groups-rings-fields

[strong valve]

what you gonna do next?

[chilly ocean]

Idk probably imma take a break for a week cos I’m rlly burnt out

[strong valve]

valid

[chilly ocean]

And then maybe topology if I’m up to it

Or catch up with school I haven’t attended my classes since like March

[hidden haven]

Based

[strong valve]

based

[carmine fossil]

Actually don't catch up

It's not worth it

[strong valve]

based and schoolpilled

[rustic crown]

hard words

[carmine fossil]

Ok,Try this question:Prove that in a field,the multiplicative group is never iso to the additive group

[hidden haven]

iso 😳

[carmine fossil]

It's not too difficult

[strong valve]

invited

[royal matrix]

h

[strong valve]

HISO

[chilly ocean]

Well if F is finite both have different cardinalities so ofc there can’t be an isomorphism between them

If it’s infinite then Ψ(-1)=0 but Ψ(1)=0 so Ψ is not an isomorphism

And if 1=-1 then simply Ψ^-1(1)=Ψ^-1(-1)=0 so Ψ is not an isomorphism

[chilly ocean]

why not in this case?

Cos u have two elements mapped to one but an isomorphism is one one

nah

if 1=-1 then its the same element

Wait o you’re right

[carmine fossil]

Yea,That case isn't too tough tho

[chilly ocean]

How do we go about this part

I’m lost

[carmine fossil]

Hint:Look at the isomorphism from additive group to multiplicative

Phi(1+1)=1=phi(1) phi(1)

[chilly ocean]

So F has characteristic 2 and so it isn’t infinite?

[carmine fossil]

Now phi(0)=1

If we let phi(1)=x ,that implies x^2-1=0

[strong valve]

F_2[x] has char 2

[carmine fossil]

Which implies x=1 or x=-1=1

i.e,phi(1)=1

[chilly ocean]

Ohhh nice

[hidden haven]

F_2(x), you want a field

[chilly ocean]

Well clearly I have a lot to learn still

[strong valve]

right

[hidden haven]

How did you get psi(-1) = 0?

Oh wait got it

[chilly ocean]

That’s assuming 1=-1

[hidden haven]

For characteristic not 2

[chilly ocean]

Wait 1 is not equal to -1

[hidden haven]

Yeah

[strong valve]

i am getting cofusedlmao

is 1 = -1 or not

[chilly ocean]

There are 2 cases

In one of them 1=-1 and in the other one it’s not

[strong valve]

ok which one did you consider first?

[chilly ocean]

1 isn’t equal to -1

[hidden haven]

≠, in which case psi((-1)²) = 2psi(-1)

[chilly ocean]

Then I got stuck for 1=-1

[hidden haven]

= psi (1) = 0

[chilly ocean]

Yea that’s what I did

[chilly ocean]

I didn’t get how phi(1+1)=1

[hidden haven]

1+1=0

Since characteristic 2

[hidden haven]

Was responding to this

[chilly ocean]

Damn this is beautiful

I’m too shabby at math

[molten silo]

Is every Euclidean domain a principle ideal domain

[chilly ocean]

Maybe

[carmine fossil]

Prove it

[molten silo]

i forgot

[carmine fossil]

Hint:Natural numbers are well ordered

[molten silo]

so it is true

[carmine fossil]

Why so

[molten silo]

i went over this proof a week ago and cant remember a thing

[rustic crown]

what's a pid?

[small bison]

i think it was like look at the element that minimizes the euclidean function

[carmine fossil]

Yes

[rustic crown]

think about what you exactly want to show, and how being euclidean helps you

[final pasture]

I was going to answer to this question, again

[rustic crown]

lol

that's very cute lol

[molten silo]

ez

[carmine fossil]

You should actually study that proof

[molten silo]

its simple when i look at it

But say im in a test, i wont remember it

[solemn rain]

solve x^2+x+1 over GF(25)

i dont know where to start

im not good at factoring lmao]

[final pasture]

You can literally bruteforce

there's like 25 elements to test

and that's fairly fast

@solemn rain

[solemn rain]

is there another way?

i think x^2+x+1 is something simpler in F_5[x]/(x^2+2) no?

@final pasture

[final pasture]

(it's F_5[X])
no it's not ?

[solemn rain]

yea sorry

isnt this modulo a degree 2 polynomial?

shouldnt it be something simpler or is that just wrong

[delicate hawk]

is the field of fractions of a ring $R$ just the set of elements written in the form $\frac{a}{b}$ such that both $a, b \in R$ but $b \neq 0$ obv

[cloud walrus]

Fiwam

[hidden haven]

Yeah but a/b = c/d if ad = bc (equivalence relation built into the a/b notation)

[lunar coyote]

would this be enough to show it

[nova plank]

Uh

They mean a has an inverse, not a = 1

[lunar coyote]

😮

[nova plank]

And you are using equals signs inappropriately to connect your statements

[lunar coyote]

i thouht unit means 1

[nova plank]

Nope

Sometimes, but usually unity is 1

And a unit is invertible

[chilly ocean]

can somebody help me prove that a group of order 2^2 *3 * 11 is not simple?

given everything that I've tried so far, I feel like my best shot is somehow proving that the normalizers of its Sylow 3-subgroups can't have index 22, which seems a little farfetched

[rustic crown]

i think direct sylow works

look at the sylow 11-subgroup, if its normal then we're done else there are 12 sylow 11-subgroups
this accounts for 120 elements of order 11

12 remaining elements

now if sylow 3-subgroup is normal, then we're done, else there are atleast 4 sylow 3-subgroups accounting for atleast 8 elements of order 3

This leave precisely enough room for just 1 sylow 2-subgroup hence that is normal.

i just proved that at least one of the sylow subgroups is normal, hence the group isn't simple.

[chilly ocean]

can we also rule out the possibility of 4 3-sylow subgroups because, assuming simplicity, the group would somehow fit into S4?

