groups-rings-fields

[unique juniper]

didnt understamd

[carmine fossil]

$a_1a_2^{u_2}..a_k^{u_k}=a_2^{v_2}...a_k^{v_k}$

[cloud walrus]

MoonBears-D-

[carmine fossil]

If u_1!=v_1

[unique juniper]

but

G is a set

[carmine fossil]

So?

One side will have a factor of a_1 and the other side will not

If the factor of a_1 is not on lhs,multiply both sides with a_1

[unique juniper]

ok

[cyan marten]

If you have seen Cauchy's or Sylow's theorems, you can quickly prove that |G| = 2^m. If p is any prime dividing |G|, there's an element of order p. But every element has order 2, so p = 2.
Proving that G is abelian is easy: We have (ab)^2 = abab = 1. Multiply both sides by ba to get ababba = ba = ab (because bb = aa = 1).

Actually you don't need the full Cauchy theorem. Since we know G must be abelian (by the second part), you can use "Cauchy's theorem for abelian groups," which is easier to prove.

[golden pasture]

wiki provides a counterexample (https://en.m.wikipedia.org/wiki/Artin–Tate_lemma), more concretely we have

let A be the ring Q[x_1,x_2,...], clearly not noetherian and let I=(x_1,x_2,...)
Let C=A[y]/y^2 and let B=A+Iy be a subring of C
verify the conditions but B cannot be finitely generated as an A algebra cuz you need x_1y, x_2y,... as the generators of the algebra

theres another counterexample that this resembles a lot trying to recall what isit rn

oh ye theres one thing in groups where subgroup of finitely generated group may not be finitely generated:
Take the group generated by the matrices $\begin{pmatrix}1&1\0&1\end{pmatrix}$ and $\begin{pmatrix}2&0\0&1\end{pmatrix}$
now consider the subgroup with $1$s in its diagonal

[cloud walrus]

a cute cat ٩(˃̶͈̀௰˂̶͈́)و

[golden pasture]

the general intuit behind finding nonfinitely generated things inside finitely generated things is somehow restricting enuf such that all your generators are missing and no new finite set can be found

[cyan marten]

Nice! I didn't know what it was called.

[cyan marten]

Oh, yeah. Also the free group on two elements contains free groups of all ranks.

[golden pasture]

yea

[cyan marten]

True!

If we forget about the finite/finite type distinction (maybe we shouldn't...) then all we have really is that C is fg over B and A. It's not obvious now that B must be fg over A.

[golden pasture]

yee

[unique juniper]

if x is the only element in G why is it hinting to consider yxy

[carmine fossil]

yxy^-1 is an element of order 2

[unique juniper]

how do you know

[chilly ocean]

ness

[carmine fossil]

Try it

[unique juniper]

well ok, it is

what im confused with is

x is the only element is G

[carmine fossil]

No

x is the only element of order 2

[chilly ocean]

to be fair

[unique juniper]

-_-

[chilly ocean]

it is worded a bit strangely

[unique juniper]

-_-

hm

yxy-1 is also an element of order two ?

so x isnt the only one?

this is weird

[viscid pewter]

no, x is the only element of order two

and yxy-1 is an element of order two

soooo

[unique juniper]

?

[viscid pewter]

how can x be the only element of order two when yxy-1 is also of order two

there is a way, what is it

[unique juniper]

maybe im missing a theorem

[viscid pewter]

...

it's if they're the same element

it's if x = yxy-1

[unique juniper]

lmao

[viscid pewter]

so how do you use that

[unique juniper]

we can uhh

take a y in G

and we need to show something like

x * y = y * x

[viscid pewter]

ok

[unique juniper]

so just replace x for that when we need it

and done ig

i might actually switch books

[golden pasture]

exactly

it's a good lemma

good name

good proof

good result

[inner acorn]

I understand how to prove this not using the ping-pong lemma

xD

but I don't know how to use the ping-pong lemma for this

[golden pasture]

try to stare extremely hard at it

it shouldnt be too too hard to find 2 regions that satisfies the ping pong lemma i think

lemme stare at it

[woven delta]

This is generally good advice

[wispy scaffold]

Proof by stare at it

Btw what book is?

[inner acorn]

Clara Loh's text "Geometric Group Theory"

[wispy scaffold]

Ty

[inner acorn]

fismo be googlin'

[golden pasture]

ok so

lets just call the matrices say

A B

There has to exist some region R such that BAR = R

(take the principal axis, and expand to a cone)

then select region r such that Ar is not in r but BAr=r

then apply ping pong lemma

[inner acorn]

[golden pasture]

a lot of "visualize this"

kinda painful but writing it explicitly is more painful

i believe this construction shows any 2 rotations that are on diff axis and not by a rational multiple of 2π should be free

[inner acorn]

thank you for the explanation, will keep working at this, and if I come to any explicit conclusions I'll ping ya

[wispy scaffold]

There is some similar content in the second chapter of The Banach Tarski Paradox book

[inner acorn]

the one by Wagon?

I'm aware of how to prove this using a more.. number theory method, as well as the whole "pick 2 rotations of two axes that are irrational multiples of 2pi"

there's some pretty interesting theory

I was unaware this proof could be done using a straight forward application of the Ping-Pong lemma, since I've only seen it applied to SL(n, Z)
otherwise the regions get a little higgledy-piggledy

I see where Ariana is coming from o: (pretty sure she meant BᵐAⁿR = R?), it's just finding such a region xD

[wispy scaffold]

yes

[golden pasture]

ah yesh oops

mbmb haha

[golden pasture]

yee

[smoky cypress]

What are the axioms for A-algebra?

[golden pasture]

this

have a ring homomorphism A->B

and then like

it satisfies both axioms for an A-module mixed in tgt with those of a ring

so kinda like multiplication distributes over addition

what are the other ring axioms idk

f(a)1=f(a)

f(a)b=bf(a)

[smoky cypress]

Ok

So it’s ring B with an A module structure

[golden pasture]

really just define an A-algebra as (B,f) where f:A->B is ring morphism and B is an A-module with multiplication defined by ab=f(a)b

[smoky cypress]

Ok

Well I asked because algebra over a field has a different definition for some reason

Well it’s still equivalent

But different

[golden pasture]

i mean it kinda is the same

just :s/field/ring/g

[smoky cypress]

Fair

[chilly ocean]

Speaking of

B is a flat A-algebra means, B is an A-algebra (hence also an A-module) and B is flat as an A-module

?

