groups-rings-fields

[coral shale]

and 'if the image is a field, then it is surjective'

[oblique leaf]

i think this is true because the image contains a spanning set of K1K2, and since fields are closed under everything, this spanning set is gonna give us all of K1K2 in the image no?

[sharp sonnet]

the image of this homomorphism is a field, since the domain is assumed a field

but the composite is the smallest field containing K_1 and K_2

and the image certainly does via image of (1, k) and (k, 1)

[coral shale]

ohhhhh

[oblique leaf]

ok yeah, i think that makes sense

[coral shale]

nothing much about tensors after all

[oblique leaf]

yeah seems like it doesn't matter

[sharp sonnet]

well, you need that the tensorproduct is a field

[oblique leaf]

thanks @coral shale @sharp sonnet

[sharp sonnet]

you can also construct preimage of an element explicitly

[coral shale]

x + y 👀

[sharp sonnet]

K_1K_2 is K_1(K_2)

so the elements are like quotients of polynomials in K_1[x_1, ..., x_n] evaluated at elements of K_2

[oblique leaf]

wait last question, this statement is true only if K1 and K2 are finite right?

[sharp sonnet]

no, why?

[oblique leaf]

I am talking about this statement

the statement in problem 5

[coral shale]

finite extensions dont necessarily indicate finite fields being involved
the degree of the extension

[sharp sonnet]

finite extensions, not finite fields

if you had finite fields, you would get surjective just for set theoretic reasons

[oblique leaf]

yeah thats what i meant, K1 and K2 are finite extensions

so for infinite extensions, it wouldn't work right?

like the statement wouldn't be true?

[coral shale]

in the infinite case we would want to claim cardinalities are equal iff ... I presume?

[sharp sonnet]

i imagine it becomes more subtle

just search "when is the tensor product of fields a field"

there is e.g. https://mathoverflow.net/questions/82083/when-is-the-tensor-product-of-two-fields-a-field

[coral shale]

im following, btw

[oblique leaf]

it is just that i don't know how to handle the index, because it seems to make little sense to me to be multiplying infinity dimensions

[coral shale]

is there an easy explicit example?

Let F = Q, K1 = Q(rt2), K2 = Q(rt3) works I think?

degree 4 vs deg 2, deg 2

[oblique leaf]

Q(rt3) has deg 3 no?

[coral shale]

sqrt3

[oblique leaf]

also i want to know if this is true for infinite dimensions

[coral shale]

x^2 - 3 is min poly

[oblique leaf]

oh root 3

not cube root 2 sorry

[coral shale]

so how can we represent rt2 + rt3

as a product was my q

hmmm

Well everything in the compositum is of the form a + brt2 + crt3 + drt6 = a + bx + cy + dxy

Oh wait not quite - quotients in this

[sharp sonnet]

i think this works? no need to take quotients

[coral shale]

yes it turns out this works

since we have finite extension

I'm just thinking about this apriori

[coral shale]

It's still not obvious to me how we can factor this

more think

[sharp sonnet]

represent where?

[coral shale]

something from Q(rt2) times something from Q(rt3)

[sharp sonnet]

it should live in the tensor product right

rt2 \tensor 1 + 1 \tensor rt3

[coral shale]

But don't we have this

Supposing the map is surjective

[coral shale]

we should be able to explicitly write anything in Q(rt2, rt3) as a product like this?

(k1, k2) -> k1k2 is surjective

[sharp sonnet]

we should be able to write it as an element in the tensorproduct

[coral shale]

ohhhhhhhhh

The original map isn't necessarily surjective

But the induced map is

[sharp sonnet]

ye, k_1 \times k_2 certainly isnt a field

[oblique leaf]

what do you mean shuri?

[coral shale]

Right right I get it

[sharp sonnet]

it has zerodivisors

[coral shale]

I kept thinking this was surjective

with the original domain

But no, its the induced map with the tensor product as a domain that is

[oblique leaf]

oh ok yeah

yeah

[coral shale]

That makes much more sense

https://images-ext-2.discordapp.net/external/NCEu9iMUJ9jxb8BMC9Vj-ERDRwC2O6k7uqc1hCTSJj8/https/i.imgur.com/PX56SkB.png

[oblique leaf]

This argument though, does this even depend on the extensions being finite at all?

thats the thing i want to know

because to me it seems that nowhere in this proof do the use the fact that K1 and K2 are finite extensions

[coral shale]

Perhaps those equalities, implicitly?

[oblique leaf]

you mean dimension equalities?

[coral shale]

yes

[oblique leaf]

but it does make sense to say \infty = infty*infty though

[coral shale]

If it is known/given/proven before these equalities hold for cardinalities in the infinite case then sure (but idk)

I also don't know if they hold in the infinite case for sure, but I suspect they would.

[oblique leaf]

what do you mean? it doesn't seem that there's anything in here that depends on cardinalities

[coral shale]

when we write = for things that are infinite, we are referring to equality of cardinalities, aren't we?

Otherwise it doesn't make sense

[oblique leaf]

but the dimensions are more like cardinality of bases?

i dont know

[coral shale]

it would be imo

[oblique leaf]

do we distinguish between a dimension being aleph null or aleph 1

even

[coral shale]

yes we probably should/would.

[oblique leaf]

but either way, the proof would work no?

[coral shale]

not that I really have studied this

tbh I'm not confident one direction of the statement holds for the infinite case (just a feeling)

[oblique leaf]

yeah ...

but i also don't know what could be wrong

[coral shale]

I think the reverse implication might come across problems

[oblique leaf]

you mean assumptions with the basis and all?

[coral shale]

If we just pick a random example, I don't think it will work out

[oblique leaf]

but infinite dimensional vector spaces have bases too

you think we can find counterexamples? like in the statement isn't true?

