groups-rings-fields

[dull ginkgo]

(finitely generated ideals are principal)

which is just like, induction on bezout identity

[south patrol]

Yeah lol

One nice application of this theorem is with separability

You can call a polynomial f over a field separable if f and f' are coprime

[dull ginkgo]

[south patrol]

And this shows that that is preserved under field extensions

Sometimes different definitions are used tho

[dull ginkgo]

wait lol

[south patrol]

Well one thing is that x^2 is always separable according to some definitions and never according to others

[dull ginkgo]

aE + bE = dE, a = ux, b = vx
dE = (ux)E + (vx)E = (u + v)xE subset of xE so x | d

so if d | x | a,b then x | d thus they are associated, d is still GCD

[south patrol]

Yeah

Hopefully this shows the benefit of the notation aha

[dull ginkgo]

ax + by = d. Assume a = uk, b = vk
ukx + vky = (ux + vy)k = d so k | d

[cobalt heath]

Oh

What is GCD domain?

[dull ginkgo]

just that there's a unique maximal divisor

[dull ginkgo]

There is a unique minimal principal ideal containing a and b

In GCD Domains: $(a) + (b) \subseteq (\gcd(a,b))$ while it's an equality for Bezout domains.

[cobalt heath]

Ah, so only minimal among principal ideal?

[cloud walrus]

Arzela-Açaí Theorem

[dull ginkgo]

principal ideals containing (a) + (b)

[cobalt heath]

Yeah

[dull ginkgo]

ax + by = d, Assume a = uk, b = vk
ukx + vky = (ux + vy)k = d so k | d

this implies that every common divisor divides all elements of (a) + (b)

it's just with the assumption that (x) + (y) EQUALS (d) then we have the funny

Okie so

(a) is maximal in a pid if a is prime, since if (proper) J contains (a), then (a) must strictly be in some (b), so b | a and is therefore not prime

so it's a field from there

otherwise we have some xy = a where neither x = 0, y = 0 (mod a), and xy = 0 (mod a)

(a)/(a) = (0) is not a prime ideal lmao

Assume x and y have degree leq 0, then d(x + y) = max(d(x),d(y)) <= 0, so the set of degree leq 0 elements is additively closed

By euclidean-ness, for x of degree 0 , there must exist a q, r such that 1 = qx + r, d(r) < d(x) = 0, so r = 0. Thus 1 = qx. Thus every nonzero degree 0 element is a unit.

idk how to show the multiplicative identity is degree 0

Oh my fucking god I skipped over d(ab) = d(a)d(b)

d(1) = d(x)d(x^-1) = 0

So the set of degree 0 elements is a field

[dull ginkgo]

Assume there exists x, y such that d(x) and d(y) are greater than 0. Let y = qx + r, d(r) < d(x)

if d(x) >= d(y), then d(q)d(x) = d(y - r) <= d(x), so d(q) <= 1

might try this tomorrow morning

[crystal vale]

Let N be a normal subgroup of a group G. Suppose that | N | = 5 and that |G| is odd. Prove that N is contained in the centre of G.

Let G act on N by Conjugation, because N is the normal subgroup of G.

Then by the class equation we get |N | = sum of |G|/| stab(n) | , n runs over conjugacy class.

Since |G| is odd so |stab(n) | must be odd then |G|/|stab(n) | be odd.

Now we have only two ways to write 5 as sum of positive odd numbers , 5 = 1 +1 +1 + 1 +1 and 5 = 1 + 3 + 1.

If we have the first one then N is contained in the centre of G. How can I eliminate second case?

[lapis latch]

So in the second case you’ve confirmed there would be 2 elements from the centre in N, N is order 5 and so can be generated by almost every one of its elements, so try and use that to show all of N must be in the centre

[crystal vale]

|N| has prime order so it is cyclic so it will be generated by all non-identity element, right?

[lapis latch]

Yep

[crystal vale]

So we have shown there are two elements that are in the centre of G so one of them is the non-identity element in the centre of G. Hence N is in the centre of G.

But I doubt if N is contained in the centre of G then how N have conjugacy class of order 3?

[lapis latch]

Well really you’re just showing that isn’t possible, if it helps you could view it as a proof by contradiction

[crystal vale]

Yes got it, thank you

[quartz wind]

where did i mess up? my conclusion is clearly false since then every Z/pZ for prime p is a multiplicative group

[vapid vale]

firstly i think this notation of considering Z/nZ as residue classes of Z is quite eccessive. it leads to an abundance of overlines and +sn / +tn, but we don't really care about Z itself anymore. if youre being taught to do this, then fair but i really think its not a great way to think of cyclic groups

also notationally alpha and a together are difficult to read

[quartz wind]

oh sorry about that. yeah i really wasn't thinking with the notation choice i guess

i csn rewrite if need be

[vapid vale]

all good!

[quartz wind]

but im not sure how else to define Z/nZ

[vapid vale]

ok actually nvm i think its fine here because its pretty instructive

but re: your error, consider a = 0

or $\overline{0}$ lol

[cloud walrus]

hk

[quartz wind]

okay so that one doesnt have an inverse i see

but its not the only residue class without an inverse for prime n, right?

[vapid vale]

you did not make any logical missteps

you just have to interpret what it means for gcd(a, n) to be 1 for all a

[quartz wind]

it means that a and n have no common prime factors, right? but then example $(\mathbb{Z}/7\mathbb{Z})^{\times}$ does not include $\overline{2}$

[cloud walrus]

esca | এস্কা

[vapid vale]

why doesnt it

[quartz wind]

oh wait hold on

[vapid vale]

whats 2*4

[quartz wind]

sorry i totally misread the thing i was reading

[vapid vale]

lol

[quartz wind]

yeah youre totally right i just skimmed over this part and interpreted it wrong lmao

sorry about that

then would it be sufficient to say that $a = \overline{0}$ contradicts the assumption?

[cloud walrus]

esca | এস্কা

[vapid vale]

sure if you show that

[quartz wind]

that there exists no 0 + mn congruent to 1 mod n?

[vapid vale]

well thats trivially true

[quartz wind]

okay i think i got it now. tysm!

[vapid vale]

cheers

[edgy plaza]

i feel like im missing some notation info - im seeing $U(\mathbb{Z}_p) = \mathbb{Z}^\star_p$ in reference to the subset of elements with a multiplicative inverse and $\mathbb{Q}^\star = \mathbb{Q} \setminus {0}$

what does the star next to the set indicate o . o i see from google that it has a certain meaning in discrete math and that it has meanings as an operator (kleene star) that im familiar with, but those don't seem to apply here

[cloud walrus]

haru~

[edgy plaza]

does it just mean nonzero

[lusty marlin]

For any field F, F* generally denotes the unit group of F

That is, the multiplicative group formed by all elements of F with a multiplicative inverse

[quartz wind]

this definition of symmetry isnt making sense to me. specifically, i dont see how rigid motion in 3-space permits all permutations of the vertices. i am conceptualizing rigid motion in 3 space as rotation about 3 linearly independent axes that intersect at the n-gon's center. but then the permutation, say $\sigma(2) = 1$, $\sigma(1) = 2$, and $\forall n > 2\space(\sigma(n) = n)$ for some n > 3 cannot be achieved by just rigid motion in 3-space? or am i misunderstanding something

[cloud walrus]

esca | এস্কা

[lusty marlin]

Rigid motion does not permit all permutations of vertices

Only some permutations can be achieved by rigid motions, obviously

Since we need to preserve the manner in which the vertices are connected

[quartz wind]

right

oh i see i misread the definition

thanks

[edgy plaza]

i thought the unit group was denoted U(F)? and here the sets arent necessarily fields (like Z), so they present that U(Z_n) = Z*n iff n is prime

[rocky cloak]

Both of those notations are used

[edgy plaza]

then why does it say that U(Z_n) = Z*_n if and only if n is prime

[rocky cloak]

Then they probably use R^* to mean the nonzero elements

[edgy plaza]

ah

yeah thats what i suspected

wish they said it somewhere 😔

thanks :D

[crystal vale]

Are the number of elements of order 5 in a group of order 20 is 4?

Because by the Sylow theorem we have subgroup H of order 5. And using the Sylow third theorem we get there is only one Sylow 5-subgroup.

Since H is a group of order 5 so all non-identity elements have order 5. And there is no other Sylow 5-subgroup therefore only 4 elements have order 5.

Is it correct?

[lusty marlin]

Yes

[crystal vale]

To prove that no group of order pq, where p and q are prime, is simple.

First, if p=q then the order of the group will be p^2 and it is abelian so it cannot be simple because the order is p^2.

Second, if p≠ q then there is the result that if there p is the smallest prime number which divides |G| and H is a subgroup of index p then H is normal in G.

Let p<q, So by Sylow there is subgroup of order q which has index p, and using the result we have normal Subgroup of order q.

