groups-rings-fields

[prisma ibex]

If you’re considering rational points on the projective elliptic curve then you include the point at infinity, if you’re considering rational points on the affine elliptic curve you don’t

Usually papers will be a little annoying about conflating the two but it’s usually possible to figure out which one is meant from context

Usually elliptic curve means the projective one, while names like affine elliptic curve or punctured elliptic curve are reserved for the affine one

[vestal snow]

I see

This was really helpful

[prisma ibex]

[vestal snow]

One last question. Is the function field associated to this curve just the fraction field of k[x,y]/y^2-x^3-1?

[terse crystal]

yes

Since D+(z)=spec(k[x/z,y/z]/((y/z)^2-(x/z)^3-1)) and the function field doesn’t change restricted on a non-empty open subset (therefore dense)

[vestal snow]

What's D?

[terse crystal]

different symbols perhaps, principle open subset

[storm turret]

Hi, I don't have colleagues to talk to right now but I want to know if the following proof is sound and I'm not using any wrong theorems for Sylow-groups. \
\
I want to show first that any group of order $15 = 3 \cdot 5$ is cyclic. From the Sylow theorems, there can only exist only one 3-Sylowgroup and one 5-Sylowgroup as 1 is the only number of the form 1+3k or 1+5k that is a divisor of 5 or 3, respectively. Since 3 and 5 are coprime, the Sylowgroups only have the neutral element in common. Now, there are exactly $15-3-5+1 = 8$ elements which are not inside a Sylowgroup. Taking any element from these 8, they must generate a cyclic subgroup whose order must divide 15, so either 1, 3, 5 or 15. Since it's not the neutral element, the order cannot be 1. It can also not be 3 or 5 because any subgroup of order $p^n$ must be a Sylowgroup and we chose these 8 not to be in one. Hence it must generate a cyclic group of order 15, the group itself and the group must be cyclic. $\Box$.
\
\
The above proof can be generalized a bit for groups with order $n = pq, p < q, p \nmid q-1$ since there are $pq - p-q+1 = (p-1)(q-1) \neq 0$ elements that aren't in either Sylowgroup. \
\
Also a group of order $21 = 3 \cdot 7$ doesn't have to be cyclic because it may have 7 3-Sylowgroups ($7 = 1 + 2\cdot 3$) (and you can actually construct it but I'm not interested in that)

[cloud walrus]

T0lgi01

[novel parrot]

If a module is finitely generated

are all submodules finitely generated then also

or is this generally not true

[stone fulcrum]

I don't think that's generally true, let me look for an example

[novel parrot]

so here, why are K and L finitely generated?

[stone fulcrum]

From Roman's advanced linear algebra

[novel parrot]

oh damn ok

ty

[stone fulcrum]

Maybe there's extra assumptions on your thing?

[novel parrot]

heres some context

[chilly ocean]

pid is probably enough for that to be true

[novel parrot]

i see

[novel parrot]

actually, wdym?

[chilly ocean]

for this

[novel parrot]

oh ok

any idea whats going on above?

[dusty river]

is A a noetherian ring?

[novel parrot]

yes

[dusty river]

ok cool

finitely generated modules over noetherian rings are noetherian

and submodules and quotients of noetherian modules are noetherian

[novel parrot]

oh

thanks

why does the fact that x_s annihaltes K and L mean we can remove it from A?

[dusty river]

This is what zef and I were telling you about yesterday

If you have an A-module M, and ann(M) contains I, then M is also an A/I module

where the scalar multiplication is given by [a]*x = a*x

[novel parrot]

lmao

[stone fulcrum]

How did you guys memorize all the module stuff? I find I just can't get it to stick

[dusty river]

or you can verify this as an instance of the first isomorphism theorem using what I said about modules being homomorphisms into end(M)

[dusty river]

You can get some intuition for many of the results. For others, knowing nicer language sometimes helps a lot

[novel parrot]

what does the brackets mean

[dusty river]

equivalence class

[a] = aI

[novel parrot]

oh

[dusty river]

coset of I

[hot lake]

in End(M) you mean endomorphisms as abelian groups ?

[dusty river]

a+I sorry

yes

with multiplication = composition

[hot lake]

yeah

[stone fulcrum]

Nicer language? Sorry if I'm interrupting oof

[novel parrot]

so if xs didnt annihalate then the "restriction" wouldnt be well defined

?

[dusty river]

cat theory but also just knowing some language/results/equivalent ways of phrasing things from other fields can help

[dusty river]

yep

[novel parrot]

got it

actually

does it need to be exactly I

or just need to contain I

[dusty river]

contain

[novel parrot]

alrighty

[dusty river]

thats what I said lol

[novel parrot]

i was just making sure

[hot lake]

R has only 4 elements so you can just brute-force who is in R* and then brute-force that statement

or you can also give an explicit inverse for 1+2r

as a polynomial in r

yeah

or their equivalence classes

[novel parrot]

when multiplying by t^n+ks how do we get that result?

[hot lake]

well in the sum you recognize all the P(M,t), P(L,t), P(M,t)*t^ks, P(L,t)*t^ks, except maybe a finite number of missing terms

[novel parrot]

wdym

[hot lake]

those missing terms you group and call them g(t)

what is the definition of P(M,t)

[hot lake]

you start with $\lambda(K_n) - \lambda(M_n) + \lambda(M_{n+k_s}) - \lambda(L_{n+k_s}) = 0$

[cloud walrus]

Zef Klop 🍃 🌿 🌻

[novel parrot]

yes

[hot lake]

you multiply by t^(n+ks)

$t^{n+k_s}\lambda(K_n) - t^{n+k_s}\lambda(M_n) + t^{n+k_s}\lambda(M_{n+k_s}) - t^{n+k_s}\lambda(L_{n+k_s}) = 0$

[cloud walrus]

Zef Klop 🍃 🌿 🌻

[hot lake]

you sum with respect to n

[novel parrot]

what does this mean

we do it for every n?

[hot lake]

it means $\sum_{n \ge 0}$ of that expression = 0

[cloud walrus]

Zef Klop 🍃 🌿 🌻

[novel parrot]

ok

[hot lake]

$\sum_{n\ge 0} t^{n+k_s}\lambda(K_n) - \sum_{n\ge 0} t^{n+k_s}\lambda(M_n) + \sum_{n\ge 0} t^{n+k_s}\lambda(M_{n+k_s}) - \sum_{n\ge 0} t^{n+k_s}\lambda(L_{n+k_s}) = 0$

[cloud walrus]

Zef Klop 🍃 🌿 🌻

[novel parrot]

ok

[hot lake]

you replace $\sum_{n\ge 0} \lambda(K_n) t^n$ with $P(K,t)$

[cloud walrus]

Zef Klop 🍃 🌿 🌻

[novel parrot]

ok

[hot lake]

$t^{k_s} P(K,t) - \sum_{n\ge 0} t^{n+k_s}\lambda(M_n) + \sum_{n\ge 0} t^{n+k_s}\lambda(M_{n+k_s}) - \sum_{n\ge 0} t^{n+k_s}\lambda(L_{n+k_s}) = 0$

[cloud walrus]

