groups-rings-fields

[sturdy mirage]

but there exists no polynomial q(x) such that x+1 = q(x)*f(x)

[zinc cloud]

...

but we need it of the form f(x)*p(x) +[1]

which we can then do

to get x+1

[sturdy mirage]

remind me again, why do we need that? for what?

i thought you asked in general why [0] was exactly equal to I

[zinc cloud]

.

yes why did u chose this to be [0]

u keep saying g(x)*p(x) +[0] = [0]

im so confuse

[sturdy mirage]

because he said we need to find an h(x) such that x*h(x) + 1 belongs to the ideal. and i said that "belonging to the ideal" means the same as "being equal to [0]"

thats all i talked about after that. being in the ideal is equivalent to being equal to [0]

[zinc cloud]

then did u mean for it to become h(x)*p(x) +[0] =[0]

if so why did the [1] become [0]

[cloud walrus]

ProphetX

[cloud walrus]

ProphetX

ProphetX

[sinful mirage]

How does this follow from the above?

[chilly ocean]

howcome $1+1\neq{0}$ in fields larger than 2 elements?

[cloud walrus]

Reaper

[chilly ocean]

Like a field with 3, 4, 5 elements

[hidden haven]

It can be 0 in a field larger than 2 elements

[gritty sparrow]

In the field of 4 elements, it is 0

[hidden haven]

Just not in prime cardinality fields

[gritty sparrow]

Moldi based opinion on two?

[hidden haven]

How can an even number be prime bro

@chilly ocean try to prove this is the case for 3 elements, then try to make a similar argument work for 5

[chilly ocean]

I figured out the tables

at least

[hidden haven]

ohh

Addition and multiplication both?

[chilly ocean]

Yeah

I used what you said

it helped me a lot

1+1=a

[hidden haven]

Yay

[chilly ocean]

but idk if i did something weird

[hidden haven]

Send table

[chilly ocean]

i said 1+a=b, 1+b=c and 1+c=0

then was able to find all the rest

[hidden haven]

Right, so you're trying to see to what extent those choices are forced?

[chilly ocean]

well

Idk if I saw that

but

I did find that

the elements in each row and column

have to be like unique

so it became a sudoku kinda puzzle

[hidden haven]

Yess

[chilly ocean]

Also I used it to show multiplication elements aswell

for instance

[hidden haven]

How did you prove that each element appears in each row and column exactly once?

[chilly ocean]

Cancellation laws

ab = ac \implies b=c

but that contradicts uniqueness

[hidden haven]

Yes

But for multiplication you have to be careful because cancellation only works when a≠0

[chilly ocean]

The only exception is multiplication table where it has a row and column of zeros

yessss

[hidden haven]

Yes

[chilly ocean]

Everything worked really well

[hidden haven]

Yeah, I didn't think about solving it like a sudoku

That's pretty nice

[hidden haven]

Yess

[chilly ocean]

I used addition table for multiplication

[hidden haven]

Yes this is exactly what was needed

[chilly ocean]

I chose to find

all the squares first

like a^2, b^2, c^2

then find ab, ac, b*c because they helped with distributivity

[hidden haven]

Elegant solution

[chilly ocean]

Damn

yeah this makes so much more sense now

[hidden haven]

I guess the only thing that remains is to see why 1+1 can't be 0 or 1, and a+1 can't be 0,1,a and so on

[chilly ocean]

Yeah

[hidden haven]

For that try to see what adding 1 does to the remaining elements

[chilly ocean]

oh damn

i forgot about that

it can't be 0, 1, a because of uniqueness

in all cases i think

[hidden haven]

Uniqueness says it can't be 1 or a

[chilly ocean]

Ohhh

hmmm

[hidden haven]

So that solves those 2 cases

[chilly ocean]

Hmmm

[hidden haven]

And similarly the for b+1 not being 1,a,b

But for all these cases you have to show that they can't give 0

1+1, a+1, b+1

[chilly ocean]

oh a+1!=0 since (1+1)+1!=0

cause then 1=0?

No wait

[hidden haven]

Then c+1 will have to be 0 because thats the only thing that will remain in the row

[chilly ocean]

thats just wrong

[hidden haven]

! as in factorial?

[chilly ocean]

! as in not

[hidden haven]

ah

[chilly ocean]

ohhhhh crap

[hidden haven]

Yeah so a+1 = 1+1+1 part is correct (once you have proven that 1+1 ≠ 0)

But then you need to show that 1+1+1 ≠ 0

[chilly ocean]

Ohhh

[hidden haven]

So 3 things to show, 1+1≠0, 1+1+1≠0 and 1+1+1+1≠0

Go in order

[chilly ocean]

suppose 1+1=0 then a+1=0 => 1=0 which is a contradiction

er

[hidden haven]

That proves that 1+1 and 1+1+1 both cannot be 0 simultaneously

(you are assuming both 1+1=0 and a+1=0 to get the contradiction)

[chilly ocean]

Yeah

[hidden haven]

So just start with 1+1 = 0

And see why this gives a contradiction

[chilly ocean]

You can do the same for 1+1+1+1 != 0

Ok ok

[hidden haven]

Yes but we need to show that none of the 3 happen on their own

[chilly ocean]

right.

[hidden haven]

Which is slightly harder to prove

So if 1+1 = 0, look at the row of the table corresponding to adding 1

You have 0+1 = 1 and 1+1 = 0

Try to complete the row

[chilly ocean]

Hmmm

Ok

wait which row?

the second?

[hidden haven]

Yes

[chilly ocean]

1+a would have to be b or c

[hidden haven]

The +1 row

[chilly ocean]

similarly 1+b a or c and 1+c a or b

[hidden haven]

Right. So can you pick 1+a to be whichever one of b and c or is a choice forced?

[chilly ocean]

There is no choice forced yet.

[hidden haven]

Yes, because so far b and c behave exactly the same with regards to addition

[chilly ocean]

Yeah

[hidden haven]

So you can just pick a+1 = b

[chilly ocean]

but we might get a contradiction down the line

[hidden haven]

Yes

That's what we want

Can you now try and fill the remaining 2 things in the row?

[chilly ocean]

Yeah so

a+1 = b, b+1=c and c+1=a

[hidden haven]

Right, and that is forced by the uniqueness

Try to see why this is problematic

[chilly ocean]

wait how so?

oh

[hidden haven]

c+1 = a or b, but b is already a+1, so c+1 ≠ b

Similarly the other equation after this

[chilly ocean]

Ohh yess

hmmm

well

this is problematic because

(a+1)+1=c => a+(1+1)=c => a=c

[hidden haven]

Yess

So this is like 0 and 1 form a cycle of 2 elements under +1

while a,b,c form a cycle of 3 elements under +1

And that's not possible

Because +1 twice should already give you 0

(I'm not saying this very precisely, I hope it makes sense)

[chilly ocean]

I don't understand the terminology too well but I get what you mean

Is all of this learned in abstract algebra?

