groups-rings-fields

[crystal vale]

i asking here because it is in ring problem set

[lone niche]

let p/q be the zero of f, then both must be odd since coefficients are odd.
I think you need to elaborate more on this

[crystal vale]

since p/q is zero of f, so let f = a_nx^n +...+a_1x + a_0, then p divides a_0 and q divides a_n, since both are odd therefore p and q must be odd

[lone niche]

since q is odd and all coefficent are odd and even degree so it gives 1 = 0mod2
This part is also hard to follow

[lone niche]

Yes, you definitely need to include that.

[crystal vale]

q is odd so and p is odd, pq^-1 = 1 mod 2

and even degree implies there odd number of terms

and all coefficient are odd

therefore there are odd number of 1

[lone niche]

how are you defining p/q^-1 mod 2?

ah nvm

yes I think that works

But yeah include allll of that.

[crystal vale]

okay thank you

yes

[tardy hedge]

If you have R (+) R, whats a basis for that as a free R-module? R has 1

Kind of silly question but im confused about something but im not yet exactly sure what it is

Im confused by saying its 1+1 ..?????

Oh maybe im confused because i havent even seen R being written as a direct sum like that anyway

Right, it does NOT even make sense to write R^2 = R x R as an internal direct sum R (+) R right?

No way

[lone niche]

Can we assume R is commutative to simplify things?

[tardy hedge]

Yea

[lone niche]

Ok then what about writing (r,s) as an element of R oplus R

Then you can check you can write (r,s) as a linear combo of (1,0) and (0,1)

[lone niche]

As an internal direct sum of what?

Oh of R^2

[tardy hedge]

Yeah i suppose thats what i was thinking

But then R cap R is obviously not 0 so

[lone niche]

(R,0) cap (0,R) is (0,0)

[tardy hedge]

Ok i kind of see where i was getting confused now

So really I should write (R,0) (+) (0,R) if i wanted to be i guess super careful with it

And writing it like that i understand what is happening now

Thanks

(R x {0}) (+) ({0} x R)

Lol

[lone niche]

Yeah

[tardy hedge]

If you just have a free R-module, not knowing if it has a finite basis or not, is it still isomorphic to a direct sum of R’s?

But could be arbitrary # of R’s?

Say R is commutative because then every free R-module M has the same size basis right

[lone niche]

Yes, that's why I restricted to commutative, I am not too familiar with the non-commutative case

[tardy hedge]

I see

[lone niche]

Yes

that is one of the alternative definitions of free module

[tardy hedge]

Thank you. Tbh, lots of little details for free modules have gotten me a little weary

[lone niche]

Aluffi Chapter 0 goes in very deep detail about free modules.

He explicitly writes the equivalences of all the common definitions.

[tardy hedge]

Wow I should read that. Ive heard lots of good things about that one

My exam is tmr morning so ill prob do it after lol

[lone niche]

good luck!

[tardy hedge]

Thanks!

[rocky cloak]

Fun fact, there are noncommutative rings R such that R is isomorphic to R^2 as a module.

So yeah you need commutativity (or some other assumption like R being Noetherian for example)

[rotund aurora]

It is impossible that R be isomorphic to an infinite direct sum of R (as R-modules), right? (R non-trivial of course)

[rocky cloak]

The only examples I know have R isomorphic to an infinite direct sum of itself

[rotund aurora]

oh xD

You can have funny examples if you take R=L(V) where V is a vector space of large dimension, L(V) denoting the ring of linear transformations V-->V

[lone niche]

Yeah, I remembered the invariant basis theorem before I answered and googled some examples, but decided I didn't have the energy to understand what went wrong lol

[crystal vale]

let f in Z[x] such that f(0) and f(1) are odd. Prove that f has no zero in Z.

since f(0) is odd so if a in Z and a is a zero of f then a must be odd.

and since f(1) is odd so f(1)-f(0) is even.

let f = a_nx^n + ...+a_1x + a_0, then f(a) mod 2 = 1, which is contradiction.

it is not proof but is idea correct?

[rocky cloak]

Yeah, this is the easiest example. But I'm sure you can have even funkier stuff

[rotund aurora]

like?

[rocky cloak]

Well like R not being isomorphic to R^2, but R^2 being isomorphic to R^3 or weird stuff like that.

Or something like that, idk

[lone niche]

since f(0) is odd so if a in Z and a is a zero of f then a must be odd.
Can you elaborate more on this?

Don't answer for him croquetaOh I guess you weren't lol

[rotund aurora]

I may be clowning, but say R is spanned by (v_i)_{i\in I} as an R-module. Then 1 is expressible as a finite linear combination of the v_i, and hence is spanned by a finite subset of the v_i. No?

[crystal vale]

Same reason which I gave in previous question

[bitter rover]

yeah refer to the LITERATURE you dingus

[crystal vale]

Or since f(a) mod m ≠ f(b) mod m implies a ≠b mod m .

f(a) mod 2 ≠ f(1) mod 2 but a = 1 mod 2.

I used a is an odd

[lone niche]

I have the memory of a goldfish sorry

[crystal vale]

Okay, a is a zero of f, so a must be divide a_0 which is odd so a must be odd

[lone niche]

Can you elaborate more on

let f = a_nx^n + ...+a_1x + a_0, then f(a) mod 2 = 1

[crystal vale]

a_na^n +...+a_1a + a_0 = a_n+...+a_1 + a_0 mod 2, because a = 1 mod 2

Now it is given f(1) is odd so f(1) = 1 mod 2

I don't understand why you are asking me for elaboration?

I know its a good thing but not every time

[lone niche]

Because these things are not immediate to me. For example you are citing the rational roots theorem implicitly above at some point.

[crystal vale]

Yes I know

[lone niche]

I would take points from your answers.

I am thinking you are submitting these as HW for some class. And as a grader it's very hard to know what one specific student knows and doesn't know.

[crystal vale]

This is not HW, I was verifying the idea

[crystal vale]

I can understand

[lone niche]

What is obvious to you is not obvious to me. You might have been playing with the idea for a while, I just saw the problem the moment you showed me your solution.

[crystal vale]

My mistake

[lone niche]

It's okay, I also do it.

[bitter rover]

The idea can work but it's roundabout. Just get at the heart of the matter, which is that if x ≡ y (mod n) then f(x) ≡ f(y) (mod n)

[lone niche]

How did you format it like that?

[tardy hedge]

How would I do (b)? I was trying to do it like how the proof worked for P being a direct summand of a free module -> P is projective

[bitter rover]

If x is odd then f(x) ≡ f(1) (mod 2)

And if x is even then f(x) ≡ f(0) (mod 2)

So, if both f(1) and f(0) are odd then...

[bitter rover]

https://support.discord.com/hc/en-us/articles/210298617-Markdown-Text-101-Chat-Formatting-Bold-Italic-Underline

[tardy hedge]

P1 (x) P2 (inclusion) -> F1 (x) F2 , and by (a) F1 (x) F2 is free so u can do that diagram kind of thing to show p1 p2 projective

[lone niche]

thanks

[lone niche]

Sorry, I made the assumption you were submitting it for grade. Perhaps just checking the idea indeed would've been more beneficial in this case. 😔

[lone niche]

I think they want a more direct approach

[tardy hedge]

Hm ok, but does what i did work?

[lone niche]

I am sorry but I would need to see the diagram. The idea I have in mind is somewhat different.

[lone niche]

Well, okay probably yes.

[tardy hedge]

If I is a principal ideal in the domain R then I (x) I has no torsion elements

Not really sure on how to approach this. I know saying an element is 0 in a tensor product is kind of tricky

Or more subtle than i am thinking

[rocky cloak]

You're right yeah. It's infinite product not infinite direct sum, so I'm the one whose clowning

[tardy hedge]

Jagr is the goat

[elfin wraith]

Do you know about free modules?

[tardy hedge]

Yea

Oh I is a free R-module and tensor product of free modules are free

[elfin wraith]

I could also be mistaken here because it’s not something I’m super comfortable with either admittedly, but I think the argument is, principal ideals generated by a regular element are free, tensor product of free modules is free, free modules are torsion free?

[tardy hedge]

Yeah makes sense i think

[elfin wraith]

Yeah this was what I had in mind

[void cosmos]

yes

got this in an exam b4

[tardy hedge]

Yooo

[void cosmos]

yeah

[elfin wraith]

I’m so upset reading over the solutions to the first homework for my noncom class, the first question was honestly so difficult because it felt so trivial. There was such a clear isomorphism between these rings and I wanted to just claim send x to x and y to y but like the different x and y

Turns out that is the correct argument you just also throw some universal properties at it

Frustrating

[tardy hedge]

Hi friends

[dull ginkgo]

I need to look into projective and injective objects

[glad osprey]

If R is a commutative ring and I is a maximal ideal, then R/I is a field. What if R is not commutative? Will R/I be a division ring?

[lapis latch]

Consider a matrix ring over a field

(0) is the only maximal ideal, yet isn’t a division ring when the dimensions greater than 1

[glad osprey]

I see, thanks

I seem to remember that the proof that a commutative simple ring is a field uses commutativity in some important way, so I guess that is kind of the reason

But if you have some noncommutative ring R, and R/I turns out to be a field, then you can conclude that I is maximal, right? Just by the correspondence principle?

[lapis latch]

yeah

[rocky cloak]

So one thing that's true is that R is a division ring if and only if 0 is a maximal left ideal.

Factoring out a maximal ideal makes 0 a maximal ideal, but you can still have bigger left ideals.

[dull ginkgo]

i keep forgetting the morpism form of matricies reverses multiplication

aaaaaaa

[rocky cloak]

Guess you need to switch to right modules

[dull ginkgo]

I thought that would still have the same problem

[dull ginkgo]

No I mean like if M* is the induced morphism from matrix M on the free finite-rank modules, then (AB)* = B* \circ A*

[rocky cloak]

Yeah, so that problem goes away by using right modules.

[dull ginkgo]

So uhm, if it’s commutative wouldn’t that mean matrices commute

Which isn’t true

[rocky cloak]

No, it doesn't mean that.

It just means that for a matrix if you think about it as acting on the right on row vectors or on the left on column vectors, those define different homomorphisms on R^n

For R commutative both of those make sense, but for noncommutative only one of them makes sense depending on whether R^n is a left or right module

[dull ginkgo]

I need to revisit this

Typically we use the column-basis approach

[rocky cloak]

In which case it makes the most sense to think about right modules.

Otherwise you have to take the transpose multiply then take the transpose again

[dull ginkgo]

I need to ponder this

[rocky cloak]

I mean it's easiest if you just think about 1x1-matrices

[dull ginkgo]

So which does our standard matrix multiplication definition abide by

Why did we chose that

Is that through the canonical End(M^N) and M_N(End(M)) isomorphism

[rocky cloak]

So historically f(x) is how we wrote application of functions.

So converting a function to a matrix you might want
f(x) = Mx

So now the left side of x is "busy", so if you want to define a compatible action of the ring on x it should be on the right.