Zef Klop 🍃 🌿 🌻

[hot lake]

same with M

$t^{k_s} P(K,t) - t^{k_s}P(M,t) + \sum_{n\ge 0} t^{n+k_s}\lambda(M_{n+k_s}) - \sum_{n\ge 0} t^{n+k_s}\lambda(L_{n+k_s}) = 0$

[cloud walrus]

Zef Klop 🍃 🌿 🌻

[hot lake]

the two sums left are like a sum for a P(M,t) or P(L,t) except the first k_s terms are missing

you can do a change of variables to get
$t^{k_s} P(K,t) - t^{k_s}P(M,t) + \sum_{n\ge k_s} t^n \lambda(M_n) - \sum_{n\ge k_s} t^n\lambda(L_n) = 0$

[cloud walrus]

Zef Klop 🍃 🌿 🌻

[hot lake]

there are missing terms to recognise P(M,t) and P(L,t) so you add them and remove them

$t^{k_s} P(K,t) - t^{k_s}P(M,t) + \sum_{n\ge 0} t^n \lambda(M_n) - \sum_{0 \le n < k_s} t^n \lambda(M_n) - \sum_{n\ge 0} t^n\lambda(L_n) + \sum_{0 \le n < k_s} t^n \lambda(L_n) = 0$

[cloud walrus]

Zef Klop 🍃 🌿 🌻

[hot lake]

then you recognise P(M,t) and P(L,t) again

$t^{k_s} P(K,t) - t^{k_s}P(M,t) + P(M,t) - P(L,t) - \sum_{0 \le n < k_s} t^n \lambda(M_n) + \sum_{0 \le n < k_s} t^n \lambda(L_n) = 0$

[cloud walrus]

Zef Klop 🍃 🌿 🌻

[hot lake]

all those leftover is just a polynomial that you call g(t)

[novel parrot]

sorry where did the t^ks go

[hot lake]

it's still there ?

[novel parrot]

should there be t^ks there?

[hot lake]

no they were absorbed in the change of variable

I replaced n+ks with n

maybe I should have used a new variable name like m

[novel parrot]

so its summing starting at ks

[hot lake]

you turn $\sum_{n\ge 0} t^{n+k_s}\lambda(M_{n+k_s})$
into $\sum_{m\ge k_s} t^{m}\lambda(M_{m})$

why is this so big

[novel parrot]

ok

[cloud walrus]

Zef Klop 🍃 🌿 🌻

[novel parrot]

how are we applying the inductive hypothesis tho

[hot lake]

you apply it to K and L

they are A[x1...x(s-1)]-modules

[novel parrot]

even for the base case

its a polynomial

but how does it have the desired form

[hot lake]

how is it not ??

if s=0 the product is empty so it is 1

and a polynomial is in the form of a polynomial divided by 1

[novel parrot]

i see

ok

[hot lake]

R[X] is the ring of polynomials in the indeterminate X with coefficients in the ring R

R[X]* should be the ring of invertible polynomials

[simple mulch]

I have a question as well

[hot lake]

the polynomials P in R[X] for which there exist Q in R[X] such that P * Q=1

[novel parrot]

sorry still slightly confused: $P(M,t) = \frac{P(L,t) - t^k_sP(K,t) + g(t)}{1-t^k_s}$

[cloud walrus]

ActiveChapter

[novel parrot]

then we cann apply to P(L,t) and P(K,t)

[hot lake]

yeah something like that

you apply the induction hypothesis to those two

[novel parrot]

got it

[simple mulch]

I am trying to solve (b)

I have this

but I am not sure how to finish the first inclusion

[novel parrot]

that w maybe not be same

[hot lake]

x may not be equal to y

[simple mulch]

ok, so assume w_1,w_2 in W such that x = v_1 + w_1 and y = v_2+w_2

[novel parrot]

yeah that works

[hot lake]

if you want to show that v1+W is included in v2+W you have to show that for every w1 in W there is a w2 in W such that v1+w1 = v2+w2

[simple mulch]

Oh yeah, just as $v_1 + W \subseteq v_2+W$ and $v_2 + W \subseteq v_1+W$

[cloud walrus]

Mns98

[hot lake]

I think your converse is needlessly complicated

[simple mulch]

hum

[hot lake]

if you're closed under multiplication you can multiply things by -1

and then if you're close under addition you can make a + (-1) * b to do a - b

oh wait I was thinking about ideals

it also depends on whether you work with rings with unity or not

but to be a subring you need to contain 1, be closed under addition, taking opposite, and multiplication

but saying closed under substraction is like saying closed under addition and taking opposite

what does the book say exactly ?

it seems mighty suspicious that they don't talk about addition/subtraction so maybe that one is implied by something else

[hot lake]

well additive subgroup implies it's closed under addition/subtraction

so your book is correct

[novel parrot]

in the converse of this proposition, it says it holds for every N module

But in the proof

they take specific N modules to prove it

[hot lake]

they are doing 4') => 4)

they can pick N to be whatever they want at any time

if you do 4) => 4') then you would need to have a proof that forall N, the sequence is exact

[novel parrot]

ive done 4 -> 4'

for all N

but

We are assuming it holds for all N yeah?

[hot lake]

we assume it holds for all N yes

we can deduce from that that it holds for M/Im(u)

[novel parrot]

how do we know that there isnt another N

[hot lake]

[novel parrot]

that contradicts the condition on u and v

[hot lake]

so you are asking what happens if from 4') you can deduce a contradiction ?

[novel parrot]

i mean ...

N = N/Im then v is surjective

[hot lake]

[novel parrot]

:/

the first part of the proof where they show that v is surjective because v bar is injective

[hot lake]

ah I haven't actually read the proof besides "next take N = M/Im(u)"

I don't know how they do that

[novel parrot]

you gotta take N = M/Im(u) to show v is surjective

but

[hot lake]

that's not what it looks like to me from what they wrote

but maybe I haven't thought about that proof much

[novel parrot]

"it follows that v is surjective"

that part

[hot lake]

yeah and then they do something else

[novel parrot]

no

then they show the next part of exactness

but my question is - How do we know that there isnt some other N, that makes v not surjective

[hot lake]

v doesn't depend on N

you start with a sequence M' -u-> M -v-> M'' -> 0

[novel parrot]

but N depends on v

[hot lake]

we show that if (forall N, the sequence (4') is exact) then the original sequence (4) is exact

the collection of A-modules that N can be does not depend on v at all

[novel parrot]

but we only see that 4) is true for specific N?

[hot lake]

4. does not have an N appearing anywhere

[novel parrot]

but we use a specific N to show that exactness

[hot lake]

and that's perfectly fine

do you know how it follows that v is surjective ?

was that explained somewhere earlier ?

[novel parrot]

ive done this theorem before

but always had this issue with understanding it

logically

but it follows because take N= M/im

follows from there i think

[oblique river]

Or maybe M’’/im?