[rustic crown]

1, 4, 22 are the divisors of 44 which are 1 mod 3

so you'll need to use the argument with 11 to rule out first that there could be 22 sylow 3-subgroups

[chilly ocean]

so I'm back at square one

[rustic crown]

your argument would then say either the sylow 11-subgroup is normal or the sylow-3 subgroup is normal.

wait actually no

[rustic crown]

because the kernel could be something arbitrary.

are you trying to see exactly which sylow subgroup is normal?

[chilly ocean]

I'm only trying to prove that it's not simple

[rustic crown]

yea so do you not like the above argument by counting elements?

[chilly ocean]

I do

but I've never seen it before even if it seems very natural

[rustic crown]

maybe i'll rephrase

Assume the group is simple, then sylow 11-subgroup isn't normal and so isn't sylow 3-subgruop. Number of sylow 3-subgroups can't be 4.

This leaves us with 22 sylow 3-subgroups and 12 sylow-11 subgroups

well this is all i can think of right now 😛 that then there are 44 elements of order 3 and 120 elements of order 11 and we just exceeded the size of groups

[chilly ocean]

I got it

[rustic crown]

i thought in this direction because once i looked at the number of sylow 11-subgroup, i got it should be 12, and that's quite a lot

when you get like a small number you think of stuffing in S_{small number} like you said

since 12 was huge, it practically filled up my entire group. So intuitively it was not hard to see that room for other sylow groups is very contrained. And indeed this was the case.

[chilly ocean]

alright, thank you for the explanation. I'm the one at fault here since I didn't know these counting arguments before

[old lava]

Sylow exercises are a painful amount of counting arguments

[chilly ocean]

and somehow we've never seen one in class

[rustic crown]

i first saw this kinda argument for a group of order 12. So if the sylow 3 subgroup isn't normal then we get 8 elements leaving only enough elements for one sylow 2-subgroup.

[chilly ocean]

I was struggling with groups of order 12 earlier today lol

[old lava]

1. counting arguments
2. use sylow subgroups to construct a subgroup of small index k, which implies a homomorphism from G to S_k, which has kernel in G (which is a normal subgroup). You can often find contradictions if you assume G is simple (i.e., the kernel is trivial)
3. can do the above and identify G as a subgroup of some symmetric group S_k if G is simple, and then show that either S_k contains no subgroup of order |G|, or you can construct a subgroup of G that can't possibly be in S_k
4. do some fuckery with intersections of sylow subgroups
5. can look at normalizers and centralizers of sylow subgroups

these are usual methods of finding contradictions

using sylow

to show a group isn't simple

[rustic crown]

4. lol

[old lava]

it just takes too much time to explain over discord

makes more sense if you just look at examples

[lunar coyote]

is this even possible?

[old lava]

yes

[rustic crown]

like 1 and -1 divide every integer

[lunar coyote]

oh ok

[old lava]

suppose u is a unit in R. Then, u v = 1 for some v. Let r be arbitrary. Then, r = 1 r = (u v) r = u (v r), so u divides r

it's just a one liner

[unique berry]

Isn't that like the definition of a unit?

Hold on lemme check

[rustic crown]

yep, a unit divides 1 and hence anything

(but you need to be careful, if the ring isn't commutative, cause both left and right units satisfy that)

[lunar coyote]

attempted this question and the solution is something completely different

they ended up doing an expansion instead :/

[nova plank]

Your solution could be right too

[lunar coyote]

idk if it's right

[chilly ocean]

b^2=0 doesn't imply b=0, eg consider z mod n for n divisible by a square

[lunar coyote]

bruh

so what do i do about that b^2 =0

[nova plank]

You seem to think the problems are totally trivial, when it was a*b, you just made b = 1 and now you make b = p 😂

I thought you had a solution

[lunar coyote]

well i'm kinda retarded

so i'm an exception

[nova plank]

Don't say that

Just don't say things you can't prove

There is no reason b^2 =0 would imply b = 0 except that it feels right.

If you are working with unfamiliar structures, your intuition can be misleading

[lunar coyote]

feels right in the integer field 🙂

[nova plank]

But also, if they meant b = 0, they would have just said that

[lunar coyote]

true

idk if the questions from contemporary abstract algebra are too hard for me

[nova plank]

You just have to get used to it

Think about what you want

You want to multiply a+b by something that gives you 1

You know a is invertible, so you can handle an a by multiply by a inverse

But you need to get rid of the b's first

So can you multiply a+b by something that will get rid of the b's completely using b^2 = 0

[lunar coyote]

was thinking b^2 mod p = b^2 * q + r

actually nvm

that wouldn't get rid of the b^2

[nova plank]

Remember, b^2 is 0. So if the only b's you have is b^2, that would get rid of it since you could replace it with 0.

[lunar coyote]

ah

so i can multiply by b then

[nova plank]

Well, then the first term would be ab

So you would still have a b there

Sorry, watching tv

[lunar coyote]

b/b

[nova plank]

Will check back in when I'm done before I sleep

[lunar coyote]

can't i just multiply by b/b

and get

(ab+b^2)/b = (ab+0)/b = a

[old lava]

division isn't exactly an operation in rings (with more work)

[nova plank]

What is b/b?

You don't know if b has an inverse

[lunar coyote]

oh damn

felt good for a second there

[nova plank]

If it did, b * b inverse would be 1. So that wouldn't do anything.

[old lava]

hint: use difference of squares

[chilly ocean]

Ah, I just solved it, neat problem!

hint: iterate through every possible polynomial dimension (in a and b) until you find one that works. if one exists you'll eventually find it

My original thought was to try to use x is a unit iff x^2 is a unit, and try squaring (a+b)

But this didn't seem to help, but....

[nova plank]

Yeah, objective 1 is getting rid of b and there is really only one way to do that

[old lava]

my thought process was

1. think of concepts that have a b^2 term isolated from a term with a
2. need an expression that divides a + b
3. thought of squares of terms, and difference of squares
4. made the connection with difference of squares

and then it's immediate

[final pasture]

my thought process =>

1. That's quite close to a similar problem I've seen about nilpotent endomorphism
2. The sol generalizes

but not sure that helps

[lunar coyote]

is slader.com a reliable source to get answers?

