groups-rings-fields

[copper kestrel]

ye thats what i planned on doing, i think Z[i]/<3> would be {(ai) + <3>, (1 + ai) + <3> , (2+ai)+<3>}, where a is any integer

but the any integer is throwing me off

[delicate orchid]

i is a unit in Z[i] so this is just {(3), i+(3), 2i+(3)}

[wraith cargo]

this isn't correct

[copper kestrel]

not surprised

[delicate orchid]

but yeah where is (3) itself

[desert verge]

i think from what u did earlier,
Z[i]/<a> = { <a> + r : N(r) < N(a) }

[copper kestrel]

oh true

[copper kestrel]

wait but then wouldnt we have 0, 1, 2, i, 2i, 1+i?

bc <3> is just 3Z

[delicate orchid]

you need everything with a norm less than 9

[copper kestrel]

oh true

[desert verge]

im not sure about this

[copper kestrel]

not less than 3

[delicate orchid]

so you're missing 2+i, 1+2i

[delicate orchid]

yeah it's 3Z[i]

[copper kestrel]

rats

[desert verge]

yeah

[copper kestrel]

:')

[delicate orchid]

oh and 2+2i lol

[desert verge]

it's not the subgroup generated by 3

viewing Z[i] as just a group

[copper kestrel]

eh?

so thats: 0, 1, 2, i, 2i, 1+i, 2+i, 1+2i, 2+2i

[wraith cargo]

note that this is the same as taking mod 3 in both coefficients

[copper kestrel]

yep

so then the order is 9

[wicked patio]

this a^2+b^2 < 9 condition is strange

it works by coincidence here but it doesn't in generality

[copper kestrel]

maybe i should've done 3^2?

[wicked patio]

no the problem is like

[copper kestrel]

o

[wicked patio]

in Z[i] / <5>

well

ok

i'm assuming a and b are nonnegative too?

[copper kestrel]

i didnt think of that

[desert verge]

it's because of division with remainder

[wicked patio]

like here you can have 4+4i, for which a^2+b^2 > 25

[desert verge]

and the fact that Z[i] is a ED.. i think

[copper kestrel]

it is

[desert verge]

which is the same as -1 - i + <5>

or umm even -1 + 4i + <5>

and here a^2+b^2 < 25

[wicked patio]

ok so it works if you allow a and b to be negative as well

it's just a strange condition

[desert verge]

^

[copper kestrel]

yaya

[desert verge]

maybe this helps

[copper kestrel]

i would think the order is 9 up to units

[desert verge]

wdym upto units

i think the order is 9 yeah.. not too sure

[copper kestrel]

like we have that multiplying by units is an equivalence relation

idk im just trying to reason why we dont need the negative values

bc if we need them the order is then 17

[desert verge]

isn't Z[i]/<3> a field

[copper kestrel]

it is

[velvet hull]

saying that the order is 9 up to units is not meaningful

because you are either counting the units or you are not

[desert verge]

oh. u just mean different representatives of the same coset probably, which also satisfy a^2 + b^2 < 9

[velvet hull]

the order is not something that you can ignore the units for

you can mean coset, as succ said, but not unit

[copper kestrel]

probably

sorry yall

[desert verge]

all good

[copper kestrel]

and now i have to do this two more times

[delicate orchid]

another way of doing this problem is to just add 1 to itself in each ring until you get 0. This immediately gets you that the characteristics of these fields are 3, 2, and 5

[copper kestrel]

i can do the characteristics fine its just the orders

[delicate orchid]

where does it ask you for the orders

[copper kestrel]

i wrote the question wrong

[wicked patio]

"orders of the characteristics"

whatever that means

[south patrol]

I have characteristic 6 or 7

[copper kestrel]

it should be orders and characteristics

[delicate orchid]

yeah which doesn't mean anything so I ignored it

[copper kestrel]

i forgot to correct it :')

[delicate orchid]

well once you have the characteristic pick a power of the prime that makes sense and you're done

[copper kestrel]

wdym

[delicate orchid]

you can estimate this based on the area of a square

[wicked patio]

i drew <1+2i>

[azure cairn]

is there anything i can say besides that |G| is a multiple of 10 and 6

[velvet hull]

you can clean that statement up a little bit more but yeah, there's not much more to be said

[azure cairn]

umm

multiple of 30?

[frigid epoch]

it could just so happen that these stipulations somehow imply that the group must have an order of 600

[velvet hull]

as a bonus exercise, you can always find a group that satisfies these requirements for every multiple of 30

[azure cairn]

would that be the converse statement

noooo i cant use lagrange's thm anymore

[proud vigil]

you don't need any fancy machinery really

[azure cairn]

(ur saying that every group whose order is a multiple of 30 has elements of order 10, 6?)

is that true

it doesnt feel true to me

[proud vigil]

no, they're saying there exists a group of size 30k that has an element of order 10 and order 6 for every natural k

[azure cairn]

oh

[rapid cave]

can you give such a group of order exactly 30? If G is such a group then Cn x G is a group of size 30n that has elements of order 6, 10

[azure cairn]

cant u just construct the product group of a group comprising an element with order 6, 10 and a group with order 30

i would need to verify on paper but shouldn't (g, e) and (g', e) (where g is order 6 and g' order 10) have orders 6 and 10 as well

[proud vigil]

i think you mean group with order 30 containing the desired order elements x some group with size k for whatever k

[proud vigil]

which is what this is

[azure cairn]

no i'm saying, take some group G_30 with order 30 then a group G with an element that has order 6 or 10

then take their direct product

[proud vigil]

okay, how do you get an order 60 group that way

[azure cairn]

|G x H| = |G| |H|?

so let |G| = 30 and |H| = some integer k

then |G x H| is a multiple of 30

[rapid cave]

this is not what you said earlier

[azure cairn]

what

[proud vigil]

you're saying one group is order 30, and the other group has an order 6 or 10 element

[azure cairn]

yes

then take their direct product

[proud vigil]

how can either group have order 2

[azure cairn]

where did order 2 come from

[proud vigil]

how would you make an order 60 group

using your methd

[azure cairn]

oh for every natural k

[azure cairn]

but i dont think this is true for a group of order 30

wait

lemme think

no it is

OOps

[proud vigil]

yeah cyclic groups get everything

[azure cairn]

wait but it is possible

G = Z_30
H = Z_2

[proud vigil]

H doesn't have an order 6 or 10 element

[azure cairn]

(5, 0) has order 6

(3, 0) has order 10

[proud vigil]

right, this just isn't what you said before

this is correct though, Z_30 x Z_k works for every natural k

[azure cairn]

ah wait i see

you choose the order 6, 10 elements from Z_30

then you dont care about what Z_k is

ok i was like 90% of the way there 😊

[proud vigil]

you actually can prove that every group of order 30k for natural k has an element of order 6 no i made an error

[azure cairn]

interesting

also for clarity every cyclic group of order n has every element whose order divides n right

|x^k| = |x|/gcd(|x|, k), so then we can choose k specifically to make the right side any divisor of |x|

if you let x be the generator then |x| = |G| = n

which is why cyclic is needed in the hypothesis

when it says "generated by these elements" what does it mean

<xy> or <x> U <y> or what?

[south goblet]

<x, y> I think

[azure cairn]

what is that

[south goblet]

A subgroup can be generated by 2 or more elements right?

That's what I mean

[azure cairn]

so like x^ay^b for every a,b \in Z?

[proud vigil]

oh this is true for order 15 elements

[azure cairn]

i dont think ive ever seen the notation <x,y>

[south goblet]

oh

that's what I use

What do you use

[azure cairn]

let me go back to the cyclic group section but i can bet money that i've never seen notATtion for the cyclic subgroup generated by two elements

[proud vigil]

that's not really a thing

cyclic = 1 generator

[south goblet]

it's not cyclic

[south goblet]

Is this sylow theorems or am I stupid

[azure cairn]

right

so then i've definitely never seen <x, y>

[azure cairn]

idk havent done that yet

[south goblet]

oh

[proud vigil]

it's the smallest subgroup that contains the elements in the brackets

equivalently the intersection of every subgroup that contains them

[azure cairn]

ah ok

[south goblet]

wait snowflake is it sylow theorems

specifically the fact that they exist

[proud vigil]

without context it can also be considered as the free group on 2 elements

oh let me scroll up one sec

[proud vigil]

i think that's just cauchy?