[hidden haven]

Yeah it's not proper terminology

[chilly ocean]

Like properly?

[hidden haven]

Yes

[chilly ocean]

Table questions like this?

[hidden haven]

No there are more efficient ways to do this :P

But we can build up to that using what you learn from the table

[chilly ocean]

Damnn

Where did you learn this from?

Any sources?

[hidden haven]

There are lots of books on abstract algebra. A famous one is dummit and Foote but I haven't read it, you can try asking in #book-recommendations

I learned this in a course in my college using a different book

[chilly ocean]

gah damn

[hidden haven]

Now can you try to prove this more general thing: if 0, 1, and a few other elements form a cycle of m elements under +1, then any other cycles under +1 should also have m elements

So here you got a cycle of 2 and a cycle of 3 and that gave a contradition

But you want to prove this general thing that all cycles should have the same size

[sullen island]

hey, im trying to prove this boxed equality for derivations of $\mathbb{F}-$algebras, and i typed out my attempt here. i got stuck at where i left off and stopped typing, and dont know how to continue to obtain the desired result. would appreciate any help and pointers!

[cloud walrus]

xy

[hidden haven]

xy, could you wait a bit? I'll tag you when the channel is free, and you can repost

[sullen island]

yes, sure! sorry for interrupting

[hidden haven]

[chilly ocean]

Oh its ok!

Yeah ill see if i can prove it

[hidden haven]

Cool

[chilly ocean]

Ill check out the textbook

Abstract Algebra seems sort of daunting

[hidden haven]

If it's done in order it usually isn't

But jumping to constructing finite fields directly is a bit hard

[chilly ocean]

Yeah

this course is really weird

out of no where we did fields

[hidden haven]

oof

So if you want to prove this theorem using the textbook then it will be done in a more general setting using some group theory

It's called Lagrange's theorem (this is a specific easier case)

But here you can use the same trick you used for cycle of 2 and cycle of 3

[chilly ocean]

Hmmm

[hidden haven]

The cycle of 2 tells you that 1+1 = 0, but then for any x in the cycle of 3, x+1+1 = x which means it's really only a cycle of 2

[chilly ocean]

Ohhh

[hidden haven]

Same thing works for cycles of any m and n

[chilly ocean]

Alright alright

ill take a look

I have go now

[hidden haven]

ok cool

[chilly ocean]

thank you moldilocks

[hidden haven]

Welcome

[chilly ocean]

ill come back and ask more questions eventually

🙂

[hidden haven]

Sure

@sullen island

[sullen island]

thanks moldi 😄

hey, im trying to prove this boxed equality for derivations of $\mathbb{F}-$algebras, and i typed out my attempt here. i got stuck at where i left off and stopped typing, and dont know how to continue to obtain the desired result. would appreciate any help and pointers!

[cloud walrus]

xy

[sullen island]

for context, $\delta$ is a derivation of an algebra and $\delta_s$, $\delta_n$ are its semisimple and nilpotent parts respectively

[cloud walrus]

xy

[sullen island]

<@&286206848099549185>

[gritty sparrow]

Here is a hint: you are taking $\delta - (a+b)$ on a sum of products, so distribute it on each term and use what we know in the n-1 case. Once you do that collect the like terms together

[cloud walrus]

saketh

[gritty sparrow]

I think this should do the trick, but I haven’t verified too carefully

[sullen island]

i split it up into $(\delta - a) - b$ and distributed it, then one of the terms will have powers summing up to $n$ instead of $n-1$

[cloud walrus]

xy

[sullen island]

which is as close as possible to getting something of the form ${{n}\choose{i}}$

the inductive case is $$(\delta - (a+b))^{n-1}(vw) = \sum\limits_{i=0}^{n-1} {{n-1}\choose{i}} ((\delta - a)^{n-1-i}v) ((\delta - b)^{i} w)$$

[cloud walrus]

xy

[gritty sparrow]

It is probably possible, by your way, but the term collection looks quite tedious

Should i just send a pic of what i’ve done?

[sullen island]

sure thing

[sullen island]

im hoping somewhere along the way i can use pascal's identity: ${{n}\choose{i}} = {{n-1}\choose{i}} + {{n-1}\choose{i-1}}$

[cloud walrus]

xy

[gritty sparrow]

Yeah i used that

[sullen island]

oh but you're doing $(\delta - ab)$

[cloud walrus]

xy

[sullen island]

not $\delta - (a + b)$ 🤔

[cloud walrus]

xy

[gritty sparrow]

Typo

(Or writo?)

[sullen island]

hmm, im not understanding the second equality

[gritty sparrow]

Did you see what i defined as vi’ and wi’?

[sullen island]

yeah

[gritty sparrow]

Then i just applied the operator on them

[sullen island]

shouldnt it just be $\sum\limits_{i=0}^{n-1} v_i 'w_i'$?

[cloud walrus]

xy

[sullen island]

oh okay i see it now sorry

[gritty sparrow]

Cool

[sullen island]

right, lemme continue reading

[vestal snow]

Let B be a basis of a topological space X and P a presheaf on the basis. Let F_Psh and F_Sh represent the functor which turn P into a presheaf and sheaf on the topological space X respectively. Let G_B and G_X represent the sheafification functors on the presheafs of B and X respectively. Is it true that G_X(F_Psh(P)) = F_Sh(G_B(P))?

[sullen island]

i cant derive the third equality you have @gritty sparrow

[gritty sparrow]

Use the n=1 case on vi and wi

[sullen island]

holy shit

okay i got the desired result for my proof

ur a genius 😄

[gritty sparrow]

Thx

[snow flint]

they were talking to me

[minor badger]

If i have a connected matrix group G with Lie algebra g, and i have shown that two representations of g are isomorphic, is there an easy way to show that the corresponding representations of G are isomorphic? i know that there exists an morphism by a theorem but is the bijectivity preserved?

[vestal snow]

How do I prove that this presheaf on a distinguished base is a sheaf?

More specifically, how do I prove the gluability axiom

[chilly ocean]

i dont know

but what book is that

[next obsidian]

I think you can do it in a slick way using that Prod A_{fi} is a faithful flat extension

But in general idk, I don’t wanna think about it

[vestal snow]

The Rising Sea by Ravi Vakil

[sturdy marsh]

lmao

you would still need to prove that faithfully flat descent works

[next obsidian]

Well...

“Slick” in the sense that it’s like... cool

[cloud walrus]

squirtlespoof

[next obsidian]

Is the polynomial degree n already?

If so, if it wasn’t irreducible then the extension just can’t be degree n!

[hot lake]

which polynomial are you talking about

[next obsidian]

Something along those lines. I can’t think about it yet at the moment since I need to drive, but it’s something like that I think

[hot lake]

galois extension => separable extension

separable polynomial => the polynomial has distinct roots

?

