groups-rings-fields

[somber bramble]

I mean a module over a field, I guess

(but not a given field, obviously there are many groups that couldn’t be vector spaces over ℝ or 𝔽₂)

idk what a group ring is, but I mean essentially a vector space over some field F, where the elements are all elements in G and addition is the group operation of G

oh, G has to be commutative of course

otherwise it obviously won’t work

so that’s definitely a necessary condition

[solar wyvern]

then you can just take formal linear combinations of G over a ring R and you have a vector space if your ring is a field?

[somber bramble]

will that automatically work?

I mean you have to define what the scalar multiplication is

[solar wyvern]

sorry just waking up not sure if there's more to your question

[somber bramble]

in order to be compatible with the addition

and it’s obvious this won’t always work - ℤ/3ℤ cannot be a vector space over 𝔽₂

since the former has 3 elemenst but all finite dim. vector spaces over 𝔽₂ have cardinality 2ⁿ

but in this case there is a field where it is a vector space over it, namely 𝔽₃

but what about e.g. ℤ/2ℤ ⊕ ℤ/4ℤ?

can that possibly be a vector space?

my gut says no, but idk how I would verify that

I mean I guess since it’s got 8 elements, it would have to be a vector space over 𝔽₂ so you could just exhaustively test all possible ways to define the multiplication and see if any of them work out? but that seems like a bad way to go about this

[solar wyvern]

elements are the form $\sum_{g} r_g g$ so $F_2[C_3]$ would be iso to $(F_2)^3$?

[somber bramble]

…well there are not very many ways to do that come to think of it :P

[cloud walrus]

flimflam:

[somber bramble]

what does the notation $F_2[C_3]$ mean?

[cloud walrus]

Sascha Baer:

[thorny slate]

that's the group ring

that was just mentioned

F[G] is F^|G| as a vector space

[somber bramble]

so you just take every element as a separate unit vector or what? but then the addition wouldn’t make any sense, would it?

and also that wouldn’t make G the vector space, F^|G| would have way more elements than G

[thorny slate]

you take the free vector space with basis elements of G

and you define an F-algebra structure by (f1 g1) (f2 g2) = f1 f2 (g1g2) where g1g2 is the group operation

what does making G a vector space mean?

a vector space is an abelian group with an F-action

do you mean a natural F-action on a given abelian group?

[somber bramble]

when I say making G a vector space I mean making a vector space over some field (which we’ll have to find) such that the elements of G are the elements of the space, and the operation of G is the addition in the space

so yes, I guess with that definition, for a given group abelian G, does there exist an F such that there is an F-action on G?

[thorny slate]

okay so you need to start with an abelian group with only has factors a given prime

or only Z as factors

I mean Q

I mean

very few groups work

[somber bramble]

okay

^^

I figured

[thorny slate]

they're the additive groups of the rings F^k for some field F

[somber bramble]

mhm, makes sense

[fringe nexus]

Lagranges theorem implies that a group of finite order p where p is prime is abelian right?

[somber bramble]

idk if lagrange implies it, but it is true

I guess it implies it via like

|G| prime ⇒ no nontrivial subgroups ⇒ cyclic ⇒ abelian

[chilly ocean]

so it does imply it

the 2nd step was lagrange

[fringe nexus]

Let a be an element of order 24 in a group G. Determine all the generators of <a> .

Would the answer to this just be a^n where n is coprime to 24?

because if it has order 24 -> a^24n is the smallest power to get an identity, as long as lcm(n,24) =24n

[random crag]

yes

the mumber of generators of a cyclic group is phi(n) (the number of elements less than n and coprime to it)

this is because the order of an element g^m (g a generator) in a cyclic group of order n is n/(n,m)

the n/(n,m) is n iff (n,m)=1 ie n and m are coprime

[somber bramble]

yea @chilly ocean I know, but it doesn’t like, directly imply it

[chilly ocean]

ah ok

[cloud walrus]

Creaver:

[covert vector]

oof

creaver !!

[thorny slate]

@surreal sonnet

[fringe nexus]

If we're finding unique left cosets of <4> in Z_8, then there are only 4 unique cosets right? 0<4>, 1<4>, 2<4> and 3<4> right?

[chilly ocean]

Yes, assuming you are talking about Z8 as an additive group or a ring
Though they should be denoted 0 + <4>, 1 + <4>, 2 + <4>, 3 + <4> in this case

[fringe nexus]

Suppose that k divides n. How many elements are there of order k in Z_n?

is the answer n/k?

[chilly ocean]

No

Check k = 1

[fringe nexus]

wait is it just 1?

cause let G = {a^0,a^1,a^2 .... a^(n-1)}

then if k | n then you want z^k = e where z = a^x for some x so you get a^xk = e, and since k divides n, n/k exists and is a integer, so then a^xk = a^n -> x = n/k

[copper ferry]

You can't really conclude that.
n = 6

a^(20*3)=a^6 but 20 certainly isn't 6/3

Also the order of z is specifically the least positive integer for which z^k=e

[somber bramble]

consider G = ℤ/12ℤ. Then 4 has order 3, but so does 8
so 1 is not the right answer

because for k=3 there are at least two elements with that order

(note I don’t actually know the answer it’s just a counterexample that came to mind)

[copper ferry]

I think an easier question to figure out is what's the order of a^x in C_n (cyclic group of n elements). You want a^2x,a^3x, ..., a^(k-1)x to all not be the identity and then also for a^k to be the identity

Also there won't always be an element of order k for any k in Z

[somber bramble]

yea but it was specified that k divides n

in which case n/k is an element of order k

[copper ferry]

oh yes, oops

[chilly ocean]

The answer involves the Euler's function
Because element of order n <=> coprime with n

[fringe nexus]

phi(n) ?

[somber bramble]

the quewstion is elements of order k

not order n

[fringe nexus]

o

[chilly ocean]

That's a special case

[somber bramble]

oh like that yea

I see what you mean

n divides n too

[copper ferry]

Your proof was a good way to start thinking about it

Suppose a^x has order k. Then a^xk = e which means that xk divides n. That's all you can conclude

but you can use that to guess what what form x will take

[chilly ocean]

For an element of order k there is a Z_(k) inside Z_n generated by it
So elements of order k come in piles of phi(k)

[copper ferry]

^

That's a simpler proof than where I was going with this

[chilly ocean]

disregard the edits, I'm stupid

[copper ferry]

wait so am I

Yeah, that that is simpler.
Another way to see it is that a^(rn/k) has order k whenever r and k are coprime

and there are phi(k) of those

[daring wolf]

Anyone know anything about block cypher with substitution and permutation?

