groups-rings-fields

[chilly ocean]

You can try them by hand

[hidden wind]

sort of, i.e. i know the definition and i will have to derive its consquences during the exam

[dull ginkgo]

A commutative ring quotiented out by it’s maximal ideal is a field

[hidden wind]

[knotty badger]

yeah i know

though there can be many maximal ideals

[chilly ocean]

Minor thing but you should say "a" rather than "it's" since there can be multiple maximal ideals

[dull ginkgo]

Well it’s up to iso y’know

For this case

[knotty badger]

wait really

that seems strange

[chilly ocean]

Finite fields are special

[knotty badger]

right

im not an algebraist so I wouldn’t know ig

[dull ginkgo]

There’s an exercise in Jacobson basic alg in the intro ring theory section to find a closed form for the amount of irreducible polys in a finite ring using dirichlet convolution lmao

Less difficult than you’d think tho ironically

[hidden wind]

are there multiple irreducble pols of degree 1 over Z/2Z ?

[chilly ocean]

Actually this is not true for F_p[x], is it? You could have an irreducible poly of degree > 2 couldn't you?

[knotty badger]

hm surely every poly of degree 1 is irreducible

unless im misremembering

[hidden wind]

yes

[dull ginkgo]

The point is the field that is the quotient by the principal ideal of the irreducible polynomial of equal degrees are iso

[chilly ocean]

Yeah but there are still multiple maximal ideals

[dull ginkgo]

That’s what I meant

[hidden wind]

actually nvm i can easily check this myself once i’ve some paper in front of me

[dull ginkgo]

Try to calculate how many there are ;3

Actually rather easy… consider how many quadratics there are, and how many are the product of two linear terms ;)

Not now tho

[hidden wind]

thanku i will if needed during the exam (which is likely)

[dull ginkgo]

Later as an exercise

[hidden wind]

god i’ve only had a first look at the majority of the curriculum yesterday

i’m not usually this stupid with my studying i swear

[dull ginkgo]

Also quick question chat

A ring is a division ring iff it is BOTH left and right simple right

since xR and Rx would both have to be R and thus both contain 1

[chilly ocean]

I think that works

I found this question which claims something stronger https://math.stackexchange.com/questions/1821467/division-ring-if-and-only-if-it-has-no-proper-left-ideals#1821470

[dull ginkgo]

Unfortunately I don’t think this is true at a glance

An element in a ring can have two left inverses

(Giving zero divisors)

In fact if there is two, there are at least countably many (this is an exercise in Jacobson)

Actually it can’t have 0 divisors

So I guess it doesn’t matter

And one more thing

So $M$ being a left $R$-module means there’s a map $R \rightarrow \mathrm{End}(M)$, and for the right analogue, $R^\mathrm{op} \rightarrow \mathrm{End}(M)$. For bimodules is it $R \otimes_\mathbb{Z} R^\mathrm{op} \rightarrow \mathrm{End}(M)$?

[cloud walrus]

The Library of Babble

[rocky cloak]

That's correct, but it is in fact enough to just be one or the other (either left or right)

[rocky cloak]

That's correct

[rocky cloak]

The key is that if your left inverse has a left inverse, then it's a two sided inverse

Well, there are some key facts that are nice to keep in mind:

If you just look at the subgroup (under addition) generated by 1, it will look like Z/n. So any finite field has Z/n as a subring for some n. And since a field doesn't have zero-divisors this n must be a prime.

Also by Lagrange, n must divide the order of the field, so in this case it must be 2.

So your field has 0, 1, some other number x and then you'll also need x+1, that's 4.

x+x = (1+1)x = 0

So the only question is how things multiply. You can't have x^2 = 0 obviously, because then dividing by x would entail x=0.

x^2 = 1, doesn't work either, because it would mean (x+1)^2 = 0, and same problem.

x^2 = x leads to x=1, no good.

So x^2 = x + 1.

In general you construct finite fields by starting with Z/p and adjoining a new element. Then for it to not have zero-divisors you need the new element to satisfy an irreducible polynomial.

[glad osprey]

I hope rødbet is allowed to check discord during the exam

[dull ginkgo]

I didn’t know simple rings usually referred to no nontrivial/proper two sided ideals

I assumed it was a left/right/two sided kinda idea where it depends on the context and which side you are thinking of

More Miz rambling time

[chilly ocean]

Usually a simple object is defined as one that has exactly two quotients

And quotients of rings are in bijection with two sided ideals

[dull ginkgo]

I guess so

I just thought like “oh ideals are left/right/bi modules of the ring inside of the ring” so

[chilly ocean]

yeah rings are special so two sided ideals of rings turn out to be equivalent to quotients of rings

its not true for other algebraic structures

[dull ginkgo]

Yep

[rocky cloak]

Yeah, it's a bit weird. Like a ring is called semi-simple if it's semisimple as a module, but then a ring being simple is about the ring structure, not the module structure

[dull ginkgo]

When in doubt, blame the French

[mighty kiln]

Isn't R simple iff it is a simple R-module

[rocky cloak]

No

[dull ginkgo]

well for noncommutative rings there’s like

3 types of modules

Ergo there’d be 3 types of simple rings lol

[rocky cloak]

You got your division rings, which are simple as a module.
You got your nxn matrix rings over division rings, which are semi-simple as a module.
Then you got your non-noetherian simple rings, we don't talk about those

[next obsidian]

Why do you all do this to yourselves

When ring is commutative, all is nice

ab - ba ≠ 0

Statements dreamed up by the utterly deranged

[dull ginkgo]

matrix rings of division rings I just assert are doubly simple

[next obsidian]

Don’t say some shit like “oh noncommutative rings show up in nature”

No that is not nature

[dull ginkgo]

Endomorphism rings

[next obsidian]

Nature has inherent beauty

This is an abomination

[chilly ocean]

Nature has ugliness too

[rocky cloak]

Mfw linear algebra isn't in nature

[next obsidian]

🙉 lalalalalallaa I can’t hear you I can’t hear you

[chilly ocean]

🪑🙉

[dull ginkgo]

I think you may be able to generalize Jacobson Density Theorem a bit.

Assume we have two sets X, Y in simple left R-module U whose union is linearly independent over D = End_R(U).

If A is in End_D(U), and A(x) = rx on X, then it does on Y too

Because every set is a generating set on U, EVERY R-linear map from R^S (DIRECT SUM, not texing this shit on mobile) to U is an epimorphism

[rocky cloak]

I mean A could be given by multiplication by r' such that rx = r'x, but ry =/= r'y

[dull ginkgo]

Hm

I get the proof of Jacobson Density Theorem, which is actually super simple when you think about it, I just wonder if it generalizes in a simple way

To generating sets but I guess it doesn’t. It depends critically on U being simple and having no nontrivial/proper submodules

[shell pilot]

I have a very simple silly question but genuinely want to make sure I understand this topic...

A lot of examples for an equivalence class use the set of Integers Modulo 3.

$Z_3={0, 1, 2}$

[cloud walrus]

Soap_Opera

[shell pilot]

The definition of an equivalence class is $\bar{a}={b\in A | b \sim a}$ if we let A be $Z_3$ and a be an element in $Z_3$

[rocky cloak]

\sim not \tilde

[shell pilot]

Thank you

[cloud walrus]

Soap_Opera

[shell pilot]

So then if we write [0] out by that definition, why does it include elements that are not 0, 1, or 2?

[0] = {...-9, -6, -3, 0, 3, 6, 9,...}

But it says that b must be in A and only 0 is in A technically

[chilly ocean]

If A is {0,1,2} then you are right that only 0 is in [0]

[chilly ocean]

But if A is Z (and the equivalence relation is equal modulo 3) then this is true

[shell pilot]

Ok, but when they say "integers mod 3" are they saying only the set Z3? Or are they saying the set of all integers and then mod 3 is just the relation

[chilly ocean]

They are saying the set Z_3

And they are defining Z_3 as the set of equivalence classes of Z modulo 3

[rocky cloak]

Typically the set Z3 refers to the set of equivalence classes

[shell pilot]

Ok, I think that is where I need help with this topic

When a text or when someone says Z_3, I think about the set {0, 1, 2}.

Are you saying the set is actually {[0], [1], [2]}? Then doesn't that equal the integers?

[chilly ocean]

The set {[0], [1], [2]} only has 3 elements so it doesn't equal the integers

[shell pilot]

Oh ok

So I'm just wondering how that definition works with the set notation

[0] = {b \in A | b ~ 0}

The only "elements" in A are 0, 1, and 2

So shouldn't [0] be {0}?

[chilly ocean]

Depends on what A and ~ are

[chilly ocean]

what is ~ here?

[shell pilot]

Oh...

