groups-rings-fields

[rustic crown]

like consider the divisor lattice that's the sup of 6 and 9?

an upper bound is somethign that is divisible by both 6 and 9, so by 18. and so 18 is actually the sup

[novel parrot]

whats divisor lattice?

[hidden haven]

m <= n iff m | n

[novel parrot]

and normal lattice?

i didnt realise lattice was another structure

[hidden haven]

poset with all binary sups and infs

[novel parrot]

from what set?

[hidden haven]

what

[novel parrot]

you said powerset?

but dont we need a set to take power set from

[hidden haven]

poset is partially ordered set

[novel parrot]

oh

what does binary sup and inf mean

[hidden haven]

sup and inf of any 2 elements exist

[summer falcon]

So i had a problem in category theory (which is trivial, but I don't know much about category so can't solve it)

It says find a category where product of rings is a universal object

I saw definition of Universal objects but i dont dont clearly understand what they actually are

[lethal dune]

@summer falcon #category-theory

[summer falcon]

Oh

[novel parrot]

what does it mean for ^ distribute over the other?

[hidden haven]

same thing as * distributes over +

[novel parrot]

ok

[amber gorge]

Hi everyone. I'm practicing some problems for my abstract algebra class, and I have a quick question. I'm running into a few questions involving Aut(U(n)). I know that Aut(Z_n) is isomorphic to U(n), but I don't know what I can say about Aut(U(n)). If there is a fact about Aut(U(n)), I will certainly try to prove it myself, but I'm not sure what I should attempt to prove. I'm basically wondering if there is an isomorphism between Aut(U(n)) and another nice group and then I will attempt to prove it. I'm just not sure what that isomorphism is, so I'm not sure what I should prove. Does that make sense? Thanks!

[novel parrot]

are you asking if Aut(U(N)) is isomorphism to anything?

[south patrol]

This seems tricky as U(n)'s structure depends on properties of n

Like there almost certainly isn't one universal thing thing for this I believe (in the sense of oh Aut(U(n)) is a member of this family)

What sort of questions were coming up?

[amber gorge]

yes exactly. Once I know what isomorphism to look for, I can try to prove it. But I agree, it does seem tricky. The problem I was doing where this arose was the following: write Aut(U(25)) in terms of Z_n \oplus Z_m. I know a relation from Aut(Z_n) to U(n), but I need to get out of the Aut(U(25)) in order to use that.

25 not 35 hold on let me edit it

[south patrol]

Ah, okay, well this is somewhat easier as |U(25)|| = φ(25) = 20

for example

i.e. we've already narrowed it down to two possible groups

Perhaps that's the sort of thing they had in mind...

[amber gorge]

phi(25) = 20, not 4. phi(25) = 5^2 - 5^(2 - 1) = 25 - 5 = 20

[south patrol]

my bad, yup

[amber gorge]

but still, we're dealing with U(n) here, not Aut(U(n)). Also, what does CRT mean?

[south patrol]

chinese remainder theorem

[amber gorge]

I haven't learned that

[south patrol]

but yeah, we've not shown it's cyclic so that doesn't apply yet

oh sure, interesting

the idea is that if n,m are coprime then Z_(mn) is isomorphic to Z_m x Z_n

[amber gorge]

that fact I have been given, yes. They didn't call it the chinese remainder theorem, but if that is what it is, then I do know it

[novel parrot]

ive never used the chinese remainder theorem for anything

[hasty sorrel]

Really? The Chinese remainder theorem is awesome.

[amber gorge]

now, I do have formulas to turn Z_m \oplus Z_n into U(k). They gave a few in the book. I suppose I could turn that U(k) into Aut(Z_k), and then I would have to show Aut(U_s) is isomorphic to Aut(Z_k). I wrote s to avoid confusion with the n in the direct product

[hidden haven]

I remember you have done AM exercises which use it

[novel parrot]

lol really

[south patrol]

lol i forgot but there is a theorem from gauss that kills this i think lol

but yes let's not just nuke it yet

[hidden haven]

Z_m ⊗ Z_n is the 0 module I think it was

[novel parrot]

ohhh

i ended up showing that 1 x 1 was 0

with elementary number theory

[south patrol]

But basically you can show that U(25) is cyclic

and then you can just use the same theorem once more to find Aut(U(25))

[novel parrot]

messing with bilinear maps was confusing

[south patrol]

:)

[hidden haven]

CRT is elementary number theory given m and n coprime, there are x and y such that mx+ny=1

This is just another way to say CRT

[south patrol]

bezout = crt?

based

but yes

[novel parrot]

lol

[amber gorge]

wait, what theorem are you referring to? So yes, I could show U(25) is cyclic. I'd write out all of the terms and find the generators. But what does this tell me about Aut(U(25))? Am I missing something?

[novel parrot]

i never called that chinese remainder theorem

[hidden haven]

Well CRT is more general because you can replace 1 with gcd(m,n) and Z with any PID

[south patrol]

well if U(25) is isomorphic to Z_n, then Aut(U(25)) is isomorphic to U(n)

[amber gorge]

yes I also called that the Bezout identity. The one listed above

[hasty sorrel]

I really think about the chinese remainder theorem as a homomorphism between rings

[amber gorge]

C_n being a cyclic group? Sorry I haven't seen that notation. Or is that the complexes?

[south patrol]

Sorry, i use that notation for Z_n

[hidden haven]

Before this server the only bezout I'd heard of was bezout's theorem from AG

[amber gorge]

ok, no problem, I just wanted to make sure

[south patrol]

But yeah, you get that? Now you just need to find said n

[hasty sorrel]

which mx + ny = 1 defines that the ideals generated by m and n are homomorphic to the parent ring which has the identity.

[south patrol]

Well, we know n = 20 as we mentioned before

So all we need to do is show U(25) is cyclic

[novel parrot]

yeah this is how i learnt it

[amber gorge]

so if U(25) is isomorphic to Z_n (yes n is clearly 20 from Euler-phi), then we can say Aut(U(25)) is isomorphic to U(20). I will have to prove that to convince myself of it, but it can certainly try that on my own. It will be a good exercise

[novel parrot]

and then the one from elementary number theory is just a corollary of that

[south patrol]

Sure

I'm not actually sure off the top of my head how to show U(25) is cyclic, but yup

[hidden haven]

You have to use CRT plenty to prove the structure theorem for modules over PIDs

[amber gorge]

I wonder if I can even say if U(n) is isomorphic to Z_(phi(n)), then Aut(U(n)) is isomorphic to U(phi(n)).

[novel parrot]

havnt learnt that yet

[amber gorge]

I can just write out the terms and find a generator. That will show it is cyclic

[south patrol]

Oh well that is just easily true from the theorem you had earlier that Aut(Z_n) is isomorphic to U(n)

So yeah

[hasty sorrel]

oh. Is it useful for studying tensor products on modules as well?

[south patrol]

And yup, good idea

[hidden haven]

I don't think so but I haven't worked with tensor products much beyond the first 5 chapters of AM

[amber gorge]

Oh yeah, that is true. That is a good fact to know then. Alright, I think I'm good to go. Thanks for your help!

[hasty sorrel]

Which book is AM? I have only read Dummit and Foote

[novel parrot]

what makes PID modules better than normal ones

[novel parrot]

atiyah mcdonald

but AM is so much better than dummit and foote

[hasty sorrel]

Nice thanks! I am always looking for new books to read.

[novel parrot]

depends on where u are in dummit and foote

like if u are past chapter 9 in dummit and foote

i recommend switch to AM

[south patrol]

Isn't AM com alg so for different purposes?

[hasty sorrel]

I have read quite a bit of Dummit and Foote. The only section I haven't really touched is 15+

[novel parrot]

ohh chapter 15 is the worst

[south patrol]

Hm how does df compare to artin

I only have artin

[hasty sorrel]

But I find chapter 14 is pretty tough already lol

[novel parrot]

galois theory in DF is done very good

[hasty sorrel]

I understand the automorphisms, but it is challenging to understand the proofs.