[sturdy marsh]

yes

[chilly ocean]

Do you have a hint of what I should look at for ii

I know there are solutions online but I’d like just a hint

I’m not sure how to invoke the property of B being an A algebra

[sturdy marsh]

try to write the thing as a composition of exact functors

The reason why we want B to be an A algebra is so that you can consider modules over B as modules over A

[next obsidian]

I just used rewrite rules

Altho I think you might have needed to sneak in a (x)_B B in there somewhere

[golden pasture]

just play with the properties of tensor AM gives you and have a long series of trivial equivalences and it magically falls out

[molten silo]

How would i show this

[chilly ocean]

hint: Q is a field

[uncut girder]

Q is a FIELD!
'

[cyan marten]

Assume the ideal J is nonzero, take a nonzero element of it and use the definition of an ideal.

[chilly ocean]

This is the thing that confuses me lots

[uncut girder]

Q is my favorite field

[chilly ocean]

Say M can be considered both an A module and a B module

[molten silo]

Is a field just a group under multiplication?

[next obsidian]

This is called restriction of scalars

B being an A algebra is the sam thing as a map phi: A -> B

Such that the image of A is a subset of the center of B

[chilly ocean]

Yeah so if we have a a ring morphism from A to B we can do this

Which we do

[next obsidian]

Okay so

The way you consider a B module as an A module isn’t where you’re confused?

[chilly ocean]

What I am confused about is, it does not seem that these module structures are interchangeable. But we seem to use them as if they are. For example phi from A to A that sends everything to 0 is ring morphism so we can view M as an A module in two ways and they don’t seem related

[next obsidian]

Well A isn’t flat over A in that case

Err

I guess it is

They are related tho like

Consider M as an A module like under the second version

And then if you apply restriction of scalars you’re now saying that am = 0 for all a, all m

Like...

The module structures are different

And they’re built up on the same set

[chilly ocean]

So when people say ‘now view M as an A module’

[next obsidian]

Well

[chilly ocean]

We really have a map

[next obsidian]

In this case it’s a bit weird

Since also B = A

[chilly ocean]

From B-modules to A-modules sending M as B module to M as an A module

[next obsidian]

Right

Yeah it’s functorial

[chilly ocean]

Call such a thing a bi-module

If M and N are bimodules, it’s not nesccairly true that M tensor_A N is isomorphic to M tensor_B N

[next obsidian]

Yes

Also a bimodule is a bit different haha

Altho I guess for commutative case it’s all sorts of weird

The thing is tho

You have the following result

If N is an (S,R) bimodule, M an S-module and L an R-module

Then (M (x)_S N) (x)_R L = M (x)_S (N (x)_R L)

So it’s associative even tho you’re changing what you’re tensoring over

[smoky cypress]

For some reason these two propositions don’t assume nor conclude the full short exact sequence

But is it true that a short exact sequence is exact iff the hom sequence is exact

I meant a short sequence

If that makes sense

[chilly ocean]

I was just asking myself this earlier

The tensor product is always right exact

Oh sorry that’s a different question all together

[chilly ocean]

This is because Hom isn’t an exact functor

[chilly ocean]

This is not true

The common example is the exact sequence 0 to 2Z to Z to Z/2Z to 0

[smoky cypress]

Wait the tensor product isn’t exact either

[chilly ocean]

Hom(Z/2Z,2Z) and Hom(Z/2Z, Z) are both zero. The thing you have to be careful about is to remember that our morphism of commutative rings my definition is unital

[smoky cypress]

Yeah

[chilly ocean]

Yeah the tensor is only right exact

[smoky cypress]

Makes sense

What does right exact mean?

[chilly ocean]

F being whatever functor

[smoky cypress]

Oh

I see

[chilly ocean]

I am also working through atiyah atm

[smoky cypress]

Nice

Ok so both hom functor (one is reversed) are left exact

[chilly ocean]

Feel free to @ me whenever

[smoky cypress]

Alright

[chilly ocean]

Yeah

[smoky cypress]

This chapter of modules is just pain

[chilly ocean]

I might not know the answer but it’s nice to work through things together

Yeah I’m just trying to focus on doing the exercises

[smoky cypress]

I kind of just finished the chapter

I’m trying to review

Before going into exrercises

[chilly ocean]

Oh I might be wrong about the terminology of left vs right exactness since Hom since one of them is contravriant (flips the order)

[next obsidian]

Did you ever figure out how to show the thing about a flat B-module being A-flat?

I can give you a proof but it sounded like you were trying to understand why it’s true intuitively

Imo it seems kinda intuitive to me, it’s just like a transitivity sorta thing

[sturdy marsh]

both homs are left exact

[loud root]

Hi, I'm doing an assignment and one of the questions asks me to prove two groups (in particular a subgroup of GL(3, R) and R^2) are isomorphic and I'm using the first isomorphism theorem, and I think I have the solution. The problem is that I don't have my head around it, I have used the definitions and it's worked out but I don't have a firm grasp on understanding what exactly all of it means. Does anyone have a good way to visualise the first isomorphism theorem?

[next obsidian]

I always find it odd when people want to visualize things that are purely algebraic

You might get some amount of headway thinking about real vector spaces or something, but I don’t really see how visualizing comes into play when you deal with GL(3,R), the isomorphism you make let’s you port this over to R^2 and then for the particular subgroup you can start to visualize things maybe

[old lava]

I mean I don't find it odd, humans like visuals

and I think visuals are pretty important to help motivate certain things

or to help understand things

but after a point, it's also important to remember when things can't be visualized

in nice, intuitive ways at least

[loud root]

i feel like without some visualisation i can't remember it or properly understand it

[old lava]

that might be a bit of an L considering the whole "abstract" part of abstract algebra

[loud root]

fuck

although

generally i'm good with concepts in abstract algebra so far, except for statements where there's a lot going on

here i get a bit confused

[sturdy marsh]

'abstract' doesnt make it less geometric tho

in fact more in many cases

[loud root]

G/K is the quotient group , i.e. all left cosets of K in G (and K must be a normal subgroup) ?

[old lava]

I mean it's still important to be able to abstract things out

[loud root]

but is this equivalent to G minus K ?

[sturdy marsh]

no

assuming you meant setminus

[loud root]

yes

set minus

G and K are sets (groups*)

i wouldnt subtract them as numbers...

