groups-rings-fields

[somber bramble]

@chilly ocean what are the 0 and 1 elements of that ring

[tame bear]

why even include -

[fading wagon]

Maybe that's the inverse operator

[somber bramble]

and what does it even mean for those things to be 9, 9 isn’t in your “ring”

[chilly ocean]

@somber bramble I mean 9 is the result

[somber bramble]

but what is 9

[fading wagon]

The operators need to be closed

[somber bramble]

your thing only has three elements

which are 1, 4 and 5

[chilly ocean]

I do not have 0 in the set????

[fading wagon]

You can't jump out of the ring

[somber bramble]

what is a 9

I do not understand this symbol, the only numbers that exist are 1, 4 and 5!

[chilly ocean]

@fading wagon The operators need to be closed yes of course I am just checking ONE property of the ring. my ring is not a ring

[fading wagon]

Actually, just 1, 4 and 5, even 5! isn't there

[somber bramble]

obtuseness is not a virtue

[fading wagon]

Well, then state all the elements in the ring

[chilly ocean]

{{1,4,5}, +, X} is a a personal reminder I nmade for {R, +, X} it is not a real noitation

[tame bear]

why?

[somber bramble]

what

[fading wagon]

To check one property of a ring, you need to check one property of the ring, like remember what the property says exactly with the "for all"

[chilly ocean]

@fading wagon of course. it is not a real ring, it is an example

[somber bramble]

if it’s not a ring, then it’s a really bad example of a ring

don’t you think

[fading wagon]

It's a non-example of a ring

[somber bramble]

“yes, yes I know that this is only a cardboard wall and not a house, but it’s an example of a house!”

[chilly ocean]

i am just trying to understand somethin gvery specific

[somber bramble]

and you’re confusing yourself in doing so

[fading wagon]

"yes, yes, this cardboard box is a great example of a house"

[somber bramble]

no, a box would actually be reasonable

we’re talking only a single wall

a box even has like, rooms

well, room

[chilly ocean]

so all what I remember of this roon is if I want to check this distributivity of a ring, I should not focus on the set but check the operators

lol I told room

[somber bramble]

I’m… not sure that sentence makes sense

[fading wagon]

well, you need to check that the operator does what it needs to do on all elements of the set

[chilly ocean]

ok

it is also what I mean by if I want to check this distributivity of a ring, I should not focus on the set but check the operators

[somber bramble]

and I think you need to stop making up your own examples and look at some made by people who actually understand the subject

well, no

you need to look at everything

[odd swan]

hey, I am not sure if this is the correct channel for this but here we go

[somber bramble]

there’s a convo ongoing you know

again, consider verifying that the integers modulo 5 are a ring

[odd swan]

I got stuck in a really simple math calc while programming and I said why not seek for help

[somber bramble]

you’re definitely in the wrong place then

[odd swan]

I am making a progress bar and trying to get percentage of 2 values

[somber bramble]

are you even reading what I’m saying

[odd swan]

yes I am but it's simple math can you listen? lol

[somber bramble]

you’re interrupting a conversation and you’re in the wrong channel

[fading wagon]

@odd swan Wrong channel

[somber bramble]

this is a channel about abstract algebra

[odd swan]

where do I go

[somber bramble]

read #❓how-to-get-help

[fading wagon]

use #prealg-and-algebra

[tame bear]

#❓how-to-get-help

[fading wagon]

that also works

[somber bramble]

anyway, as I was saying

[odd swan]

thanks

[chilly ocean]

ok let's take Zn as a REAL example:
Zn = {0,1,2,3,4.,5... n-1}
so for the commutative ring {R,+,X} with the set Zn or Z4 I have to do:
if 0x(1,2,3,) = 0x1+0x2+0x3 then the ring ensure preporty of distributivity.
0x(1,2,3,) = 0
0x1+0x2+0x3 = 0
then the ring has this property

[somber bramble]

if 0x(1,2,3,) = 0x1+0x2+0x3
this doesn’t even make sense

what does 0x(1,2,3,) even mean

and you’re still making examples

why are you multiplying stuff with 0 specifically

0 is not an arbitrary element

[chilly ocean]

In my book I have:
a × (b + c) = a × b + a × c
(a + b) × c = a × c + b × c

[somber bramble]

yes, but you can’t just take a=0

[fading wagon]

you need to show for all

[somber bramble]

0 is not all

0 is just one of them

[chilly ocean]

oops I mean:

ok let's take Zn as a REAL example:
Zn = {0,1,2,3,4.,5... n-1}
so for the commutative ring {R,+,X} with the set Zn or Z4 I have to do:
if 0x(1+2+3) = 0x1+0x2+0x3 then the ring ensure preporty of distributivity.
0x(1+2+3+) = 0
0x1+0x2+0x3 = 0
then the ring has this property

[somber bramble]

that requires understanding of ideals and quotients

[chilly ocean]

0 is not all
AHHHHHHHHHHHHHH OK I HAVE REALLY UDNERSTOOD NOW!!!!

[somber bramble]

of course

once you have the theoretical background

[mild laurel]

I don't feel like you've understood at all

[chilly ocean]

if 0x(1+2+3) = 0x1+0x2+0x3 AND if 1x(0+2+3) = 1x0+1x2+1x3 AND if 2x(0+1+3) = 2x0+2x1+2x3 and all !!!!

[somber bramble]

well

we’re getting closer

it’s something

I’m not entirely sure why you’re doing sums of three numbers tho

[mild laurel]

I'm not sure why you're doing distributivity over three numbers

[somber bramble]

how you actually prove it is:
•take the fact that ℤ is a ring
•let [x],[y],[z] ∈ ℤ/nℤ
•then x = [x(y+z)] = [xy + xz] = [x][y] + [x][z]
details to be filled in as much as you desire

well, no

that’s the wrong advice

go read about how to prove things

first

[chilly ocean]

@somber bramble you tolds we are going closer. cool. what is the next step?

[somber bramble]

then do what blitz said

[mild laurel]

No the right step is

[somber bramble]

like, you can prove that ℤ/nℤ is a ring without knowing any real ring theory if you know a bit about how remainders behave under division

[mild laurel]

To go back and read a book on how to do proofs

[somber bramble]

oh shit #books-old

I curate that

and I haven’t looked at the list in so long

I should probably update it huh

I just completely forgot about it

tomorrow

wait no that might not be opportune. well, I’ll get to it

[tame bear]

Z/nZ just inherits ring structure from Z

qed ez

[somber bramble]

prove it in such a way that ilove can understand the proof

[chilly ocean]

ok so to conclude I have the example:
if 0x(1+2+3) = 0x1+0x2+0x3 AND if 1x(0+2+3) = 1x0+1x2+1x3 AND if 2x(0+1+3) = 2x0+2x1+2x3 and all !!!!
that permit to check if a ring has the property of distribution. but this way is not the best because it requires to check any element of the set even when the size of set of the group is inifinite. for the moment it is all what I want. In the future if I want a more complete way to check not matter the size, I can check the operators. I have all what I want.