[crystal vale]

To find a sylow p-subgroup of GL_2(F_p).
If I let the subgroup generated by [ a 0; 0 1 ] , where a is non - identity element of F_p , F_p denotes the field of order p.

Is it correct?

[delicate orchid]

That group is order p-1

[crystal vale]

Yes🥲

[delicate orchid]

So you’ll have to try a different subgroup, try finding a matrix of order p

[crystal vale]

[1 1; 0 1 ] ?

[south patrol]

That sounds good

This is the transformation f: (a,b) | -> (a +b, b) so you can easily se it has order p

[crystal vale]

Oh

[cobalt heath]

I find it quite mysterious how we can directly see how multiple action behaves here, while matrix power is more difficult to compute.

[delicate orchid]

Ermmm… what the Aff_1…

[south patrol]

Yeah

I guess philosophically it's like, matrix multiplication is essentially always the same difficulty, whilst composing linear maps varies ig lol

[cobalt heath]

Hmm

Maybe it all boils down to my stupidity. Idk

[south patrol]

and here the map is so simple that it happens to be that the latter is easier lol

[cobalt heath]

Hmm, simplicity..

Maybe some kind of 'operation cost' is involved

[crystal vale]

Is {e} is the proper normal Subgroup?

[lusty marlin]

Yes

[crystal vale]

I think {e} not considered as a proper normal subgroup because there is a question, if G has order n = p^e × a where 1<=a < p and e>=1, then G has a proper normal subgroup.

If yes then this question is trivial

[lusty marlin]

Maybe the question means a nontrivial proper normal subgroup

But {e} is certainly both a proper and a normal subgroup of any group with more than one element

[crystal vale]

But if a = 1 = e then G has order p then how does it have a non-trivial proper normal Subgroup ?

[west nova]

to prove that the field Q doesn't contain any nontrivial subfield is it enough to show that Q is the field of quotients of Z, hence Q is the smallest field that contains Z, there doesn't exist any field properly containing Z and contained Q

[crystal vale]

And to prove if G is a group of order 385 show that it's 11- Sylow subgroup is normal and it's 7- Sylow subgroup is in the centre of G.

To prove the first part, let s be the number of Sylow 11- subgroups but s=1 mod 11 and s divides 35 so s has only one choice , s=1. Hence, Sylow 11- Subgroup is a normal subgroup.

And
Second part, let G act on H, where H is Sylow 7-subgroup of G then by Class equation we have |H| = sum of order of orbits of a, where a runs over conjugacy class. Since order of orbits divides order of G.

Hence order of orbits can be 5,1,7 because it cannot be greater than 7. Thus there are only choice to write 7 = 1+1+5 or 7 = 1+1+1+1+1+1+1.

So the centre of G has a non-identity element of H and H has order of p so it is cyclic hence H is in the centre of G.

Is it correct?

[kind temple]

you can also take any non-trivial subfield K of Q. it contains the integers, so you can show that Q is a subfield of K directly

[west nova]

yea that also works thanks

i have another question, it's about showing that every subfield of R must contain the rationals. i think the multiplicative Identity 1 of R is contained in any subfield, and 1 isn't irrational. so This subfield must contain rational numbers as well. but the thing is does the subfield contain all of Q?

[mighty kiln]

A field of characteristic p contains Z/pZ

p = 0 included

And if it contains Z, it contains Q

Oh that was mentioned

[olive granite]

A curiosity which arised from some other exercise: what is an example of a non abelian group G with an element x of infinite order such that G is the normal closure of x?

[lusty marlin]

Yes, it contains all of ℚ because one can obtain the integers from 1, and therefore the multiplicative inverses of all nonzero integers must also be contained in the subfield

[west nova]

holy shit how did i forget about inverses 😮‍💨 thanks

[crystal vale]

Normal closure ?

Any hint? To prove that if G is a group of order p^2q, p and q are distinct prime numbers then it is not a simple group. I have shown for when q< p but when p < q, how can I show that?

[loud merlin]

Have you tried counting the elements

[crystal vale]

Yes

If p< q and number of the Sylow q-subgroup is p^2 then there are p^2(q-1) elements of order q

[lusty marlin]

The smallest normal subgroup of G that contains x

[crystal vale]

Intersection of all normal Subgroups that contain x ?

[lusty marlin]

Yes

[wraith swan]

I got an exercise to prove the additional groups Z, Q, and R are not isomorphic. Now I think proving that Z is not isomorphic to Q and R is because Z is generated by 1, while Q and R do not have such elements. But can someone give a hint on what to do with Q and R?

[lusty marlin]

We can't have a bijection between ℚ and ℝ, immediately ruling out the possibility of an isomorphism

[wraith swan]

Oh right, let's see if i can prove that

[lusty marlin]

Or you could consider a homomorphism from ℚ to ℝ and prove that it cannot be surjective. That way you don't need to deal with cardinalities directly

[wraith swan]

Ok thank you Knight

[coral spindle]

The cardinality method is straightforward, but here's a hint that I think provides a cool way to prove this.

In Q, consider the subgroups <1/2> and <1/3>. What's their intersection? What about <x> and <y> in general?

In R, consider the subgroups <sqrt(2)> and <1/2>. What's their intersection? What about <x> and <y> in general?

[wraith swan]

Ah i think i know where you are going

But I'm not really sure what the implications are. Subgroups are in the chapter after free groups :p

[delicate orchid]

one of the more backwards methods of teaching I've heard

[wraith swan]

Haha

[south patrol]

What on earth

I don't know how you construct free groups without mentioning subgroups lol

[wraith swan]

Categorically

[south patrol]

How are you doing stuff categorically without having touched subgroups 😭

Is this following aluffi or something

[wraith swan]

Yeah

[south patrol]

Feels very odd

Fair

[vapid vale]

interesting

[coral spindle]

I tend to find it cringe when people rib on category theory by saying it gets in the way of actual maths.

However, I think this categorical introduction is getting in the way of actual maths.

[south patrol]

Lol

Yeah I find it odd introducing category theory alongside one of the first maths courses people take lol

[coral spindle]

If you don't already know either basic category theory or basic group theory, this is a bad approach.

[south patrol]

like to me it should more be introduced once you have actually done more stuff to motivate it

[coral spindle]

And I think I am not alone in that belief

[south patrol]

i agree and i am generally very pro-categories

[coral spindle]

I quite like category theory too.

[wraith swan]

I did read about some category theory, mathematically logic, HoTT :P, but never abstract algebra. So when I read about a book that teaches CT and AA, I though let's read that. It sounded fun.

I am not studying mathematics or something

[coral spindle]

I am not studying mathematics
What do you mean by this? You are studying mathematics rn

[wraith swan]

Right, right I meant not at a university or something

[coral spindle]

I'm glad though that you're not treating this as your introduction to both.

[south patrol]

I guess you are an exception in that case aha

[wraith swan]

It is my introduction to abstract algebra 🙂

But the categorical stuff is still pretty basic. It didn't even touch natural transformations yet

And of course I had my linear algebra in the past

[agile burrow]

Yeah, I suppose I only like aluffi as much as I do because I already had some prior experience with group theory

[dim widget]

I am very ind-categories

[south patrol]

Nearly made that joke lol

[south patrol]

Lol that's so odd

Doing stuff categorically-theoretically but not defining one of the most basic notions

[agile burrow]

Lol yeah functors and natural transformations aren't introduced until chapter 8

[south patrol]

What is the point of using categories then lmao

[wraith swan]

I don't know, I am reading it for fun

[agile burrow]

I mean I guess you learn to recognize functors even if you don't know what they are

[next obsidian]

Indnouns is one of my favorite jokes 🙂

[sonic flint]

Hi guys, if I have a generator (a,b). This genereator is equivalent to (g). Where g = gcd(a,b), furthermore if I can find $c \in R$ s.t. g = a - cb then (a,b) = (a-cb). Suppose I have two vectors now $v_1, v_2$. Let $g = gcd(v_1), g'=gcd(v_2)$. If I can find a $c \in R$ s.t. $g - cg' = gcd(g, g') = gcd(v_1, v_2)$ does this imply that $(gcd(v_1, v_2)) = (gcd(v_1 - cv_2))$?

[cloud walrus]

18岁的反应81岁的操作

[south patrol]

Sorry but this doesn't really make sense to me

How are you taking a gcd of a single thing?

And what is (a,b) a generator of?

what is R? etc

[sonic flint]

gcd of a vector, say the vector is (1,2,3), then gcd(1,2,3) = 1.

R is a ring say Z/(s).

Where s is some intger.

What I wanna achieve is given two vectors say $v_1, v_2$ of same size, imagine this as a matrix of 2 columns with n rows (call it M). Apply a unimodular column transformation (say V) s.t. MV = M'. Where the first columns of M' captures all the gcd infromation of the whole matrix. i.e. $gcd(M'_1) = gcd(M_1, M_2)$ where $M_i$ denotes the ith column of M. Is this achievable if we are in Z/(s) for some integer s?