[chilly ocean]

The definition they are using here for [0] is not the same as yours. They aren't using a set A with only 0,1 and 2 as elements. Their A is the set of integers

[shell pilot]

So the set is Z and the relation is mod 3

[chilly ocean]

In that case [0] is indeed {...,-6,-3,0,3,6,...}

[shell pilot]

Ok, now I get it

I was thinking the "set" was Z3 and I wasn't even thinking about the relation

Z3 means the set of integers with the relation of mod 3

[chilly ocean]

Z_3 means the set of equivalence classes of integers for the relation of mod 3

[chilly ocean]

this is one of those equivalence classes

[shell pilot]

Ah, ok, that helps a lot! Thank you

[mystic arrow]

[0] \subset Z

it is very common for people however to treat Z_3 as {0,1,2} and offload the equivalence class properties on the algebraic operations

[chilly radish]

{0,1,2} would be more common

[mystic arrow]

with very common I mean, by starting students

[mystic arrow]

yes true

fixed

I didn't write that bit with thought

although honestly, there is an argument to be made, that most people also think about Z_3 in the above sense

its not particularly rare to think of equivalence classes as their representatives

despite the pitfield of issues that might cause

[distant summit]

Can I get a hint (in addition to the one given which was fairly obvious)? $\mathrm{GL}^+ (n, \bR)$ means the set of invertible matrices with strictly positive determinant.

I've tried $\phi: \mathrm{GL}^+(n,\bR) \to \bR^+,\ \phi(A)=\det(A)$. Showing that this is well-defined, a homomorphism and surjective is fairly straightforward, but this doesn't satisfy the kernel condition. Consider $A=\mathrm{diag}(2,2,2)$, then $A$ is in the centre as a multiple of the identity, but $\det(A)=8$.

[cloud walrus]

Douglas

Douglas

[quiet pelican]

Does A -> A/det(A)^(1/n) work?

[distant summit]

So is this GL+ to SL+?

[quiet pelican]

Yes

[mystic arrow]

is 1 the matrix of ones?

[distant summit]

It looks right, lemme write it down

Well

Identity

[mystic arrow]

well that's slightly different

[distant summit]

1=diag(1,1,..,1)

[distant summit]

Hence I deleted yes

[mystic arrow]

ok here's a hint: view the matrices in their jordan normal form

and work from there

[distant summit]

I think Micose's hint was sufficient

[quiet pelican]

That’s massive overkill for this

[distant summit]

But I need to check it works

[quiet pelican]

Just a sanity check to make sure you’ve got everything when you’ve done it: why doesn’t this work for even n?

[mystic arrow]

probably, true, but any problem like that can be understood from first applying it to diagonal matrices, then jordan form and then all matrices

[mystic arrow]

a reasonable question here is "how did you come up with this map"

[quiet pelican]

I looked for something that sends every diagonal matrix to the identity

Division by λ = det(A)^(1/n) does

[mystic arrow]

nice!

[distant summit]

I can see that the centre is in the kernel of the map you suggested, but how do you know that the kernel is in the centre?

$A=\mathrm{diag}(2,2, 1/4)$ has $\det(A)=1$, so $A\in \ker\phi$, but $A$ is not a multiple of $I$.

[cloud walrus]

Douglas

[mystic arrow]

whats itching me is the n odd condition

the projective linear group is defined for all n

[quiet pelican]

How is A in the kernel?

[mystic arrow]

i guess the homomorphism is easier then

[quiet pelican]

A/det(A)^1/n = A =/= id

[distant summit]

Uh yes I was forgetting that we are going to SL+ and not R+

What's a rigorous way of proving that the kernel is in the centre then?

Contradiction?

Assume there is an element of the kernel that is not in the centre. This means it has at least two distinct eigenvalues, but then it cannot be a multiple of the identity matrix?

Something like that

[mystic arrow]

$A/det(A)^{(1/n)}$ is just A divided by a scalar

[cloud walrus]

Stef

[mystic arrow]

Divide a matrix by a scalar and get identity, what does the matrix look like

[distant summit]

Oh I am being dumb again

[distant summit]

.

You can just take any A in ker, then A=(det A)^1/n I which is clearly diagonal and therefore in the centre

@quiet pelican Why is n odd?

[quiet pelican]

How did you prove the centre is a subset of the kernel?

[distant summit]

.

Actually it should say a multiple of I

rather than diagonal

[quiet pelican]

Other way around I’m asking

[distant summit]

Well $Z={\lambda I | \lambda>0}$, so if $z\in Z$ then $z=\lambda I$. Hence $\phi(z)=\frac{\lambda I}{(\lambda^n)^{1/n}}=I$

[cloud walrus]

Douglas

[quiet pelican]

So what have you implicitly assumed there?

[distant summit]

So Z is in ker

[distant summit]

That $(\lambda^n)^{1/n} \neq -\lambda$

[cloud walrus]

Douglas

[quiet pelican]

Actually that (λ^n)^1/n = λ

When is that true?

[proud ice]

What is a nice way to think about finding an group $G$, and subgroups $H,N$ such that $H \trianglelefteq N \trianglelefteq G$ but $H$ is not normal in $G$. I know there are such examples but is there a way to think about it in terms of field extensions. For example we can easily find extensions like $Q, Q(\sqrt 2), Q(2^{1/4})$, where quadratic extensions are galois, but $Q(2^{1/4})$ is not Galois over Q. On the group side, these quadratic extensions reflect index two subgroups which are always normal, and I was wondering if this can be used to obtain such groups with the property,

[distant summit]

Que?

we're working with real matrices

Oh I think I misread what you asked

Yes we're assuming what you've said

But...

[distant summit]

If n is even, then (l^n)^1/n=l since that is how ^1/n is defined, and since the centre consists of positive multiples of I, you still get l/l=1 (not l/|l|=-1 depending on the sign)

Oh I think I see your point now

[cloud walrus]

SoloBolo

[distant summit]

@quiet pelican if n even, then you could have ((-3)^2)^1/2=3≠-3

Is that right?

But then -3I isn't in the centre of GL+ because it's not a positive multiple of I

We're assuming that l is positive to begin with

There isn't actually a contradiction there. It would be a contradiction if we found an element that was in the centre but not in the kernel, but -3I isn't in the centre. negative*I isn't in the centre

[mystic arrow]

Why what?

Wait

Whats GL+(n,R)

Is it not the general linear group?

Regardless -I commutes with everything so its in the center

[quiet pelican]

It’s the general linear with positive det

[dull ginkgo]

I think positive determinant matrices

Because positive reals form a group under multiplication

[mystic arrow]

Ah

Then n odd explains itself

[delicate orchid]

ermm... what the sigma?

[mystic arrow]

Its not a group for n even

[rocky cloak]

It's a group in either case

[hidden wind]

thanku i’ll have a thorough look at this when i get home

[distant summit]

det(AB)=det(A)det(B), if det(A),det(B)>0, then so is their product

[hidden wind]

exam went well

the only finite field that showed up was F5 but prime orders are easy

[distant summit]

Inverse is obviously fine.

Identity is just the n-by-n ID matrix

Associativity follows from associativity of matrix multiplication

The det of the inverse will be 1/det(A), and this is positive since det(A) is

[hidden wind]

only question i didn’t manage write up a (at least) halfway decent answer to was the one about galois theory :S

which is not surprising since the exam was also the very first occasion i ever tried to compute a galois group ehehe

[rocky cloak]

But the center is different

[distant summit]

I still don't see where the issue is. If the centre is ${\lambda I | \lambda>0}$ then the issue you were referring to doesn't seem to be an issue

[cloud walrus]

Douglas

[mystic arrow]

Det(I) = 1, det(-I) = (-1)^n <0

[rocky cloak]

You don't need to contain -I to be a group

[hidden wind]

one of the five questions was one i thought of and solved myself yesterday mwahaha (are there any simple groups of order 30?)

[rocky cloak]

Do you use sylow theorems?

[mystic arrow]

Oh right multiplication

[rocky cloak]

Or something else

[mystic arrow]

Mb

[delicate orchid]

classification

[rocky cloak]

"excuse me, invigilator? I'm gonna need some more paper"

"just gonna prove this lemma real quick"

[urban geyser]

Here’s an interesting question I saw—what are your thoughts?

For a binary operation $\alpha : X^{2} \to X$ and a set $B := { X : (X,\alpha)$ is a group$}$ what is the cardinality of B?

[cloud walrus]

thimg

[dull ginkgo]

good luck

[rocky cloak]

If I'm not mistaken some early values are
0, 1, 2, 3, 16

[hidden wind]

mhm

[urban geyser]

how’s that?

[rocky cloak]

Well, X empty or |X|=1 are clear.

For |X| = 2 you just pick which element is the identity. Similarly for |X|=3.

For |X|=4 the group is either cyclic or Klein 4. For cyclic you pick the identity and the order 2 element, after that the rest is determined. For the Klein 4 you just pick the identity. In total that's 4*3 + 4 = 16

For |X|=5, you pick the identity, then for a fixed nonidentity element you just determine its multiples, so that should be 5*3! = 30

[urban geyser]

I think I see

[rocky cloak]

https://oeis.org/search?q=1%2C2%2C3%2C16%2C30 id:34383

Found it

[urban geyser]

there’s a follow up: “regardless of if it is possible i have two main problems: the possibility of the set becoming paradoxical, and the question being void due to the rules of ZFC and such” which im not sure if either are really “valid” questions

[urban geyser]

oh wow, wonderful

thank you!

