, though I would have to check the details.

Unfortunately this gets complicated due to non-commutativity, but at least for commutative monoids this most certainly works.

[delicate orchid]

if it's a commutative monoid this is the Grothendieck completion/group

if it's not then I'd take the free group on each element of the monoid then quotient out by whatever relators are in the monoid

[coral spindle]

I think I have a less brute-force approach. Take the free (monoid) product M \ast (M^op), where M^op is the opposite monoid, and quotient by relations that make the terms in the second copy inverses of the things in the first. This preserves noncommutativity and ought to still contain M. Details, as ever, must be checked.

This should satisfy the obvious universal property, i.e. making cancellative monoids a reflective subcat of groups

[small bramble]

anybody know what remarks Dummit is referring to and/or can directly explain corollary 33?

meantime guess ill just reread 15.2 to try to find said remarks

[rocky cloak]

The remark is the thing about radical of ideals in integraj extensions. In the middle of your second picture with R and S and I

[small bramble]

oh ye

ok p obvs from there

[gilded frost]

i just read the wiki definition and i smell a universal property

is there a good category theory definition for that?

[delicate orchid]

there is a universal property, any monoid homomorphism from M to an abelian group factors through it's grothendieck completion

[gilded frost]

ty

[cloud walrus]

wew ladz

[rocky cloak]

Apearantly this is not the case https://link.springer.com/content/pdf/10.1007/BF01571659.pdf

[barren sierra]

$h \cdot (a_g g) = a_{hg} hg$

and then distribution is a thing ofc

[coral spindle]

Hint: the action of G on itself is transitive. That is, for any g, g' in G there is an h in G such that hg = g'.

[rocky cloak]

The example is the free monoid on a, b, c, d, x, y, u, v modulo the relation ax=by, cx=dy, au=bv

[cloud walrus]

spamakin

[barren sierra]

fixed @chilly ocean

[coral spindle]

I feel like this doesn't make sense, spam

is a_g, a_hg meant to be the coefficient? If so, why does it change index? It shouldn't.

You can simply say that ZG has a basis {v_g | g in G} and the action of G is h . v_g = v_(hg)

You don't actually need to mention the coefficients at all – you only need to define the action on the basis

Yes we require linearity to hold. We call it "extending linearly"

The action of G is defined by extending the formula h . v_g = v_(hg) linearly

[barren sierra]

I feel I always see the change of coeff in notation

that's the only reason I did it

[coral spindle]

Yeah, but it should be h . (ag) = a(hg)

I don't understand why you changed the name of the coefficient

For the avoidance of doubt I gave a more explicit basis.

[barren sierra]

fair

[coral spindle]

The coefficients do not change.

They are carried along with where the basis element goes.

No.

You've contradicted yourself there

You say that there is only one element in the invariant!

namely \sum_{g \in G} g

But then you say it forms a ring. What gives?

Right. In particular, it consists of \sum_{g \in G) zg, for some fixed integer z. These are exactly the invariants.

Indeed, but if G is infinite then ZG is not a unital ring, so we have some issues anyway. way wrong!

I don't know what you mean by that

[rocky cloak]

The action of G on ZG is not actually an action on a ring (G doesn't act by ring homomorphisms), so it's a bit weird to look for the invariant ring. In this case it happens to be closed under multiplication, but doesn't contain 1.

[rocky cloak]

ZG is always a unital ring. The unit in the group is the unit in the ring

[coral spindle]

True, it's the invariant Z-module rather. Good catch.

[coral spindle]

Oh you're right. I was confusing it with a quiver algebra. Sorry putlica!

[delicate orchid]

the 1 in question:

you are right in the more general context of category algebras over non-finite (small?) categories though

over Z[G]? good luck

[rocky cloak]

Yeah the covariant module here is just Z

[delicate orchid]

What’s the kernel of the map given by multiplying by two

and 0, right?

And what’s the image of the same map

Yup! So we have the kernel of the map is equal to the image of the map

Can you see why the sequence is exact now?

Exactly

Is that your definition of exact or do we actually need to compute the homology groups

[long nebula]

based

[delicate orchid]

Uhhh hold on let me draw the triangles in my head

Oh is this just a regular homotopy or a chain homotopy (I know they’re the same on some abstract level but shush)

[delicate orchid]

I also mainly focus on finite group reps, welcome to da club

[glossy crag]

Can perfect char p fields be classified in any reasonable way? I know that in char p perfect \iff Frobenius surjective, but that doesn't really say much.

[barren sierra]

I like Rotman's treatment

[white oxide]

does anybody know what this notation represents? i tried looking in the earlier parts of the book but i couldn't find anything

[barren sierra]

in Advanced Modern Algebra

I'm pretty sure he also has a homological algebra text tho

[glossy crag]

that's kind of meaningless, because then it doesn't have any algebraic extensions (so it's vacuously perfect)

[glossy crag]

It's just the notation Kx, the 1-dimensional subspace generated by the vector x in the K-vector space V. In this particular case the vector space is C^n and the vector x is e_1+...+e_n.

[frigid lark]

All of the finite ones are, but that's trivial

[glossy crag]

I know, I was looking for something more encompassing

[white oxide]

ohhh ok thanks

yup

[frigid lark]

Lang has a statement that if a field is perfect every algebraic extension is separable and perfect

[white oxide]

oh so just the subspace generated by e1 + e2 + ... + en over the complex numbers

[glossy crag]

that's basically a rewording of the definition (one i use is "every algebraic extension is separable")

[white oxide]

wait i forget my lin alg that's just the eigenspace spanned by (1, 1, ... 1) right

no i'm just working through a text over the summer

thought it seemed interesting

oh damn

yea looks like just some algebra and a good knowledge of lin alg

[frigid lark]

So you want something beyond the fields contained in an algebraic closure?