[cloud walrus]

18岁的反应81岁的操作

[delicate orchid]

I still don't know what your gcd notation means

but this reminds me of smith normal form a bit

[sonic flint]

Yes, I'm working on smith normal form.

[delicate orchid]

you can do smith normal form over any PID, which includes Z/sZ

[sonic flint]

say my vector is v = (2,4,6). then gcd(v) = gcd(2,4,6) = gcd(2, gcd(4, 6)).

[delicate orchid]

yes but then you took the GCD of two vectors

[sonic flint]

gcd(v1, v2) = gcd(gcd(v1), gcd(v2)).

[wraith swan]

The notation C^{>0} is that different from C\{0} ? Is the first the set with elements in polar coordinates (r, theta) with 0 < theta < pi/2? Or just |z| > 0?

Nevermind the exercise was about Q^{>0}, i misread

[long geyser]

Is 5/2 and 5/3 not a counterexample to this claim?

[celest furnace]

Yeah ur right you could just take each bi to be 1 and any non gcd 1 sequence of ais. You def need that gcd of the ais is 1

[long geyser]

ye I considered that first, then I thought what if author intended all bi =\= 1 and still found a counterexample so

[sonic flint]

This seems to be true. I implemented in maple and tested on few thousand random vectors.

[barren sierra]

How do I rule out n_5 = 16 for (iii)?

[dull ginkgo]

Uhh maybe consider normalizer? Idk man I have been thinking about this for a bit

Or intersection of P and Q being trivial

implying that any two 5-Sylow subgroups have trivial intersection

W a i t

@barren sierra I think I know

[barren sierra]

I arrived at this but got nowhere

[dull ginkgo]

assume there are 16 Sylow 5-subgroups, who’s intersection is pairwise trivial, thus we can consider their union. The union would be of size (5 - 1)*16 + 1 = 65?

hm

[barren sierra]

yea

[sonic coral]

isn’t this immediate from them being order 5

[barren sierra]

yea

[dull ginkgo]

Yeah it’s cyclic

[barren sierra]

that's how I got that at least

[dull ginkgo]

Anyway uh, 480 = 2^5 * 3 * 5

[sonic coral]

have you ruled out any possibilities the number is sylow 3 or 2s

[dull ginkgo]

Assume there are 16 5-Sylows, so 64 elements get “eaten up” by the Sylow 5 groups

We now have Sylow 3 subgroups to consider

Disjoint (except for identity) from the Sylow 5’s

Oh but there’s a lot of possibilities for that

[sonic coral]

yeah i thought about that and didn’t get very far

[barren sierra]

yea me either

[dull ginkgo]

n_3 = 1 (mod 3)
n_3 | 2^5 * 5

[barren sierra]

n_3 = 1, 4, 10, 16, 40

_>

[dull ginkgo]

Sylow 2’s - all of order 32,
n_2 = 1 (mod 32)
n_2 | 3 * 5

omfg

There’s 1

Just one

it’s normal but who cares

So uh there’s 31 nontrivial in that subgroup so uh that’s 385 remaining

[barren sierra]

oh so uh know how many Sylow 2's there are and assume number of Sylow 5s

[dull ginkgo]

Yeah but only a rather small number of elements are actually in a Sylow 2 or 5

[barren sierra]

yea and there can be at most 119 elements in a Sylow 2

[dull ginkgo]

what

[barren sierra]

wait no

I fucked up my calcs

[dull ginkgo]

480 = 2^5 * 3 * 5
There is only one Sylow 2

But it’s normal

[barren sierra]

I meant Sylow 3s

but then I messed up other calcs

[barren sierra]

tryna rule these out

[sonic coral]

probably look at the normalizer of the intersections

[delicate orchid]

groups of order p_1^ap_2p_3...p_n moment

[dull ginkgo]

they are cyclic so the union of non-identity elements of Sylow 5’s is 4*n

But we run into no issues

[dull ginkgo]

I forget how that helps tbh

[dull ginkgo]

Of what?

There’s one Sylow 2, and the other Sylow groups are cyclic

[sonic coral]

i don’t think this is helpful actually that would be to show it’s not simple

[dull ginkgo]

So every Sylow subgroup is pairwise disjoint

[delicate orchid]

I think just explicitly computing the normaliser is far easier

[barren sierra]

n_p = [G : N_G(P)]

ugh

[delicate orchid]

should be things that look like
(a b)
(0 c)

[dull ginkgo]

Mmm true

[delicate orchid]

a, c \neq 0 b whatever you want

[barren sierra]

this blows

[dull ginkgo]

Very strong should

[delicate orchid]

not really it's obvious

[barren sierra]

groups are ugly man

yea that makes sense tho

[delicate orchid]

upper triangular matrices preserve the flag

[barren sierra]

the what?

[delicate orchid]

oh look I win again

so this is uhhh 4*4*5 = 80

480/80 = 6

[barren sierra]

ya

ty

[delicate orchid]

but is this what it wanted you to do?

I don't know what "B2" is

[barren sierra]

I'm 99% sure they meant (C)

or eh I guess to compute normalizer you use (B2)

who cares which is which tho

[delicate orchid]

size of orbit then burnside's perhaps

[barren sierra]

hmmmm

[delicate orchid]

or sorry not burnsides, orb-stab

[dull ginkgo]

This problem is more of a pain in the ass than I thought, let d be delta because I refuse to LateX this right now

This is the shit I have thus far

First, assume d(0) = n > 0. Then n = d(0) = d(0^2) = d(0)^2 = n^2. This can only happen if n = 1. Thus d(0) = 1
But then d(x * 0) = d(x)d(0) = d(0) = d(x) implying d(x) = 1 forall x

But by euclidean division: if we have x, y nonzero
y = qx + r, d(r) < d(x)... but d(x) = 1 forall x so that's a contradiction

Thus d(0) = 0

Now assume we have a nonzero x such that d(x) = 0. Then 0 = qx + r, where d(r) < d(x) = 0, which also can't happen

so d(x) = 0 iff x = 0

Now let K_1 = d^-1({0,1}) i.e x such that d(x) <= 1

if x and y are in K_1

then d(x + y) <= max(d(x),d(y)) <= 1 so x + y is in K_1

d(x)d(y) <= (1)(1) = 1 so xy is in K_1

d(x) = d(1 * x) = d(1) d(x) for each x, so d(1) = 1 and is therefore in K_1
Likewise d(1) = d((-1)^2) = d(-1)^2 implying d(-1) = 1 and -1 is in K_1

lastly, assume x is in K_1. then there exists a q and r such that 1 = qx + r, d(r) < d(x) = 1, thus d(r) = 0 implying r = 0. Thus 1 = qx, and x is a unit as d(1) = 1 = d(x)d(q) = d(q) so x^-1 is in K_1.

These properties imply K_1 is a field

If K_1 is D itself, then it's a field

Lets assume it isn't. Let k = min(d(x), x not in K_1)

Now fix a of degree k . Allow the valuation map from K_1[X] to D fixing K_1, but sending X to a

A major observation is that, for all b in D, we have euclidean division:
b = qa + r, d(r) < d(a) = k, then r is in K_1.

This map is onto and this is through strong induction. The hypothesis is:
for d(b) = h then b = f(a) for some polynomial f(X)

Base case is where d(b) = 1, then b is in K_1, so it's just the constant polynomial for b.

The inductive step is assume 1 < d(b) < h => d(b) = f(a)
Let d(c) = h, then c = qa + r for r in K_1, thus
d(qa) = d(q)d(a) = d(q)k = d(c - r) <= max(d(c),d(r)) = d(c) = h. Thus d(q) < h, so q = f(a) for some a
thus c = f(a)a + r = g(a), where g(X) = f(X)X + r is a polynomial in K_1[X]

[cobalt heath]

Ohh non-Archimedean field

Yeah I guess it could be complicated

[dull ginkgo]

It's like an "ultra euclidean function"

[cobalt heath]

You can treat d here as a norm, I think

[dull ginkgo]

still typing out the last bit gimme a sec

Checking out the kernel: K_1[X] is a euclidean domain, and is thus a PID. So the kernel is generated by a single nonconstant polynomial, call it f(X).