END.

[tame bear]

the elements of Z/nZ are the classes of nZ
[0], [1], ... up to [n-1]

seriously

[chilly ocean]

ok I read

sorry

[tame bear]

why do you keep multiplying by 3

its a( b + c)

not a( b + c +d + e +f)

dont need to go to ideals to explain the Z/nZ notation

[chilly ocean]

I am very tired. qwick. can I determine by the operations of the ring if the ring is distributive regardless on the values of the set?

@chilly ocean so if I have 1 set and 2 operator, I necesary have a ring no matter the values of the set???

[tame bear]

no

thats not how it works

[chilly ocean]

ok so I should have told can I determine by the operations of the presumed ring if the presumed ring is distributive regardless on the values of the set?

https://gyazo.com/179baf7d52a05d7eebd656ae40bea231

Have to use another operative

please, help me.

its urgent..

[tame bear]

you have to use the set to determine if the operations are distributive over the set

you cant just ignore them

[chilly ocean]

how would the answer be

[tame bear]

not talking to you

[chilly ocean]

because
a × (b + c) = a × b + a × c
(a + b) × c = a × c + b × c

XD

@tame bear ok thank you. so the mathematical formula with
if 0x(1+2+3) = 0x1+0x2+0x3 AND if 1x(0+2+3) = 1x0+1x2+1x3 AND if 2x(0+1+3) = 2x0+2x1+2x3 and all !!!!
seems good? nothing to add?

[tame bear]

you cant simplify 0x ( 1 + 2) and 0x ( 2 + 3)
into 0x ( 1 + 2 + 3)

stop putting 3 terms in the ()

[chilly ocean]

ok

cool

then all is fine

everybody thank you

thank you @tame bear

thank you @chilly ocean

idem for @somber bramble

and thank yoy @mild laurel

XD

[mild laurel]

I feel like you still don't really get this but

[woven raptor]

i would think not, take a quadratic and factor it

its two binomials arent constant

oop im retord then

[smoky cypress]

he said can not have to be constant

[bleak abyss]

Under most people's definitions, yes

[mild laurel]

@chilly ocean if these constants are units in your ring, then I think most people wouldn't count this as a factorization

Yeah I mean

You wouldn't really call x = x \cdot 1 a factorization of x

Yeah I mean I'm not sure merriam-webster is the best place for math definitions

What are you even trying to say

Like unique factorization in UFD's is only up to a unit

So generally people consider the equivalence class of associate elements

Like, would you call 2 = 2 \cdot 1 a factorization of 2?

At least in my opinion, I think it'd be weird to call that a factorization so

[oak coral]

facrotization of n is the smallest multiset such that its product equals to n

[mild laurel]

@oak coral that seems it implies that factorization would be unique

[oak coral]

@mild laurel yeah, that’s what we want, dont we?

[mild laurel]

@oak coral factorization is not always unique

[oak coral]

@mild laurel it’s meaningless to make it not unique

[chilly ocean]

no it isn't

[mild laurel]

What

There are plenty of non unique factorization rings where thinking about factorization into irreducibles is super helpful

[brisk granite]

Is this over?

[mild laurel]

@oak coral

[brisk granite]

ok, so, given that $N/G_{i}$ is normal in $G_{i+1}/G_{i}$, do I know that $N $ is normal in $G_{i+1}$?

[cloud walrus]

∀ScoopityPoop, Scoop ≠ Poop:

[fading wagon]

hmm

@brisk granite I think you notated wrongly

Did you mean $G_i/N$ normal in $G_{i+1}/G_i$?

[cloud walrus]

Element118:

[fading wagon]

wait

[brisk granite]

no

Didn't mean that

[fading wagon]

you use the same normal subgroup to divide

Okay, I'm really not familiar with normal series

$G_{i+1}$ has a normal subgroup $G_i$ such that $G_{i+1}\geq N\geq G_i$, given $N/G_i$ is normal $G_{i+1}/G_i$, prove or disprove $N$ is a normal subgroup of $G_{i+1}$

[cloud walrus]

Element118:

[brisk granite]

uh, yep. that's the question

Do you know how I could start?

Is this obvious?

[fading wagon]

I don't see how this is obvious yet

[brisk granite]

ok

I can ping helpers now, right?

15 min is the wait time?

Or is that only for the questions channels?

[fading wagon]

Probably only questions channels

The aim of pinging helpers is as a reminder

but you already have me thinking about it

[mild laurel]

So you have that G_i is contained in N is contained in G_{i+1}?

[fading wagon]

yes, that seems to be the case

[hot lake]

you have two canonical surjective group morphisms G(i+1) -> G(i+1)/Gi and G(i+1)/Gi -> (G(i+1)/Gi) / (N/Gi), N is the kernel of their composition so it is normal in G(i+1)

[fading wagon]

Hmm, so if f is the composition, then we need to show f(n) for n in N is the identity and f(g) for g in G(i+1) not in N is not the identity

[chilly ocean]

I am reading a book about gcd and I find a page ambigous:
"If b m = 0 and b m−1 exceeds 1, then there does NOT exist a multiplicative inverse for b 1 in arithmetic modulo b 2"
if bm-1 does not exceed 1, then has b-1 NECESSARY a multiplicative inverse?

[fading wagon]

???

I think we are missing some parentheses?

[chilly ocean]

my question is mostly about english

[fading wagon]

Is there even more context?

[chilly ocean]

ok I am going to tell you mo0re

[fading wagon]

Oh, it's $b_m$, and $b_{m-1}$

[cloud walrus]

Element118:

[fading wagon]

Subscripts?

[chilly ocean]

yes

absolutely!

[fading wagon]

Is this number theory?

[chilly ocean]

exactly

no

maybe actually. it is about gcd. I do not know if it is in number theory

[fading wagon]

Are you working on integers?

[chilly ocean]

yes

[fading wagon]

It's number theory

[chilly ocean]

then yes

do you want I ask in the next room?

[fading wagon]

^read channel description

[chilly ocean]

with galois and all

[fading wagon]

okay then #advanced-number-theory might actually be a better fit

[chilly ocean]

thank you

🙂 bye

[mild laurel]

We have that (G_{i+1}/G_i)/(N/G_i) is isomorphic to G_{i+1}/N @brisk granite

[fading wagon]

whoa yeah, as a consequence of that surjective map

[brisk granite]

um, why?

Also, the kernal argument looks good!

[fading wagon]

try understanding what that surjective map does to elements in N and elements not in N

[brisk granite]

which surjective map?

There are two, right?

Do you mean the composition?