[dull ginkgo]

Actually this might explain it

[dull ginkgo]

Yeah it just fried my brain because Jacobson uses Row vector convention

But still has that (AB)* = B* \circ A* theorem

[rocky cloak]

Yeah, some fields of math do use that, so
f(x) = xM
and then composition of matrices is opposite of composition of functions.

[dull ginkgo]

I don’t think he does that

[rocky cloak]

The main stickle is that if R is a left module over itself End(R) = R^op.

So I guess you could think of M as a matrix with coefficients in the opposite ring. So then matrix multiplication would work as normal except you reverse all multiplications of elements

[dull ginkgo]

Yeah that’s what I was pondering

But it still has that “reversing composition” issue

[rocky cloak]

Well, now the composition is correct. It's the multiplication of elements that is reversed

Again the 1x1 case is enlightening

[dull ginkgo]

I need to review how matrix multiplication is defined vs the standard way of doing it

this always messes with me

[rocky cloak]

The easy solution is of course to forget about matrices and just work with homomorphisms

[dull ginkgo]

Pain

[elfin wraith]

I’ve done the proof that End_R(R) is isomorphic to R and it’s still very surprising to me honestly, it’s not something that feels at all obvious

[dull ginkgo]

*antisomorphic

a_l(b_l(x)) = a(b_l(x) = b_l(ax) = bax

[elfin wraith]

Right module endomorphisms

[dull ginkgo]

oh, pain

[dull ginkgo]

then R^op maps into End(M)

Bimodules are R (x) R^op into End(M) iirc lol

[elfin wraith]

No that’s my bad actually that’s some bad notation I chose to write that with tbh

[dull ginkgo]

End_R(M) is actually the commutator of the ring's embedding into the ring of abelian group endomorphisms

literally like the first exercise jacobson gives with rings because he's jacobson

[glad osprey]

I'm supposed to prove that the left modules over R/I are exactly the modules M such that IM = 0, but I don't understand what the multiplication in IM means. If M is an R/I module then we have a multiplication between R/I and M, but how do we get a multiplication between I and M?

[elfin wraith]

That’s wild, i can’t remember what Wibels first cat theory problem is but I remember it being similarly fucked for someone who’s literally just seen the definition

[dull ginkgo]

it's like the orbit basically

orbit of the module under that ideal

[glad osprey]

The proof I'm reading takes x in I, m in I, then
0 = (0 + I)m = (x + I)m = xm
which proves that IM = 0

I understand everything except the last equality

[dull ginkgo]

Also images and preimages of modules under homomorphisms aka module morphisms are modules still

consider the preimages of the modules of R/I

[glad osprey]

the homomorphism I'm thinking about is the projection pi : R -> R/I, but that's a ring homomorphism

[dull ginkgo]

It's also a quotient module

R/I for left ideal I is still a left module quotient

[glad osprey]

I'm not seeing the big picture, how does it relate to M?

[dull ginkgo]

I'll respond in a few mins

[dull ginkgo]

Here's another approach

Like how the bread and butter of group actions are orbits and stabilizers

annihilators are the bread and butter of modules

[dull ginkgo]

left modules over R/I implies we have a canonical left multiplication map of R/I into End(M), right. It's what makes it a module.

Thus we can compose the quotient map of R onto R/I onto End(M) to give it the structure of an R module, but what happens to the kernel (annihilator)? Show this kernel contains I

In general for preimages, (f \circ g)^-1(S) = g^-1(f^-1(S)) and the kernel literally is the preimage of 0

[vapid vale]

is there a general word for the terms A and C in a short exact sequence 0->A->B->C->0

[dull ginkgo]

not really, I call A the inner and C the outer terms

[sly crescent]

C is called the quotient object

[glad osprey]

Ah, I see my confusion now: I was thinking a ring hom phi : R -> S allows you to define an S-module structure from an R-module, but it's the other way around 🙃 So the projection pi : R -> R/I allows you to multiply I and M, just pass I through pi

[dull ginkgo]

silly consequence of this:

If we have a ring homomorphism of A to B and we have a B-module M then it admits an A-module structure from that through that composition

[errant wedge]

JDT as a first exercise wtf 💀

Oh wait

nvm

[dull ginkgo]

It's like 2 steps

if that

Also the R-module structure through the map R or R^op to End(M) is such a silly way to blast through intro ring theory problems

[errant wedge]

fair i just feel like itd be tough to see how to start if ur just starting noncomm alg

[dull ginkgo]

it's like a very slight variation of how to deal with group actions

instead of group G to Aut(S)

it's ring R to End(G) {abelian G}

[errant wedge]

fair enough yeah

[dull ginkgo]

more sillies

Let M be a left-R module

For any subset S of R, let Tor(S) = {x in M: Sx = 0}. Then if K is a right ideal of R, Tor(K) is a submodule of M

Tor and Ann form an antitone galois correspondence between the left submodules of M, and the right submodules of R (the right ideals)

this isn't really impressive since it's due to the heterogenous relation where rRx for r in R and x in M iff rx = 0

[dull ginkgo]

It's intro algebra

literally expected to be a first intro to rings

[errant wedge]

i misremembered how much of the proof of Wedderburn-Artin was JDT lol

[dull ginkgo]

JACOB SON DENSITY THEOREM MY BELOVED

[errant wedge]

JDT was not in my intro to groups/rings fwiw

[dull ginkgo]

oh no not the density theorem

[errant wedge]

Is that not what we're discussing wait

I might be stupid

[dull ginkgo]

though it is a fun exercise with simple modules and annihilators, imo could be with a small little guide

This was my first fiddling around with it

oop order is swapped

[dull ginkgo]

@errant wedge honestly if I ran a reading group id make this an intro guided exercise or maybe split it into parts

[rustic rapids]

How would you approach classification for these problems in general? I know i want to find a homomorphism from to the group to the symetric group of whatever order, but im not sure really what to actually do from there. Specifically, would i need to manually check every possible orbit size? Even then, im not too sure when i can exclude some said possible orbit

[lusty marlin]

You need to describe all homomorphisms from the given group to the symmetric group of the set being acted on.

[fading summit]

Which of the following sets are subrings of the field Q of rational numbers? Justify each answer.
a. { m/n | n is even }.
b. { m/n | 4 does not divide n }.
Also, assume gcd(n, m) = 1.

do i just have to check these three properties for each of them:

• if addition is closed
• if multiplication is closed
• if additive identity 0 is contained
  ?

the professor did not clearly explain the subring criteria...

[rocky cloak]

Yes, the criteria for subring is contains 0, and 1 and closed under addition, subtraction and multiplication

[fading summit]

do i have to check closure of addition and subtraction separately?

[rocky cloak]

It's enough to only check subtraction

[fading summit]

okay thanks

[crystal vale]

To show an integral domain with the property that every strictly decreasing chain of ideals I_1 \superset I_2 \superset .....must be finite in length is a field.

Let x≠0 in R and x is not a unit, then take a sequence of ideals (x) \superset (x^2) ...., it contradicts that it has finite length.

So x is a unit.

Is it correct?

[tough raven]

You should elaborate why the chain is strictly increasing, i.e., why (x^n) ≠ (x^{n+1}) for all n.

[crystal vale]

You mean strictly decreasing?

[crystal vale]

Let (x^n) = (x^(n+1) ) then we can write x^n = x^(n+1)s, now since ring is integral domain implies that xs = 1 which contradicts x is not an unit.

[crystal vale]

I don't want to memorize which ring is UFD but not ED, which ring is PID but not ED, or some counterexample.

Is there any way to understand this so I don't need to memorize

[vapid vale]

what do you want to understand then

[dim widget]

It’s hard to show that a ring is not a Euclidean domain if it’s a dedekind domain which is a PID

But for a general ring for instance a polynomial ring in more than one variable is a UFD but not Euclidean

[crystal vale]

Yes because if it is the Euclidean domain, say R[x_1,...,x_n], n>1 is the Euclidean domain then R[x_1,...,x_n-1] will be a field which is not possible, right?

[crystal vale]

I want to understand the structure say Z[ √d ], for some d it is ED and for some d it is not.

Is there any bigger picture?

[crystal vale]

Here I took a finite variable

[dim widget]

There isn’t a bigger picture

There are not many Euclidean domains, and there’s not much of a pattern for which d are Euclidean

For instance the ring of integers of Q(sqrt(-19)) is not Euclidean but it is a PID

[crystal vale]

Yes

So I need to go through their proofs?

Individually

[dim widget]

I don’t think it’s worth while unless you have a reason to

And I think for some of the imaginary quadratic fields which are Euclidean they are only Euclidean for a different function than the norm function, or a different algorithm than the obvious greedy algorithm

So it’s just a mess

[crystal vale]

Okay thank you

[serene dune]

hypothesis
group of order 60 and simple

im trying to find number of sylow-3 subgroups
after useing sylow theorem i get {1,4,10}

1 gets cancelled due to hypothesis, but** 4 gets cancelled tooo**

i need the argument for the bolded part

[rotund aurora]

With GRH you can prove that if a number field with infinitely many units is a PID then it is an ED. The euclidean function need not be the norm function, but you can define one that takes the same form for all such number fields (assuming GRH), but it's complicated. The only possible number fields excluded here are imaginary quadratic fields.

See "On Euclidean rings of algebraic integers" by Peter J. Weinberger

Also, Z[sqrt d] is not always integrally closed, you should look at the ring of integers of Q(sqrt d) instead.

[crystal vale]

i will see that

thank you

let G be a group and G' be the commutator subgroup, and N a normal cyclic subgroup of G. Prove that gn = ng for all g in G' and all n in N.

let N = < n >, then i have to show xyx^-1y^-1nyxy^-1x^-1 = n for all x in G and y in G.

because G' = < xyx^-1y^-1 >, any hint, i know N is normal so xyx^-1y^-1nyxy^-1x^-1 = n^i, but how can i show i =1?

[crystal vale]

What is GRH?

[rocky cloak]

I guess a hint for a slightly different approach would be that conjugation defines a homomorphism G -> Aut(N). Now think about what Aut(N) is

[rotund aurora]

Generalized Riemann hypothesis

[crystal vale]

I got it, Aut(N) is abelian because N is cyclic, thank you Jagr ❤️.

But what is the thought process for this question? Like how this idea comes to you?

[rocky cloak]

Well, anytime you have a normal subgroup it's natural to think about conjugation.

So you have this condition
gn = ng
which you can rewrite as
gng^-1 = n
so g fixes N under conjugation, means g is in the kernel of the map G -> Aut(N).

[crystal vale]

Yes

I am not sure but say a is n-cycle element in S_n, then any other n-cycle element is in < a > ?

[coral spindle]

Let n = 100

Wait no lmao

Much simpler

n = 5

a = (1 2 3 4 5)

What about the cycle (1 2 3 5 4)?

[crystal vale]

Yes I am thinking (1234)

[coral spindle]

But that's not an n-cycle so not sure what you mean

But yes you can also do this with n=4.

[crystal vale]

I mean 1234 in S_4

I know (12) and (123...n) generates S_n. Now I am thinking that any 2 cycle and n cycle can generate S_n?

