groups-rings-fields

[tribal moss]

I repeat, I do not know or remember what your goal was beyond showing directly that Z[x,y] satisfies the universal property.

[void cosmos]

bro i did not ask about remembering i just told you what my goal was

.

[delicate orchid]

they're putting Z[x,y] in the bottom right of the triangle

[void cosmos]

yesa

as my C

[delicate orchid]

that might be the missing piece of the puzzle

[void cosmos]

i said that 3 times ;d

[tribal moss]

No, I do not know what your goal was, and I'm not going to try to scroll up an hour to find it!

[void cosmos]

@tribal moss yea i am not telling you to scroll up

f i were to find an inverse to the map induced (from the tensor product to Z[x,y] ) from finding a bilinear map Z[x] x Z[x] to Z[x,y]
would that be the same as showing its isomorphic to the tensor product?

just say yes or no ;D

[tribal moss]

Here you were asking me about a different approach to an unknown goal. I have no opinion about whether that would have reached the goal THAT I NEVER KNEW WHAT WAS.

[void cosmos]

wait you never knew i was trying to show that Z[x] tensor Z[x] is Z[x,y]?

[delicate orchid]

that was the first goal too tropo

[tribal moss]

I thought that was a strategy towards something else.

[tribal moss]

Then this still doesn't make sense to me.

[void cosmos]

basically taking Z[x,y] as my C in the diagram

and ik the tensor product exists , so it has the universal rpoperty, so it has a homomorphism from it to Z[x,y]

now whats missing is an inverse

is htat correct

[cloud walrus]

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

[tribal moss]

That would be showing Z[x,y] is isomorphic to the tensor product, not that Z[x,y] is itself a tensor product.

[void cosmos]

got you

your amazing

tysm

for your patience

u make an excellent tutor

[delicate orchid]

although, wouldn't that follow quickly?

[void cosmos]

it would but he says it in a way so that it gets to my intuition

he deserves a medal

[tribal moss]

You can show in general that anything that is isomorphic to a tensor product is itself a tensor product too, but that step would need to be there.

[void cosmos]

ty all guys

and for the problem that started it yea it worked

Z[x] tensor_Z Z[x] is just Z[x,y]

while Z[x] tensor_Z[x] Z[x] is just Z[x]

[delicate orchid]

huzzah

[vagrant zinc]

I'm looking at examples 2 and 4 of the contemporary algebra book and it tells me that if G is a cyclic group of order n, then G is isomorphic to Z_n, then it gives me example 4, I've tried to interpret it but I don't understand it, could someone explain it to me in more detail?

gives me this function
a^k->kmodn
I have tried to apply the function but it is not given, in detail

[void cosmos]

whats example 4

do you mean 2

yo @south patrol in the proof of tensor product is right exact

why exactly is the sequence Hom(M,Hom(N,P)) --> ... is exact?

like after u simplify things using "adjoint associativity"

[south patrol]

Well

[south patrol]

this sequence is obtained just by applying Hom(-, Hom(N,P))

and we know that that functor is uh left exact

[void cosmos]

right yeahh

cuz N --> N' --> ... is exact

cool

tysm

[south patrol]

np

[remote nymph]

I'm trying to understand orderly algorithms on graphs and I'm a bit confused by this function:

where G_X is the setwise stabiliser of X in G

If G_X stabilizes X, wouldnt the orbit of G_X always be X?

there is also this paragraph:

[untold turret]

in Z/(3), 2' and 5' are representatives of the same coset (the ' indicates they're representatives). how do i formalize the fact that the canonical quotient map applied to 2 sends it to the "natural" representative 2', while this is not the case for 5?

this is not the case 8, 11, etc, either

[summer path]

the quotient map sends 5 to the equivalence class [2] = {2,5,8,...}, so in some sense they already are the same

as you said, they are representatives of the coset, so you already have that [2] = [5] ...

[untold turret]

right, i'm trying to say, if we mod 2 by 3, it doesn't change, but if we mod 5 by 3, it does

[summer path]

what

[untold turret]

2 mod 3 is 2, while 5 mod 3 is 2

5 has changed, 2 has not. this is informal, how do I make this formal?

well, what I really want to say is, considering instead ideals of Z/(3), we have (2) + (3) = (2), while (5) + (3) = (2)

so, in another sense, if A is an ideal of a ring R, and considering the quotient map R to R/A, if all ideals I in R have I mod A = I, we have that the quotient map is a bijection

i want to get at what is in a sense a minimal representative

[tribal moss]

Umm, what does "I mod A = I" even mean? A priori the two sides are ideals of different rings ...

[untold turret]

yeah that's the question

suppose an ideal B contains A. then the quotient map R to R/A sends B to B + A. but notice, 2B + A = B + A, and I want to be able to say 2B+A is not minimal

uh, i'm not sure we have 2B+A = B+A actually

but we have 2B+A equals another ideal, but that certainly isn't the minimal representation of that other ideal

the minimal representative of (7) + (18) in Z/(18) is (1) + (18), for example

[vagrant zinc]

The example 4 is U(10)≈Z_4

[tribal moss]

It's a quite strong property of Z that there even is a concept of canonical representative for each residue class. Even "Euclidean domain" in itself doesn't quite guarantee it.

[untold turret]

hm that makes sense

thanks i'll think about this for a while

[untold turret]

let A,B,C be ideals in R and A \subset B \subset C, so the generators of B are a linear combination of the generators of C. let f be the canonical quotient map R -> R/A. what's a way to show that the generators of f(B) are the same linear combination of the generators of f(C) as the one between B and C?

[next obsidian]

Write x in f(B) as f(b) for b in B. Write b as a linear combination of stuff in C, then apply f to that linear combination and you get x. Now x is the same linear combination as b was, but with f’s everywhere

Apply that to a set of generators

[round hull]

this seems generally true for any ring homomorphism no

[next obsidian]

Only thing to note is that f(I) is not in general an ideal

[round hull]

oh right

[untold turret]

i see, thanks for the help!

[next obsidian]

It’s true up to completing it to an ideal. So doing f(I)B for a map f:A -> B

Like you’re just taking the image and looking at the ideal it generates, so statements about generators are just the same

[ivory trail]

f(B) is an R-submodule of S for a map f: R -> S and ideal B of R

[untold turret]

wait but say the generators of C are g1, g2, etc., and we write b = r1g1 + ... + r_ng_n, then f(b) = f(r1)f(g1) + ... + f(r_n)f(g_n), right? but we have no guarantee we have the same linear combination in the sense of the "action" between f(B) and f(C) as we have between B and C

[untold turret]

oh this might make more sense, am parsing

[ivory trail]

well it's probably not very helpful (sorry)

[untold turret]

so this is where we might use specifically the fact we're dealing with the quotient map

[ivory trail]

what do you mean "action" between f(B) and f(C)

if you want to think about the linear combination angle strongly then you might want to consider how a map of rings f: R -> S makes S an R-module

in which case everything works out

[untold turret]

what I want is not only f(r1)f(g1) + ... + f(r_n)f(g_n), but r1f(g1) + ... + r_nf(g_n)

what's why i'm thinking your comment on modules makes more sense

[ivory trail]

r_k are elements of R, so you can't just immediately multiply them with elements of S

but the map f: R -> S induces an R-module structure on S ("restriction of scalars") whereupon you can think of r_k as scalar coefficients in a linear combination

but at the cost of talking about modules (possibly unnecessarily)

so if r in R and s in S, the R-module structure on S is given by r \cdot s := f(r)s

[untold turret]

yeah but this might be what I want more specifically, the closest thing to "conserving the action"

i.e., the closest thing to: that ideal is built the same way from this other ideal in both spaces

will keep thinking about your comments, thanks!

[ivory trail]

you should look up restriction of scalars in more detail then

[untold turret]

noted, tyvm!

[untold turret]

let $R$ be a ring with ideal $A$. let $\phi: R \to R/A$ be the canonical map. the property can be expressed, i think, as the existence of some sort of canonical set map $\psi:R/A \to R$ such that, for $x \in R$, we have $x = \psi \circ \phi(x)$. in $\mathbb Z$ the map $\psi$ can be defined formally, i think, as sending an element $\overline x$ that is $n$ unities away from unity to $n$

[cloud walrus]

nrs

[remote nymph]

sorry to reask, but could anyone help me understand what is meant by "an orbit of G_X on X" where G is a group acting on a set and G_X is the setwise stabilizer of X in G. Perhaps if someone could express this in set notation?