Of F_p

[barren sierra]

I'm learning them on the fly in an REU 💀

undergrad wasn't taught

grad algebra class assumed we already learned it

Texas State University

[white oxide]

ooh i'm gonna apply for reus this year

and gonna get fucked

[barren sierra]

just apply to a ton

[white oxide]

how was the application process for you

[barren sierra]

honestly

uh

[white oxide]

here they're super fucking competitive lol

[barren sierra]

I'll DM you something

[white oxide]

ok thx

[barren sierra]

Research Experience for Undergraduates

basically the NSF funds undergraduate summer research programs

where you can go to a different university to your home one, do research, get a small stipend and such

so I'm doing some group classification stuff

I would absolutely be the wrong person to ask lol

sorry

it is not easy

I know like some basic definitions and that's it

we talked about it for like a under a week

[karmic moat]

could just ask in this channel lol

[barren sierra]

also that

[karmic moat]

thats more of a #point-set-topology i think

[prime sundial]

what 0 map are you referring to

but what space are you talking about

in Z_4?

can you send the exact question

how are you defining a homotopy?

[dreamy plume]

Im pretty sure it means chain homotopic

Like show that f-g\neq dh+hd where f ,g:Z/4Z->Z/4Z are zero map and identity respectively and h is some map called the chain homotopy

The way I would show this is that you let g be the zero map and f be identity

So then we suppose that f-g=dh+hd

And say x is some point in Z/4Z

Then x=2h(x)+h(2x)

||We use the fact that h is homomorphic and get 2h(x)=h(x)+h(x)=h(2x)||

@chilly ocean can you go from here?

Woo

Our choice of x was an arbitrary point though?

So this seems like it contradicts that there exists a chain homotopy

Where are you getting the exercises from?

Oh cool

I think i could explain how this definition of homotopy connects to the one you see in topology

Yeah dont worry I dont care

I dont remember

I dont know if we use simplicial complexes to show the connection either

No the connection/analogy between chain homotopic and homotopy in topologies

[void cosmos]

are u guys done

i need mega help

[dreamy plume]

“The name "homotopy" comes from the fact that homotopic maps of topological spaces induce homotopic (in the above sense) maps of singular chains.” Says wikipedia with no sources

[dreamy plume]

🫵

[void cosmos]

i can undersatnd 0 of this proof

like

i understood the proof of

0 --> A --> B --> C exact gives 0 --> Hom(D,A) -_.... exact

i have no idea why this is any different from the proof

and i think the statement that A-->B-->C-->0 exact gives 0-->Hom(D,A) -->. is false

[dreamy plume]

Oh nice you play league

[void cosmos]

yea

[undone bay]

Let G be a group and U, V finite dimensional vector spaces acted upon by representations of G. I'm interested in the ring of G-equivariant polynomial maps U -> V: I think there's an argument where you can use Hilbert's finiteness theorem to argue that the ring of G-invariant maps U -> R is finitely generated, from which you can argue that these G-equivariant maps are also finitely generated. But Hilbert's finiteness theorem only holds for certain groups, and I'm unsure how constructive either of these proofs are.

Say I want to get my hands dirty and start computing the bases for the ring of G-equivariant polynomial maps: anyone have any recommendations on resources that might discuss this, and for what types of groups its possible? Some groups I'm interested range from things as simple as the symmetric group to (the group of permutations + rescalings in each components of n-vectors), SL(n, C), or groups as "bad" as SL(n, Z)

[dreamy plume]

Have you tested any examples?

[void cosmos]

yea i willa fter i finish this game

i just have to figure out how Hom(Z_n,Z) looks like

which is easy

but anyways

the proof is whats bugging er

me

[dreamy plume]

I dont know if thats the best example

For homomorphisms generators of domain map to generators of subgroups of codomain

Its kind of what first isomorphism theorem says

[void cosmos]

yea i see that

[dreamy plume]

To elaborate the projection map pi depends on what the kernel of our homomorphism phi is. Kernels are always subgroups. Pi always maps generators in G to its representatives in the quotient group. In a way this tells us that homomorphisms are uniquely determined by how they send their generators

So Z/nZ->Z maybe not the best choice since there arent many places to send your generator(1) such that the image is a subgroup/subring

[void cosmos]

yea

good point

so it might work as a false-positive ig

[void cosmos]

anyways

anyone knows how to explain this proof for me?

[void cosmos]

.

[dreamy plume]

Which part

[frigid lark]

I think you are trying to prove that A --> B --> C --> 0 exact => Hom(A, D) <-- Hom(B, D) <-- Hom(C, D) <-- 0 is exact

Cause Hom(_,D) is contravariant

[frigid lark]

This statement is in fact false as that implies that D is a projective module

[void cosmos]

no

or yeah

yea thas what i want to do

i cant understand the ker is a subset of the img part

[frigid lark]

Ok, let f be in the kernel of theta (p) hat, then fp = 0

That means that f[Im(p)] = f[Ker(Z)]. Then there exists a map, a in Hom(C, D), where a(x) = f(z^(-1)(x)). a is well defined as if y, y' are in the pre-image of x under z, i.e. z^(-1)(x), then y = y' + c such that z(c) = 0 hence f(c) is 0.

Furthermore zhat a(x) = f(z^(-1)z(x)) = f(x)

Hence for every f in the kernel of that hat, there exists an a in Hom(C,D) such that zhat(a) = f

[median pawn]

suppose a^4 = a for every a in a field K. i want to find the characteristic of K. obviously we have a^2 + a + 1 = 0 for every a in K - {0,1}. maybe a hint for what's next would help

[rustic crown]

replace a with something nice ||-a||

[median pawn]

ahhh crap i see it

thanks lol

[rustic crown]

(or yea just plug a = -1 lmao)

[delicate bloom]

y not ||a=1||

[rocky cloak]

Plugging in various integers for a, is a good hint in general

[coral spindle]

We obtained a^2 + a + 1 = 0 only for a \notin {0, 1}

[median pawn]

you get 1 = -1 if you put it in a^4 = a. 2 = 0

[coral spindle]

And? Merosity suggested a=1, not a=-1.

I did indeed notice the other thing.