[nova plank]

I don't recommend going to look for answers

Try yourself

If you get stuck, ask here for hints

Sometimes solutions can be helpful, but most of the time, it's just not as effective as struggling until you get it

[old lava]

slader solutions are often also not exactly correct

for non-calculus things

I've seen

I've never used slader for calc, although I know the stewart solutions are extremely popular

on slader

it's also decent for some intro engineering textbooks from what I've heard

if you have the clear isomorphism

than all you need to do is show it's surjective and injective

[viscid pewter]

show it satisfies the presentation, show the orders are the same?

[old lava]

I mean you can just show everything directly

[wild sapphire]

hmm so @old lava that makes total sense, but like, the injectivity is so clear that I feel like I'm just saying "well look at it"

[old lava]

it's not particularly difficult to show that the kernel is trivial

or that the map is surjective

it's almost by definition

once you've constructed the homomorphism

[wild sapphire]

Yeah

so I have this mapping, and like basically by the property of group powers it has to be injective. I feel like that's all there is to say

[rustic crown]

by the way you have defined, it is very clear that this is a bijection. but why is it a group hom?

[wild sapphire]

Hmm yeah I need to show that it preserves the respective group operations, like the structure. That's where I'm stuck because I can't think of how to do it other then by cases/examples

It's pretty easy to evaluate for say, D4 y'know. I just need to almost generalize an example further

[rustic crown]

i would have done it the way Kaisheng21 suggested

[wild sapphire]

I don't really understand what he meant by presentation.

Maybe I haven't learned that

[rustic crown]

Dn = <s, r | s^2 = r^n = srsr = e>

[wild sapphire]

Oh! Yeah that makes sense

[rustic crown]

so the idea is Dn is characterised by a rotation and a flip. So looking at the analogue for them in the matrix group, we have S = [-1, 0; 0, 1] and R = [1, 1; 0, 1].
its just basic checking that S^2 = R^n = (SR)^2 = [1,0;0,1]

can there be more relations between these though?

the answer is no, because if there were more relations then we would have less than 2n elements!

[wild sapphire]

aren't all other relations characterized by the presentation?

I don't quite understand the last message, but I fully understand how the proof should go now

:)

[rustic crown]

okay so look at the trivial group and take s = r = id
indeed s^2 = r^n = srsr = 1

but does this mean that this is Dn?

[wild sapphire]

oh hmm

[final pasture]

det is saying that if there was other relations not deducible from the ones we already know, then there would be some elements that are in Dn but that aren't in the group that answers the relations enhanced with the new relations

[wild sapphire]

its just like such a trivial Dn that it can't really be Dn

ah yeah I think I get it

[rustic crown]

if you want to write this formally, search for how presentation is actually defined.

[wild sapphire]

I think I was thinking that he said we would need less than 2n elements to have an extra relation. Which was confusing.

When what he was really saying was that having an extra relation would imply that we started with under 2n elements, which is false

[solemn rain]

is x=1 a solution for x^2+x+1 over GF(25)?

what i did was umm

GF(25) = F_5[x]/(x^2+2)

x^2 +2 is 0 here so

x^2 is -2

so x^2+x+1 is -2+x+1 = x-1 = x+4

so x=-4 which is 1?

am i bullshitting here or what

[chilly ocean]

Can't you just do 1^2+1+1=3 != 0?

[solemn rain]

ye ai did that but idk why what i did is wrong

[chilly ocean]

F5 is a subfield of GF(25), so there is not even anything crazy

[chilly ocean]

I think this should be using a variable other than x

[solemn rain]

why

[chilly ocean]

Because you already used x as a variable in the polynomial

When we write GF(25)=F5[y]/(y^2+2), y is a (representative of) an element of GF(25)

[solemn rain]

yea

[chilly ocean]

Basically you are conflating two different variables, I think

[solemn rain]

okay so

anyways

how do i solve x^2+x+1 over this then

[next obsidian]

Wtf is GF(25)

[solemn rain]

galois field of 25 elements

[chilly ocean]

25 yo girl friend

[next obsidian]

???

[solemn rain]

?/

[next obsidian]

Is this just the field of 25 elements

[solemn rain]

yes

[next obsidian]

Lol okay

But no 1 is not a root of that

[solemn rain]

okay

i see that

now what should i do

[chilly ocean]

How to factor it: try every element of GF(25)

[solemn rain]

thats alot

25 is alot

is there another general way

[chilly ocean]

Maybe rational root theorem?

[next obsidian]

I mean how are you even gonna write down the elements of that field lol

[solemn rain]

i can do that

actually

[next obsidian]

Besides the obvious 5 elements

[chilly ocean]

Ah wait, does RRT work in arbitrary fields?

[solemn rain]

idk

[chilly ocean]

Mo2men already introduced y variable for it

Just write down 0, 1, 2, 3, 4, y, y+1, etc...

[solemn rain]

yea

anyways

how do i factor 😦

[chilly ocean]

I could think 25 is not that big of a number

[solemn rain]

okay say im in GF(64)

now what

is there a more general way

to solve these quadratics

should be easy but i just dk how ig

wait

lmao

i got it

quadratic formula

this isnt char 2 so yea

[chilly ocean]

Ah right

[solemn rain]

okay so for another problem

how do i factor x^12-1 over GF(25)

[chilly ocean]

I think if you factor over eg Z you get a polynomial that has all linear or quadratic factors except one 4th degree thing right?

So you can do quadratic formula on those quadratics and then worry about just the 4th degree one?

[solemn rain]

okay

and just for checking

x^2+x+1 has no solutions over GF(25) right?

anyone?

[chilly ocean]

,w factor x^2+x+1 over GF(25)

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by 88ddda#7872

[solemn rain]

so/

?

[chilly ocean]

So it looks like it does have a root

[solemn rain]

i just used quadratic formula

it has -1+sqsrt(2)/2

and -1-sqrt(2)/2

those wouldnt be in F_5[x]/(x^2+2) no?

[chilly ocean]

i don't know, does F_5[x]/(x^2+2) have a sqrt of 2?

[solemn rain]

idk

how do i know

x^2?