[kind dust]

@daring wolf
I'm vaguely familiar with Block Cyphers in general, but this link may help you out.

https://crypto.stackexchange.com/questions/8487/the-exact-difference-between-a-permutation-and-a-substitution

[daring wolf]

@kind dust ooooh gonna go read it up

This is the problem i am having

Just don't understand how does it go from an alphabet to a number

sorry

what that was dumb... I meant how does one permute from 16 numbers with an operation of 8 numbers permutation

NEVERMIND SOLVED IT!

thanks!

[daring wolf]

@kind dust by any chance you also know how to decrypt CFB?

[chilly ocean]

in what 3 orders can I write the axioms of a group where 1 is associative, 2 is identity and 3 is inverse.

[desert plover]

I wanna ask some question related to history of quaternion,
What is the reason behind Hamilton coming up with the infamous i^2 = j^2 = k^2 = ijk = -1
History stated that he somehow couldn't came up satisfactory multiplication of triples (i.e. 1 + i + j), but he finally came up with multiplication for quadruple (i.e. 1 + i + j + k)

I wanna know why can't he came up with multiplication for triples, and why does introducing the 4th element solve this problem

cant I just define i*j = -1 or something like that and come up with multiplication for triples?

why does i*j*k = -1 make sense while i*j = -1 does not

[somber bramble]

@desert plover I wrote up sth about where some of the problems come from in #math-discussion. I can’t really show you the rigorous proof though

(cause I don’t know it)

[thorny slate]

ij = -1 implies j = -1/i = i

[somber bramble]

assuming you still want to be able to divide

which, well, you want

[kind dust]

@daring wolf Sorry! I'm late, but I'm glad you were able to solve it. 😃

I might know a bit. In what context are you referring to decrypting CFB though?

For the record, this is a little source that you could use for writing your code,

https://encode-decode.com/aes-256-cfb-encrypt-online/

[daring wolf]

@kind dust I actually ended up understanding it too XD

so I got it~ 😄

[kind dust]

Glad to hear.

[daring wolf]

since you are around :3 was wondering if you are familiar with applied algebra

[idle moth]

If I'm supposed to disprove B being a basis for P_n(x) (polynomials up to degree n), does proving that B_k is not a basis for P_k suffice for any k of my choice?

Basically, does this hold?

<@&286206848099549185>

[kind dust]

@daring wolf I am a bit familiar with Applied Algebra. 😅
Are there any specific questions that you need help answering, or any context?

[stone fulcrum]

@chilly ocean
You can't write inverse before identity as an inverse requires an identity to make sense. Otherwise, the order is free

[blissful yarrow]

Can anyone explain free monoids? I've been told it's the most "general" monoid, but I'm not sure in what way

[thorny slate]

they work just like free groups

if you are familiar with those

any monoid is a quotient of a free monoid

which is copies of N

[stone fulcrum]

@blissful yarrow
Still looking at it?

[blissful yarrow]

So a free monoid imposes no relation between its elements

And a non-free monoid with a relation is a quotient of a free monoid under that relation

[thorny slate]

yeah

so every monoid is a quotient of the free monoid on some generating set

[craggy robin]

I am doing a lot of diagram chasing at school this semester and I realized I suck at it, how do I get better? we have a bunch of exact sequences sprinkled with some epimorphisms and monomorphisms, there's not really any other information being used normally right? is the salamander or snake lemma or whatever worth figuring out so I can reduce to it (like some people on the internet seem to suggest)? what's a good source for info?

[thorny slate]

lol that happens

there's no great source

you just have to chase enough diagrams

so that it becomes second nature

do the five lemma and the snake lemma

bredon (topology and geometry) does them in a very nice way imo

along with a very sober explanation of how arrow chasing works

but general wisdom is you just get used to it

and there's no way of making this easier

[craggy robin]

okay thought so, typical math

practice some more then I guess

I'll check out that book

[final sonnet]

what would be the minimal generating set of a group for a rectangle under rotation?

[thorny slate]

r180

[final sonnet]

thanks bro

wouldn't it be 2 (r180 and identity) because i need I so the group is closed?

nvm

am stupido

[indigo scroll]

Hey u guys

Im trying to proof something and im kinda stuck

[oblique river]

what is it

[indigo scroll]

Prove that for all integers x, x^2+3x is even

[timid cipher]

take the case where x is odd and the case where x is even

[oblique river]

this probably belongs in #elementary-number-theory

[indigo scroll]

Okay ill move

Thank woo

[chilly ocean]

What does a Cayley table for a Klein group look like?

Hi @chilly ocean do you mean the Klein 4group?

If you know the operation and elements it is easy to construct the Cayley table

[chilly ocean]

yeah is it the same as R4 but just replaced with V, H, VH

[stone fulcrum]

@chilly ocean
No, very specifically not that

[fringe nexus]

Is A_3 abelian?

I think A_3 is just S_3 so i dont think it is?

[somber bramble]

A₃ is not just S₃, here are the elements of S₃ with the ones in A₃ bolded:

()
(12)
(13)
(23)
(123) = (12)(23)
(132) = (13)(32)

so A₃ is the (only) group of three elements, which is an abelian group

@fringe nexus

[fringe nexus]

Im so sad

Midterm exam was so hard :/

[somber bramble]

ugh don’t remind me, I have a “mid””term” coming up too

[fringe nexus]

There was a question about G being a group under multiplication , and definin a new operation a*b = ag^(-1)b where g is the identity

[somber bramble]

(it’s actually a “let’s make the first class of the second semester an exam” exam)

[fringe nexus]

G is the identity under *

It asks to find inverse of x under *

I kept getting x^(-1) equals to itself..

[somber bramble]

I’m confused

if g is the identity, then that’s literally just ab

[fringe nexus]

Yes

Like its unsolvable is it not

[somber bramble]

no, I mean it’s literally just the exact same structure?

[fringe nexus]

Is 1/x the right answer

It pretty much is

Cause g is the identity implies g inverse is the identity

[somber bramble]

the inverse of x under this new operation is whatever the inverse is under the regular multiplication, since it’s the same operation

[fringe nexus]

Yea so i kept getting x inverse equal to itself

[somber bramble]

I get the sneaking suspicion you misinterpreted the exercise somehow though

[fringe nexus]

Would 1/x be right

[somber bramble]

I mean, I wouldn’t use that notation, but yes

[fringe nexus]

It was pretty much what everyone got

Also what is the index of a permutation?

[somber bramble]

dunno

[simple valley]

In a finite group $G$, is $\prod_{g \in G} g$ well defined regardless of in what order we multiply the elements?

[cloud walrus]

mniip:

[chilly ocean]

No

[covert vector]

yeah no

counterexample Q_8

[chilly ocean]

If it was independent on the order, it would be identity

and check S_3

[covert vector]

swapping i,j would give the negative product

so it's not necessarily well defined

I suck at computing products in S_n lol

@chilly ocean are you sure it would be identity necessarily?