[fading wagon]

yeah the composition

the composition of surjective function is another surjective function

the composition of injective function is also another injective function

[brisk granite]

yea, ik

[oak coral]

@mild laurel oh lol i obviously forgot to note that numbers should be prime

@mild laurel took few hours to figure it out

[mild laurel]

What does that even mean? Numbers should be prime?

[oak coral]

@mild laurel tbh Im kind of lost rn. i forgot to notice every number in the multiset should be prime

[mild laurel]

Every prime element is irreducible

So I'm still not really seeing the point

[oak coral]

yeah

[mild laurel]

There are rings without unique factorization

[oak coral]

oh

well cool

[bleak abyss]

If only Zopherus were wrong

Kummer would've proved FLT way back when

[clear obsidian]

so the answer is B I think, & I know that you can see commutativity by symmetricality, but is there a way to see associativity visually (ie without having to check individual examples)?

[potent lynx]

just disect every table

see if there is an inverse for every element

and an identitiy element

identity*

so for example 1

a*a = a

a*b=b

a*c =c

a*d=d

a is identiity here

[tame bear]

that doesnt check associativity

[potent lynx]

inverse for b is d

[clear obsidian]

inverse is easy to see visually, just check for the identity in each column

in this case a

[potent lynx]

inverse for c is c

inverse for d is b

[clear obsidian]

like I get that part

[tame bear]

lights associativity test is the only thing that ive heard of, but im not sure if theres any better way

[potent lynx]

check associativty

[tame bear]

mo2, that doesnt check associativity

[potent lynx]

is (ab)c = a(bc) ?

[tame bear]

which is what he asked about

[clear obsidian]

yes I know

[potent lynx]

for all a b c in group

ok

ye sigma got it

[clear obsidian]

I know things about abstract algebra, Im just trying to figure out the best strategies for the mgre

[potent lynx]

never heard about that tbh sorry

[clear obsidian]

mgre is the mathematics subject graduate requirement exam

[tame bear]

https://math.stackexchange.com/questions/168663/is-there-an-easy-way-to-see-associativity-or-non-associativity-from-an-operation

[clear obsidian]

thanks sigma

yeah I got that

looking at it now

on wiki

says that commutativity implies associativity? I feel I should know this if its true

not sure why that should be true though

[potent lynx]

never heard of it

[mild laurel]

Think the other thing that might be helpful here

Is knowing that there are only two groups of order 4

So you can just check it against those multiplication tables

[clear obsidian]

yeah V4 and whats the other one? Z4?

[mild laurel]

Yep

[clear obsidian]

nice

and z4 is cyclic so thats easy enough

v4 I think is just a^2 = e, b^2 = e, ab = ba = c I think

yeah I thought not ^^^ thanks

[stone fulcrum]

There's really no easy way to check associativity with Cayley tables

However, if there's only four elements, you don't need to

[clear obsidian]

its easy with low group order now based on what Zopherus said

of course will get more tedious as we increase order I think

[stone fulcrum]

Also, if you happen to know the (two) groups that are of order 4, this is immediate

[clear obsidian]

like obviously order 1 is trivial , order 2 is still trivial, order 3 I think is unique structure iirc so also trivial, order 4 theres 2 up to iso, order 5 isnt bad since 5 is prime, then up from there might get a bit more tedious

ah yes

bc you take the powers in the prime factorization

to get the albelian ones, and those happen to be all of them for order 4

[stone fulcrum]

And yeah, Cayley tables are useless after a certain point

There's other means to find groups that are larger

[clear obsidian]

to be super helpful it really needs to work well at order 6 and up

yeah like I was saying too, if we have prime order its trvial since its cyclic

and thus albelian

and yeah I need to look up more about finding nonalbeilian/general groups order n

yeah this too ^^^ is what I mentioned with v4

uniquely determines group structure by a set and a few equations

[stone fulcrum]

nonabelian can be a hard question, there's lots of tricks to finding those

[clear obsidian]

mhmm, yeah I need to look up more about this ^^^

e.g. number of groups of order 48 or something

if were asking for albelian its trivial but yeah

[thorn delta]

To use the isomorphism theorem, is this question wanting me to find a group G', and find a surjective homomorphism between G and G' which has one of the normal subgroups as the kernel?

[mild laurel]

Yes

[thorn delta]

that sounds like a lot of guess work

[mild laurel]

Not at all

[thorn delta]

The determinant homomorphism is the only thing that comes to mind when I think of matrices, but I don’t think that would work here.

Oh wait do I need to know the cosets to come up with one?

[fading wagon]

let's see, the kernel of the determinant homomorphism are the matrices with determinant 1

yeah, not too useful here

[bleak abyss]

Oh I missed upper triangular

This is actually very relevant

I thought it was just GL_n

Take the product of two general such matrices

This will cause the answers to scream in your face

[thorn delta]

The product of a generic matrix in the normal subgroup and in G?

[fading wagon]

I suppose

both ways

[bleak abyss]

No no no

I just mean

[fading wagon]

two matrices in G?

[bleak abyss]

Yeah

[fading wagon]

I'm still pretty lost

[bleak abyss]

Have you done it?

[fading wagon]

yeah

I'm still trying to find out what it means

[bleak abyss]

I mean first you'll notice the diagonal entries just do not depend on what happens in the top right corner

[fading wagon]

yes

[bleak abyss]

And so sending a matrix to either diagonal entry, or to the ordered pair of the two, is a homomorphism

[fading wagon]

oooh

[thorn delta]

Oooh... I have a class that is about to start, so I’ll do this later, but thanks

[fading wagon]

oh, it's just $\mathbb{R}\setminus{0}\times\mathbb{R}\setminus{0}$

[cloud walrus]

Element118:

[bleak abyss]

$\mathbb{R}^{\times}$ is the notation

[cloud walrus]

Dami:

[fading wagon]

oooh

So that resolves (a) and (d)

[bleak abyss]

I actually had a similar problem on my algebra qual

[fading wagon]

haven't done a lot of this yet

[bleak abyss]

Part (d) in particular, you had to come up with this homomorphism to prove that the group is solvable (though with R replaced with F_p)

[fading wagon]

so what remains is showing (b), (c)

I'm guessing the best way to solve this is to try about 10 random examples of g^-1hg to see if they belong in the H, then when it suggests that H is normal, then we try to prove it

wait, maybe we can multiply it in general

hmm, I guess there's only doing that to prove a subgroup isn't normal

[tender mist]

Mm, also (c) is normal, one can send each matrix to a_11/a_22

[fringe nexus]

is the dihedral group completely determined by the relation sr=r^(-1)s ?

im trying to show that in a group with order 2n which has 2 distinct elements x,y of order 2, then it is isomorphic to D_2n

[mild laurel]

That statement isn't true

[fringe nexus]

wait

rs = sr^(-1)

[mild laurel]

For example if n is even, then you could take two copies of Z/2Z and a cyclic group with the rest

[fringe nexus]

crap i might have forgotten the question

is dummit and foote 1.6 q 24

But i dont have the book with me rn

[mild laurel]

The two elements x,y have to generate the grup

That's the condition you're missing

[fringe nexus]

oh

ok

well so basically if i map xy -> r

and say x -> s

and i show that the relation above is satisfied, then i am done right?