[coral spindle]

Well any 2-cycle and (123...n) generates S_n, this is not hard to argue

and then you can conjugate to get the result you need

[crystal vale]

Okay

[crystal vale]

Yes

So any n - cycle is conjugate to (123..n)

Conjugation has the same cycle length structure

[crystal vale]

I don't get it how conjugation shows any 2- cycle and n-cycle generates.

Let a is n-cycle, yes a is conjugate to (123..n)

[coral spindle]

any 2-cycle and (123...n) generates S_n.

[crystal vale]

Yes

[coral spindle]

So choose any 2-cycle and n-cycle

[crystal vale]

Yes

[coral spindle]

this is then conjugate to some 2-cycle and (123..n)

QED.

[crystal vale]

Oh conjugation generates same group

[coral spindle]

In the finite group case, to be clear

[crystal vale]

Oh

[coral spindle]

For infinite groups it is different.

[crystal vale]

Can you tell me more?

[coral spindle]

Uh

No I was wrong

It is true for infinite groups too

I mixed up a couple ideas

[crystal vale]

I am wrong conjugation doesn't generate the same group maybe there are different thing

https://math.stackexchange.com/questions/2958978/let-sigma-in-s-n-be-an-n-cycle-and-let-tau-in-s-n-be-a-2-cycle-then?rq=1

[coral spindle]

Exercise: Let $g \in G$ and let $X \subseteq G$ be some subset. Prove $g\langle X \rangle g^{-1} = \langle gXg^{-1}\rangle$ and infer the result.

[cloud walrus]

$\mathbf{Boytjie}$

[coral spindle]

So to put it more directly, indeed you are wrong

[crystal vale]

it is not hard but on mse there is a post that (1234) and (14) does not generate S_4

[coral spindle]

Ha, drat. I did make a mistake.

I see what happened

Not every 2-cycle works with (123...n)

All 2-cycles work if n is prime

[crystal vale]

there is a question on mse that for which k, (1 k) and (123...n) generates S_n

[coral spindle]

Should be when gcd(k-1, n) = 1

Or no, just gcd(k, n) = 1

[crystal vale]

but why?

[coral spindle]

It's just a guess, idk

I'll look at this again later, I need to do other things

[crystal vale]

okay thank you

[coral spindle]

Nevermind, I thought about it quickly

The condition is this: the only k that does not work is k=n/2 in the case that n is even. All other cases work

The way to argue this is by looking at the size of the orbit (by looking at the stabiliser) of (1 k) under the conjugation action of <(1...n)>. Then you can choose an ordering and conjugate to get to the generating set (1 2), (2 3) etc

[crystal vale]

okay

[serene dune]

may i know, what result are we talking about ?

[serene dune]

tho you are correct in your thoughts, better term here would be cycle type, no ?

[crystal vale]

yes

how do i calculate the number of subgroups of (Z/pZ)^n, is it same as number of subspace of (Z/pZ)^n over Z/pZ?

[serene dune]

#groups-rings-fields message

[crystal vale]

thank you

[tardy hedge]

In the tensor product Z/nZ (x) Z/mZ, why is the element d(1(x)1) = 0?

Where d is gcd

How do we know that 1x1 has order d?

[south patrol]

Tbh this is a good exercise to try like to understand some properties of tensor product

[tardy hedge]

U right

Exam in 30mins tho 😜

Lol

[south patrol]

Oh lol

[tardy hedge]

😭

[south patrol]

Well the easiest way to show this (which is harder than what you originally said) is to use the fact like

M (x)_R R/I = M/IM

So here you get Z/(n,m)Z

And (n,m) is generated by the gcd

[crystal vale]

S_3 times Z/5Z has 3 elements of order 2 and D_30 has 15 elements of order 2, right?

[tough raven]

Write d = mx + ny for some x, y in ℤ.

[coral spindle]

When you see the GCD, the answer is probably Bézout. It's almost always Bézout. The GCD is easy with Bézout so use Bézout when you see the GCD it's Bézout it's always Bézout

Hope this helps

If I ever teach an elementary number theory course I will have every student tattoo the name Bézout in three different places on their body

[tough raven]

When the ℤ[X]

[serene dune]

haha involutions

[crystal vale]

let G be a group of order 56 having at least 7 elements of order 7. Prove that G has only one Sylow 2-subgroup P, and that all non-identity element of P has order 2.

first one is proved now i don't have idea how to show second one

[tough raven]

I think you should look at the (transitive!) action of P on the set of Sylow 7-subgroups by conjugation.

[serene dune]

exploiting the normality !

tho not completely same, may i get a hint on this

[crystal vale]

it is non trivial homomorphism from P to S_8

[languid trellis]

Hi all, I haven't been able to verify that r'=r^2-2 is also a solution, but given it is true, I intuit that Gal(E/Q) should be C_3. We have the identity map, which keeps the roots in place, the map taking r -> r^2-2 permutes the roots, as does r-> (r^2-2)^2 -2. But, tbh, I have no idea what to do and I would appreciate guidance on how to determine Gal(E/Q) 🙂

[lone niche]

Yeah, basically since r'=r^2-2, you know that all the roots are in E.

So the polynomial splits completely after adjoining one root.

[languid trellis]

Yes, E being normal over Q is clear

[lone niche]

Yeah and |Gal(E/Q)|=3

[languid trellis]

oh of course yeah

[lone niche]

That answers everything?

[languid trellis]

because E is a splitting field, |Gal(E/Q)| = [E:Q] = 3

[lone niche]

yeah

[languid trellis]

How do we get to Gal(E/Q) being the cyclic group of order 3 definitively? I feel as if I'm missing something obvious

[lone niche]

Hmmm there is only one group of order 3

is that what you are asking, or is it another thing?

[languid trellis]

god i am so silly

Thank you

[lone niche]

np

[tough raven]

Oh, there are 8 Sylow 7-subgroups. I don't know what I was thinking.

[tough raven]

Actually I think you look at the action of one of the Sylow 7-subgroups on P by conjugation (which is a group automorphism). Sorry for the mix-up.

[jovial prawn]

If (R,m) is a local ring is 1 + a for a in m always a unit? Doesn't this follow from (1+a)(1-a+a^2-a^3+...) = 1? I feel like the local ring condition is irrelevant here

[tough raven]

Well, what if (R, m) = (ℤ, 2ℤ)?

Your inverse for 1+a is a series, which has no reason to converge (indeed, convergence is not even defined) in an arbitrary ring.

It is true that if a is nilpotent, then 1+a is invertible (and m is now irrelevant) because that series becomes a finite sum.

[jovial prawn]

I see, so for a general local ring 1 + a need not be invertible unless we have nilpotence for a?

[glad osprey]

Let M be an R-module and N a submodule of M. If N and M/N is finitely generated then M is finitely generated. I proved this by writing an element of M as the sum of generators of N and M/N. Could you just claim that M = M/N + N, or does that not make sense?

[crystal vale]

No problem

[tough raven]

No, locality is the key property. If R is a ring and m a maximal ideal, 1+a need not be invertible for a in m, but if m is the unique maximal ideal of R, then it does have to be invertible.

[crystal vale]

But both structures are different, I think you mean in the sense of isomorphism

[glad osprey]

Yeah, M/N is not a submodule of M, so it doesn't quite make sense, but maybe it makes sense under some isomorphism

I think the idea I'm thinking about is that for a projection P in a vector space V, we have V = im(P) (+) ker(P), just the equivalent for modules with the projection M -> M/N

[tough raven]

No, what you are saying is simply false in general. If it is true one says that "the short exact sequence 0 → N → M → M/N → 0 is split".

[jovial prawn]

Could you elaborate on this, what does the uniqueness imply that makes this work? I don't see directly why uniqueness would force a to be nilpotent.

[tough raven]

It doesn't force a to be nilpotent, it forces 1+a to be invertible.

Specifically, a in m, 1 not in m implies 1+a is not in m. But that means 1+a is not in any maximal ideal (since m is the only one), which forces it to be invertible.

[glad osprey]

I see Does M/N and N being f.g imply that the SES splits?

[jovial prawn]

I see! If 1+a was in m, then so would 1+a-a = 1 be also and m = R?

[tough raven]

Nope. General rings are waaaay messier than that.

It does split if N is free though - you just pick arbitrary preimages of the elements of a basis of N (you could say it splits "purely due to set theory").

Splitting of short exact sequences (or dealing with non-exact ones) is pretty important in module theory IMO. For example, the definitions of projective, injective and semisimple modules are all about certain short exact sequences splitting, and they're all very important.

[rocky cloak]

I guess some things to notice is that this is the semidirect product of C7 and a group of order 8. The semidirect product must be nontrivial, otherwise there would be only 6 elements of order 7.

From there I guess you could just go through all the groups of order 8, but there's kind annoyingly many.

Something more clever:
||So the question is just whether there is an element of order 4 in the group.||

||Notice that a group of order 8 must contain at least 1 element of order 2 and at least 1 of order 1.||

||That means there can be at most 6 elements of order 4.||

||But C7 is supposed to act nontrivially on the group, hence nontrivially on the elements of order 4. But C7 can only act nontrivially on a set with at least 7 elements. Hence there are no elements of order 4||

[tough raven]

meanwhile me trying to classify all groups of order 8 ad-hoc and compute their automorphism groups to see if they had order divisible by 7

[glad osprey]

Looking forward to learning more about it we haven't covered it in this course, but maybe in a later course like homological algebra or something

[rocky cloak]

Something vaugely similar, M/N being finitely generated means you have a surjective map R^n -> M/N, now you can always (non-uniquely) lift this to a map R^n -> M. Then M will be the sum of N and the image of this map.

So this image is like your "approximation" of M/N inside M.

[tough raven]

Oh wait did you just want a sum and not a direct sum? Then everything I said was irrelevant. ☠️

[glad osprey]

Thanks anyways, I'm curious about both cases

[tough raven]

@rocky cloak do you know how the injective module I(i) = FreeVectorSpace(paths into i) at a vertex i of a quiver is defined when it's infinite/has cycles? Is it a direct product of k's over paths into i instead of the free vector space?

Because I'm not sure my argument about the simple module V(i) being an essential submodule remains true then.

[rocky cloak]

I think free vector space still works fine.

I don't think the product one will be indecomposable

[tough raven]

How do you show that it's injective then?

It seems difficult to do explicitly because the ideals of the path algebra might be messy, and the abstract construction as Hom_k(projective right module at i, k) gives the direct product.

As a check, for a loop (so k[X]-modules), V(i) = k[X]/(X), whose injective hull should be(?) ∪_n k[X]/(X^n) = k((X))/k[[X]]. Which is a free vector space on a countable set, so you're probably right.

Maybe we can replace the exact functor Hom_k(-, k) with (+)_{i vertex} Hom_k(- e_i, k)?

Actually, that would only help with infinitely many vertices.

[crystal vale]

Sorry C7?

[barren sierra]

C_n = cyclic group with n elements

[rocky cloak]

Hmm, actually I'm not sure.