[chilly ocean]

conjugation with any g permutes elements of the group

[covert vector]

in Z_6 the sum of its elements is 3

mod 6

and it's abelian

[chilly ocean]

oh

yeah

i can prove it lies in the center

[covert vector]

that would make sense haha

[simple valley]

I am asked to prove that if $f \in G$ is the only element with order exactly 2 then $\prod_{g \in G} g = f$

[cloud walrus]

mniip:

[simple valley]

but I'm struggling to make sense of what this says for a nonabelian G

[covert vector]

does it even work for nonabelian G?

[simple valley]

it's an exercise so it has to :p

[covert vector]

wait can you have exactly one element of order 2 in a nonabelian group?

[chilly ocean]

can a nonabelian group have only one element of order 2?

I'm not sure

[covert vector]

jinx :p

wait yes

Q_8

and it's still a counterexample

because if you had such a product with i,j adjacent

even if the product was -1

[simple valley]

what's Q_8

[covert vector]

swapping i,j in that product would give a product of -(-1)=1

[chilly ocean]

quaternion group

[covert vector]

{±1, ±i, ±j, ±k}

with i²=j²=k²=ijk=-1

and (-1)²=1

you get the following cyclic equations
ij=k, jk=i, ki=j
ji=-k, kj=-i, ik=-j

so the product is dependent on the order you multiply in

[simple valley]

hmm

well the proof for abelian G is trivial

so I'll just assume that's what was meant

[covert vector]

Ok

[onyx mirage]

question

Fields typically have 2 operations defined for them, namely an additive one and a multiplicative one

Are there commonly used mathematical constructs where only having these 2 operations and their inverses defined just isn't enough?

[somber bramble]

as in, you need more operations?

in general, one thing that comes to mind is that only having the elements and the operations does not yet give you a topology

so you can’t talk about stuff like continuous functions yet

e.g. on the reals you introduce a concept of ordering (e.g. the ≤ relation)

[onyx mirage]

🤔 aha

[somber bramble]

I’m not quite sure what exactly you asked about so I jsut gave an example of sth where we demand more structure than just that of the operations

the reals are the (unique) complete ordered field

[onyx mirage]

Was wondering about whether the overloadable operators in C++ would be enough to describe higher maths better

[somber bramble]

uh

[sharp sonnet]

if you're not just interested in fields, hilbert spaces have addition, scalar multiplication and an inner product

[oblique river]

@somber bramble you also need to assume archimedian to make the reals unique

[somber bramble]

I meant that version of complete

not cauchy-complete

but supremum exists complete

[oblique river]

ah okay. also don't you need separable? or is that not necessary?

like with things like the long line

(separable as in has a countable dense subset)

oh wait but the long line isn't a field

[somber bramble]

our definition of the reals did not mention the word at all (nor any other part of the analysis course)

[oblique river]

ah okay, I think you can also characterize the reals as the unique [something only involving topological axioms and no field axioms]

[somber bramble]

we characterized the reals as the ordered field (15 axioms) which further fulfilled the following version of the completeness axiom (later proven to be equivalent to existence of the supremum of bounded sets):

For all $X, Y$ nonempty subsets of $\mathbb{R}$, if $\forall x\in X, y \in Y$ the inequality $x \leq y$ holds, then there exists a $c \in \mathbb{R}$ with $\forall x\in X, y \in Y: x \leq c \leq y$

[cloud walrus]

Sascha Baer:

[somber bramble]

or in purely symbolic form

[thorny slate]

15 axioms wtf

[oblique river]

I think you can also characterize R by saying it's totally ordered, has no first/last point, is connected, and has a countable dense subset

[somber bramble]

1-4: addition on the reals is abelian group
5-8: multiplication on the nonzero reals is abelian group
9: distribution
10-13: axioms of ≤ (reflexivity, antisymmetry, transitivity, linearity)
14-15: compatibility of ≤ and the operations

[covert vector]

can you have an ordered field with a maximum

[oblique river]

no

you can prove that 1 > 0

[somber bramble]

nah, try adding it to itself

[oblique river]

and then x + 1 > x

[covert vector]

then why do u need to say "has no first/last point"

[oblique river]

because we're giving different characterizations of R

mine made no reference to R being a field

[covert vector]

o snap

[oblique river]

just R as an ordered space

[covert vector]

that's cool

[thorny slate]

biggest totally ordered archimedean abelian group also works

[somber bramble]

btw, are there complete ordered Fields?

as in, fields but on a proper class

[oblique river]

maybe the surreals

although I'm not sure about completeness

[somber bramble]

they wouldn’t be archimedean

[oblique river]

you didn't specify archimedian

[somber bramble]

again, I meant the complete-in-that-sense complete

and not cauchy-complete

sorry, different terminology

[oblique river]

wait, is that really equivalent to archimedian?

[somber bramble]

from our completeness axiom, you can prove archimedean

from cauchy-complete you can’t

specifically we followed it from the existence of the supremum

[oblique river]

o

[somber bramble]

the proof is short but ugly

[oblique river]

you dont need to give it

[somber bramble]

I don’t feel like typing it out

I only have it in german :p

[oblique river]

good cuz if you did type it out then I would feel compelled to read it but I don't wanna read it :P

[somber bramble]

but the way it’s done is in three steps
(i) every nonempty bounded from above subset of ℤ has a maximum
(ii) for every x∈ℝ there exists exactly one n∈ℤ with n≤x<n+1
(iii) for every ε>0, there exists an n∈ℕ with 1/n < ε

you need supremum for the first

and then you can follow the others in sequence

[chilly ocean]

Rigid Linear transformation R^2 -> R^2 represents a complex number, right?

And additive group of an angle represents multiplicative group of complex numbers

[pearl mural]

🤔

[chilly ocean]

Is there some field related with these?

Why emily

[stone fulcrum]

What's a rigid linear transformation?

[chilly ocean]

Err somewhat made up word for something which preserves the shape

Like distance ratios between points

@stone fulcrum ..?

[stone fulcrum]

What is a "shape?"
What are distance ratios between points?

[chilly ocean]

Uhm, distance ratios are literally ratios of distances

Given many points, this transformation preserve relative distances among them.

There should be a better term, but I forgot.

(Kinda like similarities in geometry)

[stone fulcrum]

Isometry?

[chilly ocean]

Isometry but for ratios

Well simple isometry transformation might fit as well

It covers rotation anyway

Isn't this common and elementary one? @stone fulcrum

[stone fulcrum]

I have a lot of questions

A linear transformation is a function from one vector space to another such that
f(u + v) = f(u) + f(v)
f(ax) = af(x)
This concept doesn't consider distance or ratios. Where are you pulling these concepts from?

