groups-rings-fields

[normal tree]

both statements are true

they just combine to give you a real constraint

thats not impossible

constraint: M≅End_R​(M)^2

it just means M cant be arbitrary if R≅M⊕M

u dont need anything fancy

if S is simple pick nonzero s ∈ S, then r -> rs is a surjection R -> S

semisimple means every submodule is a summand

so the kernel splits off

which gives R≅ker⊕S

[rocky cloak]

Consider for example R = M2x2(C) the ring if 2x2 matrices and M = C^2 the unique simple module.

Then R = M(+)M and
HomR(M, M) = C
HomR(R, M) = M = C(+)C

Note that the R-module structure of HomR(R, M) comes from R being a bimodule, and you cannot have R = N(+)N for a bimodule N.

Because then
Z(R) = End_bimod(R) = End_bimod(N^2) = M2x2(End_bimod(N))
the left being commutative and the right not being commutative

[brisk tendon]

Hom(R, M) = M = C(+)C ? @rocky cloak ? fr simple r-mods ?

[rocky cloak]

M is a simple R-module here.

Not sure I understand your question

[brisk tendon]

@rocky cloaki think HomR(M, M) = C and HomR(R, M) = M are true but M \notiso C ( + ) C fr r-mods

is not

[rocky cloak]

C is not an R-module, but M = C(+)C

HomR(M, M) is not an R-module, so this is to be expected

[brisk tendon]

sure, i guess so, i jsut meant it's kknd of true fr spaces but in genral know , sorry just chatting @rocky cloak 🙂

[crystal vale]

Say, I define the normal extension in the following sense:

F \subfield E, E is said to be a normal extension of F, if E is algebraic extension and for each a \in E, its all conjugates in E.

Okay this is equivalent to saying every F-morphism E -> C, where C is algebraic closure of F, is an automorphism of E.

Okay i got it, I was trying to extend this notion to any embedding F -> C, but yeah if it is not identity mapping then we can't talk about automorphism of E

[tardy hedge]

Notknow is back

[copper kestrel]

yayyyyy

[hidden cairn]

oh i see okay this makes sense.

[hidden cairn]

ah this is a simple proof

thank you!

[lofty garden]

all this time thinking Z/nZ is just a weird notation but finding out it's actually a factor ring over Z of the principal ideal generated by n

https://giphy.com/gifs/boruto-tbv-sasuke-its-so-peak-6YIqtuwLN4fVJ42hoM

[south patrol]

Or quotient group by the subgroup nZ

[rocky cloak]

I guess the notation nZ is more reminiscent of an ideal than a subgroup

[south patrol]

Yee maybe

[jaunty bear]

If an Abelian group G contains cyclic subgroups of order 4 and 6 then what other sizes of cyclic subgroups must G contain?

[marble hinge]

Definitely 2 and 3 :)

[quiet pelican]

And 1 s

12

[normal tree]

g contains a cyclic subgroup of order 12 and a cyclic group of order n contains a unique cyclic subgroup for every divisor of n.
so {1,2,3,4,6,12}

beyond the given orders 4 and 6, G must also contain cyclic subgroups of orders 1,2,3 and 12

so

[jaunty bear]

My professor gave me this to help me solve this
However the gcd is not 1 so I'm stuck

[normal tree]

to get around that, from your cyclic subgroup K of order 6, you can extract a cyclic subgroup of order 3 (every cyclic group has subgroups of every order dividing its size)

if b has order 6 then b^2 has order 3

H = <a> with |H| = 4
K' = <b^2> with |K'| = 3

since gcd(4,3) = 1

now your professors theorem should apply

and you get a cyclic subgroup of orderr 4*3=12

the divisors of 12 are what we said earlier

so G must contain cyclic subgroups of all those orders

[jaunty bear]

So tats the key move
Thanks I get it

[normal tree]

[candid patrol]

Nope ?!

G = C2 x C6

has not cyclic subgroup of order 12

[rocky cloak]

Does not have any element of order 4

[candid patrol]

Whoops, I red 2 and 6

yeah ok good proof

[normal tree]

💔

[normal tree]

❤️

[slow dagger]

hey so im unsure if this is the right chat to ask since this is coding theory, but could anyone give me a hint for this question or maybe tell me where im going wrong?
cause tbf my answer feels verry wrong since i think i coulda done this also without my initial claim.

(first pic is exercise, second is my answer)

[copper kestrel]

abstract algebra midterm assignment has officially been created in canvas but is currently ungraded

68% FAWK!!!!

almost a C though

best exam score i've gotten in abs alg

average was a 35.5/50 and i got a 34

[tardy hedge]

Canvas

[copper kestrel]

real

[maiden crater]

kind of silly. How is this a direct conseqeunce of lagrange's?

Sure, we have o(HK)|o(G)

We also know o(HK) = o(H)o(K)/o(HnK)

oh ,that's why?

[cursive spindle]

yes

[maiden crater]

Don't need help with this, just wanted to say this is a really pretty proof

[cursive spindle]

Looks similar to the classical one

But yeah I guess more linear algebra

Cool!

[twilit citrus]

I am trying to understand what $\mathbb{F}_q^{\times}/\mathbb{F}_q^{\times n}$ looks like

[cloud walrus]

al

[ivory anvil]

in what structures can a*a not equal -a*-a?