[south goblet]

oh

[proud vigil]

we know x is either order 2, 11 or 22 by lagrange, and it cant be 22 since y is not a power of x, so x is order 2 or 11, and the subgroup it generates can't cover everything

[south goblet]

bazooka for a mosquito type of situation...

[proud vigil]

i never properly learned the sylow theorems tbh :-( i just learned how to apply them 😭 i should probably internalize the proofs

[south goblet]

oh

[chilly radish]

The proofs aren't particularly enlightening or important to the applications of the theorems

[proud vigil]

thats kind of the vibe i got but it feels a bit disappointing

[rocky cloak]

I think they can be quite enlightening, at least for a good proof.

But not particularly relevant to how you would apply the theorems

[azure cairn]

could i get a hint? i know that for phi to be a homomorphism G must be abelian

[rocky cloak]

Write out what it means for phi(xy) to equal phi(x)phi(y) and try to manipulate the expression

[azure cairn]

yes i did that

unless there's something else i should get besides G is abelian?

i was talking about the injective and bijective parts

[rocky cloak]

Oh, I misunderstood what you were asking then

[azure cairn]

also, like, i know that x^2 != 1 means none of the group elements can be involutory (for phi to be injective)

but i think there is something bigger im meant to get from phi being injective

[rocky cloak]

Do you know when a group doesn't have any element of order 2?

[azure cairn]

when the order of the group is odd?

oh

[rocky cloak]

Because...

[azure cairn]

counting formula

[rocky cloak]

So that covers injectivity

[azure cairn]

ok great ty that's good

[azure cairn]

wait but

ah yes ok

the problem was that i didnt jump from "x^2 != 1" to "no element of G may have order 2"

i will proceed from here thanks

wait a minute

isnt phi already surjective lol

like if it's injective and maps from itself to itself then it must be surjective no?

[rocky cloak]

The magic of finite sets

[azure cairn]

idk how i forgot about that

ok so this is only trueif G is an abelian group of odd order

ty

[azure cairn]

oh wait yeah that's only true for finite sets

[south goblet]

wait does it have to be abelian

[azure cairn]

for it to be a homomorphism yeah

phi(xy) = phi(x)phi(y) = x^2y^2 = (xy)^2 = xyxy

x^2y^2 = xyxy

multiply both sides by something (too lazy to figure out what) to force yx = xy

x^-1 * x(yx)y * y^-1 = yx

x^-1 * x * x * y * y * y^-1 = xy

[azure cairn]

yeah premultiply by x inverse and postmultiply by y inverse

[south goblet]

Oh

oh that does make sense!

[proud vigil]

a related problem

if f(x) = x^3 is an automorphism on a finite group, then the group is abelian

[south goblet]

xyxyxy?

and not divisible by 3

I think

[rocky cloak]

Notable that for the x^2 problem you only used the homomorphism part to show it's abelian, but here that's not enough

[azure cairn]

wait actually

i used the converse of lagrange's thm fuck

just cuz the group doesnt have elements of order 2 doesnt mean the group's order can't be a multiple of 2

cauchy's thm works

but i havent learned that in the book yet

[rocky cloak]

Note you already have that the group is abelian, so then Cauchys theorem is much easier to prove or follows from the clarification of finite abelian groups if you have that

[azure cairn]

what is the clarification of finite abelian groups

do you mean characterization

[rocky cloak]

Classification*

[azure cairn]

is that that all finite abelian groups are isomorphic to Z/pZ?

i dont have that yet either

:p

[rocky cloak]

They are products of cyclic groups

[azure cairn]

dont have that either idt

wait what

all finite abelian groups are products of cyclic groups?

[south goblet]

yes

[rocky cloak]

Yup

[azure cairn]

damn

[south goblet]

most confusing one of the finite classifications for me is simple groups...

why 26

[rocky cloak]

You can even say that they are products of cyclic groups of prime power order in a unique way.

Or a product
Z/a1 x Z/a2 x ...
such that a1 divides a2 which divides a3, etc. Also in a unique way

[azure cairn]

yeah i dont know any of that

idk if this will come up later

[rocky cloak]

Even the sporadic simple groups didn't make it to the 27 club

[south goblet]

THE 27 CLUB?

what

is that?

Oh i remember now

[rocky cloak]

Just a joke

[south goblet]

If you include that one other group which is barely of Lie type, it is 27!

not factorial

[karmic moat]

lines on a cubic

👀

[rocky cloak]

Let's include Tits, why not

[azure cairn]

so the classification of abelian groups as u put it is a corollary of another thm in chapter 14

and cauchy's theorem is a corollary of sylow's thm(s?)

[south goblet]

yeah

but it's usually used in the proof

[rocky cloak]

I usually use Cauchys theorem as a lemma for the sylow theorems, but you could prove it without

[azure cairn]

this is all much past ch2 though so 😅

[south goblet]

interesting

[azure cairn]

ch3-5 are easy ones though cuz theyre just linalg, same with ch8

[south goblet]

what book are you reading

[azure cairn]

artin

[rocky cloak]

Anyway, it's pretty easy to prove that an abelian group of order a multiple of p has an element of order p without using any of these heavier theorems

Just induction and Lagrange

[south goblet]

i first read dummit/foote 💀

[azure cairn]

here's the toc

[azure cairn]

ill just keep cauchy's thm in my toolkit

[south goblet]

chapter 8/9/13 look pretty interesting not gonna lie

But i'm biased because i love number theory

[azure cairn]

ch14 looks interesting

and ch10/11

[south goblet]

see the book i read has ch10 way later

[azure cairn]

lol

d&f is huge no?

like 1000 pages

[south goblet]

yeah

💔

[azure cairn]

this is pretty big too, 600 pages 😭

[south goblet]

Abstract algebra is foundational though

...

right...

I mean it

is

[novel star]

lang is the best one to build pain tolerance with

just read something else for homology

[warm badge]

huge? lmao if i drop it on someone's head, they're not surviving 😭

[azure cairn]

LOL

[swift root]

i guess academics will have one use in the apocalypse..

[warm badge]

our abs algebra prof is so fkin bad, he singlehandedly makes people leave maths and choose different discipline. i sometimes find it way easier to throw this on him in class😭

[feral slate]

Did he go “Damnit my foot!”?

[azure cairn]

wut

isnt 2's multiplicative inverse 1/2

oh wait

i just solve 2k = 1 (mod n) right..?

then 2 * (that) = 1 in Z/Zn

[warm badge]

yes

[azure cairn]

is it for odd n only

gimme a sec

yea

[azure cairn]

can you not just set up the system $x = u + ta = u + tb$ to show that $x \equiv u \pmod a$ and $x \equiv v \pmod b$?

[azure cairn]

and this syseq has no solutions only when a = b, which only happens if a = b = +/- 1 (since gcd(a,b) = 1)

oh wait

integer sols

oops

ok abck to drawing board

[crystal vale]

What is exactly a Kummer extension?

[sonic coral]

i think it’s just when you adjoin an nth roots from the base field

like the field extensions where you adjoin roots of x^n-a for some a and n

[crystal vale]

I think that's called a radical extension?

[azure cairn]

ok wait i think i got it?

for $x \equiv u \pmod a$ AND $x \equiv v \pmod b$, $x = ra + u = sb + v$ for $r, s \in \bZ$. wlog, let $s$ be negative then rearrange to yield $ra + sb = v - u$.
note $r', s'$ can be chosen s.t. $r'a + s'b = 1$ (since $\gcd(a, b) = 1$), so simply choose $r = (v - u)r'$ and $s = (v - u)s'$. then $ra + sb = (v - u)(r'a + s'b) = v - u$. note that if $d = \gcd(a, b)$, then $d \in (\bZ = \bZ a + \bZ b)$ and so $d = 1$ is some integer combination of $a, b$

[azure cairn]

    \begin{problem}[Chinese Remainder Theorem]{Let $a, b, u, v \in \bZ$ be integers, and assume $\gcd(a, b) = 1$. Show there is some $x \in \bZ$ for which $x \equiv u \pmod a$ and $x \equiv v \pmod b$.}
        For both conditions to hold, $x$ must take the form $x = ra + u = sb + v$ for some $r, s \in \bZ$. Without loss of generality, let $s$ be negative, and so rearranging yields $ra + sb = v - u$. Note that there is a choice of $r', s' \in \bZ$ for which $r'a + s'b = \gcd(a, b) = 1$ (since $\gcd(a, b) \in (\bZ(\gcd(a, b)) = \bZ a + \bZ b)$. Then let $r = (v - u)r'$ and $s = (v - u)s'$, yielding
            $$ra + sb = (v - u)r'a + (v - u)s'b = (v - u)(r'a + s'b) = v - u$$
        as desired, completing the proof.
    \end{problem}