[prime sundial]

take an equivalence relation over the set given by $x ~ y \iff x = g\cdot y$ for some $g\in G$. an orbit here is an equivalence class of this relation

[cloud walrus]

maximofs

[prime sundial]

oh an orbit of G_x

[remote nymph]

yeah I understand orbits of specific elements, but wouldn't an orbit of the stabilizer of X on X itself just be X?

this kind of language is used in multiple places and i dont understand it

[prime sundial]

well it’d probably be a singleton

but i’m thinking that’s probably not what they mean here

[remote nymph]

X is a set, so just the entirety of X?

[prime sundial]

if we restrict the group action to the stabilizer of X we just get that the orbits are singletons

[remote nymph]

ah gotcha

[prime sundial]

since everything in X is a fixed point

but i still don’t think that’s what they mean

[remote nymph]

definitely not, but i dont see any other clear interpretations

[prime sundial]

where is this snippet from

[remote nymph]

https://www.sciencedirect.com/science/article/pii/S0012365X97001672?ref=cra_js_challenge&fr=RR-1

section 2

[prime sundial]

lol they don’t even define what X is

[remote nymph]

yuuuuuup

i just assumed its a subset of V

[prime sundial]

yeah i think in that case them referring to theta(X) as a singleton may make sense

i’m like it seems to be for an arbitrary subset of V

[remote nymph]

here is another snippet talking about the same thing from another paper

[prime sundial]

did they give any examples?

[remote nymph]

of course not!!!!!

😦

the same kind of orbit lingo is used in the algorithm that follows

[prime sundial]

also what is 2^V

[remote nymph]

the power set of V (all possible subsets)

[prime sundial]

oh that’s odd

ah they’re using P(X) elsewhere

fair enough

then the notation makes sense

[remote nymph]

sorry, so what would the orbits that theta outputs look like?

[prime sundial]

well that i’m still unsure about, i just understand what X is now haha

[remote nymph]

oh lol

also they write $X + x$ instead of $X \cup {x}$. are they the same or is this some weird set operation?

[cloud walrus]

srhoosteen

[prime sundial]

okay i think the confusion is in what the setwise stabilizer is

the orbits of G_X aren’t necessarily singletons

they are just orbits of the restriction of the group action to G_X

what i was thinking of was the point wise stabilizer

[prime sundial]

also maybe i am still guessing that it’s this but i can’t see any other sensible interpretation

[remote nymph]

aaahh so maybe it means to pick an arbitrary $g \in G_X$, which may take any $x \in X$ to some $y \in X$, thus creating a cyclic subset of $X$?

[cloud walrus]

srhoosteen

[remote nymph]

yeah that makes sense

oh and one more question:

an orbit representative is just an arbitrary element of the orbit, right?

[prime sundial]

also i can’t see any other interpretation here either, but not sure

[prime sundial]

yes

[remote nymph]

thank you so much, the algorithm makes sense now

maximo goat fr

with good music taste

[dense raven]

does anyone know if geometric group theory has any applications

other than obscure cryptography stuff

[balmy belfry]

what is topology a pre-req to

[tender wharf]

wrong channel

[summer path]

#advanced-lounge or #point-set-topology

[tender wharf]

alg top

[balmy belfry]

ah sorry

[warped fable]

I have never heard of geometric group theory having applications in cryptography

GGT primarily has applications in geometric topology

Unless you mean real world applications which is something mathematicians mostly do not care about

For instance the proof of virtual Haken conjecture was settled by applying heavy geometric group theory machinery

[coral steeple]

What is the analogous statement asked for in b)? My guess is something like if H is a subset of a group G, where H is a group under the induced map, and x \in H, then the inverse of x in H is the inverse of x in G

[pastel yew]

Ya I think so too

[coral steeple]

Well that's a bit nicer to show than a)

thanks

[chilly ocean]

yeah.. a bit wordy though

[chilly ocean]

hello austin

[glossy crag]

If L/K is finite, is L(x)/K(x) (function fields) finite as well? If L/K is separable I suppose you can use the primitive element theorem, what about the general case?

[rocky cloak]

If L is finite then L = K(a,b,c, ...) for some fintie set of algebraic elements. But then L(x)= K(x)(a,b,c,...), so is finite as well.

[glossy crag]

Lol, true

I suppose the only point is to note that in a function field K(x) both in terms of notation and adjunction (smallest field containing K and x is K(x) itself).

[rocky cloak]

More explicitely if f(x)/g(x) is in L(x) then there is a polynomial h(x) such that g(x)h(x) is in K(x). Then f(x)/g(x) = f(x)h(x)/g(x)h(x), so it is enough to express f(x)h(x) as a linear combination of a, b, c, ... which you can do coefficient wise.

[glossy crag]

How do you get such an h(x)?

Just bring the coefficients of g to a common denominator and multiply by that?

[rocky cloak]

Let $\alpha_i$ be the roots of $g(x)$. Let $f_i$ be the minimal polynomial of $\alpha_i$ over $K$. Then $g(x)$ divides $\prod f_i(x)$

[cloud walrus]

jagr2808

[glossy crag]

Huh... Roots in an algebraic closure of L, right.

[rocky cloak]

Yes

[glossy crag]

Neat, didn't cross my mind.

Thanks.

[glossy crag]

R is the valuation ring of a complete discrete field, pi is a uniformiser. Why should a=a_n mod pi^n hold?

[solemn dew]

Do you guys think that my proof is valid?

[celest furnace]

You have the right idea but the middle line: for all x,y in G Hx = Hy is wrong

[hot lake]

{xh | x in G, h in H} = GH = G

[celest furnace]

You basically need to argue that Hx is not H and you are done

[solemn dew]

Gotcha

Just a question:

if a\in\H, is {ah, h\inH and a\inH} a left coset?

where H is a subset of G

Or is this only true if a\inG and a not in H?

[delicate orchid]

gonna assume you mean subgroup instead of subset, and yes that's a left coset it's just trivially equal to H

[solemn dew]

That's what I meant

Thnx

I think I'm done with my proof then!

Feedback anyone?

[dense raven]

i was talking real world. i personally dont care about real world applications but its nice to have one so i can give non math people an idea of what i do

[tender wharf]

this is fine

[solemn dew]

@tender wharf Alright thanks. Anything weird about my writing that I can improve?

[tender wharf]

nah looks fine

quite easy to read too

which is good

[dense raven]

@pastel yew lol its you

[cloud walrus]

jagr2808

[rocky cloak]

For example if q(x, y) = (x-y)^2 then the roots (n, n+1) converge to (1, 1)/sqrt(2)

[white oxide]

no way ultimate chad in #groups-rings-fields ???

[celest furnace]

I know very little but I found one I can answer!!!

[indigo ridge]

Is my proof correct? we want to show $f(x+y) = f(x) \cdot f(y), f(x+y) = 2^{x+y} = 2^x \cdot 2^y$, therefore f is a homorphism

[cloud walrus]

jayzsparrow

[coral spindle]

Yup

[void cosmos]

for 4c why cant i just take Z/(n,m)Z tensor Q over Z

then like 6 tensor x = 5 tensor y = 0

[coral spindle]

who says you can't?

[void cosmos]

cool

ty

[void cosmos]

son of sam

[coral spindle]

Sure.

[delicate orchid]

1 \tensor_Q 1 and 2 \tensor_Q 1/2 le troll face

[formal ermine]

beat my wife to it

[delicate orchid]

please stop saying that

[void cosmos]

no i think the tensor must be over Z

[delicate orchid]

it said any ring

in that case 1 \otimes 1 and -1 \otimes -1

[hidden haven]

Or 0 ⊗ 0' = 0' ⊗ 0 in R ⊗ R where R is a ring with 2 zeroes

[coral spindle]

Also Z (x)_Z Q is cute imo

[static yew]

okay I'm still trying to process everything I read yesterday

the elements of Q[sqrt(2)] are all (a + b sqrt(2)) for all a,b in Q, right?

[coral spindle]

Yes

[static yew]

which is why R[i] is C

[coral spindle]

We would write $\bQ[\sqrt 2] = \build{a + b\sqrt 2}{a, b \in \bQ}$.

