@gian just map everything to $0$!

I mean, to $0$, not to $0!$.

@AlexClark Memes are my life now. It's awful, I know

Monomorphism

Oh yeah Leaky has a point you need "nontrivial"

So there is a morphism?

@gian there's a very trivial one

There is but it's a troll

so trivial that you would bash your head against a wall

@skullpatrol Dave? Is that you? It's been so long!

Huh, so the question is poorly worded.

Thanks.

The "map everything to the identity" map always exists between two groups

There aren't any other maps here though

ugh

Well why is that?

Nice that 0! = 1 and 0 != 1 are both true

@gian if $f(1)=1$, then $3=f(1)+f(1)+f(1)=f(1+1+1)=f(0)=0$

and

you know

Yeah $f(1)$ satisfies $x+x+x=0$ in $\Bbb Z_4$ so it's determined, similarly for $f(2)$

@AkivaWeinberger that's much better

2! = 2 and 2 !=... Shit

Can I random for a sec

@AkivaWeinberger sure

yeaboi

this is dank

wait

octahedron is dual of cube right

so rotation of octahedron is also S4!

> Give a man a fish and feed him for a day. Don’t teach a man to fish, and feed yourself. He’s a grown man. And fishing’s not that hard.

now the 4 vertices of the tetrahedron can permute but alternatingly: so it is A4

How are you taking the factorial of a group now?!?

@Daminark $\Bbb Z!!!!$

(Dragon ball)

@Daminark That's nothing, I can take the factorial of an entire sentence!

But yeah so tetrahedron is A_4, cube/octahedron/cuboctahedron are all S_4

There's this amazing piece of geometry that does the icosahedron/dodecahedron

Well first there's gotta be like 5*12 (or 3*20) elements to this thing, so 60

@Daminark ?

Langlands is this thing to relate algebraic number theory and rep theory/algebraic geometry

Specifically Galois groups and algebraic groups over local [REDACTED]

I was really just making a thing about "Tetrahedron is A_4", as if they're the same object

Homomorphism of Q-vector spaces is homomorphism of Q-rings

Homomorphism of Q-rings is homomorphism of Q-vector spaces

A Q-ring is a bicycle part innit

who am I kidding, of course not

TIL $2\Bbb Z/8\Bbb Z = 2/8 = 1/4$

qed reacs only

...

reacts

TIL $3\Bbb N < 6\Bbb N$ because $3<6$

oh notations do you really have to reverse everything

See you might think our notations sucks, but really [knocks Leaky over the head awkwardly and runs away]

what is the connection between topology and abstract algebra?

@Balarka I summon you

I'm too busy listening to Shooting Scatstorms

You can give each topological space a bunch of algebraic objects. The most famous example is the fundamental group.

@LeakyNun Well, there are multiple connections.

@BalarkaSen I'm all ears

There's also the homology groups (these are all abelian) and the cohomology rings.

And higher homotopy groups, but I'm 89% sure they're evil

Akiva just explained some

Wait hold on I thought you knew about fundamental groups @LeakyNun

The connection is fine af too. Affine connection, you may say

Jesus I'm too tired for this shit ^

@AkivaWeinberger I know some fundamental group

Higher homotopy groups are abelian as well

But they're pretty obscene

so what are the other connections?

and connection from algebra to topology?

So the point of a lot of these algebraic structures is that they're invariant under homotopy equivalence

But their constructions out of topology can be abstracted away

@LeakyNun I think by "multiple connections" he just meant "there are multiple ways to assign algebraic objects to topological objects"

i.e. fundamental group $\pi_1$, homology groups $H_n$, etc

what is the symmetry group of beehive?

there's two translation generators

one rotation generator

one reflection generator

$\Bbb Z^2 \times D_6$?

There's the power generator

@AkivaWeinberger ?

of the flamethrower I'm pointing at the beehive

loool

imgur.com/rDClx10

i think Gromov once wrote that the apparent flabbiness of topology is just the presence of homotopies, and that everything becomes as rigid as a crystal once you mod out by it

...Flabbiness?

yeah like how everything is all stretchy stretchy

because the instructor is a freaking normie

@LeakyNun Maybe he wants to see $a+H$ and $b+H$? I dunno

@AkivaWeinberger I'd use multiplicative notation here

since additive notation usually is for abelian groups :P

Nah, it's taking place in $G$

oh wait, $G$ is abelian.