[hidden haven]

How did you show isomorphic to S_n implies that there are some n roots that it is acting on?

[hot lake]

isn't that the whole point of the problem

[hidden haven]

yeah but I think they said that was done?

so I was confused what the problem was

ah i see

[hot lake]

but you don't have roots

since you don't have a polynomial

[hidden haven]

you can say that things in the same orbit will have the same min polynomial

yes

[hot lake]

no because K is already given

but you can say "let f in F[t] be some polynomial such that K is the splitting field of f"

[hidden haven]

every finite normal extension is the splitting field of a polynomial in F[t]

[hidden haven]

how is that different?

[hot lake]

it starts with "let f be" instead of "let K be"

[hidden haven]

bruh I dont think what you start with matters

but ig better phrasing

yes

what was your argument?

[hot lake]

f could have degree up to n!

but you want a polynomial of degree n

[hidden haven]

yes

[hot lake]

you can't because f might have degree n!

[hidden haven]

yes exactly

[hot lake]

unless you take more care about how you choose it

but if you define f by just saying "let f be some polynomial such that K is the splitting field of f"

then f might have degree n!

[hidden haven]

for example the extension is simple since its finite separable

which would mean that its also the splitting field of a degree n! polynomial

I'll have to think

[hot lake]

maybe you can look into how you proved "there is a polynomial f such that K is the splitting field of f" and see if you can change the process

[hidden haven]

my approach would be to look at the orbit of a primitive element and see if I can pick out an n element set on which the galois group acts exactly like S_n

The action on that orbit is transitive and faithful, but idk yet what else I could say

kinda important for galois theory lol

but might be a different way to do this

[hidden haven]

this actually makes no sense, nvm

every element in that orbit will be primitive so it wont give anything

[hot lake]

also maybe it helps if you know the fundamental theorem of galois theory

[hidden haven]

yeah thats what im trying to use

I have an idea

Let r be a primitive element

Then orbit of r under S_n has n! many elements (all the conjugates of r)

Let H be a subgroup of S_n

Let O be the orbit of r under H

This is a subset of the original orbit

take the product or sum of everything in O

(I think product or sum wont matter, both should work)

lets say sum

Then think of the size of the orbit of that sum

under all of S_n

H acts trivially on that sum

ah shit give me a sec

[hot lake]

yeah you're on the right track

if you could explicit which H to take a little

[hidden haven]

yeah im just not sure, aren't stabilizers normal?

But i think Im getting that H is the stabilizer?

of that sum

so im just seeing what else can be there

[hot lake]

so you define H by saying it is the stabilizer of the sum of the orbit of r under H ?

[hidden haven]

No, I wanted to prove that the orbit of the sum under S_n would have size the same as index of H

and then I could just take H to be any of the (n-1)! order subgroups of S_n

but I think I dont get equality

Because H is only contained in the stabilizer of the sum

[hot lake]

yeah it can happen that you are super unlucky and the stabilizer is bigger, whether you take the sum or the product or any other method

[hidden haven]

[hot lake]

that's why the fundamental theorem of galois theory is nice

[hidden haven]

does that give something here?

[hot lake]

well... yeah ?

[hidden haven]

I dont see it

[hot lake]

you want an element x such that stab(x) = H

[hidden haven]

oh god

take fixed field of H

then a primitive element

will that work?

[hot lake]

just anything in the fixed field of H that is not in the fixed fields of bigger subgroups

[hidden haven]

Ah right

[hot lake]

I think primitive element of K^H works too

[hidden haven]

hmm let me see how this helps now

[hot lake]

and then it leaves the choice of the subgroup H of Sn to make the rest work

[hidden haven]

Oh wait

I already said it makes the rest work

[hot lake]

I wouldn't be so sure

[hidden haven]

Stabilizer = H → orbit has index H many elements. The minimal polynomial of those index H elements is an irreducible deg index H polynomial over F and now my intuition suggests that it's splitting field is K

[hot lake]

you need to show that K = F({gi(x) for gi in G/H}), and irreduciblity of the minimal polynomial of x should be uuh easy

[hidden haven]

Yeah

[hot lake]

also, how do you know Sn has a subgroup of size (n-1)!

[hidden haven]

The subgroup which fixes 1

When acting on {1,...,n}

[hot lake]

okay

to show the splitting field is K you need to show that the intersection of all the conjugates of H is {id}

[hidden haven]

Well that is true

Because conjugates of H are the subgroups fixing a k in 1,...,n

[hot lake]

yeah

[hidden haven]

And if each k is fixed that's just id

[hot lake]

idk how you would do it without an explicit choice of H

so that's why I said to get an explicit H earlier

[hidden haven]

Ah I see

[carmine fossil]

What have you tried?

[pine patio]

im not sure how to begin

[hidden haven]

Just finished going through the proof again in my head squirtle did you follow any of the discussion?

Ok I'll continue after this ig

[carmine fossil]

Recall what a conjugacy class is

More importantly recall what the number of elements in one conjugacy class is

[pine patio]

|{x in G : g = hxh^-1 for some h in G}|

[carmine fossil]

Your attack plan is to group elements of G into conjugacy classes and note sum of |C(g)| as g varies over elements of one conjugacy is |G|

[pine patio]

is C(g) the conjugacy class of g?

[hot lake]

I don't think so

[hidden haven]

That doesn't seem right? Since C(1) has 1 element

So the sum is just 1 for that conjugacy class

[pine patio]

do u mean C_G(g)

[carmine fossil]

Yes

[pine patio]

then for C_G(e) i see why the sum is |G|. let me convince myself for others

i mean i see why |C_G(e)| = |G|

[carmine fossil]

There is one element in the conjugacy class containing e

And everything commutes with e

[pine patio]

C_G(e) is the centraliser of e, so it contains all of G

[hidden haven]

oof

I was thinking C_G(g) means conjugacy class of g

[pine patio]

yeah. theyre similar so confusing

[hidden haven]

Usually centralised of S is Z(S) lol

[pine patio]

isnt that the centre of the entire group?

[hidden haven]

Z(G) is the centre

[pine patio]

so suppose we have 2 elements in a conjugacy class {a,b}. then ur saying that C_G(a)+C_G(b) = |G|

[hidden haven]

Z(S) is centraliser of S for a subset S

[pine patio]

oh ok thats a better notation

[carmine fossil]

Yes

[pine patio]

why is that true

[low surge]

generally, the orbits partition the set and can use orbit stabiliser

||so you can write the sum as |g| times sum of 1/|orb(x)|. but sum of 1/|orb(x)| counts each distinct orbit once, as each 1/|orb(x)| is counted precisely |orb(x)| times for all of its representatives and they are disjoint so no overlapping||

can some1 check thats right

[pine patio]

oh so if G acts on itself then number of orbits is same as the number of conjugacy classes

right?

