groups-rings-fields

[topaz solar]

Suppose you have a bijection I -> B, then how do you handle the P(I) many indicator functions

[marsh scaffold]

Wait is it true for any vector space that a LI set can be extended to a basis?

[topaz solar]

Yes

(With choice yada yada)

[marsh scaffold]

Okay pls continue

[topaz solar]

You’ve seen diagonalization yeah?

consider I as the basis of (+)_I K

And let’s look at f: I -> B

[marsh scaffold]

Now that you've asked me again I only know the part where they get a decimal number in [0,1] not previously in the list

[topaz solar]

Oh aight

Well, you can do this more generally to show that |X| < |P(X)|

Which is the size of the indicator functions

You get the idea though, take a “diagonal,” get an element you missed

Ye?

[marsh scaffold]

I see the set {x : x \not \in p(x)} was basically selecting things different in diagonals

[topaz solar]

Ye!

[marsh scaffold]

That just blew my mind

[topaz solar]

So, you believe when I say $\bigoplus_{i\in I} K = K^{\oplus I}$ has its dual space as $K^I$ yeah?

[cloud walrus]

Sharp-Malliaris regularity lemma

[topaz solar]

Since finite nonzero indices allow us to just sum f(i)a_i

Right?

[marsh scaffold]

I don't understand the notation

What is big O

Here

Functions?

[topaz solar]

Oh, the K vector space with a basis I

[marsh scaffold]

Okay

[topaz solar]

Basically, functions with only finitely many nonzero values

Right, this is using the suggestion in the problem that this shows we don’t have isomorphism with the duals

[marsh scaffold]

This is believe is just (isomorphic copy of the vector space with a basis of cardinality I)?

[topaz solar]

Let’s suppose we have a bijection $g:I\to \mathscr B$, then since $g(i)$ is an element of the dual of $K^{\oplus I}$, we can evaluate it at $i\in I$ right?

[cloud walrus]

Sharp-Malliaris regularity lemma

[topaz solar]

So what do you suggest we do, in the spirit of diagonalization

[marsh scaffold]

A function h such that h(i) is not sth at that point

Uhhh

[topaz solar]

Well

What about g(i)(i)

This is a function I -> K, so it’s in K^I too

[marsh scaffold]

So h(i) = g(i)(i) ?

[topaz solar]

Yep

[marsh scaffold]

Okay

Let me sink that in

Can you give me a moment

I'll be back in like 2 mins

[topaz solar]

I think this should work anyway

Been a while since I’ve done this I just know a diagonalization trick should work

[marsh scaffold]

Ok back

So sps a bijection from I to B exists we construct a function that is not in the dual list

Right?

[topaz solar]

Well, you’d wanna verify it’s not in the span since vectors

But that’s the idea

[marsh scaffold]

Wait ,g(i) is an element of the dual because?

I'm confused again

[topaz solar]

(Or maybe we wanna do like, g(i)(i) + smth)

[marsh scaffold]

Hmmm

[topaz solar]

g(i) is now a function I -> K, which we can extend linearly to a vector space w/ basis I

[marsh scaffold]

Alright so it's extension is a dual

[topaz solar]

ye, and every dual element arises in this way because linearity

so we can evaluation g(i) at i

Let’s do +1 too

[marsh scaffold]

How do you do a +1 o_0

[topaz solar]

Call h(i) = g(i)(i)+1

That’s a function I -> K

[marsh scaffold]

Wait

Okay

It makes sense

It's going to a field

[topaz solar]

And, h(i) \neq g(i)(i) for any i

So it’s certainly not in B

So the trick is doing something to exclude it from being the span

(So instead of +1 we’d wanna do something that isn’t in B at all, so might as well be g(i)(i) ig)

If you consider mapping I to the singletons in the obvious way, g(i)(i) gives you the indicator for the whole set, for example

So let’s take B as being a basis from the indicator functions so that we get indicator functions out, and we basically get the original cantor diagonal

(Up to doing 1-whatever instead ig?)

[marsh scaffold]

If I am getting this correctly
We are taking all the functions with all but finite many non zero from I to K
Then we FTSOC assume that there is a bijection from I To B where B is basis of functions from I to K

Now we use this bijection to construct a new element in the dual

This new element is having the property that ......uhhh

It's not in the span of ...(I think I need clarity here)

[topaz solar]

I am very scattered in an impromptu explanation

[marsh scaffold]

That's fine

[topaz solar]

1. let’s take a basis for the subspace of K^I spanned by the indicator functions

Call it B’

[marsh scaffold]

@topaz solar they have given this hint

[topaz solar]

So the size is bounded by basis sizes for K^I

(Bad service)

There’s an extension of B’ to a basis B of K^I

So, if we show we can’t map to it, we’re done

This is where we use the diagonalization trick

Cantor says use {x | x not in p(x)}, so let’s biject p:I -> B’, and let’s look at the indicator function we get from 1-p(i)(i)

Clearly that’s not in B’

However, we’re dealing with a basis, so we have to show there’s no finite linear combination which hits it

But uhh, since the only coefficients to consider are 1 and -1, and some cardinal arithmetic, there’s only |I| many sums

But we have 2^|I| > |I| many indicators

So it can’t possibly span the subset of indicators, much less the space they span

So we can’t surject I into B’ < B, so we can’t surject it into B

[topaz solar]

Oh ok that hint is basically what I did with the counting those sums

Cardinal arithmetic

(I’m a logician or analyst moreso usually)

@marsh scaffold this make sense?

[marsh scaffold]

I am reading it

[topaz solar]

That said, this might not be the intended way

[marsh scaffold]

I'll ping you on any doubts

It might take me some time since I am very slow

[topaz solar]

Same so understandable

[marsh scaffold]

Okay so you've proved sth stronger than we can't surject I into an even smaller B'

But one thing why is B' < B

Interms of cardinalities

@topaz solar

[topaz solar]

Well, we can extend B’ to a basis B

So it’s a subset

[marsh scaffold]

Uh?

Subsets can have same cardinalities?

Like uh R-{1} and R?

Or maybe I am not understanding what you mean

[topaz solar]

Cardinality is \leq

But B’ < B is a literal subset

And anyhow, we show |I| < |B’| \leq |B|

So |I| < |B|

[marsh scaffold]

Oh okay

[topaz solar]

Ye

[marsh scaffold]

So it's leq

I think I get it now

[topaz solar]

Cardinalities do be nice under ordering

assuming choice

But doing things without it is not your problem for now

[topaz solar]

They might’ve wanted a cardinal arithmetic argument that’s a bit nicer, but we can do this anyway

[marsh scaffold]

Alright then thanks a lot ...
I may return incase I get any doubts popping up

[marsh scaffold]

Just some after comments:
Isn't the dual vector space of a vector space V with basis of size B is just of the size |K|^|B|
And our indicator vector space is of size |big O thing|

And there was 2.30

Which gives us the size of V which is the size of our big O thing

[topaz solar]

Well 2.30 solves 2.31 when |B| = max(K, B) things, so you could use 2.30 + that 2.31 hint about countable fields

[marsh scaffold]

Oh hmmmm?

[topaz solar]

Well, it’s what I was meaning by here

But that solves it in a base case then reduces

Whereas we do everything at once

So it’s basically the same

[marsh scaffold]

Wait...what's the problem with saying that the size of the dual space is K^|B| (K is field) which is greater than equal to as a field has atleast 2 elements 2^|B|

[topaz solar]

Just remember things like |R| = |R^n| = |R^N|

So

If K is big, we can’t just look at the cardinality of the vector spaces

Where R is reals, N is naturals

[marsh scaffold]

Isn't|2^B| strictly bigger than B?

It's just the power set?

[topaz solar]

So yeah it’s gotta beat out that 2^|B|, but for very large K we won’t see differences over K

Also:

[topaz solar]

We can have countable vector spaces over Q with countable dimension

So this is wrong actually

[marsh scaffold]

I see

[topaz solar]

Remember, basis B makes a vector space by functions which are 0 except on a finite set

Not all functions

so, countable dim vector space over R vs R^N might be hard to look at cardinality-wise

So you have to actively look at basis trickery

[topaz solar]

I am walking on a sidewalk and misread this for a sec oop

[south patrol]

Isn't this fine

Oh lol

Tbf idk what you wanted to prove

[marsh scaffold]

I am actually loosing my mind

[marsh scaffold]

https://media.discordapp.net/attachments/496784958430380033/1259273109538865273/Screenshot_20240707-035116.jpg?ex=668b14f9&is=6689c379&hm=9f49aca478c24fe220ab8ef0be105b33a26ecad4465c213162e06d45e11f8817&

2.31

[topaz solar]

I made an argument about looking a basis for the subspace of simple function inside the spanning set of indicator functions, extending it to K^I basis, then by sum counting on indicators + Cantor + cardinalities ordering we win

[south patrol]

Yeah sure

[marsh scaffold]

The funny thing is: this is a textbook in topology and I haven't done any topology so far(maybe)

[topaz solar]

(maybe)

[hoary estuary]

I have no clue how to start with the blue part of this question. (I have attempted the rest)
Any help? Thanks.

[tribal moss]

Suppose g and h are two elements of G\A, and a is an element of A that is not its own inverse.
Then, because of (i), agh = ga^-1h = gha.
But if gh is in G\A too, then agh = a(gh) would also be (gh)a^-1.