[chilly ocean]

Geometry

Thinking about it, It is more about isometry than linear transformation

As old greek geometricians thought, I was thinking of something which doesn't change the 'shape'. Which is about distance ratios - this might be encoded in other words

@stone fulcrum are you there?

You aren't. =/

<@&286206848099549185>
Might be able to help me right?

Or not.

[somber bramble]

@chilly ocean you're thinking of similarity transformations and yes, there is a field which studies this and it's complex analysis. Needham's Visual Complex Analysis looks at functions in precisely this way

specifically the first chapter will interest you, but probably also stuff later on

(analytic functions are locally similarity transforms)

(there aren't that many which are so globally, it's pretty much just translations, rotations and scaling, ie addition and multiplication with complex numbers)

[chilly ocean]

Oh lol

I guess complex analysis is one field which is deeply related

Btw I think one might generalize this for n-th dimension

Anyone now?

<@&286206848099549185>

[simple valley]

orthogonal group

over R

[crimson mango]

Define f: z-> z and g: z-> by

Is f injective? Is it surjective? Explain.

Is g injective? Is it surjective? Explain.

Compute and . Are either of these surjective? Injective? Explain.

f(n)=2n and g(n)= n if n is even or 2n-1 if n is odd

[thorny slate]

nice

[crimson mango]

so i understand that f(n) is injective but not surjective but could anybody help me with g(n)? and the composition of both mappings

[worthy kindle]

you can talk about surjectivity when you know what the output set is

here, there's nothing about it

for example

$\fun\varphi\bbR{[0,+\infty[}x{x^2}$

[cloud walrus]

Tuong:

[worthy kindle]

is surjective

but

$\fun\psi\bbR\bbR x{x^2}$

[cloud walrus]

Tuong:

[worthy kindle]

isn't

[crimson mango]

so phi isnt because not ever codomain has a connection to the domain

[worthy kindle]

no, phi is surjective, but psi isn't

[crimson mango]

i ment psi

[worthy kindle]

okay

ψ isn't surjective because there does exist some y in **R ** that can't be output by ψ

for example -3

[crimson mango]

ya -3 has no preimage?

[worthy kindle]

Yeah

[crimson mango]

so i know that f(n) would be injective but not surgective

[worthy kindle]

you can't talk about surjectivity if the codomain isn't specified

[crimson mango]

f is defined by z to z

Thanks for the help ill be back later have class

[worthy kindle]

You're welcome

[uncut girder]

If you have 2 subgroups F1 and F2 of abelian group G, and T is another subgroup of G,
Can we say (F1 + F2) n T = F1 n T + F2n T

I.e does intersection distribute over set sum

[pearl mural]

is that supposed to read
,tex $(F_1 + F_2) \cap T = (F_1 \cap T) + (F_2 \cap T)$ ?

[cloud walrus]

emily:

[uncut girder]

NVM it's false

New question

If F1 contains no torsion elements and F2 contains no torsion elements does that imply F1+F2 contains no torsion elements?

[thorny slate]

yes

[uncut girder]

Why @thorny slate

[thorny slate]

who knows

I mean

abelian groups are direct sums of Free and Torsion

so F1 and F2 lie in the Free part

so their sum lies in the Free part

[uncut girder]

I'm trying to prove that the free part is unique

So using that here would be cyclic reasoning

[thorny slate]

oh

assume n(f1 + f2) = 0
then n f1 = -n f2 is a common element

[uncut girder]

How does that help

[thorny slate]

dunno

im thinking

[uncut girder]

I already knew that much

[oblique river]

@thorny slate sums of elements of infinite order need not have infinite order

[thorny slate]

in abelian group?

oh

you are right

[oblique river]

yeah, let t be a torsion element and g have infinite order. then g+t also has infinite order, but t = (g+t) + (-g)

[thorny slate]

yeah

wow mind blown

so this is false

[oblique river]

I dont think there is a canonical "free" part of an abelian group. For example, if you take Z + Z/2Z (direct sum) then the torsion part is canonical, but you could also take the "free" part to be generated by (1,1)

and that group also has a direct sum decomposition as <(1,1)> + <(0,1)>

[thorny slate]

yeah that's what this implies

that's weird

[oblique river]

is it? The free part of a group is canonically a quotient

namely G / T where T is the torsion subgroup (which is canonical)

but there's no reason to expect that the map G --> G/T splits in a unique way

[uncut girder]

Interesting

Thanks

[thorny slate]

hmm

[uncut girder]

So it's not unique and I'm going to find a counter example

[thorny slate]

I mean what what non uniqueness means is the non uniqueness of the splitting

surely it's unique up to iso

[uncut girder]

Yes

[thorny slate]

and as a quotient

I hadn't thought of these things

I need to read the structure thm for modules over pid again

[oblique river]

the decomposition is G = T + G/T, but there isn't a canonical way to embed G/T into G as a subgroup

[white hull]

Hi, I wonder if there is any formula for making something expand exponationelly yet add simmular to this: 100+200+300+400+500+100*n

[thorny slate]

too advanced

[elder valley]

Sounds like you want something that grows both expoentially and linearly?

[somber bramble]

which does exist, it’s called the 0 function

(and taht’s it)

[elder valley]

Anyone have any idea what it would mean for an imaginary quadratic number field to be locally maximal at a prime number s?

Ugh sorry, that's not quite right. I'm too confused to even formulate the proper question hah

[cedar gate]

If R is a commutative ring and $R^n\cong (R^m \bigoplus A)$, can we conclude that A is free? (In particular that it is isomorphic to $R^{n-m}$?)

[cloud walrus]

nathan_lowry23:

[uncut girder]

@cedar gate yes

Take the quotient

And use the first isomorphism theorem

[cedar gate]

gotcha, thanks

[thorny slate]

@uncut girder how

quotient of free modules need not be free

example: let R = End V, V countably infinite over F, then R = R + Hom(V,F) but is Hom(V,F) free over R?

@cedar gate

[uncut girder]

R= R+ F?

What's the isomorphism

[thorny slate]

shift the matrix diagonally one entry

and add the missing entry

does that work

I think so

I mean

V = V + F
R = End(V) = End(V + F) = End(V) + Hom(V, F) = R + Hom(V,F)

waaaait

that doesn't look right

what's Hom(V,F)

it's not F

[uncut girder]

Idk

[thorny slate]

looks pretty big

it might be just R again

hmmmmm

oh well the point is you need to say more

quotient of free modules isnt free

[uncut girder]

But these free modules have finite rank

In the original question

[thorny slate]

R has rank 1

over itself

[uncut girder]

I se

[thorny slate]

but anyway consider Z/2Z

quotient of free modules

that isnt free

[uncut girder]

What about R' = R^infty
Then R' is isomorphic to R' (+) R via
(r1, r2, ...) |-> ( (r2, r3, ... ), r1)
And R is not free over R'.