[chilly ocean]

i guess x is a sqrt of 3, x^2 = -2 = 3

[solemn rain]

wait

x^2-2 is irreudicible in F_5[x] right?

should be

[wild sapphire]

can two normal subgroups overlap?

non-trivially i mean

[chilly ocean]

Sure, consider abelian groups

[wild sapphire]

hmm what if it's non-abeliean?

[solemn rain]

alot of groups are the product of their normal subgroups

[wild sapphire]

im trying to show that the product of two normal subgroup's elements always commute, but I only know how to show this if they don't overlap 🤔 So I was wondering if there were any facts about overlapping normal subgroups i am missing

[solemn rain]

if they overlap non trivially they do not have to commute

the proof actually uses the assumption that they intersect at the identity

@wild sapphire

[wild sapphire]

Oh, so my question is wrong? It doesn't specify the non-trivial overlap

[solemn rain]

ig

can u show me the question

in ur textbook or source

[wild sapphire]

Suppose H and K are normal subgroups of group G, show that $\forall h \in H$ and k \in K, hk=kh$

like that

[cloud walrus]

darkninja175
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[wild sapphire]

the forall is for both, sorry

[solemn rain]

take G = D_2(4), N = {e,r,r^2,r^3} and K = {e,r^2,s,r^2s} , and take h = r^2 , k = s --> r^2s != sr^2

verify N and K are normal subgroups

@wild sapphire

[wild sapphire]

Hmm interesting

[chilly ocean]

(x-1)(x^2+x+1)(x+1)(x^2-x+1)(x^2+1)(x^4-x^2+1) that is the factorization of x^12-1 over Z right? @solemn rain

[solemn rain]

yes

i agree

[chilly ocean]

So the only tricky one is the last factor that's degree 4? (I'm assuming you will figure out the degree 2 ones)

[solemn rain]

why would it be the only tricky one?

i just wanna understnad

[chilly ocean]

Because you have quadratic formula

[solemn rain]

yea i can still factor it further ok

[chilly ocean]

And presumably you can figure out which square roots exist in GF(25)

[solemn rain]

how

[chilly ocean]

Well, it is only sqrt(2) and sqrt(3) that must be considered. And sqrt(3) was already shown to exist

[solemn rain]

sqrt(2) shouldnt exist

[chilly ocean]

And apparently wolfram alpha already told us sqrt(2) must be in there too

[solemn rain]

why

[chilly ocean]

Because x^2+x+1 factorable is equivalent to sqrt(2) in GF(25)

[solemn rain]

yea lmao

what could it be

ik it isnt in F_5

right/

[chilly ocean]

Yeah, it is not in F5

Hm

[solemn rain]

omg i hate math

[chilly ocean]

I think 2x and 3x are both square roots of 2

Yeah

So all quadratics with coefficients in F5 can be factored in GF25

[solemn rain]

whats sqrt(2) in GF(25)

[chilly ocean]

I just wrote it, 2y and 3y in F5[y]/(y^2+2)

[solemn rain]

so

the original problem i had

x^2+x+1

has solutions what

(1+2x)/2

[chilly ocean]

2 +/- 3sqrt(2), where sqrt(2)=2y, say

Yeah

Er

[solemn rain]

how do u get 4 solutions

in a field

for a quadratic

[chilly ocean]

There are only 2 solutions, 2y and 3y are negative of each other

[solemn rain]

oh yea

magic lmlao

[somber harbor]

can someone give turn me in the right direction on this question?

[solemn rain]

can you describe for me this group

@somber harbor

how is it defined for u

[somber harbor]

(Z/mZ)* = {a in Z/mZ s.t gcd(a, m)=1}

[solemn rain]

okayy

so ur this specific group

u want the count the elements a such that gcd(a,35) = 1

[somber harbor]

i just go through

[solemn rain]

u can do that yes

[somber harbor]

all 35 numbers

brute force

or is there a better way?

[solemn rain]

well

[somber harbor]

i guess that is more of my question

[solemn rain]

okay

u can try to have fun ig

lets define a function

call it phi(x) or whatever

and let it be this specific number actuall

like

define phi(x) to be the number of numbers coprime with x

okay?

[somber harbor]

ok

i am following

[solemn rain]

okay so

lets do some digging

so lets for example ask

what happens if x is prime

is it easy to find phi(x)?

[somber harbor]

if x is prime then ur only co prime with other primes

[solemn rain]

if x is prime

how many numbers are u coprime with

tyt

( try examples )

please no spoiler

haha

let him/her try on his/her own

[hidden haven]

oh sorry I didn't read the last few messages

[solemn rain]

its ok

@somber harbor anything?

[somber harbor]

i am uh

ok so let say x is 12 and 4

[solemn rain]

wdym x is 12 and 4

x is only 1 number

u want to find how many numbers are coprime with this number

this is phi(x) as we defined it

and now im asking u to think about phi(p) if p is prime

[somber harbor]

ok

[solemn rain]

what do you think phi(p) would be

[somber harbor]

i am sorry

hold on

[solemn rain]

for what

tyt

hint

if u struggle to do something for the general case

try a small example

saay p is 7 or any small prime

how many numbers are coprime to that 7

and keep going with more examples untill u see something u can generalize later

so what do u think is phi(7)

[somber harbor]

@solemn rain

6

[solemn rain]

yea

okay

what about phi(11)

[somber harbor]

10

[solemn rain]

okayy

do u see something here

and if so do u see why

[somber harbor]

phi(x) is x-1 if p is prime

[solemn rain]

iff

okay

cool

so now

thats ur first theorem ig

so

can u think of phi(pq)

where p and q are primes

( hint : think of the elements that actually have a common divisor with pq )

[somber harbor]

so phi(2 and 3)

[solemn rain]

for example

[wild sapphire]

@solemn rain did you say above that sr^2 != r^2s, because that's not true

[somber harbor]

6 ur only co prime with 2

2 numbers

phi 10 is 4

2 and 5

so phi(pq) is phi(p)*phi(q)

[solemn rain]

so sorry , i meant rs

[wild sapphire]

yeah, that one works 👍

[solemn rain]

phi(q)*phi(q)?