[coral shale]

for the converse

Just intuition - might be wrong

idk - let F be Q and come up with 2 non-algebraic fields over Q whose tensor product isn't a field

I just felt this should be possible

Ok how about just Q(x), Q(y)

What will their tensor product look like

If it is Q(x, y) then nvm for this

[oblique leaf]

i have no idea lol

[sharp sonnet]

does this basis thing work?

products of basis elements being a basis of the composite

i dont really want to think about it lmao

[coral shale]

$\frac1{x+y}$

[coral shale]

I have a feeling we fail here
(1/x)+(1/y) = (x+y)/(xy), looks like a deadend

https://i.imgur.com/TmK6QgU.png

Does this mean we don't even know if a basis is formed

If we don't even know if e.pi is rational, then I don't think we can do the 'same thing' as the finite extension case

That induced map from the proof won't necessarily be surjective (let F = Q, K1 = Q(pi), let K2 = Q(e))

'it is clear phi is surjective' must use the fact we have finite extensions

[sharp sonnet]

choose like Q(e) and Q(1/e), then their composite is still Q(e), the dimension statements work (unlike in the case e = sqrt(2) for example)

how about this?

[coral shale]

I agree the statement might still be true (although I'm not sure) --- but was pointing out the proof method probably doesn't extend.

[next obsidian]

I think that equality holds as cardinals

[coral shale]

https://i.imgur.com/rAlXaTC.png

What about the reverse implication?

I feel that doesn't necessarily hold (just a hunch)

[zealous flame]

i've gotten rather far with this question but just wanted some external feedback to make sure i am not going insane. So what i've gathered is I am almost certain Z3 is not a subgroup of Z9 as there exists numbers that are divisible completely by 3 but not by 9 such as 6. But using lagrange's theorem I was able to show that a subgroup order 3 is possible and Z3 has order 3 because it's just (0,1,2) .

[coral shale]

I am almost certain Z3 is not a subgroup of Z9 as there exists numbers that are divisible completely by 3 but not by 9 such as 6

Can you elaborate

[zealous flame]

well I construced a group table for mod 3 and for mod 9 and found no overlap between the 2 other than the first horizontal and verticle lines

[coral shale]

I think you are mistakenly thinking that

0, 1, 2

have to correspond exactly to 0, 1, 2

When we say Z3 is a subgroup of Z9, what we really mean is:
There is a subgroup of Z9 which is isomorphic to Z3

[zealous flame]

oh we haven't done that yet 😳

[coral shale]

then I don't get how you're meant to even do this question

Without that in mind, it doesn't make sense

[zealous flame]

what does it mean for Z9 to be isomorphic to Z3?

[coral shale]

no no

(subgroup of Z9) which is...

isomorphic means the 2 groups have exactly the same structure

To understand this formally, you need to know what a homomorphism is

[zealous flame]

oh i have that in my notes somewhere

[coral shale]

An isomorphism is a bijective homomorphism

2 groups are isomorphic to each other if there exists an isomorphism mapping one to the other

[zealous flame]

so a group homomorphism from G to H is a function from G to H such that f(g g')= f(g)*f(g') f(eg)=f(eh) and f(g^-1)=f(g)^-1

[coral shale]

it is a one line definition

[zealous flame]

i dont know what g' is tho i jsut wrote it down thinking it will be important some day

[coral shale]

f : G -> H is a homomorphism if for all a, b, f(ab) = f(a)f(b)

This is precisely the definition

Everything else you wrote is merely a consequence (that can be proven quite easily - try, if you haven't)

[zealous flame]

ah so is it commutitive ?

[coral shale]

no, certainly not

[zealous flame]

or is that associative

im confusing

[coral shale]

'commutative' has a very specific meaning

no neither.

[zealous flame]

oh

[coral shale]

If for all a, b, we have ab = ba, then the structure is commutative ('abelian' for groups)

If for all a, b, c, we have (ab)c = a(bc), then the structure is associative

[zealous flame]

oh I understand now, what is being said is that multiplying one mapped value with another will give you another mapped value ?

[coral shale]

It is saying you can do f first then multiply

or multiply then do f

[zealous flame]

ah

[coral shale]

I suppose this is similar to the idea of commutativity

but I would hesitate to use that term - at least, not without clarifying what exactly you mean

[zealous flame]

ah right

[coral shale]

'The group operation and the group homomorphism commute'

but no this isn't quite true

The group operation in G is different from the group operation in H

So yes there's this kindof idea, but not quite

[zealous flame]

so associative shows you can re group and communitive means you cna move numbers around ang get the same answer

[coral shale]

yes that is the idea.

Back to this then

hmmm

I think the best approach is to identify the subgroups of Z9

And then figuring out if any could be isomorphic o Z3

[zealous flame]

e,9,3

[coral shale]

There's actually 3 subgroups

[coral shale]

This isn't a subgroup.

[zealous flame]

oh

oh no i was stating the order

that i got from lagranges theorem

[coral shale]

Yes, but explicitly do you know what the subgroups are?

2 of them are easy

[zealous flame]

i knwo Z9 and e are individual subgroups were Z9 contains (0,1,2,3,4,5,6,7,8) and e contains (0)

[coral shale]

Right

What would the non-trivial proper subgroup be though

From lagrange, it must have order 3

[zealous flame]

a subgroup with 3 elements

ah so is this about finding the specific 3 elements from Z9 that map to give us a similar sequence to mapping out mod9?

[coral shale]

You need to find a subgroup in Z9 which has the same structure as the group Z3

So the first step is to find the subgroup with 3 elements

[zealous flame]

ah ok

[coral shale]

I say 'the' because I am telling you there is only 1 of them

but this should be something you convince yourself of later

[zealous flame]

ok i'll start getting to the subgroup with 3 elements

[chilly ocean]

Basically integers mod n are cyclic groups
And every cyclic group is an abelian group.
And every subset of that group is a subgroup

[zealous flame]

yeah ive started to realise doing this that its cyclic

[chilly ocean]

Yes

[fossil shuttle]

careful

not every subset of a group is a subgroup

[coral shale]

'every subset of that group is a subgroup' =...=

[chilly ocean]

But for abelian it's true

[fossil shuttle]

what?