$a+H$ is a subset of $G$ (being an equivalence class)

13 chapters later, the author introduces ideals and basically refers back to normal subgroups for like every theorem and exercise

Can you mod out by freaking normies

Normie subgroups

Section 27: 20. Let $R$ be a commutative ring with unity of prime characteristic $p$. Show that the map $\phi_p:R \to R$ given by $\phi_p(a)=a^p$ is a homomorphism (the Frobenius homomorphism).

But yeah they're essentially the same thing, they're "things you can quotient by"

this is cool

Yerp

Doesn't do much on $\Bbb Z_p$, though, does it

oh, the frenshman exponentiation :P

"Frenchman"? Wait 'til @Astyx hears about this

Section 19: 41. (Freshman exponentiation) Let $p$ be a prime. Show that in the ring $\Bbb Z_p$ we have $(a+b)^p = a^p+b^p$ for all $a,b \in \Bbb Z_p$. [Hint: redacted]

@AkivaWeinberger oops :P

Oh lol

correction: it is section 26, not 27.

mmm, $p|\binom{p}{k}$ and all that

@Semiclassical right

Ah yeah there's two reasons for that one, one of which is monumentally more stupid than the other @LeakyNun

(unless k=0,p of course)

For $\Bbb Z_p$ specifically

@AkivaWeinberger I wanna hear the stupid one

$x^p=x$ always

@AkivaWeinberger what

It literally just says $a+b=a+b$ there

@_@

What's the name of the thing

the cake is a lie

This is Fermat's little theorem, isn't it?

Fermat's little theorem?

right

right

Snaped!

why haven't I thought of that @_@

But yeah the other reason generalizes to other rings of characteristic $p$, where the Frobenius map is not the identity

@AkivaWeinberger the binomial expansion?

Yeah

Like in rings of characteristic $3$, we have $(a+b)^3=a^3+3a^2b+3ab^2+b^3$

$U(8) \cong V_4$

$U(10) \cong \Bbb Z_4$

which is $a^3+b^3$ 'cause the middle two suffer from $3=0$

and they die

Rest in kill

@AkivaWeinberger can you show me a finite ring with char. $3$?

nvm, RG. where G=Z_2 and R=Z_3

:39714659 lol, apart from that

$\Bbb Z_3[x]/\langle x^2+1\rangle$ I think

@LeakyNun What?

@AkivaWeinberger that's interesting

$\Bbb Z_3[i]$

Yeah

was about to complain about that [x] being missing, lol

@AkivaWeinberger group ring

apparently only my book uses the notation RG

Yeah mine's a field

@Secret hi

@AkivaWeinberger a field extension

Yeah that

What book are you using?

quick proof that there's no field with 6 elements (trick question)

The Good Book

@Daminark a first course in abstract algebra, John B. Fraleigh, seventh edition (2003)

@AkivaWeinberger not

@AkivaWeinberger you in?

I dunno that book but if you don't like it, try Herstein or Hungerford

@Daminark when did I say I don't like it, lol

@Daminark no, The Good Book is referring to Ted's book

Ah

@Daminark you in?

Assume you have a field with 6 elements, its additive group is $\mathbb{Z}_6$

@AkivaWeinberger the multiplicative group is $\Bbb Z_8$?

@Daminark there's a quicker proof :P

wakes up wha-What?

@AkivaWeinberger of $\Bbb Z_3[x]/\langle x^2+1\rangle$

What?

Quicker proof that it's a what

Oh OK

I'm asking two questions at the same time

Oh I mean what's the characteristic?

2. multiplicative group of $\Bbb Z_3[x]/\langle x^2+1\rangle$ is $\Bbb Z_8$?

You're screwed there

wait, I messed up in my quick proof. It's not a proof at all.

Rip, what'd you have?

$x$ generates it doesn't it

Wait no

@Daminark it's very stupid

I was gonna say there's no group of order 5 (because the multiplicative group has order 5)

x+1 then?

@AkivaWeinberger $x+2$ is generator

Ah OK

$x+1$ works as well

What's its inverse

ya I just looked up my table above

Arright yeah

to find $a^2=i$

Yeah that'd work

simple proof :P

I think I gotta go

Bye

Section 26: 7. The kernel of a homomorphism $\phi$ mapping a ring $R$ into a ring $R'$ is $\{\phi(r)=0'|r\in R\}$.