[hot lake]

yeah take f to be the projection from M'' to M''/im(v)

you apply the exactness of the sequence (4') with N = M''/im(v)

[oblique river]

Yes i just didnt want to give away the whole thing :P

[hot lake]

if f is nonzero then f°v is nonzero because °v is injective

but f°v is 0

so f must be 0

and so M''/im(v) = 0, and so im(v) = M''

[novel parrot]

yeah

but i mean

if N was an arbitrary A module

we cant show that v was surjective

[oblique river]

v does not depend on N

You are choosing a specific N

For the express purpose

Of showing that v is surjective

Which you can do because you know (4’) is exact for all N

v is fixed. Given to us by god. It does not change. It is one specific function

[hot lake]

and if v were not surjective then for N = M''/im(v), the sequence (4') would not be exact because v bar would not be injective

so it would be false that (4') is exact for all N

[novel parrot]

we should be able to take any A module and show that v is also surjective then

[oblique river]

No

If you take the A-module 0

You get absolutely nothing

Some A-modules will give you useful info

And some will give you trash

But it doesnt matter because we are assuming that (4’) is exact for ALL A-modules

[hot lake]

forall x, (P(x) => Q) is logically very different from (forall x, P(x)) => Q

you are confusing the two

[oblique river]

So we can pick the A-modules which are convenient to us

That’s a good way to get right to the core of the issue zef

[novel parrot]

i see, we have the second case yeah?

[hot lake]

yes

if I have an integer n and I know that forall prime number p, p divides n

then n = 0

so it's true that (forall x, x divides n) => n=0

but forall x, (x divides n => n=0) is super false

and I can prove that (forall x, x divides n) => n=0

by picking some x that's larger than |n|

then I get n = 0 or |n| >= x > |n|

so n=0

it's the same kind of proof

I picked an x for the express purpose of showing n=0

I don't need to look at what happens for the other x

and I agree it can be feel mind-boggling at first

like, my x depends on n !!

because I want it to defeat n

[novel parrot]

So we know it for every N, the sequence was exact

so then if N was something, whatever we deduce about v and u must be true for allN

yea?

[hot lake]

yeah

[novel parrot]

how do we know there isnt some N

that says that v was not surjective

[hot lake]

because we have irrational faith that ZFC is consistent and you can't prove a contradiction inside it

or if you assume that (4') is exact forall N and derive that v is both surjective and not surjective

then it must be false that (4') is exact forall N

[novel parrot]

ok ..

this is messing with my head

[hot lake]

in fact if you go, pick N = N1 and assume (4') is exact for N1 and show that v is surjective, and then pick N = N2 and assume (4') is exact for N2 and show that v is not surjective, then (4') must have not been exact for one of N1 and N2

one of the assumptions must have been false

[novel parrot]

ok

[next marsh]

hey! im super stuck on this problem:

[novel parrot]

C5 means cyclic with order 5?

take 1 to be identity, all automorphisms act the same on it

and count the rest

do commutative diagram mean

u'f' = fu

etc

the homomophisms to get to one module composed are the same

no matter what "path" u take

like that?

[next marsh]

ahhh gotchaa

[novel parrot]

ring of all units

[vocal wolf]

thats right

[novel parrot]

ok good

[cursive temple]

is the ring finite or infinite

yeah i doubt it

but consider the map from R into R of the form x -> ax for some a in R

[somber marsh]

what does likee

algebraic structure ; algebraic structure mean

[paper flint]

An L-structure where L has no relation symbols

[hidden haven]

Group of units the set of units is not a subring

[hidden haven]

To expand on this slightly, it is a set with some special constants, and functions from some power of it to itself.
Eg groups are (G, *, 1), 1 is the special constant and * is the function from G x G → G (you could remove 1 from this "signature" since it's determined by * anyway)
Rings are (R, +, *, 0, 1)
Example for what he says about relation symbols: ordered sets (S, ≤) are not algebraic

[mild laurel]

It follows from the fact that the map x -> ax is surjective

That map being surjective means that 1 is in its image, so there exists some x \in R such that ax = 1

[next obsidian]

That isn’t a ring, it’s a group

The operation on it will be multiplication, units are not closed under addition

[hidden haven]

[next obsidian]

There’s no need to apologize

I just wanted to clarify that you don’t have a ring structure

Only a group structure

[hot lake]

what's R

[golden pasture]

isnt it given explicitly for you already 🤔

are you having issues with the first or second equality

[hot lake]

yeah you have to show that those two groups are isomorphic

it shouldn't be too hard to come up with a map form R* x Z into the other thing

[golden pasture]

think like what are the elements of R* x Z
how do they multiply

[hot lake]

into {uX^k : u in R* , k in Z}

[golden pasture]

Say you have 2 groups G and H

what is G x H

[hot lake]

yeah, shockingly, 0 is not an invertible polynomial

[golden pasture]

may want to figure what it is first then :)

[dull root]

Let $R$ be a commutative ring and $K$ its field of fraction.. If $f,g \in R[x]$ such that $f$ is primitive and $f \mid_{K[x]} g$, why is it that $f \mid_{R[x]} g$?

[cloud walrus]

Mr.Hahn-Banach

[dull root]

I tried lining up the coefficients, but I'm not seeing how to get that implication

[next obsidian]

This is Gauss’s lemma

I think you wanna show first that the product of primitive polynomial is primitive

And then you do some shit from that

[novel parrot]

wht to do after the hint?

hmmmmmmmmmmmmmmm

i guess all i need to do is show a tensor M is aM

[small bison]

honestly i'd prob just do it directly and try to write down an iso

idk how helpful tensoring would be since you don't know if M is flat or not

[novel parrot]

we dont need M to be flat though?

this does work right?

@small bison

[small bison]

yeah maybe that's enough then

[novel parrot]

[potent briar]

i'm not sure what this means

If, in the set C of all complex numbers, addition is defined as usual and multiplication of a complex number by a real number is defined as usual, then C becomes a real vector space

which are scalars and which the vectors?

[mild laurel]

the scalars are R and the vectors are C

[potent briar]

ok thanks

[prisma thunder]

Bet

[next obsidian]

It is over the 0 ring

[dire bramble]

0[x] moment

[urban acorn]

0[x] is cursed

just like -0

[languid moss]

Can anyone help with 4?

[quaint ivy]

do you understand the hint

[kind temple]

the columns (and/or rows) of every matrix in GL_n have to be linearly independent. you can pick p^n - 1 vectors for the first column (or row) , since u just can’t pick the zero vector. how many choices are there for the second column (or row)?

(i say row in parentheses because u can use either rows or columns. i just prefer columns)

[languid moss]

Is the top rows just 1?

I meant 2

[kind temple]

there are p^2 - 1 suitable choices for the first row in GL_2(Z/pZ)

[languid moss]

so

( 1 0
0 0 )
(0 1
0 0

(0 0
1 0

(0 0
0 1)

[kind temple]

are u doing this for GL_2(Z/2Z)?

[languid moss]

I have to compute GL_2(Z/2Z) first then do GL_n(Z/pZ)

[kind temple]

u have to compute GL_2(Z/pZ) first

[languid moss]

I thought it was the other way around

I'm confused

[kind temple]

lol reread the question

[languid moss]

ahh

ooops

I see

[kind temple]

just refer to this hint

[languid moss]

2 choices for second column ?

[kind temple]

for GL_2(Z/pZ), there are p^2 - 1 choices for the first column.

the second column just can’t be a scalar multiple of the first column.

so there are p^2 - p choices for the second column after u have chosen the first column.

[languid moss]

2 choices as in
(1 0
0 0

(0 0
1 0 ) ?

[kind temple]

where are u getting those from?