Still working on it

[novel parrot]

did u read chapter 13

also very good

[hasty sorrel]

Yeah, I love that chapter actually

[novel parrot]

but chapter 13 has little problems

and then chapter 14 has like 40 problems each section

[hasty sorrel]

what do you mean?

oh, you mean the problems in the end lol

I thought you meant it has issues

[novel parrot]

yeah end of section

[hasty sorrel]

I am not sure... I haven't read artin

[novel parrot]

just by looking at contents page of artin

df and artin look similar

but df has some extra stuff

commutative rings

[hidden haven]

Artin has ring theory

[novel parrot]

yeah but not stuff that DF has

i dont think atleast

but artin has group representation

ohhh no DF has group representation too

but the last section

DF even has category theory

[trim grove]

While proving quotient group of a cyclic group is cyclic

They are saying $Ha^{n}=Ha....a$\$HaHaHaHa..=(Ha)^{n}$

I want to confirm that last step is beacuse H is normal so Ha=aH thats why

$HaHaHaHaHa...=(Ha)^{n}$

[cloud walrus]

Algebra

[hidden haven]

I don't think so

looks like just (Ha)^n expanded out

like just the definition of ^n

[thorn delta]

wtf am i looking at hahahaha

[hidden haven]

haha

[novel parrot]

HaHa

[trim grove]

Laughing in Math

[trim grove]

I think i made a mistake i should ask for

$Haaaaaa..a=HaHaHa..Ha$

[cloud walrus]

Algebra

[hidden haven]

Yes this is how multiplication is defined in the quotient group, and requires normality of H to be well defined

[wooden ember]

Lmao DF has like a 8 page appendix on cat theory and it really sucks

[next obsidian]

If you learned category theory from D&F rip to you

[novel parrot]

its not that bad right?

[hidden haven]

In 8 pages you can barely cover the most basic definitions and examples That really doesn't count as having cat theory lol

[novel parrot]

lol youre right

hahaha

[hidden haven]

It would be like saying this book defines a ring homomorphism, so it has commutative algebra

[novel parrot]

[errant sierra]

how do find what group this is isomorphic to

i tried messing around with the abc=e and acac=e but im left with a mess

[novel parrot]

what groups are there

not familiar with that notation

is a,b,c order 4,2,2?

and abc = e

acac = e

[south patrol]

it's a list of stuff that are e

[novel parrot]

right

[errant sierra]

its supposed to be isomorphic to some familiar group

maybe Zn x Zm n Zo but not commutative obviously

so not sure

[novel parrot]

what groups are you familiar with

it shares some relations with D4

[errant sierra]

Dihedral groups, Z Q R C Z_n M_n(R) and R[x]

general ones

how is abc=e in d4

[novel parrot]

rsrs = e

-> acac

i dont know for sure

or maybe two groups

how many elements in ur group?

[errant sierra]

ok true with the acac

would it be 16 elements in the group bc we have a^4 b^2 and c^2 ?

[novel parrot]

idk

non commutative

and other identity relations

D8 x F2?

dont think so

[errant sierra]

what about symmetric group in combination

[novel parrot]

maybe

but is there not more than 16 elements

in ur group

ababababab ...

we could multiply ab forever?

[errant sierra]

or add

but i think so?

and we get e every time

[novel parrot]

from which

[errant sierra]

maybe not. only if its multiple for 4 of a

[sharp shale]

(ab)^n=c^(-n) so ab of order 2 because c is?

[wooden ember]

ab = c^-1 so abab = c^-2 = e

[novel parrot]

yes ur right, mb

[errant sierra]

right

we need for sure order 16 group right

[wooden ember]

So you automatically know a and b don’t commute ig

[novel parrot]

i dont feel like it could have some part of symmetric group bt idk

[wooden ember]

Not necessarily

cac^-1 = a^-1

Might be able to do some semidirect product construction

And similarly bab^-1 = a^-1

So maybe $Z_4\rtimes (Z_2 \times Z_2)$

[sharp shale]

must be related to D8 in some way

[cloud walrus]

𝓛ittle ℕarwhal ✓

[novel parrot]

how about D8 and F2

[wooden ember]

Or something around those lines

[wooden ember]

What’s F2

[novel parrot]

Z/2Z sorry

[errant sierra]

F_2=<a,b>

all powers of a and b combinations

[wooden ember]

Yeah but clearly they didn’t mean that

[errant sierra]

a^9b^11

[wooden ember]

I’d think about this more but I have a class recording to watch for tomorrow and it’s almost 22:00

Gl

[errant sierra]

Z_4\times (Z_2 \times Z_2)
so let a=(1,0,0) order 4
b=(0,1,0) order 2
c=(0,0,1) order 2
abc isnt necessarily identity though right

[errant sierra]

thanks for your input! and see yaa

[novel parrot]

what does this cross mean

[errant sierra]

semidirect product symbol

[novel parrot]

oh

[past temple]

im struggling to understand what an algebra over a ring is

there are two definitions

one being that B is an algebra over A if B is the image of A under some ring hom

the other is that B is an A-module equipped with a bilinear operation A x A -> A

im confused as to how these definitions are equivalent

also i cant think of any concrete examples

[hidden haven]

not image, but codomain

R/I is always an algebra over R

C is an algebra over the reals which are an algebra over the rationals which are an algebra over the integers, all by means of the inclusion homomorphisms

The idea is that B is a module over A which is also a ring at the same time, and the 2 structures are compatible

[hidden haven]

This is the concrete realisation of this while the other definition is saying the same thing abstractly. You can try proving that both are the same

instead of saying bilinearity of that operation, you can try writing down all the equations that that gives

and this will give you the algebra axioms similar to the usual form you see group or ring axioms in

[past temple]

im confused as to how an A-module can have an operation A x A -> A tho

like the elements of the module are not elements in a

*A

[hidden haven]

similar to how a set S can have an operation S x S → S to give extra structure

It is a new operation that is being added to the already existing operations of the module

[past temple]

im confused tho bc ur not doing anything to the elements of the module

[hidden haven]

wait hold on

that is incorrect, it should be B x B → B

because you are adding a multiplication to the A-module B which makes B into a ring and the bilinearity says that this B is then a ring with module addition and new multiplication, and that this new multiplication is compatible with scalar multiplication by elements of A

it is A-bilinearity, which means that it respects both addition and scalar multiplication by elements of A

bilinearity says exactly that the new operation distributes over addition

A-bilinearity says a(mn) = (am)n = m(an)

this is what compatibility means

[past temple]

ok its a little confusing bc the operation is multiplicatoin right?

like in the case of a quotient ring

multiplying the representatives

is the bilinear map

[hidden haven]

yes

so you have both a scalar multiplication and a ring multiplication

[past temple]

ok the quotient ring example def helps for understanding

im still confused as to like

how it relates to being the codomain of a ring hom tho

i mean for a quotient ring its the natural projection

but what if its not surjective

[hidden haven]

Look at the other examples

The algebras induced by inclusions

since any homomorphism may be broken into a surjective map then an injective one, all algebras would be formed by inclusion of a quotient of A into B

(first isomorphism theorem)

Also for the first definition, keep in mind that B is not an algebra in that case, but the algebra is like (B, that homomorphism)

different homomorphisms induce different algebras

eg Z[x] → Z[x] could happen by identity or x → 0

both give non isomorphic algebras

one of them being finitely generated as a Z[x] module and one of them not

[past temple]

hmm okay

i think this all makes a bit more sense now ty

[terse crystal]

How to calculate Gal(k(x_1,…,x_n)/k) where x_1,…,x_n are algebraically independent. I mean automorphisms of k(x_1,…,x_n) that is identity map restricted to k. What should the image of x_1,…,x_n look like ? I only know that when n=1 the image of x should have the form (ax+b)/(cx+d)

[golden pasture]

i dont think theres a simple way but you could construct it as say a tower of normal field extensions but then you only know that your group is some extension of galois groups so nothing much

[terse crystal]

I see. Thanks

[frank fiber]

if i have a quotien map $\pi:A \rightarrow A/I$ and an idea $J \subset A/I$ then is it true that $\pi \sqrt { \pi^{-1 }J}= \sqrt J$ ?