[rocky cloak]

You mean B? Or what is "the set"

[urban geyser]

I suppose B is indeed the set

[rocky cloak]

B seems completely well defined in terms of ZFC to me

And for X infinite I think B will just have size the power set of X.

[urban geyser]

Makes sense, interesting that there’s already material for this question—thank you

[distant summit]

$GL^+(n, \bR)={A\in GL(n, \bR) | \det(A)>0}$

$Z(GL^+ (n,\bR))={\lambda I | \lambda>0}$

[cloud walrus]

Douglas

[distant summit]

Why does n need to be odd?

I know the isomorphism (Micose helped with that), but I don't get what is special about odd n

[rocky cloak]

If n is even then the center will be bigger

Then you can drop the lambda>0 requirement

[distant summit]

Right

that makes sense now

[chilly ocean]

Can there be a non-surjective morphism f: R->S of (commutative) rings where R has a nonzero proper ideal and for every non-(1) ideal I of R, f(I) is an ideal of S?

[urban geyser]

Is there a way to do this in reverse? Pick a general operation and find the sets which are a group under it?

[sage lodge]

not sure if I am misunderstanding you but the singleton set is a group under any operation (there is a unique one) but the collection of all singletons is not a set

[urban geyser]

I think that would still be the first question—like, for example, taking standard multiplication—how many sets under it form a group? singleton set is always in there, and so are the integers, reals, etc etc

would it even be possible to enumerate this?

[lone niche]

Let H N G be a chain of normal subgroups. And let's suppose that G is the Galois group of some field L/K.
Then N being normal to G is equivalent to L^N being Galois over L^G=K.
H not being normal to G is equivalent to saying L^H not being Galois over L^G=K.
Finally, H being normal to N is equivalent to L^H being Galois over L^N.
Letting L^N=E, L^H=F

So this is equivalent to finding extensions K < E < F < L where F is not Galois over K, but the rest are, AND F is Galois over E.
How hard is this? One can abstract your example by finding polynomials where adjoining a single root does not make it fully split, and then find E in some special cases.

But I think it's generally easier the other way around: to use group theory to help find subextensions with the given property and not the other way around.

[proud ice]

I can take the Galois group corresponding to the splitting field of $x^4 - 2$ over Q. Then from my example, there exists group with such property, but I don't write it explicitly in the most common form, but it shows existence of groups with such property

[cloud walrus]

SoloBolo

[distant summit]

Is it correct that $GL(n, \bC) / Z(GL(n, \bC)) \cong SL(n, \bC)$?

[cloud walrus]

Douglas

[chilly ocean]

Hint: use first isomorphism theorem

[distant summit]

Well yeah I think it is but some website is saying it's not

[chilly ocean]

What website?

[distant summit]

http://at.yorku.ca/b/ask-an-algebraist/2010/1524.htm

I assume this is a university chat room or smth

I used the same map suggested here (also Micose's one from earlier)

[quiet pelican]

For even n this is false
The right isn’t centreless, the left is

For odd n, it’s true by the same map

[distant summit]

Hmmmm. How to show this then?

Because I was assuming you would do something like LHS=G and RHS=G, therefore LHS=RHS

[sage lodge]

Random question that occurred to me:

Any Rubik's cube permutation also permutes the corners of the Rubik's cube. Is there a way to calculate the expected number of disjoint cycles (i.e. the number of separate cycles when writing in cycle notation) in the corner permutation induced by a random Rubik's cube scramble?

[chilly ocean]

Oops I'm sorry, I thought the center would be multiples of the identity because that's true for M_n(K), but it's apparently not true for GL_n(K)

[lone niche]

Did you read what I wrote?

[distant summit]

Um, isn't GL a subset of M_n tho?

So it won't be true for M_n either

In general

[quiet pelican]

Yes
But subgroups can have larger centres than the whole group

Although I did think that the diagonal scalar matrices were the centre of GL_n?

[chilly ocean]

M_n is a monoid rather than a group but yes

[distant summit]

Uh yes that's true but we're talking about GL/Z=PGL (not) being isomorphic to SL

So Z(GL) are the multiples of the identity

But I don't think that helps

[rocky cloak]

The center of GLn is multiples of the identity yes

[chilly ocean]

Why doesn't this argument show GL_n(K) / Z(GL_n(K)) is isomorphic to SL_n(K) then?

[rocky cloak]

How would it

[chilly ocean]

Consider the map from GL_n(K) to SL_n(K) given by A-> A/det(A)

[rocky cloak]

The image is not in SLn

[quiet pelican]

Because for even n, det(- Id) = 1

So that doesn’t get mapped to Id

[quiet pelican]

(Also it should be det(A)^1/n, but that’s pesky scaling stuff)

[rocky cloak]

Like det(A/det(A)) = det(A)^1-n

[tribal moss]

SL(2,R) contains an element of order 2; does PGL(2,R)?

[rocky cloak]

[0, 1; 1, 0]

[tribal moss]

Oh bah. Right.

[rocky cloak]

Even [0, -1; 1, 0] in PSL(2, R)

[chilly ocean]

Ohh ofc

[rocky cloak]

If you're trying to distinguish them I think the only normal subgroup of SL2 is it's center, while PGL2 maps onto C2 by the determinant

[tribal moss]

I was.

[rocky cloak]

So I guess you can show that SL2 has no map to C2, by showing that every element has a square root

At least I think that's true

[distant summit]

Apparently G/Z(G)=Inn(G)

[tribal moss]

Hmm, how about Diag(-2, -1/2)?

[rocky cloak]

Yeah, it doesn't quite work, but it's the product of two things that have square roots. So either way, no map to C2

I guess alternatively prove that the commutator subgroup of SL2 is itself. That one I know is true

[distant summit]

Is the correct surjective homomorphism to use here f(A)=(det A / |det A| )^N?

[quiet pelican]

No
It’s A -> Det(A)

I think

Yes

Recall that U(1) is just the unit complex numbers

[distant summit]

I'm a muppet

Idk why I assumed it would be more complicated than det

[distant summit]

Hang on I think I know why I thought it wasn't det.

If the homomorphism is det(A), then won't it send matrices to re^it instead of e^it, so you need to "normalise" it

[quiet pelican]

U(N) are all matrices that have unit modulus determinant

(As UU^dagger = I => det(U)det(U)* = 1 => |det(U)| = 1)

Not every matrix with unit det is unitary

But every unitary matrix has unit det

[distant summit]

Oh ok

I had not appreciated that unitary matrices had |detU|=1

[sage lodge]

maybe this question would be better suited for stackexchange?

[gaunt bone]

Hi

I have a question

My goal is to use Category Theory, Algebra, and Graph Theory to try to branch together different Types of Math (Homotopy Type Theory Types). Does anyone have any suggestions?

Similar to Langlands Program

I need to narrow my focus.

If I'm going to study something new/old, I need some proof to understand how it would meet that goal, I'm way to spread out in my studies.

[lone niche]

Or one of the combinatorics channels might work too.

[gaunt bone]

it's in there too

that's where I began the question

[chilly radish]

They weren't responding to you

[barren sierra]

If N is normal in G, and N contains it's centralizer, then |G| ≤ |N|!

How do I prove this? My thought is show that G acts on N and this induces a morphism phi: G -> S_|N| and then idk how to show this is injective

[quiet pelican]

Well what is the condition for conjugation by g to preserve N pointwise?

(Hint: it is closely related to one of your given hypotheses)

[barren sierra]

gn = ng for all g in G and n in N

Which is the centralizer yes

[quiet pelican]

Ah right

[barren sierra]

So I'd want to show the centralizer is trivial??

[quiet pelican]

There is a slightly subtle point

[barren sierra]

Which doesn't sound right

That was my first thought but that didn't work

[quiet pelican]

Your morphism can equivalently be taken as one G -> S_(|N| - 1)

Why?

[barren sierra]

🤨

[rotund aurora]

what if Z(G)=1?

[quiet pelican]

||every element fixes the identity||

[rotund aurora]

oh, oops, I misnunderstood the problem

[barren sierra]

Ah

[barren sierra]

So then kernel is centralizer

Damn that's slick

[quiet pelican]

Yes
Which has size <= |N|

[barren sierra]

Ya

Damn

[cerulean swan]

I am looking to prove the first Sylow theorem, but I keep getting confused around the last step. I have followed arguments presented in several books and videos, though I have made the most progress with this https://math.uchicago.edu/~may/REU2016/REUPapers/Idelhaj.pdf. I understand how we obtain a candidate subgroup using the stabilizer of an element whose orbit is not divisible by p, but I am lost after that... could someone please explain this to me

[hollow imp]

If $\omega$ is the element whose orbit is not divisible by $p$, and $H$ is the stabilizer of $\omega$, then since $\omega = \qty{g_1, \dots, g_{p^n}}$ is a subset of $G$, you can consider the group action of $H$ on $\omega$. This is well-defined because $H$ is a stabilizer of $\omega$, meaning that $\qty{g_1, \dots, g_{p^n}} = \qty{hg_1, \dots, hg_{p^n}}$ for all $h \in H$, which means the action of an element $h \in H$ is just a permutation of the $g_i$'s.