[rocky cloak]

So one possible way to prove this woulde be

0*x = 0*x + 0*x - 0*x = (0+0)*x - 0*x = 0

a*b + (-a)*b = (a-a)*b = 0

and similarly for x*0 and b*(-a).

Then (-a)*(-a) = -(-(a*a)) = a*a

The properties being used is that addition has (unique) inverses, and that multiplication distributes on both sides.

Removing the first property it might be less clear what -a means, unless maybe just some elements have additive inverses...

You could remove distributivity, but then there's no connection between multiplication and addition anymore.

[rocky cloak]

Do you know the structure of (F_q)^x?

[twilit citrus]

it's a cyclic group of q-1 elements?

[ivory anvil]

can ring multiplication just be left distributive/right distributive?

or does it have to be both sides

[rocky cloak]

Right, so do you know the structure of subgroups and quotients of cyclic groups?

[rocky cloak]

It needs both to be a ring, you could define something with only one side I guess

[twilit citrus]

subgroups would be just of order divisible to q-1 I guess

and the quotient would be another cyclic group of order gcd(q-1, n)?

that's all there is to it?

is there an algorithm to find the generator?

for each of these cyclic groups?

[twilit citrus]

this makes sense intuitively but while working out explicit computations in case of say F_5

I am not so sure

there the gcd is just one

so it's just trivial???

I don't see it

[rocky cloak]

Once you have a generator g for F_q^x then a generator for the subgroup would be g^n or g^gcd(q-1,n).

In general g^d for d a divisor of q-1 gives a choice of generator for each subgroup.

[twilit citrus]

huh so just raising the generator to the order of the subgroup gives me the generator? that's wild

[rocky cloak]

Depends what n is, but yeah for example if n=3 then
1^3 = 1
2^3 = 8 = 3
3^3 = 27 = 2
4^3 = 64 = 4
So F5)^3 = F5

[rocky cloak]

Index of the subgroup*

[rocky cloak]

I mean it should make sense, every element is of the form g^k, so to get every element of the form x^n you just do
x = g^k
x^n = (g^k)^n = (g^n)^k

[twilit citrus]

Right, I didn't mention n, it was 2 by default in my head lol. But I don't get it, what are you doing here?

are you taking a particular q?

here?

[rocky cloak]

Yes, 5. You said F_5

[twilit citrus]

right, right, as in order of group divided by order of subgroup

[rocky cloak]

Well if n=2, then it wouldn't be trivial. gcd(5-1, 2) = 2

[twilit citrus]

ohh holy shit

why tf did I not see that lmao

wow I am so dumb

[twilit citrus]

hmmmm

[twilit citrus]

oh right, my bad

thank youuu

[crystal vale]

Let f be a polynomial over F[x] such that it has multiple roots in its splitting field, can I get counterexample such that its splitting field is not separable extension.

where separation extension definition is an algebraic extension such that for every a in extension its minimal polynomial is separable

[velvet hull]

Take F = Z/pZ, and let f be any irreducible separable polynomial in F[x].
Then f(x^p) is an example of what you are looking for
Not exactly true, see below

in general this is kind of the "only way" that inseparable extensions can occur, there is always some kind of Frobenius map involved; and in the characteristic 0 case irreducible polynomials cannot have repeated roots

[rocky cloak]

If f is a polynomial over Z/p, then f(x^p) = f(x)^p

[velvet hull]

Ah, a bigger field is needed

[rocky cloak]

Let F = Fp(t) be the field of rational functions over Fp, then f(x) = x^p - t is an irreducible polynomial with only one root

[crystal vale]

Yes

[rocky cloak]

If you want it to have multiple roots you can do similar to what HChan said. Take some polynomial g over Fp and consider
f(x) = g(x^p) - t

Or I guess just take the product of two polynomials, but I guess you want it irreducible

[crystal vale]

Yes

[rocky cloak]

So that would work, or you can cook up lots of other examples.

x^2p + tx^p + t

is irreducible by Eisenstein for example

Like HChan said any irreducible nonseparable polynomial will be off the form g(x) = f(x^p^n) for some seperable irreducible polynomial f.

The key to make g irreducible is for one of the coefficients of f to not have a pth root.

So for example any irreducible polynomial over Fp(t) where one of the coefficients involves a single t should do

[astral ivy]

What are some steps to do when examining if 2 groups are isomorphic? I'll usually get a group product and then have to figure out if it's isomorphic to some other group product. I know some rather basic things are:

1. Group order - if not equal, they're not isomorphic (usually this is the trivial part so all group orders are made to be the same).
2. Cyclic or Abelian - only works for sorting out specific groups such as dihedral, symmetric and alternating group.
3. What's next? Their elementary/normal forms? Counting elements of certain order?

How do I do that and what should I do after those? I know how to find all elementary and normal forms but not sure what exactly am I supposed to compare. As for elements of certain order, how do I pick one and count all of them? Do I use the elementary form for that and how so? The demonstrator randomly picked elements of order 2 and turned out to be correct as the numbers mismatched although everything before passed. What if I picked 3 or 5 or something else? They could potentially match there but not on the others.

Sorry if this is too many questions I am just trying to understand the background of it, why and how we're doing what we're doing.

[rapid cave]

If your two groups are abelian use the fundamental theorem

Otherwise its harder

[chilly ocean]

Do you mean abelian and finitely generated?

[rapid cave]

Yeah... It seems like in this context the groups are finite

[thorn jay]

proving isomorphisms for infinite groups is a whole other beast lol

[charred iris]

well that's undecidable so

[thorn jay]

finite groups are too, no?