[cloud walrus]

Or x1

[next obsidian]

yes

sqrt(J) = intersection of primes containing J

Let's change notation a bit

say your ideal is J/I and J> I

then sqrt(J/I) = intersection of p/I where p is a prime containing J

this is defined as intersection pi(p) for p primes containing J

then this is the same as pi(intersection p) cuz intersection commutes with pushing stuff through functions

and now intersection p = sqrt(J)

(note that J = pi^-1 J/I)

[frank fiber]

thanks

[hidden haven]

Doesn't algebraically independent imply transcendental? In that case isn't that just an automorphism group rather than Galois group?

[terse crystal]

yeah… turned out it’s quite complicated, cremona group

So never mind

[sharp shale]

Any ideas how to do this?

[wooden ember]

formal power series can only be taken on commutative rings right?

cause DF only says ring but you dont get a commutative ring out of it if R isnt commutative itself

[hot lake]

well if R isn't commutative then R[[x]] isn't commutative, but it's still a ring

[dusty sapphire]

Possible map for homomorphism?

[hot lake]

since you have a presentation for D12 already, you just need to find the ways to pick r,s in G such that r^12 = s² = (rs)² = 1

[dusty sapphire]

i.e. r map to x and then what about s

[hot lake]

G only has 3 elements

just go through all of them and see if things work

[dusty sapphire]

Ok thanks i try

it is correct?

[hot lake]

yeah

also I think x^3 would want to be called 1

[dusty sapphire]

means

[hot lake]

G = {1 ; x ; x²}

the presentation says x³ = 1

[dusty sapphire]

so it true no

[hot lake]

yes but it's just that x³ is a weird name for 1

[dusty sapphire]

aha got it

all map to identity trivial

[hot lake]

yes

the only group morphism from D12 to G is trivial

[chilly radish]

what's the reason that the notation for formal power series is R[[x]]

[frank fiber]

i think that is for diferentiate it from the polynomial ring R[x]

[chilly radish]

i'm wondering if there's any meaning to the notation tho

[next obsidian]

You get it from R[x]/(x)^n taking an inverse limit and inverse limits add an extra [] didn’t you know

[past temple]

by the second isomorphism theorem

(S + I)/I is isomorphic to S/(I \cap S)

im kind of confused as to why its not just isomorphic to S/I

if we take an element s + i in S + I

then in the quotient that immediately just becomes s + I

[small bison]

I might not be a subgroup of S

[hidden haven]

Who said you are?

[past temple]

not sure if im correctly understanding the definition of a graded ring

is R_mR_n = {x + y | x in R_n, y in R_m}

?

bc they're just groups right

so they dont have a well-defined multiplication

or is it multiplication

[hot lake]

RmRn = {xy | x in Rn , y in Rm}

it's multiplication from R

R is a ring

if you forget about multiplication and look at R as an additive group, it decomposes into the direct sum

but also, multiplication from the ring has to take elements from Ri and Rj to some element in R(i+j)

[past temple]

ah okay that makes sense

ok back to my previous question about like the second isomorphism thoerem

what does an element of (S + I)/I look like?

because an element of S+I is going to be s + i

but then the i is killed by quotienting so that just becomes

s + I

[chilly radish]

Yes, you just need to add all the elements of I to S so that taking the quotient actually means something

But S+I/I ={s+I| s\in S}

So it's the same thing

[past temple]

ok

so the product of homogeneous elements in a graded ring is homogeneous

but not necessarily the sum?

[hot lake]

the sum of nonzero elements from different graded parts of R won't be homogeneous

[lime ivy]

can someone clarify for me why exactly the permutations of polynomial solutions is important in galois theory and proving abel-ruffini theorem? the set of polynomials forms a ring and is commutative in multiplication so i don't understand why changing the ordering of the factorization would make any difference

[terse crystal]

Any field F whose characteristic is 0, consider this polynomial f=Π(x-x_i) in K=F(x_1,…, x_n) now f is also a polynomial in E=F(t_1,…,t_n), and in E f=x^n +Σ(-1)^k t_k x^(n-k) where t_k are elementary symmetric polynomials.

And there exists a radical extension tower E=E_0 <= E_1 <=…<= E_m such that K is contained in E_m if and only if the Gal(K/E) (which is isomorphic to S_n) is solvable

I think any algebra text book has it . I read Jacobson

[upper cape]

Hey, I was asked to prove that for a ring A, nil(A)+U(A)=U(A), but this seems trivially false, since no nilpotent element is a unit. Just want to check if I am being stupid, or if the question is wrong, thanks!

[terse crystal]

If you are asking whether {x+y: x is from Nil(A) and y is from U(A)} = U(A) then I think it’s correct

Since any x from Nil(A) y from U(A), x+y=y(1-(-y^-1 x))=y(1-t) where t=-y^-1 x is also nilpotent

And 1-t has inverse Σt^k

k from 0 to a large enough integer

[digital yoke]

hm I have a thought

let f:G \to H be some group homomorphism

if G/ker f is isomorphic to G, what does that actually say about ker f?

like I know that G/ker f is isomorphic to im f, but still

[hidden haven]

G will be infinite or ker f will be 0

Hmm idk much else you could say, maybe you'll get even stronger infiniteness conditions on G than just cardinality, something like rank maybe

[gritty sparrow]

A very trivial example of ker(f)/=0 would be g= Z^|N|

and ker(f) being some copy of Z

[hidden haven]

Maybe you can say that G ≈ G ⊕ ker f through the splitting lemma somehow, but I'm not able to finish this argument

[hidden haven]

In this case I'd imagine you could say that G must have infinitely many summands isomorphic to ker f

[digital yoke]

hm I get that if ker f isn’t 0, G has to be infinite

like, hand-wavily

something like inf/finite = inf

but idk how to rigorously wrap my head around that

[terse crystal]

Not necessarily

let G=Π{G_j: j is from N} where all G_j are Z. Then we have a homomorphism from G to itself mapping (x_0,x_1,…) to (x_0,x_2,x_4,…) clearly this is surjective

Kernel has also infinite rank

[hidden haven]

If G were finite and ker f non zero then G and G/ker f have different cardinalities

[digital yoke]

right

[hidden haven]

I don't think this happens nvm

But can't see counterexamples

[gritty sparrow]

I think I have one

consider Z/2Z\oplus countably many copies of Z, and take Z/2Z\oplus2Z as the normal subgoup (sitting in the first two factors)

[hidden haven]

orz

[gritty sparrow]

G\oplus(ker(f) has a different two torsion subgroup (Z/2Z\oplusZ/2Z) hence they are not isomorphic

[gritty sparrow]

what is orz?

[gritty sparrow]

I see

[upper cape]

Thanks! for some reason I was reading it as a union of sets

[woven obsidian]

Only the bottom is relevant for my question, but I attach the rest for context

The problem is that I cannot show that the left hand side of 6.23.a is intertwining, $$\pi_W \circ \pi_V^{*} : End(V) \to End(W)$$

[cloud walrus]

AoiKunie

[warm holly]

I have and exercice which asks me to find all the equivalence relations over Z such that for all x, (x,x+2) is in equivalence relation. My idea was to look at 2 different partitions of Z (first one containing all pair number subset and all even number subset and second Z it self). Clearly equivalence relation defined on these partitions verify the condition (x,x+2) in R. But, how am I sure that there is no other partition which verifies that? Should I suppose that one subset of a partition has some even and odd numbers at the same time ?