[hoary estuary]

OK, got it, thx. (The red part below is my own work)

[shell pilot]

Suppose that a, b, and c are elements of a dihedral group. Is a^2b^4ac^5a^3c a rotation or reflection?
The answer says that even powers are rotations, but why is that? Why can't it be an even power of a reflection and that just returns it back to the identity?

[dark wave]

I think the identity is technically a rotation

So what are some interesting properties about cyclotomic fields?

[prisma umbra]

Correct?

[nimble folio]

Assume $R$ is commutative. Prove that $R$ is a field if and only if 0 is a maximal ideal.
\begin{proof}
Suppose $R$ is a field. Then $R \cong R/0$, hence 0 is a maximal ideal.
\
Suppose 0 is a maximal ideal. Then there does not exist an ideal $I$ such that $0 \subset I \subset R$. Hence the only ideals of $R$ are 0 and $R$. Thus $R$ is a field.
\end{proof}

[cloud walrus]

clubsoda14

[nimble folio]

i dont really know if this is valid

[clear fiber]

I think it is valid.
The first part depends on "If R/I is a field, then I is a maximal ideal."
The second part depends on "If the only ideals in a commutative ring R are 0 and R, then R is a field."

The second statement I wasn't immediately sure if it was true, but then I found this page which shows how to prove it https://math.stackexchange.com/questions/101157/a-commutative-ring-is-a-field-iff-the-only-ideals-are-0-and-1

[south patrol]

Lol I mean you have a characterisation of fields which is basically the same as what you have to prove

[south patrol]

You can do it much more directly though

Suppose 0 is maximal and x is a nonzero element of R. What can you say about the ideal generated by x?

[dim widget]

It’s an ideal

what do I do now?

[south patrol]

Sylow theorems

[languid trellis]

What are those again?

[south patrol]

Sylow theorems state that every ring has a maximal ideal except for the empty ring

I should say that the last couple messages have been joke responses to tteg

[rustic crown]

empty ring

[warped shadow]

I found a paper on the topic that characterizes the cases when a geodesic exists for GL(R,n), when it is unique, and how to compute it if it exists and is unique:
https://arxiv.org/abs/1412.4565

I know it was forever ago when we discussed this, but I thought you might find it interesting also

[nocturne bone]

So if R[X] is a UFD. And f is an irreducible monic polynomial in R[X]. Then is R[X]/(f) a field?

[rocky cloak]

No, but it's true if R[x] is a PID

Which I believe is true iff R is a field

[nocturne bone]

Thanks!

[south patrol]

Yes, the hammer being that dim R[x] >= dim R + 1 lol

[rocky cloak]

Or I guess, since x is prime, (x) is maximal, so R = R[x]/(x) is a field

[south patrol]

Yes :)

I just find what I said amusing lol

[rocky cloak]

Consider me amused 🐼

[south patrol]

Although I am disappointed in myself for using R

[jolly pendant]

This is supposed to prove that for an a in some group (G, *), (a^-1)^-1 = a… is this just saying “if you look at it hard enough, you can see it’s true!”, or is it actually saying something??

[coral spindle]

No this is a proof

It's elided details, but it is a proof

[fading field]

there is a unique b such that ba^-1=e (and by definition this b is (a^-1)^-1), and also aa^-1=e

so b=a

by uniqueness (i find the paragraph to be a bit wordy, and that might serve to obfuscate what’s actually being said here)

[tribal moss]

It's probably a typo that the proof says "(since by part (2) a has a unique inverse)" instead of "since by part (2) a^-1 has a unique inverse".

[jolly pendant]

Oh, I see what they mean now

Thanks for that 🙏🏿

[dark wave]

Can someone explain why this is true?

p is prime and M is an integer not divisible by p

[lone niche]

The polynomial on the right is the pth cyclotomic polynomial which is the minimal polynomial for the pth roots of unity. So Z[T]/P(T) is isomorphic to Z[zeta_p]. Notice that zeta_pM generates both zeta_p and zeta_M. Since p and M are coprime, then zeta_p\cdot zeta_M is a primite pM root of unity.

[swift tundra]

Hey, I am trying to teach myself some group theory over the summer and I wanted to know if anyone has textbook recommendations? I am reading “An introduction to the Theory of Groups” by Joseph Rotman right now since it was free on springer, but I am curious to hear other suggestions.

[mighty kiln]

Artin's algebra is good for a first viewing though it sometimes goes on tangents

Aluffi is good for a second read on abstract algebra

[topaz solar]

I do not know your prior knowledge, Lang is maybe a bit too encyclopedic-styled and brutal

Rotman covers some interesting topics but idk how “important” they are these days

[surreal sluice]

Could someone explain why the image of the map is closed?

[rocky cloak]

Notice that if s: Omega -> Omega is an automorphism, then for each finite Galois subextensions M we get an automorphism s_M on M.

Now if you have a family of automorphisms, such that s_N is the restriction of s_M whenever N is in M, then you can define s by s(x) = s_M(x) for M containing x.

Hence the image are exactly the things with these compatibility relations. So the only way for something not to be in the image is if there are extensions N < M such that s_N is not the restriction of s_M. Just fixing those two elements gives you an open set in the product topology district from the image.

[surreal sluice]

Nice, thank you!

[lean sail]

Hi everyone. Reading through Gallian's 6th ed. abstract algebra book. He does a preliminary section on mod arithmetic. Just trying to understand in this example using an error checking application as an example, why is the dot product equal to 5i and not -5i? I think this is because since there is an error in the 2 position, the mod is calculated to be "5 over"... is this thinking correct? Thanks!

[delicate bloom]

I feel like there's some missing context here needed to answer this

[lean sail]

there is, i tried to narrow down the scope, but i could see where maybe more info is necessary

[delicate bloom]

I dunno about how this encoding scheme thing works, it seems to be whatever that's about explains why it must be 5i vs -5i

[lean sail]

basically the idea is when you mod the code, it should be 0 (divisible by the mod number)... if it's not equal to 0 then, that's the magnitude of the error (how much you're off by). make more sense?

i could put the entire page from Gallian, but want to avoid if possible

[delicate bloom]

the way I'm reading it is the blue box is the left side and the red box is the right side of the equation:

the left side ends up being 10 and the right side ends up being 0+5i

which is why it's + not -

[lean sail]

you're right! i see it now, i was misreading that, it's kinda early, everything is blurring together

[delicate bloom]

it's sort of confusing because it almost looks like they've "factored out" the left into the dot product on the right

but I'm guessing that dot product on the right was precomputed as an error correcting number that we already have or something

[lean sail]

yes, factored it out, and we already know it mods to 0

[delicate bloom]

how do we know it mods to 0, just was calculated and happened to be that before the digit was flopped to 2?

I don't think it should be assumed to be 0 always at least if that's what you're suggesting

oh well we're done I guess, maybe if I saw the algorithm I'd know lol w/e have fun

[lean sail]

on second thought, i dont think this evaluates to 10

i'll post more context when i get a chance

[lean sail]

okay, here is the full context:

i can post as pictures too

@delicate bloom

[crystal vale]

Since r and s are relatively prime numbers so there exists c and d such that rc + sd = 1.

Now 1a = rca +sda, thus a = b + c, where b = sda and c = rca, right?

[tardy hedge]

Not really sure how to think about “this Z mod hom factors through H …”

What is the significance of elements in H mapping to 0 ?

[mighty kiln]

What's H

[tardy hedge]

All elements of the form

[mighty kiln]

Are R and S related

[tardy hedge]

R is a subring of S

[mighty kiln]

If you have a Z-morphism A → B and the image of the submodule C ⊂ A is trivial then you induce a map A/C → B

[tardy hedge]

Ok i can see that in that new map the 0 in the quotient gets mapped to 0 because C gets mapped to 0 in the original map

This induced map just takes a coset and maps it to where the original map sent any representative ?

[mighty kiln]

Yea

Slightly more generally if you have submodules A ⊂ B, C ⊂ D then a map B → D that restricts to a map A → C will induce B/A → D/C

This is a common trick to define a map between two groups / rings / modules

[tardy hedge]

Oh. Yeah. Seems important

[mighty kiln]

E.g. if B, D are the free groups / modules on generators and A, C are the relations

[tardy hedge]

Yea this is the first time ive seen the construction of an algebraic structure by first starting out with some free object and then quotienting by the relations u want satisfied

Thats neat

This section is about tensor products of modules and i thought the way they built it was cool

[tardy hedge]

Ok so just to be sure on details , the point of everything in H being mapped to 0 is necessarily only so that the map from F\H to L is well-defined

If you have two different representatives but the same coset, their difference is in H, so then they map to 0 and u can show that then they map to the same thing by rearranging equation etc

Sort of thing right

[coral spindle]

Watch out, you mean F/H not F\H!

[tardy hedge]

Ohh yeah haha

After this section on tensor products the book covers exact sequences

Big topic!

[mighty kiln]

The fun part of algebra

[tardy hedge]

No way

NOW it gets fun?

So actually upon reflection i am really enjoying modules

More than group theory

[mighty kiln]

That's cuz R-Mod is pog

Homological algebra go brrr

[tardy hedge]

Das cool

Although i do feel a bit naughty for not having studied topics in group theory like sylow theorems

At some point i will go back

[languid trellis]

As someone who 'studied' them, don't.