[chilly ocean]

Only for abelian groups

[coral shale]

??????

bruh

[fossil shuttle]

Hmm... I don't know what you mean by that

[coral shale]

(Z, +) is abelian

{0, 1} is certainly no subgroup

[zealous flame]

yeah Z is cyclical

[fossil shuttle]

I wonder if you're mixing up this with the idea that every subgroup of an Abelian group is normal

[coral shale]

its abelian

[chilly ocean]

Wait

[coral shale]

Your statement is absurd as it stands

[chilly ocean]

Bruh I mixed with that 🤦‍♂️

[coral shale]

A subset need not even contain the identity

[coral shale]

that's not true

[zealous flame]

Ah rip

[coral shale]

The <2> you've written isn't closed under addition.

what is 8 + 2

Similarly for the rest.

[zealous flame]

10

Ahhh

[coral shale]

1

[zealous flame]

I c

I only did the first 6 this time

As for the last 3 I had already gotten 9 the previous time

So <6> contains the exact same elements as <3>

[coral shale]

construct an isomorphism between <3> and Z3

(and show it indeed is an isomorphism)

[zealous flame]

so when i constructed a table for <3> under addition+ and used Z3 on it, i got a table with 0's returned back to me

[coral shale]

No......

I don't quite understand what you have done

But first understand what it means for 2 groups to be isomorphic

It means there exists an isomorphism between them

This means you have to construct a function mapping one group to the other

And show it is a bijective homomorphism

[zealous flame]

ah

[coral shale]

<3> and Z3 arent meant to combine or something as you seem to be doing

[zealous flame]

This is what I had constructed

And did 2 little examples at the bottom of the table to show what I was doing

[coral shale]

3+3 isn't 0.

Be clear to yourself where elements live

0, 3, 6 are referring to the elements within Z9

What you can do to these elements is ONLY the operation within Z9

That is addition modulo 9

They have nothing to do with the group operation in another group, like Z3

[zealous flame]

ohh so i'm not supposed to use addition under Z3

[coral shale]

Firstly, are you aware of what a function between the 2 groups might look like?

Can you construct an arbritrary one please?

[zealous flame]

ok i'll try

[coral shale]

I'm worried there is some missing knowledge that is often taught in an intro to proofs class, here.

[coral shale]

good, but explicitly we might write

wait

[coral shale]

,rotate

[coral shale]

🤔

I'm confused

[zealous flame]

im trying to map Z onto 2Z under mulitplcation

[coral shale]

1. Your 2 and Z are confusing - write a horizontal bar across your z's in writing

[zealous flame]

yeah sorry i've got dysgraphia

[chilly ocean]

Can't we try making cayley tables for both groups

[coral shale]

2. sure, but we don't want that here??

We are talking about Z3 and {0, 3, 9}

I dont see how Z and 2Z have any relevant here

[zealous flame]

ohhh i thought u said to construct a function between 2 groups not THE 2 groups

😩

[coral shale]

people usually write their z's like this

[zealous flame]

i thought u were trying to see if i could write out a domain and codomain with a mapping between the 2

[zealous flame]

i'll use that in future

[coral shale]

But anyways - I see you understand what a map

or a function is

Now, you need to create one that will be a bijective homomorphism

between the 2 groups

[zealous flame]

so its one to one and onto

[coral shale]

{0, 1, 2} under addition modulo 3
{0, 3, 6} under addition modulo 9

[zealous flame]

so am i mapping the mod 9 one onto the mod 3 one ?

[coral shale]

it doesn't matter which way round you do it

[zealous flame]

oh yeha

[coral shale]

because a function which is bijective is invertible

[zealous flame]

its nijective

bi

how would 3 mod9 map onto 3 mod 3 ?

i keep thinking it will be 0 but i know thats wrong

[coral shale]

You have these 2 sets

There are 3 elements in each

[zealous flame]

oh right

i just map between those

[coral shale]

yes

Not sure what you were attempting earlier

These 2 groups happen to both use numbers

[zealous flame]

so i assign f(0)=0 f(3)=1 and f(6)=2

[coral shale]

sure, and then show this is a homomorphism

(its clearly bijective)

[zealous flame]

alright

so (3+1)=f(3)+f(1) is an example of what i'm trying to show?

[coral shale]

f(3+1)

note the two + signs you use are different

one is addition mod 3

the other is addition mod 9

To emphasize this, it would be good to write this

[zealous flame]

ahhh right yeah i'll bear that in mind

[coral shale]

$+_3$

[coral shale]

$+_9$

[coral shale]

subscript

The key is your 2 groups might not even be numbers in general

one could contain something completely different from the other

One group might contain numbers
The other group might contain vectors

The groups might still be isomorphic, though

[zealous flame]

aight imma have to wrap my head around this over summer while i do real analysis

[coral shale]

The 📌 in this channel is a bit of a joke

but gives some intuition

[zealous flame]

ah yeah i've taken a look at it before

[coral shale]

italians want to add mod 3, while they do it mod 9 in nyc. So the italians just stick to 0, 3, 6

[chilly ocean]

Lol

[coral shale]

no not there

The + subscript should be under the operations

[zealous flame]

oh right i'll fix that

[chilly ocean]

$f(3+_31) = f(3) +_9 f(1)$

[cloud walrus]

Pencil

[coral shale]

For example, this is a possible statement $$f(1 +_9 2 +_9 3) = f(1 +_9 2)+_3f(3)$$

[coral shale]

Right. You showed it for the specific case

But try showing it algebraically

[zealous flame]

ok

[coral shale]

Rather than case by case

To do this, perhaps define f algebraically

But hmmm 🤔

yh ok.

[zealous flame]

wait could i show that the order of both sets are equal and as such bijective through that ?

as D>=G and G<=D

where D is under Z9 and G is under Z3

oh wait

no thats not what im trying to show

im trying to show its homomophorphic

This is what I got when I showed it algebraicly

[coral shale]

In order to show it algebraically, you need an algebraic definition of f

[coral shale]

Bijectivity should be fairly clear - check it is injective and surjective

(no 2 things map to the same thing and everything is mapped to)

[zealous flame]

I thought I had created an algebraic expression of f

[coral shale]

I can't see it

state it?