[solemn rain]

yea so sorry i fucked up

[somber harbor]

i meant phip * phiq

[solemn rain]

do you mean phi(p)phi(q)?

yea

okay so can u tell me why this works

convince me

with a more 'general' argument

rather than showing me a couple examples

[somber harbor]

ok

[solemn rain]

and also can u smplify this more

whats phi(p) when p is prime again?

[somber harbor]

p-1

[solemn rain]

okay so

phi(p)phi(q) = (p-1)(q-1) okay cool

now can u convince me again why phi(pq) equals phi(p)phi(q) where p and q are primes

[somber harbor]

ii think i can convince

you but let me think about it in english

[solemn rain]

tyt

[somber harbor]

um

[vital fossil]

how can i express this kernel as a Z6 module, like Z6[ ??? ], in otherwords, how do i find the basis vectors for this module?

[somber harbor]

so we are checking the the gcd of (pq, q) which is q, but the gcd of (pq, of lets saay 2q) is also q, and this also works for (pq, q)

[solemn rain]

so like

what do u think

are the numbers

[somber harbor]

we figured out that p-1 and q-1 but i don;t know why it is multiplicative

[solemn rain]

that do not have gcd of 1

okay

think about the what numbers that are not coprime with pq

what are they

[somber harbor]

they are numbers that are multiples of p or q

[solemn rain]

okay

so we want multiples of p or q that are less than pq

right?

[somber harbor]

yes

[solemn rain]

how many of those?

so

how many multiples of p that are less than pq

and how many multiples of q that are less than pq

[somber harbor]

q and p

[solemn rain]

pq is not less than pq

so no not quite but close

[somber harbor]

pq is not less than pq?

if i want the multiples of p less than pq i just divide pq by p which gets us q

O

but its less than

so we dont include pq

[solemn rain]

yea

[somber harbor]

so its q-1

[solemn rain]

yes

and how many multiples of q that are less than pq

[somber harbor]

we get p-1

[solemn rain]

exactly

so how many numbers that are NOT coprime with pq?

[somber harbor]

p-1)(q-1

[solemn rain]

no

why multiply

[somber harbor]

wait

[solemn rain]

we agreed the multiples of p less than pq are THOSE not coprime with pq

and we agreed the multiples of q less than pq are those not coprime with pq

so this one has q-1 and this one has p-1

so how many numbers not coprime with pq all around

[somber harbor]

um

[solemn rain]

so for example

3*5

the number of elements not coprime with 3*5 is the number of multiples of 3 and thenumber of multiples of 5

less than 3*5

[somber harbor]

we ar eadding it?

[solemn rain]

yup

its just both

so now

we now the number of numbers NOT coprime with pq is (p-1)+(q-1)

how many numbers all around less than pq

[somber harbor]

pq-1

[solemn rain]

how many numbers not coprime with pq

[somber harbor]

p-1 + q-1

[solemn rain]

yes which is p+q-2

so

we have pq-1 positive integers

we know p+q-2 of them is not coprime

how many are coprime

?

[somber harbor]

we take the difference?

[solemn rain]

yes

which is what

[somber harbor]

pq-1 -(p+q-2)

[solemn rain]

simplify this

[somber harbor]

pq-1-p-q+2

pq-p-q+1

[solemn rain]

okay

thats it

if u look at this long enough

u see its just (p-1)(q-1)

which is phi(p)phi(q)

thats

so now

[somber harbor]

i see

ok

[solemn rain]

phi(x) as we defined it is just the order of |Z/nZ|^x right

so now

can u answer ur question much more easily now

[somber harbor]

yea

[solemn rain]

the order would be just phi(35) right

[somber harbor]

so its 5*7-5-7+1

[solemn rain]

can u simplify this 35

or (5-1)(7-1) but yea

[somber harbor]

24

[solemn rain]

exactly

good job

thats it

and btw

[somber harbor]

i see

[solemn rain]

this phi(x) is actually a p famous funtion that has more cool properties

called the euler totient function

u can look it up if u want

[somber harbor]

I seee

I seee

wow. thank you u just walked me through a pretty big problem in my head

thank you!

[solemn rain]

no problem!

[somber harbor]

you were so patient

jesus. thank you

[solemn rain]

idk about that

u should see me getting helped lmao

[somber harbor]

[vital walrus]

Hello everyone, I wanted to ask if my argument for this proposition seemed sound. The claim is:
“Show that the unity element of in a subfield of a field must be the unity of the whole field”

Proof:
Let F be a field, and let K be a subfield. Denote 1_F the unity of the field F, and denote 1_K the unity of the subfield K. Since both 1_F and 1_K are in the field F, it follows that
1_F * 1_K= 1_K = 1_K^2
Using our additive property of the ring, we have
1_F*1_K - 1_K^2 = 0; then we factor out 1_K by the distribute laws of the ring, hence
1_K ( 1_F - 1_K ) = 0 ; so either 1_K = 0 or 1_K = 1_F ; but 1_K cannot be 0, so the unity of the subfield K must be equal to the unity of the field F

[solemn rain]

what is 1_K^2

@vital walrus

if you mean (1_K)^2

then idk how

unless u assumed 1_F = 1_K

which is what u want to prove in the first case

im sorry but i have to say

this is not a proof

[vital walrus]

Oh yes sorry; (1_K)^2 ; because 1_K=(1_K)^2 by idempotence of unity

[solemn rain]

okaay

yea thats on me lmao

gj

this is a not a proof

haha

[vital walrus]

Okay thank you, I wasn’t sure if I misstepped somewhere along the way 😅

[solemn rain]

np

wait

hold on 😠

how is (1_K) * (1_K) = (1_K^2)

again

?

[chilly ocean]

that is how squares are defined

lol

[solemn rain]

yea was just seeing if he/she is focusing or not ..