[low surge]

i dont think thats true

im using orbit stabiliser then it falls out by just a counting argument really

[pine patio]

so how does this relate to the number of conjugacy classes

ik this formula

oh

[carmine fossil]

Let's say there is an element a in a conjugacy class

[small karma]

Ques 14

[low surge]

you dont need burnside's lemma however you can use the thinking behind it

[carmine fossil]

Then there are |G|/|C_G(a)| elements in the said conjugacy class

[pine patio]

so if G acts on itself by conjugation then isnt it true that |fix(g)| = |C_G(g)|

[low surge]

btw number of conjugacy classes = number of distinct orbits i just prefer using the words orbit and stabiliser because its easier for my brain lol

[carmine fossil]

And if b=gag^-1 |C_G(b)|=|C_G(a)|

So, sum over a conjugacy class just becomes |C_G(a)|+|C_G(a)|
... |G|/|C_G(a)| times

[low surge]

actually i think this is true

then it would be immediate

but doesnt that defeat the purpose of the q?

[carmine fossil]

I think the point of this question is to prove burnside

[pine patio]

theres so many terms and results to remember that it gets very confusing

[low surge]

ye thats why i use orbit and stabiliser

[pine patio]

idk what the purpose of the qu is

[low surge]

as in using burnside makes it too easy

because its too powerful

[pine patio]

fun fact

[low surge]

but they already have lots of stuff attributed to them

i think calling it burnside makes it less confusing

idk

lol

[pine patio]

why is this true

[low surge]

isnt that just orbit stabiliser?

[pine patio]

oh so again we're thinking of G acting on itself by conjugation right?

[low surge]

yes

C_G(a) = stab(a)
ccl_G(a) = orb(a)

ye the cs are confusing

[pine patio]

ok i think i understand it now. but idk how to get used to all this terminology. it's that the lecture notes that i use are too succinct and dont discuss different ways of writing the same things

and when i read other texts then it gets confusing

[low surge]

i think just using it helps lol

[pine patio]

what text did u use?

[low surge]

none

i just sat in lectures and never consulted any books

[pine patio]

ur lecturer must be good, and u must be very attentive

[low surge]

i guess im lucky because my lecturer is good

well its a recorded video

[pine patio]

my lecturer is online and she just reads a set of hand written notes

[low surge]

o ok

my lecturer writes as she talks, so its a good pace

[pine patio]

but i havent seen any of her lectures cause it seems like she reads the exact notes

maybe she writes them on the go

ill check them out lol

thank you very much for ur help!

[low surge]

np

the rest of the q i think is fairly straightforward

[pine patio]

and @carmine fossil too

[pine patio]

yes so if G is abelian then the classes are singletons, so m = |G| and it fits with the probablilty of 1

[low surge]

ye

and the probability itself is just noting g, h commute iff g is in stab(h)

[small karma]

<@&286206848099549185>

[low surge]

an even length cycle is odd and an odd length cycle is even

so alpha is odd and beta is even

can you solve the rest yourself?

note btw that the sign of the permutation is the same as that of alpha^9 times beta

even though the elements dont commute

(ig technically coz sgn is a homomorphism into C2 which is an abelian group)

[sturdy mirage]

I have an exercise here.
To show that $Z_{(p)}$ is a subring, I'm first trying to check if it is closed under addition. To do that I want to check that if $a,b \in Z_{(p)}$ with $a = \frac{a_1}{a_2}$ then $a + b \in Z_{(p)}$ right? But how can I check that $gcd(a_1b_2 + b_1a_2, a_2b_2) = 1$ ?

[cloud walrus]

reking

[sturdy mirage]

And.. do I even need to check this?

[celest mantle]

we don't really care about gcd here, your ring is about p not dividing the denominator, so you can just check whether or not p | a2b2

[sturdy mirage]

yes, that just followed by euclid's lemma, but the gcd thing confused me a little.

Thanks

[pine patio]

ive done the first bit, but im not sure about the second

[low surge]

doesnt the group of symmetries of the tetrahedron include reflections?

i dont think you can conjugate a rotation to get a reflection

because det(XRX^-1)=det(X)^2det(R)=1 where X is some matrix in O3

@pine patio

@rustic crown i like ur name

lol

[rustic crown]

[pine patio]

oh yeah i think i read the qu wrong. i thought they meant the group of symmetries of cube and S4, and it wasnt making much sense

[low surge]

o i think the group of symmetries of a cube isnt S4

i think its S4 x C2 iirc

u certainly have more elements anyway

[pine patio]

rotational i meant*

[low surge]

yeye

[pine patio]

wait ur saying the group of rotational symmetries of a cube isnt S4?

[low surge]

no it is

i meant all of the symmetries are S4 x C2

i think

[pine patio]

how can u say stuff like this so quickly

itd take me like 15 mins to figure it out

[low surge]

because my exam is next week

lmao

[sturdy mirage]

where you at bobles?

[pine patio]

mine is soon too

[low surge]

luckily its paired with vector calculus so i can rely on the other if the qs are bad

[low surge]

university

UK

[sturdy mirage]

oh it just sounded like the same thing as me, female lecturer writing as she talks, exam next week

[low surge]

o lol

[pine patio]

mine too lol

[low surge]

i think its just exam season for everyone

r u guys US?

[sturdy mirage]

norway

[pine patio]

no uk too

[low surge]

o ok

first yr?

[pine patio]

yeah

[low surge]

epic

[pine patio]

u?

[low surge]

same

lol

[pine patio]

xDDD

[sturdy mirage]

this is like supposed to be 2nd year for me i believe, but i took courses all in a messed up order

[low surge]

its mandatory for us i think

[pine patio]

same

[viscid pewter]

i'm not sure it's S4 x C2 precisely? i just feel like they shouldn't commute

and with a direct product you will surely get them commuting

[rustic crown]

same... but ig they do commute somehow

[viscid pewter]

are you sure?

[rustic crown]

yea

[viscid pewter]

why

[rustic crown]

fit the cube in R3 centered at 0

then you can take the odd element to be x --> -x

(if you wish call it -id)

and definitely that commutes

[low surge]

S4 isnt abelian

if thats what u mean?

[viscid pewter]

no, i know, but i feel like the rotations shouldn't commute with the reflections

[rustic crown]

yea they probably don't

[viscid pewter]

which they would if it were a direct product?

[rustic crown]

but that C2 doesn't correspond to reflection about a plane

its reflection about 0 which is reflection about a plane composed with a rotation

[viscid pewter]

well ofc

[low surge]

does the cross product imply they commute?

[viscid pewter]

but sometimes that rotation will be 0

[low surge]

hmmm

[rustic crown]

didn't get you

[rustic crown]

so like if you have two group G and H then the elements (g, 1_H) and (1_G, h) commute... that's what we mean by commute here i think

[low surge]

oh wait i see the problem now

lol

[rustic crown]

reflection about the point 0 isn't same as reflection about a plane... you need to multiply it with a non-trivial rotation to get reflection by a plane... and since center of S4 is trivial this won't commute

[viscid pewter]

wtf is reflection about the point 0, enlargement by scale factor -1 or something?