Then f(X) = g(X)X + r. Under the valuation map, f(a) = 0 = g(a)a + r implying r = -g(a)a, so d(r) = d(g(a))d(a) = d(g(a))k <= 1, which implies d(g(a)) = 0, so g(a) must be 0, but g(X) must have degree less than f(X), and is in the kernel. This contradicts f(X) being the kernel's generator

Thus the kernel is trivial. We have an isomorphism by FIT

holy FUCK

that was a LOT OF WORK FOR ONE PROBLEM

[cobalt heath]

I think you can simplify this solution by writing less >.>

[dull ginkgo]

I was typing it out not as a proof by itself but my train of thought explicitly so I could keep my train of thought

[cobalt heath]

Ahh

Good that you worked out handling the nonArchimedean norm yourself

Maybe after establishing that you have embedding K_1 -> D, you could use K_1-algebra argument

[dull ginkgo]

oh wait it is the nonarchimedean norm

[cobalt heath]

Yes

[dull ginkgo]

For general fields I forget if ultranorm-ness and archimedeanness are dual lol

d(x + y) <= max(d(x),d(y)) being dual to there existing an n where d(n) > 1, from Z's canonical map into D

[cobalt heath]

You mean they are complement to each other?

[dull ginkgo]

yeah thanks

[cobalt heath]

Yea

[dull ginkgo]

actually being able to do this myself without help and in a single sitting actually is very satisfying

but it's kinda obvious where to go when you just kind of ask basic questions like "oh shit what's after degree 0"

time to write it down (hell)

[dull ginkgo]

The main bit is realizing K_1 itself is a field

which frankly is NOT OBVIOUS at first

[cobalt heath]

Yeah, it's one of the problem where knowing about things could help

Cool that you managed to pull this off without knowing about non-Archimedean!

[dull ginkgo]

I saw the map into N and was like "lol we have minimums for sets" because well ordering and kinda knew where to go

Wait

euclidean map into an ordinal, we just need minimality to basically get the "next elements after K_1"

we could use transfinite induction instead to do the same proof

lmao

actually

[south patrol]

this is a very useful result i have used time and time again

[dull ginkgo]

limit ordinals

I could add a funny extra step to see

[cobalt heath]

Hm I wanna know if the only PID F-algebra is F[x]/I

[dull ginkgo]

What is I

[south patrol]

Well like

for it to be a domain you need I prime

and then I is zero or maximal

so you either get F[x] or another field

not too exciting

[cobalt heath]

Why maximal I mean - Ah

So only F[x]

[dull ginkgo]

wait in a PID is every ideal prime

[south patrol]

Well there are other F-algebras which are PIDs

[cobalt heath]

Not really

[south patrol]

Definitely not

[dull ginkgo]

wait no

[cobalt heath]

6Z \subset Z

[dull ginkgo]

having a moment

Every MAXIMAL prime

and it's an iff

[cobalt heath]

I wanted to ask about this.

[south patrol]

wdym

[dull ginkgo]

prime <=> maximal for PID

[south patrol]

no

Non-zero prime <=> maximal

for PIDs which aren't fields

[dull ginkgo]

i should've mentioned that lo

[cobalt heath]

I mean, this also holds for fields vacuously

[south patrol]

One way to say this is "PIDs have dimension 1 or 0"

[cobalt heath]

Wait

[south patrol]

No

[cobalt heath]

Is (0) maximal in Q

[south patrol]

Yes

[dull ginkgo]

That's a krull result

[south patrol]

Yes

Lol

Idk how you are pronouncing krull if that is a joke

[dull ginkgo]

don't be krull to the joke bro

[south patrol]

actually tbf it works

in the german pronunciation

my bad.

!

[dull ginkgo]

chat is slowing to a krull bro

[cobalt heath]

Hmm, how does being Krull dimension 1 specify K-algebra

[dull ginkgo]

there we go

now all cases are exhausted

[south patrol]

wdym

[cobalt heath]

How to characterize such K-algebra?

[dull ginkgo]

i have to write down the entirety of this fucking exercise now god damn it

[cobalt heath]

Like, can we classify them?

[south patrol]

It's hard

[cobalt heath]

Meh

[south patrol]

Well like you get a whole family from curves

well this is essentially by definition but yeah

like k[x,y]/(x^2 + y^2 -1) i guess

[cobalt heath]

I guess K[x1, .., xk]/I is the easiest reduction and then you need proof after that

[south patrol]

and things

[cobalt heath]

Ah, right

[south patrol]

and k[x,y]/(x^3 - x - y^2) or something lol

lots of stuff

[cobalt heath]

So is proof for Euclidean domain field-algebra being k[x] or k bound to be complicated?

[south patrol]

That isn't true afaik (well you also need to allow for field extensions but also beyond that)

What I was saying was for stuff of the form k[x]/I, like domain <=> euclidean domain fairly immediately

[cobalt heath]

Well, yeah I noticed that problem

But you can take different field and it holds
..ah wait

[dull ginkgo]

okay tbh this is probably one of my favorite ring theory exercises along with local artinian rings being products of finitely many local rings

[cobalt heath]

How beyond field extension?

[south patrol]

Well I don't see why it should be true tbh, maybe it is possible

[dull ginkgo]

Hey potato, in a few do you want to see the absolute god awful thing I realized

[cobalt heath]

miz's proof does not seem to use non-Archimedean property after showing K_1-algebra.

[dull ginkgo]

It's used for the inductive step for the epi part

but that is a finitely generated K_1-algebra part though I guess lmao

showing it's into does not use the ultranormness no

[cobalt heath]

Hmm

[dull ginkgo]

The proof is basically that the kernel is principal, but that polynomial is g(X)X + r for some X. Taking the image, we have g(a)a = -r

but the norm's multiplicativity is what gives the contradiction

and d(x) <= 1 iff x in F

[cobalt heath]

I mean, I guess showing the epi part is important (- showing that it has one generator)

[dull ginkgo]

yes

crazy question for an intro ring theory textbook

[cobalt heath]

Intro ring theory (not really)

[dull ginkgo]

Literally

like the ring theory section is uh

[cobalt heath]

I know of "Introduction to number theory" which handles Adele rings and such, currently taking a course using it as a textbook

[dull ginkgo]

pages 85 to 156 are about ring theory

though there would probably be problems like that on an exam in a first year abstract alg course though

@cobalt heath want to see something silly and funny

[cobalt heath]

Ohh what is it?

[dull ginkgo]

For a general integral domain $D$, we can define $\mathrm{GL}(D,2) = { M \in M_2(D) : \det(M) \in D^\times }$ right, where $D^\times$ is the group of units

[cloud walrus]

Arzela-Açaí Theorem

Arzela-Açaí Theorem

[dull ginkgo]

I need to write this out formally first

Mistated, sad

SL(D,2) IS

stuck on this again lol

[south patrol]

UwU

[dull ginkgo]

By Euler criterion, (-1 | p) = (-1)^((p - 1/2) = (-1)^2n = 1.
Thus there exists 0 <= n < p such that: n^2 = -1 (mod p)

i.e n^2 + 1 = 0 (mod p)

so n^2 + 1 = kp for some natural k

[celest furnace]

Don’t.

Oh sorry I thought this was a different server

[dull ginkgo]

the problem is that idk where to go from here

[celest furnace]

Hint: norm of n + i

[dull ginkgo]

i mean ye

N(n + i) = kp

[celest furnace]

Don’t worry about k

What happens if p is prime? Then p divides n+i, right?

[dull ginkgo]

not sure how I'd prove that

[celest furnace]

What is the norm equal to

[dull ginkgo]

x^2 + y^2

[celest furnace]

Z times Z bar right?

[dull ginkgo]

for x + yi

[celest furnace]

And x^2+y^2 = (x+iy)(x-iy) right?

[dull ginkgo]

p is prime in Z not necessarily Z[i]

so idk what we'd do then

[celest furnace]

Yeah it’s a proof by contradiction

You’re trying to show p is reducible so assume otherwise

[dull ginkgo]

oh yeah i'd have to divide (x \pm iy) which can only happen if p | x and p | y but then p | 1 which is a contradiction

[celest furnace]

Yep u got it

Nice job sir

[cobalt heath]

Hmmm, I don't see silly stuff here

LU decomposition on general GL(D, 2)?

[celest furnace]

So you know it’s reducible

[dull ginkgo]

Extended euclidean algorithm

[dull ginkgo]

x | p implies N(x) | p^2 implying N(x) = 1, N(x) = p, or N(x) = p^2

[cobalt heath]

Ah, yeah euclidean algorithm is kind of like alternating-ly applying this kind of X=L and Y=U

[dull ginkgo]

First case can't happen except for the units, ignoring that

[dull ginkgo]

yeah that's what I realized

was weird at first but it proves extended euclidean algorithm and how it computes a pair of bezout coeffs

[celest furnace]

can the last?

[dull ginkgo]

Trying to think

You can prove x | p iff N(x) | p using xy = n implies (x - yi) | p i think?