[charred iris]

checking if two finite groups are isomorphic are absolutely decidable

just go through every possible function

(this is stupid, but it is a working algorithm!)

[thorn jay]

ah i was thinking of another definition then

[chilly ocean]

it doesn't really make sense to talk about decidability for infinite groups in general

But for finitely presented groups it is indeed undecidable

[charred iris]

yes I did mean finitely presented

I would be surprised if there were a more practical way to check isomorphism of finite groups either though

[chilly ocean]

i'd be surprised if there's not a way that's much more efficient than this

but would probably still be quite inefficient

[charred iris]

We know that a lot of properties are preserved under isomorphism (we don't care about the ones that aren't, and as long as the property is phrased only using things like "there is g in G such that g is P" and not specific like "3 is P" then it is preserved), and so if we find that there's a mismatch between the groups, they're not isomorphic. In terms of figuring out what to check, it's basically either intuition about the groups (+ experience), or start with the easy stuff (like order of the group) and work your way up to the more complicated things.

So as you say, if you decided to check number of elements of order 3 or 5 they could have matched and you wouldn't have an answer, but 2 is smaller so it is more reasonable to start there. Of course, you could have exactly the same number of elements of every order and still have non-isomorphic groups (they can't both be Abelian in this case though), so this still won't work in general, but if it's easy to count it's worth trying.

There are sufficient conditions for two groups to be isomorphic though, (e.g. if two abelian groups have the same number of elements of each order then they are isomorphic) and so if you know theorems about them then trying to apply them is also a good way to go about things.

[novel star]

what is the point of towers of groups

(or fields or rings or modules or whatever)

[thorn jay]

things like subnormal series encode very important properties of groups

towers of fields are chains of field extensions and can be a tool to study field extensions. For example, to study K[α, β] / K it might be useful to study the tower K ⊆ K[α] ⊆ K[α, β]

filtrations of groups/modules also give rise to natural topological structures on them, and completions of these

also, constructions of certain objects can involve recursively constructing a larger object, yielding an infinite tower X0 ⊆ X1 ⊆ X2 ⊆ ..., and the union of all Xi is then the object youre looking to construct

[novel star]

hmm

so are they there to chunk problems into smaller ones

[thorn jay]

exactly

[novel star]

ok

ty

[thorn jay]

mathematicians like to do that a lot lol

[novel star]

how are these two conditions different?

oh wait

they extend different things

and 3 follows from 2

[tulip otter]

dont you also need 1 for 3 to follow?

you are given that E/k and F/k are in C, so EF/F is in C by 2. Now since EF/F and F/k are in C then EF/k is in C follows from 1. Is there a way to do it only using 2?

[novel star]

yes

[tulip otter]

ohh i see

[tulip otter]

btw its nice that you are using lang :chad:

[novel star]

lang is the only book that i could actually comprehend

[tulip otter]

I am using it and milne too

[novel star]

what is milne

[tulip otter]

milne has notes on field (and galois) theory

[novel star]

ohh

that may be helpful

[tulip otter]

he also has notes on alot of other stuff too like algebraic number theory etc..

[novel star]

gem alert

[tulip otter]

https://www.jmilne.org/math/CourseNotes/index.html

here you go, i posted the link because milne himself posted these notes and made them freely available as far as i know

[astral ivy]

Are you talking about the isomorphism theorems? I'm not sure I know what you mean when you mention theorems around Abelian groups.

[charred iris]

oh my edit earlier didn't go through (it did now)

I meant to give an example of such a theorem, and then say that if you know general theorems like "if G and H share these properties then they are isomorphic" then trying out those properties is not a bad shout

[rocky cloak]

How many elements of a certain order is usually my go to.

Otherwise comparing abelianization / center / derived series might be tractable to compute.

[astral ivy]

Yup we didn't do centers etc

So what's the general algorithm for this?

Group order, abelian, cyclic, elementary/normal form, count elements of certain order?

[rocky cloak]

Wouldn't really say there is a general algorithm. Do the ones that are easiest to check first or go for the one your gut tells you will work

[candid patrol]

Does someone have good references about capable groups ?

[supple ice]

on capability of finite abelian groups it’s a preprint of Sunic Z

[astral ivy]

Okay thx.

Can you tell me one more thing: How would I compare their elementary and normal forms. Should I even do it or is it just a waste of time?

Or maybe if I write out the elementary form would it be easier to count all the elements of certain order?

[rocky cloak]

I think just splitting into product of cyclic groups of prime power order is the easiest, but I guess it depends how the group is given to you

But if your group is finite abelian, just counting how many elements of each order is sufficient, so just doing that would be fine.

Whether it be easier to compute some standard form in order to do that or not might depend

[astral ivy]

Okay

Thank you

[white oxide]

Could I please have a hint for 7? Why is it that $r + s \sqrt{D}$ is not the root of a monic linear polynomial in $\mathbb{Q}[x]$, i.e. so that $r + s \sqrt{D} \in \mathbb{Q}$? I know this implies that $r + s \sqrt{D} \in \mathbb{Z}$, but how exactly is this a contradiction?