[zealous flame]

From the F:D->G to the |G| part was supposed to set up the function

Prop 19.2 part 3 was what I was going to use to prove the bijection

*19.1

[coral shale]

Yes but that is likely overkill

(I would not expect anyone to do anything like it in a proof)

Anyways - to clarify what I meant

This is slightly informal, but you can let f(x) = x/3

But hmmmm this is slightly crappy notation

Without justification

[zealous flame]

Ah i think I know where you are going with this

[coral shale]

Are you familiar with equivalence classes?

[zealous flame]

Yup a equivalence relation within a group onto itself?

[coral shale]

You define an equivalence relation on a group

So we consider the integers mod n

We do not write

0, 1, ..., n-1

formally

But we write bars or square brackets

on the numbers

$\bar{1}$ or $[1]$

[zealous flame]

ohhhhhh

[zealous flame]

thats what the bars meant

[coral shale]

To indicate these are equivalence classes, not numbers

[zealous flame]

on the question

[coral shale]

So working modulo 9 for example

1 bar represents a set.

$\bar1:={1+9k:k\in\bZ}$

[coral shale]

By definition, this is what it really means

[zealous flame]

ah ok

[coral shale]

All the numbers with the same remainder upon division by 9

forms an equivalence class

I think I prefer the notation $[a]_n$

[coral shale]

$[a]_n:={a+kn:k\in\bZ}$

[coral shale]

Does this definition make sense to you ^

[zealous flame]

yes

i cant explain why

but it does

in my head

[coral shale]

You are familiar with set builder notation?

[zealous flame]

a+kn is the notation used for mod

so a is the remainder

[coral shale]

Expanded, this set is
{a, a - n, a + n, a - 2n, a + 2n, ...}

So if n = 9 (we are working mod 9)

[zealous flame]

never heard of set builder notation

[coral shale]

[1] = {..., -17 , -8, 1, 10, 19, 28, 37, ...}

set builder notation is exactly the notation used here

Have you been shown this formally?

'This set contains a + kn for all the possiblek in Z'

[zealous flame]

not by the uni but i have been using my own textbook that does

[coral shale]

Right if your uni didn't go through this

I think it is perhaps best to quickly browse an intro to proofs set of notes

which goes though notation/mathematical language that is very commonly used at uni level

[zealous flame]

ok i'll get onto that asap

[coral shale]

#proofs-and-logic

The first pin

loch wrote a set

[zealous flame]

ok

[coral shale]

and it isn't too long (you should know quite a bit of it already)

But anyways, so modulo 9

$\bar1$ and $\bar{10}$

[coral shale]

These are equivalence classes

And they coincide

[zealous flame]

yeah

[coral shale]

(you have exactly the same set)

[zealous flame]

i noticed that

when i was doing the question

[coral shale]

So earlier, I suggested for f to 'divide by 3'

we have 0, 6, 9 bar

and when I say 'divide by 3'

I mean divide every single element in the class by 3

[zealous flame]

ah ok

[coral shale]

Like how we write $3\bZ$

[coral shale]

Which means all the multiples of 3

[zealous flame]

would you write :3Z

[coral shale]

You take the set Z. And multiply everything in it by 3

[zealous flame]

oh right

[zealous flame]

ignore this

[coral shale]

$3\bZ:={3n : n \in \bZ}$

[coral shale]

In fact, there is another way of writing the equivalence classes

And this notation will be important later on

when you come across cosets

$$a + n\bZ :={a+kn:k\in \bZ}$$

[coral shale]

So if I wanted to write the equivalence class corresponding to 2 mod 9

..., -7, 2, 11, ...

$2+9\bZ$

[coral shale]

This is another way of writing it

[zealous flame]

ah ok yeah

[coral shale]

So back to the question - you can define f to divide everything in the equivalence class mod 9

by 3

Or alternately make f map the other way round

[zealous flame]

for the proof i think i have to do it both ways either way

[coral shale]

{0, 1, 2} under addition mod 3 to {0, 3, 6} under addition mod 9

And define f as multiplying everything in the equivalence class by 3

[coral shale]

no no

You only have to define 1 function

And show it is a bijective homomorphism

[zealous flame]

i got marked down the last time i didn't show it for both ways

[coral shale]

I'm not quite sure what you mean

As long as you show it is a bijective homomorphism (isomorphism), that is enough

Once you show f is bijective, so must f inverse

[zealous flame]

in a feedback question a while ago i showed a set being bijective for one way but they marked me down for not doing it the opposite way

[coral shale]

And if f is a bijective homomorphism, so must f inverse also be a bijective homomorphism (this can be proven).

That is odd 🤔

If possible - show us what exactly it was

it could be some other problem

[zealous flame]

i'll try and find it although this was last sememster so it might not still be up

[coral shale]

But ok, try proving this perhaps

And that might convince you only need to construct f

[zealous flame]

alright

[coral shale]

To show f is bijective, you just need to show it is injective and surjective

[zealous flame]

this is a bit of a silly question but how do I do that, all the proofs so far i've had to do for injective and surjective I could get away with using the proposition

[coral shale]

lets see

I will give an example

$f : {0+5\bZ, 1+5\bZ, 2+5\bZ, 3+5\bZ, 4+5\bZ}\to{0+15\bZ, 3+15\bZ, 6+15\bZ, 9+15\bZ, 12+15\bZ}$

[coral shale]

Using my previous notation, does this make sense

The domain has 5 elements, and so does the codomain

(but note each element itself is a set)

[zealous flame]

yeah that makes sense

so the function maps a 3F

[coral shale]

I will now define f to map by multiplying everything in a class by 3

$$f(a + 5\bZ) := 3a + 15\bZ$$

[coral shale]

for a = 0, 1, 2, 3, 4

Firstly, this function is well-defined (I have said what everything in the domain maps to, and each maps to exactly one thing)

For injectivity, suppose f(a+5Z) = f(b+5Z)

Then
3a + 15Z = 3b + 15Z

Hmmm this is a bit hard without knowing what you have/have not proven about cosets

[zealous flame]

well we've only gone over the definition

of left and right cosets

[coral shale]

Well alright so from first principles, we have

$${3a + 15n : n\in \bZ} = {3b + 15n : n\in \bZ}$$

[coral shale]

And we want to show a = b

[zealous flame]

yup

[coral shale]

Remember a must be between 0 and 4

[coral shale]

From how I chose to define the function

So what I could do is suppose a and b are not equal for contradiction

Then show 3a cannot be a member of the set on the right

Hence a contradiction of the equality of sets

(I'm just using contrapositive tbh)

Hmmmmmmmm I feel I can do this smarter

$${3a + 15n : n\in \bZ} = {3b + 15n : n\in \bZ}$$
$\implies$
$$3a \in {3b + 15n : n\in \bZ}$$
$\implies$
$$(\exists n\in\bZ)\quad3a = 3b + 15n$$
$\implies$
$$(\exists n\in\bZ)\quad3(a - b) = 15n$$
$\implies$
$$(\exists n\in\bZ)\quad a - b = 5n$$

[zealous flame]

rip

[coral shale]

Have you seen logical quantifiers?