[mild laurel]

No not quite

The generators for D_{2n} are usually r and s as you note

But s is the rotation

So that has order n, not 2

[fringe nexus]

oh

i thought r is rotation

s is flip

[mild laurel]

Either way, whatever you use

[fringe nexus]

at least thats how we defined it in our class

okie

[smoky cypress]

What is the map v here? And what does commutative mean?

[mild laurel]

Commutative just means that $\overline{\eta} \circ \nu = \eta$

[cloud walrus]

Zopherus:

[smoky cypress]

ok

[mild laurel]

They tell you what nu is

[smoky cypress]

Yes

[smoky cypress]

Nvm I see it now the v is defined before the theorem

🤔

But that’s kinda weird

[fringe nexus]

can anyone give a hint for c?

Not sure where to start

[fringe nexus]

nvm i got it i forgot i proved part a already

[analog oracle]

A paper I'm reading mentions the "fact" that the # of involutions in a coset H = the order of the subgroup formed by the elements of H that are commutative with one of the involutions. Is this a standard fact I don't know? If anyone knows it I'd appreciate some references

[obtuse coral]

I'm watching a video https://www.youtube.com/watch?v=9V7VMd_sMmU

[cloud walrus]

myrkraverk:

[topaz solar]

There's an explicit exception for 0

0 neither needs nor has an inverse

[mild laurel]

This is for all fields

0 never has an inverse in any field

[obtuse coral]

Ok, fair enough.

[obtuse coral]

Oh I see, I think. In any case, the lecturer in the video didn't mention special treatment of zero, so it struck me as odd.

Right.

[magic owl]

wait what

no

R is a field

[topaz solar]

isn't a field by itself
R is a field

well, "by itself" might mean "without the addition, multiplication, inverse"

[magic owl]

Sure, but thats not R

[topaz solar]

fair

[magic owl]

R is defined by its field isomorphism class

or as a construction on Q

but you get the operations for free

In fact, you can't even reliably construct R up to isomorphism without them (it's undecidable in ZFC if you can do this)

[topaz solar]

what you mean when you said "R isn't a field by itself" i think

[magic owl]

You said "also, R isn't a field by itself, which is a common thing people forget"

this is at best misleading and at worst totally false

R is a field

it's not just a set

Q isn't just a set

When you talk about the set of reals you're "forgetting" the operations

but every construction of R is inherently based on a smaller field (from which it inherits operations) or as an isomorphism class in the category of fields

Ordered fields are fields

I agree R is an ordered field

You said something that was just wrong, and I pointed it out

there's not a debate

[topaz solar]

I mean, stating "R isn't a field by itself" i guess might be whether R refers to the set you get after it gets forgetful'd or to the field structure from construction

[bleak abyss]

Okay wait wtf

[topaz solar]

but, it's kinda constructed as a field so it's kinda dumb to say it's not a field "by itself"

but who the fuck makes the distinction between R and the underlying set in most things anyway

[bleak abyss]

That's my question

[next mango]

Proof by triviality

[bleak abyss]

Okay okay everyone just hold up

[magic owl]

You were stating something wrong as if it were fact

[bleak abyss]

Define "by itself"

[magic owl]

if "by itself" means "forget every quality that makes R, R" then sure

[bleak abyss]

Alright so your statement is worthless then

[magic owl]

lmfao

just admit you were mistaken

[bleak abyss]

Blitz you're just talking shit at this point

[topaz solar]

flexing your knowledge on me

[next mango]

apparently teachers are flexing on me all thetime

[bleak abyss]

No it was pointless once you jumped in

[mild laurel]

If you think your opinion is right, argue against his points and argue for your opinion

[bleak abyss]

Okay Blitz I'm actually just gonna have to make this a rule since this is a recurrent problem

If you're not involved in a conversation from the start in one of these channels, stay out

Especially in abstract algebra I have never seen you actually add you anything worthwhile to any of these conversations

[mild laurel]

Wait I'm not sure you understood what happened

I think you should scroll up and read all of it

[topaz solar]

might need a slight clause for clarifying things/correcting to the rule, should the conversation actually need it

[bleak abyss]

So in my mind the discussion started off with myrkraverk saying "The lecturer states R is a field"

[mild laurel]

If you think your position is defendable, then you should defend it

[bleak abyss]

And then you guys explained that 0 doesn't have an inverse

Blitz decided to say "R by itself isn't a field"

And refuses to clarify what "by itself" means

[bleak abyss]

I mean MaxJ already explained why R as only the set is a bad definition for "by itself"

[magic owl]

for the record you jumped into the conversation 15min after it ended with this pointless remark

[bleak abyss]

You think it wasn't pointless

And you're wrong

Nah not always just happens to be the case that you're stupid

[next mango]

almost surely this chat is trivial

[mild laurel]

Max made an argument that what you said was pointless

[bleak abyss]

Boom he's banned

[mild laurel]

And you haven't seemed to defend it so

[bleak abyss]

This was too much of a waste of time

[solemn hollow]

Wait so what does $x\in\mathbb R$ mean? Cuz if $\mathbb R$ isn't just a set...

[cloud walrus]

EpicGuy4227:

[solemn hollow]

Oh wait $g\in G$ and such is used in group theory a lot

[cloud walrus]

EpicGuy4227:

[magic owl]

lmao

[solemn hollow]

ah

[clear obsidian]

If I can ask for clarification about what happened there, so am I correct that what we mean by R is the unique (up to iso) ordered field with Q as a subfield (up to iso) with completeness (/supremum property), so just the set isn't actually R right? Because we need 2 operations to have a field (and thus 2 operations to have R) And I suspect its kind of pointless to talk about just the set unless we're comparing cardinalities or something like that. Is this all correct or am I misunderstanding what happened there?

I guess when we introduce |R| cardinality notation then what we mean is clear as opposed to something like (R, +, x) <-- Idk if we'd ever write it this way though as opposed to just R

[solemn hollow]

how about if it's like R is a set that is or isn't equipped with +, x, <, depending on context 😮

[clear obsidian]

Well I think the issue is how we define R though

[solar vessel]

R is weird

[clear obsidian]

I guess all I was saying is just that stuff like |R| will be clear what we mean in context, but I don't think you can separate R from it's operations

[solar vessel]

R (note we refer to the set here)

that should do it

[clear obsidian]

something like that yeah to clarify as needed

[solemn hollow]

yes veri compact

[clear obsidian]

because you can define the iso class, then once you do that you can "talk about" the sets I guess

[untold musk]

sorry guys for dumb question, but how can you prove that any dihedral group has 2n elements using permutations?