[cloud walrus]

boytjie

[static yew]

if I said "z3 = cube root of 2"
would the elements of Q[z] be all a + bz + cz^2

[coral spindle]

They would

It's not totally trivial to see why this is true from the definition

but it is true

[static yew]

Which means that if I have a field F and a transcendental t that is not the root of any polynomial over F[x]
then the elements of F[t] would be a0 + a1 t + a2 t^2 + a3 t^3 + a4 t^4 + ... for all a_i in F

[coral spindle]

They would be finite combinations of that

[static yew]

how so

[coral spindle]

e.g. 1 + t + t^2 + t^3 + .. is not in F[t]

[static yew]

when does it stop though?

[coral spindle]

It never stops

[static yew]

it has to stop if it's finite

[coral spindle]

That's why it's not in F[t]

[static yew]

no no

what's the highest degree in F[t]

[coral spindle]

There is no highest degree in F[t]

You can have polynomials of arbitrarily high finite degree

[static yew]

Oh, you mean all finite polynomials

[coral spindle]

This is like asking "what's the biggest number in Z"

[white oxide]

Here, do they mean $\lambda(K_H)$ instead of $\lambda(E_H)$?

[static yew]

But the ring itself is still infinite

[cloud walrus]

okeyokay (analysis is aight ig)

[coral spindle]

Yes ofc

You can write down (in some sense) infinitely many elements

[static yew]

Also, applying Cantor's diagonal argument, it's uncountably infinite

[coral spindle]

No

Typically it is not

It will be uncountably infinite iff F is

[hidden haven]

Wait yes

[coral spindle]

Cantor's diagonal argument does work for F[[t]] tho

[static yew]

Wait what's F[[t]]

is this premium math

[coral spindle]

the ring of possibly infinite polynomials

so e.g. 1 + t + t^2 + ... is in F[[t]]

[static yew]

X-treme polyomial rings

[white oxide]

wait i'm confused because later on they have this in the proof:

[hidden haven]

The notation confused me

[wraith cargo]

Ring of power series in t with coeffieicnts in F

[hidden haven]

Mb

[white oxide]

oh okay

thx

[static yew]

anyway, that's not what I was focused on while laying in bed in the late hours last night

the elements of F[t] look a lot like the polynomials of F[x]

[coral spindle]

Yes

I'm just writing t instead of x

The symbol I choose does not matter

[static yew]

so when we say F[x] it's basically transcendental x

[hidden haven]

They be isomorphic

[coral spindle]

Yes, but it gets very difficult to work with this transcendental idea.

It is way easier to just think of it as a symbol.

[static yew]

Q[sqrt(2)] isn't different because it's a specific number
it's different because it is specifically a root of a polynomial with coefficients in Q

[coral spindle]

For example, is Q[pi, e] = Q[x, y]? I don't think this is known

by = I mean isomorphic ofc.

[static yew]

Q[pi] would be a ring much like Q[x]

[coral spindle]

Q[pi] is indeed isomorphic to Q[x]

[static yew]

Q[pi] sounds like a great way to troll somebody. possibly engineers and mathematicians at the same time

[coral spindle]

I don't think it would be very effective

[hidden haven]

ℚ[t][π]

[coral spindle]

Holy shit it's a picture of....

um....

[coral spindle]

idk I can't think of anyone for whom it wouldn't be creepy to say

so

we'll leave that there

[hidden haven]

[coral spindle]

https://math.stackexchange.com/questions/456097/are-pi-and-e-algebraically-independent source btw

[alpine island]

What is the story behind this? I know what it means and I see it posted a lot, but where is it from?

[coral spindle]

That's not what's being asked lol. Look at the name of it: clearly there's some story

[delicate orchid]

uinviersal pplpropertyy of the uqotient!!

[hidden haven]

The time when all of #discussion united to explain the first isomorphism theorem to 🅱️etalninja and still failed

[delicate orchid]

how long ago was that lol

[hidden haven]

A year or 2 probably

[delicate orchid]

yeah

metal doesn't strike me as someone who's just recently learnt first iso

[hidden haven]

Metal doesn't strike me at all

Ninja in hiding

[karmic moat]

omg stoooop got me kicking my feet

[coral spindle]

Have you thought about why diagonalisation doesn't actually apply here?

Because of course as I said, R[x] is uncountable iff R is uncountable. So for example Z[x] is countable

It is worth thinking about why it doesn't work.

[static yew]

diagonalization would only work if you could have a polynomial with an infinite number of terms

[coral spindle]

That's right. Typically the 'polynomial' produced is in fact a power series.

So it doesn't lie in R[x]

[static yew]

Although it seems suspicious to me that if you were to follow the procedure described in the diagonalization argument, at each step along the way, you would have a polynomial that doesn't match any existing element and never will

[coral spindle]

I don't find that suspicious at all

That is basically saying that R[x] is infinite lmao

[white oxide]

how does 50.3 imply that E is a splitting field? I understand that we can take Fbar = K, but we don't know if the isomorphisms map E onto itself right

[delicate orchid]

anything involving infinity is sus

[coral spindle]

discussion

[alpine island]

do you have a link to that? that sounds funny

[delicate orchid]

how do you not have access to discussion what did you DO

[coral spindle]

studying role lmao

[alpine island]

how many mods do you have to piss off to get locked out of discussion

[rocky cloak]

It's a long sentence. You have to read the last part as well

[white oxide]

oh right oops

thanks

[hidden haven]

Its very old lol I'm not gonna look it up

[formal ermine]

oh

[chilly radish]

he has the studying role which blocks access to the social channels

[white oxide]

oh shit i didn't know that

that makes sense then

that's hilarious

bruh here they mean $G(E/F)$ right

[cloud walrus]

okeyokay (analysis is aight ig)

[hidden haven]

Yes 💀

[white oxide]

my guy fraleigh speedrunning the galois theory section

mf has typos everywhere

[delicate orchid]

can't say I blame him

[hidden haven]

Death

[white oxide]

true

[pastel yew]

🙈

[pastel yew]

🙈🙈

Hello guys 👋

[celest furnace]

Howdy 🤠

[pastel yew]

Lol

Hi

[pastel yew]

My opinion as well

[white oxide]

here do they mean f(x) is equal to the expression pointed to with the arrow

[karmic moat]

yeah probably

[untold turret]

A,B coprime ideals in ring R, f:R -> R/A is canonical map. how do you "see" that f(B) = R/A?

[chilly ocean]

there are elements a and b s.t a+b = 1, so f(b) = 1 and then f(B) = R/A

[karmic moat]

is ur pfp untitled unmastered

[untold turret]

real recognize real

[untold turret]

this makes sense, but how come f(b) = 1? obviously we have b + A = R, but why should we have the analogous "b + A = R/A"?

[chilly ocean]

Because if B+A=R there is an element b in B and a in A s.t b+a=1. You don't have b + A = R/A. But if you know f(b)=1, then f(rb)=r, so f(B)=R/A

[round hull]

take f on both sides of a + b = 1

[untold turret]

ah! yes that makes sense! ty

[untold turret]

thanks for the help

[untold turret]

in a sense, this is what we get from the fundamental homomorphism theorem, no? movements in a ring are in a sense the same movement in the subring induced by the image of a ring homomorphism

[indigo ridge]

Im trying decompose these permutations into tranpositions am I correct? $(1 3 2 5 6)(2 3)(4 6 1 5 2)= (1 6)(1 5) (1 2) (1 3)(2 3)(4 2)(4 5)(4 1)(4 6)$

[cloud walrus]

jayzsparrow

[indigo ridge]

and $(12345)(413) = (15)(14)(13)(12)(43)(41)$

[cloud walrus]

jayzsparrow

[coral spindle]

Yup

to both

[indigo ridge]

lol thanks

[chilly radish]

Kind of an open-ended question. The very innocuous statement that G/Z(G) cyclic implies G abelian is unreasonably effective (especially in finite group theory). It got me thinking if there's some deeper/more general principle/statement hiding behind here

somehow this almost seems like it could generalise to a cohomological statement

[coral spindle]

So this is saying Inn(G) Abelian → G Abelian. Is that the kind of path you're thinking about?

[chilly radish]

I was actually thinking about this saying something about extensions of cyclic groups by abelian groups

[coral spindle]

Right, OK

[chilly ocean]

Hello can someone help me with a problem. We have a finite group G with order n. And we know that n is not divisible by 4 and we have only n-1 cyclic subgroups. Find the group G.