[azure cairn]

hopefully this is correct 🥰

|(x, y)| = lcm(r, s) right

[proud vigil]

yes

[azure cairn]

do youmind checking the proof above as well

[proud vigil]

your step of letting s be negative is kind of weird

[azure cairn]

how so

[proud vigil]

it seems like you're just swapping s for -s

[azure cairn]

i am

[proud vigil]

letting s be negative would be like "assume the value of s is negative"

that isn't the same as replacing s with -s

this isnt a structural issue but its kind of unclear as written

[azure cairn]

how would you avoid it then

or well

how would you rephrase

[proud vigil]

honestly i doubt you actually have to do that step, but otherwise i would either say "for convenience replace s as -s" or just add a third variable

[azure cairn]

wtvr

[glass sinew]

you could have posted this in #elementary-number-theory honestly lol

[azure cairn]

ig

[proud vigil]

beyond the negative wording issue i think this is fine

[fossil beacon]

arent there regular talks

[copper kestrel]

im a bit confused on how to solve x(a,b) + y(c,d) = (e,f) and seeing where x and y lie in Z, like am i solving for x and y on their own and making a system of equations?

question 8

[rocky cloak]

With some linear algebra you should be able to express x and y in terms of e and f.

Then the question becomes when does every integer e and f yield integers x and y (as opposed to arbitrary real numbers)

[copper kestrel]

hmmmmm

would it be like we have x = (e-yc, f-yd) * (1/a, 1/b)?

and then do the same thing for y?

[rocky cloak]

Well, now you still have x depending on y. But the idea is to solve the system of equations yeah. Do you know how to solve systems of equations using linear algebra?

[copper kestrel]

yep i do! i was just confused on how to get the system of equations lol

bc theres only one equation

[rocky cloak]

I mean you're given two equations
xa + yc = e
and
xb + yd = f

[copper kestrel]

OHHHHH

omg im so sorry

i forgot thats how you do it for coordinates, thank you so much 😭

[copper kestrel]

okay i got this, and now i just have to find when this is an integer, or 1 preferably

or i guess solving when x = 1 and y = 1

bc (1,1) is a basis for Z x Z iirc

it should be

[delicate orchid]

the question seems to be asking for a Z-basis, so your basis will need 2 elements

[copper kestrel]

ah sick!

so then i can do (1,0) and (0,1)

[copper kestrel]

i forgot about that in the original question xD

[delicate orchid]

I mean there's an infinite amount of such bases so you'll have to find them all not just one

[copper kestrel]

o

right

then i'll do (n,0), (0,n)

for n \in Z

(not equal to 0)

[rocky cloak]

So you want
(eb - fa)/(cb - da)
to be an integer for all e and f.

I'd start by picking e and f so that eb - fa is as small as possible. How small is that?

[copper kestrel]

i ended up setting that equation equal to n, and then trying to solve from there

unless thats useless

[rocky cloak]

Not exactly sure what you mean by that

[delicate orchid]

(1,1), (0,1) is also a basis

[copper kestrel]

rats

[delicate orchid]

my hint is that the matrix you get by considering these pairs as the rows are all determinant 1 or -1 and integer valued

[copper kestrel]

nevermind then

[copper kestrel]

maybe e = f?

[rocky cloak]

Maybe consider some concrete examples.

How small can you make
2e - 3f?

What about
4e - 6f?

[delicate orchid]

6e-7f

[rocky cloak]

Banned!

[copper kestrel]

i dont know how small i can make those honestly, bc you can provavly just choose e = 1 and f = something large to get a very very large negative number

[rocky cloak]

Small in absolute value

[copper kestrel]

ah

[copper kestrel]

well since i think we want it to be small in the integers maybe we can set e = 3 and f = 2 to get 0

[rocky cloak]

So my point is:

If you want m/n to be an integer and m is very small, then n must also be very small

But m=0 doesn't help there

[copper kestrel]

yeah

shoot

[rocky cloak]

So let's go small but positive

Or nonzero at least

[copper kestrel]

then i guess we can make m = 1 and get 2e = -3f + 1 to get e = {-3f + 1}/{2}

im sorry if this is wrong i know im being really frustrating by not knowing what to do

[rocky cloak]

So, yes.
2*2 - 3*1 = 1 for example

[copper kestrel]

yeah

[rocky cloak]

What about the 4 and 6 case

[copper kestrel]

i think the smallest we can get is 2 with 4(2) - 6(1)

or maybe theres some higher multiple where the difference is 1

[delicate orchid]

oh we're doing bezout nonsense?

[copper kestrel]

whats that

[rocky cloak]

So what's the relationship that would make 2 and 3 give 1, but 4 and 6 give 2?

Or why shouldn't we be able to make 1 with 4e - 6f?

[delicate orchid]

what jagr is about to help you discover. This would not be my approach at all fyi. After he's finished I'll tell you how I'd do it. My method has a geometric interpretation

god damn sentence fragment heaven in ts post whatt am I smoking

[copper kestrel]

because theres no integer multiple of 4 and 6 whose difference is 1

[rocky cloak]

I guess when I started this approach I assumed tt knew elementary number theory. But best not to diverge now

[copper kestrel]

i havent taken a number theory class

and its not a prerequisite to this class

for some reason

[delicate orchid]

agreed

[copper kestrel]

sorry yall

[delicate orchid]

twin ur currently doing elementary number theory

[delicate orchid]

yes this is a restatement of what jagr said, but why isn't there

[rocky cloak]

And why wouldn't that be possible.

How can I manipulate the equation
4e - 6f = 1
to make nonsense?

[copper kestrel]

i know but like :')

[copper kestrel]

maybe like e = {6f + 1}/{4} = {3f}/{2} + {1}/{4}, which doesnt have an integer solution?

[rocky cloak]

Well, why is it clear it doesn't have an integer solution?

[copper kestrel]

bc getting a common denominator requires f to be n/2

[delicate orchid]

by jove they've got it./.......!!!

[copper kestrel]

:')

[rocky cloak]

Indeed, so maybe how I would put it is that if we divide by 2 we get
2e - 3f = 1/2

[delicate orchid]

anyway yeah you just divide by 2 lol

[copper kestrel]

that makes sense

im guessing im overcomplicating this

[rocky cloak]

Similarly for an a and b
ae - bf
will always be a multiple of any common factor of a and b

[copper kestrel]

interesting

[rocky cloak]

And in fact Bezouts theorem/ Euclids algorithm tells us we can achieve
ae - bf = gcd(a, b)

[copper kestrel]

ohhhhhhh i remember seeing that a long while ago

i know what the euclidean algorithm is yeah

[rocky cloak]

So then all the way back to
(ae - bf)/(cb - da)
needing to be an integer.

Then we know
gcd(a, b)/(cb - da)
must be an integer

Then what does that tell us about cb - da?

[copper kestrel]

uhm

im not sure, maybe that it has to equal 1?

for it to be an integer

[delicate orchid]

cb-da is also an integral combination of a and b

[rocky cloak]

Well how would it be related to gcd(a, b)?

[copper kestrel]

i want to naively say -(gcd(a,b)), but that doesnt seem right, i thought it would be negative bc we're subtracting b from a instead of a from b

[rocky cloak]

Let me ask this.

Would it be bigger or smaller or equal to gcd(a, b)?

(in absolute value)

[copper kestrel]

i want to say smaller or equal to

[delicate orchid]

we just established that the smallest integer comination of two integers a,b is gcd(a,b)

so how would it be smaller

[copper kestrel]

shoot thats right

so it would be equal to

[delicate orchid]

or larger than

but you're right if we assume that fraction is an integer

[copper kestrel]

wouldnt that not make it an integer then?