[zealous flame]

yes

there exists

for all

[coral shale]

Yeah this would be a neater way of showing it

a - b is a multiple of 5

So they must be equal since a and b are between 0 and 4 inclusive

[zealous flame]

ok

[coral shale]

Like as you move on with the course

at a higher level you would not bother writing this out

(because of some results you see)

So we have f(a + 5Z) = f(b + 5Z) => a + 5Z = b + 5Z

Hence we have an injective function

Now surjectivity --- you can either use that theorem you had

The sets are finite and of the same size, so an injective function must also be surjective

Or you could say 'by inspection everything in the codomain has a preimage in the domain'

(since the sets are so small, surjectivity can be checked by brute force inspection, essentially)

[zealous flame]

aight

imma just say that surjectivity of the function has been left as an excercise to the reader lol

[coral shale]

👀

[zealous flame]

😎

alright so i'll get cracking on showing the injective part now

[shadow coyote]

studying bilinear forms rn

is the proof of this very simple? i thought about it for a couple minutes but am stumped

[zealous flame]

bruh they really jsut left the proof as an excercise 💀

[shadow coyote]

if my memory serves me right, this is the first thing that artin left as an exercise in the bilinear forms chapter

but it isn't obvious to me :/

[zealous flame]

for defining my 9Z and 3Z should i use the same variable a such that . a+9Z and a+3Z or should i use a different one for each

[coral shale]

What do you mean

I have to define the function this way

I am saying the class 'a + 5Z' maps to the class '3a + 15Z'

[zealous flame]

alright ok

[coral shale]

also, you don't have to use this chonky notation

that's up to you

[zealous flame]

aight

[coral shale]

this is probably more compact

But later on, you will see the notation I used being useful for other contexts

[zealous flame]

This is what I ended up getting

For the homomorphism and injection part

As that's really the part I found challenging

[fallow plume]

ngl i suggest typing it up in TeX @zealous flame

[zealous flame]

Aight I'll try it when I'm back home

[coral shale]

https://i.imgur.com/Hm4Uis9.png

I'm writing up notes and I

on how it 'looks'

idk - I feel like I could have structured this better, but 2nd opinions appreciated

[latent anvil]

second opinion: don't worry about it

[gilded gull]

I've checked the conditions to make sure it's an equivalence relation

but what would the equivalence classes be?

[coral shale]

have you an image in your mind

[gilded gull]

what do you mean?

[coral shale]

hmm

Let f(x) = x

Now what exactly are the functions related to f

graphically what might you draw

[gilded gull]

all functions that go through 3?

y=3

I mean

[coral shale]

exactly

Go through (3, 3)

All the continuous functions that do

So I might vaguely sketch something like this

This is one particular equivalence class

[gilded gull]

But thats just the equivalence class for x

right

ok

I understand what it is now

but I'm not sure how I would phrase that

[coral shale]

What is the defining feature

[coral shale]

of this class

[gilded gull]

All of the functions go through the point where f(x)=3

[coral shale]

you mean f(3) = 3

[gilded gull]

yes

sry

[coral shale]

Or another way to think of this

is the intersection of the function

with the line x = 3

If you want to be more precise

[coral shale]

but this is perfectly fine too

[gilded gull]

So each equivalence class is the set of functions that go through a particular point on x = 3

[coral shale]

yes

all functions which share the same intersection

[gilded gull]

Ok

tysm that was really helpful

[abstract nimbus]

Hey so I think this is the place to ask my question about rings:

I'm working on the question "for r in the set R with R being a non-zero commutative ring, if r is nilpotent then 1+r is a unit of R".

I started by saying r^n=0 for some least positive integer n => 1 + (-1)^n*r^n = 1 and I'm trying to factorise 1 + (-1)^n*r^n into (1 + r)(1 - r + r^2 - ... + (-1)^(n-1)*r^(n-1)) but when I was thinking of doing this, I realised it relies on the fact that (-a)(-b) = ab. But I'm not sure how to think of this property intuitively;

(-1) is the additive inverse of the multiplicative identity. But then how does (-1)^2 = 1 ?

It might be a simple property that I'm overthinking but I can't explain it to myself intuitively

[coral shale]

This is a property to prove

(-1).(-1) = 1

Where -1 is the additive inverse of 1

It comes directly from the axioms - try

[abstract nimbus]

I have a proof for it:

0 absorbs multiplication so we can write 0 = 0*x for any x in your ring.

so 0 = 0*0 = (1 + -1)(1 + -1)

Since multiplication distributes over addition we get
0 = 1(1 + -1) + -1(1 + -1) = -1 + (-1)^2

and rearranging gives you 1 = (-1)^2

[coral shale]

yes that's the way.

[abstract nimbus]

idk though I can't say it feels intuitive

[coral shale]

what is your intuition for -1 . -1 = 1

in the integers

It is by definition, yes? From a hs/elementary schooler perspective

So I doubt there is intuition to be had - it is a consequence of how we have defined rings.