@cedar gate

[cedar gate]

Hmm, the original context was that I have a surjective r-module homomorphsim R^n-> R^m and I want to show that m<=n

So we have the short exact sequence $0\rightarrow \ker\phi\rightarrow R^n\rightarrow R^m \rightarrow 0$

[cloud walrus]

nathan_lowry23:

[cedar gate]

Then since R^m is free, phi splits and we can work out that R^n is isomorphic to R^m + ker(phi)

Maybe in this context it holds?

[oblique river]

@elder valley I might be able to help you

[elder valley]

Why not just take a preimage of each basis element generating R^m

[oblique river]

if you can say a little more about what you were looking for

[elder valley]

I don't want to interrupt their discussion here

[oblique river]

sure thing, we can also try to help them figure it out haha

[thorny slate]

@cedar gate this is false

you can have R^n = R^m for any n,m

for some rings R

in particular End(V)

[oblique river]

is there an assumption on commutativity?

if the ring is commutative then I don't think that can happen

[elder valley]

Yeah commutativity implies invariant basis number property

[thorny slate]

yeah

https://math.stackexchange.com/questions/106786/am-hookrightarrow-an-implies-m-leq-n-for-a-ring-a-neq-0

I guess he did say commutative ring

my bad

[oblique river]

whelp, there's the answer then haha

[cedar gate]

So in conclusion, my approach is flawed and I should approach it differently?

This was a 2 part problem, and I never used the first part, but this was definitely not the intended approach. Thought it might work though

[elder valley]

What was the first part?

[cedar gate]

$R/I\otimes_{R}M\cong M/MI$

[cloud walrus]

nathan_lowry23:

[cedar gate]

Show that this is true

I'm gonna try and figure out a new approach on my own, but if I get stuck I might come back. Thanks for the help though

[elder valley]

Alright

[oblique river]

oh cool

yeah I can think of a clever solution to the second part using that

good luck! (and as a hint: yes, you can use the first part to help you)

@elder valley did you wanna get back to your other question?

[elder valley]

@oblique river there's a massive amount of context I can give but I think it's mostly irrelevant for my question, though I could be wrong. But I'm trying to make sense of: for an order $\mathcal{O}$ of an imaginary quadratic number field $K$, what does it mean for $\mathcal{O}$ to be locally maximal at a prime number $\ell$?

[cloud walrus]

Auvera:

[oblique river]

are you familiar with the p-adics?

[elder valley]

Somewhat yeah

[oblique river]

usually "locally blah" has something to do with that

[elder valley]

Hmm I assumed it was something with localization

[oblique river]

lemme do a quick calculation, one sec

yeah so my guess is that you could check something like, if you take the completion of your order at a prime above \ell, do you get all of the local ring of integers or not

but maybe it's simpler than that

can you give a little more context?

[elder valley]

Okay that sounds more in line with everything else here. $\mathcal{O}$ is the endomorphism ring of an elliptic curve $E$, and $\ell$ is interpreted as the multiplication by $\ell$ map, $[\ell]$, sending $P \mapsto \ell P$

[cloud walrus]

Auvera:

[elder valley]

The paper was also talking about how the ideal $([\ell])$ factors as the product of two prime ideals with very simple representation, so you could be right

[oblique river]

hmm, okay

I don't know as much about that situation

[cloud walrus]

Auvera:

[oblique river]

although without more information I am a little worried because I don't see why it's necessarily true that \ell always factors as a product of two primes in O

like what if your curve has CM by Z[i] and \ell = 3

[elder valley]

Everything is over F_p

Curve is supersingular

[oblique river]

oohhhhh

well then your O is an order in a quaternion algebra

not in an imaginary quadratic field

[elder valley]

Well I guess this theorem I'm reading is stated without the supersingular condition

[oblique river]

and then yeah "locally maximal" means is the order maximal in the local quaternion algebra

oh

[elder valley]

But it's restricted to the F_p rational endomorphisms, which is an order in a quadratic number field. My bad

[oblique river]

oh

so back to quadratic number fields haha

so let's see, we know that frobenius is generating the "non-Z" part

(I think, right? Like frob isn't a square of something else I don't think)

I got my hopes up there for a sec because I know what "locally maximal" means for orders of quaternion algebras

but I've just never seen that applied to orders in number fields because that picture isn't as interesting I don't think

I mean I guess it's the same thing though

[elder valley]

Right, $\pi^2 = -p$

[cloud walrus]

Auvera:

[oblique river]

although if \ell is split then I think it's automatically locally maximal

[elder valley]

What does split mean there?

[oblique river]

\ell is split in O if it's the product of two primes

as opposed to not

so like 3 is nonsplit in Z[i] but 5 is split because of (5) = (2-i)(2+i)

[elder valley]

Okay makes sense

[oblique river]

although I am still slightly worried because random orders in quadratic fields need not have unique factorization of prime ideals

[elder valley]

Yeah there's a restriction on \ell that makes it split into two. So I guess this theorem is just stating it splits under those restrictions and so it's locally maximal

[oblique river]

I see

[elder valley]

I've never seen completion before so I'll have to look at that

[oblique river]

don't trust me super hard on this but I have a feeling this is the right idea

yeah I kinda believe that it has to do with Z_p here

or like, local fields in general

usually when people talk about "local" things, they mean the p-adics (or extensions of p-adics)

(oh yeah, "local field" basically just means "p-adics or an extension of them")

[elder valley]

Okay cool, that helps a ton. I know local fields play a role in elliptic curve theory but I haven't looked at it at all

[oblique river]

I dont know too much about elliptic curves beyond what one might find in silverman for example

I just think about number fields and forget about the rest hahaha

[elder valley]

Haha, I'm working my way through Silverman but it's brutally hard for me

[oblique river]

even though I shouldnt because abelian varieties are like, a big deal (even for number fields!)

🤷

[elder valley]

Thanks for the help though!

[oblique river]

np, good luck

[uncut girder]

Tfw emoji

[somber bramble]

tfw bad font

[spark plank]

[elder valley]

I'm having a derp moment. Need clarification

I thought a quadratic extension K of F_p contained square roots for every element in F_p

I was sure uniqueness of finite fields would ensure it but I'm seeing F_3[i] as a counterexample

2 has no root

Nevermind I'm retarded, 2=-1

[teal ivy]

Can someone help me with this problem?

[teal ivy]

I think there can be 14 ways for ab together
and 14 for ac together

[viscid canyon]

why 14?

if B(4) = 15 is the number of partitions of a 4 element set

[teal ivy]

c,d,e,cd,ce,de,cde each have two possibilities either with ab or outside the ab partition

[viscid canyon]

i'd think you take all the partitions of {a,c,d,e} then just jam b into all the ones that have a in em

i don't follow your argument there

[teal ivy]

how many will have a? 2^3?