[sturdy marsh]

yeah they don't have anything to do with each other

[loud root]

i see

[sturdy marsh]

quotients take a while to 'get'

[loud root]

I remember learning about this theorem last year at the end of my linear algebra course without being taught normal subgroups

but quotient groups are defined by normal subgroups right?

[old lava]

I personally found lattices nice to help me understand quotients

[loud root]

what's a lattice

[old lava]

a way of visualizing group structure

[sturdy marsh]

yeah or you could define them using a group operation on the fibers of a map

[old lava]

in some sense, the quotient groups appear at the "top of the lattice"

[loud root]

this has me thinking of structure of chemical compounds and salads

yeah i'm lost

[old lava]

a lattice basically partially ordering every subgroup of a group by inclusion

for example. this is the lattice of D_8 (or D_4 depending on which notation you like)

suppose you wanted to know what the quotient group of D_8 by the subgroup <r^2> is

what that does, is it "collapses" <r^2> to 1

[loud root]

this is the symmetries of a square?

[old lava]

yes

so the lattice of the quotient group is the red part

because everything else gets collapsed to the identity

that's how the lattices of quotient groups are contained in the main group, near the "top" of the lattice

(the lattices of subgroups on the other hand are contained near the "bottom")

it's a neat way of visualizing things

it does get too much to handle as your groups get more complicated

but it's what helped me to develop some intuition for quotients

[loud root]

ok i kinda get it

[old lava]

you can also sometimes see how certain groups are contained in other groups as isomorphic copies

like this

you can see that an isomorphic copy of the klein 4-group is in D8

[loud root]

so these tiers are decreasing with order

[old lava]

yes, each row are subgroups of the same order

[loud root]

and a subgroup is contained within the one on linked above it

and this lattice shows every possible subgroup ?

[old lava]

yes, well you can sometimes make a partial lattice

if you only care for some of the subgroups

[loud root]

okay

[old lava]

but ya, that's what you can think of happening when you take quotient groups or talk about isomorphic copies of groups in other groups

[loud root]

so this shows that <r> is the normal subgroup?

[old lava]

no, <r^2> is the normal subgroup here

[loud root]

oh

[old lava]

this is D8/<r^2>

<r^2> becomes the identity

in the quotient group

so <r^2> is the "new 1"

and becomes the bottom of the quotient group's lattice

[loud root]

i see

[old lava]

by the fourth isomorphism theorem, every subgroup that contains <r^2>

is reflected in the lattice of the quotient group

[loud root]

wait wait i haven't reached the second theorem yet x')

but thanks, this is super neat

i wonder my lecturers didn't show us these lattices

[old lava]

they're mostly a visualization tool

not really a necessary thing to learn group theory

so I guess for time reasons or something they might've skipped past it

[smoky cypress]

Is it true that $\frak{a}\otimes A=\frak{a}A$ where $\frak{a}$ is an ideal of a commutative unital ring $A$

[cloud walrus]

Whoever

[loud root]

does this tie in with number theory? so far i've only come across ideals and rings in number theory

[smoky cypress]

Well idk

Rings and ideals are definitely not only limited to number theory

[chilly ocean]

It’s true for and A-module M that M tensor_A A is just M

[smoky cypress]

Oh wait

[chilly ocean]

For any *

[smoky cypress]

Yeah

Lmao

It’s just a

Brain dead moment

[chilly ocean]

Well aA is just a

But it is often written like aA

To emphasis you are viewing it as an A-module not just an ideal of A

[smoky cypress]

You mean small a right?

[golden pasture]

(a) is an ideal, also a module as well, (a)=aA

[smoky cypress]

Yeah

Just realized

[simple valley]

is there an elementary reason for why the additive group Q is not iso to QxQ

the simplest argument I came up with is that there is a way to quotient QxQ by Z, such that there exists a monic Z -> QxQ/Z

whereas in any quotient Q/Z (they're all the same), all elements are of finite order

[viscid pewter]

this might be really dumb but

you can quotient QxQ by Q and get Q out right?

so it can't be the same

[simple valley]

how do you should you can't do that in Q?

[viscid pewter]

because if G/G = G, G = 1?

[simple valley]

you're running into an issue with set theoretical notation

[viscid pewter]

i don't understand the issue

[simple valley]

if you're talking about things being invariant under isomorphism

then having specific subsets is not invariant under iso

[viscid pewter]

i am very confused

[simple valley]

when you write G/H where H is a "subset" of G, the isomorphism invariant version of the statement is that you have a monomorphism H -> G which you're quotienting by

[viscid pewter]

look: QxQ has a subgroup Q

oh wait

no, yeah

[simple valley]

Z has a subgroup Z

in multiple ways

[viscid pewter]

QxQ/Q is iso to Q

ah, bugger

yeah really dumb

[simple valley]

ok so set theoretically, Q is not a subgroup of QxQ

[viscid pewter]

whaaat

[simple valley]

something like {1}xQ is

or Qx{1}

[viscid pewter]

oh jesus

yeah ok fine

[golden pasture]

kinda like saying Z isnt subset Q at this point

[viscid pewter]

there is a subgroup of QxQ isomorphic to Q

[golden pasture]

but interesting qn

[simple valley]

maybe you can show that there is a proper subgroup iso to Q

whereas in Q that's not the case

[solemn rain]

Q cant be the direct product of any two non trivial groups

[viscid pewter]

'is to to'

[solemn rain]

suppose Q = H x G

then like take an element and then like project it into G ; f:H x G --> G

[simple valley]

and then?

[solemn rain]

this has H as a kernel

the kernel*

f:Q-->G

[simple valley]

and then?

this is all trivial so far

[golden pasture]

you know whats a very slick way instead

find the endomorphism ring

[simple valley]

Q vs Q x Z2?

no...

I'm thinking Aut

[solemn rain]

H must be zero

^

[simple valley]

why?

[solemn rain]

cuz this homo must be injective

[latent anvil]

from Q?

[simple valley]

why?

[latent anvil]

are you thinking of ring homs?

[simple valley]

you can most certainly project Q onto things with nontrivial kernel

[golden pasture]

theres only one noninvertible endomorphism on Q

[latent anvil]

what's Hom(Q, Q)?