[devout crow]

I never studied them and they've never come up

[tardy hedge]

Oh … i see 😅

[coral spindle]

Depends on what you study later on

[tardy hedge]

In winter semester next year i am taking galois theory

Is it relevant there?

[coral spindle]

No

[tardy hedge]

Where DO they come up then?

[coral spindle]

In group theory

[tardy hedge]

I would like to eventually study algebraic topology, algebraic geetry

Geometry

[tardy hedge]

Ah. Lol

[coral spindle]

One place where they come up is in my research area, rep theory. There are several important conjectures regarding the relationship between the reps of a normaliser of a Sylow and the whole group.

[tardy hedge]

Ohh yeah actually i really want to learn representation theory especially

More than alg topology

Is there research that goes on purely in group theory?

[coral spindle]

Yes

[barren sierra]

Yea but I've heard many mathematicians say they're surprised by that fact 😭

I did an REU at a university which has a few people working on group theory. Community seems small but very nice from the couple conferences I went to

[devout crow]

lots of important research is being done in geometric group theory, which studies groups using topology

[coral spindle]

I'm talking about just group theory though, not GGT

GGT is a big field that even crosses into non-group stuff. I know semigroup theorists who do 'GGT'

[rocky cloak]

I guess it comes down to what counts as "purely in group theory". I'm sure there is lots of research being done on algebraic groups for example

[devout crow]

fair enough, I guess maybe it comes down to whether the topics of research concerns the questions or the methods. a lot of the topology of GGT is in the techniques, but the (overarching) questions are often purely group theoretic

[lone niche]

By the fundamental theorem of Galois theory, there's a one-one relation between intermediate field extensions of some extension, and the subgroups of the Galois group. So some field theory questions get translated into group theory questions, which is where things like Sylow groups show up.

[sly crescent]

What does count as “purely group theory”?

[rocky cloak]

That is indeed the question

[languid trellis]

I'm in the midst of doing D&F exercises, and thee are a whole lot of them. How do I decide which exercise is worth doing?

and at what point can I safely move onto the next section.

for the record, I've on 10.3 and I've done most exercsises up to 18

Please ignore everything I've just said. I am moving on because most of the exercises from this point on require things I haven't done

[tardy hedge]

Yea dude like if youve done that many exercises i think u can move on

Also lets help each other on 10.4 man this shit hard 😭

Ive been literally on this page for days

I do understand the overall idea but i dont like moving on until i really understand each detail of the proof and why everything is the way it is yanno how it is u guys study math too

[languid trellis]

I've just started the chapter, I'll get there probably by tomorrow or the day after

[tardy hedge]

Ok cool, maybe lets ask each other questions. Could be helpful to work it out together

There is something unique about working thru it with another student vs just getting help from the experts all the time

Tbh ive been super sleep deprived so im thinking at like 45% capacity anyway

[languid trellis]

for sure. i'm quite busy tomorrow and I'll working quite a lot over the next few weeks but I'll definitely find some time to get through this myself and then we can go through this in here]

[barren sierra]

Don't the higher number exercises tend to be harder

Get a good mix of questions throughout the set

[languid trellis]

Perhaps I'll go back to the exercise about modules over noncomm rings not having a well-defined rank (I think? essentially that # of the minimal generating set isn't well defined). that seemed interesting

[topaz solar]

Non-IBN

[sterile condor]

what's the lifting property?

[wraith cargo]

what's the context?

[sterile condor]

lifting maps

like from simplices to P^n

[sterile condor]

might be better to say lifting map but idk

[wraith cargo]

if this is in the context of covering spaces then you can image lifting maps as being a right inverse to the covering map
So they "lift" objects from the base space to a layer of the covering space

[sterile condor]

thanks

[barren sierra]

Working on (3) right now. What's the relevance of this isomorphism $L \otimes_K U_0 \xrightarrow{\sim} U$? So let $\sigma \in \Gamma, {f_i}{i \in I}$ a basis for $U_0$, and $x = \sum{i \in I} \alpha_i f_i \in U$ where $\alpha_i \in L$. Then $u_\sigma(x) = \sum_{i \in I} \sigma(\alpha_i) f_i$ which is in $U$ still since $\sigma(\alpha_i) \in L$ still. But then for the reverse direction I want to use that $u_\sigma(U) \subseteq U$ to show that $L \otimes_K U_0 \to U$? is an isomorphism but I don't even see where the isomorphism is used in fact that $U_0$ is a $K$-structure?

[cloud walrus]

Spamakin🎷

[nimble folio]

Let $R$ be a commutative ring with 1. Prove that the principal ideal generated by $x$ in the polynomial ring $R[x]$ is a prime ideal if and only if $R$ is an integral domain.
\begin{proof}
Suppose $\langle x \rangle$ is a prime ideal of $R[x]$. Let $p(x),q(x) \in R[x]$ such that $p(x)q(x) = 0$. Since $0 \in \langle x \rangle$, $p(x)q(x) \in \langle x \rangle$.
\end{proof}

[cloud walrus]

clubsoda14

[nimble folio]

Does this imply that either p(x) = 0 or q(x) = 0?

not too sure where to go from here

[barren sierra]

If you wanted that p(x)q(x) = 0 implies p(x) = 0 or q(x) = 0 then you need that R[x] is a domain

[nimble folio]

right thats what im trying to show

[barren sierra]

Do you know that for a ring $S$ and ideal $I$, if $I$ is prime then $S / I$ is a domain?

[cloud walrus]

Spamakin🎷

[nimble folio]

Yes

[barren sierra]

and of course the converse holds

so it's an if and only if

ok so try that direction for suitable (obvious) choice of S and I

[nimble folio]

Yea is there any way to do it directly with the fact that <x> is a prime ideal

my question was if p(x)q(x) = 0 and p(x)q(x) ∈ <x> (a prime ideal), does this somehow imply that either p(x) = 0 or q(x)=0

[barren sierra]

no, it just implies that x divides p(x)q(x)

[nimble folio]

Darn

Thanks I'll try the other way

[lone niche]

I think the idea is that $V$ can be recovered from $V_0$ by an extension of scalars. In particular, if we have $v\in V$, and a basis of $V_0$, what this isomorphism says is that we can express $v$ as an $L$-linear combination of the basis elements of $V_0$.

[cloud walrus]

kibocchi

[lone niche]

I didn't read your proof, but when you took $x\in U$, you implicitly made use of this fact.

[cloud walrus]

kibocchi

[barren sierra]

Yea

Ok that makes some more sense

[sterile garden]

If I'm given the additive group $\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z} \times n\mathbb{Z},$ is there a formula for the number of elements of maximal order? For instance, I've shown that if $n$ is prime then this is just $n^2 - 1.$ I had initially thought it was $\varphi(n)n + \varphi(n)(n - \varphi(n)),$ but some computations showed this to be incorrect because I miss a bunch of elements.

[cloud walrus]

IAmDerek

[rocky cloak]

If n = ml with m and l relatively prime, then
(Z/n)^2 = (Z/m)^2 x (Z/l)^2
and an element has maximal order if each entry has maximal order.

So if n is square free it's the product of p^2 - 1, where p divides n. And in general it would reduce to the case n = p^k

[sterile garden]

Product of p^2 - 1 and what?

So if it's p^k, it seems that my idea of it being phi(n)(2n - phi(n)) is correct. I'll take a look at this m,n coprime case then

[rocky cloak]

Just the product of p^2 - 1 for the primes p.

So if n = 30
(2^2 - 1)(3^2 -1)(5^2 - 1)

[sterile garden]

Ahhh, okay, so it seems it would be something like
$$ \prod_{p \mid n}(p^{2k}-p^{2(k-1)}) $$

[cloud walrus]

IAmDerek

[sterile garden]

Errr, I guess that would be p^k dividing n

Thank you 🙂

[serene dune]

damn, have u solved all of part 1, 2(i.e. group and ring) ?

[kind temple]

suppose we have a group $G = \langle S_G | R_G \rangle$ and a set map $f:S_G\to H$, $H$ a group.$\newline$

Does there exist a unique extension of $f$ to a homomorphism $\varphi:G\to H$?

[cloud walrus]

c squared

[kind temple]

i have tried looking at the following diagram:

and showing that the kernel of p_G is a subset of the kernel of tilde(f) to use the universal property of the quotient map (viewing G as a quotient of the free group on the set of generators S_G)

but i can’t quite show that

[median pawn]

could i get a hint or two for this?

[kind temple]

F^x is cyclic (of order 2^r)

[void cosmos]

isnt like

the multiplicative group of any finte field cyclic

[kind temple]

yes, it is like that

[chilly radish]

No, it's generally not going to extend. If the map respected all the relations it would though, by the first Isomorphism theorem

[fading field]

I could give you something really dumb like G = <1,1* | 1 = 1*>, take f(1) = 1 and f(1*)=0, to Z, and then there is no group homomorphism extending f

[kind temple]

ok thanks

[chilly radish]

Yea, this is the condition basically

It doesn't always happen, as smay showed

[chilly radish]

In fact any finite subgroup of the mult group of a field is cyclic! Even if the field is infinite

[median pawn]

yeah i have that

[kind temple]

so if f(r) = f(r’) for all r,r’ in R_G, then it would be okay?