[chilly ocean]

lol

i believe you

[solemn rain]

how is 1_K = 1_K^2 again

@vital walrus

[solemn rain]

okay

[hidden haven]

Bruh

I keep thinking that you're asking the question

Sorry lol

[solemn rain]

hahaah np lmao

[vital walrus]

Ooooh yes it’s the identity 😅

[somber harbor]

@solemn rain

i am once again requesting guidance

[junior moth]

order(2) divides phi(35)

so you just have to check for factors of phi(35)

[chilly ocean]

Yea cos Lagrange’s theorem

[somber harbor]

so i got that phi(35) is 24

so the order(2) have to divide 24

so order of 2 is just 24?

cuz only 24 in the 35 intergers divides 24

[junior moth]

check 2^12

[chilly ocean]

No it can be any of the factors of 24 check for them all

[somber harbor]

oh

oh

righ tright

wait does lagranges thm work here? 35 is not prime

[chilly ocean]

I don’t think Z/35Z is a finite group so we can’t apply Lagrange’s theorem nor can we apply the corollary cos 35 isn’t prime

I suppose it has 0 order cos 2^n never loops back around cos Z/35Z is infinite

[somber harbor]

ok i just listed all 24 factors of 35

and brute forced for the answer

[chilly ocean]

what

[somber harbor]

i got 2^12=1 mod 35

[valid bridge]

How do you show that finite order elements of the free product of two nontrivial groups G and H are just conjugates of finite order elements in G and H

[carmine fossil]

Find 25 distinct elements in (Z/35Z)^* in that case

[valid bridge]

nvm i got it

[wild sapphire]

Is it possible for two abelian normal subgroups to have a non-trivial intersection? I feel like it's not

[rustic crown]

If you have an abelian group, then just pick two non-trivial subgroups, one contained in the other

Another example is D4, we have the 2 subgroups {1, r, r², r³} and {1, f, r², fr²} (r is the rotation, f is the flip)

[vital walrus]

Does the inverse of an element of a group always form the same cyclic subgroup? For example if a is in G, is it always the case that <a> = <(a)^-1> ?

[rustic crown]

how do you define <a>?

[vital walrus]

Oh yes sorry;
<a> = {(a)^n : n is an integer}

[rustic crown]

and then what is <a^-1>? do you see why they have to be the same?

[vital walrus]

<(a)^-1> = {(a^-1)^n : n is an integer}
So it’s just because n can vary freely? As in they “swap” based on whether n is positive/negative but they are still yielding the same set?

[rustic crown]

yep. if you have to write it down, you'll just need to show every element in <a> is also in <a^-1> and vice versa. These follow precisely from your "swap" intuition.

[vital walrus]

Okay thank you!

[rustic crown]

[next obsidian]

What the hell happened here. Someone said Z/35Z isn’t a finite group?

[rustic crown]

[scarlet estuary]

(Z/35Z)^x

was i think the convo?

[chilly ocean]

Prof: ah, I will give an easy exercise, "write down an infinite group", no one should get this wrong.
Student: Z/35Z.
Prof:

[scarlet estuary]

all numbers > 8 are basically infinite

[rustic crown]

i have infinite fingers >.<

[scarlet estuary]

for all practical purposes you do

i mean think about it

have you ever been like

"man"

"i ran outta fingers"

[rustic crown]

if we had 12 fingers, i could keep my index fingers on G and H >.<

[scarlet estuary]

but [G : H] is already an index

[rustic crown]

[snow flint]

👉👈

yubi yubi

[scarlet estuary]

who the fuck can use their toes to count

what

[snow flint]

sorry you arent skilled nami

[scarlet estuary]

how do you move your second toe without moving like 3 others

[rustic crown]

you can go till 3^10-1 if you can control your fingers >.<

[snow flint]

wait wydm 3^10

is this a meme

[scarlet estuary]

retracted

extended

in between

[sturdy marsh]

use your hair to count

could count up to 100000 if you do that

[scarlet estuary]

youre assuming our hair can be well ordered

[snow flint]

you're assuming just because you cant, that we cant

[sturdy marsh]

are you anti-choice?

[scarlet estuary]

recall that any number > 8 is basically infinite

we conclude that they are indistinguishible

even with choice

[celest brook]

axiom of name convention that makes sense

[unique juniper]

$D = {m + n\sqrt{2} | m,n, \in \mathbb{Z}}$

[cloud walrus]

Yes

[unique juniper]

trying to find the quotient field from this ring

not sure how tho

[carmine fossil]

Field of fractions of that is Q[√2]

Map (a+b√2,1) to a+b√2

and (1,a+b√2) to 1/(a+b√2) which can be rationalised

[unique juniper]

ahh right

ty

[carmine fossil]

I mean be more explicit as to why (a.1)(b.1)=(ab).1

[lunar coyote]

distriutivity?

[carmine fossil]

a and b should also be in N

Just write b.1-a.1=(b-a).1 the b-a step is kinda unnecessary and weird

[carmine fossil]

Yes

[chilly ocean]

A ring

[delicate hawk]

this seems self evident

but how do you formalize this

R is a subset of Q containing all the integers, so ofc it can written as a/b such that a is in Z and b is in a subset of Z that does not contain 0

[chilly ocean]

So what are you saying is S?

[delicate hawk]

say R is just Q but without 1/2

idk how to write S

would S just be Z without 2?

[final pasture]

that wouldn't be a ring

[delicate hawk]

that cant be possible since 2/4 would still be in R

[final pasture]

I mean, R is supposed to be a ring here right ?

[delicate hawk]

yes

[chilly ocean]

This sounds close to what I'm guessing is the correct answer

(I'm not completely sure what S should be but I have a guess)

[delicate hawk]

so say R does not have an element of Q in it, say 1/2, that means it'll also not have all multiples of said element

1/2 , 1/4, ... so on, and also -1/2, -1/4 and so on

since it has to be closed under multiplication

but idk how that manifests itself in the restriction on S

[chilly ocean]

I guess 2,4,... Would not be in S, I'm not sure what else can be said

Maybe it is fruitful to think of the example of the ring R that is just Z with 1/2 added, so Z[1/2]

[rustic crown]

that is just Z localized at the multiplicative set {1, 2, 2^2, ...}

I think taking S = {s in Z\{0} | 1/s in R} should do the job.

you can check that this is a multiplicative set

[delicate hawk]

@rustic crown what about when we have 2/s in R, but 1/s isnt, or vice versa?

i feel like this S doesnt account for that

maybe im missing something

[rustic crown]

it does actually, because if a/b is in R with a and b coprime, the we can find c in Z such that ac = 1 (mod b)

hence if a/b in R then ac/b - some integer = 1/b in R

so b in S and then a/b in S^-1 Z. This shows one inclusion.
other one is even easier.
an element of S^-1 Z looks like a/s for s in S. But then 1/s in R and hence a/s in R.