I think it should be determined by where 1 goes

[golden pasture]

but there are many noninvertible endomorphisms from GxH to itself

[solemn rain]

Hom(Q,Q) is functions of the form f(x) =ax for some a in Q

[golden pasture]

ye

[latent anvil]

yeah, that sounds right

[simple valley]

multiplicative Q

[latent anvil]

it's linear stuff

okay so homming into Q is not helpful

[solemn rain]

this is bijective

[latent anvil]

since we get Hom(Q, Q) vs Hom(Q, Q) x Hom(Q,Q)

[golden pasture]

here you get your contradiction already

[latent anvil]

oh yeah good point ari

I agree with this argument

oh wait

[golden pasture]

i.e. End(GxH) isnt a domain for nontrivial G,H

[latent anvil]

here's an elementary way, I think

any abelian group map between Q vector spaces is Q linear

dimension is isomorphism invariant

[golden pasture]

mm yes lin alg spam

[latent anvil]

idk maybe this does not count as elementary

but I think "this is linear algebra"

[simple valley]

probably not

[latent anvil]

is a nice way to prove things

[simple valley]

I think I'll stick with "has a proper subgroup iso to Q"

[golden pasture]

it is more elementary in the sense it is just lin alg

[latent anvil]

I mean like, the claim becomes k^2 and k aren't iso as k vector spaces

[golden pasture]

but just takes more pain

[latent anvil]

wait sorry how do we know that's impossible for Q?

[simple valley]

by looking

[viscid pewter]

aw is that a thing you can say

[latent anvil]

proof by inspection

[viscid pewter]

i was so close to that

[simple valley]

so you're looking for a monic Q -> Q, look and the image of 1 and observe it's a surjection

[latent anvil]

ah gotcha

so this is

injective = surjective for linear maps

[simple valley]

what

[latent anvil]

yee

[golden pasture]

yee

lin alg cute

[latent anvil]

abelian group maps Q -> Q are Q-linear

so a monic map Q -> Q is a linear injection

[solemn rain]

im sorry f is from Q to Q

[latent anvil]

so automatically surjective

[simple valley]

that's not what I consider lin alg

[latent anvil]

wait what

what do you consider lin alg?

[simple valley]

maybe I'm just a brainlet

[latent anvil]

linear algebra for me is very broad

[simple valley]

all of my lin alg happened over R and C

[golden pasture]

lin alg is basically modules over a field

uh

[latent anvil]

ah yeah I would go more general than that

for "basic linear algbera"

[simple valley]

I'm vaguely familiar with vector spaces over other fields

[golden pasture]

that is called physicists lin alg

[solemn rain]

if Q is iso to GxH --> we have a map from Q to Q where we send each element to G --> this is a homomoprhism from Q to Q

[simple valley]

but nowhere ready to translate things to them

[solemn rain]

as some1 said

[latent anvil]

that's fair

[solemn rain]

endo(Q) is Q

[golden pasture]

most lin alg intros goes more generally for modules over fields

[solemn rain]

so bijective --> H is 0

is this wrong

[simple valley]

well

that's supposed to be this class that I'm taking right now

[latent anvil]

linear algebra is very broad

includes riesz representation, the open mapping theorem, Tor, and nakayama's lemma

😌

[golden pasture]

at some point it becomes rep theory

[solemn rain]

isnt nakayama ag

[golden pasture]

idk where

[latent anvil]

in my brain all of this is linear algbera

[solemn rain]

CA*

[latent anvil]

oh true ari it also includes artin wedderburn

good point

[prisma ibex]

Nakayama is commutative algebra yea

[latent anvil]

it's linear algbera

it;s just

cayley's theorem

:^)

[solemn rain]

how lmao

[prisma ibex]

[simple valley]

[latent anvil]

lol

[golden pasture]

pushout
rep <- la -> modules

[prisma ibex]

I mean Lie theory is closer to linear algebra than is most of (commutative) algebra imo

[golden pasture]

yea

[latent anvil]

oh true!

[solemn rain]

yes

[latent anvil]

lie theory is also linear algebra

[golden pasture]

lie is very much linear

[simple valley]

just say all algebra is linear because reps exist

[solemn rain]

naive lie theory by someguy

is just linear algebra groups

which is c0o,

col\

[latent anvil]

lie groups, banach spaces, finite group representation theory, homological and comm alg

all of this is linear algebra

[golden pasture]

all of math is secretly lin alg

[simple valley]

well

I'm also taking lie theory right now

[solemn rain]

why is homological algebra linear algebra

[latent anvil]

K theory? did I hear you say *vector* bundles?

sounds like linear algbera

[golden pasture]

[latent anvil]

okay anyways I have an reu app due in < 3 hours and need to start my research statement

bye yall

have fun not knowing linear algbera

[golden pasture]

rip

[cyan marten]

Can we reason by trying to show that there's no way to impose a field structure on Q x Q? But I am not even sure that's the case..

[old lava]

it is the case

[golden pasture]

End(F^+)=F^+

hence yes (no field structure)

[old lava]

in general, the direct product of any two (nonzero) rings is never a field

[chilly ocean]

tfw zerodivisors

[old lava]

there's a bunch of ways to reason it

direct product always has a nontrivial ideal

[chilly ocean]

you're a nontrivial ideal

[old lava]

[latent anvil]

How do you know the ring structure is the product ring?

[chilly ocean]

called "ideals"
literally the obstruction to being a field
hence, not ideal

[golden pasture]

[latent anvil]

This doesn't seem to rule out an arbitrary field having underlying additive group Q × Q

And in fact Q(sqrt(2)) does have that as underlying additive group, right ?

Since it's a degree 2 extension of Q

[golden pasture]

uhh

[latent anvil]

degree 2 extension => underlying vector space is Q^2

[golden pasture]

hm sounds right

yea

[old lava]

oh true, he was asking specifically about the additive group

in that case my answer is not that helpful

[latent anvil]

This is not the case, take any degree two field extension of Q. Eg Q(i)

[cyan marten]

Exactly

[cyan marten]

Ugh

[latent anvil]

also nobody bulli me for coming back to chat, I submitted my application

u mich better accept me or else

[sturdy marsh]

good luck

what did you apply for?

[latent anvil]

Their program is weird

You decide on a topic with a faculty member

If they let you in

[sturdy marsh]

so do you need to mention any faculty members you would like to work with or something?

[cyan marten]

Nice

[vestal snow]

What is the module that represents this matrix?

[1 x 0]
[1 -1 -x]
[0 x x]

over k[x]

for some field k

Can someone give me a hint?