[median pawn]

oh okay! misread

[median pawn]

this is nice, it means if a^n = a for any even n, then the characteristic of K is 2. gotta see what happens for odd n

[rocky cloak]

For K the field with two elements a^n = a for every n. But there can be other fields for some n

[median pawn]

yup

[delicate bloom]

yeah I was careless

[rustic crown]

(i thought you meant plugging a = 1 in a^4 = a )

[rocky cloak]

If you want more hints:
||If it's true for K it's true in the prime field, so enough to consider fields of prime order||
||What does Fermat's little theorem tell you about when a^n = a mod p?||

[coral spindle]

||I think you need a slightly stronger fact than simply Fermat's little theorem to prove what you're suggesting. Specifically, you need the presence of an element of order p-1.||

[rocky cloak]

||fermats little at least gets you for which characteristics it is true. Then you need primitive roots for which it isnt true||

[coral spindle]

Well no, I don't think it does automatically. ||A typical element of F_p has order much less than p-1, so without more information we cannot say p-1 | n-1.||

[delicate bloom]

I guess a shortcut is, ||if you know a^4-a is 0 for all a in K, then the field has at most 4 elements. Then just brute force through all the fields with <= 4 elements lol||

[coral spindle]

Yeah I had the same thought Mero, but I guess the ||element-of-order-(p-1) approach is stronger||

[delicate bloom]

whats strong mean

like it gets it without brute forcing

[coral spindle]

yeah

[rocky cloak]

||Fermat tells you that a^p-1 = 1, thus if n = k(p-1)+1, it works in characteristic p. This doesn't tell you that it won't work in other characteristics. Then you need primitive roots||

[coral spindle]

Right, this is a sufficient condition on n, but we have not proven necessity yet. Fair enough.

[delicate bloom]

I think here's a bit more preciseness while skipping brute force ||look for double roots with the derivative 4a^3-1. If it's got 2 or 4 elements, we see that -1 is nonzero so there are no double roots. If it's got 3 elements, then it's a^3-1=0 for all elements by fermat, but that's not possible, so we're done||

[lethal dune]

|| why is everything censored ||

[coral spindle]

To prevent the government from tracking us

[void cosmos]

anyone? :()

still stuck

[rocky cloak]

The two proofs should be almost exactly the same, so it's weird you would understand one and not the other. Any particular step you're confused about?

[void cosmos]

are u sure?

the first one when proving Ker(phi bar) subset of img( other map bar ) uses the fact that it has an ivnerse

restricting the monomorphiusm to its img

the other seems to use first iso

@rocky cloak

[hidden haven]

Yeah those are like duals of each other

In that sense it is the same proof

[rocky cloak]

Yeah, so the use of the first isomorphism theorem is the only difference, right? Is that the step you don't understand?

[void cosmos]

yes

how is it quoteitn ker phi

not quotient f bar

[rocky cloak]

$C \cong B/ker\zeta$

[cloud walrus]

jagr2808

[rocky cloak]

Because $\zeta$ is surjective

[cloud walrus]

jagr2808

[void cosmos]

the map he is using is f

not even zeta

im so confused

what is zeta

[rocky cloak]

We have an exact sequence $A \to B \to C \to 0$

[cloud walrus]

jagr2808

[rocky cloak]

The first map is called $\theta$, the second $\zeta$

[cloud walrus]

jagr2808

[void cosmos]

okay

now he wants to prove ker( zeta bar ) is the same as im ( theta bar )

right?

walk trhough it with me please

[rocky cloak]

Say f is a function B -> D

[median pawn]

oh hi moldilocks, long time

[rocky cloak]

We want to prove that the kernel is contained in the image, so if $f\theta = 0$, we must show that there is a map $g\colon C -> D$ such that $f = g\zeta$

[cloud walrus]

jagr2808

[rocky cloak]

The fact that $f\theta = 0$, means that f of the image of theta is 0.

[cloud walrus]

jagr2808

[rocky cloak]

But the image of theta is the kernel of zeta

Thus f factors as $B \to B/ker\zeta \to D$

[cloud walrus]

jagr2808

[void cosmos]

how

[rocky cloak]

Say $f = \hat f \pi$ where $\hat f(b + ker\zeta) = f(b)$

[cloud walrus]

jagr2808

[hidden haven]

Hullo

[rocky cloak]

This is well defined because $f(ker\zeta)=0$, and is the defining feature of a quotient

[cloud walrus]

jagr2808

[rocky cloak]

From there you just use the isomorphism between $B/ker\zeta$ and $C$, which your picture calls $\varphi$

[cloud walrus]

jagr2808

[rocky cloak]

Then $\varphi\pi = \zeta$ so $\hat f \varphi^{-1}$ is the map you want

[cloud walrus]

jagr2808

[void cosmos]

yea

i got it now

or

just 1 thing

how did u go from image of theta is 0 to B --> B/ker

[rocky cloak]

Image of theta equals the kernel of zeta

By exactness

So B/imtheta = B/kerzeta @void cosmos

[void cosmos]

yea

ohh

lmfao

its the coker

so B--> B/im theta = B/kertheta --> C

[rocky cloak]

Indeed

[void cosmos]

by first iso

yea got it

tysm man

also one more question

when proving the reverse implication

why cant i choose D = R

( the whole ring as an R-module

and then iget it for free

cuz Hom_R(R,A) is just A

f --> f(1)

[rocky cloak]

If you're considering Hom(D, -) you can choose D=R. Here we're looking at Hom(-, D), so D=R would give you Hom(A, R) which is not that helpful

[void cosmos]

oh fuck yea

thats how i did the first one

tysm

ur so good

[rocky cloak]

Np

[delicate orchid]

cracked at the craft

does a similar result hold for general left/right exact functors or is Hom a special little snowflake

[void cosmos]

isnt this stating excatness

[frigid lark]

I think if the sequence is split exact stuff works

[void cosmos]

left

[delicate orchid]

the forward direction is yeah

[frigid lark]

Oh and the functor is additive

[hidden haven]

Ye it's not true for general left or right exact functors

[delicate orchid]

I didn't think it would be but it's worth a shot

[hidden haven]

The zero functor is exact but it also takes non-exact sequences to exact ones

[frigid lark]

Oh, both ways

Shit

[delicate orchid]

very good point

[hidden haven]

[void cosmos]

whats the 0 functor

[delicate orchid]

but yeah I meant when both ways hold

[void cosmos]

maps to 0

in the 0 category

hehe

[hidden haven]

Yeah constant zero

Like Ab → Ab

Everything maps to 0

[void cosmos]

yea

[frigid lark]

Derived functors in a world where functors are exact

[delicate orchid]

also just to double check, additive functors are ones that commute with coproducts right

[hidden haven]

wew why is this sad

[delicate orchid]

it's sad for the derived functors

[hidden haven]

explain yourself

[delicate orchid]

I am an empath moldi

I have telepathically moulded my consciousness into that of a derived functor

[hidden haven]

Commute with finite biproducts

[delicate orchid]

what the FRICK is a biproduct

a product and a coproduct at the same time?