[rustic crown]

problem with taking other 2/s and all the in the definition would be that s could be even. So say R could be just Z, and 2/2 in R. But i can't have S contain 2. else that would force me to have 1/2 in S^-1 Z.

and if you force coprimality, then ig it would work, but showing that S is multiplicative would require that a/b coprime => 1/b thing.

does it make sense?

[delicate hawk]

@rustic crown i think i understand now, tyvm ❤️

[rustic crown]

[chilly ocean]

yo det

yo $\sum_{i_1,i_2, \dots , i_n=1}^{n} \varepsilon {i_1 \dots i_n} a{1,i_1} \dots a_{n,i_n}$

[cloud walrus]

Godel

[rustic crown]

lol

[maiden ocean]

Since the prime ideals of of S^{-1}A (ring of fractions idk how standard this notation is) are S^{-1}p for p prime and p cap S = 0, if the 0 ideal is prime in S^{-1}A does this imply that 0 is prime in A?

im pretty sure this is not correct and maybe im misunderstanding the correspondance here, but my book writes it as p <-> S^{-1}p

or more precisely

im dealing with A_p is an integral domain for all prime p, which i know implies that the nilradical of A_p is zero and so the nilradical of A is also zero

and we also have that N_p is prime for all p

does this imply that N is prime?

t minus 3 minutes until chmonkey arrives and talks about sheaves

[oblique river]

I dont think that you can pass the property of being an integral domain backward. For example, Q x Q is not an integral domain and has 2 prime ideals, and if you localize at each prime you get Q I think

[maiden ocean]

is the correspondance as written not actually correct

like does S^{-1}p prime not imply p prime

[oblique river]

I think the issue is that you can have multiple ideals I which become the same under localization

What is true is that every prime of S^{-1}A is of the form S^{-1}p for some prime p of A

[maiden ocean]

yeah i think i g et it

the contraction isnt necessarily p

[oblique river]

But just knowing that S^{-1}J is prime doesnt mean J is

[maiden ocean]

yea

i thought this seemed extremely wrong lol so im glad i asked

[golden pasture]

it works if p intersect S = empty

it works for prime and primary ideals

for other ideals funnier things can happen

[bright pasture]

I have no clue how to define the function

[rustic crown]

Do you know about freegroups and presentations?

[bright pasture]

no

this is from an introductory course

[rustic crown]

okie so we might need to verify some things by hand.

we want to find a group homomorphism from D_n to S_n.

What do you think is a good choice?

[old lava]

was this exact question not asked yesterday

[bright pasture]

i was thinking of defining f(x) = s^u r^k

[rustic crown]

yep. with the same notations.

[bright pasture]

some1 from my class im guessing lol

[rustic crown]

But we used presentations yesterday

[bright pasture]

nvm then

[rustic crown]

basically define this map, it would be easily a bijection. but you might need to do some work to show it preserves the operation.

[bright pasture]

yea so basically have this f(x) = s^u r^k

and i have proved bijection

i just cant show that it preserves the operation

[rustic crown]

or lets just do the way you're doing...

y = [v, m; 0, 1]

compute xy

then see where xy is sent to

[bright pasture]

[uv, um+k; 0, 1]

[rustic crown]

yep, and this is sent to s^(uv)* r^(um+k)

[bright pasture]

yep

[rustic crown]

take cases on values of u and v to show this is same as (s^u r^k)(s^v r^m)

[bright pasture]

thats basically what I have

maybe my problem is simplifying (s^u r^k)(s^v r^m)

[rustic crown]

yea you just need to take cases on value of v

[bright pasture]

actually we probably need to map it to s^(u+1)/2

[rustic crown]

wait there is a mistake in this...

[bright pasture]

yes

[rustic crown]

yep

[rustic crown]

more like (1-u)/2

u = 1 should map to s^0 and u = -1 should be s

[bright pasture]

wait maybe thats my mistake

is there a difference ?

[rustic crown]

well (1+u)/2 is 1 when u = 1

[bright pasture]

is there a problem with having u=1 mapped to s^1

and u = -1 mapped to e

[rustic crown]

okei where will [1,0;0,1] go?

[bright pasture]

s

[rustic crown]

its the identity on the left group

[bright pasture]

omg

has to identity on right

[rustic crown]

yep

[bright pasture]

my eyes are opened

thank u

[rustic crown]

when you have -1 in the first entry, then the matrix sends the column vector (1, 0) to (-1, 0). which is like flipping.

[bright pasture]

I see

yep that was my problem

[bright pasture]

@rustic crown

is this not wrong ?

[rustic crown]

yea look wrong

[bright pasture]

x = [u, k; 0, 1] y = [v, m; 0, 1]

[rustic crown]

okie so i want

[-1 0]
[ 0 1]  |-->  s

and

[1 1]
[0 1]  |--> r

[bright pasture]

yes so thats basically what i have

[rustic crown]

i'm just checking f(x) = what we gave indeed works with matrix multiplication... should have made sure of it earlier 😦

ye so you see the problem is

[-1 0] [1 1]^k  =  [-1 0] [1 k]  =  [-1 -k]
[ 0 1] [0 1]       [ 0 1] [0 1]     [ 0  1]

so the group map would send this matrix to s r^k

but what we defined makes it go to s r^{-k}

the fix is easy

either change your f(x)

or notice

[1 k] [-1 0]  =  [-1 k]
[0 1] [ 0 1]     [ 0 1]

so the group map wants to send this matrix to r^k s

instead of s r^k

So the map is,
send the matrix

[(-1)^a   b]   -->   r^b * s^a
[  0      1]

[bright pasture]

wait if we swap the order its fixed ?