[low surge]

yeah i think the C2 is due to a point reflection

it has to be

like an inversion

x to -x as det said

[viscid pewter]

yeah i don't actually know what that means

[rustic crown]

ye

[low surge]

acc i dont think inversion is the right word

lol

[rustic crown]

i said to fit the cube in R3 centered at 0... so symmetries of cube will be just linear maps on R3

[viscid pewter]

i don't suppose you could just give me the transformation matrix for a point reflection, i'm just lost

like (-1, 0, 0; 0, -1, 0; 0, 0, -1)?

[rustic crown]

yea -id

[viscid pewter]

ok

right, that's what i thought

[low surge]

i think for the cube it swaps vertices that lie on opposite ends of a long diagonal

[viscid pewter]

yes

[low surge]

but hard to visualise

lol

[rustic crown]

so this has @rustic crown -1, which means not an element of the rotational group

[viscid pewter]

heh

[viscid pewter]

oh, yes

[low surge]

ye so that group includes the point reflection

even though we cant acc carry it out

[viscid pewter]

well we can tho

with maths and logic

[low surge]

epic maths moment

[rustic crown]

lol

you mean with preserving orientation right

[viscid pewter]

how come no one talks about preserving orientation for 2d shapes

but suddenly when you want to rotate a cube in 4d everyone's like 'nooo'

[low surge]

ig orientation is just determinant

so u can still talk about it in a way

idk

[rustic crown]

well we sadly live in 3d space 😦

[viscid pewter]

whack

[low surge]

maybe we live in 4d

but we cant see the 4d

[rustic crown]

i can only visualise 2d tho

[viscid pewter]

i wanna see a 24-cell

[rustic crown]

3d is too hard

[viscid pewter]

i can visualise a tetrahedron

[low surge]

flex

[viscid pewter]

it took me many years of practice

[low surge]

lol

[rustic crown]

well technically i can't even visualise a square... i can just draw it on a paper and say i'm visualizing it

[sturdy mirage]

How would I go about finding the smallest field containing $\mathbb{F}_5$ and roots of both $x^2 - 2$ and $x^2 - 3$? The exercise here sort of gives away the primitive polynomial $f(x) = x^2 + x + 2$, but by which method would I find this myself?

[cloud walrus]

reking

[sturdy mirage]

is there some algorithm or something to find a primitive polynomial like that?

[golden pasture]

F_{25} will contain both x^2-2 and x^3-3

there is a unique field of size p^n

[sturdy mirage]

My first instinct was that this field would have to be of size 5^4, not 5^2.. but appearently the priomitive polynomial only has to be deg 2

[golden pasture]

$$\frac{\mathbb F_5[x]}{x^2-2}\cong\frac{\mathbb F_5[x]}{x^2-3}\cong\frac{\mathbb F_5[x]}{x^2+x+2}$$

[cloud walrus]

ari 十年生死两茫茫，不思量，自难忘。

[sturdy mirage]

yes, since they are all of the same order, they have to be isometric, right?

[golden pasture]

isomorphic yes

which means

[delicate bloom]

helps to recognize 2=-3 mod 5 and that sqrt(-1) is contained in F_5

[golden pasture]

sqrt(3) is in F_5[sqrt(2)]

every quadratic polynomial has both roots in F_5[sqrt(2)]

[delicate bloom]

just to clarify what I meant, square root both sides of that eqn you get sqrt(2) = sqrt(-1)*sqrt(3) so if you have one, the other is just sqrt(-1) times the other, and ofc by sqrt(-1) I mean either 2 or 3 since these all square to -1 mod 5

[sturdy mirage]

right, right, because 3 * sqrt(2) = sqrt(3) in F_5... i think i see it now

[golden pasture]

but you have this anyways since the polynomials are irreducible

[sturdy mirage]

is it generally hard to find a primitive polynomial?

[golden pasture]

The most brute force way is to factor x^{p^n}-x

apparently as long as char isnt 2 we will have some primitive polynomial of the form x^n-ax-b

so one way is to just guess a,b and then see if you can factor

[delicate bloom]

I guess that does factor kind of nicely, $$x^{p^n}-x = x*(x^{p^n-1}-1) = x (x^k-1)\frac{x^{p^n-1}-1}{x^k-1}$$ then expand out the geometric series, assuming k divides $p^n-1$ ofc

[cloud walrus]

Merosity

[gritty sparrow]

can’t we just say that x^4=-1 so x^5\=x but x^8=1 hence x^25=x for both of the roots

[golden pasture]

it doesnt factor nicely

it literally gives you

every possible poly

[delicate bloom]

I was trying to be optimistic lmao

[golden pasture]

lol

[delicate bloom]

idk maybe you could reason on the degree with how you split it apart with the divisor k or something to make some kind of argument but... yeahhhh... lol

[golden pasture]

lemme

look at how sage does ti

https://github.com/sagemath/sage/blob/6aa4ecee167d8152ad17370f674b5e778ba7cbc6/src/sage/rings/finite_rings/finite_field_constructor.py#L589

does anyone

feel like r

eading

OL

so

sage

does it

randomly

[delicate bloom]

hmmm lol

[golden pasture]

if exists_conway_polynomial(p, n):

https://github.com/sagemath/sage/blob/6aa4ecee167d8152ad17370f674b5e778ba7cbc6/src/sage/rings/polynomial/polynomial_ring.py#L3320

https://github.com/sagemath/sage/blob/6aa4ecee167d8152ad17370f674b5e778ba7cbc6/src/sage/rings/finite_rings/conway_polynomials.py#L19

[golden pasture]

this is the key alg that computes primitive poly

R(sage.databases.conway.ConwayPolynomials()[p][n])

thanks sage

If you wish to live dangerously, you can tell the constructor not to test irreducibility using check_irreducible=False, but this can easily lead to crashes and hangs – so do not do it unless you know that the modulus really is irreducible!

lmaooo

[delicate bloom]

oh I vaguely remember hearing about conway polynomials a long time ago

Since Conway polynomials are expensive to compute, they must be stored to be used in practice. Databases of Conway polynomials are available in the computer algebra systems GAP,[1] Macaulay2,[2] Magma,[3] SageMath,[4] and at the web site of Frank Lübeck.[5]

[golden pasture]

ahh

[kindred mist]

So I am somewhat confused, according to Hungerford, any homomorphic image of a divisible abelian group is divisible

if he means epimorphism then I would understand

[hot lake]

doesn't image mean the image of the morphism

[kindred mist]

oh

yeah it does, ok nevermind. Thanks

[celest mantle]

Does every quadratic extension is built using the quotient K[X]/(X^2 - d) ? (is it even always a polynomial of that form?)