[viscid canyon]

how many of what will have a?

[teal ivy]

you said add b which have a

[viscid canyon]

yeah

what i mean is

if you have a partition of {a,c,d,e}, then some cell will contain a, and into that cell you add b

then you get a partition from Wab

so like, if the partition on acde was {ac,de}, then that would become {abc,de}

or if you had {a,ce,d} that would become {ab,ce,d}

[teal ivy]

won't it be equal to 15 as well?
all of the partition of {a,c,d,e} will have an a

[viscid canyon]

every partition of {a,c,d,e} will have a cell that has a in it, yes

[teal ivy]

so the union will be 30?

[viscid canyon]

not quite

you're doublecounting everything that is in Wab intersect Wac

Wab intersect Wac = Wabc

so you need to subtract off |Wabc| from 30

[teal ivy]

ah yeah

thanks

@viscid canyon Wabc will be 52?

[viscid canyon]

?????

that's huge

that's bigger than 30

where'd you get that from

[teal ivy]

won't it be like {abc,de}

so 5 element set

[viscid canyon]

Wabc is the partitions that have a and b and c in the same cell

it will contain things like {abc,de} but it will not, for example, contain things like {a,bc,d,e}

think about how i got the size of Wab

the case of Wabc looks pretty similar

[teal ivy]

ah so I'll just have to put bc with a in (a,d,e)

[viscid canyon]

which means that |Wabc| = ?

[teal ivy]

5

[viscid canyon]

and your final answer is?

[teal ivy]

25

[viscid canyon]

sounds good to me

[teal ivy]

thanks

I really need to understand the concept better

[timber bay]

having trouble figuring out where to go with this one

[viscid canyon]

"a splitting field"

usually when you talk about splitting fields, it should say "E is the splitting field of the polynomial p(x) in F[x]"

[timber bay]

yeah that's kind of what ive seen

we havent gone much over splitting fields but its on the hw great

[stone fulcrum]

I assume it means an irreducible polynomial in F[x] splits in Fbar and has a root in E

[timber bay]

yeah that sounds about right

[stone fulcrum]

It is irreducible, so this is the minimal polynomial for that root in E

[timber bay]

yep

[stone fulcrum]

Hmm. An interesting case is x³ - 1. It has a root in ℤ but also has roots in ℂ

[timber bay]

yeah

thats kind of what i was thinking

something like that

it has one root, but not all of them

[stone fulcrum]

Oh but that's reducible LOL

[timber bay]

oh it is in Z?

oph nvm

lol

[stone fulcrum]

Well, same idea.
x³ - 2
Is irreducible in ℚ, has a root in ℝ, but also has roots in ℂ

[timber bay]

is that not a counter example then?

[stone fulcrum]

I wonder if we're interpreting the question wrong?

[timber bay]

E is a subset of F

of F bar

R is a subset of C?

yeah

R isnt the splitting field of that polynomial

Q(2^(1/3), isqrt(3)) is

would it be fair to say that the degree of the splitting field of a polynomial in Q is the number of linearly independent elements that are roots of the poly?

[timber bay]

i also am having trouble with this one. other people showed containment of the two and the first isomorphism extension theorem but im not having any luck

@dense venture #prealg-and-algebra or one of the math helps are the right place to ask this

this is for abstract algebra, like group theory and fields and stuff

[dense venture]

oh mb

my teacher said this was abstract alegbra

[timber bay]

lol

not really, no

its kind of applicable algebra

[thorny slate]

@timber bay whats the issue

if F c E then certainly cl F c cl E

fo the other direction, since E is algebraic it is contained in the algebraic closure of F

[timber bay]

showing there's an isomorphism

[thorny slate]

oh

if you wanna do it rigorously like that you need to be more careful

[timber bay]

what if E is larger than F? then wouldnt the closure of E be much larger than the closure of F? then it can't be injective?

[thorny slate]

nope

cuz the extension is algebraic

[timber bay]

so the isomorphism is going to be an automorphism?

[thorny slate]

well they're different objects

so it's an isomorphism

fix an algebraix closure cl E of E. Then the inclusion F c E lifts to an inclusion F c cl E and so cl E is an algebraic closure of F

that's how you do one direction more rigorously

[timber bay]

well the containment seems like a better more straight forward route

[thorny slate]

it's hiding stuff

there's no actual containment

the algebraic closure isn't a unique object

it's unique up to iso

[timber bay]

but does showing there's containment prove there's an isomorphism?

oh

this class is killing me

[thorny slate]

anyway for the other direction E is algebraic over F so the inclusion F c cl F factors through E

as F c E c cl F

(these are injections, not actual containment)

etc etc

yeah it's kind of annoying

you have to keep track of elements up to iso

what one generally does

is fix an algebraic closure and work there

for example when you work over extensions of Q you just assume they're all embedded in C

in a specific way

[timber bay]

i dont get what you mean by these being injections, not containment

@thorny slate

[thorny slate]

whever I wrote c

that means an injection

containment doesn't make sense

unless you have an explicit presentation

like E = F(x1,...,xn) and the natural injection 1 -> 1 from F

[timber bay]

wait i dont know how we changed from containment to an injection?

[thorny slate]

there is no containment going on

anywhere

you were tricked

because injections behave just like containment up to isomorphism

which is always natural for field extensions

[timber bay]

then how do we know there's an injection from F to E?

[thorny slate]

because that's what it means for E to be an extension of F

[timber bay]

well i must have missed that

lol

this is news to me

[thorny slate]

I mean

there's several ways to phrase it

[timber bay]

it would make since that i was missing the big picture of it

[thorny slate]

yeah

[timber bay]

well if E is larger than F then F-> E isnt a bijection, right?

[thorny slate]

yes

[timber bay]

which is required for an isomorphism

so cl E must be the same size as cl F for there to be an isomorphism

[thorny slate]

yeah

[timber bay]

i just cant really see that since E>F

[thorny slate]

the algebraic closure of R and of C is the same

[timber bay]

okay yeah

[timber bay]

is degree of the splitting field the number of roots outside the field minus the roots in the field

kind of

like if the poly factors into 3 different polynomials, and one of the polynomials has a root inside the lower field, then the degree of the splitting field is the product of the power of the other irreducible polys?

[thorny slate]

no

given an irreducible polynomial of degree n

the splitting field can be any number divided by n

which divides n!

[chilly ocean]

Sorry for interjecting.
Geometry has its representative an in abstract algebra, right?