[cloud walrus]

Alphyte

[hollow imp]

but my favorite proof of the Sylow theorems is definitely this one: https://math.berkeley.edu/~ribet/250/Fall15/sylow.pdf

[hidden wind]

right right, that makes sense, so it's the quotient of Z/2[x] by (x^2 + x + 1), where the irreducible polynomial is the same thing as x^2 - x - 1 since 1 = -1 in Z/2

[rocky cloak]

Yeah, that's right

[hidden wind]

thank you

the part with "some other number x" is what i was struggling with i suppose, i couldn't imagine any number that doesn't look like the ones i'm used to :S like, 1,2,3, ...

[toxic zephyr]

obviously we can't consider N as an N-module, since N is not a ring. but the only axiom it seems to fail is the existence of an additive inverse. what nice properties would we lose from pretending N was an N-module? like i suppose what does the existence of an additive inverse "serve" us in terms of use? trying to find a way to motivate the existence of an additive inverse axiom for a linear algebra student and this example crossed my mind.

[chilly ocean]

The usual name for that kind of object is a semimodule btw

One of the motivations to study modules comes from rings. So maybe it would be more appropriate to ask for motivation to study rings instead of semirings

One thing about semimodules/semirings is that the kernel doesn't determine whether a morphism is injective

A morphism can have kernel {0} and still not be injective

So it's more complicated to study them

[toxic zephyr]

interesting. what would be an example of a semimodule homomorphism on N as an N-semimodule?

[distant summit]

Suppose $G=\bZ_n=\langle a \rangle$, where $n$ is even integer greater than $2$.

Is it true that $G$ is never simple, since ${e, a^{n/2} }$ is always a proper normal subgroup?

[quiet pelican]

Yes

Similarly, it’s not simple for composite n

[charred iris]

false technically

but only because of n=2

where that subgroup isn't proper

[cloud walrus]

Douglas

[charred iris]

otherwise yeah that's right

[distant summit]

Ye ok

[chilly ocean]

Every semimodule homomorphism from N to another N-semimodule S is of form nx for some x in S

[arctic trail]

If the group is abelian you only need to show that you have a subgroup

[distant summit]

How would it work for general composite n?
Is it just a case of "factorising" the group into $\bZ_{nm}=\bZ_n \times \bZ_m$?

[cloud walrus]

Douglas

[toxic zephyr]

that sounds very familiar. isnt the same thing true of any module homomorphism from Z as a Z module?

[arctic trail]

For coprime n and m this is true

[quiet pelican]

Yes

[chilly ocean]

The same is true for any R module homomorphism from R

[toxic zephyr]

oh neat

[distant summit]

Uh what if ord(G)=p^2 and p is prime

[quiet pelican]

Then the centre is non-trivial so works

(Similar for other p groups)

[toxic zephyr]

yeah that proof is pretty simple. so this is apparently also true for any S semimodules homomorphism from S (S a semiring)?

[chilly ocean]

Should be

[toxic zephyr]

i dont see why the proof shouldnt carry over. interesting

thank you @chilly ocean

[distant summit]

How do you know it's non trivial?

[delicate orchid]

it's a p-group

it's always non-trivial, use the class equation and prove it

[quiet pelican]

Consider the orbits of the action of G on itself by conjugation

[distant summit]

Yeah ok this I think goes a little beyond what I've studied of group theory

But not too much

I will take a look when I've got more time on my hands

[celest cairn]

Hello, my mentor was saying that $G = Gal(\mathbb{Q}(\sqrt{6}) / \mathbb{Q}) \cong V_4$, but the only permutations are $e: \sqrt{2} \mapsto \sqrt{2}$, and $\sigma: \sqrt{2} \mapsto -\sqrt{2}$ right?
Thus $G \cong C_2$.

[cloud walrus]

Sapphire

[lapis latch]

The order of the extension is 2 so the Galois group is gonna have order 2

[celest cairn]

Thought so. Thanks. 🙂

[uncut girder]

Bro sapphire what are you doing

[rocky cloak]

Could be that it was meant to be Q(sqrt2, sqrt3) and not Q(sqrt6)...

[dull ginkgo]

So let’s say we have a module $M$ over commutative ring $R$. If there exists an $r$ such that $r \bigwedge_{n = 1}^{N}{m_n} = 0$ then are the $m_n$ linearly dependent

[cloud walrus]

The Library of Babble

[surreal dagger]

What does the upside down V denote?

[dull ginkgo]

Wedge product in/of the exterior algebra

[cobalt heath]

What is m_n here

[dull ginkgo]

Module elements

[cobalt heath]

Yeah so I am confused since you are taking wedge product of elements.

[arctic trail]

that's confusing

[surreal dagger]

Why are they nice and where do they show up naturally and how essential are they where they show up?

[dull ginkgo]

They are essentially the free-est antisymmetric algebra you can put on the module

[arctic trail]

this is a genuinely interesting question

I'll assume the ring is unital

Maybe studying the case N = 2 first is better and naturally leads to arbitrary N

[dull ginkgo]

I only think it works with M being finitely generated

M is a module not a vector space

[arctic trail]

what I meant is

You can absorb r into a module element, this won't be the same question however it'll be similar

[dull ginkgo]

I’ll come back this in a bit

[arctic trail]

if I remember correctly a wedge product is null if the terms are linearly dependent

so absorbing r into m1, you have that rm1,...,mN are linearly dependent

This implies that m1,...,mN are linearly dependent if there exists a linear dependence a1 (rm1) + a2(m2) + ... + aN(mN) = 0 with at least one an != 0 for n >= 2 or a1 r != 0

This sheds light on what can fail

[dull ginkgo]

The other direction is easy to show

I am trying to see if it’s an iff

[rain grove]

Suppose that there are a nonzero polynomial $f(X)\in \mathbb{R}[X]$ and a nonconstant polynomial
$g(X) \in \mathbb{R}[X]$ in the field $\mathbb{R}(X)$ connected by the equality [
f(X) + \frac{1}{f(X)} = g(X) + \frac{1}{g(X)}
]
Show that then $f(X) = g(X)$

[cloud walrus]

OHHELLNAH

[rain grove]

Why does it say nonconstant?

[chilly ocean]

Consider the function h(x)=x+1/x from R-{0} to R. Is h injective?

[rain grove]

yes from $h(x) = h(y) \Rightarrow x = y$

[cloud walrus]

OHHELLNAH

[chilly ocean]

Why?

[rain grove]

oh

well I can just calculate that and I get x-y = 0

[quiet pelican]

What about 2 and 1/2?

[rain grove]

Yeah ok 2 and 1/2 are not good for injectivity

Ok and what if its not injective?

[chilly ocean]

What does it mean for it to not be injective?

[arctic trail]

Is injectivity even important in this question?

[chilly ocean]

Yes, they want to show that if x+1/x = y+1/y then x=y whenever x and y are nonconstant polynomials

And they asked

Why need to assume they are nonconstant polynomials

[chilly ocean]

this is injectivity for the function f(x)=x+1/x

[arctic trail]

that notation you used is cursed

[chilly ocean]

Sorry for that

[arctic trail]

cause f is a function IR[X] to IR(X)

Not IR

[lean sail]

Quick question… given a group G under multiplication, and a subset of G, H… does anyone know if using the two-step subgroup test to prove H is a subgroup of G, requires knowing in advance that H is closed under the operation?

[chilly ocean]

IR is naturally a subset of R[x]

[arctic trail]

yes

that is indeed factual

[rain grove]

Its not bijective

[chilly ocean]

Something can be not bijective and still be injective

[chilly ocean]

What is the two-step subgroup test

[chilly ocean]

If it didn't say nonconstant it would be false by putting f(X)=2 and g(X)=1/2

[rain grove]

Omg right my brain is not working rn... ty

[lean sail]

you probably know it by a different name, https://proofwiki.org/wiki/Two-Step_Subgroup_Test

[chilly ocean]

I knew what it is, I'm asking because the definition should make the answer clear

[arctic trail]

this turns out to be (f(X)^2+1)/f(X) these two expressions are coprime so they can't be simplified further (except maybe multiplication by a constant).
so for that equality to hold you must have g(X)^2 + 1 = R(f(X)^2 +1) and g(X) = Rf(X)
This gives R = 1 and so g = f

[lean sail]

Oh you’re asking a rhetorical question?

[south patrol]

Well there being closed under multiplication is in the definition

[lean sail]

Yes, this is what is throwing me a little bit

[arctic trail]

@rain grove

[lean sail]

Having trouble reading the def

[chilly ocean]

Are you asking if it can fail if you don't require H to be closed for multiplication?

[lean sail]

Not exactly, let me try again another way: is it possible to apply two-step subgroup test, if we don’t know H is closed under the operation?