[languid moss]

you said two choices. I'm not sure where you are getting that. So I assume you are talking about a 2x2 matrix

and from there assume I mean 2 pivot points

[terse crystal]

In GL_n(Z/pZ) first you have p^n-1 choices of the first row, (any non zero vector in (Z/pZ)^n) next you have p^n-p choices of the second row (there are p vectors being linear dependent with the first row which you shouldn’t choose) next you have p^n-p^2 choices of the third row….go on and on

[quaint ivy]

they mean choices for the columns. So e.g. if you want |GL_2( Z/3Z )| there's the following choices for non-zero vectors: (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,1), (2,2). That's 3^2-1 = 8 possible choices. Suppose we picked (1,2). Then the second column can't be a multiple of (1,2), so any vector other than (0,0), (1,2) or (2,1) (=(2,4) ) works. That's 3^2-3 = 6 possible choices, so |GL_2( Z/3Z )|=8*6=48.

if we go with our choice of (1,2) then e.g. (1,1) is a valid choice for the second column, giving the matrix
(1 1)
(2 1)
(too lazy to LaTeX rn)

a matrix is invertible iff the columns are linearly independent

[next obsidian]

Over a fieldddddddd

[magic root]

Reposted from questions, I'm having some trouble with this question

I just tried to show some facts about an additive homomorphism, but I have no clue how to show that the homomorphisms are limited to just that
Like I was thinking trying to prove by contradiction, something like suppose \varphi \notin \Phi and show its a homomorphism
but like there's nothing much I can show for \varphi except it's not of the form \varphi(x) = nx lol

[terse crystal]

bφ(a/b)=φ(a/1)=aφ(1)

[signal remnant]

Hm... Suppose I have a group G that is a subgroup of another group H and also that G is isomorphic to some other group K. Does this say anything more about how H and K relate?

[gritty sparrow]

It says that there is an injective homomorphism from K to H, of course you may not consider this as “saying more”. Other than this, I don’t think we can say much.

[signal remnant]

you mean surjective from H to K, right?

[gritty sparrow]

No

[signal remnant]

oh wait i see.

ty

[gritty sparrow]

If you want an example of a situation where a surjective homomorphism is impossible, consider Z as a subgroup of Q. There is no surjective homomorphism from Q to Z

[terse crystal]

I think there doesn’t even exist non-zero homomorphism from Q to Z since nf(1/n)=f(1)

[signal remnant]

i'm almost up to homomorphisms...

[novel parrot]

how does that part work?

$(M \bigotimes_A N)_k = M_k \bigotimes_k N_k$

[cloud walrus]

ActiveChapter

Coldilocks ✓

[dusty river]

specifically k is the (A, k)-bimodule

[terse crystal]

Construct morphisms from $(M \bigotimes_A N)_B$ to $M_B \bigotimes_B N_B$ and the other direction respectively then prove that are the inverse to each other: $(m \otimes n) \otimes b$ mapping to $(m \otimes b) \otimes (n \otimes 1)$ and $(m \otimes b) \otimes (n \otimes c)$ mapping to $(m \otimes n) \otimes bc$ respectively

[cloud walrus]

Cogwheels of the mind

[novel parrot]

but like how do we know this?

[dusty river]

use the bimodule isomorphisms given in AM to simplify the right side in the equation I wrote (if you want to follow that solution)

ex 2.15 in AM

[novel parrot]

but isnt $(M \bigotimes_A N)_k = k \bigotimes_A (M \bigotimes_A N)$

[cloud walrus]

ActiveChapter

[dusty river]

yes

[cloud walrus]

Coldilocks ✓

[dusty river]

This will eliminate one of the k's after applying the isomorphism in 2.15

[novel parrot]

tensoring by k is an extension of M,N A-module structure yeah?

[dusty river]

Yeah it makes them into (A,k) bimodules

[novel parrot]

but k doesnt contain .. A ?

[dusty river]

k is an A-algebra

[hot lake]

k = A/m

[novel parrot]

x in m isnt in k tho?

[hot lake]

k is the quotient of A by the maximal ideal m

so uh

no ?

[novel parrot]

hmm

so u

uh

f : A -> A/m and thats how we extend scalars?

[hot lake]

restrict ?

so yes, a k-module automatically becomes a A-module via that projection map

[novel parrot]

yes

but we want the other way

[hidden haven]

Then tensor_A with k

[terse crystal]

I already sent the answer I don’t know whether you want to take a look

[hidden haven]

Of any A-module becomes a k-module by scalar multiplication on the k coordinate

[novel parrot]

i dont know about morphisms

[hidden haven]

That just means homomorphisms

[terse crystal]

Sorry homomorphism

[hidden haven]

In this context

[novel parrot]

i see

i dont think i understood extending scalars tbh

[hot lake]

extending scalars turns A-modules into new k-modules

M -> M tensorA k

and restricting scalars says that a k-module is already a A-module

[novel parrot]

but what happends if we wanna multiply by something in m

is it well defined

[hot lake]

yes

you multiply by f(m)

but f(m) = 0

because f is quotienting by the maximal ideal m

[novel parrot]

so M as an A - module wont be contained in K tensor M

?

[hot lake]

so uh you want to start with a k-module M, then restrict then extend ?

[novel parrot]

no

[hidden haven]

Yes

Extending scalars gives you new relations

[novel parrot]

i see

[terse crystal]

Given an homomorphism of ring A—>B and an A module M then B tensor M becomes a B module by b(b’ tensor x)=(bb’) tensor x

[hot lake]

you have a map from M to k tensor M given by x -> 1 * x

[novel parrot]

got it

[novel parrot]

injective?

[hot lake]

dear gods no

[novel parrot]

[hidden haven]

In this example extending by scalar of k quotients your M by mM

And this map in this case is not injective

(kernel is mM)

[hot lake]

yeah that

[hidden haven]

You might want to explicitly work this out lol

[novel parrot]

ok

oh i think i see it

elements with mx -> mx * 1 = x * m = x * 0

[hidden haven]

Yep

Harder part is to prove the converse

You actually asked some questions about the converse kind of question before

[novel parrot]

hm

[hidden haven]

And once you have this kernel you use first isomorphism theorem to get that M tensor_A A/m is M/mM

Which is mentioned in the problem

[novel parrot]

oh i just used excersise 2

to show that

but now im thinking if i did it correct or no

i just tensored with M and first iso thm

[hidden haven]

Right yeah

[hot lake]

uncrop

[novel parrot]

why do we still have A under tensor

[novel parrot]

ok

[hot lake]

ah yeah

[novel parrot]

we still have multiplcation by A and also B yeah?

[hot lake]

well back then you only that M is an A module

so what else can you take tensor products over

[novel parrot]

wdym?

[hot lake]

what would you have wanted to write instead of A

[novel parrot]

B

[hot lake]

you only know that M is a A-module

[hidden haven]

What does _A represent in the tensor?

[hot lake]

you need M to be a B-module for tensor_B to be possible

[hidden haven]

It doesn't mean that your result is an A module

[novel parrot]

multiplication is defined over A?