[wet terrace]

Dividing

[fading wagon]

@untold musk not sure what you are asking about there

are you trying to show that any dihedral group has an even number of elements?

[untold musk]

https://proofwiki.org/wiki/Order_of_Dihedral_Group
I’m trying to prove this but with permutations.

[fading wagon]

Well, just mention permutations and you should be fine

By definition ... consider the permutations of the vertices. Let's not do anything with them. By inspection, we see that...

[untold musk]

Permutation of vertices is n!, no?

[fading wagon]

well

yeah, but now we only want the permutations that preserve distances

[untold musk]

@fading wagon and how can i find which ones are preserving distance?

[fading wagon]

well, you need numbers which are adjacent stay adjacent

that's a necessary condition

[untold musk]

k

thanks for the help.

[fading wagon]

(it's also sufficient, but proof of that is probably not required)

[fallen bluff]

Exact sequence A -> B -> C -> 0, where f:A->B, g:B->C implies C iso to coker(f) riiight?

[final gulch]

yes that looks correct

since C = img g = B / ker g = B / img f =: coker f

using exactness at C, first isomorphism theorem, and exactness at B

[fallen bluff]

👍🏿

[thorn delta]

this problem.... i mean, would the correct argument be: let there be a homomorphism G to G' whose fibres partition G into P and apply first isomorphism theorem?

[tender mist]

@thorn delta I would suggest to prove that the equivalence relation associated to the partition P is compatible with the operation on G, this implies that the class of 1 is a normal subgroup, it being N.

[thorn delta]

so... I know to prove that N is normal you need to show that gN = Ng for all g in G. Im pretty sure Im missing something simple, but how can you come to any conclusions without knowing the other elements of N?

[tame bear]

take one of the partions, say H, and N. Then what can you say about the elements of HN

[thorn delta]

they are the products hn of h in H and n in N

1 is in N, h1 is in HN. Since cosets are disjoint, and we have h1 = 1h = h, HN = NH = H. Something doesn't seem right about this hmmm.

[tame bear]

so given any g in G, and g in H
gN = HN = H = NH = Ng

actually, that might not be right

[thorn delta]

why do you think its wrong?

[tame bear]

youd need to show that $HN = gN$

[cloud walrus]

Sigma:

[thorn delta]

HN = C another partition of G.
for g in H, and 1 in N, g1=g in HN = C, but g is in gN, so gN = C = HN.
Like that?

[tame bear]

youre only given that HN is contained in C
and you arent sure that gN is a member of the partition

[thorn delta]

huh tru tru.

[tame bear]

but we do know that NN = N
and so N is a subgroup

[thorn delta]

right... that makes sense

[bitter mauve]

can sb recommend me a book on galois theory pls? should contain exercises. the smaller it is the better.

[final gulch]

@bitter mauve someone answered you in the other channel

[bitter mauve]

oh thanks, @final gulch.
i deleted there cuz it was getting buried and this is more relevant channel

@scarlet estuary thanks. i needed one for a semester course on galois theory.

but that one's quite big

:(

[magic owl]

big books are good

idk what you hope to get out of a shorter book

[bitter mauve]

idk either

one more question. does anyone have a syllabus on galois theory taught at their school?

if so, can u pls give it to me

[clear obsidian]

@untold musk hey if you still need this result: I think you can prove that the dihedral group has 2n elements using induction

start with the triangle, listing it out we have the identity map, 2 other rotations, and 3 reflections

then assume the (regular) n-gon has 2n distinct permutations of veritces

then the (n+1)-gon has 1 extra rotation and 1 extra reflection, as in, it has 2n + 2 permutations of vertices

then the result holds by the PMI

[tame bear]

actually, i didnt even try until now
but yeah, the first iso might be how they want you to do it

[brisk granite]

um, quick question. Why is Z/nZ/Z/mZ = Z/(n/m)Z

I can see why their orders are the same but I don't see why the group on the right is cyclic

Actually, I think I see it now

The coset 1 + Z/mZ is the generator?

[fading wagon]

Okay Z/nZ is the integers mod n
Z/mZ is the integers mod m
m divides n so the quotient group exists...
Z/(n/m)Z is the integers mod n/m

[brisk granite]

yea, the last bit confuses me

1 + Z/mZ is a generator of Z/(n/m)Z, right?

[hot lake]

how is Z/mZ a subgroup of Z/nZ ?

[fading wagon]

Well, it's just isomorphic to a subgroup of Z/nZ if m divides n

I guess

and that subgroup would contain multiples of n/m

[hot lake]

yeah

[fading wagon]

So, elements in Z/nZ/Z/mZ would be k+<n/m>

and this clearly puts equivalence classes on elements k in Z/nZ by combing k which are the same mod n/m

[languid moss]

Whenever I try to check i always get the@wrong value. Can anyone help me with this

This a class called classical algebra . Need this class in order to take abstract algebra

I got it

I got it for base 8 but for base 2 I’m not sure what I’m doing wrong

[thorn delta]

Where is your work for binary? @languid moss

[languid moss]

Which one?

[thorn delta]

Idk you said converting to base 2 was the one you had trouble with

[languid moss]

Oh.

Hold on. I’m out now eating

[languid moss]

@thorn delta

,rotate -90

[languid moss]

,rotate -90

[thorn delta]

wait is this how you were checking you answer?

it should be 1*2^0 + 1^2^1 + 0*2^2 + 0^2^3 + ....

[languid moss]

Yea

I wasn't sure

DOes the power increase or decrease?

[thorn delta]

its just depends on the order you add. As numbers are written, the powers get smaller as you go from left to right

[languid moss]

Okay

So what's the common way of doing it?

So its always left to write?

[thorn delta]

402 = 4*10^2 + 0*10^1 + 2*10^0
^ its the same as how we do it in decimal.

[languid moss]

Okay. Thank you so much!

So for base two

which power do i start from?

[thorn delta]

np. one more thing. the last computation is the digit with the highest place value in your number. so if ur looking at the digits, it reads 11111000011

[fading wagon]

oh, you are doing it by calculating using Horner's rule

just that you aren't writing brackets?

[thorn delta]

wdym by "which power do i start from"? The order you add doesn't matter as long as the correct digit is associated with the correct place value.

personally i would just do 1*2^0 +1*2^1 * 0*2^2 + 0*2^3 +...

[languid moss]

Okay. because I was taught going backwards

like highest power to lowest power

[thorn delta]

yea, that makes more sense since we read numbers left to right, but I'm not counting all of the digits in that lol

[languid moss]

12^3 +12^2 * 02^1 + 02^0 +...

I will just keep practicing. Thanks anyways

[thorn delta]

np c:

[clear obsidian]

Naive question, but do we need to specify (fix) a subfield to take a Galois group because then it wouldn't be well defined otherwise?