But k[x, x^-]/(x) is an injective k[x] module, so it checks out in that case...

[tough raven]

Maybe show that the direct sum is a summand of the direct product?

[crystal vale]

Yeah

[tough raven]

I think I'm starting to intuitively believe it at least. For any path a: v → w, the corresponding map I(i)_v → I(i)_w is surjective (with section given on the basis by precomposing a), so I(i) is "divisible".

It would be interesting to show (if true) that such "divisibility" characterises injective representations of a quiver.

[rocky cloak]

So I'm thinking you have an injective map I(i) to M and you want to show that it splits.

So you just take a linear splitting of V(i) -> I(i) -> M and extend this to a homomorphism using this divisibility and some zorn maybe

[tawdry plover]

(d) uses (c) here I think but I want to avoid it

Is there a way to proceed from the fact that every ideal should be finitely generated and baer's criterion

Also tbh I find the wording of c confusing

[rocky cloak]

I mean, that is basically what they're doing in c.

The point is that the if you map from R to a direct sum the image will land in some finite direct sum.

But if you take an infinitely generated ideal an are able to create a map so that each generator maps to a different summand, then the image can't land in a finite sum. That's what they're doing in (c).

[tawdry plover]

Okay the let's not do c

How do I construct such maps

Oh nvm maybe to say an ideal is not finitely generated it's easier to do stuff by ascending chaims

That is I think the simplest such condition

[rocky cloak]

Well, say I is generated by x1, x2, ...
Then consider a map from I/(x1, .., xn) to an injective Qn.

Then show that the image of the product map I -> Prod_n Qn lands in Sum_n Qn

(this is again just what they're doing in c)

[tawdry plover]

But to say sth is not finitely generated doesn't mean it's generated by infinite elements

Finitely generated things can also be expressed as such

It's imo more like one can't find a finite generating set

So hence the ascending chain doesn't terminate

[rocky cloak]

Yeah I'm assuming xi+1 is not contained in (x1, ..., xi)

[tawdry plover]

Oh ok

Yeah then I guess this could work

But as you said itsjust c

I_n=(x_1,x_2,...,x_n)

Okay I guess that answers my queries

Thanks

[crystal vale]

Prove that every prime ideal of a Boolean ring is maximal.

I proved this, but if we remove that unity from the ring then how can I show that?

[bitter rover]

If R is your ring and P is your prime ideal, what does R/P look like?

[crystal vale]

Since R is not a unital ring so R/P is a domain

[bitter rover]

How'd you prove it for rings with unity?

(I'm now doubting myself after a quick sketch haha. The idea I had in mind tacitly assumes a unit, but I don't think it's necessary.)

[crystal vale]

Then R/P is integral with Boolean ring so it will be isomorphic to Z/2Z which is field so P is maximal

[tardy hedge]

There are 4 ideals of C[x]/(x^2-4) right?

[bitter rover]

It's still true. Say an ideal I strictly contains P. Look at an element contained I\P

And then multiply it it by something clever which shows that any element of R must be in I

(Using the fact that P is a prime ideal, so if the product of two things is in P then at least one is in P)

[crystal vale]

I thought this way, I let J be the strictly ideal contains p, then I take j which is not in P but in J, then I take r in R and thinking about to show r-j in P

[bitter rover]

Well, what about j(r-rj)?

[crystal vale]

Got it, thank you ❤️

[elfin wraith]

Could you reason why?

[tardy hedge]

By the lattice theorem there is a 1-1 correspondence of ideals of C[x] containing (x^2-4) and ideals of C[x]/(x^2-4). C[x] is a PID so all ideals are principal. any principal ideal of C[x] containing (x^2-4) means its generator must divide x^2-4, so the only options are (x-2) and (x+2)

[elfin wraith]

Nice

Also TIL the correspondence theorem is also known as the lattice theorem

[tardy hedge]

Yeahh ive heard it being called lattice theorem. I think they call it that in dummit and foote?

[tawdry plover]

How is the coinduced module hence obtained cohomologically trivial?

And the H_n are obtained like this...

[glad osprey]

Is the annihilator of an R-module M just the kernel of the action R -> End(M)?

[tawdry plover]

I mean how do you define this map from R to End(M)

It's by r mapsto a mapsto ra right ?

And what is the definition of the anhilator

It's precisely those r such that ra=0 forall a in A

Now can we make inclusions both sides

[tawdry plover]

Yeah I misread it's like when H is trivial then it's cohomologically trivial

[glad osprey]

yep, I see it now trivial when I use noggin

[toxic zephyr]

if H is a group is Aut(H) all bijective group homomorphisms from H to itself or just all permutations of elements of H? i get confused i feel like sometimes it depends on the context

[glad osprey]

It depends on the category AFAIK, Aut_A(H) is the group of automorphisms of H in the category A, so in Set it's permutations while in Grp it's group automorphisms

So if H is a group it's probably group automorphisms, for clarity you could write Aut_Grp(H)

[tawdry plover]

What's STfFGMOaPID

[rocky cloak]

I guess you can probably prove it by just writing out the definition of H^n, but with a little machinery:

H^n(G; M) = Ext^n_ZG(Z, M)

So take a free ZG resolution of Z and can it P*, and notice that this is also a free resolution of Z as an abelian group.

The coinduced module satisfies Hom_ZG(-, A^G) = Hom_ZH(-, A).

So Hom_ZG(P*, A^G) = Hom_Ab(P*, A)
taking homology yields Ext_Ab(Z, A) which is 0.

[toxic zephyr]

structure theorem for finitely generated module over a principal ideal domain lol

ya know, like CRT = chinese remainder theorem. trying to popularize STfFGMOaPID

[toxic zephyr]

okay yeah that explains why i thought it could be either one. i'm supposed to define a group action of Aut(H) on H, so i assume isomorphisms is what we're talking about

[rocky cloak]

And I guess the same argument gives
H^n(G; A^G) = H^n(H; A)
In general

[candid dove]

Oh I mean when H is trivial ZH is just H?

[rocky cloak]

ZH is just Z

[candid dove]

Yes Z

So basically their the resolutions should be uhh

I mean how bad can Hom_Z(ZG,A) be

It's just abelian group maps

Maybe uhh

[rocky cloak]

Yeah it's just the |G| fold direct product of copies of A

[candid dove]

Ah yea

So a resolution is just it

Right?

Like it's it free then

[rocky cloak]

ZG is a free abelian group yeah

[candid dove]

No like hom_Z(ZG,A)

I mean it's uhh

Very large basis but it should be free ig

Well idk

[rocky cloak]

I mean, A need not be free, and if G is infinite then it won't be free even if A is.

But it's not really relevant...

ZG is free, that's what matters

[candid dove]

Yeah okay sorry how does that lead to a resolution

[rocky cloak]

Like you take a ZG resolution of Z, P*.

Since ZG is a free abelian group, P* is also a Z-resolution of Z.

That's the resolution

[candid dove]

Ah okay yeah

It should also be exact at Z module morphisms

[rocky cloak]

Then if you apply Hom_Z(-, A) you get a complex that computes Ext(Z, A)

[candid dove]

Uhm but you know

I think there should be a simpler cochain

Like uhhh

One that has a lot of zeroes

[rocky cloak]

Simpler than what?

[candid dove]

Does this have a lot of 0's

[rocky cloak]

I'm not sure what you mean exactly.

Like Ext^n(Z, A) is 0 for all n>0.

But the resolution is probably big.

[candid dove]

Oh then how did u compute this ext

This makes me wonder if there's some LES stuff that can be done

[rocky cloak]

Well Ext(Z, A) is easy to compute, since Z is projective.

[candid dove]

Oh

Yeah right

Right

[rocky cloak]

And then the argument is that this equals
Ext_ZG(Z, A^G)

[candid dove]

Okay uhhh

[rocky cloak]

Showing that that must be 0 as well

[candid dove]

Wait

Oh cause of the isomorphism

[candid dove]

.

[rocky cloak]

Yeah
Hom_Z(ZG, A) = A^G
and
Hom_ZG(ZG, A^G) = A^G

They're the same

[candid dove]

Yeah right

You know

I feel like I see it

But i really need to write out the details

But tbh it wasn't even my question lol

This cohomology stuff is kind of weirdly cool

[tawdry plover]

Not exactly what youve said but I guess they do give some lead ontohowto proceed with these things that come later

[tawdry plover]

Oh I see

[tawdry plover]

Okay also there's sth called as shapiro's lemma

Which makes this deduction almost immediate

Yeah jagr they basically did what you said

Took a projective resolution of Z and pretty much whatever you said

Maybe by machinery you meant shapiro's lemma

---

Whats the adjoint associativity theorem hinted at in the proof tho

I wouldn't mind spoilers

[rustic rapids]

for b) I said that n must be geq 4 as n! must be geq 12. I then showed that it couldnt be n = 4 as D6 has an element of order 6, and S4 does not. I dont know where to go from there tho. Im p sure itsn=6. but idk how to show it, if thats even the case (i want to say only since its D6...)

[viscid pewter]

ok it's been a long time since i did group theory but iirc this is basically equivalent to finding the smallest n so that D_6 is isomorphic to a subgroup of S_n

so it's just like... i mean you can just sorta go for it

get generators that work out

i mean ok the order 6 idea is good

start with an element of order 6, use that as one generator, then find another one that works the way you want

oh i forgot to mention the key part, so there are elements of order 6 in S_5

so - big hint - try stuff in S_5

[white oxide]

can somebody help me see why |Y| <= n?

[rustic rapids]

hmm alright. so i want to find two elements of order 6 in S5 that have like the same orbits or something as the rotations and reflections of D6?

[viscid pewter]

yeah

like do you know the relations for D_n

it's just that

[rustic rapids]

I know its isomorphic to A4 where A4 is just even permuations of 4 elements?

[viscid pewter]

uhhhh

like, generators and relations

ok wait

[toxic zephyr]

Suppose that R is a ring with identity. Let I be an ideal in R. Prove that it is contained in a proper maximal ideal.
doesn't this just follow straight from Zorn's lemma? since R is an upper bound for any chain?

[viscid pewter]

have you seen this before

[viscid pewter]

i guess it's just like. for any r less or equal than m, r <= n. so m <= n

[rustic rapids]

i think i may have seen something like this for diheadreal group when i first learned it

[viscid pewter]

yeah so you just want to find r and s

[rustic rapids]

ooh i think i kinda get it. So in the context of like a dihedral group r and s are reflection and rotation (other way), but it would be fine to say its some more "abtract" operation that preserves those relations. So we want to find a subgroup of S5 that has these relations as well and then we can conclude they're isomorphic?

[viscid pewter]

yup

that's what generators and relations can do for you

different groups have different presentations, it's worth looking at the ones for the most common types

[rustic rapids]

I see. So these types of questions are really dependent on the groups themselves rather than some overarching "theorem" about group actions if that kinda makes sense?

[white oxide]

so we just take r = m then right

[viscid pewter]

i don't see why not

most of the argument is way way over my head

[white oxide]

lol ok

[viscid pewter]

i mean the overarching theorem about group actions in this context is just like

if you have a faithful action then it's basically like a subgroup of S_n

i think?

homomorphism etc.