[cloud walrus]

okeyokay

[white oxide]

Oh never mind

If r + s \sqrt{D} \in Z then r + \sqrt{D} = n = n + 0 \sqrt{D} where n and 0 are in Z duh

[wise ember]

Is there a typo here...? I am not aware of the notation $M^{-T}$

[cloud walrus]

Varixiuqlhfbgraijbzjnqghppxnqmvw

[azure iron]

i would interpret that as inverse of transpose (or equivalently transpose of inverse)

[wise ember]

Works, thanks!

[novel star]

is the order of the galois group G of a finite galois extension K/k just the dimension of the extension

bc the primitive element theorem tells us that K=k(α) for some α in K

and every automorphism of K over k is an embedding of K into itself (and vice versa)

[rapid cave]

Yes

[novel star]

so then |G| is the separable degree

and K is separable so its just the dimension

idk y i said prim element thm

[ivory anvil]

why does distributivity need commutative addition?

[novel star]

wdym

[earnest hatch]

I’m pretty sure the other ring axioms force addition to be commutative

[next obsidian]

I don’t… think so?

I mean with a 1 you do get this

a + a + b + b = (1+1)(a+b) = a + b + a + b

Now subtract to get a+b = b+a

I have no idea if you can have a non-unital ring with nonabelian addition tho

It is incredibly non-obvious to me at the very least

Actually no it is easy

Let G be a nonabelian group and define the product to always be e

This is associative and kind of trivially distributes

[ivory anvil]

The axioms other than distributivity?

That doesn’t seem right

[next obsidian]

I just proved it when you have a unit and provided a counterexample when you don’t

lol

[ivory anvil]

With distributivity yeah

[next obsidian]

Oh Hurb

Yeah yeah my b

I think they’re saying when you have distributivity it forces commutativity

Not like, that it needs it, moreso it just makes it true

[earnest hatch]

Yes, sorry I was half asleep I definitely didn’t word that right

[ivory anvil]

if m is an axa matrix over a finite ring, does the probability that m is invertible decrease as a increases?

[noble nexus]

you'd want to know the size of the unit group I guess and look at the determinant function

provided the ring is commutative

[south patrol]

Over a finite field you can do a complete computation of the # of invertible matrices and the answer is yes

But also okay if you go from n x n to (n+1) x (n+1) then you can take any nxn matrix which is non invertible, chuck it into e.g. the top left corner and then make some non invertible matrices using this

[rocky cloak]

I guess for a general finite commutative ring, it's the product of local rings and then a matrix is invertible iff it is modulo the radical, so it reduces to the finite field case

[south patrol]

Nice!

[storm cipher]

hello , the question asks us to determine all subgroups of (C,+) which are connected space,

I'm really stuck, i first want to prove (idk if it is the way to solve it) that if G is a subgroup C which is not {0} then it contains a line : here is a potential starting point

[chilly ocean]

What's the simplest subgroup you can think of that isn't {0}?

[storm cipher]

vR with v in C \ {0}

[chilly ocean]

There are simpler ones

[storm cipher]

C ?

[chilly ocean]

can you think of some nontrivial subgroup that is smaller?

[storm cipher]

i don't think there is smaller if it is a connected space but idk

[chilly ocean]

oh ok, I thought you were claiming that if G is a subgroup of C which is not {0} then it contains a line

did you mean to say connected subgroup?

[storm cipher]

i don't know if it's the same definition for connected subgrouop and space but connected space A is a space in which for all x,y in A there exists an continuous application f [0,1] -> A such that f(0) =x and f(1) = y

Ok I made some researches i don't think there is even a solution to this question if there are not more hypothesis ...

[vapid dew]

Not sure if this fits here but it seemed to be the most suitable channel:

Suppose I’m given a finite set of known points in ℝ², and I’m allowed to perform only straightedge constructions: drawing lines through known points, and marking intersection points of such lines. I may repeat this process arbitrarily many times (even infinitely i.e. the ability to iterate the process without bound, so we can always do one more construction if needed, but each individual point is reached in finitely many steps), but every point I generate must be a deterministic result of previous constructions.

1. If I begin with 4 points with rational coordinates in a non-rectangular (or similar degenerate) configuration, can I generate all rational coordinate points i.e. ℚ²?

2.What if I include a fifth point (on top of the prior 4 rational ones) with at least one irrational coordinate, how does that affect the set of points I can reach?

3. More generally, what is the smallest set of initial data (and also operations, like introducing new ones such as a compass, leaving this deliberately vague as I'm interested to see what operations people mention) that would be sufficient to generate all of ℝ² (i.e., every real coordinate pair)? If the answer is none, then is there a way we can represent or express the coordinate pairs we can reach?

[warped fable]

I'm only allowed to draw new lines which pass through two old points?

In that case what is the second iteration for your 4 point configuration? I draw all 6 lines. A new fifth point is born.

I can't join it with anything else to produce any more interesting points

[rocky cloak]

So the definition you have here is actually path-connected which is stronger than just connected.

The only path connected subgroups should be things like vR, but you can have some weird connected ones

[vapid dew]

If you're asking “Am I only allowed to connect dots that already exist?”, the answer is yes.

[vapid dew]

I think you're confusing line segments (finite length) and lines (infinite in both directions).

[warped fable]

A, B, C, D are your four points. I draw bi infinite lines AB AC AD BC BD CD. Then AC and BD intersect, I mark that E.

Now what do I do. How do I generate more points

I am out of moves

[vapid dew]

AC and BD aren't the only intersections?

[warped fable]

Oh, because your configuration is sufficiently nondegenerate

AB and CD intersect somewhere far. They're not parallel

[vapid dew]

Yep

[warped fable]

Interesting question...