What's the problem with this?

@AkivaWeinberger ok bye

Wait isn't that right?

@Daminark In exercises 5 through 7, correct the definition of the italicized term without reference to the text, if correction is needed, so that it is in a form acceptable for publication.

So it might be right

I mean if you want to be annoying you have to swap $r\in R$ and $\phi(r) = 0'$

@Daminark oh!!

$\Bbb Q$ is locally cyclic: any finite subgroup is cyclic

@Semiclassical my equations, don't recall posting anything like that within the past 3 months???

@LeakyNun That looks like the equations of quaternions

@Secret yes

they are the group presentation of the quaternion group

@LeakyNun It seems you are back to the terrible avatar.

@WillHunting right

Hmm, I think I will delete my account soon.

again?

Yeah.

@Secret you can, it just takes more effort

you might like level 2:

oops

careless mistakes on line 3

@Secret Maybe the next time I come back, you will no longer be Secret, LOL.

@Secret?

nice

In general for the above "cyclic permutation like structure in groups", the words for the identity is found by picking a starting element (say a), and then follow the equations given in sequence by keep acting elements on the left (or right), eventually it will map back to the starting element. The word that accummulate on the left (or right) must be the identity if this is a group (or more generally if the algebraic structure has a two sided identity)

interested in level 2?

I am now on mobile. Expect slow typing

ok

a, ga, aga, baga, cbaga, dcbaga, edcbaga, fedcbaga. Thus fedcbag is the identity. To be simplified in terms of a

interesting

My account will be deleted in 24 hours, bye guys.

Why?

fedcbag=fedabbag=fedagagaag=decddagagaag=bccdabbcbcagaggaag=gaababbcagagaabgaaba‌​gaggaag=gaagaagagaabagagaagagaagaagaggaag=gaagaagagaagaagagaagagaagaagaggaag

Hey @Tobias!

And oh god @Secret

@Daminark Hi

How's everything going?

Good

Oh actually I'm wondering, how much background would be needed for homological algebra?

@Daminark Depends on what the aim would be

Also depends on how general one would like to do it.

The description: An introduction to Grothendieck's six functor formalism, aimed at second year students.

I have no idea what that is :)

Same

Sounds like a finishing move in a fighting game

I'll just ask the guy teaching

LOL

second year undergraduate or second year graduate?

Grad

ok

Yeah gonna audit at most

if you do it you're gonna have to take notes

for me

lol

Lmao

@AlexClark The name doesn't ring a bell

@AlexClark Well, given a vector space and a basis, there is a unique inner product which makes that basis orthonormal

i'm having a bit of a hard time wrapping my head around the concept of a "face" in a planar graph. are faces basically the boundaries of the regions of a planar graph?

@Max No, the faces are the connected components you get when you embed the graph in the plane and then remove the graph

given a graph like number I (the triangle one) here tutorialspoint.com/graph_theory/images/cycle_graph.jpg

how many faces are there? i would expect 2

but aren't they the same cycles? abca for the inner and abca for the outer?

@Max Yes, those all have 2 faces each

@Max A face is not actually a cycle in the graph

ok, from my understanding, faces are defined by their cycles?

ah, it isn't?

Only almost, as can be seen by these examples

Essentially, they are defined by the cycle and an orientation

to tell you which side of the cycle you are on

Of course, in many cases, there is only one possibility for a given cycle (or none)

ah yeah i see, that makes sense to have the orientation

but what does "connected components" then mean?

a component is a region?

yes

and they are connected via the faces?

so in the triangle case, you cut out the lines to get a region shaped like the triangle and a region shaped like the rest, and they are not connected to each other because you have removed those lines

@AkivaWeinberger Make a man a fire, and he'll be warm for a few hours. Give a man a box of matches and he'll be warm for a few days. Set a man on fire and he'll be warm for the rest of his life :^)

(you can think of all the lines having an actual thickness if you wish)

@Steamy Copyright violation (minus the matches)

steamy got sniped

awww, rip

ok i see. in my text, when they talk about faces, they usually refer to them as "closed walks" defined by the vertices along that walk

i'm a bit confused how to define the faces then in the example of a planar graph with two faces, since the outer and inner face cross have the same walks

Is there a name for objects X st. for a functor F there are no higher derived images L^i FX?

