groups-rings-fields

[past temple]

how do i find the splitting field of x^4 - 7 over F_5

[golden pasture]

attach every root?

[next obsidian]

compute how big it is

[thorn delta]

Looking for confirmation on this: so F/K really doesn't need to be Galois? I could prove this when F is galois over K, but when F/K is not galois, it seems like this changes things quite a bit

when $G = \operatorname{Gal}(F/K)$ and $F/K$ is galois, you can do $\operatorname{Gal}(LM/M) \simeq G_M/G_{LM} = G_M/G_L\cap G_M$ and apply second isomorphism theorem.

[cloud walrus]

kxrider

[thorn delta]

also, to show that $LM/M$ is finite galois, i showed that $\operatorname{Gal}(F/LM)$ is normal in $\operatorname{Gal}(F/M)$. But this relies on $F/M$ being finite Galois, which would (usually) rely on $F/K$ being finite galois

[cloud walrus]

kxrider

[magic owl]

Anyone who explains why the first sentence under the chart is true

I will buy you a drink or coffee of your choice should we ever be colocated

[magic owl]

I found it and am now ordering myself a drink of my choice thanks me

[latent anvil]

Aww

I wanted to think about it and claim my drink

But there was analysis chatter happening

[chilly ocean]

shamrock prefers analysis over u max

[magic owl]

it ends up being pretty technical

[river nebula]

Hi, I'd just proved that every Noetherian module is finitely presented, I wanted to check if the converse is true but I can't get anywhere.

Are all finitely presented modules Noetherian?

[latent anvil]

No

Consider your favorite non noetherian ring

as a module over itself

It's finitely presented

but not noetherian

[river nebula]

thanks, i should have thought of that

[thorn delta]

Suppose F/K is a field extension and L and M are intermediate fields with L/K finite Galois. How do I show LM/M is finite Galois?

[chilly ocean]

I think you can say: write L=K(a1,..., an). Then LM is generated over M by a1,..., an, so LM/M is a finite extension. Now to show that it is galois, let us use the definition that an extension is galois if it is the splitting field of a separable polynomial. We can just use the same polynomial that makes L/K galois to show that LM/M is galois.

[thorn delta]

Makes sense. But It seems kind of nontrivial that LM is generated by M and the roots of the separable polynomial

[chilly ocean]

Ah wait yeah maybe there is a hole there

Actually, I think it works

[thorn delta]

ok yea i see it. If u1, ..., un are the roots of the separable polynomial f, we see that L = K(u1, ..., un) \subset M(u1, ..., un) as well as M \subset M(u1, ..., un) so LM \subset M(u1, ..., un). The other inclusion holds since the ui are elements of L.

thanks!

[untold sapphire]

can anyone who is good at Homological Algebra help me understand this paper:

https://arxiv.org/abs/1906.08837

[thorn delta]

im trying to show that no finite field K is algebraically closed. Suppose |K| = p^n. If f = x^{p^n} - x + 1, then f has no roots in K, but also f(x) = 1 for all x in K. So does f count as a nonconstant polynomial?

[final pasture]

no it doesn't, your proof work

another polynomial would be product(a € K, (x-a)) + 1 (and it also work with finite rings, since it doesn't require any multiplicative order existence )

[thorn delta]

"no it doesn't" as in f = x^{p^n} - x + 1 is nonconstant?

[delicate bloom]

it's not "another", they're equal $\prod_{a \in K} (x-a) +1 = x^{p^n}-x+1$

[cloud walrus]

Merosity

[final pasture]

yes it's nonconstant

[final pasture]

(yh you're right, bad wording, they're the same polynomial but ""my presentation" makes it obvious that it works in any ring)

[thorn delta]

thanks!

One more question. For a field extension $F/K$ and intermediate fields $L,M$ with $L/K$ finite Galois, I have been trying to show that $$Gal(LM/M) \simeq Gal(L/L\cap M).$$ At first I tried a second isomorphism theorem argument, but I could only make it work assuming that $F/K$ was finite Galois. Now, I have the following set-up. There is a homomorphism $$Gal(LM/M) \to Gal(L/L\cap M)$$ given by restricting automorphisms of $LM/M$ to $L$. Its not hard to see this map is injective, but I have been having some trouble proving surjectivity. Since the base field shrinks in the image, i don't really see how you're supposed to be able to extend automorphisms of $L/L\cap M$ to $LM/M$. Any ideas?

[cloud walrus]

kxrider

[thorn delta]

Oh yea, and LM/M is finite Galois (also follows from L/K being finite Galois). That is probably needed somewhere.

so... from the proof that LM/M is finite galois, I know that LM is the splitting field of the same separable polynomial as L/K. Also, I believe the separable polynomial for L/K is a separable polynomial for L/L cap M. Therefore we can extend an automorphism L/L cap M to an automorphism LM/L cap M, but i don't know if this extension fixes M a priori

[thorn delta]

hmm no what i've said isn't quite right. We can't extend these automorphisms at all as far as i can see

[vestal snow]

How are exact sequences defined in categories where the objects might not be sets and the morphisms might not be set functions?

[latent anvil]

is the category abelian?

in an abelian category if you have f : A -> B and g : B -> C such that gf = 0 you can juggle universal properties to get h : im f -> ker g such that A -> im f -> ker g -> B equals f. Exactness of A -> B -> C says h is an isomorphism

[vestal snow]

So I get what ker means, but what is the notion of im in arbitrary categories?

[latent anvil]

The image is the kernel of the cokernel

[vestal snow]

Actually

Oh wow

[latent anvil]

I think aluffi's chapter on homological algebra goes through all this in depth

[vestal snow]

I never though of it like that

We haven't reached that yet

[latent anvil]

(not telling you you should know this, just giving a reference)

[vestal snow]

I'm trying to prove that right adjoint implies left exact

[latent anvil]

In an abelian category?

[vestal snow]

Yes

[latent anvil]

nice

[vestal snow]

Is this true?

[latent anvil]

Yes

so the thing to note here

Is that left/right exactness really say you preserves kernels/cokernels

[vestal snow]

Is it still true if I replace abelian with "coker and ker" exist

[latent anvil]

Yes

Well actually hang on

I think you need F to be additive

just to back up

I think you'll need F(0) = 0

otherwise things could break down

[vestal snow]

Wait I have a question

[latent anvil]

Sure

[latent anvil]

yup

[vestal snow]

I'm trying to generalize this

But it doesn't specify that the right adjoint functor is additive

[latent anvil]

This feels slightly incorrect...maybe I'm missing something

Oh I am missing something

Adjointness implies you preserves 0

[vestal snow]

Okay gotcha

[latent anvil]

Because it implies you preserve either the terminal or the initial object

Sorry, wrong name

The right generalization of this is that left adjoints preserve colimits and right adjoint preserve limits

and kernels and cokernel are particular kinds of limits and colimits

[vestal snow]

I've proved both of those

There's only one thing that's bothering me

[latent anvil]

sure

[vestal snow]

Let 0 -> A -> B -> C be exact

It's easy to prove that a right adjoint functor maps 0 to 0

Using the kernel thing

[latent anvil]

Sure

[vestal snow]

So 0 -> F(A) -> F(B) is exact

[latent anvil]

Although the proof I have in mind of the kernel thing relies on 0 mapping to 0

F(0) = 0 because 0 is both the limit and the colimit of the empty diagram

[vestal snow]

But I can't show that F(A) -> F(B) -> F(C) is also exact

[vestal snow]

I proved it differently

[latent anvil]

Don't think about it like this

0 -> A -> B -> C is exact iff A -> B is a kernel of B -> C

prove this lemma

now your statement about kernels applies directly

[vestal snow]

Shouldn't it be 0 -> A -> B -> C is exact iff A -> B is a kernel of B -> C and A -> B has kernel 0?