[vapid dew]

It's been stuck on my mind for a while, math stack exchange didn't give me an answer for 1. either. (though I did have another part answered, which I omitted here).

[warped fable]

I think it's true.

Instead of working in R^2 work in RP^2, the projective plane. Now the problem of parallelism goes away

Without loss of generality by a projective transformation you may choose your points to be [0:0:1], [1:0:1], [0:1:1] and [1:1:1]

This is a nongeneric configuration on the plane (vertices of a unit square) but you're working in the projective plane and there's no issue of parallelism

I don't have a clean proof yet but I'll think about it. It seems much of the troubles should go away by projectivizing, so that's the first thing I'd try

Say you have produced two points [a:0:1] and [b:0:1] on the X-axis. Can you produce [a-b:0:1] and [a/b:0:1]? I think that would be the question to think about

Then you can get all the rationals along the X axis

Very cool question though. I like it a lot.

I don't think you can ever generate the entirety of RP^2 by your infinitary construction, though, with any finite choice of initial points. Even if the initial points have irrational coordinates the new coordinates are Q-linear combination of these at worst. Dimension of R as a vector space over Q is infinity

But QP^2 from a choice of rational points should definitely be possible. I'd not be surprised if 4 is enough

[warped fable]

So just as a quick proof of impossibility (3), at every step you're getting a finite collection of points. You're performing countably many steps. A countable union of finite sets in R^2 can never be all of R^2; cardinality is wrong.

[vapid dew]

I follow this part, but I’m not very familiar with projective geometry, so the rest is a bit alien to me. I originally came up with this just from randomly drawing dots and lines on paper; I didn’t expect it to have such deep connections. Thanks 🙏

[novel star]

why is solving polynomials difficult

even ones with solvable galois groups

and does it have anything to do with the fact that factoring in general is difficult

[next obsidian]

Holy shit this post made me Jimmy Neutron remember like, sapphire-gaming or whatever

The guy who didn’t like, know any algebra except he just kept trying to solve Galois groups of random polynomials

And then asking about them in this channel

[novel star]

like how little algebra

on a scale from american alg 1 to all of lang

[next obsidian]

idk

[azure cairn]

for anyone who's read artin front to back, how long would you guys expect for a full reading o fit to take (assuming pre-existing linalg knowledge)?

im aware it varies by person and whatnot but im just wondering

[hollow dust]

can anyone check my proof of this question? im new to tensor products and honestly im just trying to make sure i didnt make any easy to miss mistakes

[rocky cloak]

All good 👍

[hollow dust]

thx 😁 the tensor products hard to wrap my head around sometimes

[knotty badger]

Their universal property helps a fair bit

[south patrol]

Oh yeah lol

[south patrol]

I wouldn't say all of lang is the end of the scale aha

Well I guess I mean something like being a prof lol

[craggy oar]

You sure being a prof is the end?

[copper kestrel]

how does someone go about showing an ideal is maximal? i'm very confused on how to do that

[rapid cave]

Depends on the context

[copper kestrel]

yea that makes sense

[rapid cave]

Generally you would assume there is a larger proper ideal and show its equal to your ideal

❗ Equivalently you can show the quotient is a field

[copper kestrel]

so show it equals our ideal or if it contains a unit then it equals the ring?

[copper kestrel]

yeah i saw the proof of that, quite strange

[rapid cave]

If it contains a unit its the entire ring always

[rapid cave]

Its quite useful

[copper kestrel]

it is but this homework is a lot of finding prime and maximal ideals and showing quotients are fields (by showing that the ideal is maximal) etc etc

[rapid cave]

I see

[copper kestrel]

very difficult for me LMAO

[rapid cave]

If you want any help you are welcome to ask

[copper kestrel]

sure! i'm trying to find the prime and maximal ideals of Z_{12}

i know {0} is a prime ideal

since Z_{12} is an integral domain

and i know any ideal containing 1,5,7 and 11 is not maximal since they equal Z_{12}

[rapid cave]

Is it?

[copper kestrel]

it should be bc 3 times 4 = 0 mod 12

and 6 times 2 is 0 mod 12

[rapid cave]

So it not an integral domain

Because you found zero divisors

[copper kestrel]

omg i mixed up the definitiin

youre right

wait hold on

nvm

i made another mistake

i really dont understand working with prime and maximal ideals at all

well we have a corollary that says every maximal ideal is prime if R is a commutative ring with unity, so i believe we just need to find maximal ideals

[rapid cave]

But not all prime ideals are maximal

So you need to verify there are no prime ideals which are not maximal

[copper kestrel]

oh right

my lord

thank you, im sorry

[rapid cave]

You should look into the prime ideals of Z

[copper kestrel]

i know the prime ideals of Z are pZ bc that was an example

[rapid cave]

These are exactly those generated by prime numbers and 0

[copper kestrel]

yeye

[rapid cave]

This case is very similar

[copper kestrel]

i dont see how it is unfortunately other than we have Z's

[somber goblet]

i do not understand why $|\operatorname{Gal}(K/F)| = [K : F]$

[cloud walrus]

lexi

[copper kestrel]

like its clear Z/pZ is a field, so clearly pZ's are maximal

[rapid cave]

Seperability gives you that there are [K:F] different embeddings into an algebraic closure, and normality gives that these are all automorphisms of K/F

[copper kestrel]

i think all ideals of Z/nZ are of the form mZ?