Since there are n-1 cyclic subgroups we have two distinct elements a and b such that a and b generate the same subgroup

And from here I am stuck

[coral spindle]

Do you mean to say that you have exactly n-1 cyclic subgroups?

When you say "only" you confuse me.

[chilly ocean]

Yes

Sorry

[coral spindle]

Hint: we can count the number of distinct generators of any cyclic group; there is an explicit way of doing so.

[chilly ocean]

isn't phi(order of some generator)?

So every cyclic subgroup that has not the generator a/b must have order 2?

So the elements that are not a/b have the proprety x^2=e.

So b=a^-1

So the order of a is 3, 4 or 6.

The order cant be 4 because the 4|n by lagrange and this is false

And order 6 is not good because then we have an element that is not a/a^-1 which does not have the proprety: x^2=e. So ord(a)=3 so the subgroup generated by a is {e,a,a^-1}.

And then the order of G is 2^k × 3. But k cant be >=2 so the order of G is 3 or 6.

And I think from here is easy

Thx for the hint!

[coral spindle]

I think you did that all on your own

[untold turret]

in commutative algebra we encounter the notion of a ring R-algebra in passing, is there somewhere where it is taken as an object of study?

[round hull]

you get this for free, from how functions work

or maybe i can't quite understand what you're saying

[untold turret]

yeah i realized afterwards, we can think of it in exactly the same way as group homomorphisms. a + b thought of as the action of b on a under multiplication/addition will be in a sense mirrored in the action of f(b) on f(a) under multiplication/addition

[round hull]

the fundamental homomorphism theorem says that the image of a map is basically the original structure without the kernel

[untold turret]

hm need to think more on that

[round hull]

you should do exercises instead of thinking about it

[untold turret]

that's good advice, will go do some exercises heheh

[warm wyvern]

54.1

[open sluice]

be very afraid

[white oxide]

nah i skipped some sections and fraleigh has sections for everything lmao

he had an entire section for the field of quotients

which was mainly just the construction but he goes into a lot of detail

[delicate orchid]

lol?

like just the field of fractions not even localisations in general?

[white oxide]

given that i don't know what a localisation is yes

[delicate orchid]

weird

[white oxide]

it's probably the friendliest introductory algebra text i've seen

but he explains things really intuitively

[white oxide]

how did they get from the first equality to the second equality? it obviously has to do with the properties of the inner product but im kinda rusty on that and struggling to work out the equality

[round hull]

linearity

[white oxide]

nvm got it

yea

[rocky cloak]

Representation theory of algebras typically studies such things. Also in algebraic geometry one often restricts to schemes over a base ring. Which then is the same as R-algebras when restrict yourself to rings.

[untold turret]

very cool stuff, didn't know there was representation theory for things other than groups!

[delicate orchid]

wait till you find out about quivers

[rocky cloak]

People do representation theory of groups, rings, algebras, quivers, posets, Lie algebras, and probably other things

[untold turret]

very cool

[hasty sinew]

Was wondering what a representation of a quiver would be; for some quiver Q and ring R, it's a functor FreeCat(Q) -> RMod

Very cool

[sweet echo]

you also usually identify them with modules over the path algebra

[delicate orchid]

new cool representation theory
look inside
a functor into R-Mod

everytime

[sweet echo]

propaganda dreamed up by Big Category Theory

[open sluice]

category nii-san

[wraith cargo]

This is what I thought too
Until I read some rep theory
Categories didn't prepare me for this shit

[delicate orchid]

Trolled

Also doing categories before group rep theory

Based alert

[wraith cargo]

I don't know shit rep theory
The only rep theory that I know is what I learned by reading some Tate thesis nonsense regarding like Gelfand transforms

[wraith cargo]

Wew stop pretending like you know rep theory
It's widely known that rep theory is a joke invented by analysis to troll algebraists

[untold turret]

C is nontrivial ideal, and A \neq B nontrivial ideals. If A+C = B+C then A,B coprime with C. why?

[rocky cloak]

Doesn't sound true. Just have A, B, C equal to (8), (4), (2)

[untold turret]

hm, right. thank you

I think we have: C is nontrivial ideal, and A \neq B nontrivial ideals, then if A+C = B + C, C not in A,B

[hasty sinew]

The statement basically reduces to saying that a representation associates each vertex with an R module and each edge with a linear transformation from its source vertex to its target vertex

[untold turret]

we know ideal addition and intersection correspond to join and meet in the lattice of ideals

is there an interpretation of product?

[frigid lark]

The product of 2 ideals, I, J defined as IJ = {sum i_r × j_r | i_r in I, j_r in J} (a finite sum), is actually contained in the intersection of I and J and can be equal to the intersection, if I + J = R

[lusty marlin]

Let A[x] and B[x] be polynomial rings such that A[x] ⊆ B[x] and B[x] is a UFD. If we want to show that a polynomial p(x) is irreducible in A[x], then would it be sufficient to demonstrate a factorization of p(x) in B[x] and then show that this factorization cannot be equal to a factorization in A[x] up to multiplication of factors by units?

Eg. We want to show that p(x)=x²-√2 is irreducible over ℤ[√2]. So we show that (x-2^(1/4))(x+2^(1/4)) is a factorization of p(x) in ℝ[x], which is a UFD, and then show that this is not equivalent up to multiplication by units to any factorization over ℤ[√2].

[frigid lark]

I think if the degree of p(X) is larger than 2 (3?) you might have to consider more cases then just the irreducible elements that constitute p(X), specifically their products

[lusty marlin]

What do you mean by this?

[frigid lark]

Like for example X^4 - 4 over Z[X]

If you took it to R[X], and split it into irreducible components you would observe that none of them are in Z[X]

But X^4 - 4 obviously has a factorisation in Z[X] consisting of 2 quadratic factors

Which you would need to account for in your argument

[lusty marlin]

Yeah, of course. I should have been specific, I was asking about the validity of this method for polynomials of degree 2 or 3. A degree greater than that leads to this method falling short in the way you described.

[frigid lark]

Yeah I think your method should work

[lusty marlin]

Thank you. For some reason I felt like I was making a logical leap somewhere while applying it. Seems like that isn't the case.

[frigid lark]

Formally you would assume p(x) = q(x)r(x) for some q(x), r(x) in Z[sqrt(2)][x]. Since p(x) is monic, WLOG assume q(x), r(x) are monic. Then since R[x] is factorial, you would conclude that WLOG q(x) = (x - a), and r(x) = (x + a) for a = 2^(1/4).

That's your contradiction

[lusty marlin]

Yes

[median pawn]

nice way to show that x^5 + 3x + 1 is irreducible over Q?

[lusty marlin]

Let p(x)=x⁵+3x+1
By Gauss's Lemma, it is sufficient to show that p(x) is irreducible in ℤ[x]
By the rational root theorem, any rational root of p must be equal to ±1, and one can check that neither are roots of p.
Therefore, assuming p is reducible over ℤ, it must have a factorization consisting of a degree 2 and a degree 3 polynomial with integer coefficients. Let them be x²+ax+b and x³+cx²+dx+e. Equating the coefficients and using the fact that they must be integers, it is fairly simple to reach a contradiction and hence show that it is irreducible over ℤ.

[median pawn]

cool! thanks

i did the same, i just feel like "equating coefficients" is not particularly nice

[lusty marlin]

be=1 places strong restrictions

that lead to the conclusion following easily. Although I would agree that it isn't an 'elegant' method

[rocky cloak]

But -1 is a root modulo 3

[lusty marlin]

mod 3, this polynomial becomes x⁵+1

[south patrol]

Yes I forgot about -1 and stuff lol

[chilly ocean]

Does a "p-extension" of a field K just mean some extension adjoining p-power roots to K?

[wind parrot]

It usually means a Galois extension whose Galois group is a p-group (or a pro-p-group if you're dealing with infinite extensions)

[chilly ocean]

ah ok thanks

[void cosmos]

yo

let R be a ring and I be an ideal , M be an R-module

so i wanted to prove that R/I tensor M is the same as M/IM

IM = {im | i in I , m in M}

and one way i thought was just to use the known short sequence

0 --> I --> R ---> R/I-->0

then tensor then by the left and i get a homomorphism from M to R/I tensor M

and now i got stuck

any hints or is that not the approach at all

[formal ermine]

first iso

[void cosmos]

yea is the kernel IM

[elder wave]

Remember that M tensor R is identified with M by r \otimes m = rm

Then go from there after tensoring

And use exactness ofc

[summer path]

i could use some chicken tenders right now...