[hidden haven]

Yes

[delicate orchid]

oh so direct sums in all of the categories that matter

[hidden haven]

[hidden haven]

👎

[frigid lark]

No Tor, no Ext

[hidden haven]

Good.

[frigid lark]

Damn

[void cosmos]

how do you guys enjoy this

homological aglebra stuff

[delicate orchid]

I don't enjoy it

[frigid lark]

No employment for homological algebraists

[void cosmos]

no proofs no good problems no nothing

[hidden haven]

I enjoy it

[void cosmos]

no intuition

[frigid lark]

I go to Galois theory

[delicate orchid]

I don't know how you enjoy that either

[hidden haven]

I would enjoy it more if the universal coefficient theorem were H(X; A) = H(X) tensor A

[frigid lark]

My birthday is 25/10/some year

[hidden haven]

so true

[delicate orchid]

I understand the left/right exactness functor nonsense because I have to for my research

[hidden haven]

Why are you doxxing me

[delicate orchid]

but I have no intuition for like

homology/cohomology

[frigid lark]

Wait, fellow reincarnation of Galois?

[delicate orchid]

which is a problem because I have to for my research

[hidden haven]

Is Galois also a bro?

[void cosmos]

i dont think there is intuitoin man

i only know 1%

[delicate orchid]

or will do very soon - Mackey functors may start being a thing I have to worry about

[void cosmos]

but like

its just some rigged shit l;ike everything is defined for some thing to work and thats it

[delicate orchid]

yup that's how maths is

[void cosmos]

😠

[frigid lark]

I was giving a presentation on why Tor is iso to tor, in a book club and that was the dryest math I have ever done.

I do not find homological algebra fun

[delicate orchid]

need my moist maths

[frigid lark]

In general math makes me moist

[hidden haven]

Homologicalgebra is the best left as an exercise subject

[frigid lark]

True

[void cosmos]

yea

the proofs are literally sarcasm

like its not even trickery

[hidden haven]

What does that even mean

[void cosmos]

its like

think of any algebra proof

[hidden haven]

💀

[delicate orchid]

I don't know but I agree with him

the vibes align

[hidden haven]

Based

[void cosmos]

there are details when u have to prove this is well defined or this is an acutal homomorphism

[frigid lark]

Lang section 4 homological algebra, make Lang section 4

[void cosmos]

now turn this shit into a whole subfield

lmfaao

thats all what u do

[hidden haven]

Yeah that's because you are doing homie alg before cat theory 💀

[void cosmos]

im not doing it man 😦 i have a grad algebra exam

and its only 1 section

in hungerford

[delicate orchid]

you WHAT

[void cosmos]

homie

[frigid lark]

Are you doing it after homology?

[void cosmos]

no idk AT yet

which is supposedly the math where insane fuckery makes sense

[delicate orchid]

you should do category theory before AT as well

[frigid lark]

Isn't homological algebra motivated by homology?

[delicate orchid]

it is

[void cosmos]

i think it is

[frigid lark]

I don't know as I haven't done homology

[delicate orchid]

most of AT is homology lol

it's like homology and the funny wholesome pi_n's

[hidden haven]

Some of the most of AT

pi_n's are not funny wholesome

[delicate orchid]

agree to disagree

[hidden haven]

Agreed.

[void cosmos]

AT is the coolest shit ever honestly but i cant learn it

cuz im too dumb for it

[delicate orchid]

same

[void cosmos]

it requires imaging things

[frigid lark]

Last time we did that we got 0=1

[void cosmos]

imagining *

[delicate orchid]

it requires imagining continous maps of continous maps of continous maps of con-

[hidden haven]

Bro what's stopping you from imagining RP^3

[void cosmos]

i can imagine that

from the sphere

but its so cool that

[delicate orchid]

brohomology of the nth configuration space

[void cosmos]

alot of the problems are done b

by

[frigid lark]

$C^{C^{C^\ddots}}$

[void cosmos]

like they define map by "okay so we lift this cylinder and then attach to it a carbon fiber body then engine ..

[delicate orchid]

\ddots you fool

[void cosmos]

like they define it visually

lmfao

and then give u an exercise to actually define it

and its the worst shit ever

[delicate orchid]

post-rigour

[cloud walrus]

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

[hidden haven]

Don't tell me you aren't using the lifting properties of fibrations and cofibrations to define those maps

[delicate orchid]

pranked

[void cosmos]

i wont

cuz idk what are fibrations are

:(*

[delicate orchid]

preimages for nerds

[frigid lark]

I'll dddots you

[hidden haven]

Lmao parrotty on his own character arc

[void cosmos]

so like fibers?

[south patrol]

Nice

[void cosmos]

anyways the only good application of algebra is ANT so im learning that after i finish this test

[hidden haven]

Nah it's a more complicated thing but makes a lot of existence arguments for maps easier

[void cosmos]

ty again for helping me with the proof

[delicate orchid]

computed that one as an exercise lol

[south patrol]

Which one? Lol

I didn't read chat properly

[void cosmos]

can u define them here or does it take 200 years of math learning

[delicate orchid]

the cohomo of the configuration space of C^n

[void cosmos]

to just understnd the definitoin

[south patrol]

Sullying this lol

[frigid lark]

The only good application of abstract nonsense is more abstract nonsense that has a really low floor?