[rapid cave]

1 - ideals of Z12 are subgroups of the additive group (Z12,+), and since its cyclic you have a small amount of ideals

2 - Z is PID so Z/12Z is also a PID, again you have a small number of ideals to check

3 - you have the correspondece theorem

[somber goblet]

we know the automorphisms of the splitting field of $p(x) \in F[x]$ fixing $F$ permute its roots because the automorphisms preserve the field operations leading to $p(x)$ being zero at a particular value of $x$...

[rapid cave]

This is exactly the correspondace theorem... a small correction: it should be m(Z12)

[cloud walrus]

lexi

[rapid cave]

Yeah

[copper kestrel]

ah

[copper kestrel]

we have not learned about PID's yet

[copper kestrel]

whats that?

[rapid cave]

The ideals of A/I are of the form J/I where J is an ideal of A containing I. Also, J/I is prime if and only if J is prime in A.

[copper kestrel]

i have not learned that

[rapid cave]

Ok

[copper kestrel]

sorry

[rapid cave]

Its fine...
What we want to do it to check wheter m(Z12) is prime/maximal for 0<=m<12

But to do this we need to say somehow every ideal of Z12 is of the form mZ12

[copper kestrel]

that makes sense! thats what i planned on doing

[copper kestrel]

yeaaaaaa

[rapid cave]

Which can be done by what I said above.

[copper kestrel]

wouldnt i need to prove that though? if it wasnt given to me in the book?

[rapid cave]

Also this is slightly different

[rapid cave]

Yes

[copper kestrel]

right

i need to see some other example problems bc im not understanding this at all, i'll try to find some on youtube

[rapid cave]

Sure

[copper kestrel]

i saw someone make a subgroup lattice which was helpful (of a diff problem), rhat seems helpful in finding maximal ideals

im giving up on this problem

i'll come back to it

[novel star]

can you give me a hint for this exercise:

whoops

Let $f$ be a homogeneous polynomial of degree $d$ in $n$ variables, with rational coefficients. If $n>d$, show that there exists a root of unity $\zeta$ and elements
[
x_1,...,x_n\in\mathbb Q [\zeta]
]
not all 0 such that $f(x_1,...,x_n)=0.$

[cloud walrus]

qchs

[wise socket]

It looks a lot like Chevalley Warning theorem, maybe try to look at the proof ?

[vapid vale]

have you tried just computing ideals to see what happens?

[copper kestrel]

if i knew how to construct ideals i would

[vapid vale]

every ideal is generated by one element. so you can multiply that element by all the other elements and see what you get

the first statement is maybe not trivial, so perhaps try constructing an ideal generated by two elements and see what it looks like. maybe this exercise is a little tedious but once you construct your ideals see if you can find patterns

[copper kestrel]

i can only make principal ideals

but idk if every ring/field/whatever has one or not

[vapid vale]

has what?

[copper kestrel]

has a principal ideal

[vapid vale]

you can always make a principal ideal by just doing it

if you take an element, you can multiply it by every element in the ring and that is the ideal it generates

you just have to identify what they are

[copper kestrel]

okay, i also dont understand how to verify if ideal are maximal or prime, like it makes 0 sense

especially maximal

prime i kinda get

[south patrol]

{0} is a principal ideal

[novel star]

contradiction

[south patrol]

It's sometimes hard but a major thing you'll see is that like an ideal J of a ring R is prime (or maximal) iff R/J is an integral domain (resp. a field)

So you reduce it to the question of checking if a ring is an integral domain or field

And this can often be checked by hand etc

[novel star]

isnt that the defn of a prime ideal

[copper kestrel]

right but idk how to do that when i'm quotienting like, Z_{12}/Z{6}

[south patrol]

It is a definition but I wouldn't even say it's the standard one

[copper kestrel]

and finding out what theyre isomorphic to

[south patrol]

What do you mean by Z{6}

[copper kestrel]

chat honestly i'm having a terrible day im sorry

[south patrol]

If you mean Z_6 then this isn't an ideal of Z_12

[copper kestrel]

i meant Z_6

[south patrol]

Ye that isn't an ideal oop sorry

[supple ice]

every ideal is generated by one element???

[south patrol]

Or at least like

Presumably you mean 2Z_12 or smth

[vapid vale]

the problem example was Z/12Z

[copper kestrel]

yeah that makes sense i was just saying shit

[south patrol]

Otherwise not clear what you mean to me

[copper kestrel]

i've been working on all of these problems except for 1 and have gotta stuck on all of them

[supple ice]

okay sorry I didn’t have context I'll shut my mouth

[vapid vale]

good

[copper kestrel]

like i understand what i have to do but it all comes down to verifying ideals are maximal and ideals are prime

[vapid vale]

i really recommend computing the ideals and knowing what they are before getting caught up on verifying if they are prime or maximal

[novel star]

looking through the previous exercises i think its an open problem

the line above it says "the answer to this exercise is unknown"

[unkempt comet]

wth, how

[novel star]

idk

but if lang says it

theres a 30% chance its right

[tardy hedge]

This mf lang

Cheeky guy

[unkempt comet]

seems connected to it

https://en.wikipedia.org/wiki/Quasi-algebraically_closed_field

https://en.wikipedia.org/wiki/Tsen's_theorem

[novel star]

always

i think his dissertation was on an adjacent subject

[somber goblet]

we can "adjoin" elements to rings by taking the polynomial ring $R[x]$, then applying equivalence relations corresponding to how we want the adjoined number to behave
can we define a (not necessarily algebraic) field extension as the field of rational functions $F(x)$ quotiented with some equivalence?