[golden pasture]

For char not 2 yea cuz like

quadratic equation

[celest mantle]

Yea forgot that special case, im just working with Q

[delicate bloom]

maybe start with x^2+ax+b and solve for k such that (x+k)^2+a(x+k)+b = x^2 - d

[golden pasture]

K[x]/(x^2+ax+b) = K[x]/((x+a/2)^2+b-a^2/4) = K[y]/(y^2+b-a^2/4)

[delicate bloom]

I guess geometrically translation is just making it into an even function, so you can think it's just translating by where the vertex is

[next obsidian]

How could it fail even in char 2?

[golden pasture]

F2[x]/(x^2+x+1)

[next obsidian]

Wut?

Oh x^2 - d

[delicate bloom]

x^2-d = (x-e)^2 in char 2 since d=e^2

[next obsidian]

Hurb, my bad

I thought that was a general quadratic there

[golden pasture]

🇫

[celest mantle]

Thanks for your answers, as for the first question, is it always true over Q ?

Ah well i think it is, as Q[X]/(X^2-d) is isomorphic to Q[sqrt(d)] and as d is algebraic we have Q[sqrt(d)] = Q(sqrt(d)), so theorically we could construct every quadratic extension of Q using that, am i wrong ?

[delicate bloom]

thank you for thanking us for our answers, but it'd be even better if you read them

[celest mantle]

Well

Am i missing something ? Cause i just saw that you answered to the first question

Uh the second mb

Yea i just misread the first message, mb again

[daring ibex]

Question

i am not sure how the following definition of the wedge product

Is equivalent to the defn on wikipedia

i.e. the induced operation after you quotient the tensor algebra on a vector space V by the ideal generated by all elements of the form x \otimes x

They result in the same (or at least very similar) operation but i still want to see a formal proof

i haven't been able to do it on my own tho oof

i prefer the latter defn but this book does use this a lot so hm

[oblique river]

that definition on wikipedia is for the exterior algebra

which is larger than the wedge product of V with itself

[daring ibex]

Yes

[oblique river]

also i think that the definition in that book is only equivalent to the usual definition if the field doesn't have characteristic 2

[daring ibex]

I mean like i was worried about equivalence if you take the subspace

the 2-exterior power of V

[oblique river]

i'm not sure what you mean

[daring ibex]

Like if you take the subspace of the exterior algebra $\wedge^2 V$ and consider the alternating product there, is this the same as the definition given above

[hot lake]

is it even true that the set of v1 * v2 - v2 * v1 is a subspace of V² ?

[oblique river]

in characteristic not equal to 2, yes

[cloud walrus]

i.imgur.com/0ddbSVy.png

[oblique river]

no zef that's not true, the book means "subspace generated by"

[daring ibex]

how so tho

[oblique river]

i think the notation hints at how it should be true

one construction is as a quotient of $V \otimes V$. the image of an element $v \otimes w$ is written $v \wedge w$

[cloud walrus]

Buncho Bananas

[hot lake]

the definition on wikipedia comes with a projection from V² to V ^ V ?

[oblique river]

the other construction is the subspace generated by things of the form $v \otimes w - w \otimes v \in V \otimes V$

[cloud walrus]

Buncho Bananas

[oblique river]

the identification between these two spaces is $v \wedge w \leftrightarrow v \otimes w - w \otimes v$

[cloud walrus]

Buncho Bananas

[oblique river]

(i have it in my mind that maybe there should be a factor of 1/2 here as well)

(but that's more for bookkeeping)

[daring ibex]

[sturdy mirage]

can someone explain to me in clear text what it means that a group action induces an injective group homomorphism? i just can't parse this

[oblique river]

maybe after the current conversation is done

[hot lake]

are you talking to a physicist

[daring ibex]

so i was reading up about this and i think this has something to do with the alternating tensor algebra. I think what u just said is similar to the canonical iso b/w $\wedge(V) \cong A(V)$

[cloud walrus]

i.imgur.com/0ddbSVy.png

[oblique river]

I don't know what $\bigwedge(V)$ and $A(V)$ are supposed to be

[cloud walrus]

Buncho Bananas

[daring ibex]

But i was weirded out by the factor of 1/2 induced

Ah sorry one sec

[oblique river]

in particular, which one is which?

[daring ibex]

ill just ss the wikipedia one

the wedge(V) is the exterior alg

[oblique river]

so the reason for the 1/2

[daring ibex]

A(V) is alternating tensor alg

[oblique river]

oh, so AV is the subspace of TV and wedgeV is the quotient of TV?

[daring ibex]

And also it requires that the base field have char 0 but i thought that this couldn't be right bcs the author later talks about the field perhaps being finite

[oblique river]

it is correct

if you want an isomorphism of the entire alternating algebra

but if all you care about is the degree-2 piece

you only need teh characteristic to be not equal to 2

[daring ibex]

This couldn't be right as in this isn't what i am looking for

[daring ibex]

yeah

[daring ibex]

hm could you elaborate here

[oblique river]

i mean imagine that your field has characteristic 2

then 1 = -1

[daring ibex]

yeah

Oh then we have x \otimes y + y \otimes x?

[oblique river]

yes, and the problem is that when you quotient by the x \otimes x

[daring ibex]

and that goes to 0

[oblique river]

that's going to be zero

[daring ibex]

in the quotient

yeah

[oblique river]

yeah

[daring ibex]

Hrm

[oblique river]

in higher degrees, you need to be careful about more primes

in degree 3 you need p \neq 3 for example

well actually you need p \neq 2 or 3

[daring ibex]

ah i see

[oblique river]

because there is a factor of 1/n!

[daring ibex]

so in the end

when we aggregate all these cases together

we eventually have to require the char be 0

[oblique river]

that's right

[daring ibex]

okay makes sense a little

[oblique river]

but again if all you care about is the degree-2 piece then you only need to assume the characteristic isn't 2

[daring ibex]

hm, also why the 1/2 factor

sorry if im being a bit dense i just woke up

[oblique river]

in the quotient map, $v \otimes w \mapsto v \wedge w$

[cloud walrus]

Buncho Bananas

[oblique river]

which means that $v \otimes w - w \otimes v \mapsto v \wedge w - w \wedge v = v \wedge w + v \wedge w = 2(v \wedge w)$

[cloud walrus]

Buncho Bananas

[oblique river]

because of the alternating property

[daring ibex]

ohh yeah

[oblique river]

this means that $v \wedge w$ should be associated with $\frac{1}{2}(v \otimes w - w \otimes v)$

[cloud walrus]

Buncho Bananas

[daring ibex]

yes that makes sense

So all i have to do is show

The morphism given by $V \wedge V \to A(V)_2$ and $v \wedge w \mapsto \frac{1}{2}(v \otimes w - w \otimes v)$ has trivial kernel?