[south patrol]

Ah ok

Why C^n and not R^n lol but fair

[delicate orchid]

FUCK R

[south patrol]

I guess even is easier actually iirc

[frigid lark]

R is better than C

[south patrol]

Lol

[delicate orchid]

get yooo frobenius shcur indicator outta my faceee

[frigid lark]

Well more interesting than C

[delicate orchid]

yeah you can actually go around things in C

[south patrol]

Ye

[formal ermine]

biggest mistake of my life

[void cosmos]

fuck

why

[formal ermine]

i thought ant was a lot of field theory

[void cosmos]

i think its more commie algebra

like its discount AG

[formal ermine]

it's just a bunch of comm alg disguised as fields

[south patrol]

Actually yeah cohomology of config spaces of R^n is tricky enough bruh

[void cosmos]

yea hahah

[south patrol]

Don't you need to use a fibration sequence

[wraith cargo]

it's a LOT of field theory

[delicate orchid]

no clue

[south patrol]

Or is it easier for C^n

Lol

[void cosmos]

idk tbh

i took

half a semester

[hidden haven]

p: E → B is a (hurewicz) fibration is given any commutative square of the form
X → E
↓ ↓
X x I → B
there exists a map X x I → E making everything commute

[frigid lark]

Damn commie algebra influencing our fields

[delicate orchid]

for C^n you just tensor a bunch of shit together using the meyers-something sequences

iirc

[wraith cargo]

meyer vietoris?

[delicate orchid]

it was a while back and tbh I'm not even sure I did it right

[south patrol]

hm

[hidden haven]

So you get the existence of such a diagonal map without having to explicitly construct it

[delicate orchid]

yur, think so

[rocky cloak]

Just use flat modules, problem solved

[formal ermine]

yeah bro it's all fun and games when you think you just work with number fields but then the fucking ring of integers shows up and ruins everything

[hidden haven]

[delicate orchid]

like you do it inductively by partitioning the space into a neighbourhood containing just one missing point and then the rest

I think

[void cosmos]

what field of algebra solve concrete problems

by abstract shit

ANT right?

[south patrol]

Lol

[delicate orchid]

category theory

[south patrol]

Algebra?

What do you mean by concrete lol

[void cosmos]

like i wanna solve olympiad problems

easy to state

[south patrol]

Algebra is concrete enough often

[frigid lark]

I am not looking forward to the part of Galois theory that says fields are nice and all but let's use a PID

[void cosmos]

using cute amth

[south patrol]

Oh lol

[wraith cargo]

this reminded me I need to read more Weibel

[delicate orchid]

faithful functor th o the caetorgry of stdyreiufjegliref

[south patrol]

Lol

[delicate orchid]

PIDs are basically fields

[wraith cargo]

Yesterday I nearly had an aneurysm when he started talking abt why homology doesn't commute with arbitrary colimits

[delicate orchid]

structure theorem for modules over a ring is equivalent to everything being free because I don't have to think about it

[wraith cargo]

yeah good one buddy

[south patrol]

Maybe I need to make a thread so that I can be bullied into working this summe

[south patrol]

Hm

[frigid lark]

Come to Australia, it's winter here

[delicate orchid]

algebraically closed PID

[frigid lark]

ID's are contained in a field, and are also basically fields

[delicate orchid]

https://tenor.com/view/sassy-who-the-hell-cares-peter-family-guy-gif-5547786

[wraith cargo]

https://tenor.com/view/chainsaw-man-gif-26446195

[south patrol]

Oh okay lol

I thought you meant homology of spaces and was tryna think of an example there

Sure

[hidden haven]

But it commutes with filtered colims

which is better than anything you could expect

[wraith cargo]

yeah
But I never strain my pasta
So who cares

[hidden haven]

For R-modules

[south patrol]

Are there any nice examples where homology of chain complex doesn't commute w colimits

Probably smth boring

[hidden haven]

Probably try pushout

[frigid lark]

Speaking of spectral sequences, I haven't done any topology, is doing an undergrad research project on spectral sequences a bad idea?

[hidden haven]

No it is not

[wraith cargo]

you can do it purely algebraically but I think the topological perspective is interesting

[hidden haven]

Spectral sequences very cool and mostly algebraic anyway

[wraith cargo]

McCleary has a good book covering the topological side of them in depth

[hidden haven]

There are toppy applications

[wraith cargo]

Also I need to read The K-BOOK

[frigid lark]

Thx

[wraith cargo]

I wanna start talking like a K theorist

[hidden haven]

I recommend John Rognes's book on spectral sequences

He is a topologist though

So there is some top

[rocky cloak]

Cokernel doesn't commute with homology, but you get the long exact sequence which is almost like commuting I guess

[south patrol]

Awesome thanks for that example

[azure copper]

crossposting from #help-3

as i got suggested

[rocky cloak]

What is exp(2pi i)?

[azure copper]

1!

[teal vessel]

well, the definition of an injection is one where f(x)=f(y) implies x=y, so can you find some distinct x,y s.t. f(x)=f(y)?

i.e. can you break the definition with this function?

[azure copper]

well yh like f(2pi * i) and f(4pi * i)

both maps to 1

yet they're different

[teal vessel]

then you have a minimum counterexample of two valid x and y that both have the same image, therefore it cannot be a bijection

[azure copper]

so basically i can use this simple definition?

cause i remember my professor doing some long ass work every time there was to prove injectivity

like showing that Ker(phi) != {0} in my case

[delicate orchid]

they're equivalent

a homomorphism phi is injective if and only if ker(phi) \neq 0

[ashen heron]

should be equal?