[cloud walrus]

lexi

[velvet hull]

you could

[somber goblet]

is that equivalent to being the smallest field containing $F$ and $x$

[cloud walrus]

lexi

[velvet hull]

F(S) is by definition the smallest field containing F and S, where S is any set

[somber goblet]

for simple extensions, is this met by the field of rational functions over F in x

[velvet hull]

what do you mean by simple extensions

[somber goblet]

for example $\mathbb{Q}(\pi)$

[cloud walrus]

lexi

[somber goblet]

is the smallest field containing both the rational numbers and pi

it is a simple extension because it is generated by one element

[velvet hull]

well either that element is transcendental or it is algebraic

there is not much to be said in terms of galois theory

either it is a finite extension or is isomorphic to F(x)

[somber goblet]

is this isomorphic to the field of rational functions over Q?

ok thats what i thought

and then if you have algebraic $x$, with minimal polynomial $f(x)$, then $\mathbb{Q}(x) \cong \mathbb{Q}[X]/(f(X))$

[cloud walrus]

lexi

[velvet hull]

yes

[somber goblet]

in general, if you have $F \cong R$ and you know $F$ is a field, $R$ must be as well, right?

[cloud walrus]

lexi

[somber goblet]

since you can define the inverse on $R$ through its isomorphism with $F$

[cloud walrus]

lexi

[velvet hull]

isomorphism of what? rings?

then yeah sure

[somber goblet]

yes sorry

[novel star]

why does the field trace vanish on inseparable extensionsz

[velvet hull]

if you have an inseparable extension, what this looks like concretely in terms on the minimal polynomials of each elements is that there is some kind of x^p doubling of the roots

in other words, if you use the definition of trace of an elements being the sum of all the roots of its min poly, then the natrual expanation is that every single root has a factor of p appearing before it, so everything adds to 0

[novel star]

i think my confusion stemmed from the fact that i didnt realize that there are no inseparable extensions of characteristic 0

[exotic verge]

im supposed to show cayleys thm for infinite groups

the proof for finite groups is given in my book

step one is it defines a map L_g for each element g in G

if i can do this for my infinite group the rest of the proof is also basically the same

but i’m not sure its legal lol

i do have aoc, but i don’t see how to use it

i mean if i define a function F that takes every element of g into singleton containing g

then i can use choice on F(G)

but its starting to feel like since i know i can do that, its fine to just implicitly use choice and define my function

am i overthinking this or underthinking it chat?

[noble nexus]

you don't need choice you're overthinking

It is indeed the same proof

[exotic verge]

nice thanks

[exotic verge]

this is what got me, because i was expecting some gotcha

[cloud walrus]

UGOAT

[candid patrol]

@exotic verge

[kind temple]

you could also use the Yoneda lemma

[candid patrol]

haha didn't about this one

[kind temple]

yea, it is kind of cute

[glass sinew]

yoneda lemma for cayley's thm

[noble nexus]

hydrogen bomb vs coughing baby

[umbral lance]

Emily Riehl https://www.youtube.com/watch?v=mTwvecBthpQ mentions the Yoneda lemma often.

[novel star]

is the title a pun

[warped fable]

EmilyntaRiehl

[kind temple]

this talk was really good

[tidal torrent]

What are Dedekind rings, Noetherian rings, and Frobenious rings (what would happen if one of these rings acts on a vector space)

Dedekind modules just sound cursed idk

mentions that Dedekind domains have something to do with PIDs and UFDs but that’s outside the scope of the book

[rocky cloak]

A ring is Noetherian if every ideal is finitely generated.

Dedekind domains and frobenius rings are a little more complicated, but an integral domain is a dedekind domain iff every ideal is projective.

A ring is frobenius if every projective module is injective and it's socle equals it's top.

[rocky cloak]

PIDs are dedekind domains, and a dedekind domain is a ufd iff it is a pid

[rocky cloak]

I'm not exactly sure what you mean by a vector space acting on the rings, like a group action? I don't think anything particularly noteworthy happens

[warped fable]

Yeah, I don't know a unified short proof. I always go with the A_5 one which I can remember because it acts on the icosahedron.

The action is so transitive that it's sort of visually immediate that every conjugacy class generates the entire group. Once you realise this, it's not hard to write a proof

[glass sinew]

gallian gives this proof which relies on a few facts done in exercises but it's relatively short

[azure cairn]

how do you typeset this lol

[modern canopy]

\begin{tikzcd}[column sep=1em]
    1 && 4 \\
    & 3
    \arrow[curve={height=13pt}, between={0.1}{1}, from=1-1, to=2-2]
    \arrow[curve={height=20pt}, from=1-3, to=1-1]
    \arrow[curve={height=13pt}, between={0}{0.9}, from=2-2, to=1-3]
\end{tikzcd}

it's a lil ad hoc but I think this works

[novel star]

where is kummer theory used

[azure cairn]

Ty

Also where'd the flowers at the bottom come from 😭 😭 😭

[glass sinew]

why is your latex rendering beautiful

this is so pleasing to look at

[kind temple]

lmao

fr tho

[modern canopy]

it's my texit fork

you can find it in the texit support server

[kind temple]

where do you find that?