[latent anvil]

You can show that kernels are monomorphisms

more generally equalizers are monomorphisms in any category

and monomorphism = has kernel 0 in an abelian category

Let me see if chmonkey proved this in his notes

[vestal snow]

Don't worry about it

[latent anvil]

Kk

[vestal snow]

I'll try and do it myself

I think I get it now

[cloud walrus]

Yes

[rustic crown]

Say N is the normalizer of P and C is the centralizer. Then by definition, g in N means that gPg' = P

this gives you a map, N --> Aut_{Grp}(P)

the kernel of this map is, g in N such that gag' = a for each a in P, that is kernel is the centralizer

Hence N/C is isomorphic to a subgroup of Aut(P) = Aut(Z/pZ) = Z/(p-1)Z

so the index [N:C] divides both |G| and (p-1)

if a prime q divides |G| then q >= p and hence doesn't divide p-1, thus [N:C]=1 as we wanted

[hot lake]

what if |P| > p ?

[rustic crown]

in that case |Aut(P)| = p^(n-1)(p-1), and gcd(|G|, p^(n-1)(p-1)) = p^(n-1)

so ig [N:C] could be a proper divisor |P| in general?

oh wait i assumed P iso to Z/p^n Z 🤦‍♂️

[hot lake]

well they did say that P is cyclic

[rustic crown]

I read that "|P| = p" somehow... something is wrong with me >.<

[hot lake]

since P is cyclic, it is abelian so P is contained in C, so [N:C] can't be divisible by p

[rustic crown]

oh yes that works

[rigid cave]

Is there any program that can give me a description of the split conjugacy class of 5-cycles in A5?

Like a program that can print out all 5 cycles?

[golden pasture]

GAP probably

[molten silo]

Anyone know how to prove e

[viscid pewter]

find the additive inverse of -r

[molten silo]

its r

[rustic crown]

but its also -(-r)

[molten silo]

i need to show -(-r) + r =0

[viscid pewter]

no you don't

[hot lake]

no

[molten silo]

nevermind

i see

[spice lance]

hi! this might be a super basic question but how can I show that G is isomorphic to R?

[chilly ocean]

u mean G

[viscid pewter]

well can you see intuitively what the isomorphism actually is

[spice lance]

oh yes G sorry

[spice lance]

yup i was thinking of just mapping real numbers to a hahaha but i'm not sure if this is right or how to put it mathematically

[viscid pewter]

yeah that's it

so do you know how to show things are isomorphic

[spice lance]

yup i have to show that the function is bijective and that f(x * y) = f(x) * f(y)

[spice lance]

but i'm not sure how to put this into mathematical terms for the proof hahaha

[viscid pewter]

let f: R -> G be like f(a) = (1, a; 0, 1)

there's a map

now just show it

[spice lance]

okay thank you so much!!

[cloud walrus]

Fiwam

[spice lance]

hi how do i show that this is bijective? haha does it suffice to just say that for the matrices to be equal, the two variables must be equal?

[old lava]

show its injectiive, then show its surjective

[spice lance]

yes sorry i'm not really sure how to prove either really sorry

[untold sapphire]

Well, to show surjectivity, how would you show that you can get any real number using your map?

What choices do you have for the domain ?

E.g. g : G -,> R given by g(1, a, 0, 1) = a

(hint: there is only one variable you can choose the value of!)

[old lava]

for injectivity, all you need to prove is that the only element that maps to the identity is the identity. For surjectivity, show that for every element in the codomain, you can find an element in the domain that maps to it

[celest brook]

draw the cayley table

:<)

@spice lance if you can prove you can express any matrix in GL2(R) as an upper triangular matrix, you got your isomorphism

[cloud walrus]

Fiwam

[unique juniper]

sorry, i didnt completely understand

[rustic crown]

so for g in N, you get an automorphism on P, given by a --> gag'

gag' in P, as P is normal in N

[unique juniper]

ok

[rustic crown]

i misread the question, so yea Aut(P) = Aut(Z/p^nZ) = (Z/p^nZ)*

[unique juniper]

oh ok

[rustic crown]

and so |Aut(P)| = (p-1) *p^(n-1)

since N/C is (like) a subgroup, it cardinality divides that of Aut(P)

[unique juniper]

yes

[rustic crown]

but N is a subgroup of G, so |N| has to divide |G| and so [N:C] divides both |Aut(P)| and |G|

but notice that P is cyclic, in particular abelian, hence the centralizer contains P

P <= C <= N

this means that p doesn't divide [N:C]

Now if q was any other prime divisor of [N:C] then it divides |G|, which would mean q > p and it would also need to divide |Aut(P)| = (p-1) * p^(n-1)

but this this simply can't happen

[N:C] has no prime divisor! so it must be 1

[unique juniper]

why?

[rustic crown]

intuitively, the sylow P subgroup has all the p in it... so N and C have both maximum power of p allowed to them, i.e. |P| divides |N| and |C| but |N|/|P| and |C|/|P| can't be divisible by p

hence their quotient sin't divisible by p

[unique juniper]

okay yes

[rustic crown]

another way to see this is,
[G:P] = [G:N] [N:C] [C:P]

[unique juniper]

right

thank you :)

[rustic crown]

[mystic jungle]

how would I show Z/nZ is abelian? im trying to show ring axioms to be true

[rustic crown]

so what are the elements of Z/nZ?

[mystic jungle]

can we say that the sum of Z/nZ is Z/nZ so coset of p + coset of q is closed in Z/nZ

Z/nZ = { coset of 0, coset of 1, ...., coset of n-1}

[rustic crown]

what do you mean by "sum of Z/nZ is Z/nZ"

[mystic jungle]

like sum of elements in this set is also in the set

[rustic crown]

yep, you need that for to even say its a group, when we say an operation on a set X, we mean a function from X*X --> X, so being closed is already there in the definition!

[mystic jungle]

do i have to show that there exists additive identity of zero element and then negative of a to show Z/nZ is abelian

[rustic crown]

in our case, for Z/nZ, the operation is defined as [a] + [b] = [a+b]

[rustic crown]

so have you shown that Z/nZ is a group?

if not, yep you need to do that.

[rustic crown]

(you need to verify here that this operation is well defined, i'm using the notation [a] for the coset of a)

[mystic jungle]

coset of p + coset of q = coset of p + q is well defined cause (p + nZ) + (q + nZ) = (p+q) + nZ and for any a,b in Z we have coset of a = coset of b <=> a-b=nk for some k in Z

[rustic crown]

so it seems like you know how to quotient a group by a normal subgroup, and that the result is a group again.

so only thing remaining to verify here is that this is abelian, and that follows from the fact Z is abelian.
[a]+[b] = [a+b] = [b+a] = [b] + [a]

[mystic jungle]

i have to show addition is associative in the set right

[rustic crown]

technically yes, but all that is already done with you construct the quotient group

[mystic jungle]

so for additive identity of zero element we have [p] + [e] = [p] implies [p + e] = [p] but what does that say about e

[rustic crown]

e must be a multiple of n

[mystic jungle]

mmm so n | e but how would i use that to find the negative of [p]

[rustic crown]

[-p] works right!

[prime cloak]

Hello guys quick question if we have an endomorphism u how can we show that there exists a polynomial that cancels u (P(u)=0) ?

[rustic crown]

endomorphism of what? finite dimensional vector space?

[prime cloak]

yes

[rustic crown]

yep you can do that!

say V was a vector space over the field k with dimension n. Then The dimension of End(V) is n^2

do you see this? All we need is that End(V) is also a finite dimensional vector space.

using that we can say, {id, u, u^2, ......} must be linearly dependent! the linear dependence gives you a polynomial which u has to satisfy!