[thorny slate]

🤔

[chilly ocean]

❔ ❔

[thorny slate]

what do you mean

[chilly ocean]

Like there's going to be something corresponds to SAS

[thorny slate]

what's SAS

[chilly ocean]

SAS congruence of triangles

I'm quite certain that there should be an analogue of euclidean geometry (or any geometry) in abstract algebra

Like how vector spaces relates to relative positions

And linear transformations which relate to scaling, rotations and such

[timber bay]

https://en.wikipedia.org/wiki/Geometric_algebra

[thorny slate]

um

high school geometry isn't really the geometry mathematicians study

[timber bay]

is there a method of finding the degree of the splitting field

[thorny slate]

yeah

you add roots one by one

[timber bay]

i cant seem to find it online or in my book

[thorny slate]

and see what happens

[timber bay]

that sounds like a method alright

[thorny slate]

like

[timber bay]

what do you mean add them?

[thorny slate]

after you add the first root

start with irreducible f

[timber bay]

like literally?

[thorny slate]

in a field F

adding a root of f means extending to F[x]/f(x)

which is a degree n extension, n = deg(f)

[timber bay]

i still havent learned criteria for determining if a poly is irred

[thorny slate]

it's tricky in general

[timber bay]

i hear there's einstein criteria

[thorny slate]

eisenstein

anyway once you add the first root like that

you can see how f factors in F[x]/f(x)

it will split into multiple factors

maybe they're all linear and you're done

maybe there's more

let me show you an example

[timber bay]

thatd be great

[thorny slate]

take the polynomial x^3 - 3 in Q

this is irreducible, as there aren't cube roots of 3 in Q

so we'll extend to the field Q[x]/f(x) = Q(rt(3))

where we identify rt(3) with the real root of x^3 - 3

now our polynomial is gonna split as

(x - rt(3)) p(x)

where p(x) has degree 2

you can try to find it explicitly if you want

as an exercise

and you can see that this polynomial p(x) is irreducible, it has two complex roots

[timber bay]

proof will be left up to reader

lol

ill try it

[thorny slate]

you just have to divide x^3 - 3 by x - rt(3)

[timber bay]

yeah

lol

i was just typing if that's what youd do

[thorny slate]

you get x^2 + rt(3) x + rt(3)^2

so it's (x - rt(3))( x^2 + rt(3) x + rt(3)^2 )

[timber bay]

so in this case the splitting field would have degree 2 since it'd need cuberoot(3) as well as cuberoot(3)^2?

[thorny slate]

no

we have just added rt(3)

rt(3)^2 is automatically in the field

since it's a square of an element we added

[timber bay]

oh of course

whoops

[thorny slate]

rt(3) was a root of a degree 3 poly

so we have extended by degree 3

[timber bay]

ohhhhhhh

that makes sense

[chilly ocean]

Hmmmmm
Why do you think I'm referring to high school geometry

[thorny slate]

but we aren't done, this isn't the splitting field still

[timber bay]

SAS

[thorny slate]

(x - rt(3))( x^2 + rt(3) x + rt(3)^2 )

the roots of this thing are complex

[chilly ocean]

IIRC SAS congruence is one of the basic properties of Euclidean geometry

[thorny slate]

so they aren't in Q(rt(3))

@chilly ocean as you recall from HIGH SCHOOL

[chilly ocean]

Well

[thorny slate]

so we need to add a root of ( x^2 + rt(3) x + rt(3)^2 ) to get the splitting field

we are extending by a root of a degree 2 polynomial

[chilly ocean]

No, Hilbertian system for geometry

[thorny slate]

so this will be an extension of degree 2

in total, 3*2 = 6

and our splitting field has degree 6

it can also happen that the first extension already is the splitting field

[chilly ocean]

It seems like you despise geometry or something.

[timber bay]

okay i kind of get the idea

[thorny slate]

stop littering this channel with you nonsense @chilly ocean

[timber bay]

lol

[thorny slate]

all sorts of things can happen when taking the splitting field of a polynomial

in fact that's the point of galois theory

galois extensions are splitting fields of polynomials

and we want to study how they work

[chilly ocean]

nonsense as in your opinion.
So you are saying, euclidean geometry is elementary.
Including all the triangle centers and projective properties of it

[thorny slate]

that's high school stuff you dense asshole

geometry doesn't split into "euclidean" and "non euclidean" like that

geometry is about the study of manifolds

dont post in these channels again

[timber bay]

and stay out

that was a joke

im staying neutral

[chilly ocean]

here goes another burbaki

"High school" stuff which is actually much more practical and useful, lol
K I'd not ask about this here

[timber bay]

thats not necessarily true

[thorny slate]

bitch ass 12 year old talking about bourbaki

like you know anything about math

just shut up and leave

[timber bay]

bitch-ass

[thorny slate]

damn right

[timber bay]

damn straight

we dont take no shit here in #groups-rings-fields

[thorny slate]

algebra gang

[timber bay]

lol

would you like to look over my work for the splitting field of (x^2-2)(x^3-2)?

well ill post it, because apparently the answer is 12 and im getting 24

[oblique river]

I can take a look

(for the record the answer is definitely 12)

[timber bay]

is it that obvious?

lol

[oblique river]

yes :P

what did you do?

[timber bay]

so x^2 -2 is irred with x=sqrt(2) having a degree 2 poly. then you look at (x^3-2) = (x^2+2^(1/3)x+2^(2/3) which has roots of something that isn't in the field, but squared they are, and so that adds another 2

so for my roots i got stuff with i, sqrt(2), cuberoot(2), and a number that was a square root of some cube roots and stuff that was in our field

which all the minpolys add up to 24

[thorny slate]

yeah x^3-2 is similar to the example which we worked out above

so that's a degree 6 extension

all that remains to see is that x^2 - 2 is still irreducible

in that extension

[timber bay]

why 6 im getting 12

[thorny slate]

the total degree is 12

[timber bay]

yes

[thorny slate]

6 from x^3-2

and then 2 more from x^2 - 2

but you havet to check that x^2 - 2 actually gives you 2 more

[timber bay]

let me write out my work

[thorny slate]

inside the splitting field of x^3-2

[timber bay]

and take a pic

[thorny slate]

sure

[timber bay]

wait isqrt(2) is just degree 2 aspect right?

ohhh

that's my mistake

[oblique river]

so it looks like youve written out 2*3*2 which is 12

[timber bay]

They're deg 2 because they're together

[oblique river]

also just to emphasize jacobian's point

from the get-go you know the answer has to be 6 or 12

the splitting field for x^3-2 has degree 6

[timber bay]

well the two arrows for the isqrt(3) is supposed to imply degree 4

[oblique river]

isqrt(3) = sqrt(-3)

[timber bay]

well 24 is a multiple of 6

[oblique river]

but it's not 12 lol

[timber bay]

wait why 6 and 12?

[thorny slate]

the max degree of splitting field is n!