[south patrol]

nice

another way to phrase this is like

Lmao

[celest furnace]

Try to use integers more

[south patrol]

This may seem ad hoc but it actually generalises a lot lol

[celest furnace]

Regular Z is much nicer

My Potato send the smallest image man has ever seen

[south patrol]

Dedekind Koomer

[celest furnace]

Holy shit mind blown

That’s so nice lol

[dull ginkgo]

Assume (x + yi)(a + bi) = p + 0i
p = xa - yb
0 = xb + ya

xb = -ya

[celest furnace]

This is the type of thing Jagr would send

[celest furnace]

Try to use norms more

[dull ginkgo]

Still trying

[celest furnace]

Do you want a hint

[dull ginkgo]

N(x) = p^2,

xx* = p^2

[cobalt heath]

Alg geo class usually force you to memorize that A/I is domain iff I is prime

[dull ginkgo]

I still can't figure it out aaaa

N(p^2) = p^4 = N(x)N(x*) which isn't an issue

[cobalt heath]

Hmm

[cobalt heath]

Huh

Why is this

[celest furnace]

You aren’t using that p is reducible

[dull ginkgo]

asterisk for conjugate

[cobalt heath]

Is x x* = p^2 or p

[dull ginkgo]

p^2 by the assumption N(p) = p^2

[cobalt heath]

Eh? What was x?

[dull ginkgo]

x | p

[dull ginkgo]

... that's how I got x in the first place

[celest furnace]

Start by writing p = zw for z w no units

[dull ginkgo]

assuming x divides p, p doesn't divide x (not associates)

p = xy, p^2 = N(x)N(y) => p | N(x) or p | N(y)

wlog assume p | N(x)

Ah lol

[cobalt heath]

Hmm Do you know N(x) = x x* >.>

[dull ginkgo]

yes

I was gonna use that but it isn't necessary

[cobalt heath]

Welp

[celest furnace]

🔠⁉️

[dull ginkgo]

what

neither x or y have norm 1

[celest furnace]

🔠 stands for always be clearing doubts and is required to put when you have successfully performed a doubt clear

[celest furnace]

😀

[dull ginkgo]

oh lordie

[cobalt heath]

Oh, how did you know?

[dull ginkgo]

I am asserting neither are units

[cobalt heath]

Does N(x) = 1 mean x is a unit?

[dull ginkgo]

It's a euclidean domain

i think that explains it

[dull ginkgo]

Same proof as I used for the prior exercise

[cobalt heath]

Ah, my hazy memory

[dull ginkgo]

1 = qx + r, N(r) < N(x) = 1 so N(r) = 0 thus r = 0

1 = qx, unit :3

[celest furnace]

Yeah cuz x* is the inverse lol

[dull ginkgo]

that too lmao

[dull ginkgo]

I guess for this:

First check which integer primes retain primality

2 is reducible, as 2 = (1 + i)(1 - i)

Thus for odd primes, they must be of the form 4n + 1 or 4n + 3. The former case is reducible

3 is NOT a quadratic residue mod 4, you can verify that

Assume x = a + bi, N(x) = p = 4n + 3 = a^2 + b^2. Then a + b = 1 (mod 2). thus only one of a, b can be odd.

but that contradicts a^2 + b^2 = 3 (mod 4), as a = 1 (mod 2) implies a^2 = 1 (mod 4) and a = 0 (mod 2) implies a^2 = 0 (mod 4)

[west nova]

i want to show that "if U and V are ideals in a commutative ring R such that U+V = R, then prove that UV = U∩V".

UV ⊂ U∩V is true in general, i have to show the reverse and that's where i am stuck now.

[cobalt heath]

Yep in this case

Forgot general logic applying to ED tho

[celest furnace]

My sir has down too much alg geo

Forgot the easy stuff

[dull ginkgo]

So no element of Z[i] can have norm 4n + 3 in general

[cobalt heath]

Ramification behavior >.>

[dull ginkgo]

I wanted to try the same for Z[w]

cbrt of unity

i think i got 3n + 2 primes are still prime

[cobalt heath]

How about 3n + 1?

[dull ginkgo]

reducible

[cobalt heath]

Oh so you are done

[dull ginkgo]

Pain in the ass

i forget how I did it

every prime p of the form 3n + 1 has p = a^2 + ab + b^2

[dull ginkgo]

Anyway once again
Assume x | p, then xy = p so N(x)N(y) = p^2, so p|N(x) or p|N(y) but N(x) nor N(y) is 1 so N(x) = N(y) = p but no elements of norm congruent to 3 mod 4 exist

Thus p = 4n + 3 is still prime

[rocky cloak]

That U+V = R means in particular that there are u and v with u+v = 1.

Then use that x*1 = x

[west nova]

ok so let x ∈ U∩V. then x*(u+v) = xu + xv = x ∈ UV.

wait a second when did i say R has multiplicative identity hmmm all commutative ring has unity?

[west nova]

alright the claim is false if R doesn't have unity. great.

[barren sierra]

What is the difference between S and A?

definition if people need it

or is S = A the whole point of the exercise

[hollow river]

S contains x^5 y^2 from that example, for instance

Well the exponents of that, at least

[barren sierra]

Oh I see

yea ok duh now that's obvious

_> thanks

reading is hard

[ebon pine]

I need help with option c, not sure how to proceed

[celest furnace]

Was A or B true,

Try using 3rd iso theorem if so

[ebon pine]

Yeah, both are true. B is immediate from 1st isomorphism theorem and then B and A are isomorphic

[crystal vale]

Any hint? To show that if G is a group of order p^2q then it is not a simple group

[ebon pine]

Is this p^{2q} or p^{2} q

[celest furnace]

Nice . So do you remember what 3rd iso theorem is? Bijection between subgroups of G containing N and subgroups of G/N?

[ebon pine]

You mean the correspondence theorem?

[celest furnace]

Yes

[ebon pine]

Alright

[celest furnace]

So…

The question is the same as finding a proper nontrivial subgroup of G/N right

[ebon pine]

Ohh, interesting...

So N is maximal if and only if G/N is simple. Is that correct?

[celest furnace]

Yeah

[ebon pine]

Alrighty, thanks

[celest furnace]

🔠

[crystal vale]

P^(2) q

[sonic coral]

split it into cases, if q is less than p and if p is less than q

[crystal vale]

I did for q< p but for p< q, there are 1 Sylow q-subgroup or p^2 Sylow q-subgroup

If there are p^2 Sylow q-subgroup then there are p^2(q-1) elements of order q

[sonic coral]

i think it would be best to play with the congruence p^2 = 1 mod q

[crystal vale]

Then p=1modq or p= -1 mod q, right?

p = -1mod q

[sonic coral]

yeah, but remember p < q, so you know that p + 1 = q

and now you have a reduced problem where you can talk about the sylow structure of a specific ordered group in this case

[median pawn]

is part (d) true or false?

[coral spindle]

I suggest you try to find a counterexample or prove it first. I would hint to you that the quotients depends on how the subgroup embeds into the group.

Another hint: consider the group S_3 x Z_3

[median pawn]

Thanks!

[crystal vale]

And p and q are distinct prime numbers so only choice is p = 2 and q = 3, right

Hence G is a group of order 12

[crystal vale]

Let G be a group of order 30. I want to prove that either the Sylow 5-subgroup K or the Sylow 3-subgroup H is normal.

Let s be the number of Sylow 5-subgroups and t be the number of Sylow 3-subgroups,.

Thus, we have s= 1,6 and t=1,10. Now if s= 6 and t= 10 then there will be 54 distinct elements which is not possible so one of them must be 1.

Hence either Sylow 5-subgroup K or Sylow 3-subgroup H is normal

Is it correct?

[rough beacon]

I'm in the process of making a Python package to implement finite group theory. I'm trying to figure out a way to find all subgroups of a group $G$. One rigorous way would be to iterate through $\mathcal P(A)$, filter them using Lagrange's Theorem, and then apply Finite Subgroup Test to each one of the filtered subsets. This works, however with larger groups (e.g., 18 or more order), the computations become very expensive. Another basic way I've thought of is compiling all ``natural'' (?) subgroups such as $Z(G)$, ${C(a): a\in G}$, and ${\langle a \rangle: a \in G}$, then taking their union to get a list. This gives close results. However, still about 10-20% subgroups are not covered in this method. In the case of $\mathbb Z_n$, the method gives correct result. All subgroups seem to be covered (upto $\mathbb Z_{18}$ that I tried), but in $U_n$ and $D_n$, it skips some of the subgroups.

Is there any relation between this union and the set of all subgroups? I need to figure out where can I focus to reach out to the remaining subgroups, if anyone has any idea.

[cloud walrus]

n11

[kind temple]

what groups are available to users?

do you have a quick way of telling if a group is abelian or not?