[prime cloak]

what is End ?

[rustic crown]

set of endomorphisms V --> V

that set naturally acquires a k-vector space structure

[mystic jungle]

Let G be a group and H a subgroup of G,

relation 1: relation ~ on G defined by a ~ b if ab^-1 in H

relation 2: relation ~ on G defined by a ~ b if a^(-1)b in H

are these the same?

[carmine fossil]

Yes

[prime cloak]

ok thank you but concretly what does the polynomial look like ?

[rustic crown]

well not a lot from this argument, you just get that it satisfies some polynomial, and the degree of this polynomial could be upto n^2

[untold sapphire]

Very nice

[rustic crown]

you can make this better tho, using Cayley Hamilton Theorem

[carmine fossil]

Actually,This is wrong

If ab^-1 is in H,that doesn't imply (a^-1)b is in H

unless b^-1 h b is an element of H for all b and all h in H

[rustic crown]

that theorem says that if you write the endomorphism as a matrix, then look at the characteristic polynomial, the endomorphism has to satisfy it.

[carmine fossil]

@mystic jungle

[prime cloak]

ok

[mystic jungle]

hmm

[prime cloak]

but last question (Id,u,...,u^p) has a dimension of p+1 ??

and L(E) has a dimension of n**2

[carmine fossil]

If the 2 relations are same,We call H a normal subgroup of G

[prime cloak]

what makes it linearly dependant

[mystic jungle]

cant i say som,ething like this

for a,b in G since H is subgroup of G we have a ~ b iff ab^-1 in H iff (ab^-1)^-1 in H iff ba^-1 in H iff a^-1b in H iff a ~ b ?

[carmine fossil]

ba^-1 in H doesn't imply a^-1 b in H

Generally

[rustic crown]

oh we don't now if that is dimension p+1, there could already be some linear dependence

but we are not putting a bound on p

in particular

(1, u, ...., u^(n^2)) are n^2+1 vectors in a n^2 -dimensional vector space End(V0

hence they have to be linearly dependent.

[mystic jungle]

is there stronger counterexample using equivalence classes or is this reasoning good enough

[carmine fossil]

I mean literally take any group and a non normal subgroup

You will found a counterexample

Try S_5 and H=<(1,2)>

[past temple]

I'm struggling on this field theory question

suppose we have a field K and an autormophism of K of infinite order sigma

Let F be the fixed field of K by sigma

how do i show that K/F is a normal extension

[waxen hollow]

Hi guys. How I can find all composition series for D12? (Dihedral 12)

[waxen hollow]

<@&286206848099549185>

[latent anvil]

Try to find a normal subgroup of D12. Try to find a normal subgroup between that and all of D12, and one between that and the identity. Repeat

Oh wait all composition series? Not all composition factors?

More annoying

is this a group of order 12 or order 24?

[waxen hollow]

Well, order 24

[latent anvil]

Sure

[waxen hollow]

I did for C30, if you want. But it didnt help me that much for D12

[latent anvil]

so this essentially comes down to "find all normal subgroups of D12"

and then look at all the ways they fit together

Which is messy

are you 100% sure you're supposed to find all composition series? Not just one?

[waxen hollow]

Yeah, it is all composition series

[latent anvil]

hmm

Then yeah I don't know a good way to do this. The first step would be to find all normal subgroups

[waxen hollow]

I thought maybe the way was: find one -> jordan-holder then find all isomorphism and permutations between

[latent anvil]

I don't think there's a way to do that, but I might be mistaken

I mean if you find one series

This gives you information about others

So I guess to start, let's try to find one normal series

consider the element r^6

This is central, right?

It commutes with all elements of the group

[waxen hollow]

r^6 means D6?

[latent anvil]

Oh sorry the presentation I'm thinking of is <r, f|r^12 = 1, f^2 = 1, frf = r^-1>

So r is a rotation by 2π/12

Does what I said make sense?

[waxen hollow]

yeah, sure

[latent anvil]

Then D12/<r^6> will be the dihedral group of order 12, I think

So normal subgroups of D12 containing r^6 are the same as normal subgroups of D6

[waxen hollow]

i see

[latent anvil]

Idk, I don't see any nice ways to do this

Other than just very explicitly thinking about what subgroups exist

Maybe you can compute the conjugacy classes of D12 and use the fact that normal subgroups are unions of conjugacy classes?

[waxen hollow]

yeah, I understand you, I don't know why my teacher asked to look for something like that

[latent anvil]

It seems like an unpleasant problem

[waxen hollow]

i have to show all composition series for S5 too, but i'm already stucked with D12

[latent anvil]

S5 is actually easy

It has a unique normal subgroup A5, which is simple

[waxen hollow]

You're right, S5 is almost simple

Thank you

How I can proceed from A5? For me the unique composition series is C_1 < A_5

Doesnt seems right for me

[chilly ocean]

You just prove that A5 is simple right?

[waxen hollow]

I already proved it once

So, the only composition serie is C_1 < A_5 < S_5 ?

[latent anvil]

Yes

[waxen hollow]

When I saw order 120, I would never have imagined it would just be this one.

Well, do you know any method to check how many series exists for G arbitrary?

[chilly ocean]

plug it into sage

[latent anvil]

Plug it into GAP

[rose axle]

can someone help me compute

namely I have no idea what the tensor product of modules does

nvm

[oblique river]

@rose axle did you figure it out or still have a question?

[untold sapphire]

For then Tensor Product the best way to think of it is in terms of a Universal Property

I.e. it's the module through which every billinear map from M x N factors

[vestal snow]

@latent anvil A follow up question from yesterday's discussion: A sequence is defined to be exact if the (category theoretic) kernel of a map is the same as the (category theoretic) image of the previous right?

[latent anvil]

sort of

if gf = 0, there's a natural map im f -> ker g

you require that this specific map be an isomorphism

equality is way too strong, the objects image and kernel are only well defined up to unique isomorphism

[vestal snow]

Yeah that's what I meant

the (category theoretic) kernel of a map satisfied the same universal property of the (category theoretic) image of the previous

[vestal snow]

Is this it?

[sturdy marsh]

yeah that works

[vestal snow]

Okay thanks

Images can be expressed as limits right?

[sturdy marsh]

colimit

actually no

it's a limit

you're right

[vestal snow]

So since right exact functors preserve limits, wouldn't that mean that they preserve exact sequences?

instead of just left exact

which I know isn't true, but I don't see why exactly this proof does not apply

[sturdy marsh]

sorry I had to leave for a sec

right exact functors do not preserve limits

right adjoints commute with limits

you might be confusing the two

[vestal snow]

Ah my bad

I meant right adjoint preserve limits

and so they should preserve both the image and kernel

and thus, they should preserve exact sequences

instead of just left exact

[rustic crown]

but image = ker (coker)

right adjoints may not preserve coker

wait did it get it backward?

[sturdy marsh]

no you got it right

[rustic crown]

this is confusing

[sturdy marsh]

try writing this out properly

[sturdy marsh]

the image is a limit, but not quite in the way you need it to be in order to make this argument work

[vestal snow]

I see

[sturdy marsh]

it can also be written as a colimit

[vestal snow]

So is it something like Im(f) = lim(something(f)) where something does not commute with the right adjoint functor

[sturdy marsh]

actually im not even convinced that it is a limit anymore

you should be mapping out of it

[sturdy marsh]

ye so limit of a colimit

det pointed out the problem

[vestal snow]

Thanks det

This really cleared up something that had been bothering me for the past few days

[rustic crown]

i was confused why it was a limit >.<

[sturdy marsh]

image is pretty weird

ker and coker are easier to work with

as ker is just a limit

and coker is a colimit

[latent anvil]

yeah image is the limit of a certain diagram

Since it's ker coker

And ker is a limit

But the thing in that diagram involves coker

So like given f : A -> B, im f is the equalizer of B -> coker f

So F(im f) is the equalizer of F(B) -> F(coker f)

And the issue is that you can't do anything with F(coker f)

Brofib was saying this but I wanted to make it 100% explicit

[vestal snow]

Shamrock can you help me out?