We need to find what it is as "a better known entity"

[latent anvil]

Yeah, I put 4 people

That's what they ask for

Someone who does singularites in char p stuff, someone who does homotopy theory+k theory, someone who does hodge stuff, and Faye's algebraic topology prof

[vestal snow]

So we need to find a good description for R^3/Image of that matrix

[latent anvil]

(who does some kind of AT manifolds thingy)

[sturdy marsh]

cool

[latent anvil]

What does this mean?

I haven't heard this terminology before

[vestal snow]

Basically we have to find a good description for R^3/image of that matrix I just wrote

Where R=k[x]

This was the original matrix which I reduced to that using column ops

I don't know any "special" k[x] modules so I would be very curious to see what the module is

[rustic crown]

convert to smith normal form

k[x] is a pid (infact euclidean)

[vestal snow]

I don't think we're allowed to do that

That changes the image of the matrix

[rustic crown]

nope

say you have a map from R^m -> R^n

this is just a linear transformation...

once you fix a basis on both sides you can give a matrix for it

now will the image of the map depend on the basis you choose?

[vestal snow]

Yeah you're right

[rustic crown]

this is from Aluffi... right?

[vestal snow]

Yes

[rustic crown]

I just finished chapter 7 not too long ago

[vestal snow]

Oh nice

[vestal snow]

@rustic crown I got this matrix:
[1,-x,0]
[0,1,0]
[0,0,x]

I don't know how to make this into SNF

[rustic crown]

you can use row operations to killl the second entry in the first row

and you also know that presentation of block matrix corresponds to the direct sums of modules

[vestal snow]

Will that be reversible?

[rustic crown]

R1 --> R1 + x*R2

this is reversible...

reverse is just R1 --> R1 - x*R2

[golden pasture]

elementary row operations are invertible

[vestal snow]

Oh right

[rustic crown]

so we can delete a the row and the corresponding column if the only non-zero entry is a unit

[vestal snow]

that's an elementary row op

[latent anvil]

sort of

Scaling a row by a non unit isn't invertible

[golden pasture]

we shall work over fields then

[latent anvil]

[golden pasture]

people don't usually consider that a elem row operation do we?

[vestal snow]

Alright I got
[1 0 0]
[0 1 0]
[0 0 x]

[latent anvil]

¯\_(ツ)_/¯

[vestal snow]

Which means that the module is basically everything of the form (f(x),g(x),k)

Right?

Oh wait

[latent anvil]

I would think it's iso to k?

[rustic crown]

no... this is a presentation matrix

[vestal snow]

It's just k

[rustic crown]

yep

[vestal snow]

All that just to get k

[rustic crown]

lol

isn't that nice tho? working with that scary module is just working with the field...

[vestal snow]

@rustic crown Since you've done Aluffi, can you tell me what he wants me to do here exactly?

This is pg 338 for reference

Like is it something like this

Suppose blah = something where blah and something are finite sums and products of Z[x_1,...,x_r], then blah = something for every ring?

[latent anvil]

yup

[vestal snow]

Okay

[latent anvil]

interpreting the xi as universally quantified elements of a ring

[vestal snow]

Got it

Category theory time

[latent anvil]

lol

model theorem time

[vestal snow]

I kinda want to study some of that stuff

But AG 😦

[latent anvil]

Me too lol

Be like me and not really know model theory or ag

I read a book on basic fol stuff at one point but it was a long time ago and I'd like to get a bit deeper

[vestal snow]

I mean I don't know anything about cohomology and schemes

But I'm planning to speedrun death Vakil over the summer

[latent anvil]

nice

[rustic crown]

same... can i join you?

[vestal snow]

Of course

[latent anvil]

at some point I will ram Hartshorne directly into my brain

but I don't have time

[vestal snow]

We can make a study group

[vestal snow]

My advisor said I should do Vakil/Qing Lui now and EGA in grad school

The thing is EGA is in french

[latent anvil]

good thing I'm learning French now

for me it's like

AG takes a lot of time to learn

Other math takes less time

you do the math 🧐

[next obsidian]

I think that

the EGA thing is a massive meme

Would not recommend

I think that's the common opinion as well, a lot of people haven't read all of EGA (I think Vakil even said he hasn't) it's moreso something you just go "f I need a general result time to find it in EGA"

Altho at this point it seems like Stacks Project is better for that

and the stuff I care about at the moment I'm finding myself being pointed to SGA more often now lmfao

[vestal snow]

Honestly, my advisor seems like the kind of guy to say "Oh you read all of EGA? It was just a prank bro"

[next obsidian]

lol

Pick up Goertz and Wedhorn frfr

arguably runs onto being too sophisticated, but it offsets this by actually proving stuff

[vestal snow]

Does this seem correct?

Not related to our discussion Chmonkey

[next obsidian]

what are the r_i?

[vestal snow]

It's from an earlier question I asked Shamrock and det

elements of R

[next obsidian]

oh are you saying like

you have polynomials in Z

they agree as polynomials over Z

[vestal snow]

Look here for the problem

[next obsidian]

I think you can clean it up by noting that an identity is equivalent to saying something is 0

[vestal snow]

Yeah you're right

This just seems weird

Like no two different polynomials in Z[x_1,...,x_n] will be the same

[next obsidian]

But yeah like if you have a polynomial f(x_1,...,x_n) = 0 then you can map x_i to r_i using the univ property of Z[x_1,...,x_n] and then this gives you a map sending f(x_1,...,x_n) to f(r_1,...,r_n)

[vestal snow]

So there is no universal identity

[next obsidian]

and if f is the 0 polynomial then f(x_1,...,x_n) has to map to 0

so that f(r_1,...,r_n) = 0

and that can be viewed as an identity of elements in R

The universal identity is saying that if an identity holds in Z[x_1,...,x_n]t it holds in any ring

I guess you could look at it like this, if you consider an algebraic identity to be saying that

f(x_1,...,x_n) = 0

where f is in R[x_1,...,x_n]

then if you show the identity holds in Z[x_1,..,x_n] it holds in all rings

since you can just send f(x_1,...,x_n) = 0 in Z[x_1,...,x_n] to the same polynomial but viewed in R[x_1,...,x_n]

this is contrasting to say the identity

[vestal snow]

Yeah but my point is that this will never be possible

[next obsidian]

wdym?

like

consider x_1(x_2 + x_1) - x_1x_2 - x_1^2

this is mega contrived but like yeah this is the 0 polynomial

I think this is mainly useful for like matrix crap

or doing stuff with determinant or something

you might get some really trash ass polynomial in n variables and you can move it around and it happens to be 0

[vestal snow]

This is my fifth time learning matrices and though it's much better doing it from Aluffi, it is still my least favorite part of the book

[next obsidian]

I mean yeah same

[vestal snow]

Thanks tho, that explained it

[next obsidian]

I think shamrock had a good example of this

he told me about this thing

i guess the point is

you want to think of f(x_1,...,x_n) as like a formal thing

[cloud walrus]

det

[next obsidian]

Like Z[x_1,...,x_n] is equivalence classes of these formal sums

so like x - x is a different object than 0

but we call them equal

so there's times where it's like not very useful to know that r - r = 0 for any r in a ring R

but if it gets more complicated, and you ahve more variables it might actually begin to be useful

[vestal snow]

How does this look?