[teal vessel]

that is the definition as far as I'm aware. Most definitions are extremely simple, but have far reaching implications. Depending how your professor wants you to do it, you may have to show more work, but fundamentally all you have to do to "prove" that it isn't bijective is to find a counterexample if one is readily at hand. Sometimes that's harder to find than some structural argument, but when possible it's sufficient.

it's like the definition of a logarithm: f(x) is a logarithm iff f(xy)=f(x)+f(y) for all x,y > 0. With that (and the notational convention that f_a(a)=1) you can derive all the other properties of logs

[rocky cloak]

Thus is very useful for showing that something is injective. But doesnt make a big difference when showing something isn't

[azure copper]

i thought that was gonna fall under the "proof by example" thing ngl 😭

my professor is extremely pedantic

[delicate orchid]

it's a proof by counterexample

which is valid

[ashen heron]

proof by example isn't a thing

[chilly radish]

homology doesn't commute with arbitrary direct sums in general, he has a correction in the errata

[teal vessel]

proof by minimum counterexample is a valid proof, and is closely related to proof by contradiction.

i.e. "if it's of form X it should behave like X, but here it behaves like Y which isn't X"

[delicate orchid]

stinky!

[rocky cloak]

Who wants to live in a world where sums and products are not exact

[azure copper]

can I send another exercise rq

it's related

[chilly radish]

algebraic geometers

iirc categories of sheaves do not have AB4 generally

[chilly radish]

I might be wrong tho

[azure copper]

this one should be fine right?

[rocky cloak]

Damn, those wacky geometers

[chilly radish]

yea sheaves don't have AB4 but they have the dual property

wait sorry no

they have AB4 but not AB4* I think

something like that

it doesn't have AB4* which is the dual property tho

generally

[azure copper]

slightly different groups but the idea to prove surjectivity was to show that e^a = 0 has no solutions

[delicate orchid]

surjectivity is wrong because 0 isn't in R^\times

[wraith cargo]

OK what are the AB referring to?
so I can find a list of all of them

[chilly radish]

grothendieck axioms

[delicate orchid]

you could have chosen like, 2, instead of 0

[wraith cargo]

thanks!

[delicate orchid]

because there are no integers n such that e^n = 2

[chilly radish]

for abelian categories

[azure copper]

righttt

[delicate orchid]

(or just used the fact that |R| > |Z|)

[chilly radish]

https://en.wikipedia.org/wiki/Abelian_category#:~:text=Grothendieck's axioms,-In his Tōhoku&text=They are the following%3A,of exact sequences are exact.

they originated from the tohoku paper

it's also in weibel

in the appendix iirc

[delicate orchid]

I hope it's clear why 0 isn't in (R, *, 1)?

[azure copper]

obviously yh actually i think the example the professor gave us was wrong cause she always put R* to denote that its R - {0}

[delicate orchid]

yeah I agree with you, it should be (R*, *, 1)

[teal vessel]

not with that attitude (the correct one)

[azure copper]

oh yh this is cool

Cantor theorem consequence?

[delicate orchid]

yeah - you know the map can't be an isomorphism because isomorphisms are bijections

it's good that you tried to show it explicitly though

[void cosmos]

yo

any injective R-module is divisible right?

only the other way around

does it need R to be a PID

right

i have a proof :3

[delicate orchid]

it definitely holds if R is a PID

[void cosmos]

yea

[teal vessel]

meanwhile I'm over here trying to show the hard way that the rotations of a cube are isomorphic to S4

[delicate orchid]

not sure if it's an if and only if

[void cosmos]

i tihnk

divisible implies injective

[teal vessel]

I can do the assignment by hand because it's only 24 elements

[void cosmos]

is an iff

okay let A be an injective R-module.

[delicate orchid]

then why are there two different words

[rocky cloak]

What does divisible mean if R is not a domain?

[delicate orchid]

wait we gotta go all the way to non-domains?

damn

lemme think about Z/4Z

[void cosmos]

we claim that for all y in A , r in R there exists x in A such that nx = y. proof: let y in A and r in r and consider f: (r) --> A identified by sending r to y ( f(r) =y ). then by injectivity this is extended to the whole ring, and we can then let x = f(1) and see that rf(1) = f(r) = y

^ GOOD?

[delicate orchid]

oh are you just wanting to prove it for PIDs

[void cosmos]

no

thats not a PID

where did i use R is a pid

[delicate orchid]

I don't know I haven't read it yet

[teal vessel]

there's gotta be an easier way than writing down all 24 ordered pairs for this mapping. What am I missing in the structure of a cube's vertex pairs that makes it analogous to the arbitrary permutations of four objects?

mostly rhetorical

[delicate orchid]

||the funny diagonals||

[rocky cloak]

You're using that (r) is free

[teal vessel]

v. funny

ty

[void cosmos]

what :d

how

(r) is just the ideal generated byr

[rocky cloak]

You say define f(r) = y

[void cosmos]

yea

[rocky cloak]

Which might not be a homomorphisms

[void cosmos]

yea..

i got it

[rocky cloak]

Z/4 is self injective for example

[void cosmos]

Z/4 is Z/4-injective but not divisible?

[teal vessel]

wait a second... I know rubik's cubes. You can generate all orientations of the cube through a series of x and y rotations... That means that the rotation group of a cube has two generators. Therefore I should, in theory, be able to present s4 as a set of two generators and equivalent commutation relations.

[void cosmos]

but why tho

(r) is just {nr | n in R}

f(nr) = ny

like what breaks

[delicate orchid]

what if nr is zero but ny isn't, or the other way around

it's gotta have something to do with that

[void cosmos]

its weird like

i just sent n to another element

like okay

watch this

[delicate orchid]

yeah so you split it up f(n)f(r) I see what you mean, but what if n isn't in (r)

[rocky cloak]

Look at Z/4: is there a homomorphisms from (2) -> Z/4 mapping 2 to 1?

[void cosmos]

we claim that for all y in A , r in R there exists x in A such that nx = y. proof: let r in R and consider f: (r) --> A identified by sending r to y ( f(r) =y ). then by injectivity this is extended to the whole ring, and we can then let x = f(1) and see that rf(1) = f(r) for ANY f(r) in A.

^ this assumes surjectivity doesnt it

[rocky cloak]

Surjectivity is not a problem, it's that you're assuming such an f exists

[void cosmos]

no

i forgot

to delete

forget sending r to y

the thing is i can just toy with images of f

as my own y

or wait

yea nvm

is there a counterexample for an injective yet not divisible

[rocky cloak]

Z/4

[void cosmos]

okay i can see why its not divisible but im sorta new with injectivity

why is this injective

[teal vessel]

probably because for every x and y, having the same image implies the same preimage.