[void cosmos]

yeah ig its well known a subgroup of a cyclic group is cyclic

the field being infinite tho

is cool

[kind temple]

what do you know about generators of finite cyclic groups

[chilly radish]

No...

In the infinite case the big group isn't cyclic

Only every finite subgroup

[fading field]

the group of units of an infinite field need not be cyclic

[void cosmos]

oh wow

yeah

even cooler ig

[kind temple]

.

[fading field]

you can even take C for example to show this

[chilly radish]

You need f(r)=0

For all r

When here I think of f as the extension to the FREE group

[median pawn]

well there's just one generator

this is what the solution does

but i don't see why the kernel has the given size

[kind temple]

by the euler phi function, there are 2^{r-1} generators

[chilly radish]

Isn't it 2^r-2^{r-1}

[median pawn]

oh that's what you mean

yeah we have phi(2^r) choices for the generators, but one is enough is what i meant hmm

[kind temple]

same thing

[fading field]

:P

[chilly radish]

I am not sure how this helps, to be honest

[kind temple]

i didnt say it did

[median pawn]

bro

[chilly radish]

It's kind of a bad move to give hints if you're not sure of the efficacy of said hint

[kind temple]

i didn’t give that hint

they suggested it

[chilly radish]

?

[kind temple]

i just suggested to think about generators

[median pawn]

do y'all see it

[languid trellis]

Nope. I used a different book

[serene dune]

oh damn, which book was it ?

[serene dune]

i am just trying to find good problems in abstract algebra in general, any suggestion regarding that ?

[narrow shuttle]

how do I glue points on a plane? I only know what to do in 1 variable case

could it be (x^2-2, y^2-2) and (xy+2)?

[tribal moss]

(x²-2, y²-2) is not maximal.

(x+y)(x-y)=0 in the quotient, so the quotient cannot be a field.

Try ||(x-y, x²-2)||.

[swift tundra]

I am looking at one of the exercises in my group theory book, and it says

Let G be a group and H be a proper subgroup of G. Then <G -S>=G.

But I was considering the case where our group is the integers under addition and we take the subgroup of odd integers. How can the even integers generate the odd integers?

[mighty kiln]

What's the subgroup of odd integers

[wary sorrel]

odd integers are not closed under addition

1 + 1 = 2

also 0 is not odd

[midnight pasture]

i was trying to figure out the constructibility of a regular polygon and all that
btw i learned about fields and stuff just days ago so i dont know much 😭

and I don’t get how each imaginary root of unity corresponds to one additional degree of extension $deg(\frac{\mathbb{Q}(\zeta)}{\mathbb{Q}})=1$ (when there are roots labelled $\zeta^1,\zeta^2,\dots,\zeta^n$), and not two degrees or any other number for that matter

[cloud walrus]

daff

[rocky cloak]

I'm not sure what you mean be "correspond to one additional degree of extension"

Also Q(zeta)/Q has degree 1 only if zeta is rational.

[midnight pasture]

yes

oh yeah damn

i said this because

like if you extend the rationals by zeta you get in ex. $x^4-1=0$ it would be a degree 2 extension because there r two imaginary roots

[cloud walrus]

daff

[midnight pasture]

i think im misunderstanding something 😭

[midnight pasture]

thats what some uni professor said

i dont get it

[rocky cloak]

Are you asking what the degree of Q(zeta)/Q is for different choices of zeta?

In general if zeta is a (primitive) nth root of unity, then the extension has degree phi(n)

[midnight pasture]

i dont get that one either

i know gauss proved it but

wait did he

[rocky cloak]

I don't know who proved that originally

[midnight pasture]

is there an intuitive explanation?

so i was watching this

video

https://youtu.be/EIDsrNZTpTI?si=XatPdmhKae6Nd-5c&t=1710

he just says like it just happens and doesnt give an explanation

[tribal moss]

I think you need some medium-hairy algebraic argument involving cyclotomic polynomials to see it.

[midnight pasture]

medium hairy hahahaha

[midnight pasture]

where can i find the medium-hairy algebraic argument involving cyclotomic polynomials

[tribal moss]

It's simple enough that I think I've understood it once, so it can't be very hairy.

[rocky cloak]

Well, there is a little work to be done to prove it. But the basic argument is that if x is any number, then the degree of Q(x)/Q is the degree of the minimal polynomial of x.

The minimal polynomial of zeta is the cyclotomic polynomial, having the primitive nth roots of unity as roots.

There are phi(n) primitive nth roots of unity

[midnight pasture]

i downloaded disquisitiones arithmaticae and i cant get mnyself to start reading it

[rocky cloak]

For example for n=4, the 4th roots of unity are
1, i, -1, -i
and out of these i and -i are primitive

[midnight pasture]

yes i get that

wait no

is the definition of primitive like

it extends $\mathbb{Q}$

[cloud walrus]

daff

[rocky cloak]

We say something is a primitive nth root, if it's not an mth for for any m<n.

For example (-1) is not a primitive 4th root of unity because (-1)^2 = 1, so it's a 2nd root of unity

[midnight pasture]

[chilly ocean]

If V is a finite-dimensional vector space over K, and if V^E is the extension of scalars to E, then is det(A) = det(A^E) for all linear operators A on V?

[mighty kiln]

It's represented by the same matrix right

[chilly ocean]

right… thanks

[chilly radish]

Another way: by abstract nonesense exterior powers commute with tensor products, and by uniqueness of the determinant we get that det on V^E is just det (x) 1

This is morally more correct since you don't need to choose a basis

(Also for what arki said to be correct, you do need to choose a basis for V^E correctly)

[languid trellis]

I can't quite see how the universal property of free modules is being used here. I would imagine that the universal property shows only that there is a Free R-module on N (constructed in dummit and foote by considering set functions N-> R with finite support and defining addition and scalar multi as expected) through which the homomorphism can be factored. I have no idea where the free Z-Module F on S x N came from. Can anyone advise?

[tardy hedge]

I was thinking of this as being like a template that eventually induces the tensor product map that we want

[mighty kiln]

Universal property of free R-modules:
An R-module map F(S) → M is the same thing as a set function S → L

[rocky cloak]

Presumably they define the tensor product as the free group on SxN module the tensor product relations

[mighty kiln]

Here the set function sends (s, n) to sφ(n)

[languid trellis]

Yep

[mighty kiln]

And so we get a Z-module map F(S×N) → L

[languid trellis]

Wait I think I see what's trying to be done

[mighty kiln]

By mapping each "basis element" via this

[languid trellis]

We're defining a set map S x N -> L then applying the universal property of free modules

[tardy hedge]

yeah

[languid trellis]

I don't see why F(S x N) should be a Z-module?

[mighty kiln]

It's the free Z-module over the set S×N

F(S×N) is denoted F in the text

[languid trellis]

Does the universal property not sugges that F(SxN) ought to be an S-module?

[tardy hedge]

Consider L a Z-module

[mighty kiln]

We're taking R = Z

So F(S×N) is a Z-module

[languid trellis]

oh right I see

all this work is happening outside of our assumptions about L

[mighty kiln]

Yea

F(S×N) is the direct sum of a bunch of copies of Z, indexed by S×N

[languid trellis]

and clearly we can make L a Z-module because L is an abelian group already

[mighty kiln]

So you can think of each (s, n) as a basis element of F

[mighty kiln]

Yup

[languid trellis]

we can write each element of F(SxN) uniquely as c1(s1, n1) + .... + cn(sn, nn) by definiton of the free module, and becuase Z is commutative we can think of the elements (si, ni) as a basis?

[mighty kiln]

Yup

[languid trellis]

as in rank is well defined

ok

[tardy hedge]

It doesnt have finite rank right?

[languid trellis]

I'd imagine not

[tardy hedge]

Its just the collection (si,ni)

[languid trellis]

perhaps if S, N are finite sets

[mighty kiln]

I think they mean there is a unique rank since Z is a cring

Though besides the point

[languid trellis]

ok i'll carry on working through this and come back when I get stuck

[tardy hedge]

What is the rank of a free module again? Size of generating set isnt it?

[languid trellis]

thank you arki

[mighty kiln]

Yea the cardinality of a basis

Which can fail to be well-defined for general rings I think

[languid trellis]

yes

an example is given in 10.3 ex 27

[tardy hedge]

Ok, but if cardinality of a generating set is not finite then ?

[mighty kiln]

Then it's that cardinality

[tardy hedge]

Ok

[languid trellis]

oh. so we have the induced map F(SxN) -> L, and then as the subgroup H which generates the module relations is mapped to 0 by phi, we can quotient out by it to get a map F(SxN) / H = S \otimes_R N -> L mapping (s \otimes n) -> s \phi(n)

plaintext notation overload

but essentially the module relations are mapped to 0 by the homomorphism, so we can safely quotient without changing anything

and this makes the induced map an S-module hom

hmmm

I'll need to reread to get all the details but I think Iget the idea

thanks all

[rotund aurora]

what is this marked in yellow?

I assume G^* is maybe the Frattini subgroup?

but what is G^p[G, G]

[rocky cloak]

Presumably G^p are the pth powers in G, and [G, G] is the commutator subgroup.

[rotund aurora]

so is G^p [G, G] the group generated by G^p and [G, G] ?

[rocky cloak]

Yeah

[rotund aurora]

but is this ${ xy : x\in G^p, y\in [G, G]}$?