[next obsidian]

pretty much yeah

[vestal snow]

This problem just feels weird

Like I know what I want to say, but it seems hard to formalize it

[latent anvil]

the adjugate matrix

you have this matrix A

And you can build another one B

Whose entries are polynomial in those of A

And you want to say AB = BA = det(A) I

It suffices to verify this in Z[x1,...,xn]

I think aluffi does this

[next obsidian]

All of math is reducing things to statements over fields or over Z

[latent anvil]

or combinatorics

[next obsidian]

That's just statements over Z

[latent anvil]

all math is statements over Z because you can godel encode every proposition

[next obsidian]

now this is truly based

[woven delta]

Lol you're so close to rediscovering Computability

[woven delta]

This is actually a valid perspective

Except it turns out there is some math you can't encode unfortunately

But that's why we have higher order Computability

[opal osprey]

I know that the Eilenberg-Steenrod axioms describe how a homology theory of topological spaces should behave, i.e as a sequence of functors that have certain properties.

Are there similar axioms for a homology theory of algebraic structures?

Like, say that describes at the same time how a homology theory of groups, such as the Galois' Cohomology and say a (co)homology theory of certain say commutative ring (idk if that even exists tbh) should behave, both at the same time and more.

Thanks

[light tusk]

Generators and relations pop up all over the place

[prisma ibex]

Kinda. (Co)homology theories can be extremely diverse (lots of extraordinary cohomology theories for spaces, lots of cohomological constructions for algebras with funny properties...)

But the functors will generally satisfy something related to the Eilenberg-Steenrod axioms

[opal osprey]

Hm I see

I thought homological algebra described axioms for (co)homology theories in general algebraic structures.

[chilly ocean]

Can someone help me clarify the universal property of the tensor product over commutative rings?

If i understand correctly: Lets say we want to take the tensor product of two A modules M and N. Suppose we think that P is the tensor product of M and N over A. To check this we define an A-bilinear module homomorphism from the regular old product of M and N to P (call this map b). Then we let f:M\times N to Q be an arbitrary bilinear morphism into an arbitrary A-module.

If we show that this map factors over b, we know that P is indeed the tensor product

awh that diagram came out awful

[prisma ibex]

It does, in the sense that there are very general tools from homological algebra or even stable homotopy theory that give very general constructions for such cohomology theories

it's just that very general cohomology theories that come out of algebra might not satisfy some of these good axioms

you can narrow things down to nicely behaved cohomology theories and then yea there are various sets of axioms you can choose that makes these cohomology theories "behave as one would expect" in some reasonable sense

There's some cases where you can describe an algebraic cohomology theory as a topological cohomology theory: for some algebraic structure, you can sometimes give a procedure for constructing a space such that the cohomology of this space is the cohomology of this algebra

in this case you can sometimes translate the usual Eilenberg-Steenrod axioms into algebraic statements

[chilly ocean]

The tensor product is unique up to unique isomorphism

[chilly ocean]

Can someone help me with this?

If the x_i generate M over A then and m in M is a sum of a_i x_i

We need to see that we can make 1\otimes a x_i for any a in A

We allowed to multiply by b’s in B

[next obsidian]

Okay so

You want to make b (x) m for any b in B, m in M

Write M = sum a_i • x_i

Then the way that A acts on b is that a•b = phi(a)b where phi:A -> B

Now you get that

sum bphi(a_i)(1 (x) x_i) = b(sum phi(a_i) (x) x_i) = b(sum a_i•1 (x) x_i) = b(sum 1 (x) a_i • x_i) = b(1 (x) sum a_i • x_i) = b (x) m

Here I use • only to denote when the multiplication is considering something as an A-module

Which is why phi(a_i) becomes a_i • 1 because I’m thinking of how A acts on 1_B in B

@chilly ocean

The first thing is written in a way such that it’s generated by 1 (x) x_i over B

[chilly ocean]

Big ty

[noble saddle]

@chilly ocean have you read Keith Conrad's tensor product notes? If not, go now. They really make clear what the universal property is and why it is the only thing you should ever use.

[latent anvil]

I love all of his notes

They're great

[noble saddle]

lol your profile pic is my phone lockscreen

also I went to a conference at UConnn once

and Ketich Conrad tried to do standup

let's just say his math exposition is a lucky gift to us all

[next obsidian]

His pfp has tiny masks edited onto the mouths

[noble saddle]

that I would have never noticed

God bless

[obsidian sleet]

not really sure which groups to consider for this problem

gut is telling me not to use the additive groups Z_n

but uh dis is weird im not sure what they're getting at here

[chilly ocean]

We call a property local if it suffices to check that the property holds on localisations at primes

often we get even better than this and it suffices only to check at maximal ideals

is there a term for this? perhaps something like coarsely local?

[cloud walrus]

squirtlespoof

[noble saddle]

I don't think there's a word for that. no

[latent anvil]

If you want the source @noble saddle

This is probably the best image I've ever made

[noble saddle]

It's p great

How do you feel about the remaster of the record?

[latent anvil]

I like it better than the original

[noble saddle]

How could you

[latent anvil]

But also I first listened to them at the same time

[chilly ocean]

Me Big noob

What is the music

[noble saddle]

Beach Life in death itself is enough to prefer the original

[latent anvil]

@chilly ocean twin fantasy by car seat headrest

[chilly ocean]

I shall check it out

[latent anvil]

That's my favorite song on the new record 😛

It's good!