[toxic zephyr]

and doesn't I need to be a proper ideal for this to be true at all?

[rustic rapids]

ah right

[slim kayak]

For zorns you'd want an upper bound that's also a proper ideal, R isnt. Also yes, I is meant to be proper

Try the usual trick for Zorns lemma when working with set inclusion

[toxic zephyr]

yeah realized i fucked up. i decided to use
P={all proper ideals containing I} (nonempty bc I in P)
for any chain {Ja}, i showed the union of all Ja is in P and is an upper bound
then P has a maximal element M and i showed M is a maximal ideal

that's the usual trick you mean right?

just use the union of all elts in the chain

[tardy hedge]

I dont think u need to do that

Why proper ideals containing I? What is I

Oh mb

I see

[tardy hedge]

That makes sense to me

[toxic zephyr]

thanks!

[toxic zephyr]

can someone sanity check me on this that GL2(Z2)=D3 (dihedral group with 6 elts)?

$$GL_2(\bZ_2)=
\bdef{
\m{1&0\0&1}, \m{0&1\1&0},\m{0&1\1&1}
\m{1&1\1&0},\m{1&1\0&1},\m{1&0\1&1}
}$$
(s=\m{0&1\1&0}) and (r=\m{1&1\1&0})
[
GL_2(\bZ_2)=\gen{r,s
: o(s)=2,o(r)=3,sr^{-1}=rs
}\cong D_3
]

[cloud walrus]

eigentaylor (STfFGMOaPID)

[toxic zephyr]

this is what i got

[tardy hedge]

Thats cool

[scarlet estuary]

if you dont know what youre talking about, please dont offer "help"

[chilly radish]

If |Y|>n |Y|>=n+1 and the argument would imply n+1<=n which can't happen

[tardy hedge]

Woah … senior mod appears …

[ashen heron]

disturbing right

[tough raven]

Is it true that any localisation of a commutative Artinian ring is actually a quotient?

[tardy hedge]

for REAAAL. FR

[tough raven]

Indeed, for any s, there exists n ≥ 0 and a such that s^n = a s^{n+1} ⇔ (1-as) s^n = 0. So when we invert s in a localisation, the inverse becomes a. Put in other words, adjoining an inverse to s is the same as adding the relation as = 1, i.e., specifically making a the inverse of s.

[crystal vale]

I think there are only two groups, {e} and Z/2Z

Got it

My mistake

[serene dune]

an element commutes with its inverse too

so u gotta see if all elements have self inverse or not ?

[crystal vale]

Yes

[crystal vale]

Yes then x^2 = 1 for all x

Then G is abelian

[serene dune]

and even ordered

[crystal vale]

So G has at most 2 element

But it is tricky

Since identity commute with every element

So G has only identity element

[serene dune]

yezz

thats why u cant expand any further

ig

[crystal vale]

Yes

[serene dune]

in Z/2 identity commutes with the other element too, so shouldnt it be only {e}

oh u stated that, my brain confirms then freezes, then goes back

[rocky cloak]

It can be nice to think about localization in two steps. First killing any element with sx = 0, so that s becomes a nonzero divisor. Then adjoining inverses to s.

But for an artinian ring any element is either a zero divisor or a unit, so we're done after the first step.

[crystal vale]

Yes so S contains only {e}

[toxic zephyr]

if G is abelian, not cyclic, and of order 63=3²x7, then can we say there is no element of order 9? since the only possible invariant factor decomp is
Z3 x Z21

can we extend this to if p and q are distinct primes then if |G|=p²xq and G is not cyclic but is abelian, then there is no element of order p²?

I'm not feeling confident that this is the only invariant factor decomp but I think it is. if so wouldn't this imply
Zp x Zp x Zq
is isomorphic to
Zp x Zpq ?
which I think makes sense bc isn't Zp x Zq cyclic if p,q are distinct primes?

and another question:
"Suppose that R is a ring with only one non-zero proper prime ideal. Prove that R/P is a field."
does R need to be commutative for this to be true or can we not assume that? if so, then doesn't this just reduce down to showing that if there's only one nonzero prime ideal then it's maximal? if R isn't commutative this seems much harder, as I'm less familiar with what R/P is defined to be without commutativity

[lone niche]

Yes

Actually yes to all that block of questions.

You can generalize the last one further to coprime m,n

[lone niche]

There's a couple of issues you need to be careful with in the noncommutative case. I am also not too familiar with that case, so I'll just list the ones that I spot: that the definition of prime ideal you are using makes sense. That maximal ideals are prime. That the quotient of a ring by a maximal ideal is a field.

For that last one you have this: https://math.stackexchange.com/questions/4763655/quotients-of-non-commutative-rings

[rocky cloak]

In assuming P is defined to be prime iff it's two-sided an R/P is a domain.

If so you can take something like the weyl algebra which is a simple domain.

Then you can just adjoin a central nilpotent element to make P nonzero.

So yeah, commutativity is needed

[elfin wraith]

Does anyone know of any “nice” proofs that x^ay^b forms a basis of the Weyl algebra? The proof I’ve seen in class uses induction on the number of words which are of the form yux for some word u in x and y, and I don’t really love this proof

[rocky cloak]

I guess you can frame it as contradiction instead of induction.

Like take some element and write it as a combination of words so that y appears before x as few times as possible.

Replacing yx by xy + 1, reduces the number so each word must be off the form x^a y^b.

Then you just have to show these are linearly independent I guess.

[elfin wraith]

Yeah linear independence is easy enough to show, but I guess that is much the same argument really, I just found arguing about the words to be a bit of a faff but maybe it’s somewhat unavoidable

[dull ginkgo]

I wonder if you could also try to create a multilinear form from the powers of the algebra

Iirc the weyl algebra’s kernel is (v(x) u -u (x) v - [u, v])?

Maybe a similar proof to how the Symmetric and Alternating Algebras are free for free modules?

@elfin wraith do you think we could work out something like this?

Working over a commutative ring R:

Let W_n(M) be the n-th power of the Weyl algebra over module M.

If we have an N-multilinear form f from modules A to B such that swapping 2 terms introduces a term of the form [x,y], then that multilinear map should factor through W_n(A). I wonder if you can construct a canonical one (like the determinant or symmetric maps for those two algebras) and show linearly independence of the basis to show freeness

[terse crystal]

I have two finite groups N, K whose orders are relatively prime to each other, and two homomorphisms $\phi_{1}, \phi_{2}: K \to \mathord{\mathrm{Aut}}(N)$. How to prove that two semidirect products $N \rtimes_{\phi_{1}} K$ and $N \rtimes_{\phi_{2}} K$ are isomorphic if and only if there exist $f \in \mathord{\mathrm{Aut}}(N), g \in \mathord{\mathrm{Aut}}(K)$ such that for any $k \in K, \phi_{2}(g(k))=f\phi_{1}(k)f^{-1}$?

[cloud walrus]

Cogwheels of the mind

[candid patrol]

Hello guys, is there a channel with good / Interesting exercices about groups theory on this channel / discord server ?

Latex works nice !

[tough raven]

You can wait for questions to be asked in this channel and see if you like them.

[candid patrol]

Oh... A channel with interesting exercises would be good :/

[elfin wraith]

You’d be better just finding a textbook for that, try dummit and Foote, people here would be happy to help if you get stuck

There’s also a lot of problems posted here and they can be really good to think about

[candid patrol]

Ok ok

[wicked patio]

Let $G$ be a finite group and $S$ a subset of $G$ such that $|S| > |G|/2$. Show that every $g \in G$ can be written as $s_1s_2$, for some $s_1,s_2\in S$

[cloud walrus]

Dreyuk

[median fox]

Nice set of lectures on introductory Group Theory by Prof. Lothar Gottsche, ICTP.

2013-14, 20 lectures, https://www.youtube.com/playlist?list=PLLq_gUfXAnknLXjNSnKKLT4LI1AfTy9PS

2020-21, 15 lectures, https://www.youtube.com/playlist?list=PLp0hSY2uBeP9yvNkjIA_iXGy6W_OaPs-9

Prof. Lothar Gottsche, home page, https://users.ictp.it/~gottsche/
Scroll down. Lecture notes for Algebra, Algebraic Topology, and Algebraic Geometry courses.

[tough raven]

A neat fact about radicals of modules (the sum of all superfluous submodules (really a directed union, since the sum of superfluous submodules is superfluous), classically equal to the intersection of all maximal submodules): if M is generated by {m_i : i in I}, then n is in the radical of M iff <m_i + r_i n : i in I> = M for all choices of (r_i)_{i in I}. Indeed, for any submodule N, N + <n> = M iff it contains all m_i iff m_i + r_i n in N for some r_i iff <m_i + r_i n : i in I> ⊆ N for some r_i, and this forces N = M iff <m_i + r_i n : i in I> = M for every (r_i)_{i in I}.

This also allows a purely constructive proof that the radical of a finitely generated module is superfluous: if Rad M + N, then Rad(M/N) = M/N (fact: module homomorphisms map radicals inside radicals: for any m in Rad(M) with image \bar{m}, and submodule X/N ⊆ M/N, X/N + <\bar{m}> = M/N ⇔ X + <m> + N = X + <m> = M ⇒ X = M ⇔ X/N = M/N, so <\bar{m}> is superfluous). But M/N is finitely generated, and a finite sum of superfluous submodules is superfluous, so M/N is a superfluous submodule of itself, which forces it to be 0 (since M/N + 0 = M/N ⇒ 0 = M/N).

[tough raven]

Also here's a somewhat algorithmic proof that an Artinian module M with zero radical is semisimple: given N ⊆ M, inductively define N_0 := N and X_0 = M; and X_{i+1} ⊆ X_i such that N_i + X_{i+1} = X_i and X_{i+1} ⊊ X_i unless N_i = 0 (this is possible because Rad(X_{i+1}) ⊆ Rad(M) = 0), and N_{i+1} := N_i ∩ X_{i+1}. We have N_{i+1} ⊆ N_i and X_{i+1} ⊆ X_i with equality for both iff N_i = 0. But by the Artinian property, this has to happen eventually, say N_n = 0. But then N_0 ∩ X_n = N_0 ∩ (X_1 ∩ ... ∩ X_n) = N_1 ∩ X_2 ∩ ... ∩ X_n = ... = N_n = 0 and N_0 + X_n = (N_0 + ... + N_{n-1}) + X_n = N_0 + ... + N_{n-2} + X_{n-1} = ... = X_0 = M, so M = N (+) X_n.

Actually, IG the last two proofs are valid in any abelian category (although defining the radical does at least require the subobject lattice to have directed joins, we can formulate "radical = 0" as "all superfluous subobjects are 0" and "radical [resp., + N] = M" as "any suboject including all superfluous subobjects [resp., and N] includes M").

[rocky cloak]

You can do intersection and sums of subobjects simply through pullbacks and pushouts, so works in any abelian category.