[final oasis]

Is there a concise but clearly written book on rings and modules
I am trying to follow Prof. Berchards lectures on ring but he moves fast and skips a lot of the details

[potent briar]

i still can't understand this one problem ugh

[hidden haven]

||all odd numbers must be in the same position||

You need a space after \xi

[potent briar]

come on

[hidden haven]

_ after sum lol

For the index to go subscript

[potent briar]

If $(\alpha_0, \alpha_1, \alpha2, ...)$ is an arbitrary sequence of complex numbers, and if $x$ is an element of P (polynomials), $x(t)=\sum_{i=0}^{n}\xi_it^i$ write $y(x) = \sum_{i=0}^{n}\xi_i\alpha_i$.

Prove that y is an element of P' (dual space of P), and that every element of P' can be obtanied in this manner by a suitable choice of $\alpha s$.

[cloud walrus]

vik

[potent briar]

ok

i've been trying to understand this problem and its potential proof for a few days now

note: i'm not supposed to use transformations to solve it

[final oasis]

Is y in P' clear to you by verifying linearity

[potent briar]

I did verify that y is in P'

but it might be wrong

i think it was a dumb way of doing it

uhm how do i do new lines

$y(x)=\sum_{i=0}^{n}\xi_i\alpha_i$

$y(\beta_1 x_i + \beta_2 x_2)=\sum_{i=0}^{n}(\beta_1\xi_i + \beta_2\eta_i)\alpha_i$

$= \sum_{i=0}^{n}\beta_1 \xi_i \alpha_i+ \beta_2 \eta_i \alpha_i$

$= \beta_1 \sum_{i=0}^n \xi_i \alpha_i + \beta_2 \sum_{i=0}^n \eta_i \alpha_i$

and then

$\beta_1y(x_1) + \beta_2y(x_2) = \beta_1 \sum_{i=0}^n \xi_i \alpha_i + \beta_2 \sum_{i=0}^n \eta_i \alpha_i$

[cloud walrus]

vik

[potent briar]

thats proves that y(x) is a linear functional right?

[final oasis]

Yeah, only that last statement would be
$y(\beta_1 x_1 + \beta_2 x_2)= \beta_1 y(x_1) + \beta_2 y(x_2) $
As the last statem

[potent briar]

ok but that's easy

[final oasis]

Yep

[potent briar]

how do i prove that all elements of P' have that form

if i understand correctly the sequence of alphas could be anything

with repeated elements

etc

[final oasis]

Show that each basis element goes to a particular number

[potent briar]

basis element? 1, t, t^2, .. ?

[final oasis]

Yeah

[potent briar]

i mean doesn't it go from t^i -> alpha_i

by definition?

[final oasis]

Yep

[potent briar]

is this telling me there is a bijection from P to P'?

[final oasis]

Not really, but but yeh any 2 n dim vector space are iso

If over the same field

[potent briar]

ok so concretely

how do i prove the second part

[final oasis]

State the existence of the apha_i and since each poly is a linear combination of basis el you get the some from the linearity of y

ie. Each y is fully described by the aphas

[potent briar]

ooh right

but

how do the different elements of P' differ

y is just one

because it's one particular choice of alphas?

how would another y be written

i mean, all elements look like y(x), but they have different alphas because they "come from different polynomials"?

if so how do you know that all the y(x)s cover the entire P'

[final oasis]

Fix a basis of P

y works because it's arbitrary

any linear functional takes a basis element to a number. You then give a label to each of those numbers

[potent briar]

ok maybe the next exercise help me understand a bit more

[warm holly]

You mean partition ?

[hidden haven]

yes lol

[potent briar]

If $y$ is a non-zero linear functional on a vector space $V$, and if $\alpha$ is an arbitrary scalar, does there necessarily exist a vector $x$ in $v$ such that $y(x) = \alpha$

[cloud walrus]

vik

[warm holly]

Hm not sure to understand then… partition is composed by disjoint subsets of A such that its union is equal to A. If one of subsets is all odd numbers and another one is subset with all even, then equivalence relation defined on this partition verifies (x,x+2) for all x no?

[hidden haven]

so you know all even numbers are in the same equivalence class and all odd numbers are in the same equivalence class. You just need to count the number of partitions that make this happen

and check that all of them do satisfy the given condition which you pretty much showed just now

so a partition gives an equivalence relation with that property iff all odd numbers are in the same equiv class and all even numbers are the same equiv class

[warm holly]

So for example@partition X={{odd numbers}, {even numbers}} is ok, right?

The second one is juste a partition containing Z it self

[hidden haven]

yes\

yep

and you just have to prove that there is nothing else

[warm holly]

Alright. And to do that I can show that there are only 2 possible equivalence classes for first partition and only 1 for second 1?

[hidden haven]

I dont see how that proves it

[warm holly]

The set of equivalence classes forms a partition if I’m not wrong

Or another way would be to consider the first partition and modify the first sub set by considering there there exist an even number and the same with another one but with odd

Is it correct?

[hidden haven]

yeah equivalence relations are equivalent to partitions, they each give rise to each other in a unique correspondence way

[warm holly]

Or probably show that equivalence class of even and equivalence class of odd it’s union is disjoint

[hidden haven]

yeah, basically consider the situation where ||one of the sets in the partition contains an even number and an odd number. Either this situation happens or it doesn't, and the 2 cases both give you a unique partition||

[warm holly]

Nice, thank you very much

[final oasis]

Got it?

[potent briar]

no

[final oasis]

Any idea how u would get it

@potent briar

[potent briar]

i mean if i think about R^1

y(x) = x gives you every scalar

[nova plank]

You can't choose a specific transformation

It has to be true for an arbitrary one

So y a nonzero linear function on V, that means there is a nonzero scalar β and a vector v in V such that y(v) = β

Now you are given a scalar α, can you give me a vector which is mapped to α?

Also this belongs in #linear-algebra probably

I guess depends on content

@potent briar

[potent briar]

what part of the definition of linear functional helps here

"A linear functional on a vector space V is scalar-valued function y defined for every vector x, with the property that (identically in the vectors x1 and x2 and the scalars a1 and a2

y(a1x1 + a2x2) = a1 y(x1) + a2 y(x2)"

@nova plank what if the vectors aren't made up of scalars?

[nova plank]

v isn't a scalar

It's just a vector

alpha and beta are scalars

Read my description of the problem carefuly

And think how you can manipulate y(v) = beta, so that the output is alpha

[potent briar]

yeah but

why does
"y a nonzero linear function on V"
imply that
"that means there is a nonzero scalar β and a vector v in V such that y(v) = β"

the question is precisely whether you can get all scalars from one single functional

isnt it

[nova plank]

The zero linear functional is the functional that sends every vector to 0. So if the linear functional is nonzero, it must send at least one vector v to something that is not zero

Does that make sense?

[potent briar]

yeah sure

[nova plank]

Okay, so the vector is v and the nonzero scalar it gets mapped to, I call beta

Now you have to show this is enough to guarantee that every other scalar gets mapped to as well

So for an arbitrary scalar alpha, you have to find a vector maps to alpha

[potent briar]

ok

y(a/b v)

or smth like that no

[nova plank]

Yep

Perfect

[past temple]

is the direct sum of noetherian rings noetherian?

[rustic crown]

you mean direct product?

ideals of R x S look exactly like I x J, so that just translates the problem to each component

[past temple]

so it should be true right

every ideal of R should be finitely generated

and every ideal of S should be finitely generated

so the ideals I x J should be generated by the pairs of generators

of which therea re finitely many?