[cloud walrus]

croqueta3385

[rocky cloak]

Should be yeah

[median pawn]

what's happening in the last line?

it seems to only tell us that the ideal generated by x^2 - y^2 is contained in I

I ⊂ <x + y> seems weird

[chilly ocean]

This is what I was looking for, thanks 🙂

[lone niche]

I think that last line is false. I don't think x^2+1 is in the ideal generated by x+y for example. Or at least I'm finding it very hard to see how that would be.

[median pawn]

makes sense, thanks

yeah they probably wanna do <x^2+1,x+y> instead, which is a maximal ideal

[tribal moss]

The reasoning doesn't really make sense even then, I think.

"Look, this element of J is also in I, so I is a subset of J" is mad.
Even
"Look, this element of J is also in I, so I is a subset of I+J" wouldn't be much better, even though the conclusion then happens to be true.

[lone niche]

I think whoever wrote the solutions copied it wrong from math stackexchange. It looks similar to the proof that I is not prime in here: https://math.stackexchange.com/questions/4592814/find-a-prime-ideal-which-contains-x21-y21

Which is kinda funny to me

[tribal moss]

Hmm, I don't think either is necessarily copied from the other. Considering (x+y)(x-y) is a fairly standard approach for dealing with attempts to make too many ring elements with the same square, and there's not much else (other than the problem statement in common between the MSE question and the above quote.

[median pawn]

yeah i think the reasoning given is just wrong

also they did a lot of work to show that <x^2+1,y^2+1> is not prime

this is immediate from their observation that (x+y)(x-y) is inside the ideal lol

[tribal moss]

I suspect whoever wrote it didn't feel confident enough to just assert that neither x+y nor x-y is in <x²+1,y²+1>.

[median pawn]

but it's obvious from a degree argument right

[tribal moss]

Yeah -- given sufficient experience with such degree arguments.

[lone niche]

how would you prove that x^2+1 is not in <x+y> btw? I couldn't figure out a rigorous approach, except computing grobner bases, and thus decided to just use Sage for it.

[tribal moss]

Having two variables and two generators might cause someone to doubt their intuition there, especially if they have only seen those arguments carried out in detail in the single-variable case.

[tribal moss]

Every polynomial in <x+y> evaluates to 0 when you plug in x=y=0. But x²+1 doesn't.

[median pawn]

just to make sure, anything in that ideal looks like P(x,y)(x^2+1) + Q(x,y)(y^2+1), and deg(P(x,y)(x^2+1)) >= deg(x^2+1) = 2, similarly for the other term

hence x+y, x-y aren't in the ideal

[lone niche]

Ahh I think I see, by the ideal-vanishing set correspondence?

[tribal moss]

Well, more elementarily because every element of <x+y> has the form (x+y)·P(x,y) for some P, so when we evaluate that at (0,0) we get 0·P(0,0) which is zero no matter what happens in P.

[median pawn]

yep

it's sad how that's a quals solution lol

they should be more careful

[tribal moss]

I feel one needs a bit of additional footwork to convince oneself that the higher-degree terms from P(x,y)(x²+1) and the ones from Q(x,y)(y²+1) can't somehow cancel each other out.

[lone niche]

Sorry, I meant the part that follows. Since V(<x+y>) is not a subset of V(x^2+1), then <x^2+1> is not contained in <x+y>, where V is the vanishing set.

[tribal moss]

Well, yes, that too.

What I did was just to show one concrete point where x+y vanishes but x²+1 doesn't.

[lone niche]

Yeah, I was trying to see how that implied x^2+1 is not in <x+y>.

thank you

[swift tundra]

Oh lol, yeah that’s right. That’s interesting

[long obsidian]

(2) Is an ideal of the integers Z. Is (2) not an ideal of the Z algebra Z[x]? What is the ideal generated by 2 if not

[tender wharf]

well ideal generated by a thing is always gonna be an ideal

but if you mean "do the even integers form an ideal in Z[x]" then they don't because you are missing tons of stuff

well you're taking polynomials with integer coefficients and you're multiplying them by 2

to get (2) in Z[x]

[long obsidian]

Maybe the ideal of even integers in Z[x] is (2)[X]

[tender wharf]

what is (2)[X]

if you mean like

polynomials n + nx + nx^2 + .. + nx^k where n is even that's not an ideal in Z[x]

[long obsidian]

Ah shoot okay

[median pawn]

that's a valid point. how'd you fix that?

[lone niche]

You can use the same argument above. We have that the ideal $I$ does not vanish identically on $(0,0)$, but $<x+y>$ does. In other words, the vanishing set $V(I)$ is not contained in the vanishing set $V(<x+y>)$, and thus by the inclusion reversing property of $V$, $<x+y>$ is not contained in $I$. Ditto for $<x-y>$.

[cloud walrus]

kibocchi

[barren sierra]

Ok so if $L = K(\alpha)$ then clearly the only automorphism that stabilizes $\alpha$ is $\text{id}_L$ as any other non-identity automorphism must move $\alpha$ somewhere else (if it didn't then it wouldn't be non-identity). The converse direction I'm not so sure about.

[cloud walrus]

Spamakin🎷

[barren sierra]

The thing jumping out to me is that a polynomial is irreducible if and only if the Galois group acts transitively but where could I get a polynomial from. The minimal polynomial of \alpha?

[lone niche]

Consider some $\beta\in L$ such that $\beta\notin K(\alpha)$, and then consider the automorphisms that move $\beta$.

[cloud walrus]

kibocchi

[barren sierra]

wouldn't I need to know something about the minimal polynomial of beta over K?

because such an automorphism moving beta couldn't swap alpha with beta so I'd need more information surrounding beta I think

[lone niche]

For the argument I'm thinking, it would suffice to consider the minimal polynomial of beta over K(a).

[barren sierra]

ah since beta isn't in K(alpha) there exists some other root gamma and so I can always permute beta and gamma and then by irreducibility of the minimal polynomial I can extend this to an automorphism of all of L (at least iirc some theorem like that exists)

errr maybe I have to say something about how gamma should exist in L first, hm

[lone niche]

And these permutations will fix alpha, since we are over K(a), so these permutations are non-trivial automorphisms contradicting the Stabilizer condition.

[barren sierra]

right that much makes sense

[barren sierra]

just gotta figure this detail out

[lone niche]

L is Galois over K

[barren sierra]

ah so minimal polynomial splits and roots are distinct

[lone niche]

yupyup

[barren sierra]

and since degree > 1 another root does exist

ty ty

[long obsidian]

How do you check if a commutative algebra (A,f:R->A) is a sub algebra of another commutative algebra (B,g:R->B) over the ring R? If there is an inclusion map i: A->B does showing A is a sub algebra of B just amount to showing if=g commutes?

[tulip glacier]

~~Uhh, I don't know what the two functions, f and g, from the underlying ring to the algebra are.

That aside, yes, when you have any two algebraic structures A and B, you can prove that A is a subobject of B by constructing an inclusion (an injective structure-preserving map) from A to B.~~

[mighty kiln]

A commutative R-algebra by definition is a ring morphism f: R → A into a commutative ring

[tulip glacier]

Ohhh, I only knew the "algebra over a ring for which the product is commutative" definition. I see why this works too. Thanks!

[tulip glacier]

To update my answer now (thanks to @mighty kiln):

When you have any two algebraic structures A and B, you can prove that A is a subobject of B by constructing an inclusion (an injective structure-preserving map) from A to B.

Here, the structure of a commutative algebra is made out of:

• a commutative ring A
• a ring homomorphism f: R -> A.

So, our structure-preserving inclusion i: A -> B will have to:

• respect the structure of the ring A - that is, it has to be a ring homomorphism from A to B.
• respect the structure of the maps f: R -> A and g: R -> B. That is, if = g.
• be an inclusion, that is an injective function.

So yes, this is what a morphism between commutative algebras will look like.

[dark wave]

So what do wreath products do?

[chilly ocean]

What does R^2 mean where R is a ring

I guess {r+r'}

[south patrol]

The direct sum R (+) R of groups equipped with componentwise multiplication

[rotund aurora]

how can I show that if G is a profinite group and x-->x^n is surjective (for n a positive integer) then it is also injective?

[dim widget]

I think the best way is to view G as instead an inverse system of finite groups G_i. Wolog we can assume all the transition maps are finite, then multiplication by n defines a surjective map on G if and only if it induces a surjective map on each G_i. But G_I is finite so this means that each G_I has order prime to n, so G has no n torsion.

[dim widget]

^

[rotund aurora]

suppose x^n=1 for x!=1. Since G is profinite, there exists an open normal subgroup U not containing x. The nth power map will still be surjective in G/U but this is a finite group and x^n=1 mod U with x!=1 mod U

Doesn't this also work? @dim widget

I was a bit confused when looking at the diagram because it doesn't work when "surjective" is replaced by "injective", but I guess that's just totally different (eg. multiplication by p is injective in Z_p but not surjective, and it is not injective on the finite quotients)

[dim widget]

Yeah there is some duality at play, certain properties are easy to check for inverse limits, but the dual properties are easy to check for direct limits

and often it’s more tricky to check the other way around

[rotund aurora]

unrelated but what is the G-module structure here? Lambda=Z[G]. Is X taken with a trivial G action?