[copper kestrel]

ye

[delicate orchid]

it has to be equal to gcd(a,b) in absolute value

[copper kestrel]

that makes sense

[delicate orchid]

ok jagr what the fuck are we supposed to do from here? this is INCREDIBLY convoluted

[rocky cloak]

Now x and y are completely symmetric. So we can also conclude that gcd(c, d) = ±(bc-da)

Hence bc-da = ±gcd(a,b,c,d)

And the original equation tells us
xa + yc = e. In particular xa + yc = 1 should be possible

[copper kestrel]

hm

[delicate orchid]

I'm sending it. \begin{align*}
xa+yc = e, xb+yd &= f \iff \begin{pmatrix} a & c \ b & d \end{pmatrix}\begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} e \ f \end{pmatrix} \ & \iff \begin{pmatrix} x \ y \end{pmatrix} = \frac1{ad-bc}\begin{pmatrix} d & -c \ -b & a \end{pmatrix}\begin{pmatrix} e \ f \end{pmatrix} \ &\iff (x,y) = \frac{1}{ad-bc}(de-cf, af-be)\end{align*} is an integer solution if and only if $ad-bc = \pm 1$ so the bases of $\bZ \times \bZ$ are in 1:1 correspondence with the $2 \times 2$ integer valued matrices with determinant $\pm 1$.

great I'm gonna have to align it

[cloud walrus]

The Real Wew Lads Tbh

[copper kestrel]

oh my god why didnt i think to use matrices

[rocky cloak]

Yeah, so the key idea here is that systems of equations correspond to matrices and then we can use linear algebra to solve them

[delicate orchid]

the geometric interpritation of this is if you think about the submodule M of N := <(1,0), (0,1)> generated by (a,b) and (c,d), then for this to be equal to N there has to be a Z-valued linear transformation L moving (a,b) onto (1,0) and (c,d) onto (0,1). If this linear transformation doesn't have determinant with abs. value 1 then the best we can do is <(det(L), 0), (0,1)> (in other words, we need this L to have a trivial cokernel)

this is a lattice theoretic result where a sublattice is the full lattice if and only if it's volume is equal to the volume of the full lattice up to a unit in the underlying ring

that msg was mainly for jagr

[copper kestrel]

🫡

[delicate orchid]

but yeah when you're working with R-modules that look like R^n you should immediately be thinking about linear algebra

[rocky cloak]

Or if you hate geometry:
det(AB) = det(A)det(B)

[copper kestrel]

interesting, we didnt quite go over a ton of modules unfortunately

much to my dismay lol

[delicate orchid]

they're just vector spaces but with a ring instead of a field

[delicate orchid]

https://tenor.com/view/venom-bird-transform-be-careful-devilish-gif-6759896094770507840

[copper kestrel]

yeye exactly

i shall attempt to write a proof using the matrices bc that makes sense

but ty yall :]

i'm sorry if i constantly forgot stuff or made you frustrated at all i am not the best math student in the world

[delicate orchid]

it's ok dw boss

[rocky cloak]

Get some repetition while also applying the knowledge in new situations

Must be a good way to learn I think

Now in the future maybe systems of equations will scream "matrices" a little louder and
ax - by
will scream
gcd(a, b)

[delicate orchid]

but what if I'm not in a heckin gcd domain

[copper kestrel]

truth

[delicate orchid]

that's a thought actually is there a notion of a gcd domain but for ringoids

yes there is

[rocky cloak]

In AR theory you typically deal with categories where morphisms factor as compositions of irreducible morphisms

So throw in uniqueness and you'd have like a UFDoid

[delicate orchid]

yeah that's kind of what I thought

although I was thinking in a much more silly way, like an euclidean algorithm on idealoids

which seems unlikely to work

[rocky cloak]

I think non-commutativity is what kills you there

[delicate orchid]

a commutative ringoid, or as I like to call it, a commutative ring

[karmic moat]

what is ar theory 👀

[copper kestrel]

amean rvalue theorem

[karmic moat]

ohhhh

[copper kestrel]

now time to try to prove that Q under addition is not a free abelian group

[delicate orchid]

the fact that it isn't makes me cry

[karmic moat]

oh AR for auslander reiten

[copper kestrel]

im guessing its bc 0 isnt unique but

[karmic moat]

google kept giving me some stupid augmented reality whatever

[delicate orchid]

wdym, it definitely is unique

you just gotta find a Z-linear relation between elements of Q

[copper kestrel]

hashtag nevermind then LMAO

[delicate orchid]

the identity element is unique in any group

the ol e = e'e = e'

[glass sinew]

groups 101 proof

[copper kestrel]

truth

me when i forget evrything all the time

[glass sinew]

algebra moment

[copper kestrel]

idk how to work on having a better memory

[karmic moat]

this is what that little piggy said all the way home

[copper kestrel]

me when my academically challenging degree is academically challenging

[copper kestrel]

OH i realize i didnt explain correctly, i meant like 0 cant be expressed uniquely as a linear combination of basis elements

contradicting the basis part of a free abelian group

but i dont think that works

[rocky cloak]

Yeah, so you can split into two steps:

1. Q is not cyclic
2. any pair of elements of Q are Z-linearly dependent

Conclusion Q does not have a Z-basis

[copper kestrel]

ah

[crystal vale]

This question came up today in the exam

[ripe harbor]

$$\text{ev}(1) = \sum_{a \in \mathbb N^{(I)}} h(1(a))s^a = h(1)s^0 = h(1) = 1.$$ Why is $\sum_{a \in \mathbb N^{(I)}} h(1(a))s^a = h(1)s^0$?

[cloud walrus]

ILikeMathematics

[ripe harbor]

Does that look wrong @candid patrol ?

$$ev: R[X_i \mid i \in I] \to S$$$$f = \sum_{a \in \mathbb N^{(I)}} f(a) X^a \mapsto ev(f) \coloneqq \sum_{a \in \mathbb N^{(I)}} h(f(a)) s^a$$

[cloud walrus]

ILikeMathematics

[delicate orchid]

$ev$ is usually index by some $X \in \mathbb{N}^I$ and it just sends $f$ to $f(X_1, X_2, X_3, ...)$. I don't know what you're writing here, what is h? is f a polynomial? why is there no subscript on the X?

[cloud walrus]

The Real Wew Lads Tbh

[ripe harbor]

X^a is the product X_i^(a_i)

a is in N^(I)

R[X_i] is the set of all maps from N^(I) -> R

ev = ev_(h, s) with ev_(h, s)(r) = h(r) for all r in R and ev_(h, s)(X_i) = s_i for all i in I

[copper kestrel]

hey all :] im trying to prove in a euclidean domain that if nu(a) = nu(ab), then b is a unit

i was able to prove the converse surprisingly!!

[wraith cargo]

what's nu?

[copper kestrel]

euclidean norm!

just any euclidean norm

i know that if b is a unit then nu(b) = nu(1)

but i dont know if thats 1. helpful and 2. an if and only if

oh it is an if and only if!

i was trying some euclidean algorithm stuff and had there exists q and r such that

a = abq + r where 0 < nu(r) < nu(ab)

i mean maybe if ab is a unit then we have 0 < nu(r) < nu(1), which is minimal, then r = 0

but then i have to get b is a unit 😅

[delicate orchid]

well if r is 0 we have a = abq. We're in a euclidean domain, thus an integral domain. So multiplication is cancellative and we may conclude that bq = 1

[copper kestrel]

that makes sense!

but i'm not right in my justification that ab is a unit

[delicate orchid]

ab certainly need not be a unit, what if a is 0?

[copper kestrel]

exactly

[delicate orchid]

I'm not seeing where you claimed that ab was a unit

oh

mb

[copper kestrel]

but i jus need a bit more justification on why r = 0

[copper kestrel]

yayaya

thats why i said the minimal stuff

[delicate orchid]

well a definitely divides ab so eventually we will arrive at a step where r = 0

[copper kestrel]

ohhhhhhhhhh

that makes sense

idk why my brain glossed over that fact

[delicate orchid]

it's quite subtle imo

[copper kestrel]

i suppose so, i just need to use my brain

does this work?

im guessing i need some more justification but i cant seem to find where i need it

[rocky cloak]

Is the statement for a specific (nonzero) a or for all a?