[void cosmos]

the exactness gives me surjectiveness

[chilly ocean]

just write down the isomorphism xd

[void cosmos]

so the img is just the whole R/I tensor M

[void cosmos]

isnt it a homomoprhism

or do u mean the onde induced by first iso

[chilly ocean]

man

[elder wave]

Take some time to think about the hints we gave and try again

[void cosmos]

yea i got it from this

so that means that the image of the first map is just IM

which by exactness means the kernel is IM

then i get it by first iso

right?

[elder wave]

Yes

[void cosmos]

what "isomorphism" is ttepa talking about

[chilly ocean]

think

think about what my message could mean in this context

what does "just write it down" mean

[void cosmos]

you mean like just find an isomoprhism between the two modules without using sequences?

or like use the universal property?

i can find a bilinear map R/I x M --> M/IM by just (a mod I , m ) --> a*m mod IM

[chilly ocean]

warmer

[void cosmos]

hwat

[chilly ocean]

nvm

yes that is basically what i meant by "just write it down"

i'll make sure to be more explicit next time

[void cosmos]

cool man ty

[void cosmos]

let R be a ring , I and J be ideals

then R/I tensor R/J is iso R/(I+J)

just wanna check this proof

define f: R/I x R/J --> R/(I+J) ; f(a mod I , b mod J) = ab mod (I+J)

this is bilinear and it induces a homomoprhism from R/I tensor R/J to R/(I+J) , call it h such that the diagram commutes , ie h(phi(a,b)) = f

where phi is the map from the product to the tensor defined as (a,b) --> a tensor b

now define hbar: R/(I+J) --> R/I tensor R/J ; hbar(a mod I+J) = a mod I tensor 1

then those are inverse homomoprhisms hence an iso and im done?

as tropo pointed out this is not showing the tensor product IS this but rather is isomoprhic to

that is if its correct llo

[elder wave]

Yes that works

[void cosmos]

ty

[thorny knoll]

What is this discriminant, exactly?

[gritty sparrow]

It's the usual discrminant of a polynomial, except in the field k(x)[y], I suppose. So it would be an element of k(x).

[thorny knoll]

Ahh so the discriminant is usually a constant number, but here we have a function, because we always fix one x and then calculate the discriminant? Btw, any reason you are writing k(x) instead of k[x]?

[gritty sparrow]

Well, the way I was thinking about it was as follows: for a polynomial p(y) over any field , say F (ie p is an element of F[y]). We can calculate the discriminant of p by the usual calculation. This is just the case when the field F is k(x).

[formal ermine]

you can also calculate the discriminant through the resultant which should make more sense in the case of multiple variables

or over random fields/rings

[gritty sparrow]

I could have also used k[x] in the process I was talking about, but people are not usually familiar with discriminant over arbitrary rings.

[thorny knoll]

Thank you both!

[south patrol]

Well also like elements of k[x] aren't functions

[delicate orchid]

Every element of any set is a function because I said so

[solemn dew]

3

But statement is true. Where did I go wrong?

[elder wave]

what is your map phi

[solemn dew]

(123)

[coral spindle]

And how is this a group homomorphism?

[solemn dew]

Alright then I'll try to rewrite it 😅

[coral spindle]

I feel like you've ignored my question

Phi is a permutation, i.e., an element of S_3

but it is not a group homomorphism into S_3

[solemn dew]

I get what you are saying

[coral spindle]

The question required phi : G → G' to be a group homomorphism, so your supposed counterexample isn't correct. In the first place, it doesn't even define a map into S_3.

[solemn dew]

Yes

Thank you!

(i am noob)

[delicate orchid]

Also, is phi^-1 just denoting the preimage? I don’t see an assumption that phi is invertible anywhere

[coral spindle]

Yeah it ought to be

[tribal moss]

I think Jonathan was momentarily confused between maps {1,2,3}->{1,2,3} and maps S3 -> S3. His (1 2 3) is a map of the first kind, but not one of the second kind.

[solemn dew]

Yes

But I will change my strategy this time

Thank you all !

[solemn dew]

Not the cleanest proof

Any feedback?

[chilly ocean]

why did you write "suppose the pre-image is non-empty" when you're proving it right after?

[solemn dew]

You can ignore the first line ^

I was thinking that it is only true if it is not empty

Because if it were empty I could never show that it is closed, since it requires elements to exist

[chilly ocean]

if it's empty then it couldn't have an identity element and therefore wouldn't be a group

the other group operations would be satisfied vacuously

[delicate orchid]

the empty set is definitely closed under whatever binary operator you want lol

[solemn dew]

I assumed it is not empty, and because of this it was a subgroup

But if I assume it is empty it will not be a subgroup

[delicate orchid]

why do you need to make assumptions about emptyness at all

[chilly ocean]

you have proven it is non-empty, why are you assuming things about its (non)emptiness?

[solemn dew]

Oh okay I think I get it

I showed that e' must be in H' and so e must be in phi^-1(H'), so it is not empty

so I don't have to make that assumption

[chilly ocean]

your proofs that the pre-image contains the identity element and is closed under the group operation look fine logically. the proof that it's closed under inverses is a bit hard to read through. i think you can trim it down to something with a clearer logical progression

as for the very last line, i'm not sure why you wrote "gg^{-1} = e is in the pre-image" since you already proved at the start that it contains the identity. you could just write that it follows from what you showed that the thing is a subgroup

[open sluice]

assuming the statement you are trying to prove is a logical fallacy

[solemn dew]

Just wanted to show that g^-1 is in phi^-1(H'), and then phi^-1(H') has inverses, and is a subgroup

[open sluice]

there’s a difference between “a set with these properties exists” and “if this set exists then it must have these other properties”

[chilly ocean]

yeah it's obvious what you're trying to prove. no doubt there. i'm saying the way you've laid out the proof is kind of confusing

like, what's the point of "we showed that if g is in the pre-image, then \phi(g) is in H'"? you don't need to recall the definition of the pre-image

[cloud walrus]

TTeppa

[solemn dew]

Aaah okay

Yours looks so much better haha

I think I get your point

[cloud walrus]

TTeppa

[chilly ocean]

you can comfortably take something like this as known for what you're doing

[solemn dew]

Alright I don't have to write everything down then?

[chilly ocean]

well you have to write the proof of what you're trying to prove at least

but small things like this are safe to take for granted, if you're comfortable with them

[solemn dew]

I am not yet, but I will in a few days

Thank you so much 😄

[glossy crag]

If L/K is an unramified extension of local fields that means a uniformiser of K is a uniformiser of L, right? Unramified => e=1 => valuation groups are the same.

[hollow fjord]

does anyone knows how to prove Maschke's theorem for compact groups?

[formal ermine]

I don't, so no, not everyone

[hollow fjord]

xD

[rocky cloak]

The proof should be the same as for finite groups, except you switch the inner product

$\langle v, w\rangle_G = \sum \langle gv, gw\rangle$

With

$\langle v, w\rangle_G = \int \langle gv, gw\rangle$

[cloud walrus]

jagr2808

[rocky cloak]

(integral with respect to haar measure)

[coral steeple]

Is there a cleverer way to count elements of S_n of order 2 than going through them all and checking whether x^2=1? I think this reduces to counting symmetric permutation matrices but I'm not sure if that's any easier

[delicate orchid]

an element of S_n has order 2 if and only if it's cycle type decomposition looks like 2+2+...+2 for some number of 2s
so what you can do is find the number of partitions of n consisting of just 1s and 2s, then use the formula for the size of conjugacy classes of S_n to determine the number of corresponding elements

this should be faster as the number of partitions grows slower than n!

[vague granite]

I'm reading some awkwardly written lecture notes. Is the set of automorphisms on the set S not just the symmetric group of S?

[lusty marlin]

No, assuming S is a group. Not every permutation of a group is an automorphism

[vague granite]

S doesn't have a group structure

S is just a set

[wooden ember]

Then yes

[lusty marlin]

What do you mean by automorphism of a set then?

[vague granite]

automorphism in the category of sets

This is what I'm refering to

[lusty marlin]

Ah
My bad, I'm not familiar with category theory stuff

[vague granite]

I think it's written weirdly

[wooden ember]

Yep it’s the same

[delicate orchid]

yeah that is just S _|S|lol

]PGORJMLW#

you know what I mean

[wooden ember]

Maybe they’re using that notation to talk about equivariant automorphisms later?