[abstract nimbus]

no there is intuition for it i think

i think that makes sense because numbers are vectors on a number line right, and multiplication is sort of the way you scale numbers on the line

e.g. 2 * 4 is saying to take the vector that is 2 away from 0 on the right, and scale it by a size of 4 which is 8 away from 0 on the right so that's +8

2 * -4 is saying to take the vector that is 2 away from 0 on the right, flip it's direction and scale it by 4 so that's 8 away from 0 on the left so that's -8

-1 * -1 says to take the vector that is 1 away from 0 on the left, flip it's direction and then scale it by 1 so that's 1 away from 0 on the right so that's +1

[coral shale]

We can consider the R-module over R I suppose but idk about this...........

[south patrol]

basically just saying the intuition is 1 and (-1).(-1) are both additive inverses for -1, and inverses are unique

which motivates the proof strat

[spice whale]

quick question: is this really as simple as just defining an equivalence relation a~b iff HaK = HbK

[coral shale]

You want to motivate (-1)*a = -a I suppose

[south patrol]

well you need to show that that is an equivalence relation which is the work

although ye that is then easy by transitivity of equality

[spice whale]

isn't that just by equality properties

yeah

[coral shale]

Better than a simple 'Yes'

Really, a missing aspect here is perhaps the well-definedness of that notation

[gilded gull]

What does it mean for a function to be well-defined?

[coral shale]

or maybe not

and you show this indeed is a function

[delicate orchid]

x = y => f(x) = f(y)

[gilded gull]

wait so jus tinjective?

[spice whale]

nono

the other way around is injective

[delicate orchid]

injective is the other implication

[gilded gull]

Oh

[coral shale]

You want to show the function is defined on all elements of the domain too

f : R -> R
f(x) = 1/x

is not well-defined

[delicate orchid]

I feel like your coset example was better, shuri

[coral shale]

I'm too lazy to make it work

[sharp sonnet]

ye, this is still a partial function

[delicate orchid]

take f(x+2Z) = x, this is not well defined

[coral shale]

ah very good very good

[spice whale]

the function here is well defined because the group operation is well defined

[sharp sonnet]

good example is like f(x) = 0 on even, 1 on odd and 2 on divisible by 3

[delicate orchid]

you have 1+2Z = 3+2Z but f(1+2Z) = 1 and f(3+2Z) = 3, blunder...

[gilded gull]

oh ic

[coral shale]

anyways, do we care about well-definedness here or does that not even matter with regards to this equivalence relation

I feel like it doesn't matter after all

[sharp sonnet]

well definedness of what

[delicate orchid]

if a relation satisfies the properties it's an equiv. relation

[coral shale]

Yh, I don't think welldefinedness comes into constructing this equivalence relation

I thought it did

well definedness of the double coset notation which will matter for other stuff

[delicate orchid]

I guess you can kind of think of it as a function mapping two group elements to a boolean value - in which case it's definitely well defined (if it's an equivalence relation)

[spice whale]

it is

[coral shale]

(HaK)(HbK) := H(ab)K

Well-definedness comes into play when tryna do stuff like this

[sharp sonnet]

this is true

but you try to define some operation in terms of a "representative" of a double coset

this is the biggest cause of something being not well defined

wews example showed this

you define operation on some set in terms of a representative

[spice whale]

yeah

[spice whale]

double quotient 😳

[delicate orchid]

every example I've seen of funny well definedness issues is when you're mapping from some space of representitives

[coral shale]

f(alison) := im alison
but not all alisons might agree

[delicate orchid]

like quotients, tensor products, field of fractions, etc.

[coral shale]

or something silly

[spice whale]

i need to meet my alt universe selves

[delicate orchid]

alt+universe

[spice whale]

on my bucket list

[delicate orchid]

time for a lot of \Rightarrows

[spice whale]

is this valid

[delicate orchid]

reverse implication is swag tier

[spice whale]

it works if you remember they're different elements of H

[delicate orchid]

yeah it's just the set H

[spice whale]

i believe

ye

[delicate orchid]

uhhh lemme think

[spice whale]

ye no reason not to work

[delicate orchid]

it's just that

I've never seen any subgroup to the power of -1

[sharp sonnet]

[delicate orchid]

sully me all you want it's the truth

[spice whale]

if you do $h_1 g h_2 = g h_3 \leftrightarrow h_1 g = g h_3 h_2^{-1}$

[cloud walrus]

all functions are alison

[sharp sonnet]

invertibility of H?

is that a word

[delicate orchid]

ok now it makes sense

[spice whale]

closed under inverse

[delicate orchid]

and h_3h_2^{-1} runs over H as h_3 and h_2 do

[delicate orchid]

I'd write this rather than H^{-1}

[spice whale]

yeah

[delicate orchid]

I've never seen that before

[sharp sonnet]

i have

its normal notation

[spice whale]

but the swag

[delicate orchid]

true the swag tho

[sharp sonnet]

it just looks a bit weird

anyways, you have H^-1 = H

[delicate orchid]

I've seen it for sets

[sharp sonnet]

H is a set

[delicate orchid]

ah you beat me to it

yeah I meant sets that are not subgroups

[sharp sonnet]

its super common to continue operations on elements to subsets (elements of the powerset)

[spice whale]

funny matrix time

[sharp sonnet]

good exercise

[spice whale]

wait this is block multiplication isn't it

wait no

[sharp sonnet]

dont think so

unless you consider matrix multiplication block multiplication

blocks of size 1

[spice whale]

yeah

[sharp sonnet]

or size n i guess

[delicate orchid]

horrifying

[sharp sonnet]

matrix decompositions are super important

[delicate orchid]

agreed but they give me first year lin alg flashbacks

[sharp sonnet]

wew you should do linear algebra in Q_p vector spaces

lattice theory

and then teach it to me

[delicate orchid]

loch I barely even understand the construction of Q_p

[sharp sonnet]

the lattices are like the balls with respect to the metric

can you imagine that?

[delicate orchid]

yes :chad:

[sharp sonnet]

actually

[delicate orchid]

the metric on Q_p is discrete isn't it?

[sharp sonnet]

discrete?