Seems to be a specific a

[copper kestrel]

should be nonzero a bc thats what asked in the question!

i didnt specify thats what we wanted though

[delicate orchid]

well it's quite easy to prove for a = 0!

and don't give me that 0! = 1

[copper kestrel]

"you should do it as an exercise"

well if my proof works i'm done with this god forsaken assignment

and i can get back to my pokemon cards

[rocky cloak]

Proof looks good. Assuming your definition of euclidian norm has
nu(a) <= nu(ab)

[copper kestrel]

it is!

thank you for your help yall i love yall

soon i will be able to help others in my position

[delicate orchid]

no worries, it's good for me to revise my ring theory

[copper kestrel]

this class is so hard but honestly its so fun

algebra is so great

[ripe harbor]

How do you usually remember all the lemmas in ring theory? Like maximal ideal => prime ideal, prime => irreducible, integral domain then prime <==> prime ideal, PID and prime <==> irreducible and so on?

Is there some nice diagram you can draw or something that contains most of these?

[tulip otter]

quick question: Given a field F, an extension E/F and 2 elements a and b in E. Is it true that F(a,b)=F(a)(b) and F[a,b]=F[a][b] just like how things work when a and b are replaced by variables x_1 and x_2?

[south patrol]

Practice. For example these facts about maximal/prime ideal get used a lot for other things and can get drilled in from that

[tulip otter]

Like is it valid to treat a and b as variables over F in some sense so that some properties like this one are preserved?

[south patrol]

Yes, this will be a straightforward exercise

[south patrol]

But this sounds dodgy depending on what you mean lol

[tulip otter]

because I imagine this helps well with say finding the degree of the extensions ig

[south patrol]

Yes

[tulip otter]

instead of having to deal with polynomials in several variables you can find the degree of the extension over F(a) and then find that of F(a) over F

then multiply them together

[tulip otter]

i see, well I didnt have any precise meaning tbh. I was only looking for something that might help in the calculations around these extensions like the degree for example

[tulip otter]

is it just using the evaluation map on each variable one at a time vs using the evaluation map on all of them at once?

[rocky cloak]

I mean F(a, b) is the smallest field containing F, a and b. While F(a)(b) is the smallest field containing F(a) and b

So should be immediate they are the same

[tulip otter]

so for example on one hand you take f(x_1,..,x_n) to f(a_1,x_2,..,x_n) and then take that to f(a_1,a_2,x_3,...,x_n) etc.. and on the other hand you take f(x_1,..x_n) to f(a_1,..a_n) immediately and its the same thing except that the first goes F[x_1,..,x_n]=F[x_1][x_2,..,x_n]->F[a_1][x_2,..,x_n]
->F[a_1][a_2]..[x_n]->...->F[a_1][a_2]...[a_n] while the second goes F[x_1,...,x_n]->F[a_1,...,a_n]

[tulip otter]

ah right lol

[tulip otter]

does something like this do the job too even tho its overcomplicated or is there something wrong here

[rocky cloak]

Should work yeah

Guess it gets more annoying for the field part

[tulip otter]

right

[tulip otter]

on the other hand this handles it easily

tysm everyone

have a great day/night

[ripe harbor]

So you arent thinking of some diagram to remember these?

[chilly radish]

Prime means the quotient is an integral domain. Maximal means the quotient is a field. A field is obviously an integral domain

These kinds of relationships build up in your mind over time

[ripe harbor]

Alright, thanks

[ripe harbor]

?

[copper kestrel]

finals week is next week and i've just realized idk how to study for math exams besides doing the practice midterm and making a notecard :')

[swift tundra]

I mean many actually beneficial methods are best used over time. For example, spaced repetition improves your ability to accurately remember concepts; spaced repetition can take many forms, but the general idea is to review material frequently, and then less frequently as you get more comfortable with the material (for example using flash cards). To test your ability to remember, free recall is helpful. This is where you pick a topic and write down everything you can remember, right or wrong, from the given topic.

There are many more things to consider when reviewing, and I learned a lot from reading “Make it Stick” by Peter brown. Honestly I haven’t used many of these methods myself yet, but I am trying to use them more. If any one has anything to critique or add, please reply! I’m always curious to learn new study methods, or just more about learning in general.

[twilit wraith]

working with it enough tbh

in the ring theory ive done ive considered PIDs a singificant amount more than anything else, and so i just assume that PID is where all the nice stuff cuts off

[south patrol]

Not sure what diagram to draw tbh

[tulip otter]

I want to find $[K:\mathbb Q]$ for $K=\mathbb{Q}(\sqrt{18},\sqrt[4]{2})$. So $\sqrt{18}=3\sqrt 2=3(\sqrt[4]{2})^2$ so that $K=\mathbb Q(\sqrt{18},\sqrt[4]{2})=\mathbb Q(\sqrt[4]{2})$ and $[K:Q]=4$.

Is this correct?

[cloud walrus]

ali yassine

[lilac mango]

Yep

[tulip otter]

ok next is $K=\mathbb Q(\sqrt[3] 5,\sqrt{-2})$, how do I prove that $\sqrt[3] 5$ and $\sqrt{-2}$ are algebraically independent (i think thats the correct term, is it?), ie that $\mathbb Q(\sqrt[3] 5,\sqrt{-2})$ cannot be reduced to $\mathbb Q(\sqrt[3] 5)$ or $\mathbb Q(\sqrt{-2})$?

[cloud walrus]

ali yassine

[lilac mango]

Well one way is to note that sqrt(-2) isnt in the Q(cubed root of 5)

Which is immediate since the latter is a real field

[tulip otter]

I mean sure but isnt that what I am trying to prove

[lilac mango]

Yes but as I said Q(cubed root of 5) is contained in the reals

Or another way to look at it is that the degree of one is 3 and the degree of the other is 2

So one cant be an extension of the other

Since 2 doesnt divide 3 and 3 doesnt divide 2

[tulip otter]

oh yea right, but what about the other direction? I think its also immediate by noting that any power of sqrt(-2) is either a rational number or a multiple of sqrt(-2) so sqrt[3]{5} isnt an element of Q(sqrt{-2})?

[tulip otter]

ohhh

[lilac mango]

I mean what you are saying here is true but the proof of it is what I said

[tulip otter]

what you said proves that sqrt(-2) isnt an element of Q(cbrt 5) but does it also suffice for the other direction?

like shouldnt i also prove that Q(cbrt 5) is not contained in Q(sqrt(-2)) too

[lilac mango]

For that the degree argument suffices

[tulip otter]

yea fair

ok so [K:Q]=[Q(sqrt(-2)),Q(cbrt 5)] [Q(cbrt 5):Q]=[Q(sqrt(-2)),Q] [Q(cbrt 5):Q]=(2)(3)=6

[lilac mango]

Modulo your typo yes

[tulip otter]

which typo

the [Q(sqrt(-2)),Q(cbrt 5)]=[Q(sqrt(-2)),Q] part?

[lilac mango]

Should be [K: Q(cbrt 5)][Q(cbrt 5):Q]

[tulip otter]

ah yes mb

[cursive spindle]

Number theory...

[tulip otter]

is it necessary to explicitly prove that [K:Q(cbrt 5)]=[Q(sqrt(-2)):Q]?

[lilac mango]

Wdym

Thats not true

It is wait

Misread

[tulip otter]

ohh ok, np dw about it

[lilac mango]

I mean the fact that the first one is smaller than the second one is immediate

As the first one isthe degree of the minimal polynomial over Q[cbrt 5] of sqrt -2

And since the other one is 2 and the first one is not 1

You have it

Theres probably a more elegant way to do it

[tulip otter]

right

I was thinking about saying that it follows from the fact that sqrt(-2) and cbrt 5 are algebraically independent because from this the minimal polynomial of sqrt(-2) over Q(cbrt 5) should have coefficients in Q right?

it cant have cbrt 5 in the coefficients since that would contradict the fact that sqrt(-2) and cbrt 5 are algebraically independent.

or would it?

i mean yea it would

[lilac mango]

By algebraically independent you mean what exactly

[tulip otter]

I mean that cbrt 5 and sqrt(-2) dont satisfy any polynomial relation with coefficients in Q

[lilac mango]

Well I don't exactly get what your argument here is

[tulip otter]

if sqrt(-2) is a root of a polynomial with some of the coefficients in Q(cbrt 5) and not in Q then this would lead to a polynomial relation between sqrt(-2) and cbrt 5

but we know this cant happen

so any polynomial with coefficients in Q(cbrt 5) having sqrt(-2) as a root should in fact have all of its coefficients in Q

thus the minimal polynomial of sqrt(-2) over Q(cbrt 5) is the same as that over Q

which means that [K:Q(cbrt 5)]=[Q(sqrt(-2):Q]

is this correct?