[terse crystal]

yes

[rough beacon]

Right now, dihedral, Z_n, U_n, K4 and Q8. And EDPs of any of these groups

[rough beacon]

Yes. I check commutativity among all elements... and return the answer accordingly

[cobalt heath]

Making python package

[rough beacon]

The 10-20% is only upto order 18 BTW when I compared results from both methods. I used GAP to find subgroups of D_100 (order 200), it gave over 220 subgroups. The method I'm trying gave around 160

[rough beacon]

[crystal vale]

If H is Sylow 3-subgroup and K is Sylow 5-subgroup of G of order 30. And let H is normal in G then HK is a subgroup of order 15. Therefore HK is cyclic normal subgroup.

K is a subgroup of HK it implies that K is a normal in G, because HK is cyclic

[rocky cloak]

How are you concluding that HK is normal?

Because it has index 2? That would work at least I guess

[crystal vale]

Yes

[untold crag]

why did the solution not check that the expression is a left inverse too

[coral spindle]

You're right that technically we do need to check that too

Fortunately, it is easy to see it's also a left inverse since everything written there commutes.

[untold crag]

thx!

[tardy hedge]

Why are the concept of group actions introduced?

For a group G and some set S, you can define lots of different group actions on S right? So this group action doesn’t necessarily have much to do with the multiplication inside G itself?

I dunno, im just learning group actions now.

[coral shale]

probably most intuitive with practical examples

[tardy hedge]

Defn says its a function G x S -> S such that a(bx)=(ab)x and ex = x

[coral shale]

dihedral groups, rubiks cube...

The action makes your ordinary set group-like in a way... is how i might think of it. The set originally has no structure, but once you define an action on it, the action lets you kindof relate elements of it in a G-like way

[tribal moss]

There's a good argument that the main motivation for talking about groups at all is to have a language for speaking about their actions ....

[tardy hedge]

Dat shit cool

[tardy hedge]

Ok

[coral shale]

So take an equilateral triangle. That is just 3 vertices.

but we can have C3 act on it, and that gets you the rotations

D3 (D6 depending on your text) will give you the reflections as well

[tardy hedge]

Ohhh lol thats cool

[tribal moss]

Permutation groups as well as Symmetry groups of geometric spaces are important examples of groups that are literally defined by their actions on some pre-existing set.

[coral shale]

(to add)
The dihedral groups wouldnt be as interesting without the above actions

at least, it wouldnt be used as a common first example

[tardy hedge]

Thanks guys, tbh i should just continue reading the section in the booo

Book

[coral shale]

And then the rubiks cube group can be thought of to act on the cube. If you take each minicube as an element say. Or each sticker as an individual element (lots of ways to choose the underlying set)

right, cayley's theorem says every group is a permutation group

[tribal moss]

And another way to phrase this is that every group has a faithful action on something.

[dull ginkgo]

I agree with this sentiment completely

[tardy hedge]

What does a faithful action mean

[delicate orchid]

the map G -> S_n given by the action has trivial kernel

[coral shale]

no 2 elements of the group act on the set in exactly the same way

[tribal moss]

Or more concretely: the olny group element whose action is to leave everthing unchanged is the identity element.

[coral shale]

You can have C6 act unfaithfully on a triangle by making 2 elements do the same rotation if they differ by 3

[barren sierra]

How do I show that there are no simple groups of order 216 = 2^3 * 3^3? Here's what I have so far. It's not much:

• n_2 = 1, 3, 9, 27 all except 1 seem possible towards contradiction
• n_3 = 1 or 4 so towards contradiction assume 4

[delicate orchid]

trivial by classification of finite simple groups

[barren sierra]

Thanks

I'll go memorize that one rq

[delicate orchid]

honestly this problem seems awful

[barren sierra]

NVM have idea

Thanks to looking at notes 😎

suppose n_3 = 4 then G acts on the set of these 4 groups by conjugation

so get a morphism G -> S_4

[delicate orchid]

it's peak....

pure kino cinema...

[barren sierra]

And I wanna figure something about the kernel

[delicate orchid]

here's a hint: 216 > 24

[barren sierra]

I hate groups I hate groups I hate groups I hate groups

[celest furnace]

💀

[barren sierra]

Yea pretty much

[viscid pewter]

i wonder if you could automate this sorta problem

[celest furnace]

Btw @barren sierra for these type of problems you should usually start with the number of the largest p for sylow

2 usually tells you nothing

[barren sierra]

Yea it's called the SmallGrps library in GAP

[delicate orchid]

it's called the IsSimple functor in MAGMA

[viscid pewter]

optimal

[barren sierra]

There's a lot of possibilities for the number of Sylow 3-subgroups which seemed more annoying

[celest furnace]

Don’t u just have 1,4?

[barren sierra]

I misremembered whoops

[viscid pewter]

7 shouldn't work

[barren sierra]

Yea

[celest furnace]

That’s how it usually goes

[barren sierra]

I mean yes I started with the Sylow 3 subgroups

[viscid pewter]

so it's not literally the worst

[delicate orchid]

bb-0b--b-b-but what baout n_3 = 1?!??!?!?!? we don't have3 an actionerino?!?!?!??!?! Where' --__S---- the MOrphism?!?!?! 🥺🥺🥺🥺

[celest furnace]

[barren sierra]

Anyways groups suck

[delicate orchid]

they really don't

[barren sierra]

True

[delicate orchid]

literally nothing about modules makes sense

[sonic coral]

i’ve done all groups up to order 300 and the homomorphism into S_n where n is the number of sylow subgroups the most useful tool

[sonic coral]

or by burnsides theorem lol

[barren sierra]

Burnsides feels like cheating lol

Also I forgot about it

[delicate orchid]

burnside is a hack

[delicate orchid]

wait you've done WHAT??

[coral spindle]

Based

[barren sierra]

cruel and unusual punishment

[delicate orchid]

if memory serves me correctly, other than prime orders, there's only two other cardinalities of simple groups in that range

so that must have been a really boring endevour

[sonic coral]

it was actually a final project for my graduate group theory class

[delicate orchid]

but it's just the same thing 300 times... it's too based for me to comprehend

[sonic coral]

i chose that one because one of my undergraduate homework assignments was the same thing but only up until 100, but at that time i didn’t have the lemma with the honomorphism into S_n so it was much harder

[sonic coral]

no literally. i think i just like front loaded a bunch of lemmas and group order types and did it that way. there were only about 12-15 i had to do by hand iirc

i think i messed up on 240 though

[delicate orchid]

,w prime factors of 240

[cloud walrus]

Results provided by WolframAlpha

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Requested by wewladstbh

[delicate orchid]

hmm

[sonic coral]

i tried to use this result in D & F let me find what i did

[delicate orchid]

I suppose the only non-trivial ones are where there's more than 3 prime factors and the smallest prime sylow isn't cyclic

[sonic coral]

yes i think that was the case

[delicate orchid]

the cyclic thing gets you a p-normal compliment so obviously can't be simple

[sonic coral]

i tried to apply this idea for 240 but i think he said it wasnt quite right but i dont remember what exactly was wrong with my proof

oh lemma 7 was cool too

[hollow tartan]

Hello, my intuition tells me that this should hold, but idk how to show it.

[median pawn]

Do you mean G_1 = S_3 and G_2 = Z_3?

[coral spindle]

No

In fact I'm afraid that I actually was thinking of a counterexample to (e), not (d).

I will give you a different hint: look at S_3 and its normal subgroup Z_3.

[sonic coral]

i hate that you refer to it as Z_3 and not A_3

[coral spindle]

OK

[delicate orchid]

It’s the same thing

What I hate is that it’s not C_{3

[coral spindle]

I would like to call it C_3 but I fear that early group theorists know it by Z_3

[sly crescent]

But Z_3 is the 3-adic integers

[delicate orchid]

but Z_3 is the group of 3-cycles or whatever

[tribal moss]

C_{3?

[delicate orchid]

ok so I made a typo while typing on a phone whilst walking

cut me some slack jack

[south patrol]

Oh lol we were always told C_3

And yet now I always view it as Z/3 lol

[wraith swan]

C_3 = Z/3Z right?

[coral spindle]

C_n is the Cyclic group of order n

[wraith swan]

Right ok

[coral spindle]

Usually written mutliplicatively

So it is isomorphic to Z/nZ

[wraith swan]

Sorry, every time I see your name I ask myself are you Afrikaans?

[coral spindle]

Well spotted, indeed I was born there and my grandma calls me boytjie

[wraith swan]

Ah that's sweet, reminds me of my grandpa calling me kerel before he died. Well back to math 😛

[icy ore]

Well, not always. Think of additional condition that will make this statement true

[hollow tartan]

I was thinking maybe phi needs to be surjective

[rapid junco]

is this polynomial well studied? I am trying to find resources on it:

-x^2n - x^{2n - 2} - ... - x^2 + 1

[tribal moss]

What a shame about that + sign at the end.

[rapid junco]

is the latter well studied?

[rocky cloak]

With the minus sign it should be
(1 - x^{2n+2})/(x^2 - 1)

I don't know that it is necessarily well studied, or what it would really mean for a single polynomial to be well studied, but at least it has a nice formula

[old sun]

What are the irreducible polynomials of degree <= 4 in Z/2[x] and why?