[elfin wraith]

“Let p be prime. Show that the group algebra F_pC_p is not semi simple”

Any ideas how to go about this? I know the converse of Maschkes theorem gives this immediately but I don’t have that available to me

My first thought was F_pC_p \cong F[g]/(g^p-1) since a factor of the polynomial ring is probably nicer to work with than a group algebra but I’m still kinda unsure where to go after that

[rocky cloak]

Do you know what the semi-simple rings are?

You can notice that FpCp is commutative, and there aren't that many commutative semi-simple rings, so you just need to find one property they have that FpCp doesn't

Another approach is that a ring is semi-simple if every short exact sequence splits.

So you could investigate some simple maps like the projection FpCp -> Fp

[elfin wraith]

I don’t think I do, this was a bit of a rouge workshop and doesn’t seem to be too similar to a lot of the other stuff we covered

Would classifying all commutative semi simple rings be relatively straightforward or is this more of a “nice if you know it” result?

[elfin wraith]

I didn’t actually know this but that could be a nice way to do it, I’ll maybe see if I can think of something this way.

Definitely not the intended method for the course but it seems like it could be a nice proof

[rocky cloak]

I mean it's not that hard, but probably harder than showing that FpCp isn't.

Artin-Wedderburn is the name of the theorem classifying semi-simple rings

[elfin wraith]

Oh in that case sorry yes I definitely know artin wedderburn

Not sure what I was thinking there, yeah if it was semi simple we can write it as a finite direct sum of simple matrix rings

[rocky cloak]

Are you sure it's not intended. The fact that submodules of semi-simple modules are direct summands seems like the kind of thing one should know if one knows what semi-simple rings are

[elfin wraith]

Yeah it’s definitely not, there’s no expectation of knowledge of SES or anything like that, my unis algebra track is… questionable at times

If it wasn’t for my UG project I wouldn’t cover exact sequences anywhere, and if it wasn’t for this topics course I wouldn’t know anything about modules beyond the definition so yeah

[elfin wraith]

Hold on I think I’m just being dumb, F_pC_p is a P dimensional F_p vector space, so if it were semi simple, by artin wedderburn every matrix is of dimension one, so F_pC_p \cong F_p^p but then this clearly isn’t the case

[rocky cloak]

Well it will have to be a product of fields.

That can happen in a few ways. (F_p)^p, F_p^p,
F_p^n1 x F_p^n2 x ....
with
n1 + n2 + ... = p
In general

But if you can use something about these, like how many units/idempotents/nilpotents/zero-divisors or something like that.

Then you'd have a neat argument.

[elfin wraith]

Thank you jagr!

[candid patrol]

wanna try smthg ?

[rocky cloak]

What kind of challenge

[candid patrol]

Determine all the subgroups of the additive group Z^2

[rocky cloak]

So you have (0) and all the cyclic subgroups which are just given by an element of Z^2 up to ±.

The remaining ones are the subgroups isomorphic to Z^2.

If you think of it as the row space of an invertible matrix, you can row reduce it to
[p, q; 0, r]
with p>0, and r>q>=0

[tough raven]

Yes, but infinite ones might not exist, right?

[rocky cloak]

Yeah, but if you have infinite direct sums you're good

[tough raven]

This is correct. You might want to investigate when R/(r) is semisimple for R a PID (and r non-zero), then apply that to R = F_p[X].

[elfin wraith]

Yeah I’ll for sure have a think about that, my initial thought is that in a PID prime <=> maximal, so if r is prime we have a simple ring, so that’s a little too strong

Obvious candidate is then to look at semiprime ideals and I’ll need to actually work it out but I think this would work. Semiprime ideals in a PID are generated by elements with only 1 prime factor so I guess you’d get some repeated sum of simple modules from that

Yeah actually something with like the CRT should show that

[tough raven]

Try ℤ/nℤ.

[elfin wraith]

Hmm yeah maybe what I’ve said doesn’t work then, just thinking of abelian groups C_9 isn’t isomorphic to C_3 (+) C_3

I’ll actually sit down with some paper in a minute and try to have a coherent thought

[glad osprey]

To show that a ring R with no proper left ideals is a division ring, can you just use R iso to End_R(R) which is a division ring by Schur's lemma?

[rocky cloak]

That works yeah

Though I guess it should be R iso to End_R(R)^op, since you're talking about left ideals

Not that it makes a big difference

[zinc fiber]

sorry if this is the wrong channel, does anybody know what paper they are talking about here https://youtu.be/AzHWp2BV0C8?feature=shared&t=2059

34:19

[elfin wraith]

Ok yeah turned my brain back on, semi prime isn’t enough, you also need Artinian

So R/(r) is semi simple when R is an artinian PID, and (r) is semiprime (because then R/(r) is reduced so J(R/(r)) = 0)

[tough raven]

Um, R/(r) is always Artinian for R a PID and r non-zero.

Really, just do everything as explicitly as possible for ℤ/nℤ. You don't need module or ideal theoretic words to describe the answer.

[elfin wraith]

Ah yeah so it is

[elfin wraith]

If $n$ has prime factorisation $n =p_1^{\alpha_1}\cdots p_n^{\alpha_n}$ then $\mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}/p_1^{\alpha_1} \oplus\cdots\oplus \mathbb{Z}/p_n^{\alpha_n}\mathbb{Z}$ by the Chinese remainder theorem

so Z/nZ is semi simple whenever all the $\alpha_i = 1$?

[cloud walrus]

Nope

[hot pebble]

A simple (?) question that i just can't wrap my head around:\\

1. Show that $x_1x_4-x_2x_3$ is irreducible in $\mathbb{Z}[x_1,x_2,x_3,x_4]$\
2. Show that $x_2^2-4x_1x_3$ is irreducible in $\mathbb{Z}[x_1,x_2,x_3,x_4]$\\
   I see that the first one is a linear polynomial in x4 and i guess must be irreducible, but for some reason i just can't prove it whne theres all these x1 x2 x3 going on. Any help would be appreciated!

[cloud walrus]

MartinFTW

[rustic crown]

look at the ring R = Z[x1, x2, x3] all you need to know about this ring is that it's a UFD and x1, x2, x3 are "distinct" (more precisely non-associate) irreducible elements

now you have the polynomial ax - bc in the ring R[x]

with a, b, c distinct irreducibles in R.

and we want to say this polynomial is irreducible

[hot pebble]

right and that would be because a,b,c are irreducible already so any factorization must be into non-constant polynomials, which we can't by deg=1

[rustic crown]

yep :3 the important thing here is that gcd(a, bc) = 1

[hot pebble]

what does gcd mean in a general UFD like Z[x1, x2, x3]?

just factorize it and see all the maximum irreducible powers (like how we do for prime factorization in Z)?

[rustic crown]

yep precisely!

[rustic crown]

if you write ax-bc = f * g, then at least one of f or g must have degree 0

but that would force that a, bc have a common factor

which must then be 1 as they are coprime

[hot pebble]

right cuz a,b,c themselves are irreducible

so "bc" is already the factorization for bc

[rustic crown]

yee

[hot pebble]

ok!

[rustic crown]

do you know about gauss' lemma?

[hot pebble]

what about that second one? if the same logic applies i'd try doing Z[x1, x3, x4][x2]

yes i do

the primitive one and the irreducible one

[rocky cloak]

Could also use Eisenstein for both of these

[rustic crown]

it says that if you have a ufd R, then a polynomial in R[x] is irreducible iff its irreducible in Frac(R)[x] and gcd of coeff = 1 (aka primitive)

[hot pebble]

though i'm not sure how field of fractions of Z[x1,x2,x3] helps? isn't it the set of "rational functions"

[hot pebble]

what prime ideal are we taking? i thought about eisenstein for the second question

[rocky cloak]

Like (x1) for example

[rustic crown]

yee, we dont' care so much about the field, just that it's a field and any linear polynomial over a field is irreducible

[hot pebble]

ohhh a field

[rocky cloak]

I guess it gets tricky for the first one, but for the second it should work

[hot pebble]

right field of fraction is a field

[hot pebble]

considering the prime ideal (x1) in the ring Z[x1, x3, x4][x2]? does the 0(x2)^1 interfere with it?

well i guess 0 is in any ideal

[elfin wraith]

Ok so in general, for R a PID, R/(r) is semisimple iff every term in the prime factorisation of r occurs without multiplicity

So back to the original problem with F_pC_p, we look at the factor of the polynomial ring, which is a PID, so we need to find the prime factorisation of g^p-1. That’s going to be (g-1)(1+\cdots+g^p-1) no? And each of these has multiplicity 1?

And I know it’s definitely not semisimple by Maschkes theorem

[rocky cloak]

You can factor this quite a bit further

Think about ||freshmen's dream||

[elfin wraith]

Ah yes because char p

So (g^p-1) = (g-1)^p, multiplicity p, done

That was a struggle but I think a productive one, thank you both for your patience

[tough raven]

Indeed. And conversely, over a field of characteristic not p, we have (X^p - 1, (X^p - 1)') = (X^p - 1, p X^{p-1}) ∋ (X/p)(p X^{p-1}) - (X^p - 1) = 1. So X^p - 1 is coprime to its gcd, which means it cannot be divisible by any square, so F[X]/(X^p - 1) is semisimple.

[elfin wraith]

Ahh, I guess that’s sort of how the converse to Maschkes theorem goes

We only stated the special case of it over C in class, so I could be way off, all my knowledge of it just came from a brief skim of Wikipedia after wondering why it would only hold over C

[tough raven]

It holds over any field of characteristic not dividing the order of the group.

[hot pebble]

I have to simplify the ideal $(x^2+1,3+5x)$ into $(m, x+n)$ where $m,n\in\mathbb{Z}$. How should I go about this?

[cloud walrus]

MartinFTW

[hot pebble]

i guess m=3^2+5^2=34, but apparently its "doable" by only doing euclidean algorithms and finding some linear combinations, i've been trying to make that form for like an hour now

oh i read the question wrong, it might not be possible

[rocky cloak]

So if you just start manipulating like multiplying the first by 5 and the second by x and so on you'll quickly find a number m contained in the ideal.

Then you can look at the image of the ideal in Z[x]/(m) which you can decompose with Chinese remainder theorem.

So at least if m is square free this just reduces to polynomial division over a field

[south rain]

I'm given the ring $R = \mathbb{Q}[x]$ with zero constant term, i.e. no identity. Furthermore, for $H \leq (\mathbb{Q}, +)$ where $H \subsetneq \mathbb{Q}$, I have the ideal $I(H) = Hx + Rx$. Now I should show that $I(H)$ is not contained in any maximal ideal $M$. I have the feeling I should proceed by contradiction and having $M = R$ if $I(H)$ is contained in $M$, but I can't see how to start the argument. I also know that if $M$ is maximal $R/M$ is a field, but I don't see how that helps me right now.

[cloud walrus]

dellinger

[rocky cloak]

I don't think contradiction is the best way to go. Try instead to think about which ideals contain I(H).

Also R is not unital, so you shouldn't expect R/M to be a field

It might be convenient to think about H=0.