[arctic trail]

Notice cause then you have R^2f(x)^2 + 1 = R f(X)^2 +R so f(X)^2 (R^2 - R) = R-1. Of course we assume R!= 0 so f(X)^2 (R-1) R = (R-1) if f is not constant by following the degrees this implies R = 1

[chilly ocean]

Not sure what you mean by that. When you apply the two-step subgroup test, one of the steps is to test whether H is closed under the operation

[arctic trail]

They are coprime by using Euclid's algorithm

[lean sail]

This actually answered my question.

[mighty kiln]

One-step subgroup test:

• Nonempty
• gh^-1

[rain grove]

So that is the reason i can't $\frac{f(X)^2+f(X)}{f(X)} - \frac{g(X)^2+g(X)}{g(X)} = 0$ and then put on common denomitor and get g(X) = f(X)?

[arctic trail]

what is T?

nope

[cloud walrus]

OHHELLNAH

[mighty kiln]

Zero-step subgroup test:

• ∃(h∈H)∀(h'∈H)∀(h"∈H)[h'h"^-1 ∈ H]

[arctic trail]

If $a/b = c/d$ and they're both written such that a and b have no common factors and c and d also have no common factors then there exists a unit R such that
$a = Rc$ and $b = Rd$

[cloud walrus]

Trivial Lemma

[arctic trail]

units in the case of real polynomials are non-zero scalars

[grave sedge]

For a cleaner solution, solve a+1/a=b+1/b over an arbitrary field and then use the fact that a,b are elements of R[x] inside R(x)

[arctic trail]

I don't understnad

my proof works also for principal ideal domains which is nice

and the x + 1/x can be replaced with x^n + 1/x^m

[arctic trail]

solving it for an arbitrary field seems harder

a^2 - a(b+1/b) +1 = 0
If the field has characteristic 2 we are cooked.

[rain grove]

Wait At the end I just get 2 conditions f(X)-g(X) = 0 and f(X)g(X) = 1. If one of them is nonconstant then it has no inverse so the other condition can't be true and then f(X)-g(X) = 0

[arctic trail]

this must be wrong

cause f(X)g(X) = 1 not true forall non-constant polynomials

[rain grove]

Yeah that my point

[grave sedge]

It's literally (a-b)(ab-1)=0 idk why it seems harder

Which works on any domain (since a ring is a domain iff its a subring of a field)

[distant summit]

Is $\phi: K \to \bC^*, \phi(A)=(A)_{11}=\lambda$ the correct map to use for (b)?

[cloud walrus]

Douglas

[quiet pelican]

Yes

[distant summit]

Lovely

[distant summit]

For (c), is the correct reasoning as follows?

$$\phi(a^n b^m a^N b^M)=\phi(a^n a^{-N} b^m b^M)=b^{n-N}$$
$$=b^n b^{-N}=b^n b^N = \phi(a^n b^m) \phi(a^N b^M)$$

[cloud walrus]

Douglas

[coral spindle]

Not quite. You correctly tried to use the fact that $ba=a^{-1}b$ implies $ba^N=a^{-N}b$, but you incorrectly assumed that this means $b^ma^N=a^{-N}b^m$ which is not true.

[cloud walrus]

Boytjie

[coral spindle]

In the end it isn’t important to the proof, but this is a mistake.

Also you wrote n where you mean m at a certain point

[distant summit]

What is the general rule then?

I would be able to do the more "manual" method of doing the four cases depending on the value of m and M

[coral spindle]

And same for N and M

[distant summit]

Where?

[coral spindle]

Hint: b^2a = b(ba). Use induction

[coral spindle]

Wait, my bad, I misread the definition of the map. Indeed it should be n and M.

[distant summit]

How are you supposed to induction if there are two parameters?

the power of a and the power of b

[coral spindle]

Think about it.

See what you derive

[delicate orchid]

or you could just... check? there's only 8 elements

[coral spindle]

That’s what a hint is for

I want you to get something of the form where there are no bs on the left.

[distant summit]

$b^2 a^n = b(ba^n)=b (a^{-n} b)=(ba^{-n})b=(a^n b)b=a^n$?

This makes sense since $b^2=e$, but then idk how you'd write that in nice notation, i.e. as $a^n b^m$

[cloud walrus]

Douglas

[distant summit]

Also what's the kernel in (d)?

Is it Z_3?

I think the kernel is ${e, b, a^2 b}$ which is a subgroup of order 3, and there is only one such group, namely the $\bZ_3$

[grave sedge]

That's not a subgroup

[coral spindle]

Correct. So if the power of b is odd…

[distant summit]

that is true

[coral spindle]

D_4 has 8 elements. What does Lagrange tell us about this supposed subgroup?

[distant summit]

Uh yes 3 is not a divisor so it can't be

oh its ${e, a^2, b, a^2 b}$ i think

[cloud walrus]

Douglas

[distant summit]

so this is either the klein four group or the fourth cyclic group

it obviously isnt cyclic, so it must be the klein four group

[distant summit]

i think the general rule is $b^m a^n=a^{(-1)^m n} b^n$

[cloud walrus]

Douglas

[coral spindle]

That’s exactly right, well done

[coral spindle]

This is correct too

And it is indeed the Klein 4 group. You could even call it D_2

[delicate orchid]

I would cry

[spice whale]

D_2

actually that's quite interesting

the 3 different defs (that i can recall) of D_n are actually not equivalent for n ≤ 2

[delicate orchid]

Yeah they break

C_2 x C_2 does not act faithfully on the line for instance

[rocky cloak]

Then the question is, should a digon be a line or should it be this like double droplet thingy

[spice whale]

You've heard of the line with two origins

[tribal moss]

Could we say D2 is the group of isometries of the entire plane that happen to preserve a line segment?

[spice whale]

now meet the origin with two lines

[spice whale]

yeah but this breaks with D_1

[distant summit]

is inner/exterior right in the penultimate bullet point? wouldnt inner/outer or interior/exterior make more sense?

[spice whale]

because in that case D_1 is just O_2(R)

you actually get 3 different D_1 s with the 3 different defs

geometrically it's O_2

by permutations it's the trivial group

and formally it's C_2

actually if you do it via 3d rotations the geonetric case is SO_3

wait

yeah actually you can probably even argue it's like
SL_3

[delicate orchid]

It should be this definitely

[distant summit]

.

[quiet pelican]

I would naturally use internal and external

[south patrol]

UsU

[surreal dagger]

Proving this im stuck on proving the exactnes at $Hom_R(M,N)$.
I denote the top maps: f and g and the bottom maps "og" and "of".
For exactness I have to show that im(og)=ker(of).

$if x \in im(og)$ then $x=h \circ g$ and $x \circ f = h \circ g \circ f = h \circ 0 = 0$

x is a map from M->N, If $x \in ker(of)$ then $x \circ f = 0$ hence im(f) contained in ker(x).

So this gives us a unique map from M/im(f)=M/ker(g)=M´´ to N which we denote by x".

I this the candidate we want?
I think x"(m´´):=x"(m+ker(g))=x(m) per construction so this works I hope?

[cloud walrus]

eden

[hidden wind]

fun fun

so far i’ve loved everything i’ve come across by john stillwell

also ian stewart

[coral spindle]

This has no relation to anything important at all, but I love that the notation for the dicyclic groups that is consistent with Dih(2n) for the dihedral groups is Dic(4n)

[hidden wind]

is there like a the archetypical example of a group extension ?

also is the dicyclic group of order 8 the same as the quaternion group

[still spire]

I guess the first examples you encounter are simply cyclic groups of order n, i.e. $\mathbb{Z}/n\mathbb{Z}$.

[rain grove]

Let $A$ be an algebra. A linear mapping $\delta : A \rightarrow A$ is called a derivation if for all $x, y \in A $
[\delta(xy) = \delta(x)y + x\delta(y)]
Show that every derivation on the real algebra of polynomials $\mathbb{R}[X]$ is of the form
[\delta(f(X)) = u(X)f'(X)]
where $u(X) = \delta(X)$ and $f'(X)$ is the derivative of the polynomial $f(X)$ (we define it naturally as follows: if $f(X) = \sum_{k\geq 0}^{n}a_kX^k , f'(X) = \sum_{k = 1}^{n}ka_kX^{k-1})$. Derive from here that every derivation on the real algebra of rational functions $\mathbb{R}$(X) is of the form
[\delta (\frac{f(X)}{g(X)}) = u(X)\frac{f'(X)g(X)-f(X)g'(X)}{g '(X)^2}]
(even the field of rational functions $\mathbb{R}(X)$ can be treated as an algebra over $\mathbb{R}$, multiplication by scalars is defined in a self-explanatory way)

[cloud walrus]

OHHELLNAH

Eulers Lamp

[still spire]

yes, pretty much by checking the definitions

[rain grove]

I can’t get to the equality just by using that proposition

[grave sedge]

Show constants must go to zero

(it's sufficient to check δ(1)=0 and you're done by linearity)

[dull ginkgo]

I am still haunted by the problem of showing a map from R^n to R^n is injective iff it’s determinant is regular (not a zero divisor)

[stone sky]

How is Contemporary Abstract Algebra by Gallian?

[still spire]

Do you wanna talk about it? 🙂

[dull ginkgo]

In a bit

[hidden wind]

so that's...