[vague granite]

I think most people write Sym(S)

[delicate orchid]

yur

[wooden ember]

Not sure

[vague granite]

It's a course on Galois theory, if that provides any context

[wooden ember]

Ah no it’s cause it’s gonna talk about actions on a ring

I mean in general you map to automorphism groups on objects of a category

[gritty sparrow]

Keep in mind: you need the set partitions, not the integer partitions. In any case this logic gives a more or less explicit formula, by summing over number of 2 cycles, choosing the pairs one at a time.

[vague granite]

Well Galois theory is all about looking at field automorphisms, the only thing is I don't understand why they felt like calling bijection maps set automorphisms

[coral spindle]

Sometimes people write Sym(S) to mean the set of permutations that fix all but finitely many elements of S, which is typically very different from Aut(S), but it's hard to tell from this context. This smaller group has a disjoint cycle decomposition whereas Aut(S) does not.

[vague granite]

Oh really? I've seen people write Sym(\mathbb{N}) to talk about the Riemann rearrangement theorem for example

[coral spindle]

This is why I said 'sometimes'

[vague granite]

This lecturer is known for using abstract terminology unmotiviationally so I'll take it as that

[gritty sparrow]

@delicate orchid Is what I said incorrect?

Ah wait, I misunderstood what you were doing

[delicate orchid]

I fail to see the difference between the set partitions and integer paritions

[gritty sparrow]

4 = 2 + 2 is one partition

[delicate orchid]

ah yeah I omitted a bunch of 1s

[gritty sparrow]

But {1,2,3,4}= {1,2} +{3,4} and also {1,3}+{2,4}

In any case since you are using conjugacy classes, you need integer partitions

[delicate orchid]

right yeah but they would be conjugate

[gritty sparrow]

But if you don’t you get an explicit formula

[gritty sparrow]

Yeah, I missed that initially

[wooden ember]

In this case I honestly think it’s just to have a general definition of a group action on an object of any category

It’s the most general one you can get and you need that notation for it

[coral steeple]

Most of those terms haven't been defined in my book yet so I think the point of the exercise may be to just do it exhaustively (for n=4)

[delicate orchid]

oh

then yes lol

[celest furnace]

[delicate orchid]

I thought you wanted some sort of computational implimentation

[coral steeple]

I thought there would be a clever way to count matrices but I guess it's more complicated

Thanks for the help though

[glossy crag]

Uniqueness of p-adic expansion in K means that if we have two powers series with coefficients in the same pool (a representative system for the residue field of K), then equality of series implies equality of coefficients. But the coefficients a_k don't necessarily come from the same pool as the c_k do (a priori we don't even know they are in K), so how can we even invoke uniqueness of p-adic expansion?

[stuck fiber]

So I'm reading Rotmans first course in abstract algebra and one of the exercises says:

Let $k$ be a field with one $\varepsilon$ and let $R$ be the subring $R={n\varepsilon: n\in\mathbb{Z}}

I'm not sure what epsilon is though

[cloud walrus]

kenshin5334
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[delicate orchid]

is that quote verbatim

[stuck fiber]

Yes

[delicate orchid]

if so, I presume \vareps is the multiplicative identity? no clue why they'd specify that for a field though

[delicate orchid]

kind of implied

fi nte fi eld

[hidden haven]

Maybe eps is an element such that its square is 0

[delicate orchid]

is this a numerics joke

[hidden haven]

eps is the standard notation for such an element in dual numbers

[delicate orchid]

it's a field moldi

[hidden haven]

For exactly this reason

[delicate orchid]

how can it square to zero

[hidden haven]

Maybe its a field with 2 zeroes

[delicate orchid]

what's that stupid space called

[delicate orchid]

the line with 2 points or whatever

[hidden haven]

Literally the line with 2 points

Yeah sorry idk lmao

about the eps

[delicate orchid]

reading the definition of the ring, yeah - it's absolutely the multiplicative identity

@stuck fiber

[thorny knoll]

always love to ponder how such creations enter this world

[delicate orchid]

question basically seems to be "Z is initial in Ring" lol

[hidden haven]

Non unital rings Non initial Z

[delicate orchid]

I said Ring for a reason

[hidden haven]

[delicate orchid]

I will use rng for non-unital rings because a stupid construction deserves a stupid name

[stuck fiber]

Thanks I thought maybe he had used it somewhere in the chapter but I didn't see it in this one

[hidden haven]

Did you know you can talk about exact sequences of rngs? Try that with rings, hater

[formal ermine]

I only use rng when I blame the rng gods in video games

[delicate orchid]

why would I want to talk about exact sequences

[hidden haven]

gay theory

K*

[formal ermine]

What

[delicate orchid]

true slightly

[hidden haven]

Sorry similar sounds

[delicate orchid]

I know minimal K-theory

[formal ermine]

reminds me of https://www.reddit.com/r/Badfaketexts/comments/9oh7cz/big_dicks_in_his_brothers_bum/

[vernal bronze]

Hi, I have a question in ring theory.
Let $R$ be a dedekind domain and let $I$ be an ideal of it. Let $0 \neq y \in I$ be an element in the ideal, show that there is an ideal $J \vartriangleleft R$ such that $IJ = \left<y\right>$

[cloud walrus]

arielhouri

[formal ermine]

what have you tried so far

[vernal bronze]

Pretty much nothing, I really dont have an idea where to start

I thought about using the notherianity of the ring and the fact that it is an integral domain, because the other two doesn't seem helpful

But I don't know how to continue

[chilly radish]

What do you know about dedekind domains

[formal ermine]

yeah that would have been my next question

[vernal bronze]

I know the definition (the 4 rules) and we also saw some examples in class, like O_d

And I also know that the ideals separate into multiplication of prime ideals

I actually just remembered that and it looks like it might be useful

Yea I think from there it's quite simple, I just look at the prime ideal factorization of I and <y> and then create J by subtracting the powers. <y> is a subset of I thus there powers in it's factorization must be higher

That's it, no?

[white oxide]

ok i should rlly know this and idk if this is the right place to post this, but i seem to have forgot some of my trigonometry. so to see this i considered n = 4 and m = 3, and ended up with the matrix (0 1 // -1 0). i then considered the vector (2, 3) in C^2, applied the matrix to it, and got the vector (3, -2). the problem is i forgot how to plot the points LMFAO so i would appreciate some help to see how it's rotation by 3pi/2\

in other words how do i plot (2, 3) in C^2 💀

[celest furnace]

C^2 is four dimensional

U can plot it in R^2 though

Better yet derive the rotation matrix yourself

[white oxide]

oh god i really should've studied math then the way i do now

[delicate orchid]

this matrix is very clearly real - I don't see why you can't just... ignore the fact it's in C? lol

[dreamy fiber]

Now the question is - can you show that this is a natural isomorphism

[celest furnace]

#geometry-and-trigonometry

[dreamy fiber]

Jkjk

[celest furnace]

Okey did u not like math in high school?

But realized how good it was and started studying algebra

[white oxide]

no i kinda just copied all the hw answers and played games during class lmfao

but it was like

[celest furnace]

Lmao

[open sluice]

happens to all of us
unless you're terence tao or something

[white oxide]

beginning of senior year that calculus started to get cool

so i tried self-studying and i always liked learning the next thing idk

[white oxide]

lmfao true

so i should really go back to strengthen my weak-ass trig

[delicate orchid]

alternatively just like

picture it lol

or hell, draw it

[white oxide]

also my mathematical maturity last year was pretty bad, i was just getting used to math proofs and shit and i thought you could just read the textbook once and then go to lecture and do the hw problems and you would be chill and that didnt work out great

[white oxide]

ye

i drew it

lmfao

[celest furnace]

That works when the homework takes 8 hours for 5 problems

[white oxide]

lmfao definitely not for me

[celest furnace]

[celest furnace]

Did u make this yourself lmao

[delicate orchid]

yeah

[coral spindle]

wew's an artiste

[formal ermine]

wew the memelord

[celest furnace]

W

[summer path]

E

[glossy crag]

Anyone?

[real sparrow]

Group holomorph

https://media.discordapp.net/attachments/742765221956943986/1127386794556665916/215424215199776768.png

[formal ermine]

@real sparrow why do you keep posting stuff without context

[void cosmos]

hello

0-->A-->B-->C-->0 split exact gives us tensor product is exact

proof is to write B as A+C and then distribute the tensor

and then the map from R tensor A to R tensor A + R tesnor C becomes the inclusion which is a monomorphism?

and thats it?"