[rustic crown]

yeee, but enough to say generated by {(r, 0)} and {(0, s)}

[past temple]

ah right

[rustic crown]

[barren sierra]

for part a

I am struggling to show -|b| <= a <= |b| implies a^2 <= b^2

[gritty sparrow]

first without loss of generality by replacing b with |b| we can assume that b is positive. Then just split into 2 cases, one where a is positive and one where a is negative. Both of these cases should be easy

[barren sierra]

why can we replace b with |b|

[gritty sparrow]

because |b|^2 =b^2

[barren sierra]

Hmmmmm

I don't know if I have to prove that or not but that's easy to show

[gritty sparrow]

If you aren't comfortable with that, you can split into cases for b as well, I just wanted to avoid that because that sounds annoying

[barren sierra]

yea no I'd rather not

[barren sierra]

I'll try to show this and then move from there. Thanks

[gritty sparrow]

cool

[novel parrot]

what stops an arbitrary module from having linearly independant elements

[hidden haven]

what

[gritty sparrow]

what exactly do you want an explanation for

[warm holly]

I just have a question on equivalence classes and partitions, please. Given an equivalence class R we can construct all possible equivalence classes on it. Then, the union of equivalence classes forms a partition X.
So from equivalence class I can come up to a partition.
But, given a partition X( same as before) and use an equivalence relation R’={(a,b) in AxA : exist Y in X s.t {a,b} in Y}over this partition, is R’=R ?

[urban acorn]

are you asking why not all modules over rings have a base?

Exercise: Look at the proof that every (for this issue's sake, finite dimensional) vector space has a base, and find where the fact that F is a field is used. (this happens by dividing by a scalar, with the argument that it is non-zero so we can divide by it by doing scalar multiplication by its multiplicative inverse)

Example: Z/nZ as a Z-module

[hidden haven]

Yes. Assume 2 elements are related in one and show that they are related in the other

And you would probably also want to show that this process, starting with a partition and ending at a partition yields the same partition

If you are trying to show a correspondence

Because you want to show that these conversions are inverse to each other in both ways. One isn't enough

[novel parrot]

hmm i c

[warm holly]

Thank you!

[celest mantle]

Hi, i have an exercise in which we have to consider k[x1, ..., xn]/I(V) where I(V) = { f \in k[x1, ..., xn] | \forall x \in V, f(x) = 0}. Though, as I'm reading and learning algebra on my own, this quotient ring really seems hard to understand, i don't really see what it looks like...
is it like the ring of polynomial functions on V or something ? as I(V) seems to be the kernel of phi : f -> f(x)

[hidden haven]

yes it is exactly the ring of polynomial functions. You are identifying 2 polynomials iff they define the same function

[celest mantle]

alright thanks

[warm holly]

If we have a set A and a set of all partitions of A, can we construct all the possible equivalence relations over AxA by using equivalence relations over all partitions of A?

I.e {set of all equivalence classes over AxA }= {R_1,…,R_n,..} with R_i equivalence class of partition X_i of A

[gusty halo]

maybe im misunderstanding, but i dont think you need any relations on the set of partitions at all. Equivalence relations on A stand in bijection to partitions of A by R -> A/R

oh wait do you actually mean equivalence relations on AxA and not on A, i.e. relations R being subsets of A^4?

[warm holly]

Yes on A x A

It’s not an exercice just curious if it is the case

[gusty halo]

So if I understand correctly, the left set here should be the set of all equivalence relations on AxA, and the right set is the quotient of the set of partitions on A by some fixed equivalence relation

but then cardinality wise this cannot work

left set has size of number of all partitions on AxA, i.e. |A|^2-th bell number

right size is at most number of partitions on A, i.e. |A|-th bell number

[warm holly]

Right set is juste : R_i={(a,b) in AxA | exists Y in X_i : {a,b}in Y}

So set of all equivalence classes R_i over a partition X_i

So I was thinking about equakity {set of equivalence classes over AxA}={R_i | R_i equivalence class over partition X_i of A}

[gusty halo]

oh now i get it

[warm holly]

Sorry for confusion

But I guess it might hold from my last question which was answered by Moldi

Just wanted to be sure

As we can come up from equivalence class R over AxA to a partition X of A and from partition X to an equivalence class R’ (which is one over partition X) and we have R’=R

[obsidian sleet]

hello friends

i have a question about a definition

so earlier in the text he defined composition series and said that the factors are simple

this definition doesn’t say that? so i can’t assume Gi+1/Gi are simple groups, right?

or does “chain” mean composition series

pls ping me too

[urban acorn]

@obsidian sleet this definition doesn't say that the quotients are simple but

you can without loss of generality assume that they are

it's not difficult to show that every chain can be "extended" to a composition series by adding more intermediate subgroups inbetween until all quotients are simple

[obsidian sleet]

because if not i can just make a new chain

yeah

hm hm

[novel parrot]

nope

dont think they need to be simple

[wooden ember]

yeah no they dont need to be simple

here you just need a normal tower not a composition series

[obsidian sleet]

right

[urban acorn]

although as I said earlier, you can assume that they are simple without loss of generality

[wooden ember]

yeah sure

but it tends to be nicer to be general when you dont see that in the definition

i actually hadnt seen your answer lmao

so pretty useless for me to answer too lol

[acoustic pine]

Can you consider a ring endomorphism as a module homomorphism?

I'd like to make sure the notions coincide, bc I'm looking at the preimage of an ideal being an ideal, and want to prove its finitely generated

[cloud walrus]

Todd

[acoustic pine]

Perhaps a slightly different question may be helpful for me then. Consider $R$ as a module over itself, where $R$ is Noetherian, and consider the quotient $R/I$ as an $R$-module. Noting that the ideals of $R/I$ are submodules, is it true that the preimage of a submodule $N$ that is an $R/I$ ideal is an $R$-ideal?

[cloud walrus]

deathcode

[acoustic pine]

I'm trying to prove quotients of Notherian rings are noetherian

I can't imagine a situation where the above fails offhand, but the only other way it could happen is if the image of a subring of $R$ is an ideal in $R/I$

[cloud walrus]

deathcode

[acoustic pine]

It is I think

but I didn't know if that changes bc $\varphi$ is a module homomorphism

[cloud walrus]

deathcode

[acoustic pine]

We only learned to prove noetherian-ness in the case of modules, so I'm trying to view it as a noetherian module over itself

yes

I thought that was the case, but the prof is kinda omitting like, a lot of details on things

Hell, when introducing modules he just handwaved "the usual axioms", which is fine if you know what a module is, but like, ¯_(ツ)_/¯

He thought the previous course lmao

last term was ring/field theory

then group theory

this is technically our only commutative algebra course, and its half CA and half AG

its also our only AG course

perks of an engineering college I guess

Does this check out?

Sick. Not the hardest proof, but I haven't done a ton of algebra in the past 4 months

[maiden ocean]

I dont think this is incorrect but fyi its much easier to just prove that the image of a Noetherian ring under a homomorphism is noetherian

or to just use the correspondence theorem (any chain of ideals in R/I will have the form I_1 + I subset ... subset I_n + I subset ... and we know that the chain of the I_j must be stationary, hence so will this chain)

[acoustic pine]

Yeah I did it that way initially, and may go back, bc I do use that property during the proof

I also considered just showing the quotient map is order isomorphic on ideals

[maiden ocean]

By order isomorphic do you mean that the bijection of ideals of A containing I with ideals of A/I is order preserving?

[acoustic pine]

Yeah

[maiden ocean]

Are you sure you cant assume that? its a pretty key property for lots of proofs

[acoustic pine]

But that is only if the ideal contains I

[maiden ocean]

Yes

Ideals of A containing I are in order preserving bijection with ideals of A/I

[acoustic pine]

But what about ideals coprime to I

[maiden ocean]

I mean they'll be trivial in A/I because I + J contains 1

[acoustic pine]

This may help clear that up

We had this too

that definition only says ring is noetherian iff R is a noetherian module over itself, not the normal "all ideals are fin. gen."