[hidden haven]

It means you are staying the universal property of tensor product with A-bilinear maps

[hot lake]

tensorA takes two A-modules and returns a new A-module

[novel parrot]

i see

[hidden haven]

The universal property of tensor products talks about factoring A-bilinear maps for some ring A, so in fact you don't just have 1 tensor product, there's a notion of tensor product over each ring

And the relations that you quotient by in the construction depend on the ring too

[novel parrot]

ok

[hidden haven]

And in the case of the problem, you have k which is a (A,k) bimodule

So you can do k tensor_A k and k tensor_k k

[novel parrot]

ok

[hidden haven]

The latter gives just k, the former gives a non trivial quotient k/mk

[hot lake]

I think it's also k

[hidden haven]

Wait yeah lmao

oof bad example

Since mk is 0

But take R and C

C is an (R,C) bimodule

C tensor_C C is C again

But C tensor_R C is 4 dimensional

Over R

[hot lake]

over R

[novel parrot]

i see

[hidden haven]

nice

[novel parrot]

thanks for the help everyone

[hidden haven]

[novel parrot]

sorry the last part

if its 0 as a vector space, why must it be 0 set

[hidden haven]

What do you mean

The only 0 vector space over any field is {0}

But where is that even being used

[novel parrot]

tensoring with a field makes it isomorphic to a vector space yeah?

or rather is a vector space

so we can multiply the two dimensions

[hidden haven]

Yeah tensoring with a field which is an algebra over the original module makes it into a vector space over that field

And tensor product over k of 2 vector spaces over k gives you their direct product

No sorry

But I mean

Dimensions get multiplied

[novel parrot]

yea

[hidden haven]

They get added in direct product

[novel parrot]

in the question we can just multply the dimensions right? or no?

[hidden haven]

Yes

[novel parrot]

we are not doing direct products here .. yea?

[hidden haven]

Yeah

[novel parrot]

ok good

[hidden haven]

😌

[novel parrot]

XD

so yea one of thos vector spaces had 0 length basis

[hidden haven]

Yes

[novel parrot]

[cyan marten]

Is the structure of k[X, Y]-modules much more complicated than that of k[X]-modules? I mean do we have anything similar to the structure theorem?

[terse crystal]

You have to ask a specific question. Anyway, k[x] is a pid while k[x,y] is not

[cyan marten]

I did! Reread the last part.

By the structure theorem I mean the one for f.g. modules over PIDs

[terse crystal]

I see. Structures of Finitely generated modules over PID and dedekind domain are so beautiful.

[terse crystal]

I still haven’t found anything about structure of finitely generated module over k[x,y].Is there even any theory about structure of finitely generated modules over rings of dimension greater than 1?

[cyan marten]

I wonder..

[novel parrot]

if aM = 0

does a = 0?

M is finitely generated A module

[sturdy marsh]

if you restrict to projective modules, there are theorems on this

lookup the quillen-suslin theorem

[sturdy marsh]

No, for any nonzero a, M = R/(a) is a finitely generated R-module with aM = 0

[terse crystal]

I see, thanks

[novel parrot]

ok

[novel parrot]

for the backwards

if m/d != 0 then there doesnt exist a xm = 0 where x is in D

a contradiction

that works yeah?

[sour plume]

wasn't there a theorem which said something like "if we have two short exact sequences and three maps between the terms, and two of the three maps are isomorphisms, then so is the third one"

what was that called

THE SHORT FIVE LEMMA

there we go

[frank fiber]

is it true that $\langle x_1,...,x_n|x_1^2...x_n^2\rangle * Z/2Z=\langle x_1,...,x_n,x_{n+1}|x_1^2...x_n^2x_{n+1}^2\rangle$?

[cloud walrus]

Or x1

[urban acorn]

<x1, ..., xn | x1^2...xn^2 > means the quotient of the free group on x1 through xn by the normal subgroup generated by x1^2...xn^2, right?

[frank fiber]

yes

[hidden haven]

I can't see why this would be true, don't see any natural inclusions of the first group on the left into the last one. n=1 feels like a counterexample but also feels like it would be hard to prove the it is

[cyan marten]

They have isomorphic abelianizations?

[chilly ocean]

I have a slightly funny definition of support

but i want to want to know if there is relation between

say N nd M are an A-modules

and moreover M is an N-module

can we say something like the support of M bust be in the support of N?

[next obsidian]

I mean

Yeah

Legit just Supp M < Supp N

Wait

M is an N-module??

Is N an A-algebra?

[chilly ocean]

Maybe i should just say precisely the situation i have just to make sure i haven't abstracted away in correctly

[next obsidian]

I mean you can’t have a module over a module

[chilly ocean]

i think so

A is C[x1,..,xn]

i have the inclusion of A into M

and the inclusion of A into N

[next obsidian]

This doesn’t mean anything with regards to N being an algebra

If you want to consider M over N in any fashion you’ll need N to be a ring

[chilly ocean]

and algebra is just when we can also multiply things

oh okay yes

N is ring sorry

yikes

[next obsidian]

In commutative algebra being an R algebra is equivalent to having a ring map R -> S

In the noncommutative case it’s a bit trickier

[chilly ocean]

and yeah everything is commutative

[next obsidian]

Anyway, if S is an R-algebra any S module is canonically an R module

If the ring map is f:R -> S

You let rm = f(r)m

Then f(r) is in S so f(r)m makes sense

Then you could ask questions about the support of M with respect to S or to R

[chilly ocean]

this makes sense i think ive seen this before

[next obsidian]

And since S is also an R-module

You can ask about the support of S with respect to R

So now you can compare Supp_R(M) and Supp_R(S)

[chilly ocean]

and the support the non zero stalks

[next obsidian]

Umm yeah

If you want to think about it that way

I’m surprised you’d like to think about it that way tho haha

[chilly ocean]

sorry the main reason im having trouble is because

my situation is cohomology rings

but with a bigger base ring

and the definition of the support is more concrete

[ancient night]

The set of all total functions $\mathbb{R} \to \mathbb{R}$ with vector addition defined as $f+g= x \mapsto f(x) +g(x)$ and scalar multiplication defined as $kf = x \mapsto kf(x)$ is a vector space. Many of the properties that make this algebraic structure a vector space are inherited from the underlying field of real numbers. Is there a theorem/principle about inheritance of properties of algebraic structures?

[next obsidian]

The principle is just that

Maps from any set to an algebraic structure inherit that usually

[cloud walrus]

JohnDark

[next obsidian]

By just doing everything on the codomain like u noted

You can replace the first R with literally any set and it works

Maybe there’s a universal algebra way to make this precise

But it works for pretty much any structure you can think of

Ring, group, module, vector space, fucking magma

[ancient night]

I realize that but I would like to make a short proof that $\mathbb{R} \to \mathbb{R}$ is a vector space. Proving all 10 axioms is quite tedious.

[cloud walrus]

JohnDark

[ancient night]

I would like to prove at least the majority of them in one or two batches

[gritty sparrow]

Use the trick textbook authors use, the proof is obvious by inspection…

[next obsidian]

Yeah like

Just write down that everything happens in the codomain so any sort of formula you need to hold will hold

That’s all the axioms are just identities

And they hold for things in the codomain so they hold for each x which is all you need

If that’s not good enough idk, either look into universal algebra or prove them all 1 by 1 ¯_(ツ)_/¯

[ancient night]

Thanks!