[tribal moss]

Which is, up to overall sign, a quite well studied polynomial in x².

So the well-known theory would go quite some way towards factoring it, for example.

[rocky cloak]

A polynomial of degree 4, either has a root, is the product of two irreducible degree 2 polynomials, or is irreducible. The only irreducible degree 2 polynomial over Z/2 is x^2 + x + 1, and it's square is x^4 + x^2 + 1.

Things with constant term 0 has 0 as a root, and things with an even number of summands has 1 as a root. So that leaves

x^4 + x^3 + 1
x^4 + x + 1
x^4 + x^3 + x^2 + x + 1

That's of degree 4, then you can reason similarly for degree < 4

[long geyser]

isn't the p'th root of every element themselves?

[tribal moss]

This is the case for elements of Fp itself, but not for the rest of k.

[long geyser]

oh I meant (ii) not (i)

[tribal moss]

Oh right, then it's just the identity.

Which you can prove using (i).

[long geyser]

wait, you can?

oh, because it must be injective

cool

[tribal moss]

Because f(x) = x^p is a homomorphism, you have f(1+1+...+1) = f(1)+f(1)+...+f(1) = 1+1+...+1.
But every element of F_p can be written as 1+1+....+1.

[long geyser]

ah, you can prove that it is the identity

but to even show this is a homomorphism, you must show (a + b)^p = a^p + b^p which requires the very thing which you'd want to prove, so...

[delicate bloom]

you can prove that with the binomial theorem

[long geyser]

oh wait yeah, my bad

[delicate bloom]

p choose n for 0<n<p has strictly integers less than p in the factorials in the denominator, so its numerator must be divisible by p

well, "numerator" it's an integer ofc

[long geyser]

you use that same theorem for fermat which is why I got confused

at least that's what happened earlier in the book

[tribal moss]

Yeah, this is essentially Fermat's little theorem in slightly fancier language.

[old sun]

What's the kernel and result of Isomorphism Theorem of f: Z[x] -> Z[sqrt(2)] : f(x) = sqrt(2)?

[delicate bloom]

your f is not a homomorphism

but I'm guessing you meant you want to map x to sqrt(2)

[old sun]

Hm, so maybe they made a typo?

[delicate bloom]

yeah, you should prove it's not a homomorphism to see what I mean before continuing with assuming they mean x |-> sqrt(2)

[old sun]

I think correct notation is like Z[sqrt(d)] = a+bsqrt(d) right

a,b, in Z *

[delicate bloom]

I see what you're saying Z[sqrt(d)] = {a+b sqrt(d) : a,b in Z} but

that's not really related to their mistake, Z[x] is all polynomials in x with integer coefficients, they're substituting x for sqrt(2)

[tribal moss]

The "x" in "f(x)=sqrt(2)" is not a variable, but a specific element of Z[x].

[south patrol]

It's not a typo, just badly written imo

[tribal moss]

Specifying where f takes this element is enough to specify a homomophism uniquely.

[south patrol]

The unique ring homomorphism which sends x to sqrt(2)

[delicate bloom]

specifically sqrt(2) = f(xy)=f(x)f(y) = sqrt(2)sqrt(2) is not a homomorphism

[old sun]

Hm so example would be x+1 -> sqrt2 + 1

[south patrol]

Yes

[tribal moss]

Yes.

[south patrol]

You just plug sqrt 2 into your polynomial

[old sun]

x^2 + x + 1 -> 2 + sqrt2 + 1

[tribal moss]

And in particular f(x^2 - 2) = 0.

So there you at least have one nontrivial element of the kernel.

[old sun]

oh is that the kernal ?

[south patrol]

To be fair they did make a typo elsewhere

What does the theorem tells us

[tribal moss]

The theorem tell us that Z[x]/<.....> is isomorphic to ...

[old sun]

<x^2 - 2>

[delicate bloom]

oh duh, sorry for misleading you there I'm too distracted @old sun

[old sun]

not sure how to write the kernel tbh. but it makes sense that f(x^2 - 2) = 0

like i just wrote it from the notes but i dont know how to generate it

vs something like <2> in z10

but apparently Z[x] / <x^2 - 2> ~= Z[sqrt2]

???

[tribal moss]

Yes. Do you know the meaning of the notation <x²-2>?

[old sun]

Ideal generated by <x^2 - 2> = {?,....,?}

[tribal moss]

Yes.

[median pawn]

OK thanks!

[tribal moss]

Do you know that this ideal is the kernel, or is it a guess?

[old sun]

it is a guess

{x in R : f(x) = 0}

[tribal moss]

But since we've agreed that x²-2 is an element of the kernel, the kernel at least contains <x²-2> as a subset.

We need to show that every element of the kernel is in fact in <x²-2>, so assume we're looking at some polynomial p with f(p)=0.
What does polynomial division of p by x²-2 then tell us?

[old sun]

not sure how to divide an arbitrary polynomial by x^2 - 2

[tribal moss]

Hmm. Not really sure how to approach this problem if you haven't learned about long polynomial division ...

[old sun]

I do know about long division

but i dont know how to divide it if there are no numbers and just a letter p

[tribal moss]

Oh, what I means was something like: polynomial division tells us that there exist polynomials q and r, such that
p = (x²-2)q + r
and r has degree at most 1.

[old sun]

why degree at most 1?

is it because x^2 can cancel all integers?

[tribal moss]

Yes -- when you divide by a polynomial whose highest coefficient is invertible, polynomial division can always continue until the remainder has smaller degree than the divisor.

[old sun]

what does p = (x^2 - 2) q + r tell us?

[tribal moss]

Well, if r=0, then we're happy because it then tells us that p is in <x²-2>.

[old sun]

sidenote, is <> always with multiplication?

how do we tell if it's + or *

[tribal moss]

If r is not 0, then it at least tells us that r is in the kernel (since it is the difference between p and (x²-2)q which are both in the kernel).

[tribal moss]

For a single generator we always have <a> = { ra | r is in the ring }, yes.

[old sun]

but what about the ideals of Z/10Z which are (1), (2), (5), (0)

looks like this is by addition

else (1) = {1}

[tribal moss]

No, it works there too: <1> = { r·1 | r is in Z/10Z }

[minor fulcrum]

Vakil claims that if we have a ring R and an ideal I, then there's a bijection between prime ideals in R/I and prime ideals containing I in R. One way of this bijection feels clear to me (the image of a prime ideal in a quotient is prime) but the other way is less so

how does that bijection go?

[kind temple]

third isomorphism theorem

[minor fulcrum]

ty

[south patrol]

It's a standard fact that the preimage of a prime ideal under any ring homomorphism R -> S is prime. (This follows quickly from the definitions)

Also, this is a special case of the more general correspondence theorem, if you're interested in the bijection

[dull ginkgo]

Assume prime p = 4n +3, v_p(a^2 + b^2) = k

x = a + bi. p is prime in Z[i], and p | a^2 + b^2 = xx* implying p | x or p | x*

[celest furnace]

Is the product of two numbers of the form sum of two squares also a sum of two squares?

Omg. Is this multiplicativity of L2 norm!!!

[dull ginkgo]

yeah the reverse is way easy

[celest furnace]

So cool

[dull ginkgo]

Anyway, if n | a + bi then n | a and n | b so n | a - bi

therefore

p^k = p^2n where n = min(v_p(a),v_p(b))

so each prime congruent to 3 mod 4 in m's prime decomposition must have even valuation

converse is easy.

okay idk how to do this hm

wait omfg

a = 0, b = p

allow me to slam my head against my desk :3

[dull ginkgo]

weird issue, idk how to describe these "symbollically" without writing it down like with plain english

[celest furnace]

dont overuse symbols

[dull ginkgo]

The mobius function is multiplicative because if n_1 has a squared prime factor then n_1 n_2 has a repeated prime factor, thus mu(n_1) mu(n_2) = 0 = mu(n_1 n_2)

[celest furnace]

This is good

[dull ginkgo]

Otherwise, the mobius function is the parity of the number of prime factors of n_1 and n_2, so n_1 n_2 is just the sum mod 2 and (-1)^n is iso to Z/2Z lmao

so it's multiplicative

[cobalt heath]

It could be annoying to write it all down symbolically

[dull ginkgo]

That sum is the part that's easiest in plain english

We can strip out the divisors that have a squared prime factor, so it's a sum over products of n's prime factors

i sorta get stuck here showing if n != 1 then it's 0

[delicate bloom]

because it's multiplicative you can split it up into a product over sums of powers of distinct primes, prod_p (mu(1)+mu(p)+mu(p^2) +...)

[dull ginkgo]

product over sums of powers?