[lilac mango]

Not sure of how that would lead to a polynomial relation exactly

Like a polynomial relation would be when one is a polynomial over the other

But I dont see how youd isolate one of them

Like you have anx^n+...+a0=0 where the ans are in Q[cbrt 5] and x is sqrt(-2)

How do you express x in terms of cbrt 5

Or the other way around

Youd need to solve the equation

[tulip otter]

so what I am trying to say is something like the following: assume that $\sqrt{-2}$ is a root of a polynomial $f(x)=\sum_{n\in\mathbb N}a_nx^n=\sum_{m,n\in\mathbb N}b_{n,m}(\sqrt[3] 5)^mx^n$ where $b_{n,m}\neq 0$ for some $m>0$ and $b_{n,m}\in\mathbb Q$ for all $m,n$. Then $(\sqrt[3] 5,\sqrt{-2})$ is a root of the polynomial $g(x,y)=$\sum_{m,n\in\mathbb N}b_{n,m}y^mx^n$ and hence $\sqrt[3] 5$ and $\sqrt{-2}$ are not algebraically independent over $\mathbb Q$ which is a contradiction

[cloud walrus]

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

[tulip otter]

yes, and this is a non-trivial polynomial relation if at least one of the coefficients is in Q(cbrt 5) and not in Q right

[lilac mango]

Thats why I was asking what you meant by algebraically independent

Because when you say that qe proved that they are algebraically independent what we proved is that one is not a polynomial in the other

And you are claiming that that means that there isnt a polynomial in those variables that evaluates to 0

Which is not at all clear and if I understand what you are trying to say its actually false

Since like cbrt(5), sqrt(-2) is a root of g(x,y)=x^3-5

And like I know you are trying to say that you need to include sqrt(-2) in the coefficients

But I wouldn't even know how to make that precise

I dont say it can't be done just I don't see how

[tulip otter]

did you mean something like g(x,y)=(x^3-5)(y^2+2)?

[lilac mango]

No I meant what I wrote

[tulip otter]

but how is sqrt(-2) a root of that

oh wait nvm

i cant even read lol

mb

[lilac mango]

np

Honestly Id say just thinking of degrees is the way to go

[tulip otter]

True

[lilac mango]

When I was studying this for the first time I also tried to do things the way you are trying to do them

But it never seemed to work

At the end of the day what you are doing when you think of degrees is in a sense making it a question about linear algebra

Which is usually simpler

Because linear algebra is easy

While studying relationships that arent linear is tough

[tulip otter]

I see, that makes sense

I will keep that in mind. Tysm for your help

[lilac mango]

You are welcome

[tulip otter]

Have a great day/night

[lilac mango]

You too

[whole quail]

Hello guys

Did u study the sylow subgroups in groups theory?

[ripe harbor]

Let $G$ be a group of order $p^2$, $p$ prime. I want to show that $G$ must be abelian.

[cloud walrus]

ILikeMathematics

[ripe harbor]

I thought about using the first Sylow theorem, we know that there exist subgroups $${e} \unlhd N_1 \unlhd N_2 = G$$ with $|N_1| = p$ and $|N_2| = p^2$

[cloud walrus]

ILikeMathematics

[candid patrol]

That’s not Sylow

[ripe harbor]

It is the first Sylow theorem

[candid patrol]

Not at all

[ripe harbor]

For |G| = p^k * m with p !| m there exists a chain of normal subgroups so that |N_i| = p^i and N_i normal in N_(i + 1)

Thats what I know the first Sylow theorem as..

[candid patrol]

That’s not really Sylow

Sylow is useless in a p-group

[ripe harbor]

What do you consider first Sylow theorem then

The one about conjugation of a p-group being in a Sylow group?

And all Sylow groups being isomorphic?

Thats my second

[candid patrol]

The one on the wiki

[ripe harbor]

And my third is the one with n_p cong 1 mod p and n_p | |G|

[ripe harbor]

Anyways so Im using this theorem

Do you think it helps?

Let G be a group and G/C(G) be cyclic, then G = C(G) where C(G) is the center

I want to use this

We know that |G/C(G)| = p^2/|C(G)| so either |C(G)| = 1, p or p^2

We need to rule out 1 and p and we are done

[candid patrol]

no

[harsh gale]

1 is not the case as the center is not trivial

[ripe harbor]

Why not

[candid patrol]

the center of a p-group is never trivial

[ripe harbor]

Why?

[warm badge]

consider class eq

[harsh gale]

yeah, just use Lagrange theorem

[warm badge]

also we cannot rule out p, that proves G is abelian too

[ripe harbor]

Why would it?

Oh

G/C(G) is cyclic since of order p

[ripe harbor]

|G| = [G : C(G)] * |C(G)|

[harsh gale]

And if center = p^2, then obviously it's abelian

[ripe harbor]

p^2 = [G : C(G)] * |C(G)|

Why cant we have |C(G)| = 1

[ripe harbor]

So p^2 = 1 + sum

[harsh gale]

You can't have such orbits, I believe

[ripe harbor]

Hm, why not

Oh wait

[harsh gale]

I forgot the proof

[ripe harbor]

Its not p^2 = ...

S is the set, not the group

Oh nevermind

In the case of center its the group

[harsh gale]

but Z should be the set of fixed points

[south patrol]

What about the trivial group

[next obsidian]

Pwned

Pwned

Pwned

[ripe harbor]

Ok yeah no it makes sense, we just mod both sides by p

Left becomes 0 since G is a p-group, right becomes |Z|, the sum vanishes because G != G_a

So p | |Z|

So Z is non-trivial

Thanks!

[harsh gale]

Actually I had the same question, can't remember, how I convinced myself of that

[ripe harbor]

Hm, could we do this without the G/C(G) cyclic => G = C(G) lemma?

If we construct this chain, we can say N_1 = <a> and N_2 = <a, b>, no?

Maybe that helps

[whole quail]

Question 1 guys

[desert verge]

a Sylow-p subgroup of G would have order p^2, so be G itself

[ripe harbor]

Yes, thats fine, our N_i are just subgroups

[whole quail]

I knew how to prove that P exists but i don't know how to prove that P is only one

[ripe harbor]

It doesnt ask for P to be unique, does it?

[desert verge]

but it doesn't ask u to prove P is unique

[ripe harbor]

How did you prove existence?

[desert verge]

i assume HK = {hk: h in H, k in K}

[harsh gale]

I got it, it's pretty simple

[whole quail]

Existing only

[ripe harbor]

You mean to a)?

[harsh gale]

in p^n you can have only orbits of order 1 or p or p^m such that p^m | p^n

[desert verge]

how do you know S_H, S_K are p-groups?

isn't a p-group a group of order p^n for some n

[whole quail]

That's the meaning of p group

[desert verge]

oh wait

nvm me

[whole quail]

Has an order p^n

[desert verge]

im fucked up

[ripe harbor]

Right

[harsh gale]

so there cannot be nontrivial orbit of order such that mod p != 0

Whoopie

[ripe harbor]

Are you talking about the justification for Z(G) non-trivial?

Or a new proof without using the lemma

[harsh gale]

I'm talking about theorem I posted, proof of which I forgot 🙂

[ripe harbor]

Ah

[harsh gale]

I believe that's the key theorem, so no need to not use Lagrange theorem and it's extension to group actions

I wanted to not use sylow theorem here

[whole quail]

That means we have a gPg-1 intersection H belongs to Sylp(H)
in the same way we can make by conjugation with another element from G (g1) such that... g1Pg1-1 intersection K belongs to Sylp(K) then what i want is to prove that g=g1 to make P is the same with H, K

Ohh

I think i wanna solve question2 in question 1

Oh GOD!!!!!!!!!

[ripe harbor]

.. and the center is an orbit

[whole quail]

I JUST REALIZED

[whole quail]

How it makes sence in that exercice!?

Can u explain more

[ripe harbor]

Im not talking about your exercise lol

[whole quail]

Oops🥲

Please anyone i wanna solve this issue

[whole quail]

Nope... I just realized again that i'm right now i want to prove P intersection belongs to both Sylp H and Sylp K

[harsh gale]

sry, know not much on Sylow groups

I’d be very happy to help the other way

[rocky cloak]

If you know about semidirect products you can finish from here

[ripe harbor]

Actually I did it over the G/Z(G) cyclic => G = Z(G) lemma

[rocky cloak]

That works too. Just need that p-groups have nontrivial center

[ripe harbor]

Let $G$ be a group of order $55$ and $\phi: G \times X \to X$ a group action with $|X| = 39$. I want to show $X^G \neq \varnothing$. Assume it is $\varnothing$. Then $|G.x| \geq 2$ for all $x \in X$. Also, since $|G.x| \mid |G|$, we get $|G.x| = 5$ or $11$. And we also have $$39 = \sum_{\substack{1 \leq i \leq n \ x_i \notin X^G}} [G : G_{x_i}].$$ How would we continue here?