[gritty sparrow]

Also, maybe a more “theoretical” way to view this is by noticing that Homs preserve limits and most algebraic structures are just some commutative diagrams with products which are limits. So this sort of thing may work in any category with internal hom functors (I haven’t thought through this fully, but it seems legit)

[next obsidian]

This is true

But also this is an incredibly hurbed way to do it

[gritty sparrow]

Fair enough

[edgy glacier]

Struggling with this notation a bit

In Chomsky and Shutzenburger's algebraic theory of context free languages

can anybody help me with what (8) means?

its a formal power series... why are they having relation in a tuple along with the character like so:
<r, f1>f1

what does that mean? Does anybody know?

[gritty sparrow]

<r,f1> is just some integer, so basically this is a formal sum of the strings f_i with some integers as coefficients

Maybe as an example, if maps a to 5 and aa to 10, the power series is 5a+10aa

[edgy glacier]

okay, so they're abusing the fact that you can dovetail two integers here and not specifying how?

oh wait

nvm

i get what you're saying I think

[gritty sparrow]

Nice

[edgy glacier]

<r, f1> is an actual integer produced by the relation on f1

(?)

[gritty sparrow]

Yes exactly

[edgy glacier]

is that notation you've seen before? lmao

jotting that down as the millionth way I've seen notation for a relation 😆

[gritty sparrow]

No lol, it seems very weird. Also, I think r is a function not just any relation. This looks like some alternate notation to r(f1)

[tropic spade]

"For F a field and J an ideal of F[x], J has a unique monic generator", this is false for J=<0> right?

Since <0>={0} and the zero polynomial is not monic.

[gritty sparrow]

If you think of 0 as x^(-infty) it is monic

[tropic spade]

Hmm

Is that something people commonly do?

I know very little about polynomial rings.

[gritty sparrow]

No not at all

[hidden haven]

Sometimes people do define degree of the 0 polynomial to be -infty

To make degree additive

[gritty sparrow]

Yeah that was kinda why I thought it might make some sense

[hidden haven]

But not a standard thing at all

[tropic spade]

Would you even need it to be -infty for that rather than picking a negative integer?

[hidden haven]

Yeah because multiplying by x should not add 1 to deg 0

[tropic spade]

Ooh yeah I can see how that would be awkward

[hidden haven]

Only numbers stable under finite addition would be ± infty

[tropic spade]

Stable meaning?

[gritty sparrow]

a+infty=infty

[tropic spade]

Oh gotcha

[wraith obsidian]

Are split exact sequences preserved by all additive functors?

(specifically: extension of scalars from R-mod to S-mod)

My gut feeling says yes because being a section is preserved by functors

but the exactness may get lost

[terse crystal]

Yes

Since u’u=1, vv’=1 vu =0 u’v’=0 uu’+v’v=1 then apply F on them you still have those equations

[wraith obsidian]

So do you mean the property that they are chain complexes is still preserved and then we use that 0 → A → B → C → 0 chain complex + inner arrows split implies exact?

(inner arrows split means B → C splits, A → B retracts)

[terse crystal]

Yes. (u’u=1, vv’=1 vu =0 u’v’=0 uu’+v’v=1) implies that u is monic, v is epic, im u =ker v

[wraith obsidian]

very sweet, that's more intuitive than I thought

thank you

[robust pollen]

Is my proof of the following correct?
Claim: $F$ a field, $A_1, A_2$ not squares in $F$, abbreviate $\alpha_i = \sqrt{A_i}$ and look at the field $F(\alpha_1, \alpha_2)$. Then $DEG = [F(\alpha_1, \alpha_2):F] = 4$ is $A_1 A_2$ is a square in $F$ and otherwise the degree is 2.

Proof: $F(\alpha_1, \alpha_2) = F(\alpha_1)F(\alpha_2)$ is the composite field of two quadratic extensions over $F$, so $DEG \leq 4$. Since $F \subset F(\alpha_1) \subset F(\alpha_1, \alpha_2)$, we must have $DEG \in {2,4}$.
If $A_1 A_2$ is a square in $F$, then $\alpha_1 \alpha_2 \in F$, so that $F(\alpha_1, \alpha_2)$ is spanned by 3 elements, hence $DEG = 2$.
If $A_1 A_2$ is not a square, then let us assume for contradiction that $[F(\alpha_1, \alpha_2) : F(\alpha_1)] = 1$. Then $\alpha_2 = a + b \alpha_1$ with $b \neq 0$, which after massaging becomes $\sqrt{A_1 A_2} = \alpha_1 \alpha_2 = -\frac{1}{b}(a - A_1 - A_2) \in F$, and hence $A_1 A_2$ is a square in $F$, namely $A_1 A_2 = b^{-2} (a - A_1 - A_2)^2$, contradiction. Hence $[F(\alpha_1, \alpha_2) : F(\alpha_1)] = 2$, and so $DEG = 4$.

[cloud walrus]

expectTheUnexpected

[terse crystal]

the α1α2 I got is (A_2-a^2+b^2A_1)/2b, apart from this I think it’s correct

[robust pollen]

right, the a should be squared, oops. Thanks 🙂

[urban acorn]

What if F = Q, A1 = 2, and A2 = 2/9?

once you add sqrt(2) to Q, A2 also becomes a square, so we only need a degree 2 extension in order to give both of them roots

[terse crystal]

He accidentally reversed the order of 2 and 4 in his statement

[urban acorn]

oh okay

[old zenith]

continuing the discussion over in #advanced-analysis

@hidden haven so what does it mean for two elements to be separable?

[hidden haven]

Their minimal polynomials over the base field don't have repeated roots

You can ignore this condition here - everything is separable over Q

over any char 0 field in fact

[old zenith]

oh ok

i can see why both are separable

that's a nice lemma

[novel parrot]

is the definition of a flat module

a module such that tensoring with an injective map will remain injective

i didnt understand this at all lmao

[fossil shuttle]

if 0 -> A -> B -> C -> 0 is split exact by a section C -> B, then B decomposes as the direct sum of A \oplus C and the map A -> B has a retraction (the projection onto A), which will also be preserved by the functor. That extra information is part of what you need.

it's easier to present this as "additive functors preserve binary direct sums"

[next obsidian]

Yes

The two are equivalent only because tensoring is already right exact

The failure of tensoring to be exact is only in that injectivity might not be preserved

[novel parrot]

i see

[chilly ocean]

tensor product is right exact, so exactness of 0 → A → B → C → 0 implies exactness of A o M → B o M → C o M → 0 for every module M. But if M is flat then A o M → B o M is also injective and we would obtain exactness of 0 → A o M → B o M → C o M → 0

oh sorry it has been already answered

[wraith obsidian]

Thank y'all for all the different viewpoints on thhis.

[steel tusk]

I got a problem with infimum?

Is there an ordered set where all of its subsets with cardinality strictly larger than the cardinality of integer to have the infimum undefined

[kind temple]

probably best to ask in the #advanced-analysis or #foundations channel

[signal remnant]

I'm going a little bonkers on this necessary/sufficient condition for a problem... hopefully someone can help me...

this is where im at:

Provided that my condition that G ⊕ H is cyclic is correct (that is, both necessary and sufficient), I'm having a lot of difficulty understanding how to arrive at that as a necessary logical conclusion.

...To the point where i'm starting to suspect that G ⊕ H is cyclic is not the condition I am seeking at all.