I.e. what are the ideals of R/I(0)?

[south rain]

Ah, of course I need have identity for it to be a field.

[south rain]

I'm note sure I'm following, but $R/Rx$ are just the linear terms and zero, right?

[cloud walrus]

dellinger

[rocky cloak]

They will give you representatives yeah

[south rain]

Do I just have the trivial ideals?

i.e. zero and the entire ring

I can't think of any other subring, since they would all be a subset of the form ${a x + Rx }$ but upon multiplying it with any element in $R/I(0)$ I would always end up with zero.

[cloud walrus]

dellinger

[south rain]

Oh, but 0 must be necessarily in the subring of R/I, so any such subset is always an ideal or maybe I'm just stupid.

[rocky cloak]

Well it would have to be closed under addition, but otherwise yes

[south rain]

Isn't that the definition of a subring?

But I don't quite see how the quotient ring helps with determining how ideals "look" when $I(H)$ is contained.

[cloud walrus]

dellinger

[rocky cloak]

You know that the ideals of a quotient ring R/I correspond to the ideals of R containing I?

[south rain]

Oh, no I wasn't aware of that. Tbf, we just got introduced to ring theory, after group theory and didn't really derive many statements. Does this fall under one of the ismorphism theorems?

[rocky cloak]

It's sometimes called the 4th isomorphism theorem. Also goes by correspondence theorem and lattice theorem

[south rain]

So every ideal $T$ of $R/I$ is of the form $J/I$ where $I \subset J$?

[cloud walrus]

dellinger

[south rain]

If $I$ is an ideal of $R$.

[cloud walrus]

dellinger

[south rain]

Well $I \subset J$ kinda makes sense now that I think about it, it can't be a subset of $I$ since we modded out by it.

[cloud walrus]

dellinger

[south rain]

So basically, $R/I(H)$ for any $H \subsetneq \mathbb{Q}$ gives the the set ${0 + Rx, qx + Rx}$ where now $q \notin H$.

[cloud walrus]

dellinger

[tardy hedge]

How is that a ring if it doesnt have an additive identity? Its not a group under addition?

[south rain]

The additive identity is 0, this is in $\mathbb{Q}[x]$, right?

[cloud walrus]

dellinger

[south rain]

By "no identity" I meant $(\mathbb{Q}, \cdot)$ is only a semigroup and not a monoid.

[cloud walrus]

dellinger

[south rain]

So no multiplicative identity

[fallen geyser]

For a finite group, when do you have G = G/N + N?

[acoustic rose]

Could I have some help with (1)?

Not really sure how to approach it. I know I need to show that $\phi(x_j)$ is a root of $f(x)$ for any $x_j \in X$, but not sure how to do that.

[cloud walrus]

Ari

[coral steeple]

Are we missing Z8 + Z90?

[rocky cloak]

That's isomorphic to Z/2 x Z/360

[coral spindle]

(and it's not in the standard form since 8 does not divide 90)

[coral steeple]

Oh wait yeah of course

[rocky cloak]

Seem to be missing
Z2 x Z6 x Z60 though

[coral steeple]

I cut the image off there because my confusion was only with Z8+Z90

[broken pollen]

⊗ here is the tensor product of abelian groups

i might just be silly, but why is this ψ injective?

[broken pollen]

i suppose i'm struggling with going back from the equivalence classes to the group G

why is that function well-defined?

if n ⊗ g = m ⊗ h then we have 1 ⊗ ng = 1 ⊗ mh and i really want to say ng = mh here

any possibility of decategorising that and working with the pure construction? 👉 👈

[peak root]

I may be silly here. But I am trying to think about the split group extension of $C_3$ by $C_3\times C_3$. We need a group homomorphism $C_3\times C_3\rightarrow Aut(C_3)$. Here $Aut(C_3)=C_2$, so we only have a trivial homomorphism. And the entension can only be the direct product?

This is not right because I am proving the Heisenberg group over $\mathbb{F}_3$ is isomorphic to $C_3\rtimes (C_3\times C_3)$.

[cloud walrus]

Dong_Valentino

Dong_Valentino

[peak root]

$C_3\times C_3$ does not have order 2 elements, where can I send (1,0) and (0,1)?

[cloud walrus]

Dong_Valentino

[coral spindle]

Everything you say is right, except the very first thing

It's not $C_3\rtimes (C_3\times C_3)$

It's $(C_3 \times C_3) \rtimes C_3$

[cloud walrus]

$\mathbf{Boytjie}$

[coral spindle]

And indeed you can check there is an automorphism of C_3 x C_3 which is of order 3

[peak root]

thanks, I will check that out.

[languid trellis]

Semidirect product is super neato

[bitter rover]

the sequences must split!!

[coral steeple]

Why is PN/N a subgroup of Q? Why do we know anything about how P and Q relate?

(I am asking about the converse implication, one line before the last)

I really don't know where PN/N comes from. It reminds me of the second isomorphism theorem, but I don't think that's being used here.

[rocky cloak]

PN < H, so PN/N is a subgroup of H/N

[coral steeple]

Ah, so it is.

I think that answers my second question too. Thanks!

[south rain]

Alright, so I've contemplated the hint I got yesterday and reduced my problem so far (first image is the problem) , second image is how far I've got. What I'm trying to show is that $Ux + I(H)$ cannot be maximal. I know that the rng $(\mathbb{Q}, +)$ without unity has no maximal ideals. Can i somehow use that proof for showing essentially something similar here?

[cloud walrus]

dellinger

[languid trellis]

Hi all. I've done the first part of the exercise. and I've gotten that $\text{Gal}(E/\mathbb{Q}) \cong C_5 \rtimes C_4$. As $\text{gcd}(5,4)=1$, and considering the map $\phi: C_5 \rightarrow \text{Aut}(C_5)$, we get that phi only gives rise to the trivial automorphism, hence the semi-direct product is the direct product in this case. This means that we should have normal subgroups identifiable with $C_5$ and $C_4$ and hence normal intermediate fields. The intermediate field correspoding to $C_5$ is (i think) $\mathbb{Q}(\zeta)$ where $\zeta$ is a fifth root of unity. Can anyone offer any advice as to finding the intermediate field corresponding to $C_4$?

[cloud walrus]

swifteeee

[rocky cloak]

So the Galois group is the semidirect product of C5 and C4 through the isomorphism C4 = Aut(C5). There's no normal subgroup of order 4

But for example Q(fifthroot(2)) is a field corresponding to one of the subgroups of order 4 in the Galois group.

[languid trellis]

I've made a big table of automorphisms of E/Q, and I think I've found it. I don't see how the semi-direct product isn't the direct product in this case though, is the image of phi: C_5 -> C_4 not trivial?

[rocky cloak]

There's no map C5 -> C4.

The relevant data to the semidirect product is a map
C4 -> Aut(C5)
which in this case is an isomorphism.

[languid trellis]

Oh I got my semidirect product the wrong way round

I see now

Thank you jagr

[tawdry plover]

Why do endomorphisms preserve isotypic components

[tough raven]

Let f be an endomorphism and W be a simple submodule isomorphic to a given simple module V. Then f restricted to W is either 0 or it is injective, so f(W) is either 0 or also isomorphic to V. Either way, it is included in the V-isotypic component. Since the V-isotypic component is the sum of all simple submodules isomorphic to V and W was arbitrary, this shows that the V-isotypic component is mapped into itself by f.

[tawdry plover]

I see

Thanks

[coral steeple]

Why does the congruence [G:H]=[G/N:Q] \not\equiv 0 \pmod{p} matter? Is that just justifying that p must divide |G|?

[rocky cloak]

It's that |G|/|H| is not a multiple of p, so a sylow subgroup of H is also a sylow subgroup of G.

[coral steeple]

Right, so if |H|=p^km, p does not divide m, then |G|=p^kn (p \nmid n)

Thanks!

[coral steeple]

Also why is P/(P \cap N) a Sylow p-subgroup? Sure, it is a p-group, but why must it be maximal?

[rocky cloak]

Well it has the correct order

[warm dove]

How can I prove that any divisor of a zero divisor is either a unit or a zero divisor?

(just crashed out on my rings test)

[hot wadi]

Well

Let’s say a is a divisor of b, which is a zero-divisor

Let’s unpack each definition

a•c=b for some c, and b•d=0 for some d other than 0

Substituting gives a•c•d=0

If a is not a zero divisor then c•d=0

d is not 0, so c is a zero-divisor

Hm

Yeah no the statement just isn’t true

Here is a counter example:

In ZxZ, (2,2) is not a 0 divisor or a unit

But (2,2) is a divisor of (2,0), which is a 0 divisor

[warm dove]

Motherfucker the teacher lied

And it wasn’t a prove or disprove

[hot wadi]

Rip

[warm dove]

Wait this is great

I’ll get points back

[hot wadi]

Is there any chance you misread the question?

[warm dove]

No

Nobody in my class saw finite

[hot wadi]

Ohhh finite is more interesting

[regal zodiac]

Can someone tell me what is wrong with this ?
If i have a injective map f from A -> B and another module C
Why is the tensor map from A tensor C to B tensor C no necessarily injective ?

[slim kayak]

Is there anything wrong? Tensoring takes the exact sequence 0 A B C 0 to an exact sequence TA TB TC 0, so if A is zero then the tensored sequence still has an isomorphism

Tensoring is still a functor, it is forced to send isomorphism to isomorphism just from the base rules. Try to see where your example breaks down if you try to apply it to non-surjective injective maps

You will generally just factor it as A -> f(A) -> B with the right map an inclusion. You then end up just having to ask if the inclusion remains injective after tensoring

[wraith cargo]

the issue is that ker(f) (x) A != ker(f (x) A)
And vice versa for the image

[tardy hedge]

i miss algebra

i need to study topology now

Algebra best math

[elfin wraith]

Wait until you hear about algebraic topology

[tardy hedge]

thats what im tryna study 😭

I find it very hard

[elfin wraith]

Are you learning from Hatcher?

[tardy hedge]

this was mostly a point-set topology class but the prof randomly threw in the intro stuff to algebraic topology at the end

up to fundamental group of S1

we were following Janich book (which I hated) so I read Munkres the whole way through

and im reading Munkres discussions about homotopy now

[fossil pike]

Damn , even I don't understand abstract algebra at all

Now it feels scary

[cobalt heath]

I heard munkres discussion of homotopy is quite mid

You may try Hatcher instead.

[mighty kiln]

Or tom Dieck

[lime junco]

bit confused

why is this f linear?

oh wait i think im stupid and im misunderstanding

in this context, all that means is its a linear map on the vector space structure over F

nvm disregard

i confused it with F-automorphisms lmao

[toxic zephyr]

i was able to prove the operation
a*b=a+b+ab
forms a group on G=R\{-1}
does this generalize at all? can we define a group operation on R\{m}
for m any specified real number?

[toxic zephyr]

does
a*b=(a-m)(b-m)+m
e=m+1
a^-1=m+1/(a-m)
work?