[dim widget]

Yep that works too, it’s similar

[rotund aurora]

before they said this

[dim widget]

It’s the action gf(\lambda) = f(g^-1\lambda) or f(\lambda g)

[dim widget]

Okay apparently it’s the first one then

[rotund aurora]

alright, ty

[dim widget]

nw

[long obsidian]

Is it correct to say that if A is a commutative algebra over a ring R and S is a subset of A then the R-subalgebra generated by S is the intersection of all R-subalgebras of R which contain S?

I get that the generated algebra is the smallest sub algebra that contains S but the "smallest" is vague. Smallest means this condition right?

[dim widget]

Yep this condition is closed under intersections so to take the smallest X you just take the intersection of all X’s

[coral spindle]

Supposing $n = n_1 + \cdots + n_k$, the obvious subgroup of the symmetric group $S_n$ isomorphic to $S_{n_1} \times \cdots \times S_{n_k}$ has index given by the multinomial coefficient $\binom{n}{n_1, \dots, n_k}$ which has a nice combinatiorial interpretation.

\bigskip
\textbf{Question:} this suggests there should be a combinatorially nice and reasonably canonical choice of representatives for the cosets of this subgroup. Anyone know?

[cloud walrus]

Boytjie

[coral spindle]

Maybe it's actually really obvious lol

[rotund aurora]

is there a "nice" description of H_1(G, A) for an arbitrary G-module A? Something similar to H^1

[dim widget]

I think the usual choice of representatives are the minimal length elements in the cosets. Those are the permutations s such that s(1) < s(2) < … < s(n_1) and the same for the other blocks

[coral spindle]

I suppose there isn’t a nice list of these things

Yeah if I were doing coxeter combinatorics that would be the right way, alas I am not

Good to know though

[rotund aurora]

When the action of G on A is trivial I think then H^1(G, A)=G^ab otimes_Z A. But in general it seems somewhat more annoying to write down

[coral spindle]

No wait it is super nixe & I misunderstood looool

nice*

[rotund aurora]

If H is a subgroup of G, is Z[G] a free Z[H]-module?

[topaz solar]

Cosets should be disjoint for multiplying monomials, so I’d believe it?

[rotund aurora]

yeah taking representatives of G/H works for a basis

[chilly radish]

Yes, choose coset representatives

[rotund aurora]

shouldn't the left be H^q(G/H, A^H) ?

[nimble folio]

Assume $R$ is commutative. Let $I$ and $J$ be ideals of $R$ and assume $\mathfrak{p}$ is a prime ideal of $R$ that contains $IJ$. Prove that either $I$ or $J$ is contained in $\mathfrak{p}$.
\begin{proof}
Let $\sum a_i b_i \in IJ$, where $a_i \in I$ and $b_i \in J$. Since $IJ \subseteq \mathfrak{p}$, $\sum a_i b_i \in \mathfrak{p}$. So $a_i \in \mathfrak{p}$ or $b_i \in \mathfrak{p}$. Hence $I \subseteq \mathfrak{p}$ or $J \subseteq \mathfrak{p}$.
\end{proof}

[cloud walrus]

clubsoda14

[nimble folio]

Is this chill

my worry is the part where I write Σaibi ∈ p implies ai ∈ p or bi ∈ p

i dont know if thats valid

[lone niche]

I don't think it's valid.

[nimble folio]

Why

Σaibi ∈ p ⟹ aibi ∈ p?

since p is an ideal which means it is an additive subgroup?

[lone niche]

You made a big logic jump exactly where you are worrying about. Try to expand it out and see that you can't using the definition of prime ideal.

[nimble folio]

I think I need a little help understanding why

[lone niche]

it could be that a1, b2, a3, b4 are in p

why is it that a1, a2, a3, a4 are in p

or b1, b2, b3, b4

[nimble folio]

Uh

Well my thinking was that If IJ is the set of all finite sums of the form Σaibi, and since IJ is an ideal (an additive subgroup), wouldnt that imply there are two different sums such that Σaibi - Σai'bi' = aibi?

[lone niche]

simpler than that, by definition aibi in IJ

[nimble folio]

then wouldnt that mean my proof is valid

if aibi is in IJ and IJ is a subset of p, then aibi is in p

and if ai is an element of I and bi is an element of J, and since p is prime, then ai is in p or bi is in p

hence I is a subset of p or J is a subset of p

[lone niche]

that's where this example comes in

you know that a1b1 in p, a2b2 in p, and so on

so that implies that a1 or b1 in p

a2 or b2 in p

what if a1 in p and b2 in p

[nimble folio]

Uhh

[lone niche]

do you see the logic jump? i can give you a hint of where to go next

[nimble folio]

Yes please

also i dont see the logic jump

[nimble folio]

why is this bad?

[lone niche]

because then we dont have the guarantee that a2 in p

and we thus cant prove I is contained in P

alternatively we arent guaranteed b1 in p

and thus we cant prove J is contained p

[nimble folio]

Ohhhhh

Wow that makes perfect sense

Ok I see where I need to go with this

[tulip saddle]

This is probably too elementary for this channel, but my question is: is the induction here really necessary?

[glad osprey]

Depends on how formal you want to be, but I would say $a_1 \cdots a_k a_k^{-1} \cdots a_1^{-1} = e$ should be obvious without having to do the induction explicitly

[cloud walrus]

sheddow

[median pawn]

what's their explanation for $\tau$ not being able to fix $\sqrt[3]{5}$? the real subfield is $\Bbb{Q}[\sqrt[3]{2}]$ ofc

[cloud walrus]

hausdorff

[rocky cloak]

The real subfield has degree 3, Q(cuberoot(5)) has degree 3 and is real.

So if cuberoot(5) is in the field, then this will be the entire real subfield

[median pawn]

hmm

so are you saying that if we have two real subfields of the same degree, then they are equal?

[rocky cloak]

I'm saying if you have a subfield with the same degree as it's superfield, then they are the same

[median pawn]

a = cuberoot(5)
b = cuberoot(2) for simplicity.
we assume that a \in Q[b, w], so Q[a] \subset Q[b,w]. Q[a] has degree 3 over Q.
as Q[a] is real, we have Q[a] \subset Q[b], since Q[b] is the real subfield of Q[b,w].
how do you get Q[b] \subset Q[a]?

or do you not need that

[rocky cloak]

Q[a] < Q[b] and both have degree 3, hence they're equal.

Or if you like
3 = [Q(b):Q] = [Q(b):Q(a)] * [Q(a):Q] = 3[Q(b):Q(a)]
So [Q(b):Q(a)] = 1

[median pawn]

ahh right, thank you!

[barren sierra]

I've got zero ideas for (1). $\mu_m(K) = {x \in K ~|~ x^m = 1} \simeq \langle \zeta \rangle$ where $\zeta \in K$ is a primitive $m$-th root of unity.

[cloud walrus]

Spamakin🎷

[cloud walrus]

clubsoda14

[barren sierra]

I'm confused? How did you show that $\varphi^{-1}(\mathfrak{p}) \in \text{Spec}(R)$ always? If you did that then you're done (the other portion of the ``or'' would be irrelevant.

[nimble folio]

I'll type it up give me one second

[cloud walrus]

Spamakin🎷

clubsoda14

[barren sierra]

do your ring homomorphism have to be unital btw?

[nimble folio]

I think so let me check

Yes R is a ring with identity with 1 != 0

[barren sierra]

no no, the homomorphisms

does 1 have to map to 1?

or is the 0 morphism a valid ring homomorphism?

[nimble folio]

I have no idea lol

Should I assume it does not always map 1 to 1?

[barren sierra]

because if it is then you should find a map where $\varphi^{-1}(\mathfrak{p}) = R$

[cloud walrus]

Spamakin🎷

[barren sierra]

such (non-unital) morphisms exist

[cloud walrus]

clubsoda14

[nimble folio]

I think i see what you mean

[barren sierra]

because what you've shown here still holds if \varphi^{-1}(p) is the whole ring

[nimble folio]

I see

[barren sierra]

yea your proof should start "if \varphi^{-1}(p) = R we are done. So suppose not" then show \varphi^{-1}(p) is in Spec(R)

[rocky cloak]

Here's sort of a fun proof:

Let w be a primitive rth root of unity (zeta^m/r) and let c be a root of x^r - b. Then
||x^r - b factors as (x-c)(x-wc)...(x - w^r-1 c)||
||If it's not irreducible, then some polynomial dividing this will have coefficients in K. So w^something c^i is in K for some i. Hence c^i is in K, hence c^gcd(i, r) is in K.||
Now
||let d=gcd(i, r). Then (c^d)^m = a^d, which implies d=r.||

[barren sierra]

why do we know x^r - b must factor like that?

why can it not factor as say 2 polynomials or something, or 3 or whatever.

[rocky cloak]

Because a polynomial always factors into (x - the roots)

[barren sierra]

over some field yes but not necessarily K

[rocky cloak]

I'm confused what you're asking. c is not in K

Like the roots of x^r - b is c, wc, w^2c, ...
so it factors as (x-c)(x-wc)... over the algebraic closure or over K(c) or whichever field you want to work in where the polynomial splits

[barren sierra]

nvm I completely misread what you wrote so I see it now. I thought you were just showing that factorization couldn't lie in K[x] and I was lost

[tardy hedge]

When this map is not injective, how can we intuitively think is going on , as it relates to extending N (R-mod) to an S-module?