[noble saddle]

D:

it is good

but the ending of the original

[latent anvil]

I should get a new pfp

I've had this one for a while

[chilly ocean]

I always thought your image was some kind of gnome/wizards

[latent anvil]

Lmfaooooo

They're dogs boyfriends

[noble saddle]

DO YOU HAVE SOMETHING AGAINST DOGGGSSSS

granted the original came out while I was in highschool

[latent anvil]

one time I tweeted out all the lyrics to beach life in death in alphabetical order and nobody retweeted it

Yeah I think the remaster came out when I was in my senior year of high school

Depends on the time of year

[latent anvil]

oh wow

[chilly ocean]

This is what I saw

[latent anvil]

That's good

[noble saddle]

beautiful

[latent anvil]

I am loling out loud at this

Maybe I should just use my face as my pfp like max

I like using album art though

[chilly ocean]

Perhaps use this no?

[noble saddle]

what else you into, Shamrock?

[latent anvil]

I am considering...

[chilly ocean]

Custom made bespoke art

[latent anvil]

Music wise?

I like emo, skramz, and sad indie stuff

[noble saddle]

you are like a younger me

pls stop

[latent anvil]

lol

[cyan marten]

Any two involutions generate a dihedral group (with the conventions: C2 and C2 x C2 are dihedral, as is the semidirect product of Z with Z/(2)).

[chilly ocean]

Very different vibe but I would recommend a listening to this

https://open.spotify.com/track/5hHtri1T4sBpWgse0c3cnC?si=59JkKuI3Qb-aheOle55iaQ

[latent anvil]

I am listening

why am I getting d

[noble saddle]

Shamrock I just found Indian Summer's discog at a record store the other day

[latent anvil]

Not even a

[noble saddle]

if you haven't heard them

it's good for a cry

[latent anvil]

I don't think I have, I'll give that a listen too

[sturdy marsh]

anyone here like prog?

[noble saddle]

Check Angry Son for their best track

What is it 1970 gramps??

[latent anvil]

no

I haven't listened to the band Indian summer but I do like this song called Indian summer https://open.spotify.com/track/3MhOHLfjD0cGoKwgmyo4eb?si=Rtg1NhKGRweiOXPbeR-W0A

[noble saddle]

Pedro the Lion is also dope

I'll allow it

[latent anvil]

I'm a big fan

[noble saddle]

the ytoured with mewithoutYou recently

great show

[latent anvil]

In January of last year I had the chance to see either mewithoutyou or this band called glass beach

I ended up going to the glass beach concert and finding a new band I really liked

[noble saddle]

I've seen mewithoutYou like 5 times, and it's always great

but glass beach I don't know

[latent anvil]

But then covid hit and I will not see any concert in infinity years

[noble saddle]

will check out

[latent anvil]

and I think me without you broke up last year too

[noble saddle]

I had a ticket for mwY's farewell tour

so much for that

[latent anvil]

😭 😭 😭

I should've gone when I had the chance

Actually I remember why I didn't know and it's very very dumb

[noble saddle]

might be the best band I've ever seenlive

[latent anvil]

I was messaging with this guy on tinder and we said we'd go together, but then neither of us followed up and the concert was really close

And I didn't want the awkwardness of following up

So I went to a different concert instead

And never heard from him again

[noble saddle]

dude's suck

[latent anvil]

well

[noble saddle]

duedes*

wow that was worse

[latent anvil]

I am also dudes and did not follow up

[noble saddle]

dudes

[latent anvil]

So I agree

But it's symmetrical

[noble saddle]

I know the feeling

I had a a similar experience with a dude going to see Dinosaur Jr

[chilly ocean]

God banging out exercises is a great feeling

[latent anvil]

Yeah

[chilly ocean]

Is there a reason we phrase the Jac radical as all x such that 1-xy is a unit for all y

[latent anvil]

wym

Like

[chilly ocean]

This is the same as 1=xy

1+xy

Is a unit

[latent anvil]

Oh yeah I mean I've seen it defined both ways

[chilly ocean]

Oh okay

I was just wondering if there was an aesthetic choice for this or some intuitive reason why it’s true if you see it with a minus sign

Hmm I suppose it made those exercises easier

[noble saddle]

It is related to whether the ring is local, right? it's been a while

[delicate bloom]

probably because the - makes its inverse 1+xy+(xy)^2+...

[latent anvil]

ehh I guess

[chilly ocean]

Anyone have ideas on this?

,m

Ah silly mistake

[snow flint]

this

[golden pasture]

you can think of this as like
if 1-xy is a unit then the "m-adic valuation" of x has to be positive for all m, i.e. x is in all maximal ideals

im doing too much nt ig

[chilly ocean]

you can probably just brute force q1, it is only 9x9 multiplication table

[carmine fossil]

This may be a test

Probably

[cyan marten]

Is there an analogue of Nakayama's lemma in group theory?

[latent anvil]

Nothing comes to mind

Hi chm

[next obsidian]

Otrher than considering it a Z-module if it's abelian

nah

[latent anvil]

nakayama talks about modules and about rings

Which one would the group play the role of?

Idk maybe that's too reductive

I just can't think of what an analogue would look like

[cyan marten]

Well, thinking of modules as rings acting on abelian groups, I thought of something like the action of the Frattini subgroup.

[next obsidian]

[cyan marten]

I am not very sure about the terminology right now but I think we can restate Nakayama's lemma as follows: Jac R acts non-transitively on M, when M is nonzero, finitely generated R-module.

[latent anvil]

That's interesting

I'm a little uncomfortable talking about transitivity of the action of an ideal

It's not clear to me that JM = M implies J acts on M transitively

[cyan marten]

Hmmm..

[next obsidian]

the sums make it tricky

I lean towards saying it's not equivalent

[latent anvil]

I like the analogy between R-mod and G-set here but I don't think you can translate finite generation or transitivity

Those feel specific

[cyan marten]

Oh, yeah.. There's an essential difference..

For an element g of G, we have gX = X, because by definition g acts as a permutation

[latent anvil]

yup

[cyan marten]

A group action on X is a homomorphism from G to Aut X, not merely the set of all functions from X to X (i.e., "End X")

But End X is a monoid, right?

[latent anvil]

You could define it either way

A monoid homomorphism G -> M lands in the set of invertible elements of M

Where G is a group and M a monoid

[cyan marten]

We can still think of monoid homomorphisms from G to End X.

[latent anvil]

yes, and it's the same thing as the set of group homomorphisms into Aut X

[cyan marten]

As for finite generation, I was thinking of "finitely many orbits."