[cloud walrus]

i.imgur.com/0ddbSVy.png

[daring ibex]

A(V)_2 being the 2 dim part of A(V)

[oblique river]

that would do it

alternatively you could go the other way. instead of viewing $V \wedge V$ as a subspace of $\wedge V$ which is a quotient of $T(V)$, you can view $V \wedge V$ as just the quotient of $V \otimes V$ by the subspace generated by $v \otimes v$ for $v \in V$

[cloud walrus]

Buncho Bananas

[oblique river]

so you have a surjective map $V \otimes V \to V \wedge V$ and you can restrict it to the subspace $A(V)_2$

[cloud walrus]

Buncho Bananas

[oblique river]

and then ask if that's still surjective/injective

[daring ibex]

Yeah that's more economical since it restricts to stuff we actually care about atm

okay i think this is starting to make a bit of sense

One more thing, in the wiki defn, what's a homogenous and a decomposable tensor

[oblique river]

in T(V) you have tensors of all differnet degrees

and you can add them together

for example $v \otimes v + v \otimes w \otimes x$ is a sum of a degree-2 tensor and a degree-3 tensor

[cloud walrus]

Buncho Bananas

[oblique river]

a homogeneous element is a sum of terms all of the same degree

and decomposable means that ti's literally of the form $v \otimes w \otimes x$

[cloud walrus]

Buncho Bananas

[oblique river]

and not a sum of things of that form

[daring ibex]

ohh i see

[oblique river]

for example $a \otimes b + c \otimes d$ is not decomposable in general. it's a sum of two decomposable things though

[cloud walrus]

Buncho Bananas

[daring ibex]

so decomposable ones are in a sense those ones that can't be decomposed further

[oblique river]

i suppose :P

[daring ibex]

oh can you elaborate on one more thing? I'm sorry im being a bit slow today, but out of curiousity i'd like to see how this extends to stuff like $\bigwedge^k V \to A(V)_k$, and why we successively how to worry about more primes

[cloud walrus]

i.imgur.com/0ddbSVy.png

[oblique river]

the general formula for A(V)_k is: for every k-tuple (v1, ..., vk) of elements of v, define Alt(v1, ..., vk) as the sum of all permutations of v1 otimes v2 ... otimes vk, but each one is weighted by the sign of the permutation

for k = 3, there are 6 permutations of three elements

so it would look like

[daring ibex]

i see i see

[oblique river]

$Alt(x, y, z) = x \otimes y \otimes z + z \otimes x \otimes y + y \otimes z \otimes x - x \otimes z \otimes y - y \otimes x \otimes z - z \otimes y \otimes x$

[cloud walrus]

Buncho Bananas

[oblique river]

or something like that

A(V)_3 is spanned by these elements

[daring ibex]

hm yes

[oblique river]

in the quotient when you send $V \otimes V \otimes V$ to $\wedge^3 V$

[cloud walrus]

Buncho Bananas

[oblique river]

the element $Alt(x,y,z)$ will get sent to $6(x \wedge y \wedge z)$

[cloud walrus]

Buncho Bananas

[daring ibex]

this beconds 6(x \otimes y \otimes z)

[oblique river]

yes

[daring ibex]

or well the wedge

yeah

[oblique river]

so now you can see where the n! comes from

for hte degree-n stuff

[daring ibex]

yeah

[oblique river]

btw you should think of the quotient definition as the "right" definition

[daring ibex]

I think it's beginning to click

[oblique river]

and the subspace definition is equivalent (in some cases) and occasionally useful

[daring ibex]

Yeah, i prefer the first one

So the antisymmeterization defined here

having the r! there is kind of wrong?

[oblique river]

that definition only works in characteristic 0

[daring ibex]

right bcs 1/r! is in Q but who knows what field we're in? I guess it works nevermind, since if a field has char 0 it contains Q

[oblique river]

or more generally "characteristic not equal to any prime less than r"

you can define this Alt in two different ways, with the r! or without it

[daring ibex]

hm, but if we do it this way what does the 1/r! even mean? i dont think it'd be an element of ur field

[oblique river]

what do you mean?

r! is an element of any field

if r! is nonzero, it has a multiplicative inverse

[daring ibex]

Oh

[oblique river]

5! = 5*4*3*2*1 makes sense in any field

[daring ibex]

Yeah then we will define r! = (1)(1+1)...(1+...+1) right

where we don't have more terms in each sum than r

that's the only sensible defn i can think of lol

[oblique river]

I'm not sure what you're worried about

the number 3 exists in every field

you don't have to write 1 + 1 + 1

every field K admits a unique ring homomorphism from Z

this lets us talk about the integers in any field

[daring ibex]

oh yeah

[oblique river]

if the field is characteristic 0, this extends to a (unique) field homomorhpism from Q to K

[daring ibex]

Yeah

[oblique river]

so I can talk about the number 17/31 in any field of characteristic 0

[daring ibex]

Okay yes makes sense

[oblique river]

(more generally in any field of characteristic not equal to 31)

[daring ibex]

yeah

bcs multiplicative inverse

okay thanks a lot buncho this really helped

[oblique river]

👍

i'm gonna head off for a bit

so if you have more questions maybe someone else can help or you can save them for when i'm on later

[daring ibex]

sure

im probably just going to take today off, today is not a math day i was just really worried about this from yesterday bcs it's from a book i was looking forward to reading that i skimmed and if this was wrong it'd kinda mess everything up lol

[unique juniper]

just wondering why its called two dimensional

when it has triples

mmm

[delicate bloom]

because you can make lambda force one element to be 1 always, like (X,Y,1)

is one way to think of it

[unique juniper]

oh right yeah

[delicate bloom]

although not always, if that element is 0, you can still have the point (X,Y,0)

if you haven't seen it, you probably will see that you can decompose projective space into a union of affine spaces of lower and lower dimension

basically just by splitting off just like this between 0 and 1

[unique juniper]

i have not

but i have seen infinite points

[delicate bloom]

like the last coordinate can be 0 or nonzero, so we have points like (X,Y,1) and (X,Y,0)

[unique juniper]

yep

[delicate bloom]

but we could just take (X,Y,0) and force Y=1 if it's nonzero and so we again have (X,1,0) and (X,0,0)

so now (X,1,0) is one dimensional and (X,0,0) is a point, actually just (1,0,0) since we don't include (0,0,0)

[unique juniper]

yeah :D

[delicate bloom]

so there's how to break down P^2 into A^2 U A^1 U a pt lol

[unique juniper]

ohhhh

nice

ty

[delicate bloom]

yeah yw

[golden pasture]

In context of manifolds P^n(R) is a n-manifold because locally it looks like R^n

[golden pasture]

so we do this

ignore points at infinity

and you get your R^n back

do it for every component and you cover the manifold

[old hollow]

How do you intuit squaring an ideal

[hidden haven]

You take the smallest ideal that contains products of things in the given one

It's an abstract thing idk if there's a way to intuit it except in specific cases

Such as in Z, (nZ)² = n²Z

[golden pasture]

abstract thing

you intuit

by thinking of special cases

intuit com ring theory through arithmetic examples

[final pasture]

or just draw arrows and read nlab. You won't get anything, but you'll look cooler