I'm trying to prove 0 -> A -> B -> C is exact iff the kernel of A -> B is 0 -> A and the kernel of B -> C is A -> B

[golden pasture]

(thats kinda by definition right?)

[vestal snow]

Using this definition of exactness

Here's what I've done so far

[golden pasture]

ahh

[vestal snow]

Assume that the sequence is exact

[golden pasture]

yea the image is kinda finicky to play with

[vestal snow]

First I want to show that 0 -> A is the kernel of A -> B

Since we know that the kernel of A -> B is the image of 0 -> A by exactness

it suffices to show that the image of 0 -> A is 0 -> A

That is, there is a map monic map $m:0\rightarrow A$ and a map $e: 0 \rightarrow 0$ such that $me=0$ and it is initial with this property

[cloud walrus]

Have a Banana, Bitch

[latent anvil]

ah sorry banana

[vestal snow]

Obviously, m and e must both be 0

[latent anvil]

i wasn't looking at this channel

[vestal snow]

It's okay

So we have the following diagram

[latent anvil]

this is redundant

the kernel of A -> B is 0 -> A and the kernel of B -> C is A -> B

the second statement implies the first

probably easier to show that first, simplifies the problem

[vestal snow]

You're right but I'm trying to prove the first directly

Just to make sure I understand this stuff

% https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBoAmAXVJADcBDAGwFcYkRiQBfU9TXfIRTkK1Ok1bsAgt14gM2PASIiALGIYs2iEAEkA5LL6LBK0sQ0TtHbmJhQA5vCKgAZgCcIAWyQiQOCCQARhpNSR0vQxpGegAjGEYABX4lIRB3LAcACxwQGhx6LEZ2LIgIAGsjEA9vJDJ-QMQQ8S12Th43Tx9EeoDfaLiE5JNlHQzs3NCrNqqa7t7GgGYp1p0YAF52uTmkZYbg-MLinVKKvJbwkC9N2a7d-Ma-eLAoJFUATmisMGsoejgsvZbFwgA
\begin{tikzcd}
& & 0 \arrow[dd, "m=0", hook] \arrow[dddd, dashed, bend left=49] \
& & \
0 \arrow[rr, "0"] \arrow[rrdd, "0"'] \arrow[rruu, "e=0"] & & A \
& & \
& & I' \arrow[uu, "m'"', hook]
\end{tikzcd}

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

We somehow need to prove that the dotted map (which must be the 0 map) makes the diagram commute

That is, m' 0 = 0

[latent anvil]

yeah i mean

[vestal snow]

Oh wait

[latent anvil]

composition is bilinear

m' circ 0 = m' circ (0 + 0) = (m' circ 0) + (m' circ 0)

so m' circ 0 = 0

[vestal snow]

Yeah I realized that I messed something up while writing it out

I was trying to prove that there is a map from I' to 0 (which must be 0) such that 0*0=m

I drew the dotted map the other way around in my notebook

Which is why I was having problems with that

I can't believe I spent 30 mins on something just because of a misplaced arrow

[latent anvil]

it happens

[vestal snow]

@latent anvil This turned out to be much easier than I thought

Thanks for helping me out

Both directions of the proof are basically proving Im(A -> B) = A -> B

[latent anvil]

👍

Np

[vestal snow]

Just making sure I got the main idea. The proof is essentially proving that if f: A -> B is monic, then its image is itself

And when we prove the analogous statement for left adjoint functors, the proof will essentially be proving that if g: B -> C is epic, then its coimage is itself

[wooden ember]

Counter example to unit conjecture has been found

[chilly ocean]

May you teach me how make it?

[vestal snow]

If you go on https://tikzcd.yichuanshen.de you can learn it pretty easily

[chilly ocean]

Thx

@vestal snow

[vestal snow]

Can exact sequences be defined in terms of cokernels and coimages instead of images and kernels?

[unique juniper]

im trying to find out amount of finite abelian groups of order 100 there are up to isomorphism

shall i first get it into invariant factors?

and then decompose each of those into elementary divisors?

lol

why monkas me

[carmine fossil]

Ye,That should work

[unique juniper]

is this the best way tho?

[hot lake]

probably

[unique juniper]

im reading this from d&f and theres no proofs and kinda confusing

[hot lake]

it's probably as bothersome as getting the shape of the prime factorisation of every integer up to 100

[old lava]

it's not too bad, you just have to look at either elementary divisors or prime factors

d&f gives an example of doing stuff like that for certain orders

[unique juniper]

so for order 100

[old lava]

like d&f gives this example for groups of order 180

just apply the same principle

but for 100

instead

[chilly ocean]

that moment when poros did this exact exercise

[unique juniper]

yeah i did that

But now, how do i get the elementary divisors from the invariant factors?

[old lava]

you just said you wanted to find number of finite abelian groups of order 100

you don't need to get the elementary divisors

for that

you only need to use either invariant factors or elementary divisors

not both

[unique juniper]

oh

[old lava]

they're just two different approaches

[unique juniper]

right so theres 4

[old lava]

also, for future reference, to do this you just prime factor each invariant factor

[unique juniper]

ok

[old lava]

Z_50 for example, 50 = 5^2 * 2, so Z_50 = Z_2 x Z_{5^2} = Z_2 x Z_25

[unique juniper]

ok

the lecture notes dont mention invariant factors

hmmm

my teacher completely ignores this stuff

[carmine fossil]

I mean does anyone care about classifying groups?

[old lava]

just read the d&f section

then

textbooks do be for reading

[unique juniper]

d&f is kinda painful

too many words

[old lava]

[chilly ocean]

lol

it is a wordy book

[hot lake]

oh I thought the question was about the groups of order up to 100, and not the groups of order 100

[rose axle]

sanity check--if I have free abelian groups A and B a free subgroup then the rank of A/B is is rank(A) - rank(B)

yeah you can do this with like

smith normal form

I just did this

https://estrogen.fun/i/olty.png

[spiral wolf]

Is this true:

Z^2/(5Z x 10Z) = Z/5Z x Z/10Z

[viscid pewter]

i think they're isomorphic, yes?

[spiral wolf]

Is it true in general? Say AxB/CxD = A/C x B/D

Is it true in general? Say AxB/CxD = A/C x B/D

[viscid pewter]

think so

if C is normal in A, D is normal in B, as standard

[old lava]

(assuming C is normal in A, D is normal in B) you can construct a homomorphism AxB to A/C x B/D which sends (a, b) to (a mod C, b mod D). It's obviously surjective, and looking at the kernel and using first isomorphism theorem gives you the desired result

[spiral wolf]

Awesome, thank you

[sinful mirage]

how did the prof conclude that phi=0 or phi is injective given the ker phi relations?

[chilly ocean]

That is quite a chicken scratch

[sinful mirage]

what do you mean?

this is from schur's lemma

[chilly ocean]

Just that I can't read it, basically the left of "ker phi = U" I can't make out

[sinful mirage]

ahh,the statement is: phi:v_1->v_2. v_1,rho_1 v_2,rho_2 are irreducible reps. then if v_1 is irreducible(by a previous proposition), it follows that either ker phi=V_1 or ker phi={0}

my question is,how does ker phi=v_1 imply phi=0 and how does ker phi=0 imply phi is injective

[hot lake]

ker phi = {x in V1 | phi(x) = 0}

[sinful mirage]

ahhhhhh

lol

fml

and how about the second statement?

ker phi=0=> phi is injective

[hot lake]

if phi(x) = phi(y) then phi(x-y) = 0

are you doing representations of finite groups without ever doing linear algebra ?