[delicate bloom]

for instance n=12 let's say, (mu(1)+mu(2)+mu(4))*(mu(1)+mu(3))

makes sense that we can do this in general or should I say more

[dull ginkgo]

is this important for the sum?

[delicate bloom]

it is the sum

[dull ginkgo]

the fuck?

[delicate bloom]

distribute it out if you don't believe me

[crystal vale]

To show that if N is a normal subgroup in finite group G such that order of N is relative prime to index of N then N is the only subgroup of that order.

Let N have order n.
Now, let G->G/N, and if there is any other subgroup M of order n then image of M is trivial which implies that M contained in N.

By cardinality, N=M.

Is it correct?

[dull ginkgo]

how the fuck would you prove that?

[delicate bloom]

you get a mu(2^a)mu(3^b) = mu(2^a*3^b) for each term in that example

you already proved it's multiplicative

[dull ginkgo]

yeah?

[delicate bloom]

I'll be more explicit about it, if f is multiplicative then $$\sum_{d|n}f(d) = \prod_{p|n} \sum_{i=0}^{v_p(n)} f(p^i)$$, here the product is over distinct prime divisors $p|n$.

[cloud walrus]

Merosity

[dull ginkgo]

how would I prove that

it seems unintuitive as fuck

actually

induction

[delicate bloom]

each divisor d has exactly one prime factorization

[dull ginkgo]

partition d | n into sets where p^n | d

[dull ginkgo]

For each prime p dividing n, we can consider the divisor set of n, and partition it via preimages of v_p(d)

[delicate bloom]

do you understand why this example works out

[dull ginkgo]

yeah because mu(p^n) = 0 for n > 1

OOOOOH

[delicate bloom]

I mean why that sum is equivalent to the original not why after you have it it gets you the solution. I don't quite think you're grouping them in the same way as that

[dull ginkgo]

Let D(n) = {d : d | n, d > 0}. We are taking the sum of f(d) over d in D(n)

Recursively start with the smallest prime divisor of n, then we can partition D(n) into preimages of v_p(d). Just calling these preimages v_p^-1(k) in this context

thus for each d in v_p^-1(k), there is a unqiue h in v_p^-1(0) such that d = p^k h

Thus we can split the sum into a sum over subsums over the v_p^-1(k), of which we can factor out the sum of v_p^-1(0)

[delicate bloom]

maybe we should prove that the dirichlet convolution of two multiplicative functions is another multiplicative function instead

[dull ginkgo]

leaving a sum over f(p^k)

[delicate bloom]

that would make the decomposition much clearer

[dull ginkgo]

i can actually symbolically write that proof out lmao

[delicate bloom]

then we're just using the multiplicativity of the result to break it up into a product over the prime power divisors of n

makes sense?

[dull ginkgo]

I think the one I just typed out is very clear

At least to me

just factoring out the sum over divisors of n/p^v_p(n), and leaving a sum over f(p^k)

and recursively doing this until we exhaust the prime factors

[delicate bloom]

sounds off to me

[dull ginkgo]

how so?

[delicate bloom]

I only skimmed it

[dull ginkgo]

this is the tldr

[delicate bloom]

that's why I suggested this instead

[dull ginkgo]

hmm

the convolution is $\sum_{ab = n}{f(a)g(b)}$ right

[cloud walrus]

Miz

[delicate bloom]

yup

[dull ginkgo]

for 1 <= a,b <= n

that seems way harder to show

seems like a similar idea though

[cobalt heath]

Hmm what is this for?

[dull ginkgo]

Jacobson

might prove stuff about the convolution first and go from that

Anyway to do this one quickly first

[cobalt heath]

Ah, inversion formula, counting stuff

[delicate bloom]

yeah, personally I think it's criminal to introduce the dirichlet convolution without the dirichlet series to help guide it

[dull ginkgo]

... monoid rings

[delicate bloom]

it's much more obvious to guide yourself through what needs to happen when you see that this simply comes from multiplying dirichlet series and their euler products are what facilitate it

[cobalt heath]

Oo combinatorics

[delicate bloom]

doesn't matter

[dull ginkgo]

the dirichlet convolution is how you'd describe multiplication on R[N]

wait that's just R[X]

nvm

N under multiplication

There was an exercise on that

[delicate bloom]

the point is it directly encodes the convolution as multiplication of the series

[dull ginkgo]

and laurent series too

which is also cool

[delicate bloom]

no

I'm talking about Dirichlet series right now

[dull ginkgo]

wait not laurent

fuck what are those series called

a_n/(1 - q^n)

[cobalt heath]

Ah so you can consider dirichlet convolution like multiplication on R[N]

[dull ginkgo]

LAMBERT SERIES

that's what they're called

[cobalt heath]

I guess you have formal power series f(n) q^n in this case

[dull ginkgo]

oh lol

[cobalt heath]

.>

[delicate bloom]

lambert series are second rate to dirichlet series here

[dull ginkgo]

i know it's funny

also sometimes they are like related in weird ways afaik

[delicate bloom]

well if you knew we wouldn't be having this discussion

[dull ginkgo]

But yeah I was introduced to dirichlet convolutions via a monoid algebra on N under multiplication

[dull ginkgo]

i read about lambert series a while ago i just forgot the name

[delicate bloom]

I can give a quick rundown of using dirichlet series for the task at hand here

[dull ginkgo]

I know how that's done for multiplicative functions

wait.

[delicate bloom]

so you can walk through the proof of convolution of multiplicative functions is multiplicative using them as your guide

[dull ginkgo]

I realize now it's obvious lmao

[delicate bloom]

good

[dull ginkgo]

divisors of n and divisors of m have intersection 1

sets of divisiors I mean

[delicate bloom]

?

[dull ginkgo]

We are summing over pairs (a,b) where ab = nm

[delicate bloom]

are you using dirichlet series at all

[dull ginkgo]

I'd prefer not to pass to analytical means to do a combinatorics problem

[delicate bloom]

not analytical at all

[dull ginkgo]

holy shit it's 1:10 am

I want to do the irreducibles problem rq then I'll come back to dirichlet convolutions tomorrow morning

[dull ginkgo]

Alrighty so

• Every monic linear is of the form X + a, so there are q many linear monics
• Every monic quadratic is of the form X^2 + aX + b, so there are q^2 many quadratic monic polynomials by associating a quadratic to a pair (a,b)
• Every reducible quadratic is of the form (X + a)(X + b), so there are q^2/2 many reducible quadratic monic polynomials

is the last statement not correct?

oh

[cobalt heath]

(X+a)^2 is prob counted twice

[dull ginkgo]

Bijection between reducible monics and pairs (a,b) ~ (b,a)

so dividing out by 2, no?

oh shit

[cobalt heath]

(a, a) ~ (a, a) >.>

[dull ginkgo]

just realized yeah

[cobalt heath]

Now you know how difficult counting is

[dull ginkgo]

(q 2) + q

I need to go to bed, it's almost 1:30 am

[dull ginkgo]

Let L(n) = {(a,b) in N^2 : ab = n}

Assume n_1 and n_2 are coprime, then L(n_1 n_2) is in bijection with L(n_1) x L(n_2)

As if ab = n_1 n_2, then a = a_1 a_2, where a_1 | n_1, a_2 | n_2, and ditto for b
Therefore it's not hard to show a_1 b_1 = n_1, a_2 b_2 = n_2 due to coprimality. Thus (a_1, b_1) is in L(n_1) and (a_2, b_2) is in L(n_2). This is unique for (a,b)

[delicate bloom]

better off counting the other thing and then mobius inverting to get it

[dull ginkgo]

Thus: $\sum_{ab = n_1 n_2}{f(a)g(b)} = \sum_{a_1 b_1 = n_1}\sum_{a_2 b_2 = n_2}{f(a_1a_2) g(b_1b_2)} = \sum_{a_1 b_1 = n_1}\sum_{a_2 b_2 = n_2}{f(a_1)f(a_2) g(b_1)g(b_2)} = \left(\sum_{a_1 b_1 = n}{f(a_1)g(b_1)}\right) \left(\sum_{a_2 b_2 = n_2}{f(a_2)g(b_2)}\right)$

[cloud walrus]

Miz

[dull ginkgo]

Thus proving the dirichlet convolution of multiplicative functions is multiplicative

because a_1 and a_2 must be coprime as they are divisors of n_1 and n_2 (coprime) respectively

ditto for b

@delicate bloom does this work?

[delicate bloom]

sounds right to me

[dull ginkgo]

nice

[delicate bloom]

you are now one step closer to proving it's a group

[dull ginkgo]

associativity seems like a symbolic clusterfuck

though it's basically just... sum over f(a)g(b)h(c) for abc = n right

[cobalt heath]

.>

[dull ginkgo]

sotrue

[cobalt heath]

How about showing for general monoid ring R[M]

[dull ginkgo]

that was an exercise in jacobson I did already