Of course on the extreme end, when the kernel is all of N, there is basically no consistent way to “extend scalars”, but what is going on in those in-between cases?

[tardy hedge]

Probably working through some concrete examples would help

in this case, how are we even mapping something like I((mi,ni)+(mj,nj))?

based on the definition of that inclusion map?

I was trying to understand why this map is in general not a group homomorphism

[lone niche]

(mi + mj) tensor (ni + nj)

[still dew]

This is driving me mad

For the first part u(U n v)
I consider an element ug g in U n V and then ug h g^-1u^-1 is in u(U n v) somehow

But I am having trouble showing this

[prisma umbra]

Hints

[still dew]

First have u figured out the identity

[rocky cloak]

So h = xy with x in u and y in U n v.
So g xy g^-1 = gxg^-1 gyg^-1

Then you can just go down a checklist of properties. Like it's gxg^-1 still in u, is gyg^-1 still in U, is gyg^-1 still in v

[prisma umbra]

x.y=x
Y=3/2

[still dew]

Oh i forgot u is non an element

😢

[prisma umbra]

x+2y-3=3/2

2x+4y-9

[still dew]

Ah okay I'll try this now , what's the point of this excercise anyways 😭

Although Thank you so much

[prisma umbra]

y=9-2x/4

[rocky cloak]

That normal subgroups intersect nicely I guess.

Or maybe just practice in symbol pushing

[still dew]

symbol pushing 😭

[rocky cloak]

This exercise is kinda weird, since the operation given doesn't seem to form a group.

Or I must be misunderstanding something, like they can the group (G, *), but then they write x . y

[prisma umbra]

Why?

Which operstion failed?

[rocky cloak]

Well, like you said the identity should be 3/2, but
3/2 . y = 3/2 + 2y - 3

I suppose it's possible that 3/2 is the only element of G, since they didn't specify that

[prisma umbra]

Ohh it fails

[rocky cloak]

Well, it fails unless y=3/2.

So I suppose you're just supposed to realize G = {3/2}

Still kinda weird though

[prisma umbra]

How did you feel it fails for operation?

[rocky cloak]

How I feel about it?

[prisma umbra]

It has only one element to be a group?

[rocky cloak]

Yeah, you can have a group with just one element. It just seems like a kinda stupid example

[prisma umbra]

Tq very much

Can I post next doubt?

[still dew]

@rocky cloak I got through the proving normal subgroups part , I am now stuck on the isomorphisms

Any hints

[rocky cloak]

Do you know the isomorphism theorems?

[still dew]

Yea

[rocky cloak]

Then use those

[still dew]

It's not easy to make that the kernel

But I'll try

[rocky cloak]

Not the first isomorphism theorem. Notice your problem contains products quotients and intersections of subgroups

[still dew]

But it's reversed for the second iso thm (kind of)

Like H/AH kind of stuff

Idk 😭

[prisma umbra]

Given answer is A

[rocky cloak]

So for the first isomorphism consider for example N = u(U n v), S = (U n V).

Then it's asking you to show
SN/N = S/(S n N)

[still dew]

Ah okay this was easy then , my bad I didn't notice (Un v) (UnV) is UnV

Okay wait i might be able to do the second part

Thanks for the help so far

[rocky cloak]

Might be a mistake in your given answer then

[prisma umbra]

Then it will be B?

[still dew]

It should have a √2 instead of a √3 ig

i√2

Oh wait

It's 4

💀

I'm so done 😭

c

It should be c

[prisma umbra]

Why C?

@still dew

[still dew]

Uh what are the roots of x^2+4

[prisma umbra]

2i

[still dew]

||And Q(2i)= Q(i)||

The smallest field that contains both those roots is Q(i)

[prisma umbra]

I see

Why not A and B

[still dew]

They don't contain the required roots

2i is not in Q(√3i)

Can u see why

[prisma umbra]

Yes

√3i,2√3i...

[still dew]

Yeah whenever you have a i in sth there's also √3

(still a better lovestory that Twilight 💀)

[prisma umbra]

What what?

[prisma umbra]

Have a look at it?

[rotund aurora]

Bump

[untold basalt]

Prove solvability of automorphism group.

[untold basalt]

can someone help me out here please?

[still dew]

Oh i actually found why that excercise was significant it's zassenhaus lemma, which later helps in the proof of schreier refinement theorem

[tardy hedge]

How?

I mean ok sure but idk why its that just from the definition of the function alone

[prisma umbra]

Hints??

[still dew]

Wait this is a #point-set-topology question

[lean sail]

anyone here read pinter's algebra and saracino's algebra (a first course)? someone in book recommendations said pinter might be the way to go, just wondering if anyone read both.

[lone niche]

Add them before you map them

[lone niche]

I tried pinter, didn't like it since a lot of the important results were left to exercises (which is something some people like apparently).

[lean sail]

not a big fan of that, not my first time reading through something, anyway, i looked at pinter and i didn't see too many proofs in the explanations

[cloud walrus]

clubsoda14

[nimble folio]

I found a different proof online that is a lot longer, but is this fine?

they defined the natural projection S -> S/M and then showed that R/φ^-1(M) is a field

[lone niche]

why is varphi(I) not all of S? (in your chain of inclusions m phi(I) S)

[nimble folio]

if I is properly contained in R and phi is surjective then wouldnt phi(I) be properly contained in S?

Or is that not necessarily true

[tribal moss]

Why would that be?

(It would obviously be true if phi were injective).

[nimble folio]

Ohhhh

Darn

Let me think about it for a little

[rocky cloak]

Why is this a lot longer? Like you've explained the whole proof and it's quite short.

[nimble folio]

I think you're right

It does not seem that hard to show

Thanks for the help

[dull ginkgo]

What if R is the complex numbers

Wouldn’t the group-forg’d conjugation automorphism be distinct from the multiplication endomorphisms

[rocky cloak]

That's right 👍

[tardy hedge]

But the addition in the free Z module is .. not component wise though?

(m1,n1)+(m2,n2) is not (m1+m2, n1+n2) …?

Its just … (m1,n1)+(m2,n2)

[main flower]

I am having trouble with a question regarding group actions from a practice test I am working through. I will include a picture of the question along with some notes I have taken about the problem. Also for terminology, G*(alpha) is an orbit and G_(alpha) is the stabilizer. Any help would be greatly appreciated

[rocky cloak]

a) you've noticed that the stabilizer are the matrices of the form [1, b; 0, d] so that's p(p-1) matrices. The orbit has size p^2 - 1, so in total that's (p^2 - 1)p(p-1) elements. (p=3).

b) that's correct, but your missing the proof I guess.

c) [1, 0] and [2, 0] span the same one dimensional subspace. The 4 subspaces should be the ones spanned by [1, 0], [1, 1], [1, 2] and [0, 1] respectively.

[main flower]

Thank you! And dang I can’t believe I didn’t that on part C I appreciate it!

[lone niche]

lemme check, is this Dummit and Foote?

[lone niche]

I found the page

[lone niche]

We aren't doing addition in the free module. The free module is the middle step, we are doing addition in M times N (which was previously assumed to be just the cartesian product, but D&F is probably abusing notation)

[tardy hedge]

You just start with a cartesian product M x N , and then define free Z-module with M x N as generating set

I dont see where there is a direct product of modules structure going on

Where are we defining a (m1,n1)+(m2,n2) = (m1+m2, n1+n2) structure?

The map M x N -> M tensor N is mapping free Z module to the tensor product

[long obsidian]

Say M is an R module generated by a set {m_a}_{a\in A} indexed by A and another R module called N.

A hom F:M\to N is determined by its image on the generators. How do you exactly say that?

Say I had a set map f:{m_a}_A->N is this just an application of the universal property of the free product F:Free({m_a}_A)->N? This isn't exactly the original but I have a sense that it's something like this. I know if {m_a}_A is a generating set there is a map from F:Free({m_a}_A)->M but I'm not sure how to put them together

[rocky cloak]

I mean you can involve free modules and stuff if you want I guess, but it's really just every element in M is of the form
m = Sum_a r_a m_a
so
f(m) = Sum_a r_a f(m_a)

[long obsidian]

so just think of it like a linear combination okay. I guess I was worried that it wasn't obviously a hom.

To be honest I also had an interest in this question also as a statement about commutative R algebras. So the linear combination thing didn't make it clear to me that the hom would respect the multiplication of the commutative algebra

[lean sail]

how often is the concept of preimage (for functions) used in group theory?

[rocky cloak]

What do you mean "it wasn't obviously a hom", what is "it"?

[fading field]

a lot. a sort of fundamental example is that the preimage of 0 of a group homomorphism is a normal subgroup of the domain

[lone niche]

It says that we have a map f, from M x N (left ambiguously) to the free module on M x N. And then we pass to the quotient M tensor N, which means we have a map from q, from the free Z module on M x N to M tensor N. Thus we have a map which is the composition of those two maps: qf. This map is a map from the cartesian product to M tensor N, not from the free module. Now, notice that M x N is the same notation for both the cartesian product, and direct product. We can thus infer that the authors are using it as the direct product, since they are talking about homomorphisms, how the map qf does not satisfy the homomorphism condition, which would not make sense if we were only considering M x N as the cartesian set.