[latent anvil]

I don't really get how that's a similar condition

Oh hmm I guess I see your thinking

You have finitely many elements (representatives of the orbits) which "generate" the set

If you act on those you hit everything

[cyan marten]

(my internet connection is a bit slow)

[latent anvil]

still feels kind of weird

[cyan marten]

Ohhh...

Makes sense.

Yeah

So for modules, we have an algebra with two operators (?)* acting on another with one operator (2, 1), and for G-set we have (1, 0). The concept of "generation" uses the operation of the thing acted on (in this case, the operation of the module), and it no longer makes sense to speak of it in the (1, 0) case.

• I meant the universal algebra sense

[chilly ocean]

It doesn’t seem to be stated

But it seems that atiyah-Macdonald implicitly uses that if S is multiplicatively closed then 1 is in S

[next obsidian]

This is convention

when you talk about S a "multiplicative system" you want to make sure that 1 is in there

[cyan marten]

It is stated in page 36, fourth paragraph.

[old hollow]

Shouldn’t this be p^(d+1)

I’m definitely counting wrong or something lol

[carmine fossil]

The bottom thing should be a_{d-1}

[rustic crown]

I have shown (2) <=> (3) and (1) => (2). So it remains to show that if this exact functor doesn't kill non-zero modules or non-zero homomorphisms then (a) its additive (b) its faithfully exact. I don't understand this additive-ness requirement nicely.

[old hollow]

@carmine fossil ya but don’t polynomials go from a_0 to a_d if they’re degree d?

[carmine fossil]

You are doing a map a -> r +(m),where r is the remainder after divison of a with m

[next obsidian]

You don’t need to prove additiveness

An exact functor is additive

I think this is by convention...??

Or maybe it’s a result...

[rustic crown]

additive was defined by saying the corresponding map Hom_R(M, N) --> Hom_S(M, N) is a abelian group homomorphism

[next obsidian]

Go to page 495

[old hollow]

Ohhh I’m so dumb

[next obsidian]

We’re implicitly assuming the functor is additive

[next obsidian]

“Let F:R-mod to S-mod be an additive functor...”

And likewise he defines left and right exactness for an additive functor

[rustic crown]

Why do we need this additive hypothesis?

[next obsidian]

We want to know that F sends chain complexes to chain complexes

Without additiveness there’s no reason to know that F sends the 0 map to the 0 map

I think is the issue

F as a functor only needs to send the identity to the identity but this isn’t the 0 map

[rustic crown]

yea this could be it

[next obsidian]

Honestly I can’t tell you if exactness can somehow get this for you (the fact that it’s additive)

But I think when you talk about an exact functor you ought to assume it’s additive

Unless you have a really good reason not to

[rustic crown]

its doesn't make me completely happy because "being a homomorphism" is more than just mapping 0 to 0.

[next obsidian]

It’s the only reasonable property to impose which guarantees it

Imo

[rustic crown]

Thanks for that...

I'll try it again tomorrow. Time to sleep.

[sturdy marsh]

maybe you could define the notion of a short exact sequence in a weaker category

to define kernels and cokernels all you need is the notion of a zero morphism

which can be defined if you have a zero object

(and you need to assume the corresponding limits and colimits exist)

this is almost a pre-additive category

[gleaming belfry]

What's the most general algebraic structure in which "a^k = b^k ==> a = b ∨ a = -b" works (where k is a natural number)?

Clearly finite fields don't work, is an integral domain with characteristic 0 enough?

[chilly ocean]

Do you really mean if and only if?

[paper flint]

Isn't this true of groups?

[crisp badge]

Just look at C and you'll see that we have an integral domain with characteristic 0 where this doesn't hold

[chilly ocean]

If k is odd, then backwards implication is basically never true

[paper flint]

Aah, the a=-b bit is troublesome.

[gleaming belfry]

ah, yeaahh. I'm interested in just ==>, actually

fixed up my message

[crisp badge]

I just realised I said a bunch of relative nonsense

Deleted the nonsense

[gleaming belfry]

yeah, C doesn't work. So I'd need an ordered integral domain, no?

[crisp badge]

Also, do you want k to be arbitrary, or you fix k then find the structure so that this holds?

[gleaming belfry]

one structure, all k

[chilly ocean]

I guess this is sort of about roots of unity, if we are in a commutative ring, this is equivalent to (ab^-1)^k=1

[gleaming belfry]

...if b has an inverse

[crisp badge]

Or -1

[chilly ocean]

Ah true yeah

Maybe these types of rings even have a name

[gleaming belfry]

hmm, is there an unordered integral domain where this holds?

[crisp badge]

Well, if we do have roots of unity, we are requiring that we have at most 2, those being 1 & -1

[gleaming belfry]

ah, yeah, just fix b=1

[crisp badge]

Well, we will have some roots of unity (1 and -1), but we will never have any others

So we must have characteristic 0, 2, 3 or 4

But char 4 doesn't work because 2^2 = 0^2

[gleaming belfry]

why wouldn't characteristic 1 work?

[crisp badge]

I mean, sure, but who cares about char 1

Considering I'm assuming you want to have 0 \neq 1

But also you can't take any extension of F_2 or F_3

So you have F_2, F_3, and some things of char 0

[gleaming belfry]

hmm, I don't see why this is

[crisp badge]

Because if we have char = n, then we require phi(n) <= 2

Because we have the multiplicative group of Z/nZ having phi(n) elements, and these are all roots of unity

And phi(n) <= 2 -> n = 1, 2, 3 or 4

[gleaming belfry]

and Z/nZ is a subring of all rings with char n, right?

[crisp badge]

Yes

[gleaming belfry]

does it have to be a field, though?

[crisp badge]

Thinking about it, no

Uh, take a UFD with at most 2 roots of unity

That should work

So like F_2[x]

Probably other stuff too, but that is probably at least sufficient

Too tired to really think it through right now

[gleaming belfry]

ah, thanks anyway!

[vestal snow]

I had a question about Jacobian of a variety and group objects in general. My Professor likes to go off topic a lot during office hours and he described them to us, but I was a bit confused about the definition. Let $A$ be a group object over some category $C$. Then $A(X)$ is defined as $Mor_C(X,A)$ and since $A$ is a group object, this is a group. I don't understand what is the group operation in $Mor_C(X,A)$.

[cloud walrus]

Have a Banana, Bitch

[oblique river]

saying that A is a group object in C means that A comes with some morphisms