[wind steeple]

yes

you need to know over which field you take your galois group

[fickle brook]

left to write

write != right

[wind steeple]

galois group is defined for field extensions, i.e. two fields which one is included into the other

[clear obsidian]

So find Pmin of the extension over the sufield, then we get the degree of Pmin number of auto.'s, so that's the cardianlity of Gal(L/K) right?

So I guess we either need a field F and a polynomial that's not split in F, or just F and some extension of F otherwise

[wind steeple]

that's the cardinality of the Galois group iff the extension is Galois

and what do you call "Pmin of the extension" ?

Pmin is for elements of your extension

[clear obsidian]

uhh yeah I should specify

if L is a finite extension of K, L = K(a1, ... an), then Pmin is going to be the product PI(x - ai) right?

[wind steeple]

so its degree is n ?

[clear obsidian]

yep

[wind steeple]

that's not necessarily a minimal polynomial

for example Q(sqrt(2),1)/Q, P_min = (X-sqrt(2))(X-1)

[clear obsidian]

has to be monic and of least degree

[wind steeple]

I don't understand what do you call P_min here

[clear obsidian]

here lemme rephrase

Let F be a field, and L = F(a1, ... an) where a1, ..., an are algebraic over F, then I think there should be a unique monic polynomial of least degree with a1, ... , an as its roots

that polynomial should be a minimum polynomial then right?

[wind steeple]

no it's false

Q(sqrt(2),1)/Q

[clear obsidian]

but 1 isn't algebraic over Q through right? what am I missing?

[wind steeple]

it is

X-1

cancels 1

[clear obsidian]

we just need it to be the root of some polynomial in Q[x]?

my bad

lemme just look up the definition

[clear obsidian]

Sorry for the messy work, but can someone check me on my automorphism of Fp (the black marker in the middle column there)?

[fading wagon]

if only the identity map?

[clear obsidian]

I got that any field automorphism of Fp has to be the identity map, just making sure my proof is correct

[fading wagon]

Well, 0 has to map to 0.
1*1=1, so, if it maps to x, x*x=x, x(x-1) divisible by p, so x is 1, 1 maps to 1.
And 2 is 1+1, 3 is 1+1+1, and so on

yeah I think the main idea is there

[clear obsidian]

alright ty

[wide blade]

How do you find the Domain and Range of y= -2x^2 + 8

[barren delta]

$a * b = c$ (where $c$ is some integer greater than $ab$) on $\mathbb{Z}^+$

[cloud walrus]

kickpuncher:

[barren delta]

I need to determine if this is a binary operation

I think I could just let a=1 and b=1

but it's sort of written like c is just arbitrarily larger than ab always?

I dunno I keep confusing myself

if c is arbitrarily larger than ab, then this is indeed closed and well defined since there always exists some e in Z+ where c = ab + e

[final gulch]

it's not really clear how you are defining the operation

[barren delta]

exactly

[final gulch]

you mean for each pair (a,b) you arbitrarily choose some c and declare a * b = c?

[barren delta]

it's very vague

I think for each (a,b), a*b = c > ab

¯_(ツ)_/¯

where * is a binary operation (not multiplication)

[final gulch]

I will assume by that you mean what I wrote

and you're asking what?

[barren delta]

ah, yeah I guess

I'm asking if I'm interpreting it correctly

[final gulch]

I would want to see the exact source you're getting this from before I try and answer that lol

[barren delta]

it's just a homework problem

just says determine if it's a binary operation

[final gulch]

as in a function Z^+ \times Z^+ -> Z^+?

[barren delta]

yes

[final gulch]

sure looks fine to me
the condition c > ab isn't even relevant

if for any (a,b) you choose a c in Z^+, then that is basically the definition of a binary operation

[barren delta]

I mean, isn't that worded awkwardly? Or is that just me?

lol

[final gulch]

well I didn't get to see how it is originally worded
but yes based on what I read it was vague

if you choose c independenly of a and b then the condition c > ab makes no sense since it is impossible to fulfill for all a, b
if you choose c depending on a and b then it's super generic and includes a whole family of binary operations

in both cases it's not an interesting problem

[barren delta]

¯_(ツ)_/¯

I do have to determine if it's commutative and/or associative

mmk nevermind I got it

[barren delta]

oy vey I suck at this

Suppose $*$ is associative on a set $S$

[cloud walrus]

kickpuncher:

[barren delta]

Let $H = {a\in S ,|, a * x = x* a ;\forall x\in S}$

[cloud walrus]

kickpuncher:

[barren delta]

Show $H$ is closed under $*$

[cloud walrus]

kickpuncher:

[barren delta]

I've been banging my head against the wall for 30 minutes and I can't get started on this

[wind steeple]

let a and b in S

prove that ab in S

prove that for all x, a*b*x = x*a*b

[barren delta]

yeah I'm not following

prove ab in S?

you mean the product ab? how am I allowed to do that if the product isn't defined?

[magic owl]

The product is just a*b

It exists by hypothesis

[cloud walrus]

kickpuncher:

[barren delta]

what do you mean it exists by hypothesis?

[magic owl]

Like

You know * exists as a binary operation

So of course a*b exists

Because you’re just applying * to (a,b)

[barren delta]

but zak wrote ab which makes it look like they're saying multiply a and b

[magic owl]

They are

Given an operation *

It is standard to write ab for a*b

[barren delta]

but....

multiplication isn't necessarily the operation we're talking about

[magic owl]

Ok yes

[barren delta]

I don't understand fundamentally what you're saying

[magic owl]

Uh

You have an operation *

We like to think of these operations generalizing the ones we know

So we just sorta think about it as abstract multiplication

So people often just write ab

Because once you get into longer stuff

Abc*d

Fuck

A*b*c*d

Is uglier than abcd

And because your operation is associative we don’t have to worry about parenthesis

[barren delta]

ok

[magic owl]

So do you get what Zak was saying now?

[wind steeple]

yes I wrote ab for a*b sry

[barren delta]

not exactly... seems like "let a,b in S, since * is associative a*b*c is unambiguous and so a*b in S"

for prove ab in S?

[magic owl]

Uh

You want to show H is closed under *

What does that mean

[barren delta]

for a, x in H, a*x = x*a in H

i think?

ffs

[magic owl]

Not quite

The commutativity condition you wrote isn’t needed

All you want is that

If x,y in H then xy is in H

[barren delta]

right

i guess I'm struggling to see how that relates to S

[magic owl]

Huh?

[barren delta]

for x in S and a in S

[magic owl]

H is a subset of S

[barren delta]

right

[magic owl]

In particular its a subset such that if x is in H

And a is in S

Then xa=ax

So what would it mean for xy to be in H

[barren delta]

y is in S?