[void cosmos]

what

ohh ur joking

and question

in the theorem of like an injective R-module is equivalent to its a direct sum of a module B of which its a submodule

is this submodule like up to isomorphism

like Z/6 is Z/3 + Z/2

those arent submodules 😄

right

[delicate orchid]

is this the theorem that injectivity of M is equivalent to 0 -> M -> N -> N' -> 0 spliting?

[void cosmos]

yes

this comes from that

[delicate orchid]

well take N' = M

[rocky cloak]

So to check if Q is injective it is actually enough to check whether any map I -> Q can be extended to R -> Q, where I is an ideal

[delicate orchid]

there will be at least two copies of M inside N then

[rocky cloak]

For R=Z/4 there is only one ideal to check

[void cosmos]

you mean 3 ig but yeah

[delicate orchid]

it's trivial for the other two

[void cosmos]

only 1 that matters

yea

[rocky cloak]

Well, you don't need to check 0 and R

[void cosmos]

yea

just (2)

i think

cooll

ur probably chat gpt

or something @rocky cloak

ur too good

[delicate orchid]

calling them chat gpt is a massive insult

[void cosmos]

yea just figured

gpt sucks at math sadly

[delicate orchid]

[void cosmos]

i was so hyped

i thought it would solve like unsolved problems legit

[rocky cloak]

I will take it as a compliment nonetheless

[delicate orchid]

you're deluded if you ever thought that ngl

it's a language model

[void cosmos]

😠

can someone give a counter example

like

A --> B --> C --> 0 exact implies Hom(D,A) -> Hom(D,B)-->Hom(D,C) --> 0 exact?

for all R-modules D

[rocky cloak]

Let R be the integers and let A=B=Z and C=Z/2. Choose D=Z/2

[void cosmos]

so stupid but why is A --> B-->C --> 0 exact

how is ker B -->C img A-->B

[wraith cargo]

Z - x2 -> Z -> Z/2Z -> 0

[void cosmos]

oh x2

i used the identity lmfoa

[wraith cargo]

the map Z->Z is multiplication by 2

[void cosmos]

yea

yea yea

cool got it

[teal vessel]

ok so I've managed to show that the rotations of a cube can be written as <x,y | x^4=y^4=1, x^2y=y^-1x^2> (I think. There are other possible relations that hold but I'm not entirely sure if they are necessary). I've also shown that the assignment f(x)=(1 2 3 4), f(y)=(1 3 2 4) defines a homomorphism, which says that the group generated by these two permutations are a subgroup of S4. Is it sufficient now to say that since the cardinality of S4 and of the rotations of a cube are the same, that this must be an isomorphism? I could show that all assignments are unique, but since I'm already showing a mapping between generators, can't I say that the cube rotations are isomorphic to the subgroup, which then must span the group S4 because of cardinality?

[azure copper]

there's one thing that is never clear to me. If i'm asked to prove that something is a group, what are exactly the steps to follow?

show that every element of the group has an inverse?

[delicate orchid]

check that it satifies all of the properties of a group

[azure copper]

that's it?

[delicate orchid]

what else could there be

[azure copper]

associativity is assumed usually

at least in my exercises

[teal vessel]

define a group

if you have a clear definition, you simply check if all properties in that definition are met

it's like "how do I prove that something is a square?" well, the definition of a square is a quadrilateral with all congruent sides and all congruent angles. That's three conditions. If it meets all conditions, it's a square, otherwise, it isn't.

[delicate orchid]

(1234), (1324) can be shown to generate S4

[teal vessel]

I'm sure it can.

[delicate orchid]

:trollface:

what you're describing sounds like the subtitution test

[teal vessel]

I wouldn't know the name of such a test, I'm just building what I can from what I have lol

[delicate orchid]

where you extend a set map from a set into a group into a group homomorphism by showing the substituting the images of the appropriate things in the set into the relators results in the identity

but it sounds like this will be true from how you've constructed your presentation

so yeah it works

[azure copper]

an algebraic structure (a triplet) composed by a set, an operation and a neutral element. the operation must be associative and the former must be a closure.

so usually what i do is prove that e is the neutral element and that the operation is a closure.

[delicate orchid]

"must be a closure"?

also, a group is a triplet of a set, an operation, and a neutral element

[delicate orchid]

here's the actual statement

[teal vessel]

in the way my text has described it, we have four properties here, one of which I'm not sure of in this description:
a group is a pairing of a set and an operation such that the following hold:
0. (often assumed) the operation is associative over the set; for all a,b,c in the set, a(bc)=(ab)c

1. the set is closed under the operation; i.e. for any x and y in the set, x*y is also in the set
2. there is an identity; i.e. there exists an element e such that for any element a in the set, ea=ae=a
3. every element has an inverse; for all x in the set, there exists some x^-1 such that xx^-1=x^-1x=e

if you can prove that all four of these properties hold, then your set-op pairing (or set-op-identity triplet) is a group. Otherwise, it isn't.

[delicate orchid]

a thing is a thing if it satifies the definition of a thing

[teal vessel]

trying to find the most abstract, convoluted way to say that but the joke is lost

[coral spindle]

a vector is an element of a vector space

But a more general version would be "a thing is an X if it satisfies all properties that are shared by all Xs"

[teal vessel]

a square is a trapezoid if for all squares, they agree with being trapezoids (hot take, they are)

but that's an elementary geo argument, not abstract algebra

[open sluice]

a _ is an object that transforms like a _

[azure copper]

assumed 1, here the only thing I'm missing is the second point right?

wait

wrong pic

[teal vessel]

wait, I thought for sure I set associativity as property 0. Curse non-zero indices.