[fading field]

and more generally, the preimage of any normal subgroup is again normal in the domain, and you can often say things about the quotients of groups using facts like this. many exercises will reduce to finding homomorphisms such that the preimage of 0 is the “right set”

and they’re used pretty much everywhere you go in math, not just algebra, so it’s worth being really familiar with it

[lean sail]

i know what they are, but someone on the server was going on about them at length, i thought, maybe i dont know enough about preimages

[fading field]

ah lol, most of the interesting things you can say about preimages are about preimages of special sets, when you know stuff about both the map and the structure of the objects that you’re talking about

[lean sail]

what all do we need to know about preimages? i studied them briefly in a foundations of advanced math course at the graduate level

but not super indepth or anything

[rocky cloak]

Like how it interacts with intersections, unions, subsets etc is useful.

And more specifically in algebra how preimages of rings/groups/ideals/prime ideals are again that, and when their properties are carried over

[lean sail]

is that typically explained in undergraduate abstract algebra books whenever they talk about homomorphisms?

[next obsidian]

[rocky cloak]

Probably

[next obsidian]

Either proven or given as exercise. It’s easy af tho

[lean sail]

what does this mean: proven or given as exercise? like, what is proven?

[next obsidian]

It’ll either be proven in the book

Or given as an exercise

[lean sail]

whats the "it"

[next obsidian]

Everything, I suppose

[lean sail]

sorry, i dont understand

[next obsidian]

The former is maybe not because it’s just elementary set theory

[next obsidian]

🙂

[rocky cloak]

This is "it"

[fading field]

here are two exercises that you can do, and then you could probably learn everything else in an introductory algebra text (maybe even these exercises will be there, but this is my favorite exercise to give people about preimages):
Let f: X->Y be a function and let A be a subset of X and let B be a subset of Y. Show that f(f^-1(B)) is a subset of B, and that A is a subset of f^-1(f(A)) (and find examples for each where they’re not equal, eg. the subsets are allowed to be proper subsets sometimes)

[lean sail]

thanks, i'll write this down and take a look at it

[long obsidian]

I was talking about extending a set map f:S->N to a module map F:M->N where S generates M. I think you suggested if m=\sum_a r_a m_a then f(m)=\sum_a r_a f(m_a ).

But I think I should have stated that I'm actually interested in this problem about extending a set map in the case of commutative algebras where the extended homomorphism should also be a ring hom.

So that's why I wanted to understand this extension "categorically" in terms of that free object idea

[rocky cloak]

You can't in general extend maps from a generating set to the whole module. Always being able to do that is exactly the condition of being freely generated

[lean sail]

i think i got the first part:

[next obsidian]

https://tenor.com/view/music-cute-gif-10276268078873540233

I’m in London you’re in Paris

[fading field]

you don't know that f^-1(x) is just a single element, multiple things can map to x under f (or maybe nothing maps to x)

[lean sail]

ah yeah you're right, it's coming back to me...

you dont know if its one to one

[fading field]

plus, B \subset f(f^-1(B)) is not true in general, remember what you're trying to prove.

we could take this chat to #proofs-and-logic if you want

[lean sail]

sure

[dull ginkgo]

For one of the defs of a module

We have that ring R maps into End(A) where A is an abelian group

Does this have to be a monomorphism?

E.g. does the scalar operation need to be faithful?

Answered my own question, finite abelian groups are Z modules even though the order is a subideal of the kernel

jacobson says “R into End(M)” but when I see “into” I assume it’s mono

That kinda fucked with me for a moment

[dull ginkgo]

Actually I think (ord(G)) is the kernel

Since it’s the minimum n such that nx = 0

lol I look down to the textbook and my whackass pondering answered the next 3 questions lmao

Currying moment

[next obsidian]

https://tenor.com/view/butter-chicken-indian-food-rice-food-gif-18062269302545036995

[dull ginkgo]

Also I think if M is simple over R then End_R(M) is a division algebra actually

[next obsidian]

YES

This is important to rep theory

[dull ginkgo]

how do I unlearn something related to rep theory

[next obsidian]

Hammer

[dull ginkgo]

thank u

[dull ginkgo]

I don’t know much about rep theory

[tardy hedge]

Why is it kind of uncanny to see something related to real life (food) here

Its like … you guys are all … normal people, too?

[dull ginkgo]

@next obsidian so like, we have semigroup rings, like R[G] right… what if we had (left/right) modules over them?

Any use of those

[next obsidian]

Hahahahaha

This is literally rep theory

Take R = C

And then you get group rep theory basically

[dull ginkgo]

interesting

And why is this useful lol

[next obsidian]

Cuz of magical things

I honestly forget how the story really goes, but you focus on semisimple modules

Something about how C[G] is a semisimple ring tells you that you can always do this or something

Idk

[barren sierra]

at a high level we use representation theory because groups are hard and vector spaces are easy

[next obsidian]

My vector spaces are hard af

Cuz I’m not soft

You won’t ever find my VS lacking

[barren sierra]

🤨

anyways representations of groups allow us to prove nice things because honestly groups are boring

we don't care about them

we care about groups acting on things

and informations about these actions allows us to prove things about the groups

[dull ginkgo]

[barren sierra]

give us nice tools to classify them and such

by looking at their actions on vector spaces

rather than the group itself

[dull ginkgo]

Jacobson II covers some rep theory afaik

[barren sierra]

yea so burnsides theorem comes up alot in combinatorial and finite group theory

and that's proven nicely via representation theory

[next obsidian]

CHARACTER THEORY

But then had a purely group theoretic proof too

Later

[barren sierra]

yea but characters cooler

[dull ginkgo]

I really like the ring theoretic flair of 5

That R[\lambda] is a PID, and naturally T^n(x) = x so (lambda^n - 1)x = 0 for each d

Thus lambda^n - 1 is in B, but the divisors lambda - 1 and the sum of powers up to n - 1 of lambda aren’t (consider the (1,0…0) unit vector) so (lambda^n - 1) = B

[tender wharf]

is jacobson 1 worth doing or is it better to skip to 2

[dull ginkgo]

If you are a glutton for punishment

Most people don’t do 2 afaik

Just 1

[tender wharf]

I see

Okay then

I heard 2 covers the stuff 1 does though but apparently in more detail? Idk

[dull ginkgo]

I did 7 a silly way for shits and giggles

Assume a and b are homomorphisms from Q into End(M)

Consider the subfield K of End(M) consisting of x such that a(x) = b(x). Now consider that a(n) = b(n) for n in Z by Z being generated by 1 and a(1) = b(1) by ring homomorphism axioms

Thus Z embeds into K

So by universal property of Frac(Z) = Q

Q itself maps into K

I.e a(x) = b(x) for all x in Q

thus a = b.

:3

[cobalt heath]

Oh this seems interesting, but how is x an element of End(M)?

Ah maybe I misread and it's in Q

So you used that the only subfield of Q is Q?

[dull ginkgo]

Moreso field of fractions approach

[cobalt heath]

Edit: You proved subfield of Q*

[dull ginkgo]

Intuitively, if we can endow Abelian group A with a (commutative) R-module structure, then you can uniquely determine a RS^-1 module structure from the multiplicative subset S given it’s S-faithful

[cobalt heath]

Wow, seems you gained lots of algebra knowledge

[dull ginkgo]

Basically it’s that

I have basic localization stuff scortched into my head like what AM did to Ted with the HATE pillar after the Ore localization shit I did

Which now makes me wonder about modules and localizations about primes

[cobalt heath]

Localizations about primes?

Ah do you mean the local ring at prime ideal

[dull ginkgo]

Yes

Probably little instances where that matters though sadly

[cobalt heath]

I'd say, local rings are everywhere.

In AG

Now, outside of AG.. yeah.

[dull ginkgo]

Artinian commutative rings are local rings in a costume

Also apparently if we have a finitely-generated left R-module M and Jac(R) is the Jacobson radical then Jac(R)M is a sub module of M

[next obsidian]

Isn’t that just always true for any ideal

And any module

[dull ginkgo]

*Proper

[next obsidian]

Why does it need to be proper

Oh

I see what you mean

[dull ginkgo]

Nakayama generalization apparently

Just was reading through a paper about Jacobson radicals

[next obsidian]

Yeh

I mean this is how Nakayama is actually most generally stated

You prove that IM = M means there is x with x = 1 mod I such that xM = 0

For fg M

You get this by determinant trick

In the case of Jac(R) that x would be a unit, so M = 0

[dark wave]

Fun math fact: Every symmetric group except for S2 and S6 has itself as an automorphism group.
S2 has the trivial group, which is boring
Basically, if you take a set of any size except 2 or 6, and look at the permutations (ways to order things), the only maps that take a permutation, spit out a permutation, and preserve composition (automorphism) are given by conjugation
So a permutation p would define an automorphism P(x) that spits out pxp^-1 for any permutation x
The set with 6 elements has a permutation that doesn't have any analogues at different sized sets
The easiest way I could find to describe this is using graph matching
A (perfect) graph matching is a set of edges of a graph that don't have any vertices in common and each vertex is touching an edge in the matching
A graph can be factored into matchings that don't share any edges
Here is an example of a factorization of the complete graph with 6 vertices

Every permutation of the vertices defines a unique permutation of the factorizations

There happen to be 6 factorizations

No analogue of this automorphism exists for other symmetric groups