[magic owl]

...no

What does it mean for something to be in H

[barren delta]

xy is also in S?

at this point i'm literally doubting everything

[magic owl]

Just read the definition of H

[thorn delta]

dont overthink it xd

[magic owl]

H is a subset of S

So anything in H will of course be in S

The point is what does it mean for elements of S to be in H

[barren delta]

all elements of S?

[magic owl]

For any element of S to be in H

[barren delta]

it seems like you're either trying to lead me to S = H or just that S is commutative

[magic owl]

No

No

No

[barren delta]

okay then I have no fucking clue where we're going

[magic owl]

I just want you to tell me what it means

If I say

X is in H

Lower case x

[barren delta]

then x is in S

lol

[magic owl]

What

I mean yes

But that’s not what I’m saying at all

[thorn delta]

then x commutes with everything in S

[magic owl]

Do you kike

Like*

Have experience with set notation?

Do you understand the definition of H?

[barren delta]

I mean it could mean a lot of things

[magic owl]

I’m looking for the obvious thing

Literally what does it mean for something to be in H

What’s the definition

Just re-write it

[barren delta]

it's just not clear what you're asking me because I don't understand what I need to glean from the definition of H

[magic owl]

Please just do it lmfao

[barren delta]

for x in H, x commutes with a in S

[magic owl]

Ok

Now we want to show xy is in H

What does that mean?

[barren delta]

are you asking what it means for xy to be in H? or are you asking what does it mean we want to show xy is in H

[magic owl]

The first

[barren delta]

x*y*a = a*x*y for a in S

[magic owl]

Good ok

So assume x and y are in H

Can you prove that xy is in H

Using what you just wrote

[barren delta]

using associativity, = a*(x*y) = (x*y)*a -> x*y in H

[magic owl]

Ok but you have to actually prove that

How do we know axy=xya

[barren delta]

I don't know how, other than that that's just how H is defined if x and y are in H

[magic owl]

No

If x and y are in H

You know x and y commute with all of the elements of S

You want to show that this means xy also commutes with all the elements of S

This is just a simple manipulation

Start with axy for a in S

And use what you know to get xya

[thorn delta]

Just because x and y are in H doesn't mean xy is in H. But this is where associativity comes into play.

hint: a(xy) = (ax)y
we know ax = xa

[barren delta]

ok jesus christ lol

i got it now

i usually can see where i need to go and get there eventually i don't know why i struggled so much with this

trivial ass problem just fucking trashed my confidence

[thorn delta]

well if it makes u feel any better i can relate

[barren delta]

x,y in H -> xa = ax and ya = ay
a x y = (a x) y = (x a) y = x (a y) = x (y a) = x y a -> xy in H

just to make sure

[thorn delta]

yea looks good to me

[barren delta]

good because it's too late to switch my goddamn major

thanks for your infinite patience @thorn delta @magic owl

i'm gonna go rethink my life choices

[thorn delta]

np and math is the best major u should have no regrets.

[barren delta]

lol I know

but I'm in my last semester of applied and felt like taking some pure maths for kicks

clearly i'm super rusty on some fundamental stuff

[barren delta]

The map $\varphi : \mathbb{Q} \rightarrow \mathbb{Q}$ defined by $\varphi(x) = 3x-1$ for $x\in\mathbb{Q}$ is bijective. Find $*$ such that $\langle \mathbb{Q}, +\rangle \simeq \langle \mathbb{Q}, * \rangle$

[cloud walrus]

kickpuncher:

[barren delta]

so I know I need $\varphi(x+y) = \varphi(x) ,*, \varphi(y)$

[cloud walrus]

kickpuncher:

[barren delta]

and I've been messing around with algebra for a while now and feel like I'm spinning my wheels

[final gulch]

ok so if you want that equation to hold, how should you define x * y?

[barren delta]

I mean I think I have something but it's weak af

I have x * y = x + y + 1

[final gulch]

that looks fine

[barren delta]

so phi(x+y) = 3(x+y)-1 = 3x + 3y - 1 = 3x + 3y -1 + 1 -1 .... lol but that feels dumb

[tame bear]

my book wants me to determine if x^4 -3x^2 +6x +1 is reducible or not in Q[x]
but none of the techniques theyve taught apply here
what do

[stone fulcrum]

,w roots of x^4 - 3x^2 + 6x + 1

[cloud walrus]

Display results online and refine query

Requested by Kaynex#7520

[tame bear]

like, all the problems before this are easy as hell
but theres just no approach here

[tame bear]

shift it by 2 and then reduce by mod 11
then Eisensteins criterion can be applied

possibly the worst problem of my life

wait, no
you cant even use Eisenstein outside of Z

[wind steeple]

did you tried by hand ?

like writing P(X)=Q(X)R(X)

Q and R of deg 2

because P hasn't roots in Q

[tame bear]

thats cheating

ill try it when i get home

[wind steeple]

why that's cheating ?

lmao

[tame bear]

$(x^2 + ax + b )(x^2 - ax + \frac{1}{b})\$
$a$ is a solution of $a^3 + 3a - 6\$
therefore a is not rational, and $x^4 - 3x^2 + 6x + 1\$
is not reducible in Q[x]

[cloud walrus]

Sigma:

[tame bear]

[wind steeple]

Yes

[raw star]

to show that ideal I of ring P is prime is it enough to prove that P/I is isomorphic to whole numbers?

[mild laurel]

No?

Also whole numbers? That's not a ring @raw star

[bleak abyss]

"of a ring P" I should ban you just for that notation

If you're thinking of showing the quotient is Z, the answer is that this suffices but there are tons of prime ideals for which the quotient isn't Z

So that's not a strategy you can lean on

[fading wagon]

It's that P/I is an integral domain.

@raw star

[hot folio]

i'm trying to prove this

$ E, F \ are \ subspace \ of \ a \ field \ K , \ f \ is \ the \ linear \ application \ of \ E \ in \ F \ show \ that \ f \ is \ bijective \ \implies \forall \ bases \ B \ of \ E \ , f(B) \ is \ a \ base \ of \ F $

[cloud walrus]

Katydid:

[magic owl]

what the fuck is that latex lol

just put the dollar signs around the math

You can do it like $x=5$ this

[hot folio]

oh lol

[cloud walrus]

MaxJ:

[hot folio]

f is bijective that means $ \forall y \in F \exists ! x \ \in E \ st f(x) = y \ (*) \ $ let B be a base for E, that means $ \forall x \in E \exists \alpha \in \mathbb{R} \ st \ \alpha B = x \ $ by substituting in * $\ \exists ! \alpha B \in E st \ f(\alpha B) = y $ so $ \alpha f(B) = y $ which kinda looks like a proof idea

[cloud walrus]

Katydid:

[magic owl]

whats E,F

[hot folio]

E, F are subspaces of a field K

this is what i came up with

tbh i was going to look at the solution

but i figured why not give it a try

[magic owl]

Sorry a field K or a vector space K?