[delicate orchid]

like it can only take 1/p^n for some natural n

or am I talking nonsense

where did I read this

ah that's abs value

[sharp sonnet]

you get this on restriction to Q

then on Q_p as well i think

[delicate orchid]

epic

[sharp sonnet]

the cool stuff though is its an ultrametric

so you have |a+b| <= min(|a|, |b|)

a stronger form of triangle inequality

and this gives you all kinds of weird stuff

anyways, i am just thinking

[delicate orchid]

I've briefly seen ultra metrics

[sharp sonnet]

there is this computationally hard problem on computing a shortest vector in a lattice

and computers can do numerically exact computations on finite extensions of Q_p even with finite precision

so maybe i can build a meme cryptosystem based on shortest vector in some Q_p lattice

i am doing this now

[delicate orchid]

spooky

step 1: bubble sort the set of vectors

[fallow plume]

lemme think about what I'm about to say just in case I'm missing something really obvious

okay yea I'm dumb

I thought our hypothesis was for a quotient module

not a submodule

that makes much more sense

[gilded gull]

does this work?

[spice whale]

been attempting this
and the matrices aren't funny

[hybrid island]

willing to pay $ for someone to write proofs for me dm me

[spice whale]

most people will probably give you "left as an exercise to the reader" as a proof for everything

[hybrid island]

[spice whale]

do your own homework

it will help you become better at maths

[sharp sonnet]

i think the trick here is to do it in like dimension 3 and then stare at it until you are convinced it works in the general case

[fallow plume]

that's just the trick for almost everything with matrices

[sharp sonnet]

also @hybrid island we do not support academic dishonesty, so i am assuming you were just looking for a tutor but even in that case we strongly advise against financial transactions

[spice whale]

I've got it sort of now

it's related to column echelon

[fallow plume]

finitely presented <=> representable by a finite matrix?

[hybrid island]

its just hw tho

[fallow plume]

i can't tell if there's similarities or if it's something more

[fallow plume]

that doesn't mean it isn't academic dishonesty

[hybrid island]

ur allowed to get outside help for the hws

[spice whale]

then ask for help

[fallow plume]

help, not "doing proofs for you"
if that's not what you wanted, thta's why he said "So i am assuming ..."

[spice whale]

but people won't do it for you

[gilded gull]

sry I don't want to spam but I don't know if this is the right way of going ab it

[hybrid island]

ight imma be fr tho i have no idea how to even start this

lol

[fallow plume]

you're chill dw. I personally see handwritten text and ignore it though 💀 you might try TeX'ing it

[gilded gull]

If $[x]{12} = [x^{'}]{12}$, then $x^{'} = 12kx$ or $ x$. If $x^{'} = x$, clearly $f([x]{12}) = [x+1]{12} = [x^{'} + 1]{12} = f([x^{'}]).$ If $x^{'} $ does not $= x, f([x^{'}]{12}) = [x^{'}+1]{12} = [12kx^{'}+1] = [x+1]{12} = f([x]_{12})$, so the function is well defined

god damn it this is going to be pain

[fallow plume]

lmao, just remember that
$$this is a MATH statement on a new line$$
$this is an inline MATH statement$
everything else should be text outside

[gilded gull]

I'm pretty sure the morphism bit is right so i just need to know if this works to show that its well defined

[cloud walrus]

Rishi

[gilded gull]

and there's the original problem again

[delicate orchid]

this escalated quickly
https://cdn.discordapp.com/attachments/359052581022203914/940656085893799936/unknown.png

[chilly ocean]

can someone walk me thru this problem

i know the centralizer would be all elements of S4 that commute with (1,3,4)

right?

[gilded gull]

That sounds right to me but I'm not the best person to help here

Ok wait this isn't that bad I might be able to help

[chilly ocean]

can u pls

[brisk granite]

so, there are only 16 elements you're worried about so bashing this out in the obvious way isn't that bad

[gilded gull]

^^

and you can recognize a pattern hopefully after doing a few

[brisk granite]

maybe this helps: gx = xg iff gxg^{-1} = x right?

[chilly ocean]

yeah

[brisk granite]

and you know gxg^{-1} = (g(1),g(3), g(4))

[chilly ocean]

nope

[brisk granite]

x is (134) btw

[chilly ocean]

oh

[brisk granite]

or sorry

[chilly ocean]

ummm

[brisk granite]

I meant g^{-1} of all that

[chilly ocean]

all of what

[brisk granite]

okay, edited for correctness

so, now counting g's shouldnt be that bad?

[chilly ocean]

i dont get how you got this tbh

[brisk granite]

well, see if you can prove that

what is (gxg^{-1})(g(1)) ?

[chilly ocean]

g is any element in s4?

[brisk granite]

yes

[chilly ocean]

x is (1,3,4)

what is g(1)

[brisk granite]

the idea is that we're counting g such that g^{-1}xg = x

[brisk granite]

the specific number doesnt actually matter

[chilly ocean]

well woudlnt it just be

(1,3,4)g

?

[brisk granite]

wouldnt what be that?

[chilly ocean]

= (1,3,4)g

we have g(1,3,4)g^-1

yeah honestly i got no clue

[brisk granite]

okay, so, here's how I'd compute it: $$gxg^{-1}(g(1)) = (gx)(g^{-1}(g(1))) =gx(1) = g(3)$$

[cloud walrus]

polybeandip

[brisk granite]

do you see how that works?

[chilly ocean]

how did u get g(3) ?

[brisk granite]

x(1) = 3

from there, Id hope you can show g^{-1}xg = (g(1),g(3),g(4)). We want this to be equal to (1,3,4). So, there's three cases to consider.

[gilded gull]

would it be g(1) = 1 , 3, 4 and then just check if it it stays in the same order

[brisk granite]

yep

[gilded gull]

and obviously the first is just the identity so

really its just two

[brisk granite]

why would you not include the identity in the centralizer

[gilded gull]

no thats what I mean

wait i said that really weirdly

nvm lol

[gilded gull]

@brisk granite could you check this tho

[chilly ocean]

hmm

let me look at this for amin

[brisk granite]

looks good technically to me though I think there's some redundancy

or actually, I think you meant x' = 12k + x

[gilded gull]

Oh thats a better way of writing it

[brisk granite]

well, what you wrote was not right

[gilded gull]

?