[rocky cloak]

|G.x| shouldn't divide |X|

It will divide |G| though

[cloud walrus]

ILikeMathematics

[rocky cloak]

To continue, try to write 39 as a sum of 5s and 11s

[ripe harbor]

We cant

Three times 11 doesnt work, twice doesnt, once doesnt and no 11 doesnt

Ok so we are done

[rocky cloak]

Boom boom

[ripe harbor]

So basically it was just the orbit formula, everything before that was useless lol. Thanks!

[ripe harbor]

Also really quick: Is it true that $|\langle 1 \ 2 \rangle \langle 2 \ 3 \rangle \langle 3 \ 4 \rangle | = 2^3 =8$?

[cloud walrus]

ILikeMathematics

[ripe harbor]

Yeah, right?

[rocky cloak]

Is this cycle notation?

[ripe harbor]

yes

[rocky cloak]

Then no

[ripe harbor]

Do I have to reverse the entire thing?

Beginning with < 3 4>?

[rocky cloak]

To do what?

[ripe harbor]

To get order 8 (as order 2 multiplied 3 times)

[fickle dirge]

What's the motivation for studying rings? Like groups are symmetries, but why study rings? Like, why these specific axioms? In group theory, Cayley's theorem essentially shows that the axioms we have are the "correct" ones. Is there anything like that for rings?

[rocky cloak]

Like you want an element of order 8 in S4?

[ripe harbor]

In S_5, I want to get a subgroup of order 8

So I want to glue together 3 subgroups of order 2

[rocky cloak]

Well, there is a subgroup of S4 of order 8.

You can think about it geometrically

What is something with 4 points and a nice symmetry group?

[ripe harbor]

Oh, D_4

[ripe harbor]

If we didnt know D_4 though, is there a way to constuct such a subgroup

[rocky cloak]

Well, I'm not sure what "the gluing idea" is exactly.

But just looking at some elements of order 2 or 4 and check what they generate would work

[ripe harbor]

Ah nevermind, I was thinking about composing permutations but that doesnt help

Because in the end we will still be cyclic

And the order needs to be a prime, not 8

[rocky cloak]

In the same way that groups arise as symmetries of sets, rings arise as endomorphisms of abelian groups.

In particular if A is an abelian group, then End(A) (the set of homomorphisms from A to itself) is a ring where the multiplication is composition.

Just as in Cayley's theorem you always get a ring homomorphism R -> End(R).

The analog for a group action by a ring is what one calls a module and they show up everywhere.

[ripe harbor]

The other day we thought about this for quite some time, but we couldnt find a way other than looking at $\phi: G \to \text{Sym}(X)$ which I dont really want: \[10pt] Let $G = \langle a, b\rangle$ be a group with $\text{ord}(a) = 7$ and $\text{ord}(b) = 11$. Let $\phi: G \times X \to X$ be a group action and $|X| = 8$. Show that $\phi$ cant be transitive.

[cloud walrus]

ILikeMathematics

[ripe harbor]

The problem is that we can only say 77 | |G| but not |G| = 77

[placid shale]

and why do i care about the endomorphisms of abelian groups? why abelian groups specifically?

I often hear that rings are the study of "what's a number" or what behaves like a number and its properties

but that's not clear because N is not a ring, and the definition of a ring doesn't encompass all the properties of Z

what should I be looking for intuition-wise? and why do we care about this specific definition of a ring

[fickle dirge]

Thank you!

[noble nexus]

Idk I would say numbers are a good thing to think about

N isn't a ring, but when you work with natural numbers you want to do subtraction so the integers are fairly natural

[rocky cloak]

Rings are pretty broad and you can be interested in them for various reasons.

It is true that Z, Z/nZ and number rings that appear in algebraic number theory are all rings, so rings are very much a generalization of numbers. But they can do more than that.

I think the insight comes from linear algebra, where it was realized that systems of equations correspond to homomorphisms of vector spaces, and these do form a ring (the ring of nxn matrices).

[candid patrol]

Zzz

[ripe harbor]

Any ideas?

[noble nexus]

really just the best intuition is to write down a bunch of examples and abstract their common properties

this is how most algebraic structures come about

[ripe harbor]

I want to prove that for $n \geq 2$, $[S_n, S_n] = A_n$ (the commutator group). We know that $$A_n \unlhd S_n \quad \text{with} \quad [S_n : A_n] = 2$$ and that $N = [G, G]$ is the smallest normal subgroup of $G$ so that $G/N$ is abelian. Can we use these?

[cloud walrus]

ILikeMathematics

[placid shale]

Historically what was the motivation that made us reach the current definition?

Also the ring of nxn matrices has a unity, so it is not clear how we got to the standard definition without the unity?

group theory is the study of symmetries and that's clear. every symmetry is a group and every group is some symmetry, and we need exactly the group axioms for this no more and no less. Is there is something similar for ring theory

ring theory is the study of {something}. every {something} is a ring and every ring is {something} and we need exactly the ring axioms for this no more and no less

[rocky cloak]

The standard definition has unity.

Every ring is the endomorphism ring of a module, it is just a linear version of symmetries really

The way you get the definition without unity, is just taking the standard definition and dropping unity.

People relax axioms of things all the time. Not to characterize some specific thing, but because they only use the axioms they need to prove the theorems they want

[placid shale]

Thank you very much

So ring theory is the study of "linear symmetries," right?

Also another question if you would let me. If that's the intuition behind ring theory, what is the intuition behind commutative algebra. I know that it is a subset of ring theory, but why do we specifically care about commutative rings?

[rocky cloak]

So commutative algebra is motivated a little differently. It's two main applications would be in number theory (so this is very much the generalization of numbers type thing), and the other in geometry. If you have some space X then the set of (continuous/smooth/whatever) functions to R or C forms a ring, and ring properties here encode geometric properties which then motivates commutative algebra

[wraith cargo]

well clearly since S_n/A_n is abelian, then A_n contains the commutator subgroup

now you have to show that S_n doesn't have a larger abelian quotient

[placid shale]

I don't understand the geometry part of your answer.

Also if commutative algebra is motivated by number theory and geometry in the same time then what are these 2 fields have in common that commutative algebra studies?

I really liked the answer you gave for ring theory being the linear version of symmetries. Is there a similar answer for commutative algebra

[swift root]

modern algebraic geometry is the answer

it encapsulates both classical algebraic geometry and number theory

[placid shale]

so what's commutative algebra? 😅

what is it studying? I think it is different from modern algebraic geometry

[wraith cargo]

it's not

modern algebraic geometry and commutative algebra are almost the same

there are some major differences

but classical algebraic geometry is very different

while modern algebraic geometry is very deeply connected to commutative algebra

[swift root]

commutative algebra is about commutative rings and their modules

but that´s very intertwined with modern algebraic geometry which is, well, about mostly the same but viewed through the lense of geometry

[ripe harbor]

Right. Why is S_n/A_n abelian though?

[swift root]

order 2

[placid shale]

yes but why do i care about these rings and these modules specifically?