[hot folio]

it says a field K

[magic owl]

I think there might be some mistranslation or something bc I'm struggling to understand what this is saying

So you have some field K, and subfields (not sure what a subspace of a field is) E,F

you have some bijection f from E to F

I'm not sure if f has any extra structure

looks like you are saying f is linear

[hot folio]

yea f is linear

[magic owl]

Can you define field for me? maybe this is a different usage than the standard

Because this all makes a lot more sense if we are dealing with R-Vector Spaces

[hot folio]

well in french it is called corp

from wikipedia

Étant donné un corps K, un espace vectoriel E sur K est un groupe commutatif (dont la loi est notée +) muni d'une action « compatible » de K (au sens de la définition ci-dessous). Les éléments de E sont appelés vecteurs, et les éléments de K des scalaires. 

i'm also confused by this because when i google "vector spaces in fields" i get nothing

i guess you're right

[magic owl]

So a field is like

an algebraic structure where we can multiply and add

and a vector space over a field

is a structure in which we can add, but we can only multiply by th eunderlying field

OHHH

ok

just put this through google translate and combined it with my very bad french

Yes ok K is a field

and E is a K-Vector Space

we are talking about two K-Vector spaces E,F I bet

and a linear bijection E->F

[hot folio]

yes

[magic owl]

Then we have what's called a basis, but I'm not sure that matches the def you wrote earlier

A basis is a set b1,b2,....

such that if e is in E

[hot folio]

prove that for all basis B of E, f(B) is a basis of F

so a base is a basis in english yea

[magic owl]

then $e=\sum_i \alpha_i b_i$ for some $\alpha_i \in K$

[wind steeple]

e = summ alpha_ib_i for some alpha in K yes

[cloud walrus]

MaxJ:

[magic owl]

there we go

[wind steeple]

and some i in a finite subset of I

I is the indices set of the (b_i) family

[magic owl]

yeah I'm trying to make sure the definition is the one katy is using

[wind steeple]

yes, he's only manipulating vector spaces

@hot folio you only need to prove, fixing B a basis, that f(B) is a basis

[magic owl]

Uh

[wind steeple]

so that f(B) is lineary independant

[hot folio]

how would i show that

[wind steeple]

and f(B) generates F

[magic owl]

You had the right idea earlier katy

[hot folio]

there's no expression for f

[wind steeple]

this is kinda trivial since f is bijective

[magic owl]

Zak no offense

but you aren't really helping

[wind steeple]

let me finish what I am doing wtf

[magic owl]

You literally interrupted me lmao

I'm not trying to have a terf war, if you want to help go for it

[wind steeple]

ok if you want help him, I understood what he wanted to show and I'm trying to be efficient but as you wish

[magic owl]

Sure, but making sure that both people are on the same page is important

And rushing in and calling things trivial is not

Anyway, did you agree with my basis definition Katy?

[wind steeple]

I didn't finish but ok ._.

[magic owl]

You had the right idea earlier on the proof approach

[hot folio]

yea i agree

[magic owl]

Ok, so

You want to show that if $y\in F$

[cloud walrus]

MaxJ:

[magic owl]

We can write $y$ as a linear combination of things in $f(B)$

[cloud walrus]

MaxJ:

[magic owl]

You also need to show that the set $f(B)$ is linearly independent

[cloud walrus]

MaxJ:

[magic owl]

Which one do you want to do first I guess

[hot folio]

write y as a linear combination of f(B)

then show f(B) is independent

that makes sense, no?

[magic owl]

Either order works jus preference

Do you have an idea or do you want a hint?

[hot folio]

a hint would be helpful

[magic owl]

Well, you know that there’s an x in E

Such that f(x)=y right?

[hot folio]

there's a unique x yes

[magic owl]

And that x can be expressed in the basis

Try writing x in the basis and then applying f

[hot folio]

B = summ x_i for x_i in E

[magic owl]

Not quite what we want

B is a collection of elements in E

If x is in E and B={b1,b2,...,bk}

Then $x=\sum_i a_i b_i$ for some a_i in K

[cloud walrus]

MaxJ:
Compile Error! Click the reaction for details. (You may edit your message)

[hot folio]

$f(x)=\sum_i a_i f(b_i) = a_1 f(b_1) + ... + a_k f(b_k) $

[cloud walrus]

Katydid:

[magic owl]

Yep!

So y=?

[hot folio]

that sum

o

[magic owl]

Perfect yeah

[hot folio]

which means it is a basis

[magic owl]

That’s half of it anyway

Do you know the easy way to prove linear independence?

[hot folio]

i usually use a matrix and try to get to a row echellon form

[magic owl]

B is a linearly independent set if $\sum \alpha_i b_i = 0$ implies all the $\alpha_i$ are 0

[cloud walrus]

MaxJ:

[magic owl]

Have you seen this before

Basically the b_is cant “cancel out”

[hot folio]

yeah i've seen that

[magic owl]

Ok perfect

Let’s assume that $\sum \alpha_i f(b_i)=0$

[cloud walrus]

MaxJ:

[magic owl]

We want to show that all the alphas are 0

But because f is bike to earth

It must have an inverse right?

Bijective*

Lmao bike to earth

[hot folio]

you got me on that one lmfao

yeah

also all bi's can't be 0

oh

[magic owl]

Well the b_is are a basis

But try taking f^-1 of what I wrote and see if you can show what we need

Or honestly

If you’re comfortable with bijection a

You know what the inverse of f(bi) is

Which leads us to?

[hot folio]

what i meant was also all f(bi's) can't be 0

wait lemme think a lil

[magic owl]

Remember the goal

Is to show that the alphas are 0

And we know the b_i are a basis in the first space

[hot folio]

how would i know what the inverse is ?

[magic owl]

Well what’s the definition of an inverse

Well, maybe this will be a better first step

Take the thing I wrote above

And apply f^-1 to both sides

[hot folio]

if y = f(x), then the inverse f-1(y) = x

[magic owl]

So what’s f^-1(f(bi))

[hot folio]

oof i couldn't process it

[magic owl]

Huh?

[hot folio]

x/ai ?

[magic owl]

So in general if f is a function

And f^-1 is its inverse

We get $f^{-1}(f(z))=z$ for all z

[cloud walrus]

MaxJ:

[magic owl]

Does that make sense?

[hot folio]

yes

$f(x) = \sum a_i f(b_i) $

[cloud walrus]

Katydid:

[hot folio]

$f^{-1}(f(x)) = \sum f^{-1}(a_i f(b_i)) = x $

[cloud walrus]

Katydid:

[magic owl]

Let's apply $f^{-1}$ to $\sum \alpha_i f(b_i)=0$

[cloud walrus]

MaxJ:

[hot folio]

sum a_i b_i = 0