@Bubaya $F$-acyclic or just acyclic

(I assume by "no images" you mean that those are all $0$)

Also, this is probably only used when $F$ is right exact so that $L^0F = F$

@TobiasKildetoft: Thank you. Yes, I mean L^i FX=0 ∀i>0 for F right exact.

@Bubaya Then those are indeed called acyclic (or $F$-acyclic when one wants to make sure to specify $F$)

@TobiasKildetoft: I am a bit astonished b/c Weibel calls an object F-acyclic (for F a triangulated functor) if H^i FX=0 ∀i.

@Bubaya Hmm, that seems odd

@TobiasKildetoft: But I see, Weibel's convention seems unusual among other writings. Thanks!

I took the term from Jantzen's book on algebraic groups, but it seemed to be standard, as it is part of the requirements for the Grothendieck spectral sequence

Now, if Weibel had only required it for $i > 0$ that would be essentially the same as this

(up to some niceness properties or whatever, to get rid of negative cohomology degrees)

Maybe I have read over some boundedness conditions or so, dunno.

Btw, have you ever come across writing L^i instead of L_i for the left derived functor? This seems a bit odd to me.

@Tobias Nori responded and the class has been upgraded and is no longer a second year grad class...

@Daminark So it is what? A third year class?

Not sure, he just said it'll be very advanced

Hey @Tim!

I see

that sucks

Oh well

so much for homological algebra

I'll find other things to do

tru

@Daminark Hey Dami

How's it going?

Well I started college a month back, and so far its going great.

Does "Please show me your license, buddy" ring a bell by any chance?

And that's neat, what are you taking?

The answer to the first is no, and I'm doing a bachelors of math and computer science.

Nice

hi chat

hey alessandro

Hi @AlessandroCodenotti

sanity check: Z2 x Z4 has 4 prime ideals and 2 maximal ideals

the latter are Z2 x 2Z2 and {0} x Z4, and the remaining of the former are Z2 x {0} and {0} x 2Z2

@TobiasKildetoft

@LeakyNun prime ideals and maximal ideals are the same for finite rings

gg

right, coz like finite id is field

so there are 2 max and prime

@TobiasKildetoft

@LeakyNun I didn't really think much more about it. There are of course more subgroups of order $4$, but I suppose those are not ideals

I just think about the factor ring :p

(right, ideals are direct products of ideals in the factors)

coz like max ideal <=> field

Sure, and clearly the maximal ideals here are precisely those of order $4$

I have no idea how could one get eleven a s out of this:

gaagaagagaagaagagaagagaagaagaggaag

lmao

field of order 4: 0, 1, i, i+1

We need some kind of equation analogous to ii=jj=kk=ijk to simplify it further

wait, x^2+1 is reducible gg

@LeakyNun There is clearly no way to get the field of order $4$ as a quotient here

@TobiasKildetoft how clear?

right, coz like F4 isn't decomposible

@TobiasKildetoft what's your favourite F4?

@LeakyNun Basically because you would need an element to get multiplicative order $3$ in the quotient

I don't have a favorite one.

the additive group of F4 is V4 right

right

is there only one F4?

let's represent F4 by 0,1,2,3

with addition being xor

@LeakyNun Up to isomorphism, yes

that leaves us with 2x3 = 1

done

It just looks like some keyboard mashing

@SteamyRoot look up Fibonacci group. I'm having him prove that the order of F(2,7) is 11

Wut

I don't get why I cannot simplify this given it has such a nice structure

It's hard to prove something that isn't right :^)

@SteamyRoot Wait so you mean the order of F(2,7) is is not 11?

Pretty sure that's the order of $F(2,5)$

google agrees, the order of F(2,5) should be 11 and F(2,7) should be 29

Hello everyone. So I am reading this article and the author uses alot of | (pipe) symbols. Do they actually mean anything, eg in the context "A set of m| artificial ants construct solutions" (here), or is he just using them to separate math/formula from text?

$F(2,6)$ is the best Fibonacci group anyway :3

The only thing I am sure about is that, if I start with a, then fedcbag is the identity since (fedcbag)a=a. However I have trouble trying to simplify it in terms of just powers of one element. The best I got from this is fedcbag=gaagaagagaagaagagaagagaagaagaggaag

@Secret I apologize :p

you may want to show instead that a^29 = 1 bahaha