[azure copper]

if we being pedantic id have to show that A x identity = identity x A = identitity

i only did it one way

but yea

[teal vessel]

for most well understood objects like matrices I tend to just lean on "the identity matrix exists" since I already took linear algebra but then I don't have a prof to answer to

[azure copper]

lucky, that aint my case

[teal vessel]

if you can show that the matrix multiplication results in a commutative structure on a per-element basis, you can skip some of the right/left distinction

things like how a(1/a)+0=(1/a)a+0 for the two matrix multiplications

[teal vessel]

but what if I wanted to show that for each vertex pair in a cube there was an associated face pair in an octahedron with the same threefold symmetry as the vertices? dangit why must this be a problem for abstract algebra to solve rather than showing a geometric isomorphism? lol

(yes this is actually saying the same thing, just using geometric language instead)

[delicate orchid]

yeah, that's what it means for the shapes to be dual

so label the octahedron in this way corresponding to however you've labelled the cube

[teal vessel]

I know how it works lol, it's just funny to me that this is asking you to show it through the algebra side and then linking it with the geometry, rather than leaving it open for someone to use the geometric understanding to justify the isomorphism.

one of my favorite demonstrations is literally the dual inscription construction, where you place the dual inside the form such that the vertices intersect the faces at their respective centroids

[static yew]

My Elements of Algebra book finally showed up and I'm reading about quadratic closures of fields

Is there an example of a finite field other than F2 that is quadratically closed?

It seems like a prime field Fp can only be quadratically closed if p is one more than a square, like 257. Because if p=a^2+1, then a^2 = -1 and if the lower ("positive") half all have square roots then sqrt(-k) = a*sqrt(k). But GF(257) doesnt have e.g sqrt(3), and the bigger the field the less likely it seems that all x^2 - yp = z has solutions for all 1<z<p, though I'm not sure how I'd prove it

Still mulling over prime powers

Am I way off base here

[rocky cloak]

So you're looking in the wrong place a little.

Notice that squaring is a group homomorphism from F^* to F^*. What is the kernel, can it be surjective?

[rocky cloak]

It's possible for -1 to have a square root mod p, when if p is not 1 more than a square. 5^2 = -1 mod 13 for example

[white oxide]

so essentially the external weak direct product of a family of groups is the set of all tuples that have at least one identity element in any of its components?

[chilly radish]

no

ALL of the elements except a finite number need to be the identity

basically you don't allow infinite 'products'

[white oxide]

ohh ok that makes sense thanks

[wraith cargo]

I feel like I need a hint abt how to attack this problem

my first idea was to maybe extract some SES form the resolution we're given and apply the Tor long exact sequence but I don't think that pheasible

that middle multiplication map is sorta weird to me

hm could I maybe calculate the homologies of the resolution 0-> rR -> R -> R -> 0?

[wooden ember]

what does the notation rR mean here?

(subscript r)

[wraith cargo]

Elements of R that are annihilated by r

Same for rB

i.e. here our choice of r is that it's a zero divisor

[rocky cloak]

So if you just take a projective resolution of _rR, you can tag it on to get a resolution of R/r. Then tensoring with B and taking homology will immediately give you the second part. And I think if you just carefully write down what the kernels and images becomes you should get the first part, but I haven't checked

[wraith cargo]

New idea:
Since every R-module has a projective resolution i can probably extend the identity of R/rR to a morphism of resolutions and do something with that

[wraith cargo]

Hmmm will try that

[wooden ember]

ok isn't this just the start of the les on the ses 0-> R/_rR -> R -> R/rR -> 0 or am i being dumb

yeah no im being dumb

[rocky cloak]

I checked and it works

Remember the two different ways to compute homology, and that tensoring is right exact

[white oxide]

so i'm a little stuck on this question. i let sigma: F --> F be the identity isomorphism, and extended it to an isomorphism tau: Ebar onto a subfield of Fbar. i guess my trouble is showing that the image of Ebar is iequal to Fbar, and what i tried doing is considering the inverse: tau^-1: tau[Ebar] --> Ebar, and deduced that tau[Ebar] = Ebar so that tau is onto Ebar and thus Fbar. don't know if that logic works or not (tau is still an isomorphism regardless)

[wraith cargo]

Sorry what do you mean two?
I just know the ker/im way?

[rocky cloak]

You also have the dual way, i.e. the kernel of the induced map from the cokernel

[wraith cargo]

Also here
Do you mean like take a projective resolution of rR and then replace rR in the exact sequence with the sequence of projectives?

[wraith cargo]

Hm interesting

[rocky cloak]

So for A-f->B-g->C the homology at B is the kernel of the map cok f -> C

[rocky cloak]

Yes

[wraith cargo]

Ah interesting
I don't think I've heard that before (or that Weibel mentioned it)

Tho it makes sense

[rocky cloak]

I think you need it to prove the LES in homology, unless maybe you just do some horrible diagram chase

[white oxide]

any ideas?

[rocky cloak]

So you have tau^- : tau[Ebar] -> Ebar. Then you extend that to tau^- : Fbar -> Ebar, and that's it right?

[white oxide]

oh right oops i forgot tau[Ebar] lives in Fbar

got it ty

[white oxide]

can somebody help me understand why q(x) divides irr(alpha_i, F)?

[chilly radish]

irr(a_i,F) can be considered as a polynomial over F(a_1,...,a_i-1), wherein it is a polynomial with root a_i

the result follows from the defn of the minimal polynomial

[white oxide]

oh irr(alpha_i, F) = f(x) has alpha i as a zero and so q(x) divides it

duh

[chilly radish]

yup

[white oxide]

right

thx

[chilly radish]

np

[white oxide]

now im tryna see why that implies that alpha_i is a zero of q(x) of multiplicity 1 lol

oh

alpha_i separable over F implies that irr(alpha_i, F) has all zeros of multiplicity 1

[white oxide]

i have a question

why do you have a santa hat on if it's not christmas

[chilly radish]

I refuse to change

and no one was willing to make me a pride month pfp

[chilly radish]

also yes, exactly

[white oxide]

fascinating

[white oxide]

why is f(x) in F[x] if they gave the factorization in Fbar (one of the alpha_i need not be in F right)

[fossil hazel]

how do you find the 3rd derivative of this?

[formal ermine]

wrong channel?