[real sparrow]

because the context is in the other channel

[white oxide]

oh shit I just realized I misread

then yea def for me

😹😹😹

[sly crescent]

Which other channel?

[real sparrow]

what are some intresting property of holomorph of a group

[hidden haven]

Sorta yes, but recall that split exact only means that B is isomorphic to A+C and not equal. The conditions on the maps are about the commutativity of certain diagrams. So if you want to be precise, you will want to show that this isomorphism gives you an isomorphism for R ⊗ B such that the relevant diagrams still commute.

Another way to do this is to characterize split exact sequences in terms of addition of maps. If the first map in your sequence is i and the second is p, you find that the sequence is split if and only if i has a left inverse q, p a right inverse j, and jp + qi = 1_B

In this way, split exact sequences are entirely equationally described, and you don't need to check the ker p = im i condition. Then you use the fact that R ⊗ - preserves morphism composition and morphism addition to see that the equations are preserved so the result must also be split exact.

[chilly radish]

https://en.wikipedia.org/wiki/Holomorph_(mathematics)

fyi, if you want better answers, you should provide more context (saying 'it's in the other channel' is not helpful as no one knows where you were typing beforehand)

[frigid lark]

Could I have a hint for this?

Maybe just a hint to the idea that if f(w_1) = cw_1^n, and f(w_2) = dw_2^m for some w_1, w_2, and n, m minimum that n = m?

where w_1, w_2 are roots of unity, and c,d are roots of unity contained in the finite extension K

[chilly ocean]

Hm

Since you have infinitely many roots of unity mapping to other roots of unity you know it sends the unit circle to itself

And since you have infinitely many roots of unity you can find one that does not lie in the field of coefficients

Some ideas that seem helpful

[frigid lark]

Why?

[chilly ocean]

Oh oops that is actually wrong

But because the unit circle is compact and we have infinitely many roots of unity we have a convergent sequence

And so by the identity theorem we know that near where the sequence converges it sends the unit circle to itself, by continuity

[frigid lark]

Sorry, I don't know enough analysis to understand you

[tribal moss]

I don't think it works anyway. There are nice rational functions with coefficients in Z that send the unit circle to itself, but don't send 0 to 0 or infinity (so they can't have the form stated in the exercise). So a solution will have to use the specific fact that the infinitely many zetas map to roots of unity rather than just somewhere on the unit circle.

[delicate bloom]

I guess the roots of unity aren't complete so we might not even have the limit be a root of unity either, hmm

[frigid lark]

Would it be possible to disprove this statement:
For all integer m, there exists a rank N(m) such that for all n > N(m), f(zeta_n) neq wzeta_n^s for any s < m, for any root of unity w in K?

I think I can answer the question if the above is false

[tribal moss]

(For a concrete example, (2z+1)/(2+z) maps the unit circle to itself, since if |z|=1 then 2z+1 and 2+z have the same length on geometric grounds. But it maps 0 to 1/2).

[frigid lark]

Well this statement is true for f(z) = -z

So damn

[delicate bloom]

oh interesting, when you said this I assumed you meant f(z)=1/z

[tribal moss]

But that does satisfy the assumptions of the exercise -- it takes every root of unity to a root of unity -- and also the conclusion, with n=-1.

[delicate bloom]

yeah that makes sense

hmm, since it's a finite extension but infinitely many roots - that means they exist in a field L containing K, and we must get arbitrarily high degree. Then we can look at all their conjugates, which must also satisfy the equation because K is fixed, and the collection of all of these will be dense in the unit circle now

[tribal moss]

The finite extension is only assumed to contain the coefficients, not necessarily the inifinitely many roots of unity that map to roots of unity.

Consider f(z)=z, with coefficients in Q itself, but Q only has 2 roots of unity.

[frigid lark]

An extension that contains infinite roots of unity, I believe must me infinite as well

Yeah, it contains an arbitrarily high primitive n'th root of unity

And then it's order is at least phi(n)

[tribal moss]

As far as I understand the text of the exercise, it doesn't say that the zetas live in the finite extension it speaks of.

[frigid lark]

Which goes off to infinity

[frigid lark]

Yes

[tribal moss]

Does "yes" mean you agree or disagree?

[frigid lark]

Agree

[tribal moss]

Ah right, now I understand what you said here, sorry.

[delicate bloom]

yeah that was the point of my remark, the elements of K are fixed by the automorphisms of the roots in the larger field, since we have infinitely many of them

oh looks like you both worked it out, ok cool I'm sorta in and out atm

[frigid lark]

I haven't

[delicate bloom]

oh ok

so K contains finitely many roots of unity right

[frigid lark]

Ye

[delicate bloom]

well, it's just a finite extension is really all that matters

so the fact that we have infinitely many roots of unity mapping through it means they will lie in extensions of K

[tribal moss]

K is the finite extension containing the coefficients of f?

[delicate bloom]

yeah

sorry I thought it said that in the original question

but I invented that whoops

[frigid lark]

Well f(w) lies in K(w) for some root of unity w

[delicate bloom]

yeah

[tribal moss]

It could just have said "... with algebraic coefficients".

[delicate bloom]

and this will be true for arbitrarily large roots of unity

so $\sigma(f(\zeta)) = f(\sigma(\zeta))$ will hold, because K is fixed in these extensions

[cloud walrus]

merosity

[tribal moss]

Oooh.

[frigid lark]

Yeah

[delicate bloom]

yeah, so now, the p^1000th root of unity might be one of them, and we also know that all its conugates satisfy it as well

which is where I'm getting density pulled outta

[tribal moss]

Okay, I accept we have a dense set of roots of unity that each map to roots of unity.

[delicate bloom]

since we know having infinitely many roots of unity forces their order to be arbitrarily high

yeah, now I don't know what we do with that, just an observation haha

[frigid lark]

Well f(zeta) = wzeta^s for some root of unity w in K, and for some integer s

And for some root of unity zeta

[tribal moss]

If w and s can depend on zeta we can just take s to be 1 :-)

Or 0 ...

[frigid lark]

And then if f(z) = g(z)/h(z), zeta and all of its conjugates will be roots of g(z) - wz^sh(z)

So if I can bound s

I'm done

Oh, and w,s depend on zeta as observed

[tribal moss]

Oh, w in K, didn't see that, apologies.

Perhaps it would help to assume wlog that f(1)=1?

[delicate bloom]

idk if we can do that

[frigid lark]

Or show for some s, there's a primitive n'th root of unity, a with [K(a) : K] > degree r(z) = g(z) - wz^sh(z), r(a)=0

[tribal moss]

Just divide through by f(1), adjoining it to K if it's not already there.

[empty rose]

but then f(zeta) might not be a root of unity anymore

[delicate bloom]

yeah

[empty rose]

it will be a root of unity divided by f(1)

[tribal moss]

Aah, damn.

[delicate bloom]

e^i for instance 😢

[tribal moss]

But a root of unity divided by a root of unity is always a root of unity, isn't it?

[empty rose]

do we know that f(1) is a root of unity...?

[tribal moss]

Damn².

[delicate bloom]

f(z)=z with e^i sadly

[frigid lark]

Best root of disunity

[delicate bloom]

just 2pi order

i love lagranges theorem now

[tribal moss]

OK, proposal retracted.

[delicate bloom]

a good thought though

[chilly ocean]

Can anyone explain Sylow's Theorem and the notion of p-groups, p-subgroups, and Sylow p-subgroups?

[formal ermine]

the wikipedia page for it is nice

[delicate orchid]

a p-group is a group who's order is a power of p, that one's easy enough

[chilly ocean]

what is the restriction on n

[delicate orchid]

no clue what that means

[formal ermine]

none

p^n for any n

[delicate orchid]

oh

yeah

[chilly ocean]

so p^0 is a p-group? i.e. the trivial group?

[formal ermine]

yes

[chilly ocean]

and a p-subgroup is the same notion, I'm guessing?