I've filled 3 sketch pad pages in addition to this partial sketch of an answer. Apologies for the terse write-up.

I reposted over in #help-2 for if this is a bad place to inquire. Also, I'm not looking for an answer; more just a little nudge or a bit of cheeky guidance. Thanks.

[gritty sparrow]

Just to clarify, is the question asking for a necessary and sufficient condition for groups G and H to ensure that for all g in G and h in H this condition is satisfied, or is it more like we have fixed G and H and the question is asking for necessary and sufficient conditions on elements g in G and h in H such that the condition is satisfied? My interpretation of the question is the second one, but I think your interpretation is the first.

Also even in the first case, I think there is a counter example. If we let K be the klein 4 group and Z/3Z be the cyclic grp on 3 elements, their direct sum has the property that for any elements g and h the condition is satisfied (ie <(g,h)>=<g>+<h>)

[signal remnant]

Oh uh

[hot lake]

I don't think the "necessary and sufficient condition" can talk about G and H because G and H can be whatever they want when it doesn't involve <g> and <h>

so the question is really about <g> and <h>

and uuh those kinds of exercise are kinda lame

because you never know what kind of condition they want

<g,h> = <g>+<h> if and only if <g,h> = <g>+<h>
proof : trivial qed

okay usually you want an equivalent condition that talks about simpler concepts

[signal remnant]

Let G and H be finite groups and write (g,h) for elements in G ⊕ H. State a necessary and sufficient (both simultaneously) (logical) condition (ie hypothesis) for the conclusion <(g,h)> = <g> ⊕ <h>.

Idk if that cleared anything up but

[gritty sparrow]

Yeah this definitely seems like it should be the second interpretation I was talking about

[signal remnant]

I wouldn't interpret it in either of those ways tbh...

[gritty sparrow]

How do you interpret it?

[signal remnant]

reflexivity of bi-implication...

[signal remnant]

I interpret it like this I suppose? I can attempt to expound upon my perspective more though.

Using sentential logic, "Let G and H be finite groups" (logically) implies "Let G be a finite group" and "Let H be a finite group." "Let G be a finite group" implies "There exists a group G" and, you could say, a little lemma, such as "If there exists a group G, then it is finite" and "There exists a group G", to proof it.

But before I go any further...

...Is this what you're trying to ask me?

And "(g,h) in G ⊕ H" is always true. But all it means is that elements of G ⊕ H can be represented with (g,h).

Then the way I believe is the appropriate method to formulate a (single) necessary and sufficient condition for any sort of logical conclusion would as follows:

First, write H for the hypothesis and C for the conclusion of a conditional statement "H implies C."

Then, "H implies C" has a truth table it is associated with which represents the boolean values of True and False (in a grid) under each of the combinations for H and C thereof.

Now, I posit that a condition H is sufficient for conclusion C if and only if H implies C. Informally, I might say "A condition H is a sufficient condition for conclusion C if and only if for (every variable hypothesis) H => C," even if it looks a bit weird when I write it like that.

Furthermore, I posit that a condition H is necessary for conclusion C if and only if C implies H.

Does that all make sense?

[gritty sparrow]

Yes, I get what necessary and sufficient means, but I feel like you might be being overly formal here

[signal remnant]

it is a disease

[gritty sparrow]

Ok, so what do you say about the actual ambiguity here

[signal remnant]

idk how to do the "necessary" part

[hot lake]

because you chose a condition that is not necessary

[gritty sparrow]

bro what is H and C, that is where the ambiguity actually is

It feels like you tip toed past the question I asked in a way

[signal remnant]

in which part?

H is a group, or it is "the hypothesis", or even it is my guess of "G ⊕ H is cyclic."

[gritty sparrow]

The hypothesis

[hot lake]

I am not seeing any ambiguity o..o

[signal remnant]

Or, for the "necessary" part, it could be seen as the "conclusion" in the dual.

The problem doesn't state a hypothesis...

I'm sorry, are you asking me to define what the English word "hypothesis" means in a sentential-logic sense?

[gritty sparrow]

yeah so looking at the second thing I wrote in this reply, Is this not exactly what you mean @hot lake ?

but jade said that it was not what she meant, so that is what I was asking for, what she means by the question exactly

[hot lake]

it's find a predicate f(G,H,g,h) such that "forall finite groups G and H, forall g in G and h in H, ((<g,h> = <g>+<h>) <=> f(G,H,g,h))"

[signal remnant]

...I'm not sure that is the approach my book is going for.

But if that helps saketh understand, cool.

Oh! wait now I understand, predicate yes

[gritty sparrow]

Is this not what I said in my reply? It is certainly what I meant

[hot lake]

so yeah I guess your second interpretation

but at this point I wanted to make it precise

and your first interpretation was having forall g,h in the LHS of the equivalence ?

[gritty sparrow]

the first interpretation was having a sentence on the rhs that did not depend on g and h

[signal remnant]

this one? 1st interpretation seems right actually sorry

[gritty sparrow]

essentially a sentence with free variables only G and H

[signal remnant]

I mean the sufficient and necessary condition could hypothetically be on g or h I suppose but yeah.

which i believe is what Zef was saying with the predicate formulation

ie my ("base fact") necessary and sufficient conditional could be, like |g| is relatively prime to |h|.

Sure, this implies stuff about G and H, but that requires proof.

[gritty sparrow]

yeah, that is exactly the sort of sentence I think they are looking for in this question

that the orders are relatively prime

[signal remnant]

...so "G ⊕ H is cyclic" is definitely wrong?

err

[gritty sparrow]

yes see my counter example

[signal remnant]

oh the Klein group one?

[gritty sparrow]

I provided groups such that the direct sum is non cyclic, in such a way that for all elements that condition is satisfied

[gritty sparrow]

yeah that one

[signal remnant]

I do not know what the Klein 4-group is really. Z/3Z terminology is fine though.

[signal remnant]

Okay but here...

[gritty sparrow]

oh klein 4 group is nothing more than $Z/2Z \oplus Z/2Z$

[signal remnant]

oh lol

[cloud walrus]

saketh

[signal remnant]

my book must hate its readers

now hold up. you're saying <(g,h)>=<g>+<h> for K ⊕ Z3, but K ⊕ Z3 is not cyclic, correct?

[gritty sparrow]

yes

[signal remnant]

ah and that's why i can't prove the "necessary" part

[gritty sparrow]

yes

[signal remnant]

Okay. I'm happy now. Thank you!

[gritty sparrow]

Np

[cloud walrus]

datorangeguy

[eager willow]

derivations being maps $d : R \to R$ satisfying the product rule $d(fg) = gdf + fdg$

[cloud walrus]

datorangeguy

[eager willow]

I know of constructions such as $(x + y)\frac{\partial}{\partial z} + (x + z)\frac{\partial}{\partial y} + (y + z)\frac{\partial}{\partial x}$ which are defined over $R'$ but map symmetric polynomials to symmetric polynomials

[cloud walrus]

datorangeguy

[eager willow]

and similarly summing permutations from S_n over some polynomial symmetric in n-1 indeterminants, attached to the remaining nth partial derivative, will always work

[stone fulcrum]

R is all polynomials, R' are the symmetric polynomials?