[maiden ocean]

R-submodules of R are exactly the ideals

[acoustic pine]

Aren't subrings also submodules?

[maiden ocean]

No mine had more detail

No

[acoustic pine]

oh shit

Idk, I thought that there could be some submodules that are not ideals and are subrings

[maiden ocean]

i mean

[acoustic pine]

ah fuck i see

[maiden ocean]

Yea

[acoustic pine]

Time to rewrite this proof, but a lot simpler now lol

[lethal valley]

Hello, what is the difference between integral domain characteristic and its underlying ring order?

[gritty sparrow]

Characteristic of a ring is the number of times you have to add one to itself in a ring to get 0. (If you can never get 0 then the characteristic is 0). The order of a ring is just the size of the ring itself

[obsidian sleet]

hello friends

i am completely dead inside

i do not understand this

https://cdn.discordapp.com/attachments/487651208048410627/894372530201169930/unknown.png

i showed that 1 2 3 were equivalent to each other

and then that 4 of course implies 1 but then to show 1 implies 4 is so weird i don't understand it

i did what the hint said and i think i got to the point where M is abelian (it may be wrong how i did it tho) but

i think that's where i am

but then idk how this induction works

idk whats happening

why did i even prove that M was abelian

pls ping me if you have advice for me ty

[gritty sparrow]

I believe that he now wants you to quotient G by M, then as |G/M|<|G| we can use induction to get a chain of the form given in (iv) in G/M. The preimages of these groups along with M should be the required chain (well that is just what I'm thinking, I haven't checked this properly)

[obsidian sleet]

ohh okay

ty ty

[frank fiber]

if i have a quotient group A/B=C and i know what is B and C, how can i find the possibilities of A?

[gritty sparrow]

without more info, you can basically say nothing. For example let B=Z, C=Z/2Z, we can have A=Z or even A=Z \oplus Z/2Z by embedding B in various ways

[tropic spade]

Does a simple extension require the elt being adjoined to the field not already be an elt of the field?

My textbook has an exercise asking to prove any two simple extensions in R are isomorphic. But I don't see why I can't just pick 1 and i from C to get R(1), R(i) as a counterexample.

[coarse stag]

Can someone give me some intuition as to why this is well defined?

[next obsidian]

If you want a definition that doesn’t have this weird caveat for char 0 it’s the generator of the kernel of Z -> R 😎

It motivates why there’s isn’t a “char ♾ “ for me at least

[thorn flint]

Hi can someone help with 1a?

I’m not sure how to start

[next obsidian]

Just, do the definition of a ring homomorphism

And then see if the results are equal mod 6

You‘ll end up with some expression like
(Sometbinf in a,b,c) =?= (something else in a,b,c)

And then you just can plug in thevalues for c and see if they’re equal

[thorn flint]

And then I’m stuck

[cloud walrus]

Mac

[paper flint]

Mmm, use \

[cloud walrus]

Mac

[vestal snow]

I think my advisor gave me an incomplete problem:

Let C be an elliptic curve and let D be divisor of degree 0 on C. Let O represent the identity point of the elliptic curve. Prove that there exists a unique divisor Q such that Q - O = D (mod Prin(C))

The existence of this seems easy as we can just define Q to be D - O

And the uniqueness seems to be untrue unless we assume something about it (for example, unique up to an element of Prin(C))

From how he said it, this seems to be a classic problem so I was wondering if anyone knew what the correct version of it was

[hot lake]

maybe he wants Q to be supported on exactly one point

so prove there exists a unique point Q such that Q - O = D (mod (Prin(C))

[terse crystal]

You can find square root of -2=4 in Z/6Z right?

[thorn flint]

How

[terse crystal]

?

[next obsidian]

2^2 = 4

Lol

[terse crystal]

2,4

For b use the fact that Z[sqrt(-2)] is an Euclidean domain, so find an element of kerf whose norm |a^2-2b^2| is minimal

[thorn flint]

Oh that’s what you meant

[thorn flint]

Okay I’ll try that

Thanks

[vestal snow]

That might be correct

We can pretty easily conclude from this that $l(D)=l(-D)$

[cloud walrus]

Finitely Many Bananas

[vestal snow]

This is by applying Riemann-Roch

Any ideas on how to proceed would be appreciated

[oak grove]

i know this is probably an easy question

OH

nvm got it

[obsidian sleet]

lagrange

[south patrol]

gg

[obsidian sleet]

momento

[south patrol]

ye

nice how it doesn't need cauchy

[obsidian sleet]

indeeb

[oak grove]

ye

my class threw so many useless corollaries at us this week i forgot the important one rip

[south patrol]

oh that all prime groups are cyclic?

this is a cute q actually cause you have to apply lagrange twice

[oak grove]

no, just a bunch of nonsense stuff about cosets

twice?

[viscid pewter]

isn't lagrange just cosets anyway

[urban acorn]

yeah

I mean it's a very simple coset based argument

[ancient night]

How can I describe this property in the language of abstract algebra?

Here $P: \mathbb{N}_1 \to (\mathbb{N}_0)^{\vert \mathbb{P} \rvert}$

[cloud walrus]

JohnDark

[ancient night]

P - is a canonical map from natural numbers starting with 1 to the infinitely-sized tuple of exponents in the unique factorization

[urban acorn]

and $\mathbb{P}$ is the set of primes?

[cloud walrus]

Ab

[ancient night]

Exactly

$\lvert P \rvert$ is the cardinality. $ \lvert \mathbb{P} \rvert = \aleph_0$

[urban acorn]

and you're looking at it like the fundmental theorem of arithmetic says that N_1 under multiplication is equal to (N0)^|P| ?

[cloud walrus]

JohnDark

[ancient night]

@urban acorn I do

[urban acorn]

that's a very good level of understanding of the fundamental theorem of arithmetic

since you asked for an abstract algebraic way to describe this relationship

N_1 under multiplication and (N_0) ^ aleph_0 under pointwise addition are both objects called monoid

and they are said to be isomorphic as monoids

meaning there's a 1-to-1 correspondence between their elements such that the monoid operation on one side of the correspondence is exactly the same as on the other side

and this formally encapsulates the fact that they have "the same structure"

monoids are not commonly studied in abstract algebra, since there's not really much interesting about the structure of monoids the same way that there is - say - about the structure of groups - which are special cases of monoids that don't include this one

[ancient night]

@urban acorn Thank you a lot! That's a really good answer, which can help me understand it better!

[rustic crown]

i should point out that these are not isomorphic monoids... a natural number can be written as product of finitely many primes. so your "vectors" should only have non-zero entries at finitely many places.

like think about it (N0)^P should have cardinality more than 2^P which is already uncountable

for things like this people use the notation

[cloud walrus]

det

[wooden ember]

huh i never viewed the fundamental theorem of arithmetic algebraically like this

pretty neat

[wooden ember]

so is this $\bigoplus_{p\in\mathbb{P}}\mathbb{N}_0$?

[rustic crown]

yea

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

i see

neat way to write it

[urban acorn]

oh yes, that escaped my eye, indeed you need this caveat on the monoid

i.e. it's the direct sum, not direct product

[warm holly]

In linear algebra course we showed that each subgroup of Z is of the form kZ. To prove the statement we considered H subgroup of Z and claimed that H=nZ with n a minimal element in Z. I’m not sure to understand the main idea of taking the minimal element to show the statement. If someone could clarify it, please, I would really appreciate it’

[urban acorn]

if you look at any difficult number theory problem, you'll notice it involves both multiplication and addition, and the way they interact with each other.
this is because obviously N with just addition is simple enough, and the fundmental theorem of algebra describes N with just multiplication in terms of the former

[urban acorn]

it's a minimal positive element

cause you can keep taking smaller negative elements except with the trivial subgroup

it is in general a useful fact that every non-empty subset A of N contains a minimal element

that is an element x of A such that no element of A is smaller than x

that holds for arbitrary non-empty subsets

[warm holly]

Alright, thank you

[urban acorn]

it's called the well-orderedness of N

and it's actually equivalent to induction

here's a way to think about it:
induction can be justified using well-ordering of N by saying "consider the set of n where our statement fails, then consider the minimal such n"

then the induction hypothesis clearly gives us a contradiction since the statement must hold for n-1, otherwise n wouldn't be the minimal such n

@warm holly exercise: prove the other direction of the equivalence; that is prove the well ordering of N using induction

[warm holly]

Nice, thank you very much for help!