[brisk granite]

a \equiv b mod n means a-b is divisible by n so a = b + 12k

[gilded gull]

right

[brisk granite]

you wrote a product

a= 12kb

[chilly ocean]

can you explain this then

this my prof solution

[gilded gull]

that is the same thing in different notation

[brisk granite]

ah yeah, he did what we said

[brisk granite]

ohh sorry

I see what you mean now

[gilded gull]

No mine is still wrong lol

ur right

[brisk granite]

but yeah, your thing is confusing lol

[gilded gull]

Yeah I kinda mixed that up

How would I go about showing it then

[brisk granite]

sure, this looks good

Id recommend calling the operation in Z_12 + though

[gilded gull]

wait hold up im still messed up with defining x prime though

[brisk granite]

also, Z_12 is like kinda bad notation and I think Z/12Z is preferred. There's conflict with p-adic stuff

[gilded gull]

Yeah the book defines it that way for rn though so I'm just leaving it

[brisk granite]

this was what we said before

if you want to prove that equality, then check that they represent the same function on [n]

i.e. plug in \sigma(1) and show it gets sent to sigma(3)

and so on

and show that everything other than sigma(1), sigma(3), sigma(4) is fixed

[brisk granite]

I think the first section for when x = x^\prime is redundant

the next part includes this case

[gilded gull]

Right I just had to add the plus in

[brisk granite]

it's when k = 0

[gilded gull]

because then k = 0 is just

yeah

I'm writing up my answers instead of latex rn tho so I'm just leaving it

I don't think they'll be mad if its inefficient as long as its correct

thanks

[brisk granite]

you shouldn't keep redundant sections of an argument

[chilly ocean]

i dont get the highlighted part

how are those equal

[gilded gull]

I get that

I'll just cross it out then

[gilded gull]

yeah I don't know how helpful I can be with this

But all abelian subgroups have to have a conjugate g^{-1}hg= h (and this comes from the fact that all abelian subgroups are normal subgroups)

Since the centralizer has to be abelian

you get that

[chilly ocean]

fuck it

im too ass for this class lol

[gilded gull]

Honestly you just might not have learned it yet

Idk lol

[chilly ocean]

my final is on tues

[gilded gull]

O.o

[chilly ocean]

im so fucked

[gilded gull]

Ok I've found someone more screwed than I am

i'm two weeks behind

[chilly ocean]

these last 10 weeks

have been hell for me

cause idk anything that goes on in this class

im not even a math major

[gilded gull]

Rip

[tranquil parcel]

"Note that if F is the prime subfield of K then Aut(K) = Aut(K/F) since every automorphism of K automatically fixes F". I don't understand why an automorphism of K necessarily needs to fix its prime field. Doesnt the prime field itself sometimes have automorphisms?

[brisk granite]

the prime subfield is the one generated by 1

[tranquil parcel]

oh and 1 would need to go to 1

ok ok i see i was thinking of group automorphisms not field automorphisms when i said that

totally different

[gilded gull]

Is the function from $[x]{8}$ to $[x]{2}$ a morphism?

[cloud walrus]

Rishi

[gilded gull]

I don't think it is

But I'm braindead rn so I want to make sure

[brisk granite]

why not?

[gilded gull]

Honestly its the modular arithmetic at this point killing me

Isn't it that f(3 mod 8)f(5 mod 8) goes to 1 mod 2 times 1 mod 2

wait thats just 1 isn't it

[brisk granite]

your group operations are addition btw

[gilded gull]

Does it have to be?

addition I mean?

[brisk granite]

If this is all the problem gives you, then presumably you're working with a map from Z/8Z to Z/2Z

[gilded gull]

Yes

Well it wants me to verify if its well-defined and then if its a morphism

and then if its a morphism to find the kernel and image

[coral shale]

Tell us what 'f is well-defined' means.

[gilded gull]

a = b implies f(a) = f(b)

[coral shale]

where does addition and multiplication come into this

[gilded gull]

w the morphism

wait

what did I write

[coral shale]

what?

[coral shale]

there is no addition or multiplication

[gilded gull]

yes

Ok so

when i'm checking if its a morphism

[coral shale]

no.

You are checking if this function is well-defined

[coral shale]

.

[gilded gull]

Nono I already checked that

[coral shale]

You sure didn't make that clear.

[gilded gull]

My bad

[gilded gull]

that's how we got here tho

Does my group operation need to be addition to check if its a morphism

or could I use multiplication instead

[coral shale]

You don't pick your group operation...........

That's ridiculous

[gilded gull]

So I need to use addition here

[coral shale]

Yes.

[gilded gull]

Is it supposed to be that the operation is implied depending on the map?

[coral shale]

yes.

no

implied from the group

Z/nZ is viewed as a group under addition unless stated otherwise

[gilded gull]

Oh

[coral shale]

This will never even be a group under multiplication

without some adjustments

You should realise this.

[gilded gull]

Well it makes sense that my answers didn't make sense then

[coral shale]

Why is Z/nZ not a group under multiplication?

[gilded gull]

Would you not have an inverse?

[coral shale]

that's unspecific

What doesn't have an inverse

everything?

[gilded gull]

I'm not sure tbh

[coral shale]

[gilded gull]

Could we look a specific example to make it easier

[coral shale]

well have a think

[gilded gull]

Ok

[coral shale]

yes, pick any n

maybe one thats not too small

[gilded gull]

(1 , 2, 3, 4, 5) so Z_{6}

(0,

[coral shale]

k

[gilded gull]

Ok so 2 and 3 here would have inverses of each other because they give 0

1 is the identity under multiplication

[coral shale]

🤔

🤔 🤔 🤔 🤔 🤔 🤔 🤔 🤔 🤔 🤔

what?????

[gilded gull]

assuming we're using multiplication for contradiction right?

[coral shale]

Check that.

[coral shale]

.

Indeed, before even considering inverses of elements

you must first identify the identity

correctly

[gilded gull]

1

yes

Its obviously closed so thats fine

oh my bad 2 and 3 wouldn't be inverses then

Its impossible for 5 to have an inverse

so its not a group

@coral shale is my reasoning right now?

[coral shale]

very sus.