[eager willow]

other way, e_i are elementary symmetric polynomials in R

so R < R'

I have partial derivatives with respect to x_i are derivations defined over R', as are any R'-linear combination, but the right combination may restrict to a derivation R-> R

[dire shuttle]

Super elementary proof, for some reason my brain cannot figure it out. Probably because I've been staring at a textbook for the past like 5 hours lol. Direction is appreciated.

Assume that a group < G, ∗ > has the property that if a ∗ b = c ∗ a, then b = c. Prove that G is
Abelian.

[gritty sparrow]

b(ab)=(ba)b for all a and b by associativity

[dire shuttle]

DUH

Thank you lol. I knew there was some super small thing I was just not thinking off

After staring at my book for so long I think I just need a break lol

[gritty sparrow]

Yeah after a long time studying, it becomes really easy to mess up

[wild sapphire]

Hey guys, why does the subgroup test on arbitrary subgroup elements a,b need the presence of ab^{-1} and not a^{-1}b^{-1}?

I was just wondering why the subgroup test only has one of the elements inversed, and if it matters

[next obsidian]

Try proving that something’s a subgroup using both things

One of them will work, the other will not

I think the issue comes that you won’t know e is in H

[wild sapphire]

yeah, I feel like there is a hole somewhere but I can't see it

[wild sapphire]

that makes sense

hmm

[lethal cipher]

So the first part of 20 is pretty easy. <x> has 35 distinct elements and since G has order 35, x must generate G.

The 2nd part is what intrigues me. Is this still doable if all x satisfy x^33=e instead?

[next obsidian]

Uh

Why does <x> have 35 elements

[hidden haven]

Why does <x> have 35 distinct elements?

[next obsidian]

Sniped

[hidden haven]

[next obsidian]

I think the case with 33 is like 10000000 times easier

[lethal cipher]

Actually sorry, there that is not the order of x. My bad

[next obsidian]

Indeed

Haha

Oh wait

33 isn’t prime

Nvm

[lethal cipher]

I was a bit hasty there

[next obsidian]

Whoopsie

Tfw

[hidden haven]

Chmonkey prime

[next obsidian]

I’m like Grothendieck!

[scarlet estuary]

33 fails like

[next obsidian]

On a lot of counts

Like

[scarlet estuary]

2 of the easiest divisibility tests

except for "is it even"

lmao

[lethal cipher]

Okay, well one thing I do know for sure is that every x has order d where d| 35

[hidden haven]

What about divisibility by 5

How is 11 easier than 5

[scarlet estuary]

5 is a fake number

[lethal cipher]

Can we back up a bit. It seems I still need to figure out the first part

[next obsidian]

Honestly this is such a trash way to do this but

My brain instantly goes to the classification

Abelian is really important here but without just going to the classification idk a good argument for it lol

Wel

Okay I came up with one

Lol

So

How much hint

Do you want?

Or rather I’ll just lay out some facts

What are the possible orders for elements of G?

[lethal cipher]

7,5 and 35

[next obsidian]

If you have some certain combination can you show the group is cyclic?

(And 1)

[lethal cipher]

Oh true, but that's far less interesting. But fair point

[next obsidian]

But it’s very important for the rest of the argument haha

So what happens if you have an element of order 25

35*

[lethal cipher]

Then my pseudo proof from before actually works :p

[next obsidian]

Yup

So let’s assume one doesn’t exist

So now you’re stuck with 7,5,1

Can you get a contradiction if you have some combination of these as well?

[lethal cipher]

What are we contradicting here?

[next obsidian]

No element of order 35

Which is equivalent to just

Being cyclic haha

So we’re gonna show a non cyclic group can’t exist

[lethal cipher]

Oh... That is kind of the point isn't it. :P

[next obsidian]

So anything come to mind?

[lethal cipher]

Hmmm, a bit stuck

[next obsidian]

What if g is order 7

And h is order 5

[lethal cipher]

Then h^5g^7=(hg)^5 g^2=e

But I'm not sure how helpful that is

[next obsidian]

Wel what do you think the order of hg is

There’s only 4 things it could be

[lethal cipher]

Either 5 or 1

[next obsidian]

What’s wrong with 5

It can’t be 5

[lethal cipher]

Ah, cuz g^2 can't be e

[next obsidian]

Or wel

Cuz g^5 can’t be e

Rather

[lethal cipher]

That too

[next obsidian]

What about 7

[lethal cipher]

So then we just have 1

[next obsidian]

It can’t be 1

Then g and h are inverse

And have the same order

[lethal cipher]

Hmm,(hg)^7=e, this can't happen. Because g^7=e, but h^7 does not

[next obsidian]

Yeah

So…

[lethal cipher]

So the only other possibility is 35

[next obsidian]

But!

That’s bad yeah?

So as a side note:

What you showed holds more generally

The fact the only possible orders are 5,7,35 helped but

You can show in general if gh = hg and the order of g and h are coprime

Then the order of gh is the product of order of g and order of h

You cannot weaken any of those conditions

And it isn’t true that the order is always the lcm, you really need coprimality

[lethal cipher]

Oh, that's useful

[next obsidian]

So

That’s something you should prove sometime

But you don’t need

To for this application

Anyway so you can’t have an order 5 and an order 7 element

So you’re either all order 5 (and identity)

Or order 7 (and identity)

Let’s just assume order 7 for now okay

This again isn’t possible, but can you see why?

[lethal cipher]

No, I don't see this one as clearly

[next obsidian]

Okay so

We have 34 elements of order 7

Right?

[lethal cipher]

Yes

e is excluded

[next obsidian]

Now consider the sets

{x,x^2,…,x^6}

Over all x of order 7

Each of these has exactly 6 distinct elements yeah?

And all of them are order 7

[lethal cipher]

Yes

[next obsidian]

These sets partition the 34 elements

Because if two of them intersect

They’re actually equal

[lethal cipher]

True

[next obsidian]

But does 6 divide 34?

😨

[lethal cipher]

No, it most certainly does not

[next obsidian]

So it can’t be that everything is order 7

What about order 5?

[lethal cipher]

And then that has the same situation.
The partitions for order 5 would be {x,x^2,x^3,x^4}, but 4 does not divide 34

[next obsidian]

Ah

Ah

Ah

4

Does not divide 34

Is what you actually want

[lethal cipher]

I knew what I was trying to say :p

[next obsidian]

Okay haha

So…

This group can’t exist

You’re outta options

Unless you have more than one order 1 element

Which is obviously just impossible lol

[lethal cipher]

Very true.
This is a nice case by case proof. Out of curiosity, is there a different way to approach this then cases?

[sturdy marsh]

sylow

[next obsidian]

Classification of finite abelian groups

Is the easiest

It’s braindead with that

[lethal cipher]

Although this was helpful. I seriously need to think about all these nifty partitions

[sturdy marsh]

you dont even need to assume abelian

there is only one group of order 35

[next obsidian]

Anyway the same thing might work for 33

[sturdy marsh]

[next obsidian]

It is true that the abelian group of order 33 is gonna be cyclic

[dusty river]

can't you take some semi direct products of Z5 and Z7?

to get a non abelian order 35 group

[sturdy marsh]

nope

[next obsidian]

But idk if the numbers work out so that this proof works

[sturdy marsh]

there is only one 5-sylow and one 7-sylow

[next obsidian]

Namely

2 divides 32