[delicate orchid]

yes

[chilly ocean]

a subgroup of order p^n

which must mean for a group p^m n | m

[formal ermine]

if your group is p^m k where p does not divide k

[chilly ocean]

oh

[formal ermine]

then a p subgroup is p^n for any n <= m

[delicate orchid]

and a Sylow p-subgroup is a p-subgroup of maximal order

so p^m in this case

[chilly ocean]

but what is Sylow's Theorem

[formal ermine]

3 theorems

[delicate orchid]

there's more than one

[formal ermine]

that assert existence and how many sylow subgroups there are

and also extra stuff

[delicate orchid]

the conjugation one is way more important than how many there are imo

if P and P' are two Sylow p-subgroups of some G then there exists a g in G such that gPg^{-1} = P'

this proves that all Sylow p-subgroups are isomorphic in a nice way

[formal ermine]

the third one is nice for showing a group is abelian/solvable

[chilly ocean]

ah i see

[delicate orchid]

yeah I'm not saying it's useless, just that it's less useful

[chilly ocean]

gotcha guys, thanks

[formal ermine]

the motivation is like

converse of lagrange for non abelian groups

[delicate orchid]

the natural motivation comes from considering fusion systems

[formal ermine]

I hope you step on a lego

[tribal moss]

Okay, new justification. Let zeta be one of the infinitely many roots of unity that map to a root of unity. Adjoin zeta and f(zeta) to K if they're not already there, and consider g(z) = f(z·zeta) / f(zeta). Then surely g(1)=1, and the assumptions are still intact.

[delicate bloom]

that sounds good to me, and if we were to add anything to K, it'd only apply to g(z), which we'd undo when we go back to our problem on f(z) I think, so we're probably safe there too

[tribal moss]

We now have a concrete candidate for n (that is, the power in the solution): it has to equal f'(1) if the claim is true at all.

[wooden ember]

Is this the problem y’all are talking about?

[frigid lark]

yes

[frigid lark]

how can we guarantee that g(z) maps infinitely many roots of unity, into roots of unity

oh wait

[empty rose]

the same roots of unity as f

[frigid lark]

it does

[frigid lark]

after multiplying them by zeta^(-1) ?

[empty rose]

...oh that's true
yes

[tribal moss]

Since Merosity showed that the qualifying zetas are dense, we can pick a sequence of them that approaches 1 without being 1.

[wooden ember]

So if I understand the problem correctly we’re done if we can show that it maps infinitely many roots of unity to roots of unity of a bounded order right?

[frigid lark]

yep

[wooden ember]

Ooh that’s smart

[tribal moss]

As zeta approaches 1 we have log(f(zeta))/log(zeta) -> f'(1). If we can show that there's a dense set of zetas for which this ratio must be an integer, then it must eventually be f'(1), and we get the rest of the way by continuity.

Or perhaps it's at least always an integer multiple of 1/k where k is the degree of K, or something. That would be good enough.

[frigid lark]

wdym by log(zeta)?

[tribal moss]

Complex logarithm -- I'm just considering a small neighborhood of 1, so there's a single relevant branch.

[frigid lark]

okey

btw

this problem is in the Galois theory section of this book

[wooden ember]

This is probably a stupid question but how are you getting f’ here in the limit?

[tribal moss]

Something something logarithmic derivative.

Let me see if I can recall how it goes.

[wooden ember]

Oh yeah I think I see it

Didn’t really realize log(f(1)) was 0

Isn’t it just log(zeta) ~ zeta-1 as zeta->1?

Then definition of derivative + chain rule

[tribal moss]

Hmm, perhaps. It's intuitively clear to me, but my first attempt to write it down went astray.

[next obsidian]

d/dx(log(f)) = f’/f

[tribal moss]

f'/f was right, I think.

[next obsidian]

Actually I had it right the first time

[delicate orchid]

this is where I thought we were going lol

[delicate bloom]

can we make the same density argument with the fixed field in the generalized exercise

[wooden ember]

Very neat argument if it works out

[frigid lark]

Okay, it's now tomorrow, Imma go sleep

[delicate bloom]

if the finitely generated multiplicative group is, say powers of sqrt(2) then it turns out Gamma would be dense in all of C I believe

probably all examples will end up this way except the special case of roots of unity since they're all |z|=1 they can't "blow up" everwhere on us

at least my thinking is, if we work the general problem a bit, it'll help give us a new perspective maybe

maybe I'm wrong, but since he said it's in the galois theory section, I kind of think there's not any analysis involved, and this density stuff is a red herring.

[rocky cloak]

Hmmm, is a rational function determined by its value on finitely many points in the same way polynomials are? Feels like there could be some sort of argument there

[delicate bloom]

yeah that sounds likely

write f(z)=g(z)/h(z) and then make g(z)=f(z)h(z) and plug in numbers with coefficients in theory to get a system of linear equations

[tribal moss]

On the other hand, even the generalized statement does say "complex numbers" specifically, and if the intended argument was purely algebraic it could probably be generalized further.

[delicate bloom]

I think people say complex numbers when really they are being too lazy to say the algebraic closure of Q

at least this is something I commonly have seen in multiple textbooks

like for the most part the distinction doesn't matter enough

[delicate orchid]

I ain't typing out \overline{...} lets put it like that

[frigid lark]

We could write g(z) = w(z)z^{s(z)} h(z) for relevant z, well zetas

Idk if that helps

[tribal moss]

Hmm, yes, since every field of charactistic 0 with transcendence degree at most continuum embeds into C, we're not losing much by saying C :-)

My argument that we can assume f(1)=1 still seems to work in the generalized setting.

[delicate bloom]

trying to think about topological group tricks to apply now

[elder wave]

pi_1 is abelian

that's all i got

[delicate orchid]

pullback

amalgamated free product

[delicate bloom]

haar measure lol

lie group 😩

[delicate orchid]

truth group

[delicate bloom]

magic word open sesame

[delicate orchid]

basic group, perfect group, extraspecial group, clopen group

sylow B-subgroup

[delicate bloom]

Hurwitz canonical anabelianoidization of co-complete pro-normal module

[delicate orchid]

the Atyiah completion theorem for saturated fusion systems

[uneven fossil]

was told to ask here how i would prove question 9? i know that any matrix these generate would also have determinant one. do i need to prove that these generate 24 elements?

[delicate orchid]

you can literally just

multiply everything sure

or, go backwards

take an arbitary matrix in SL_2(F_3) and show it's a product of those two matrices

[uneven fossil]

hmm okay

this is a bit harder since it's not abelian

[delicate orchid]

yeah, I'd start by finding certain relations between the two generating elements

which reduces the number of words you'll need to check

[uneven fossil]

i know the order of a and b is 3 and the order of ab is 4

baba = abab

[delicate orchid]

that one's pretty handy

honestly, this exercise just seems like a bunch of tedium

the next one is basically the same thing but much faster

[uneven fossil]

ah i'll try the next one

[delicate bloom]

can we use the fact that it has order 24 to do some kind of cheap tricks like with sylow thms

[delicate orchid]

the cheap trick is knowing that SL_2(F_3) is a non-split extenstion of Q_8 by C_3 and literally cheating by using the next question

which is basically the sylow theorems

but considering how early Zwischen is in group theory I doubt they'd've done the sylow theorems

[uneven fossil]

yeah no sylow theorems yet

nor do i know what a non split extension is

in #10 it says the subgroup of SL_2(F_3) generated by (1 & 1 // 1 & -1)

but that doesn't have determinant 1?

wait

ok

F_3 goes hard

[delicate orchid]

funny

normally the trolls in this channel are a bit more high effort than litteraly just taking a massive shit on the keyboard

I'm dissapointed

[hasty lance]

Can we ask help in this channel?

[delicate orchid]

yeah sure

[wooden ember]

okay maybe this isnt very helpful but i didnt see it mentioned yet. Note we have f(z) in K(z) for any z in Qbar, so in particular for w a root of unity the order of f(w) is <= the order of w

[somber sleet]

can somebody tell me why splitting fields are finite? I don't see it

[delicate orchid]

fundemental theorem of algebra

[wooden ember]

maybe from this we can get some better bounds on things

[delicate orchid]

assuming you mean finite extenstions instead of finite finite

[somber sleet]

I neither see what they really are. Does somebody have an intuitive explaination?

[hasty lance]

For the second expression,
(D_{ij}•f_k)•f_l , I seem to be getting a 12 out front and not a 6

[delicate orchid]

polynomials have a finite degree, and thus have a finite number of roots in the algebraic closure - hence you only need to adjoin a finite number of elements to the base field to account for every root in the polynomial

full disclosure I came up with this on the spot so maybe wait for a second answer

[somber sleet]

yes I'm talking about finite extensions, my bad