$0\to \mathbb Z\overset{\times n}{\to}\mathbb Z\overset{\mod n}{\to}\mathbb Z/n\mathbb Z \to 0$

[cloud walrus]

rødbet

[hidden wind]

?

[still spire]

The first map has to be injective, so you start with $n\mathbb{Z}$ as a group and the first map is just inklusion.

[cloud walrus]

Eulers Lamp

[hidden wind]

oh

[grave sedge]

It's the same thing

[still spire]

oh yeah

nvm, my mistake

[hidden wind]

i guess my "mod n" is not quite technically correct either

mod nZ would be

[still spire]

I think its fine. you send a number to it's residue class mod n.

[kindred ice]

Hello, I have a question about rings I’m not sure how to answer

True/false: $(\mathbb{Z}/10\mathbb{Z})[x]\cong\mathbb{Z}[x,y]/(10y)$

[cloud walrus]

qianqian07

[kindred ice]

I think it’s false but I’m not sure how to prove it

my first thought was to apply the correspondence theorem to the homomorphism with kernel 10, but that won’t work because (10) is not contained in (10y)

and even if this did work it would say that $(\mathbb{Z}/10\mathbb{Z})[x,y]\cong\mathbb{Z}[x,y]/(10y)$ which is not what the question asks for

[cloud walrus]

qianqian07

[kindred ice]

could anyone help?

thanks in advance 🙏

[still spire]

One approach would be to observe that $\mathbb{Z}/10\mathbb{Z})[x]$ has characteristic 10 but $\mathbb{Z}[x,y]/(10y)$ has characteristic zero.

[cloud walrus]

Eulers Lamp

[dull ginkgo]

It’s a problem in Jacobson that I’ve been periodically working on for a while and am stuck on

[still spire]

I don't have the book. But where are you stuck?

[dull ginkgo]

Just pondering

Let me grab my laptop rq

I am sick in bed and typing this on my phone would suck

[still spire]

ok, so it's two parts. for the first part, you have to show that the $f_i$ are linearly independent.

[cloud walrus]

Eulers Lamp

[dull ginkgo]

I tried working with instead

The A being an injective map

[still spire]

whats your reasoning there?

[dull ginkgo]

If it wasn’t injective there would be linear dependence

the kernel of the map would give the linear dependence

I was going to try showing there’s a nonzero map to the kernel, and the composition would be 0, thus the determinant functor to \wedge^N would make it a zero divisor

Can’t really use the adjugate for the reason that some matrices have a 0 adjugate

[still spire]

What theorems do you have available? do you already know that $Ax=0 with x\neq 0 \iff det(A)$ is a zero divisior?

[cloud walrus]

Eulers Lamp

[dull ginkgo]

Haven’t proven that yet

[still spire]

any other things you know about the determinant?

[dull ginkgo]

Not much at all

Jacobson barely has done anything with it tbh

[still spire]

Ok, I think showing that A is injective is a good approach. You'll need it later in a potential proof.

How far did you get?

[dull ginkgo]

I’d need to find my notes

[dull ginkgo]

Jacobson hasn’t introduced the wedge product yet but I want to use it because it’s easier

[still spire]

what definition is he using for the determinant?

[dull ginkgo]

Computed one

But I prefer the functor def

And I proved the equivalence before

Namely I did some ring theoretic bullshit

[still spire]

can you give the functorial one you have?

[dull ginkgo]

Every endomorphism of M gives rise to an endomorphism of Wedge^rank(M)(M) which is isomorphic to R

That endomorphism of the wedge power is the determinant

[still spire]

M being a free R module I guess?

[dull ginkgo]

Yes

Of finite rank

[still spire]

any properties of that endomorphism you already proved?

[dull ginkgo]

Not yet

The problem is equivalent to it preserving injections/bijections because of cancellativity and def of epi and mono morphisms

I’d need to think of how to show that

[dull ginkgo]

Basically here’s how I did it

Assume we have a module M over commutative ring R. We can construct the tensor algebra T_R(M) and view it as a graded ring over R (like the polynomial ring) with a grading that is the degree of a tensor

Let’s say we have a homogenous ideal J of T_R(M) with trivial intersection with R

If we have an endomorphisms of M that fixes the intersection of J and M, then that endomorphism extends to a ring endomorphism of T_R(M)/J

Which fixes (and acts like a linear map) over the grading

[still spire]

thats a lot of machinery for a simple fact 🙂

[rain grove]

How can I show the second part then?

[dull ginkgo]

It was in an entirely different context tbf

[rain grove]

For the algebra of fractions

[still spire]

I'm not sure how this does imply that A is injective iff det(A) is not a zero divisior

[dull ginkgo]

I wanted to prove some stuff about graded rings or algebras and there’s only 3 good examples I could think of, the tensor algebra included along with the polynomial rings, and the rings over free groups equipping it with the word lexicographic ordering

[dull ginkgo]

The ideal defining the exterior algebra is a homogenous ideal of degree 2, generated by (xy - yx)

And if M is finite rank n, the nth power is iso to R

All can be easily verified

[grave sedge]

Clearly to know δ(f/g) you only need δ(f) and δ(1/g), but g*1/g=1...

[rain grove]

But δ(f/g) = δ(f/1 * 1/g) and then if i use that formula its δ(f/1)1/g + f/1δ(1/g) and idk what is δ of f/1 or 1/g, cause f/1 is element of R(X) but f is from R[X] can I just use the same formula still?

[grave sedge]

Your proof for R[x] should work the same in R(x) for polynomials, right?

Only thing that changes is that δ(x) might not be a polynomial

[dull ginkgo]

is he proving Zariski cotangent bullshit indirectly

[rain grove]

So then this happens:

δ(f/1) =? δ(f) = δ(x)f' =? δ(x)f'/1

like mid-equation im converting delta to act first on a element of the ring and then on an element of the extension field of fractions

[grave sedge]

Just take the proof for R[x] you had

Read it but instead of having f be a polynomial in R[x] you have it be a polynomial in R(x)

What changes?

[distant summit]

Best textbook for learning this stuff?

[rain grove]

this one δ(f(X)) = δ(X)f'(X)?

I don't get it

[still spire]

So thats how you show the equivalence of the definitions?

[dull ginkgo]

Well by kind of “matching up” the computation with computing the actual determinant in the functorial definition

[still spire]

I thought you where already talking about the the injectivity of A

[distant summit]

.

[dull ginkgo]

Well here was my idea

The functor from the endomorphisms of R^N to the endomorphisms of Wedge^N R^N is surjective

And being injective is basically being left-cancellative

[still spire]

But being surjective just means that for every r in R you find a Matrix with det(A)=r. Not sure how that helps.

[dull ginkgo]

Good point

I’m still unsure if the whole wedge across the basis idea would work too

[grave sedge]

Yes

[still spire]

Probably, in the end it's all equivalent.

[dull ginkgo]

I’m just unsure how to show the one direction

[still spire]

Which one?

[dull ginkgo]

Determinant being a regular element

Implies injective

[still spire]

Hmm...how about assuming it'S not and take a $x=(x_1,...,x_n)$ that gets send to zero. Than I would check where $\sum_i e_1\wedge\ldots\wedge x_i\ldots \wedge e_n$ gets send under the induced morphism.

[cloud walrus]

Eulers Lamp

[dull ginkgo]

It gets sent to 0

[still spire]

But thats just an idea...not sure if it leads anywhere

[dull ginkgo]

Wait

I should probably be sleeping instead of trying to math poorly when I’m sick lol

[still spire]

true

and i should go out instead of writing on a math channel

its friday evening

[dull ginkgo]

@still spire actually

I think if we have an injective endomorphism of M

Then it extends to an injective ring endomorphism of the exterior algebra

[still spire]

yeah, I think so too

[dull ginkgo]

Same with these

Just due to the “freeness” lol

[still spire]

and then you can interfere that since the only endomorphisms of R are multiplikations with elements of R you have your result

[dull ginkgo]

Is there an easy way to show there is an endomorphism of R^N with image within a given submodule of it

I.e the kernel of another

[grave sedge]

Not injective implies determinant being zero divisor can be done with the adjoint matrix thing

(which i think is the implication you had problems with, up to contrapositive)

[dull ginkgo]

The adjoint might be 0.

Which isn’t helpful aaaaa

[grave sedge]

It shoudlnt be, right?

And even if it is, who cares

It would imply the determinant being zero which surely is a good thing

[dull ginkgo]

hmm

[grave sedge]

If your endomorphism is represented by a matrix, you know the product of this matrix and its adjoint is determinant*identity

[dull ginkgo]

Yeah

[grave sedge]

If the determinant is regular then det*identity is injective

[dull ginkgo]

Posting

*pondering

[grave sedge]

Composition of two things being injective implies the first one (i.e. your original endomorphism) was injective

[dull ginkgo]

That’s a big jump with stuff that’s circular

Also my point is the problematic direction is det(f) being a zero divisor -> f having nontrivial kernel

If the adjugate is 0, yes det(f) = 0

But we can’t do shit with that

Because then we have 0f = 0, obviously

[grave sedge]

Which Is the opposite of what you said here

I gave you a proof for that direction

[dull ginkgo]

I haven’t proven that yet.