[sinful mirage]

i had linear algebra course taught by a physicist and now doing rep of finite groups with a mathematician

[hot lake]

ah

my condolences

usually those things are drilled into you in a linear algebra course

[sinful mirage]

wait so the definition of injective is if phi(x)=phi(y)=> x=y

where did you use the ker phi=0 condition?

[hot lake]

it's used when you go from phi(x-y) = 0 to x-y = 0

[sinful mirage]

ahh lol this is because phi(x-y)=0 implies x-y is in ker(f), but ker(f)=0 => x-y=0

ok yes,i see now,this was way easier than I thought thanks!

by the way @hot lake do you maybe know an 'intuitive' or 'physicist friendly' reference on rep of finite groups? We are following Fulton Harris but it's a speedrun/mega abstract for me, not having strong math background

[hot lake]

well fulton harris is my goto reference

I know chemists have a pretty warped notion of groups

idk about physicists

[sinful mirage]

we physicists are not used to def thrm proof rather computations

[hot lake]

also you usually do a course about groups before doing representations of finite groups

so that you have lots of example of groups

yeah they do proof theorem definition instead

and to get a feel of how groups work and how rich they can get

[sinful mirage]

how is this a consequence of schur's lemma?

the statement is that if v1,v2 are irreducible reps, phi:v_1->v_2, phi is in hom{G}(v_1,v_2), then it is either invertible or zero. I proved that,but then this should be a consequence of that( i can't see why)

[sturdy marsh]

What exactly are you trying to prove?

@sinful mirage

are you trying to prove that Hom_G(v_1, v_2) = C for irreps v1 and v2?

scratch that

[sinful mirage]

so I have proven Schur's lemma, which states let v_1,v_2 be irreducible reps, and phi:v_1->v_2 in hom_{g}(v_1,v_2). then it is either invertible or zero

[sturdy marsh]

okay

[sinful mirage]

now my professor said something further(which is a corrolary of Schur's lemma apparently). If there is another homomorphism, $\phi'$ which is either zero or invertible, then $\phi'=\lambda \phi$, with lambda in C

[cloud walrus]

ProphetX

[sturdy marsh]

umm

you can make Hom_G(v_1,v_2) an algebra over C

actually can you

[sinful mirage]

this is how she proved it,but I can't really follow

i'm ok with first line. what does it mean to form 'that map over c?'

and why does it have an eigenvalue surely?

[sturdy marsh]

any linear operator on a complex vector space has an eigenvalue

jordan form

but yeah anyway, Hom_G(v_1, v_2) is a division algebra over C

C is algebraically closed

so it is iso to C

which means that any two isos are the same up to multiplying by a constant

[sturdy marsh]

this is by Schur's lemma

[sinful mirage]

wait,so I looked up jordan canonical form and now I see why it must have an eigenvalue

why does that imply that the map is not invertible?

[sturdy marsh]

yeah idk

if it's a composition of isos

then it should be invertible

[sinful mirage]

wait its not a composition of isos

or hm

[sturdy marsh]

ok I think I see what they were going for

phi ^-1 phi' has eigenvalue alpha

i.e. phi^{-1} phi' - alpha I is not invertible

so phi' - alpha phi is not invertible (multiplying by phi)

so it is zero (by Schur)

[sinful mirage]

does that follow from eigenvalue equation being satisfied and taking determinant and that being 0?

[sturdy marsh]

it sends an eigenvector of phi^-1 phi' to zero

so not invertible

[sinful mirage]

ahh ok

makes sense

thanks

[sturdy marsh]

Perhaps a cleaner way is to prove that a division algebra over an algebraically closed field must be the field

Hom(blah, blah) is a division algebra over C by schur

and you're done

[sinful mirage]

can you please show why this is true?

i'm not familiar with the term division algebra

[sturdy marsh]

it's an algebra

and every nonzero element is invertible

[sinful mirage]

which operations would I need to define on hom to make it an algebra?

explicitly

hom_.. is just a set at that point

[sturdy marsh]

pretty much ring + compatible scalar multiplication

[sturdy marsh]

it's a C[G]-module

[sinful mirage]

would you have patience to write out the operations explicitly? I can't see it: I know that a module is a vector space over a ring, but my problem is, I do not know how to define the operations on the set itself to make it into such structure

(sorry if this is trivial,i'm coming from a physics background )

[sturdy marsh]

it might be easier to do it in general

if M and N are R-modules

[sinful mirage]

what i mean is to define + and times for hom_G(v_1,v_2) like here

[sturdy marsh]

can you put an R-mod structure on Hom(M, N)

[sinful mirage]

to see how it gives rise to a structure of module using only the +,times on v_1,v_2

so I know Hom(M,N) is a vector space by pointwise multiplication

ohh Hom_{g}(M,N) is a subset of Hom(M,N), because we allow only specific linear maps? then I can use the same pointwise multiplication and addition to get a vector space structure on Hom_{g}(M,N)?

[sturdy marsh]

I might have screwed up, this is true when blah = blah

v_1 and v_2 are different reps

nvm

this proof doesnt work

[sinful mirage]

btw,another related question: does this definition for the tensor product seem right?

are the operations defined ok?(i could not find them explicitly in books and tried constructing myself)

[sturdy marsh]

this isnt the notation im used to

[sinful mirage]

how would you define the tensor product of two vector spaces?

I can't find a satisfactory answer on the internet,where the operations would be explicitly written out such that to prove that the tensor prod inherits the vector space structure from the 2 vector spaces

[sturdy marsh]

https://en.wikipedia.org/wiki/Tensor_product#The_definition_of_the_abstract_tensor_product

[sinful mirage]

yes,I saw this,but I can't really comprehend it

[sturdy marsh]

umm

[sinful mirage]

what does 'the free vector space' mean?

F(VXW)

and how to show that the quotient vector space is a vector space,i.e. how does it inherit the structure from the other 2? what is the + and times on the quotient space?

[sturdy marsh]

okay, you should probably read up on quotients, it's essential to "define" tensor products

but very roughly, you do all the operations on representatives

if [v] and [w] are elements of the quotient

then [v]+[w]:= [v+w]

and scalar mult is given by alpha [v] = [alpha \cdot v ]

and this doesnt depend on choice of representatives

[sinful mirage]

naively am i allowed to think the + and times are inherited from the field?

or they are from the vector spaces

[sturdy marsh]

from the vector space, yes

this stuff is worth reading in detail

quotients turn up a lot

[sinful mirage]

do you have a reference?

(an intro one if possible)

[sturdy marsh]

pretty much any algebra book should do quotients

dummit and foote for example

you can get away with defining tensor products of vector spaces in a weird way but I'm not sure if that's the best way to learn about them

but if V has a basis e_i

and W has a basis f_j

then V (x) W is a vector space on the basis e_i (x) f_j

[sinful mirage]

ahh and then i add two elements accordingly, (a+_v b,c+_w d)?

and multiply lambda(e_1,f_2)=(lambda dot_v e_1, lambda dot_w e_2)?

[sturdy marsh]

it's just a vector space on that basis. If it makes you comfortable, set g_{ij} = e_i (x) f_j

then elements of the vector space are F-linear combinations a_ij g_ij

and then a_ij g_ij + b_ijg_ij = (a_ij + b_ij)g_ij

[sinful mirage]

yes this is what I was looking for

thanks! 😄

[chilly ocean]

Wow! Why do mathematicians have to make things so difficult!