[rustic crown]

b takes values in Z/nZ and a takes values in Z/2Z

[rustic crown]

right this calculation show that

the matrix corresponding to s should be on the right

Say

[(-1)^a   b]   -->   r^b * s^a
[  0      1]

[(-1)^c   d]   -->   r^d * s^c
[  0      1]

Then their product is,

[(-1)^(a+c)   b + (-1)^a d]   
[  0               1      ]
-->  r^(b + (-1)^a d) * s^(a+c)

[bright pasture]

i see

[rustic crown]

if a = 0 then
(r^b s^a)(r^d s^c) = r^(b+d) s^c

and if a = 1 then
(r^b s^a)(r^d s^c) = r^b s r^d s s^(1+c) = r^b (srs)^d s^(1+c)
= r^(b - d) * s^(1+c)

[bright pasture]

yes so the swapping works ill try writing it out

[rustic crown]

Sorry for being wrong earlier

i should have checked the details first

[bright pasture]

it was my fault for not checking it fully myself

thx for the help

[rustic crown]

[jagged dune]

I generated the ideals of Z25, but I'm not sure how to tell if they're prime or maximal

or both

so I have {0}, Z25, and <5>

just not sure how to identify it

[carmine fossil]

Z_25 is not an ideal of Z_25

[jagged dune]

ah crap

[carmine fossil]

{0} is neither prime nor maximal

<5> is maximal

[jagged dune]

How is Z25 not an ideal?

Isn't it <25/25> = <1>?

Then that's just Z25

and then there's this in my textbook

so I figured Z25 would be an ideal

[small bison]

might just be a difference in definitions

[chilly ocean]

To me, a ring R in itself should not be considered an ideal, since then the idea of "maximal ideal" becomes pointless

Then you need to say something like "penultimate" ideal, which actually, when I think about it, it sounds sort of cool too

[nova plank]

Yall are wack

Z25 is an ideal of Z25

[chilly ocean]

Ok, so what is a maximal ideal then?

[jagged dune]

lol

[chilly ocean]

"a proper ideal"

[jagged dune]

ahh

[chilly ocean]

Ok, so you really mean "maximal proper ideal" when you say "maximal ideal". But this is too long to say

[nova plank]

This is the definition I like. An ideal I of R is maximal if whenever J is an ideal such that $I \subseteq J \subseteq R$ then J is I or R. I guess I can see the motivation for not wanting to awkwardly exclude R like this, but it's what I'm used to.

[cloud walrus]

Lunasong the Supergay

[jagged dune]

so in this case

Z25 would not be maximal

based on the definition I gave

[nova plank]

Yes

[jagged dune]

hm

is it prime?

That confuses me a bit

[small bison]

depends on yoru definition also

[jagged dune]

yeah

I'm reading now and it says proper ideal again

so nvm

[nova plank]

You didn't give your definition, but I assume you exclude R there too

Yeah

[jagged dune]

I'm confused because we did proofs including R

[nova plank]

R is an ideal. It is just never prime or maximal.

[chilly ocean]

I would just say this is weird. "Maximal" and "ideal" are already defined terms.

[nova plank]

The other people are just using different definitions

[jagged dune]

so <5> isn't prime because something like 5*10 is outside of <5>, correct?

[nova plank]

Huh

[jagged dune]

or am I misunderstanding the definition

[nova plank]

Yes

[jagged dune]

I think I am then lol

ok

[nova plank]

If 5*10 were outside of <5>, would it even be an ideal then?

Luckily it isn't.

[jagged dune]

well <5> is only {0, 5, 10, 15, 20} in this case

or is it 5*10 mod 25?

because then it'd be 0

[nova plank]

Indeed

[jagged dune]

ohhh

ok

[nova plank]

5 and 10 are not the integers 5 and 10

They are the equivalence classes of 5 and 10

[jagged dune]

gotcha

so is it not prime?

or are a and b meant to be any value?

well even then, I can't think of anything where it wouldn't be right

[nova plank]

You have to take arbitrary a and b yes

And it is prime

[jagged dune]

oh lol

ok I thought I was going crazy

[nova plank]

It's also maximal

And maximal implies prime

[jagged dune]

I just remembered that

and {0} is just neither?

it should be prime

based on the def

since a or b has to be 0

oh wait unless it's like, the equivalence class thing

because something like 5*10 mod 25 is 0

but neither 5 or 10 are 0

[nova plank]

Indeed

[jagged dune]

so then it isn't prime

and then it isn't maximal either

since it's a subset of <5>

not sure if it's proper actually, but I think it is

alright all is good

thank you all for the help

and differing views

[nova plank]

Np

Proper just means not equal to R

So it's proper

[jagged dune]

oh ok

thank you

[nova plank]

@chilly ocean question

If you define R not to be an ideal. Do you not define R/R as a quotient ring?

[chilly ocean]

I guess not

[nova plank]

[nova plank]

Do you know of a text that doesn't define R as an ideal?

[chilly ocean]

Ah, my favorite algebra book, d and f, defines ideal in a way that R is allowed

[nova plank]

Blegh

Good thing I stopped using df

Oh, is allowed?

I thought you said isn't

[proud bear]

using lang like mirza?

[nova plank]

Damn, guess df got one thing right

Ew no

Rotman

[proud bear]

oh wow first search result is a complete pdf

[nova plank]

Good, good

Embrace good writing style and exposition.

[final pasture]

Rotman's Advanced Modern Algebra ? It's a jewel, yeah, I love it

[wild sapphire]

is it possible for a function in Z5[x] to just have no roots because they're all real or complex?

[final pasture]

X² - 1 ?

@wild sapphire

wait

Z5[X] not Z[X]

Z5 = Z/5Z

[wild sapphire]

like I mean not that it makes much of a difference right

[final pasture]

it does

X² - 1 have no roots in Z

but it does have a root in Z/5Z