Which what you gave is circular

[grave sedge]

Huh?

[dull ginkgo]

If I can just show that there is a map from R^N to the kernel then I’m good.

Like a complex

[grave sedge]

You said "i'm unsure how to show A=>B", i (think i) gave a proof for it and you said "my point is the problematic direction is B=>A"

[dull ginkgo]

You referenced a part that depends on the thing I am trying to prove

[grave sedge]

No i didnt (?)

[dull ginkgo]

I’m starting over

[still spire]

I'm also not sure about you're reasoning @grave sedge . If $A\cdot A^{ad}=det(A)\cdot I$ that only really helps you to find an inverse if $det(A)$ is invertible.

[cloud walrus]

Eulers Lamp

[grave sedge]

If det(A) is regular then the thing on the right is injective

Which implies A is injective

Just plug in something that is killed by A if you want to do it by contradiction

[dull ginkgo]

Or use that it’s in the center

[dull ginkgo]

How does that imply A is injective

Composition of two maps can be injective without the maps being injective themselves no?

[grave sedge]

One of them has to be injective

fg injective implies g injective

Do it by contradiction

(similarly fg surjective implies f surjective)

[dull ginkgo]

Ok I see

Then there’s showing f being injective implies det(f) is regular

[grave sedge]

No

[dull ginkgo]

I think I can do this

[grave sedge]

Ah no wait

[grave sedge]

I misread

Yes

[dull ginkgo]

because then the scaled adjugate would map to the kernel

So I guess it’s all due to the adjugate

[rain grove]

What is δ(1/g)?

[grave sedge]

What is δ(g*1/g)?

[rain grove]

0 = δ(g) * 1/g + δ(1/g) * g

[grave sedge]

There you have δ(1/g)

[rain grove]

yeah but g is not not necesarily invertible

its not unless its a constan

[grave sedge]

You're in R(x)

The inverse of g is 1/g

[rain grove]

Hmm ok, and then what is δ(g/1)?

[grave sedge]

g/1 is just g

[rain grove]

its δ(x)g'/1?

ok cool ty

[arctic trail]

oh lol

[kindred ice]

oh, that makes sense. thanks for the help!

[willow cipher]

is this question right? in particular the second part proving that if R is noetherian, then Ass_R(M) is not empty

ive proven the first part, but what if M is torsion-free and Ann(m) = 0 for every single m? can this not happen? if Ann(m) = 0 for each, then none of these can be prime so Ass_R(M) is empty. im pretty new to this concept and would be thankful if someone could give some clarification

[lone niche]

(0) is a prime ideal in Z for example.

[lone niche]

You need to show maximal elements of the family exist via the Noetherian property.

[next obsidian]

Following up on Kiyoshi’s example, you can see directly that R is an integral domain.

If it wasn’t, then take ab = 0, but a and b nonzero. Then am isn’t 0, but b(am) = 0, so b in ann(am) which is uh oh bad news

[dull ginkgo]

Last little bit of the zero divisor determinant problem

I am stuck showing that if a matrix A gives an injective endomorphism of R^(n) [R commutative] then det(A) is regular

The other route you can use the adjugate for, but this one you sorta can’t immediately

Starting with the assumption that det(A)x = 0, then we have the problem that xAdj(A) might be the 0 matrix

[still spire]

What about the approach we hat yesterday? That if A is injective the induced morphism on the wedge is injective?

[dull ginkgo]

I haven’t shown that the induced morphisms preserve injections

And that seems like a lot of work ngl

So otherwise I am completely stumped

[still spire]

it should be similar to the field case. somehow take any basis $e_1,...,e_n$ then $Ae_1,...,Ae_n$ is still linearly independent and hence $Ae_1\wedge...\wedge Ae_n=det(A) \cdot e_1\wedge...\wedge e_n$ is nonzero.

[dull ginkgo]

That takes a lot of computational verification

[cloud walrus]

Eulers Lamp

[still spire]

which part?

[dull ginkgo]

All of it.

Gimme a second to think

[dull ginkgo]

The problem is then showing det(A) has to be regular

Which we are back where we started

Using the exterior algebra results in circular logic

So I’m going to avoid using it

But I’m still utterly stuck which is pathetic because this should be an easy problem lmao

[still spire]

I don't think its hard, given you know that the wedges are nonzero if the vectors are linearly independent, i.e. $v_1\wedge \ldots\wedge v_n$ is nonzero iff the $v_i$ are linearly independent. Just assume you have a x in R with $det(A)x=0$, then $x\cdot v_1,v_2,\ldots,v_n$ are still linearly independent. Then you get $A(x \cdot v_1)\wedge A(v_2)\wedge \ldots \wedge A(v_n)=det(A)\cdot x \cdot v_1\wedge\ldots\wedge v_n\neq 0$

[cloud walrus]

Eulers Lamp

[long obsidian]

Is this a place to post basic representation theory questions?

Why does it follow that if p(g):V->V is a G-linear map then g is in the center of the group?

[knotty badger]

you can consider what happens for the regular representation, i think

[coral spindle]

It suffices to check just rho being the regular representation

[knotty badger]

yeah that’s what i said, right?

i guess maybe I should’ve specified that rho was the regular one

[coral spindle]

My bad

[knotty badger]

it’s ok

[tribal moss]

For representation theory, #advanced-algebra will be a better match.

[swift tundra]

If you have an isomorphism between groups, is it necessarily a homomorphism?

I can see how the isomorphism would induce a group structure on the codomain, but I am not convinced that it is the same structure given by the binary operation on that set. Do the multiplication tables of the structure induced by the isomorphism and the binary operation always match?

[dull ginkgo]

… what do you mean

An isomorphism is an invertible homomorphism

But we can have two things be isomorphic as one type of object but not as another

R^2 and C are isomorphic as vector spaces but not as rings/algebras

[chilly ocean]

Fun fact you can have two topological groups that are isomorphic as groups and as topological spaces but not isomorphic as topological groups

[swift tundra]

I’m sorry if this is a dumb question. I am relatively new to group theory. This is the same answer I have read online, but I still don’t see why.

When you read isomorphism in a group theory text book is it implied that it’s a homomorphism?

I guess I have always hear isomorphism in terms of sets. So I am wondering if you have a bijection, must it also preserve the group operation?

[chilly ocean]

Yes isomorphisms in group theory are required to be homomorphisms

Bijections needn't preserve the group operation

Consider f: Z->Z given by f(x)=x+1

It is a bijection

But f(0+0)=1 while f(0)+f(0)=2

[dull ginkgo]

forgetful/free adjunction moment

[chilly ocean]

What do you mean by that in this context? I know what those are but I don't see how it's related to the above conversation.

[dull ginkgo]

isomorphism of groups induces an isomorphism of the underlying sets but not the other way around, but it does induce an isomorphism of the free groups

[tropic spade]

I do not see how the fact about subgroups of index 2 gives us C_D8(x) has order 4?

I can see how using x not in the center gives us |D8 : C_D8(x)| is either 2 or 4 but I don't see how to rule out the case that C_D8(x) has order 2? I think I'm missing how the fact they mentioned above prevents this?

I know I could just brute force out all the conjugacy classes, so I'm really trying to just understand the specific technique they are using here.

[swift tundra]

Ok ok, thank you! This is what I was thinking but I never saw it stated explicitly.

[lone niche]

idk about this part, but one can see |C_D8(x)| cannot be 2 since x is not in the center, so CD8(x) contains Z(D8) and x, and Z(D8) has size 2.

[tropic spade]

Okay, I think I see it. Are you explicitly computing Z(D8) from the elements of D8 here to get that it has size 2?

[lone niche]

Nope, I took it as a general fact. Let me see how the proof goes for the size of the center.

[feral timber]

I have a simple query. If P is a prime ideal of R, prove that P[X] can never be maximal in R[X].

Now if I consider S={all polynomials with every coefficient in P and constant in R}, this gives me S not equal to R[X] (misses polynomials with non P coefficients), and P[X] contained in S

[tropic spade]

I was thinking it's just Z(D8) = {1,r^2} by considering presentation facts for D8?

[feral timber]

It looks pretty certain that S would be an ideal, since all the coefficients except constant (that too can be in P) are in P anyways, so multiplying it with any element of R[X] would still result in coefficients in P. Does this work?

Does this work?

[chilly ocean]

What is P[X]?

[mystic arrow]

Polynomials over P

[lone niche]

yeah this is enough to show Z(D8) has size at least 2

[knotty badger]

Suppose you have a commutative ring R such that all R-modules are free. Does R have to be a field?

[tropic spade]

Thx for the help kiyoshi!

[mystic arrow]

No

[knotty badger]

I see

What’s a counterexample?

[mystic arrow]

Oh

Commutative R

Yes

[knotty badger]

Oh!

Yeah, I’ve heard of division rings which don’t have to be commutative

Then - how do you prove this?