[urban acorn]

Solution: ||The statement is "if A (some subset of N) contains n, then A contains a minimal element." For the base case, when n = 1, then 1 is clearly a minimal element. Now assume our statement holds for any value smaller than n, then any set A containing n either contains a value smaller than A, in which case we apply to induction hypothesis to obtain that A has a minimal element, or not, in which case n is a minimal element.||

[warm holly]

But if we wanted to prove the statement for n<0 we could take a maximal element n in Z ?

But if we take minimal for n>0 then for -n would a maximal in negatif

So I think it might work

[urban acorn]

I'm not sure what you mean by "prove the statement for n < 0", there are not always minimal or maximal elements for subsets of Z

like all of Z has no minimal or maximal element

but we're proving that all subgroups of Z are of the form kZ by using the well-ordering of N

because you just look at the minimal positive element of your subgroup

note that n is in a subgroup iff -n is in that subgroup, so the negative elements of the subgroup are just a reflection of the positives

[warm holly]

Yes I was talking about element in H sorry

[wooden ember]

are you at EPFL

cause i had that exact same lesson today

holy fuck if im correct you're also in my mentor group

Tanish

whether or not you are it's an incredible coincidence

[warm holly]

Yep

[toxic yacht]

Hi, i have this question. The Question Let (G,$\star$) be binary algebraic structure .G=Z and a$\star$b=0 for a,b $\in$ G? Show that it is a group?
I know a group must satisfy the property of associativity, has an identity and an inverse. But, i am not sure how to check if it has an identity and an inverse under the defined operation a*b=0. How can i check if it has an identity and an inverse?<@&286206848099549185>

[cloud walrus]

joseph2531

[sharp sonnet]

i think you are onto something... is there an identity?

[toxic yacht]

@sharp sonnet i think it does not have an identity but, i am not sure

[eternal furnace]

why is it -1?

[wooden ember]

So let me get this straight: set wise G is Z (integers) and it has a binary operation * such that for all a,b in G, a*b = 0?

[toxic yacht]

yes

[wooden ember]

That’s not a group though: for any hypothetical e as identity, e*a = 0 so we can just take a=/= 0 for a contradiction

Or is there something I’m missing

[thorn delta]

what is g? Its acting on vectors and positive integers?

[eternal furnace]

Sn for one of the examples then a few different ones

[thorn delta]

oh wait duh i see, its a group action

[thorn delta]

$g^{-1}(g(i))= i$ for any $i$ which gives $$z_ie_{g(i)} = z_{g^{-1}(g(i))}e_{g(i)} = z_{g^{-1}(j)}e_{j}$$

for some j in {1,2,3}

[cloud walrus]

kxrider

[waxen hollow]

How to show that D4 has only 3 subgroups of order 4?

[void knot]

I didn't remember how to prove, but I think it would help:

,tex $P_1=\langle a\rangle$

$P_2=\left{1,ba,a^2,ba\right}$

$P_3=\left{1,ba^2,a^2,b\right}$

[cloud walrus]

RaD0N

[void knot]

$D(4)=\langle a,b: b^2=a^4=1,, ab=ba^{-1}\rangle$

[cloud walrus]

RaD0N

[void knot]

Correcting: $P_2=\left{1,ab,a^2,ba\right}$.

[cloud walrus]

RaD0N

[barren sierra]

in an ordered non-trivial ring

is 1 > 0

like is that always true?

cause like suppose it wasn't, then 1 < 0. But if 1 < 0 we have that the multiplicative identity flips the sign of an element x and in general x =/= -x

is that logic false?

[golden pasture]

yea

but the logic looks sus

direct proof: ||0<-1 implies 0<1 but oops||

[barren sierra]

wdym

[golden pasture]

you probably meant to say in general x and -x cannot both be positive

[barren sierra]

ah

right right I see

ok cool

is this enough?

oh I also should show order for product is maintained right?

if I show those am I good?

[urban acorn]

amogus

[golden pasture]

and this yea

[rustic crown]

say it was reducible.. and 3 = xy was a non-trivial factorization

what can you say about N(x)?

is there really an element of that norm?

yup

but what really are the elements of norm 3 in Z[sqrt(-5)]?

yea

exactly!

there are no elements in Z[sqrt(-5)] which have norm equaling 2 or 3!

yee

a^2 + 5b^2 = something
is pretty restrictive

if you know that there are no elements of norm 7, then there can't be any non-trivial factors of 7 inside Z[sqrt(-5)]

and all we need to know to find out that there are no elements of order 7 is by checking all small values of a, b in a^2+5b^2 = 7 which is pretty simple

like b = 0, doesn't work. |b| = 1 doesn't work. |b| = 2 is too much.

4+sqrt(-5) is similar as it's norm is 21 and we already verified that there are no elements of norm 3 or norm 7

[plucky flicker]

Hi! Can someone give me some hint about the solution of this problem?

Assuma that Z(A) = 1. Show that every extension of A by X can be embedded into the direct product Aut(A) x X.

Well Z(A) = 1 implies that Z(Aut(A)) = 1. Is this helpful?

[frank fiber]

what is Z(A) ?

[plucky flicker]

the center of the group A

[final pasture]

If Z(A) = 1, then since G/Z(A) ~ Int(G), you have G/1 ~ G ~ Int(G) @plucky flicker

this is probably helpful

an element isn't isomorphic to another , rings (or groups are).
But you can simply check it's a morphism by hand.
Ig you want a ring morphism, so you need to check 1 is sent to 1, that f(a+b) = f(a) + f(b) and that f(ab) = f(a)f(b)

Alright, then first check it's an homomorphism

Yeah

Yeah

I mean the f(a+b) = f(a) + f(b) is "obvious enough", you don't need to compute it to see it, but if it's not that clear for you, do the computations

Like

pick a, b € Z[sqrt(-5)]

So a = x + ysqrt(-5)

and b = p + qsqrt(-5)

now you want to check if f(a + b) = f(a) + f(b)

you know f(a+b) = f(x + ysqrt(-5) + p + qsqrt(-5)) = f((x+p) + (y+q)sqrt(-5)) = (x+p) + 3(y+q) [7]

Now compute f(a) + f(b) and check this is equal to the final expression we got from f(a+b)

Everything works as you expect it to work out

i.e addition is commutative and associative

Indeed

[topaz leaf]

(symmetric groups)
let s be a transposition of the form (i,i+1), and (j,k) be a transposition where j<k.
show that s(j,k)s = (js,ks)

is there a way to do this besides "crunching through cases"?
(please ping me/reply to me, i have this server muted)

[final pasture]

3x3s = 9xs mod 7, right @flint crater ?

But -5 = 9 mod 7 (can you see why ?)