What is the deep structure that commutativity resembles?

what are "commutative linear symmetries" exactly in that intuition

[swift root]

they're numbers

or, they're rings of functions on a subset of affine space

[ripe harbor]

Ah, right

[next obsidian]

They’re friends :)

[swift root]

in classical algebraic geometry, you have an alg closed field k (which I hopefully shouldn't have to motivate why we care about these), and consider subsets of k^n which are solutions to systems of polynomial equations. Between these sets one may look at functions which are restricted polynomial functions. It turns out that looking at this collection of sets and functions is essentially the same as looking at the collection of finitely generated commutative reduced k-algebras and k-algebra homomorphisms

[south patrol]

Classical algebraic geometry is algebraic geometry which isn't derived/spectral etc

[noble nexus]

rings generalize simultaneously number systems, functions, and linear operators

[swift root]

on a higher level, commutativity is important because it allows you to have localisation, and this in turn allows you to extend the above fact to arbitrary commutative rings

[noble nexus]

which happens to include a lot of things we care about

there is a bit of a mystery why commutative is so relevant

in some sense commutative seems to correspond to geometry

[ripe harbor]

Assume it does and S_n/N is abelian with A_n c N. Then N must contain an element with sign -1. I feel like that will help in some way

[noble nexus]

or rather "classical" geometry

[swift root]

funny, this is essentially my current research project

[wraith cargo]

okay I expressed myself incorrectly

[karmic moat]

non-commutative is hard

:(

[noble nexus]

this is because if you have some sort of geometric space, the functions on the space will form a commutative ring

[swift root]

(on a more general scale, of course)

[desert verge]

R is commutative

[noble nexus]

(functions being real or complex or values in some field)

[candid patrol]

It's not An subset in N

[desert verge]

sniped

[wraith cargo]

you want the opposite, i.e. N c A_n

[ripe harbor]

Oh

[cloud walrus]

lexi

[somber goblet]

especially if you dont have the order of the extension to begin with

[wraith cargo]

yeah when I said larger quotient I meant smalled subgroup haha

[next obsidian]

Ur a small subgroup

[ripe harbor]

Assume it does and S_n/N is abelian with N c A_n. Then, since S_n/A_n c S_n/N, we have |S_n/N| >= 3

[placid shale]

I think i will go with the intuition that commutative linear symmetries are numbers instead 😅

i didn't study algebraic geometry yet so i will be lost. I just hoped to find a motivation for commutative algebra which i am studying now like how "what's area" motivates measure theory, and that motivation helped me study measure theory really well so i wanted to do the same with commutative algebra and ring theory

the "linear symmetries" for ring theory really clicked with me and i liked it. I feel that numbers with commutative algebra is ok, but it doesn't tell me why these matter in geometry so idrk if i can have a simple motivation like "what's area" for that without needing prerequisites i don't have

[placid shale]

i would be actually interested in reading such a research

[swift root]

well, it has a prerequisite of universal algebra

I don't know if you've ever heard of that?

[placid shale]

i heard of it yes, but didn't study it

i just know that it studies what structures can exist

[slim kayak]

Linear symmetries for ring theory?

[placid shale]

^

[noble nexus]

Commutative algebra is pretty dry on its own, that's usually why people recommending studying it at the same time as studying algebraic geometry or algebraic number theory

[placid shale]

this would make me ask what's the fundamental problem of algebraic number theory or [modern] algebraic geometry

[swift root]

yeah, you study algebraic structures (over sets) in their generality, and in particular you try to say something about exactly what properties cause these classes of algebraic structures to behave the way they do

[placid shale]

that's really interesting, I might consider studying it in the future

[flint timber]

before that rings and module needed i believe

[noble nexus]

Yeah of course

[swift root]

I think it's really cool, though not as popular nowadays

to great sadness and despair

[copper kestrel]

practice final was revealed

[slim kayak]

Eh? Any module gives rise to an endomorphism ring, but the only canonical way(that I can thikn of) of getting R as the endomorphism ring of a S-module M is to take M=R and R=S

Which is completely circular. (R-)linearity requires the ring structure of R

[rocky cloak]

I mean, that's the only canonical way to make G the symmetry group of something as well

The question is more what concept is it rings capture, and they are the things that act on modules.

[slim kayak]

Sure but saying "groups are the collections of symmetries of groups" is pretty silly. Z and its extensions and k-linear endomorphisms give a rich class of structures motivating the ring axioms, I agree there.

[ruby pilot]

Do people try to discover what groups are realisable as class ideal groups over a number field like they do Galois groups?

Over Q, I mean.

[rocky cloak]

It seems people do
https://mathoverflow.net/questions/252313/is-every-group-an-ideal-class-group-of-a-number-field

[chilly radish]

It's certainly not true that S_n/A_n is a subset or subgroup of S_n/N.

[next obsidian]

Am I stupid or is S_n/A_n not just Z/2Z lol

[chilly radish]

You can define an embedding S_n/A_n->S_n/N but you should be careful with using "subset" loosely in these kinds of settings

[chilly radish]

Yes but I meant as sets

[next obsidian]

Ok

My point is like embedding Z/2Z into something is just picking an involution so it’s pretty damn arbitrary

Which I guess is also your point

[dire wren]

Is a better intuition of associative structures string manipulation or function composition

(Ik how to prove that these are the same)

Or do you have smthing better than both

[rocky cloak]

I think both are good. Why pick just one?

[dire wren]

Like let us say i want to think about associative magmas with a single sided unit, I find that the function viewpoint gives me much more information, but is more difficult to think about, so was wondering if the same applies to other things as well, maybe there is something which combines the simplicity of string manip and the information of function?

In particular the function viewpoint told me associative structures with a single sided unit is just the same as a associative structure with a unit, but with classes of objects for each object in the structure with a full unit

The string manip told me nothing special.

[vapid vale]

If you need something simpler than the function viewpoint why not just ignore it when you don’t want to think about it

[dire wren]

I don't need something simpler, just wanted to know if there was any such thing.

Because just right now, I was doing that famous exercise about showing that left inverse and left identity and associativity is enough to guarantee group.

[rocky cloak]

I'm not sure what you mean here. You can cook up collections of functions where you only have one sided unit

For example take the set of constant functions. Then every element is a right unit.

Every associative operation can be realized as some set of functions under composition in fact

Because if * is a binary operation then
f(y) = x*y
is a function.

And associative is saying exactly that composing two such function is the same as using the operation:

composition of functions applied to z = (x * (y * z) ) = (x * y) * z

[swift root]

fundamentally, these are the same thing

string manipulation, as you call it, is simply function composition when you cant assume any relations between the functions

if you extend this to functions between multiple objects you get the idea of paths in a directed graph (i.e. free category)

[dire wren]

Ik how to prove that

[dire wren]

Ik how to prove that, i am just saying which is a better intuition

[swift root]

^

[dire wren]

Ik that, but same thing doesn't mean same intuition

[swift root]

neither is better

[rocky cloak]

Then I don't understand what you mean by function composition telling you that one sided unit implies unit.

[swift root]

thats also not true

[rocky cloak]

Indeed, hence my confusion

[swift root]

the operation xy = x is associative and every element is a right-unit

[rocky cloak]

Exactly the example I gave

[swift root]

oh i may have neglected to read

[noble nexus]

more intuitions is better than less

[swift root]

that too

[astral ivy]

Can anyone help me understand the process around Wilson's theorem? I just keep getting it wrong...

I know that by definition (p-1)! = -1 (mod p) iff p is prime.

Example: Find a remainder of $(1001!\cdot 994!)^{19961} (mod 1997)$ if we know 1997 is prime.

So, 1997 = p

1996! = -1 (mod 1997)

The issue begins when I need to manipulate the original number.

[cloud walrus]

dan

[astral ivy]

1996 = 1001! * 1002...1996

but how exactly do I transform those 'free' digits into 994!

am I even supposed to ask here or this falls more into elementary number theory?

[swift root]

this seems more number theory

[astral ivy]

yee

but it's in my abstract algebra course for whatever reason

[swift root]

hm, apparently you can use sylow lol

[young solstice]

ah like write 1002 as -995

and so on

[astral ivy]

we didn't study Sylow on this course

[young solstice]

this gives that 1002\times \dots \times 1996 \equiv - 995!

[astral ivy]

ahh I see

but how did you get to that arithmetic?

this is very tricky for me

did you just 1997 (modulo number) - 1002

[young solstice]

see you have 1002 which is like very big so you reduce it mod 1997

[swift root]

🥲

[astral ivy]

and then (-1)^994 or whatever is the number

[young solstice]

yes

[young solstice]

its like -1^995 right?

[astral ivy]

yes

that's tricky

because how did you count there being 995 or 994 operations between 1997 and 1001 or 1002?

that's the confusing part for me

[young solstice]

not really obs we get -1 from 995 to 1 and from odd to odd there are always odd number of number so the prod is -1

[young solstice]

see 1002 corresponds to -995 right

and 1996 correponds to -1

[astral ivy]

well yeah, -(1997 - 1002) = -995 I guess

[young solstice]

so 1002 \times \dots \equiv 1996 \equiv (-1)^{number of numbers from 1 to 995} \times 995! then the above reasoning applies

tho obs that the exponent is 19961 which is 1 mod 1996