2!3! = 12

[oblique river]

the splitting field for x^3 - 2 has degree 6: cbrt(2) is degree 3, and then you need a cube root of unity to get the other two

and the cube root of unity has degree 2

but now that you have x^3 - 2, all you need to do is get sqrt(2)

and there are two options: either you already have sqrt(2) in which case you're done and it's degree 6

[thorny slate]

also I already said this a couple times, but you have not proven the extension is of degree 12 yet. you have shown that (x^3-2) has degree 6 over Q and that (x^2 - 2) has degree 2 over Q

[oblique river]

or you need to add sqrt(2) which is a degree-2 extension so you get to 12

[timber bay]

yeah i get that

[oblique river]

also, just to go back to jacobian's n! thing

here is the worst case scenario for a poly of degree n:

add the first root and it has degree n. Then in that field, your first poly splits as (x-a)g(x) where a was the first root

so the next root of g(x) has degree (n-1). then (again worst case scenario) in that new field the original poly factors as (x-a)(x-b)h(x) so the next root has degree (n-2)

all the way down

so you know without doing any work that the splitting field of x^3 - 2 has degree bounded by 6

and you know it's at least 3 because of cbrt(2)

so it's either 3 or 6, and you've figured out that it's in fact 6

[timber bay]

okay that makes sense

[thorny slate]

finalizing the proof can be a bit tricky, what I like to do is do all the real extensions first

so let's go back to (x^3-2)(x^2 - 2)
we add a degree 3 extension cbrt(2) and we get the splitting
(x-cbrt(2)) g(x) (x^2 - 2)

x^2 - 2 can't possibly split, because it would give a degree 2 extension inside our degree 3 extension

so we can now add sqrt(2)

and we get
(x - cbrt(2)) g(x) (x - sqrt(2)) (x + sqrt(2))

we know that g(x) doesn't factor still, because its roots were complex

so we still have a degree 2 extension to do

and then that's 3, 2, 2

in general cases it can get a bit tricky

you can use things like gauss's lemma

for irreducibility

[timber bay]

we didnt have to prove it just provide the value

[thorny slate]

but determining whether a polynomial stays irreducible is hard

yeah but you don't know whether it's 6 or 12

[timber bay]

oh I see what youre saying

[teal ivy]

So if A intersection C = B union C then it means that A = B = C, right?

[worthy geode]

Normally, it is.

[teal ivy]

and if A union C is subset of B union C then is A subset of B even if the question doesn't says anything about A intersection C?

[teal ivy]

btw I think it is also true for A being subset of C or A = C then also A union C becomes subset of B union C so A doesn't have to be a subset of B

is this interpretation correct?

[timid cipher]

if C is a subset of A and B is empty, i don't think the first implication holds

for the second, consider A subset of C and B and C disjoint

[teal ivy]

thanks!

[somber bramble]

Can someone tell me if this statement is true (and if it's not provide a counterexample)?

Let $G$ be a group and $H$ a normal subgroup. Then $G \cong H \oplus G/H$

[cloud walrus]

Sascha Baer:

[somber bramble]

(this is not homework, I'm just failing to find the right words to google and can't find it on wikipedia)

[chilly ocean]

are groups abelian?

oh wait

why would you use $\oplus$ for non-abelian groups?

[cloud walrus]

Xaositect:

[somber bramble]

direct sum, as in the group with tuples (h, gH)

that's the notation we always used

afaik

[chilly ocean]

Anyway, $A_3 \subset S_3$ is a counterexample in the nonabelian case

[cloud walrus]

Xaositect:

[somber bramble]

maybe we used sth different with nonabelian ones and I just never noticed

what's the usual notation?

[chilly ocean]

product, $G_1 \times G_2$

[cloud walrus]

Xaositect:

[somber bramble]

ah, fair

yea I've seen that, never even noticed that there were two notations for the same concept

so $S_3 \ncong A_3 \times \mathbb{Z}_2$?

[cloud walrus]

Sascha Baer:

[chilly ocean]

yes

[somber bramble]

What if both H and G/H are abelian?

[chilly ocean]

because this product is abelian

this is the case here

[somber bramble]

oh.
well darn that also is a counterexample to the midterm question I tried to justify this way then (it was multiple choice)

namely "if H and G/H are abelian, then so is G"

in hindsight, Dihedral groups would've been a good counterexample to that too

...I guess S3 is actually the dihedral group of the triangle so yea

oh, also this question on the midterm I had to guess, would be interested to knlw what it is

Let $A,B$ be comm rings and $f: A \to B$ a ring hom. If $f$ is surjective, then, then the preimage of a maximal ideal is a maximal ideal.

[cloud walrus]

Sascha Baer:

[chilly ocean]

this looks true

yeah

B is isomorhic to a factor A/ker f
f gives a one-to-one correspondence between ideals of B and ideals of A containing ker f

[somber bramble]

dammit, I second guessed myself

[fringe nexus]

Find all subgroups of $\mathbb{Z}_2 \cross \mathbb{Z}_2 \cross \mathbb{Z}_4 $ that are isomorphic to the Klein 4 group

[cloud walrus]

Victoria:

[fringe nexus]

do i just list everything out and brute force it? Or am i missing something

[somber bramble]

so the klein four group is Z₂×Z₂, right? so in particular, all its elements (apart from the identity) have order 2

so you can take all elements of order 2 in your given group

and then see which ones you can pair up

(by which I mean you take two, and multiply them together, and if that gives you a third element which is also of order 2 that’s your candidate)

[viscid canyon]

this is a cute little problem actually

[uncut girder]

Bc u like combinatorics

[gleaming onyx]

pty still in here lul

[uncut girder]

Hm?

[neon tapir]

Hi, I have a little question about rings if someone is good in algebra

if A is a ring

Do we have A+A = A

Cause i want to prove

A=Z[i]

(2-i) = A

[covert vector]

A+A as in the direct sum?

[neon tapir]

(A,+, .) is the ring

it's

the operation

[covert vector]

then yes A+A=A

[neon tapir]

Oh

[covert vector]

you are required for the ring to be closed under their operations

which means that for any a,b in A

[neon tapir]

So (2-i)A =A + A -A = A

[covert vector]

a+b is in A as well

no it doesn't work like that

[neon tapir]

Ah

Ah

I know what you said

[covert vector]

A+A = {a+b | a,b \in A}

[neon tapir]

Ah

it's true

[covert vector]

then you take a=0

and b to range over all A

to get that A+A = A

[neon tapir]

Ok

But

if A = Z[i]

(2-i)A

= 2A - iA

[covert vector]

you can't just distribute (2-i)*A though

[neon tapir]

Ah

[covert vector]

no it doesn't work like that

you're mixing up operations on sets

to operations on their elements