[sinful mirage]

they make things precise and clear 😅 if I ask a mathematician why why why he can always point me to an axiom if he has enough patience

it's just my physics background not having developed seeing clearly

[sturdy marsh]

this is not a good definition of the tensor product

for example, you seem to be interested in representations

this doesnt work

if you want to tensor representations

[sinful mirage]

yes,i'm doing finite group rep theory atm

[sinful mirage]

yes this is why I wanted to define the tensor product

[sturdy marsh]

okay you certainly should read about how quotients work

and then read about tensor products of modules

they said representations

so you certainly want to be able to take tensor products of modules

[sinful mirage]

we're gonna cover rep theory of finite groups then do ver bit of lie group rep theory and classification of complex semisimple lie algebras

so the tensor product of representations will appear often? So far the prof just defined it

[sturdy marsh]

maybe you can get away with knowing how it works for vector spaces 🤷

im not sure

[sinful mirage]

it's a vector space over a ring(as I know it)

I know there are some fancy other ways to define it (some mathematical physicist told me,but i didnt understand anything of that)

[sturdy marsh]

no that's pretty much it

[sinful mirage]

only in differential geometry

in other courses no

first heard the definition of the module $\Gamma(TM)$

[cloud walrus]

ProphetX

[sinful mirage]

basis is not always guaranteed

for vector spaces it can be proven that having a basis is equiv to zorn lemma/axiom of choice

[sturdy marsh]

the dummit section on tensor products is very good

the exercises in that section should give you some intuition on how to work with them

[chilly ocean]

@ poros

[unique berry]

I need a little help with this.
If I is an ideal of a commutative unitary ring R. Prove I is maximal iff R/I is a field

[chilly ocean]

hint: what are the ideals in a field?

🙋

When the prof doesn't call on you

[unique berry]

Ideals in a field?

[chilly ocean]

maybe a question i ought to have asked before that is, what's the relation between ideals in a quotient ring and ideals in the original ring?

[unique berry]

They're the same elements just one is a set of cosets right??

[golden pasture]

what are the ideals of R/I

how are they related to the ideals of R

[chilly ocean]

I didn’t understand the second part where gcd is d

Why is |a^k|_<S

[ocean magnet]

that comes from (a^k)^s=e

[final pasture]

The order of a^k will be the smallest x such that (a^k)^x = e

if you already know that (a^k)^s = e

then |a^k| must be atleast equal to s, and may be smaller

[cloud walrus]

Moth

Moth

[maiden ocean]

i found a proof for the reverse direction but it feels incorrect? or rather too strong

so I feel like i messed up

[final pasture]

how si S^-1A defined ? {a € A | exists s € S, sa € A} ?

[cloud walrus]

Moth

Moth

[final pasture]

oh lol

[maiden ocean]

with respect to a multiplicatively closed subset S

[final pasture]

yeah ok

[maiden ocean]

anyway then for s in S, s/1 is in S^{-1} Ann(M) so that should imply that s is in Ann(M), no?

[dapper nebula]

I think you are going 2 far

because the backward direction is just the unraveling of the definition

[maiden ocean]

i guess

i mean its not like its any harder im just kind of worried im incorrect

actually

doesnt going directly from the defn not work

because you can get like x/1 = 0/1 so xu = 0

but you dont know that u is in the annihilator

just that it annihilates a single element of M

i guess this is where you write x in terms of the generators but that seems like more effort than just using that annihilator commutes with formation of fractions

actually ok

is there a non trivial finitely generated module M with trivial module of fractions?

i guess restating this: if S is multiplicatively closed can you have a non-trivial finitely generated module M with Ann(M) cap S non-empty but not containing 1

[golden pasture]

Z/2 and {1,2,4,...} ?

[sturdy marsh]

yeah, or just any torsion abelian group localized at (0)

moth you might be overthinking this

if you have generators x_1 ... x_n

find s_i which kill x_i

and multiply them

[maiden ocean]

ik but im trying to see where im going wrong

with my other proof

like i can do it manually but id like to figure out what is wrong with what i am doing

[sturdy marsh]

which proof

[sturdy marsh]

this one?

[next obsidian]

MOTH

I literally remember you solving this exercise a few months ago

lmfao

[maiden ocean]

if S^{-1}M = 0 then S^{-1}Ann(M) = Ann(S^{-1}M) = S^{-1}A

[next obsidian]

unless you're now onto something else

[maiden ocean]

i know how to do it another way chmonkey i am just trying to figure out why this one is wrong

[next obsidian]

wtf is S^-1Ann(M) lol

[maiden ocean]

module of fractions

[next obsidian]

like localizing it as an ideal?

[maiden ocean]

wrt multiplicatively closed subset S

idk maybe

AM says localizing at p is ring of fractions of A wrt S = A - p prime

[next obsidian]

yes

[maiden ocean]

so i assume it is the same concept

[next obsidian]

Yes

it is

I don't buy S^-1Ann(M) = Ann(S^-1M) at first glance

[maiden ocean]

M is finitely generated

[chilly ocean]

Cant it be like that the order is k and then after you raise it to K then you get e and e^s is just e and s may be anything at all

[maiden ocean]

its in the book

[next obsidian]

weird

ok

[chilly ocean]

Ohh wait no I got it I mistook order of a^k for order of a

[next obsidian]

wait so what's the problem you run into?

[maiden ocean]

so my thought was that if S^{-1}M = 0 then the annihilator is all of S^{-1}A right

[next obsidian]

yeh

[maiden ocean]

then since it commutes with forming fractions

S^{-1}(Ann(M)) = S^{-1}A

[next obsidian]

sure

[maiden ocean]

ok i guess actually im not clear on the details here. do we have to do like for s in S, s/1 = x/t for x in Ann(M), t in S?

we cant just say that s itself is in Ann(M)?

[next obsidian]

well for Ann(M) you're only dealing with A

[maiden ocean]

yea

S is a subset of A though

[next obsidian]

okay?

[maiden ocean]

so s/1 in S^{-1}Ann(M) does not necessarily imply s is in Ann(M)?

[next obsidian]

right

[maiden ocean]

okay i think that makes sense

[next obsidian]

I think it does in the finite case over an integral domain

err... I guess like where you're acting faithfully

actually no

if s/1 is in S^-1Ann(M) = Ann(S^-1M), for any x in M, you know that
s/1x/1 = 0 so that sx/1 = 0 which implies there's a u in S such that usx = 0

but even if you assume integral domain, acting without zero divisors, blah blah

I don't think you can ever conclude sx = 0

[maiden ocean]

yeah i think i get it

well the thing i wanna prove still follows pretty easily from this because xu in Ann(M) and xu - su = 0 so xu in S

and S cap Ann(M) is non-empty

thank u

[next obsidian]

I didn't really do anything other than say random unhelpful things

thank shamrock instead

[maiden ocean]

:redpanda:

hi shamrock

[latent anvil]

Hello

Chmonkey what friend are you missing

[next obsidian]

Max

[latent anvil]

Ah yeah

[next obsidian]

I don't wanna cause a stir tho

You know how ppl have done gimmick names after some of the active users

[maiden ocean]

[next obsidian]

we have had a shamrock variant, an ultra variant, a buncho variant

there was a moonbears variant

We should have a Chmonkey variant where you just put Ch in front of it

so I become ChChmonkey

Moth is Chmoth

Sham is Chamrock

[latent anvil]

nope

[maiden ocean]

@next obsidian

just for u

[golden pasture]

is chow group ChCh_i then

[vital fossil]

what does it mean to say C-infinity is an algebra? does that just mean C-infinity is a ring?