I see

Ok maybe it'd be helpful to just define properly mod thingies before going further

Let's take the simplest defn (probably not the nicest but np)

We say that two numbers k, k' € Z are equal mod n (for n € Z) if n divides k - k'

That basically means that when you do euclidean division of k by n

i.e that you write k = an + r with r < n

and that you do the same for k', i.e k' = a'n + r' with r' < n

you have r = r'

So like, we can check a few things, once we defined things like that

first, we can check that if we have a = b mod n and that we pick c € Z

then we have a + c = b + c mod n

That means that we can define addition mod n

since it won't depend on the "number we choose to represent the others"

to give an explicit example

in our case we had -5 = 2 = 9 mod 7

If you pick some number k

then k - 5 = k + 2 = k + 9 mod 7

[final pasture]

We can also check that the addition is commutative and associative

And we can do the same for multiplication

crap

I forgot I was on LoL

brb lol

I'm level 13

and they are lvl 18

I think my game is fucked up

anyway so hum

We can do the same for multiplication

I.e if a = b mod n, then ac = bc mod n for any c

Alright, then the point is, that makes the set of "integers mod n" a commutative ring

(I didn't define that set, I just defined how elements behaved, but let's work with that informal set where 2 = 9 mod 7 etc for the moment, we can speak about the actual proper construction later if you want to)

So since you have a commutative ring, you can work in it as you'd do it in any commutative ring, everything simplifies as you'd expect it to

So for example, you don't need to write (a+b mod n) + (c + d) mod n as you did earlier, everything is associative, parenthesis aren't necessary

And to check if two elements are equal mod n, you can simply check if their difference is divisible by n

is this clear ?

(wow we're losing so hard, sorry team )

[green locust]

Hey everyone, I had an algebra question I put into a stack exchange post, can I post the link here for feedback or do I need to copy and paste it?

[next obsidian]

I think a copy paste is fine

Err

A link I mean

[green locust]

https://math.stackexchange.com/questions/4268148/in-f-faba-define-f-sim-f-if-and-only-if-f-f-2g-for-some-g-in-f here it is thanks in advance!

[next obsidian]

Yooo

Rushy saved my ass on this commutative algebra thing

Altho admittedly actually I was done I just misread the problem statement

☠️

[green locust]

Haha yeah his hint helped I just wanted to get a second opinion to make sure my solution makes sense

[final pasture]

Let's do it case by case

There are 7 elements mod 7

0, 1, 2, 3, 4, 5, 6

What's an antecedent of 0 ?

[next obsidian]

Antecedent

[final pasture]

man I'm just taking a french word and removing the accents

Isn't this how we speak English ?

[next obsidian]

I think an antecedent is something in grammar

I’ve never heard it in math lmfao

Do you mean like a preimage?

[final pasture]

Solve 0 = a+3b mod 7

yes preimage lol

[next obsidian]

That’s my only guess haha

Okay

Lmfao, such a useful word

Guess it does have a mathematical meaning

[chilly radish]

Antecedent is also a logic thing

[coarse stag]

How does Yoneda’s Lemma imply cayley’s theorem?

[prisma ibex]

View the group G as a category with one object

This answer on MSE explains all the details: https://math.stackexchange.com/a/1708

[coarse stag]

I didn’t understand what S denoted in that explaination.

[prisma ibex]

a set S

which is the image of the single object * of the group G, regarded as a category with one object, under the functor G->Set produced by the Yoneda embedding

[coarse stag]

Sorry, in this case what is G->S?

[coarse stag]

I haven’t had any education in category theory, but I’m assuming set valued functors map objects to sets?

[prisma ibex]

yes, and they map morphisms to functions between sets

so regarding G as a category with a single object *

such a functor maps * to a set S and maps each element of G to a function S->S

in such a way that the composition of these operations is given by the group operation on G

[coarse stag]

Are the contents of S important?

[prisma ibex]

Not really

[coarse stag]

Or is it just some arbitrary unordered collection

[prisma ibex]

Just an arbitrary set where this single object * maps

[green locust]

Anyone else have feedback on this one? 🙂

[coarse stag]

And this functor from G->S is a bijection?

[prisma ibex]

It need not be

For instance you could take S to be a set with one element and map every element of G to the identity function

Still a perfectly valid functor

[coarse stag]

I mean as in isn’t there only one morphism that takes S->S

[prisma ibex]

No, you specify one for every element of G

[coarse stag]

And when you say elements of G, you mean the morphisms?

[primal pier]

What does "truncated" in "truncated (tensor) algebra" refer to?

[prisma ibex]

To be clear with notation if you have a group G (as in a set with a group operation) you form a category called BG with one object whose morphisms correspond to elements of G

Then you’re interested in functors BG->Set

[final pasture]

Well yeah

so we want a and b such that a + 3b = 0

and then we also want another (c, d) such that c + 3d = 1

etc.

Yeah that works

That's what I meant by solving the equation

sorry that wasn't clear

I just meant find a sol

Well finding the kernel is precisely solving 0 = a + 3b

in this case by solving I do mean finding all the sols

[coarse stag]

Sorry, one more part that I don’t follow: it says that we have a functor S:*->S and hom(*,-):*->G

[cursive temple]

i dont understand how we get the "Then one has the relations" part

[thorn delta]

lambda e1 cancels out everything the first row. lambda e2 cancels out everything on the second row, etc

Because lambda ei = eta ei = sum a_ij e_j

[cursive temple]

damn

its that simple

[thorn delta]

Yup!

[cursive temple]

thanks a lot!

[thorn delta]

Np

[green locust]

Anyone willing to check out my stackoverflow question ?

On free abelian group

[thorn delta]

what's the question?

[green locust]

https://math.stackexchange.com/questions/4268148/in-f-faba-define-f-sim-f-if-and-only-if-f-f-2g-for-some-g-in-f

Just looking for a second opinion I believe it’s correct but it could be wrong

[thorn delta]

are you overloading j here?

[green locust]

Oh yeah good point, just edited it to make it more clear

[thorn delta]

other than that looks good to me

[green locust]

Awesome thanks for checking it out !

[thorn delta]

np. btw this equivalence relation is just the same as the subgroup 2F. And u could prove equivalently that F/2F is infinite iff F is infinitely generated

[green locust]

Oh ok that makes sense

[thorn delta]

and essentially what u proved is that is that distinct basis elements of F stay distinct in F/2F

[green locust]

Is this something that will show up again, this is my first brush with free groups

[thorn delta]

hmm idk if i've ever though about this particular fact before.

I'm not sure what Aluffi covers, but there is a classification of finitely generated abelian groups where free abelian groups will show up. Basically, a f.g. abelian group is the direct sum of a "torsion" part which is finite and a "torsion-free" part which is a f.g. free abelian group

so in a sense to understand all f.g. abelian groups, u need to understand free groups

[green locust]

I think that stuff is covered later on, sounds pretty cool

[dire bramble]

okay i have to ask this very stupid question or I'm going to go insane

can I just write a polynomial, prove its monic and irreducible, take a root x, and claim that the polynomial is minimal usually? or do I need to show that it is a divisor of polynomials with root x every time?

[sturdy marsh]

if it's irreducible, then it's the minimal poly

[dire bramble]

oh wait yeah

if its monic + irreducible, then if m is the minimal poly, then m divides p, p irreducible so p associate to m, but p monic. so p = m

[plush quartz]

I'm stuck on what to do after using the hint and proving the mapping is an isomorphism. Clearly y!=g

[rustic crown]

so how would you show that phi(y) = y^n is an isomorphism? what all do we need to check?

[plush quartz]

1. show phi is a homomorphism.
2. Injective + surjective

injective is straightforward (in my eyes) surjectivity we need to check when n is even, but since n is relatively prime to o(G) then we're fine.

[rustic crown]

injectivity will also use that n and o(G) are relatively prime

[plush quartz]

um

[rustic crown]

this all would fail completely when n = o(G), also notice that phi is a map from G --> G so, once you show injectivity or surjectivity other follows easily because both sides have same cardinality

[plush quartz]

suppose we have two values a & b and assume their outputs are the same. Then, taking the nth root of both sides we showed a=b

I am failing to see how we use the prime fact for injectivity.

[rustic crown]

but we can't take n-th roots

[plush quartz]

why not?