[chilly ocean]

The trivial ring is a counterexample but its the only one

[knotty badger]

Hmm, is it like

You quotient R by a nontrivial ideal?

[mystic arrow]

Ye, you need 0 neq 1

[knotty badger]

If it wasn’t a field

And then the resulting module isn’t free?

[dull ginkgo]

Sup chat

I am going to go back to my alg binge

[chilly ocean]

What are you reading?

[dull ginkgo]

Nothing

[feral timber]

?

[dull ginkgo]

Trying to understand something outside of a textbook

Graded algebra schenanigans

[mystic arrow]

Yep

[knotty badger]

oh neat!

that was easier than i expected

[mystic arrow]

Fwiw

You haven't actually shown it yet :p

You have the idea, now you need to actually show the steps

[knotty badger]

hmm, well

take a nonzero non-unit a

[chilly ocean]

I think it does

[knotty badger]

then consider R / (a)

[mystic arrow]

Why does R/I not admit a vasis

[knotty badger]

hmm, because any element multiplied by a is zero?

and that contradicts the analog of linear independence

i think

because a is nonzero

so in particular, no nonempty set can be “linearly independent”, i think

[rotund aurora]

Can Z_p (p-adics) contain Z_q for q!=p (here I'm talking only about abelian groups)?

[mystic arrow]

You are significantly overthinking this I think. A free module would mean R/I ~= R^n

But think about what happens to f(0) and f(1) and f(a)

[knotty badger]

I’m confused, did i say something wrong?

[south patrol]

I'm confused too lol seems pseudo was fine

[mystic arrow]

I guess it depends how you define free

[south patrol]

There is only one definition (up to equivalence)

[knotty badger]

has a basis, is the definition i’ve heard

[mystic arrow]

In our definition it was free iff product of Rs

[south patrol]

That only works for finite things

[mystic arrow]

Although our prof famously hated the word basis

[south patrol]

It should be direct sum

[mystic arrow]

Ye sorry direct sum

English not primary language

[south patrol]

And having a basis is the same as being a direct sum of R since the basis gives you an isomorphism (and the converse is trivial)

[rotund aurora]

sum/product is spelled pretty much the same in most european languages, probably

[mystic arrow]

Yes, but my ability to think in english is worse than in my main language

[knotty badger]

yeah that tracks

the free r-module functor preserves coproducts

[south patrol]

Well also this is just a common definition as opposed to "being in the essential image of the free functor" ig but yes that shows they agree

[long obsidian]

@knotty badger @coral spindle how can I show this with the regular representation of a finite group? It's not clear to me that if

$\sum a_z e_{ghz} =\sum a_x e_{hgx}$

then g and h commute.

Is it something like if this is the case then $z=h^{-1} g^{-1} h g x$ which implies that $h^{-1} g^{-1} hg=e$? I'm not really sure if this is the correct argument.

[cloud walrus]

HausdorffT1

[knotty badger]

I think set v = delta_e, the basis vector corresponding to the identity

[long obsidian]

Ooooh wow okay that's simple thank you

[knotty badger]

yeah often you can like

treat the regular representation really as just the regular action

its the free vector space functor applied to the regular action, after all

[rain grove]

Are generators of an Algebra the same as basis elements of its vector space?

[grave sedge]

no, they are elements such that any element can be represented as a polynomial in them

instead of a linear combination (which would be a linear polynomial)

equivalently, a set of elements generate the algebra if the smallest subalgebra containing all of them is the whole algebra

(if you want a counterexample, R[x] is finitely generated as a R-algebra but not as a R-vector space)

[rain grove]

Hmm im studying the Algebra of quaternions rn, the generator is the set {1,i,j,k}?

[long obsidian]

An algebra is a module, technically it does have a set of generators as a module. The monomials generate the polynomial ring as a module

[grave sedge]

{i,j} is also a set of generators

since your set contains it then it also is a generating set

there's no "the" set of generators

[rain grove]

But there is a subalgebra generated with just {i}

[grave sedge]

yes

[rain grove]

is there some connection with the vector space basis and the "generators"

idk exactly what Im looking for but just thats what im thinking

[grave sedge]

a vector space basis is also a set of generators

there's no "the" generators, by the way

(kind of how there's no "the" basis of a vector space)

[rain grove]

Yeah I see

[grave sedge]

and if you have a set of generators then the monomials made by multiplying those generators together are going to be a spanning set of the vector space

not necessarily a basis, but something that contains a basis

[rain grove]

Ohh nice I like that

[tribal moss]

Oh fun! I can think of G=||positive rationals under multiplication, with the usual topology|| and H=||G×Z, where Z is discrete||.
(They are distinguishable as topological groups because ||H has an open proper subgroup||).

[dull ginkgo]

Hey goobers

Lets say U is a cyclic left R-module

And we have a left ideal J of R such that

Jx = U. and we have another element y such that:
Ann(x) \cap J \subseteq Ann(y) \cap J

Then f(jx) = jy is a well-defined endomorphism of U right

[rain grove]

Im trying to show what elements are in the center Z(H). This guy wrote "commute with generators", does he mean elements of the basis?

source: https://math.stackexchange.com/questions/1595738/center-of-the-quaternions-proof-and-method

[dull ginkgo]

The irony here is that technically {i,j} form a generating set for Q_8

since ij = k

actually that's how you can give the presentation of Q_8

[grave sedge]

no, generators

[dull ginkgo]

Do you know what monoid rings are

[rain grove]

yeah just rings with identity, not neccesarily having inverses

[grave sedge]

since every element can be written as a polynomial in i,j, an element commutes with everything if and only if it commutes with i and j

[dull ginkgo]

no i mean like

If you have a ring R, and a monoid M, you can construct the "monoid ring" of R[M].

Which is finite sums rg of elements of R times elements of M

defined like a polynomial ring

[rain grove]

Ohh so then writing reals and k was unnecessary

[grave sedge]

(the hint given there is to verify it with 1,i,j,k, but it's not strictly necessary)

yes

also they are unnecessary for more or less different reasons

[rotund aurora]

If you restrict to the class of torsion abelian profinite groups and assume the GCH then this is not true, although this is just an artifact of the classification of such groups. But it is funny

[grave sedge]

k is unnecessary because it's just ij

reals are unnecessary because they commute with everything "by definition"

[dull ginkgo]

@rain grove Find a presentation for Q_8 using i and j ;3

[chilly ocean]

What is the classification of torsion abelian profinite groups?

[rain grove]

Q_8 is group of quaternions?

[dull ginkgo]

yes

[dull ginkgo]

well group of the "basis" quaternions

[grave sedge]

since "reals commute with everything" is more or less equivalent to "multiplication is bilinear wrt the real vector space structure"

[dull ginkgo]

1, i, j, k and their negatives

[rotund aurora]

$\prod_n (\Z/n\Z)^{c(n)}$ where the $c(n)$ are cardinals. The $c(n)$ here are not uniquely determined, but you can make them unique if you impose that c(n)=0 for n not a power of a prime

[cloud walrus]

croqueta3385

[rotund aurora]

and

c(n)=0 for large enough n

otherwise it is not torsion

[rain grove]

And its enough to check just for generators because of the distribution law and (xa)b = x(ab) = a(xb)

[grave sedge]

yes (assuming what you said implies x being real)

[dull ginkgo]

I wonder if you can easily compute the center of R[M] if R is commutative and M is a monoid

[rotund aurora]

that the exponent is finite uses something like the Baire category theorem (you consider the union of the G[n], n=1,2,3,..., which are closed, and whose union is all of G)

[chilly ocean]

Crazy that they can be classified. Although I'm not super familiar with profinite groups so maybe there's a simple reason why the torsion abelian ones are so simple

[dull ginkgo]

DIRECT LIMIT JUMPSCARE

@grave sedge btw are you UG?

[grave sedge]

yes

[dull ginkgo]

epic

same-ish

I wonder if this is a direct consequence of jacobson density theorem

[rotund aurora]

I don't actually know the details of the proof, but people usually take the Pontryagin dual and then it's a statement about usual abelian groups of finite exponent

[dull ginkgo]

omfg it's really easy

[rotund aurora]

bro is a scholar of the Jacobson density theorem

[dull ginkgo]

I NEED TO UNDERSTAND JACOBSON BOURBAKI IT HAUTNS ME

[chilly ocean]

I'm starting to feel that I need to undertand it too

[dull ginkgo]

i should probably focus on stuff actually on topic

[chilly ocean]

What topic?

[dull ginkgo]

intro abst alg lmao

[tough raven]

This is true whenever U is (a semisimple module over R which is) finitely generated over D (because C_E(C_E(R_U)) = End_D(U) = closure(R_U) in the topology of pointwise convergence by the Density Theorem and End_D(U) is discrete wrt that topology if U is fg over D).
But I don't see why U being simple over the right Artinian R ensures that dim_D(U) is finite...

[knotty badger]

Ok, so a 2-cycle is also known as a transposition - it’s a special kind of permutation which only swaps two elements

All 2-cycles have order 2, and products of disjoint 2-cycles also have order 2