[latent anvil]

it's a ring with a vector space structure

Where the ring and scalar multiplication interact correctly

so if c is a scalar and x, y are in the ring, c(xy) = (cx)y = x(cy)

[vestal snow]

Suppose $U$ and $V$ are universal constructions on a category $C$. Then prove that $U(A) \cong V(B)$ implies $coU(A)\cong coV(B)$ if $coU(A)$ and $coV(B)$ exist

[cloud walrus]

Have a Banana, Bitch

[chrome hinge]

Yup, it can also be understood as a vector space with an extra operation $\cdot : V \times V \longrightarrow V$

[cloud walrus]

Maikel

[vestal snow]

Here's my proof

$U(A)$ and $V(B)$ are the initial or final objects in some category $D$. $coU(A)$ must be the initial or final object in $D^{op}$ and same for $coV(B)$.

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

Is this a valid proof?

[final pasture]

(why don't you ask in #category-theory ? )

[vestal snow]

I didn't even know we had a channel for that

[final pasture]

oh lol, then now you know, ig it's more suited for ct questions than #groups-rings-fields

[vestal snow]

Thanks

[unique juniper]

$R = {m + b\sqrt{2} | m,n \in \mathbb{Z}}$

[cloud walrus]

Yes

[unique juniper]

Show $1+ \sqrt{2} $ has infinite order in $R^{\times}$

[cloud walrus]

Yes

[unique juniper]

what does this actually mean

the multiplicative order?

[chilly ocean]

Yeah (especially if it says R^*, it cannot be anything else)

[unique juniper]

ok ty

[viscid pewter]

wait, orbit stabiliser holds for subsets too? sweet

ig the same proof still holds

[rustic crown]

(the action should be closed if you expect nice things)

[viscid pewter]

In D8 since the three subgroups of order 2 are abelian, if an element x isn't in the center of D8, the order of the centraliser of x is 4

i don't quite see how the second bit follows from the first

ok so by orbit-stabiliser wrt. conjugation, |number of conjugates of x| = |G : centraliser of x|

so they say the order of the centraliser is either 4 or 8, essentially

implying the number of conjugates of any x is either 1 or 2

... but i don't see how that relates to the subgroups

so if gxg-1 != hxh^-1, uhhh (h-1g)x(h-1g)-1 != x? so what

[carmine fossil]

centralizer of x contains {1,r^2,x}. So there are atleast 4 elements in centralizer (x)

Since x is not in centre,centralizer order is not 8

So order of centralizer is 4(since centraliser(x) is a group)

@viscid pewter

[viscid pewter]

oh, right, we're just using the center to begin with

cheers

[carmine fossil]

Also, Literally every single group of order 2 is abelian

[viscid pewter]

yes ik

it was basically just a quote

... in fact i still don't see how the order 2 subgroups have anything to do with it

[vestal snow]

@latent anvil When we were trying to prove that right adjoint implies left exact, we used 0 -> A -> B -> C exact iff Ker(B -> C) = A -> B. Would the analogous statement be A -> B -> C -> 0 is exact iff coker(A -> B) = B -> C?

[rustic crown]

(looks good to me)

[vestal snow]

Are you replying to my question?

[rustic crown]

yea

[old lava]

wasn't it ker(B->C) = im(A->B)

[next obsidian]

You can make the statement with the cokernel

[old lava]

no I don't mean the cokernel part

I mean the first part where they say ker(B->C) = A->B

[maiden ocean]

yes, you can see this explicitly by using the first isomorphism theorem: if we let f: A -> B, g: B -> C then ckr(f) = B/Im(f) = B/ker(g) = Im(g)

the second equality is equivalent to exactness at B

[latent anvil]

moth, I think this is in an abelian category

[sinful mirage]

can someone give hints for c)?

i should check that it is a homomorphism, but I am not given rho(sigma2)

I can deduce it using rho(tau)=rho(sigma1 sigma2)=rho(sigma1)rho(sigma2), but then this seems to be backwards thinking,i already am assuming that it is a homomorphism 😦

[hot lake]

yeah you have to deduce rho(sigma2) from rho(sigma1) and rho(tau)

[sinful mirage]

but why am i allowed to do that?

this means that i assume that it is a homomorphism(what i want to prove)

[hot lake]

a representation is supposed to be a group morphism from G to GL(V)

you want to verify that you can fill the unknown values

so that it is a group morphism

[sinful mirage]

ah so I can deduce rho(sigma2) and verify that all relations of a homomorphism are met,given rho(sigma1) and rho(sigma2)

right?

[hot lake]

well you need to find rho of all the group elements

then check the remaining relations that need to hold

[sinful mirage]

yes,but I can find them only supposing its a homomorphism

else i cant find them

[hot lake]

btw that's faster if you know a presentation of the group

so you don't have a million cases to do

[sinful mirage]

what i mean is i am finding the rho of all group elements by supposing it is a homomorphism

is that allowed?

[hot lake]

yes

[sinful mirage]

if it would say find a homomorphism which has this elements would make sense

but it asks me to check whether this is a homomorphism or not

[hot lake]

that will force your rho to fulfill some of the constraints of being a morphism

[sinful mirage]

yes

[hot lake]

but there will still be some leftover to check even after that

[sinful mirage]

oh ok im actually not that far,i will first check those and then see what is left over

[hot lake]

here for example, you can use 4 of the constraints to find the remaining 4 values of rho

and you will be left with 36-4 = 32 cases to check if you really wanted to be exhaustive

32 is a lot but since GL(V) is associative it is not actually needed to check them all one by one, that's why having a presentation of G is nice

it lets you cut down on the number of things to check

[sinful mirage]

what is a presentation?

[hot lake]

it's a set of generators and a set of relations on them that completely describe the group

[sinful mirage]

yes,I have that

how will this help me not to check all other cases?

[hot lake]

you also have sigma1^2 = sigma2^2 = 1

and that should be enough for a presentation

[viscid pewter]

as a general thing, if things hold for the generators, they often hold for the rest of the group

so like instead of finding the matrices for each of the elements direct

you could just find the matrices for the generators, and combine them in the same way that you combine the generators to get the rest of the group's elements

[sinful mirage]

right this is what i wanted to do

supposing rho is a homomorphism

but @hot lake said there will be freedom left after this

if I have the matrix rep of all elements of the group

(supposing rho is a homomorphism)

[viscid pewter]

oh um

[hot lake]

knowing rho(tau) and rho(sigma1) is enough to find rho of all the rest but then you still have to check that it's a morphism

you oculd have some random stuff for tau and sigma1

[viscid pewter]

well if you have like two groups, and you have a presentation for one, and you have elements in the other that fulfil the presentation for the first, there's at least a homomorphism i think

[hot lake]

and things would most likely break

somewhere

you have to check that things don't break

[viscid pewter]

well if they're both the maximal order that the presentation allows, or whatever, idk

[sinful mirage]

what do you mean by having to check its a morphism?

thats only the fact that rho(ab)=rho(a)rho(b) for all a,b in G

ahhhhhhhhhh i see

there are 36 cases because the group has 6 elements, and ab can take 36 values

omg

i only checked 4 by constructing the elements

now i see your point

and what is the alternative of not checking all this?

[hot lake]

you check that they satisfy what the presentation says to satisfy

[sinful mirage]

rho(sigma1 sigma1)=rho(e)=rho(sigma2 sigma2)?

[hot lake]

and the one that they gave you

[sinful mirage]

ok great

[hot lake]

well rather

[sinful mirage]

but why does this make all conditions immediately true?

all 36

[hot lake]

rho(sigma) rho(sigma1) = rho(e)

[sinful mirage]

ah ye

[hot lake]

and rho(sigma2) rho(sigma2) = rho(e)

[sinful mirage]

yse now i know what to check

but i cant see why checking this would suffice for all 36 cases

[hot lake]

because the fact that it's a presentation means that you can deduce all the identities you want from those 3 relations and associativity

so if you can check those 3 relation on the image of rho side

you automatically have associativity because GL(V) is a group so associative

so any identity that is true in G has to be true on the image of rho side

[sinful mirage]

ok so here's a probably trivial statement i cant see

given the presentation on the group,if the presentation is true in image too(i.e. on rep)

why does it follow that all rep elements can be obtained using them?