[mighty kiln]

Yes

Exercise: any bijection S → G where S is a set and G is a group induces a group structure on S

[languid trellis]

For this exercise, I believe that $G \cong Z/pZ$ (if we denote the automorphism t-> t+1 by f, then it is clear that G = <f>, and the order of f is p).

[cloud walrus]

swifteeee

[coral spindle]

Looks right to me

Btw you can write \mapsto and use \langle, \rangle

[languid trellis]

Now to determine $F = \text{Inv}(G)$, I have so far, just by experimentation, seen that $Z/pZ$ (identified as a subfield of $(Z/(p))(t)$ is fixed, as are the polynomials $1,:t^p-t, : t^{p^2} - t^p, :t^{p^3} - t^p^2, \cdots$, their inverses and $Z/pZ$-linear combinations of them. However, I feel like I am missing something, because by artin's lemma $[E:F] \leq |G|$, so $E$ is a finite dimensional $F$-vector space. Can someone confirm my doubts and give me a hint on how to "systematically" determine $\text{Inv}(G)$?

[cloud walrus]

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

[languid trellis]

I bring up artin's lemma because it is making me feel that F should be way bigger than I am giving it credit for

[coral spindle]

Recall that in a field of characteristic $p$, the Frobenius map $x \mapsto x^p$ is an endomorphism. So in particular, where is $t^p - t$ sent by the Frobenius map?

[cloud walrus]

$\mathbf{Boytjie}$

[languid trellis]

t^p^2 - t^p

[coral spindle]

So you only need to list $t^p - t$, right? That generates the rest of the things you listed

[cloud walrus]

$\mathbf{Boytjie}$

[languid trellis]

Yes, but the point is that I think I'm missing something still

F needs to be infinite dimensional over Z/pZ

[coral spindle]

This should let you see what the degree of the extension $\mathbb F_p(t) : \mathbb F_p(t^p - t)$ is

Oh screw you texit

[cloud walrus]

$\mathbf{Boytjie}$

[coral spindle]

Hmm does it? You've stated that [E:F] <= |G| but this doesn't seem to say F = Z/pZ

[languid trellis]

[E:Z/pZ] = infty + dimensionality formula

I'm not saying that F = Z/pZ

far from it

[coral spindle]

Oh I thought you said finite dimensional

[coral spindle]

Well ok yes... but what's the issue? t^p - t is still transcendental over Z/pZ so...

[coral spindle]

But in any case, this should tell you what [E:F] is.

[languid trellis]

There is probably something obvious I'm missing here, because I'm struggling to relate it to the frob endo, but one attempt in my brian is to try construct a of F_p(t). 1 and t^p-t will be in this basis. I have a suspicion that 1,t,t^2,.... t^p-1 is a basis. As then

(t^p-t)*1 + t = t^p
t((t^p-t)*1 +t) = t^{p+1}

And so on. So this is certainly a generating set.

[coral spindle]

The frobenius endomorphism was just allowing us to see that $\mathbb F_p(t^p-t, t^{p^2} - t^p, \dots) = \mathbb F_p(t^p - t)$. Working out what $[E:F]$ is is a different matter.

Hint: find a polynomial in $F$ for which $t$ is a solution.

[cloud walrus]

$\mathbf{Boytjie}$

[languid trellis]

a polynomial in F?

Not F_p(t^p-t)?

[coral spindle]

F = F_p(t^p - t).

[languid trellis]

Okie

[coral spindle]

The notation is unfortunate I guess

[languid trellis]

Haha

(t^p-t) - x^p+x = 0

oh my god i will have a fit

everything is this polynomial

So $\mathbb{F}_p(t^p-t)[x]/(x^p +x - t^p-t) \cong \mathbb{F}_p(t)$

[cloud walrus]

swifteeee

[languid trellis]

And we know that F_p(t^p-t) is contained in Inv(G), and it has "maximum degree" (artins lemma) so it should be Inv(G)

Need to be more formal about the "maximum degree" idea

Maybe formulate it as a lemma:

Let E be a field, G a group of automorphisms of E. If F \subset Inv(G) is such that [E:F] = G, then F = Inv(G)

Wait this has already been shown

[languid trellis]

I got some signs wrong

[dull ginkgo]

Is Inv(G) your notation for the fixed field

[languid trellis]

Yes

[dull ginkgo]

This is a corollary of Artin’s Lemma

[languid trellis]

I realised

[dull ginkgo]

Artin’s lemma my beloved

[coral spindle]

You got it <3

So everything lines up

[dull ginkgo]

pretty sure you can do like, a LOT of Galois theory just from Artin’s Lemma as a base point

Since the idea of being a “Galois Extension” is basically forcing a correspondence, it’s just some mapping details and equivalences you can work out

[languid trellis]

Thanks for the help!

[coral spindle]

I think you did the heavy lifting there swift

[languid trellis]

I haven't had this much fun doing maths in a while. This subject is genuinely magic

I am becoming an algebraist

[dull ginkgo]

Iirc if you have |G| = N, then [F : F^G] = |G| and the basis is the orbit of some element under the automorphism group

[coral spindle]

Yeah Galois theory is truly lovely. I remember thinking "oh, THIS is why group theory exists"

Like it's such a good motivation

[languid trellis]

Literally yes

[dull ginkgo]

I wonder a lot about the generated Subfields of different orbits under the automorphism group, since they’re always linearly independent by Dedekind’s theorem

[languid trellis]

I'm also fascinated by how much you can learn about x object by looking at maps between other things and x. The deepest linear algebraic result without linear maps is probably invariance of cardinality of basis, for example. It's conceptually fascinating, and it's nice to see the "translation into a different language" (poly equations -> automorphism groups) in real time after hearing mathematicians say "translate it into x language" 1000 times

[coral spindle]

totally. I often think about the basis observation as turning linear algebra into combinatorics

We can suddenly just use basic counting arguments

It's very cool

[languid trellis]

Super neat

[rotund aurora]

What interesting things can you say about the Rubik's cube group?

[languid trellis]

It's massive

https://ncatlab.org/nlab/show/motivation+for+sheaves%2C+cohomology+and+higher+stacks#TheBasicIdeaOfSheaves This little section also blew my mind

[coral spindle]

Oh I think we have a basic structure result on it

like we can say it's a wreath product or sth

[languid trellis]

"We can totally determine a space by probing it with another space"

[rotund aurora]

Do you have a reference?

[languid trellis]

Have you checked out the wikipedia page https://en.wikipedia.org/wiki/Rubik's_Cube_group

[coral spindle]

This is half-remembered I'm afraid

[dull ginkgo]

There’s also neat connections to the endomorphism ring

[rotund aurora]

Also what the hell, square roots of groups

[coral spindle]

Oh it's subdirect of index 2 presumably?

Is it unique I wonder

Not the nicest description

[rotund aurora]

I wonder if you can say anything at all of God's number but for other similar puzzles. Like nxnxn cubes or similar things but in higher dimensions

like I don't expect exact values obviously but maybe you can have non-trivial bounds

[coral spindle]

Right you could ask about generating sets of any group, right?

There's nothing special about the Rubik's cube group

I guess linking these to combinatorial objects is kind of interesting

[rotund aurora]

I would say the Rubik's cube is a combinatorial object lol

[dull ginkgo]

I found this out playing around a while ago

Let R be an integral ring inside ring B generated additively as a finite basis of right units b_1… b_N

Then as a module, B ~= R^N, so End_R(B) is isomorphic to the matrix ring M_n(R^op), where the basis is the matrices that send b_n to \delta{n,m} b_k, so End_R(B) is a finite rank left-R^op module or equivalently a right-R module

But if you instead consider it as a B-module (since B-scalings are linear), you can show that the basis gets reduced to the diagonal ones, since e_n,m : b_i to \delta_{i,n} b_m is equal to e_{n,n} (b_n^-1 b_m). It’s easy to then show the diagonal endomorphisms are linearly independent, so we’ve basically cut down the dimension massively

[coral spindle]

Well yes what I'm saying is that god's number is really just a property of the group

So can we go backwards and obtain useful combinatorial objects the other way

Is it even helpful?

[rotund aurora]

Ah, I see. Unless you consider a group to be a combinatorial object, then I don't think so lol. But these questions are very complicated, I don't think you get anything from extra generality

[dull ginkgo]

@languid trellis fun little problem:
let B be a Galois extension of A.

1. Show End_A(B), the ring of A-linear endomorphisms of A, is a A-vector space of dim [B:A]^2
2. Show that End_A(B) permits the structure of a B-vector space, show the dim is [B : A] and describe the basis using Gal(B/A)
3. Describe multiplication in that ring as a finite B-algebra, and describe how it relates to the semidirect product from group theory

[languid trellis]

oh what the sigma

I'll write it down

[dull ginkgo]

That’s the fun part of noncomm ring theory when it isn’t just Jacobson radical spam

[elfin wraith]

Appologies if this is an obvious thing, but I really dont have much of a feeling for Groebner bases, but I was wondering what do you do when you get repeated leading monomials which then give you different reduction rules? Does this matter? My feeling would be no, but it might cause you to find a larger Groebner basis than needed

As a specific example of what I mean, suppose you have like K[x,y,z] and youre finding a Groebner basis for f = x+z, g = x+y, I = <f,g> (this is maybe too trivial of an example but I hope it just makes my point clear), with the lex ordering x>y>z, then we have that LT(f) = x = LT(g) which gives us the reduction rules x -> -y and x-> - z, which are different, so can we just choose either when computing S-polynomials?

[dull ginkgo]

I wanted to try to describe the Grobner basis idea using an ordered monoid for my own sanity

[dull ginkgo]

I think that point is actually why some aspect of choice is needed in proving Hilbert Basis Theorem iirc

[dull ginkgo]

If you’re lazy you can just chose it off the next largest, whichever is larger lol

[elfin wraith]

But yeah you agree it doesn't matter and youll get a valid Groebner basis either way? I mean it makes sense you just keep going until everything reduces to 0 either way but I just wanted to double check im not being silly

[rocky cloak]

Skew group algebra my beloved

[dull ginkgo]

2-cocycle jumpscare

[dull ginkgo]

Are you referring to the crossed product algebra

[rocky cloak]

I guess it is a special case of that yeah

Or they're the same thing I guess. Haven't used the word "crossed product algebra" before.

Anyway, skew group algebra is just
If you have G acting on a ring R, then you exactly like the group algebra take the free R module generated by G, but then multiplication is
rg * sh = r s^g gh
where s^g means the image of s under the action of g.

It's like the semidirect product in that
(kN) # H = k(N ⋊ H)

[dull ginkgo]

2-COCYCLE JUMPSCARE

[tough raven]

It's not true that End_B(B) ⊆ End_R(B) = M_n(R^op) consists of the diagonal matrices (for some choice of basis) unless B is isomorphic R^n as algebras and not just R-modules, since End_B(B) = B^op and the diagonal matrices form (R^op)^n as R-algebras.

But it is true that End_B(B) is a subalgebra of M_n(R^op) which is a rank-n free submodule.

[rocky cloak]

A good example might just be to look at B = C (complex numbers) and R real numbers