[golden pasture]

[old lava]

it's easy to intuit abstract concepts

just keep using them without understanding them

until your brain unwillingly adapts to thinking those concepts are right

[golden pasture]

honestly for me like my intuition of com alg kinda comes from ag memes

and ag memes intuit either comes from number theory or complex surfaces

which are actually pretty reasonable intuit

[hidden haven]

So I was really studying when I was looking at memes before exam

[final pasture]

tbh that does work sometimes, like you don't have any intuition for some stuff, you just keep using it, and then some day you realize you actually understand why everything works intuitively

[golden pasture]

my intuit for oo-memes surprisingly is from classical topology constructions so far

i still cant rmb

[free/forgetul] functors are [left/right] adjoint that preserves [colimits/limts]

[final pasture]

ariana be like: I don't do math. I do math meme

[golden pasture]

idk why but im seriously directionally impaired

[golden pasture]

maybe im a forgetful functor

[hidden haven]

You develop intuition from seeing more results because often you can intuit at least some of those

And solving problems is the best way to see results on whatever you're having trouble with

[final pasture]

yeah

[old lava]

sometimes there's just completely arbitrary definitions which have no intuition whatsoever when you first see them

and the definition was only really made so that the results following it are nice

in some way

and so you have to wait for the results

[final pasture]

definitely makes sense to me

(definition of a category in dependent-type theory )

[old lava]

[final pasture]

(to be fair in this case the only problem is the syntax, what it's saying is easy, but w/e 🐒)

[green stone]

anyone here study any sporadic simple groups before?

[scarlet estuary]

"studied" what about them

afaik theres not much interesting stuff to say about them, besides their behaviour as subgroups of the monster group

(and the 6-7 that arent)

[chilly ocean]

Let * be an associative binary operation on the set G such that the following properties hold: i) There exists e in G such that g * e = g for all g in G. ii) For each g in G, there exists g' in G such that g * g' = e. Prove that (G, *) is a group.

We have to prove that e * g = g and g' * g = e, but this is harder than it looks. It does not seem like there is enough information to do so.

[carmine fossil]

Pick a g

g g'=e for some g' and g' g"=e for some g"

you have to show g"=g

[chilly ocean]

Okay

[carmine fossil]

Ok,That requires you to show e g=g

[chilly ocean]

It does

[daring ibex]

[chilly ocean]

I have no idea on how to proceed

I have g = (g * g) * g' = g * e

How do I get e * g from there?

It seems impossible

[hidden haven]

eg = e(ge), and e = g'g" so eg = e(gg'g") = eeg" = eg" but idk how to proceed

[carmine fossil]

This feels wrong

[hidden haven]

pretty sure this is a known result

i remember working through this

eg" = gg'g" = ge = g

so eg = eg" = g

g'g = g'g(g'g") = g'gg'g'' = g'eg" = g'g" = e

I think this isnt enough when you have a right identity and a left inverse

and vice versa

[chilly ocean]

That's the problem from the book

[hidden haven]

damn, you have to construct a counterexample?

[chilly ocean]

No? It is true

Or supposedly true

[hidden haven]

oh

wait so any 1-sided identity and any 1-sided inverse is enough?

I proved it for when you have right identity and right inverse

[daring ibex]

it might just be a typo

or do they explicitly say left identity and right inverse (or vice versa)

[hidden haven]

which is the one you asked for

let me think about the other one

[chilly ocean]

I wrote it exactly as written in the book

[daring ibex]

might just be a typo from the book

[hidden haven]

yeah that one is saying that right id and right inverse exist

[lofty yew]

try taking the right inverse of g'g"=e, g'*e is the inverse of g", meaning g' is the inverse of g", so g(inverse g") =e, use right multiplication of g" to get g = g"

[hidden haven]

and then it works out

[hidden haven]

the solution is starting from here

[tough raven]

Solution obtained by doing random things until you succeed:
||for any g, g_1, g_2 such that g g_1 = g_1 g_2 = e;
g g_1 g_2 = g e = g = e g_2
For any g, g_1, g_2, g_3, g_4 such that g g_1 = g_i g_(i+1) = e;
g_2 = e g_4 and g = e g_2 = e e g_4 = e g_4 = g_2. So g = g_2 and e g = g.
Since some g_1, ..., g_4 must exist for any g, e is a left identity. Since g = g_2, g_1 g = e.||

Or perhaps a clearer rephrasing:
||1. to show that e is left identity: observe that for any g that can be written as e g', we have e g = e e g' = e g' = g. So we just need to show any element is e times something else.
For any g, there are g_1, g_2 such that g g_1 = g_1 g_2 = e; then g = g e = g g_1 g_2 = e g_2.
Thus e is a left identity.
2. Above, we have that g = e g_2, but e is a left identity. So g = g_2, so g_1 g = e.||

[tough raven]

Yes, IIRC there's a counterexample with the operation being min or something

I remember because I once tried to come up with an almost-identity that would circumvent the usual proof of uniqueness of identities by saying
f g = g f = g, except if g = e (the real identity)
then did some Googling and found this, in which every element is like an "almost identity" with "exceptions" for all the elements less than it (for max as operation)

[hidden haven]

Did you mean g g_1 = g_1 g_2 = e in the definition of g_1 and g_2?

[tough raven]

Yes, thanks

[scarlet estuary]

seems like a lame exercise

is there a better tactic for approaching it than trying stuff at random until it works

[chilly ocean]

@tough raven I don't think your proof is right?

observe that for any g that can be written as e g', we have e g = e e g' = e g' = g. So we just need to show any element is e times something else.

How do you know that g = e g'?

[hidden haven]

hes using g' for that

not for right inverse

like g' is defined as g = eg'

and then he proves that any element is of this form

[lavish gale]

egg= 🥚 ~ 0= e

[hidden haven]

[tough raven]

0 =/= e
1 = e

No wait

e = 2.718

Hence e(GG - 1) = -2.71828

[cloud walrus]

squirtlespoof

[hot lake]

you mean the discriminant ?

so if the roots are a,b,c in Fp³ then (a-b)(a-c)(b-a)(b-c)(c-a)(c-b) = -4p³-27q² ?

so I have a 50% chance of making a sign mistake

it's a square in Fp³ I guess

but now you only have to show

if something in Fp is a square of something in Fp³ then it was a square in Fp

because (a-b)(b-c)(a-c) is in Fp³

a,b,c are in Fp³ and not in Fp

because the polynomial is irreducible in Fp

no

F(p³)

the finite field with p³ elements

the degree 3 extension of Fp

in which a,b,c live their happy root lives

you have a degree 3 irreducible polynomial f(x)

so you can make an extension field Fp[x]/(f)

[hidden haven]

The coefficients of f are in the finite fields F, but it's not given that F = F_p

[hot lake]

oh dear are we using the letter p for two different things