[hot lake]

a presentation is generators + relations

[sinful mirage]

so if i am given the generators+relations satisfied in the group G

and I check that some relations hold for the rep

why does it follow that all representatives can be obtained using the relation on rep?

ohhh

is it because then the images of generators behave as the generators( the same way) thus they generate all images too? @hot lake

the thing is that they can be any objects as long as they satisfy those relations is what i understand

[hot lake]

since the generators generate the group, any group element can be written as a composition of them

so there is always some way to build rho of the group element

and if there are two ways, then there is a relation in the original group ; that relation is a consequence of the relations given in the presentation

so if your rho(gi) satisfies those relations from the presentation, then it follows that the two ways to build it give the same result

[sinful mirage]

yes,now it makes sense,thanks!

[quaint tree]

A bit of a general question, but what sorts of things are typically covered in a second undergraduate abstract algebra course?

As in, the one that's just vaguely called "abstract algebra II" and whose course description is "a continuation of the material from abstract algebra I"

[chilly ocean]

I think usually stuff like topics in groups/rings/modules/fields that were skipped in a first course, in particular galois theory is not typically covered in a first course. And maybe some commutative algebra type stuff

[quaint tree]

Oh cool, so that's basically what I want to take, then.

I'm planning out my courses for next year and didn't know whether to take a vague "abstract algebra II" or try to get an independent study in Galois theory in.

[chilly ocean]

But the contents of such a course probably depends on the school, so maybe it is good to check if it matters to you

[quaint tree]

I mean, I'm sure I'd be interested either way. I used to be the kind of person who said that I wanted to learn as much math as is humanly possible, and while I still think that's a good goal, I just didn't realize how much math there is out there and realize that I'll have to be a bit choosier about it at some point down the line.

[cloud walrus]

Yes

[unique juniper]

how do i exactly determine the elements of this?

[wind steeple]

Z_2 = Z/2Z ?

[unique juniper]

yes

[wind steeple]

you can try to see if your polynomial have roots

in Z2

[unique juniper]

ok

[chilly ocean]

Well, every coset in this thing can be represented by a degree <= 1 polynomial

[keen bloom]

I'm a Math major (Undergrad only) turned software engineer but I'm trying to apply some of the concepts that I learned in my degree to my work. One of the things we discussed in Abstract Algebra, and that I'm sure everyone in here is aware of is "idempotence", e.g. that if

if f(a) = b, then f(b) = b

Is there a concept of "partial" idempotence for functions of higher arity? E.g.,

if f(a, b) = c, then f(a, c) = c

I've been jokingly calling this "spiritually" idempotent (as in points to the spirit of idempotence), but I was wondering if there was more appropriate terminology. I've also referred to this as f is idempotent over b, but that may also be confusing and wrong.

[unique juniper]

right yes i see

so ig its just 0 1 x and x+1

[chilly ocean]

I guess "idempotent over b" sounds sort of correct in that this property is equivalent to the idempotence of the functions $b \mapsto f(a, b)$ for each a

[cloud walrus]

8da 💕

[keen bloom]

I suspected as much, I was just wondering if there was a more formal definition/terminology of such behavior of f.

[chilly ocean]

It sounds sort of similar to things like "the function (x, y) mapsto f(x, y) is said to be lipschitz in the second coordinate if y mapsto f(x, y) is lipschitz for each x" (or maybe we only say this if there is a fixed lipschitz constant uniformly over all x?)

[keen bloom]

I suspect, in my documentation, I'll refer to this as

f is idempotent over b for each a in A.

[unique juniper]

actually

why isnt x^2 also an element

x^2 isnt in < x^2 + 1 >

[old lava]

firstly, it makes no sense to think of x^2 as an element of the quotient

it's a REPRESENTATIVE of an element in the quotient

[unique juniper]

yeah

[old lava]

next, the coset that x^2 represents is the same as the coset containing x^2 + (x^2 + 1) = 1

you can always just eliminate all the powers of x higher than 1 by adding/subtracting multiples of x^2+1

to find a representative with degree < 2

[unique juniper]

ok

[chilly ocean]

Why must each cyclic subgroup have 4 elements?

Is it because a subgroup of order 10 has four generators?

[unique juniper]

wdym

[chilly ocean]

But it does, 1 3 7 9 in Z10

[unique juniper]

:o oh yh lol

nvm me

[shut halo]

I'm trying to show that the generators of a semi-simple Lie Algebra are traceless. What I know that if I have $[T_a, T_b] = f^c {ab} T_c$ then we must have that $f^c{ab} Tr(T_c) = 0$ since the tr(AB) = tr(BA) is always zero but not quite sure how to proceed from here.

[cloud walrus]

snypehype

[chilly ocean]

How did they get that x^20=1

What does an element of U(100)=Z2 x Z20 look like?

Something like (1,19) maybe

For sure, and now what happens if we take it to ^20 (or more like multiply by 20 since we are working in additive groups)?

It becomes (0,0) right?

Yeah, and this makes sense right?

Bruh😭

[old lava]

it becomes 1, 1

[chilly ocean]

Something times 20 is divisible by 2, and sometimes times 20 is divisible by 20

Yea

How does this imply 1 tho

[chilly ocean]

I guess this depends on what we mean by 1

[old lava]

I know exactly 0 people in existence who write U(100) additively

[chilly ocean]

For sure, but we are talking about additive group Z2 x Z20

Wait so x^20=(0,0) how is (0,0) 1

help pls

Poros just meant (0,0) is the identity in Z2 x Z20. And that in the isomorphism Z2 x Z20 = U(100) the element (0,0) would map to 1 in U(100)

Ohhh thank you

[vestal snow]

1. I want to avoid taking quotients because we're they might not exist in our category
2. the definition of exact I'm using is ker(f_i) is isomorphic to Im(f_{i-1}) so I don't think we can say that B/Im(f) is isomorphic to B/Ker(g)

[maiden ocean]

Yeah sorry I wasn't aware that you were working in a general abelian category

or preabelian or whatever idk there are like 12 conditions

regular??

[prisma ibex]

there is some general abstract nonsense about which kinds of categories the various isomorphism theorems are valid in

this is certainly true for Abelian categories but also true for much weaker categories

iirc this is true for protomodular categories in the sense of Borceaux/Bourne

[next obsidian]

It’s actually common to write things as a quotient for cokernels in abelian categories precisely because the first iso holds in them. But yes, in general you’d wanna avoid them

[vestal snow]

Can someone help me work this out i detail?

right adjoint implies left exact was pretty straightforward

But since the definition of exact sequences is given in terms of kernels and images and left adjoints preserve cokernels, I can't prove that left adjoint implies right exact

[latent anvil]

Did your first proof happen in an arbitrary abelian category?

Or just in eg the category of modules

[vestal snow]

arbitrary abelian category

Give me a sec. I'll type up my proof for the first one

Taking a quiz so it might be more like 30 mins

[next obsidian]

If it happened in an arbitrary abelian category I'm pretty sure the other one follows by duality

more generally, I think you should just show that right adjoints preserve limits

or I guess in this case that a left adjoint preserves colimits

But either way, the other follows by duality

[vestal snow]

Duality just seems really weird in this case

[vestal snow]

I'm sure it works, but I think working out the details of the duality argument would boil down to proving this