[terse crystal]

By the way these are covered by Kasch modules and rings. It’s the most gentle and easiest to follow book I have ever seen among all math books I have read.

[azure iron]

tried to make a less "manual" solution

\usetikzlibrary{graphs}
\tikz[every edge/.append style={bend right=45}] \graph[counterclockwise=3,phase=150,->] {[cycle] 1,3, 4};

[cloud walrus]

cloud ☁

[tidal torrent]

what does it mean for an ideal to be projective

i guess it makes sense why quasi frobenius rings/frobenius algebras arent covered in the book since they are outside the scope of an introductory class of AA

is it the socle of a group or socle of a module

is a socle a subgroup generated by the minimal subgroups of a group in the context of groups

[tidal torrent]

i meant one of the rings acting on a vector space

would a frobenius ring acting on a vector space be a frobenius algebra

[rocky cloak]

That it is a projective module. So for example of a direct summand of a free module.

[rocky cloak]

Socle of a module, so sum of the simple submodules

[rocky cloak]

So an algebra is a ring that is also a vector space. A Frobenius algebra is a Frobenius ring that is a finite dimensional algebra. And its modules would be vector spaces yes

[tidal torrent]

if i remember what differentiates modules from algebras are that algebras are equipped with bilinear form or something

i thought an algebra was a field over a field but thats an F-Algebra with more structure or something

[rocky cloak]

And algebra is like a ring over a field (or sometimes ring over a commutative ring)

[desert verge]

a field over a field is a field extension

[azure cairn]

Ah ty

[muted sparrow]

Is this a valid proof? Its obviously different from the textbook proof. Also, I clearly didn't use all the assumptions as it is recommended by the textbook.

Suppose $H,K \trianglelefteq G$ and $H \cap K ={ e }$. If $a\in H$ and $b \in K$ then $ab=ba$.

Proof: Given $H \trianglelefteq G$, we have $gH=Hg$, $\forall g \in G$ by the definition of normality.
Also $K \trianglelefteq G$ and exists $k \in K \trianglelefteq G$, so $k \in G$. Let $k=g$, then $kH=Hk$.
Since $k \in K$ was chosen arbitrarily, we have $kH=Hk, \forall k \in K$.
Thus, $kh=hk, \forall k \in K,\forall h \in H$.

[cloud walrus]

Hause

[warped fable]

kH = Hk does not mean kh = hk for all h in H

[muted sparrow]

ah I see, thanks

[candid patrol]

hkh^-1k^-1 is in H n K because the normality of both

And so hk = kh

[little ocean]

Idk if I should ask this here, or in #real-complex-analysis , but I was just wondering how many field automorphisms the complex numbers has

I know the reals only has one (the identity) and complex at least 2, but how many does it exactly have?

[quiet pelican]

If choice holds, it has infinitely many (and they’re very hard to describe explicitly)
If choice doesn’t hold, it is consistent that there are only 2

[little ocean]

Why?

What does the axiom of choice have to do with this?

[quiet pelican]

It’s possibly “more obvious” that it should be relevant via Zorn’s lemma

[little ocean]

Zorn says that every poset has a maximal element or something, right?

[quiet pelican]

Yes
The poset you want to consider is (subfield F of \C, automorphisms of F) ordered by inclusion on the first component and Extension on the second

[little ocean]

Excuse me, what?

[tardy hedge]

micoi is defining the poset to use

[little ocean]

Yeah, but I don't understand what he means here

[quiet pelican]

*she
And it’s the poset (F, g) <= (K, h) if F \subseteq K and h(f) = g(f) for all f \in F

[little ocean]

(Sorry 😔 )
And here, what are K and F and what are h and g?

[tardy hedge]

subfields of C and automorphisms

[little ocean]

Right

So how does this prove that there would be infinitely many?

And if this weren't true, how would this prove that there don't exist more than 2?

[quiet pelican]

That’s not necessarily true
It is consistent that there are not more than two
And that’s a result that I don’t know how to prove (and more properly belongs to foundations tbh)

[little ocean]

And how would you make an automorphism that is not the identity or the complex conjugate?

[quiet pelican]

Messy and scary
You prove that a maximal element of the poset I defined is such a thing

[little ocean]

Okay, but we defined this poset with tuples (F,g), where g is an automorphism

So don't you need more than 2 automorphisms for this to work?

[rocky cloak]

By Zorn's lemma there is a maximal set of algebraically independent elements, called a trancendence basis.

So C is an algebraic extension of the field of fractions in uncountably many variables. Since C is normal, then (again by Zorn) every automorphism of this subfield extends to an automorphism of C.

Hence C has at least 2^C automorphisms, and it can't have more because that's also the amount of functions from C to itself.

[little ocean]

Uhm

[rocky cloak]

g is an automorphism of F, not C

[little ocean]

So C is an extension of Q

[rocky cloak]

Yes

[little ocean]

And what does it mean for C to be normal?

[rocky cloak]

For this purpose I guess it's easier to just say that C is alg closed

[little ocean]

Cool

Right, so then we say that if F < = C has an automorphism f, then we can extend f to an automorphism g such that f(a) = g(a) for all a in F?

Okay, why does F have an automorphism?

*That's not the identity

Or do we not require it to be nontrivial?

[rocky cloak]

Well F is just any subfield of C. I'm sure you're able to come up with subfields that have automorphisms

[little ocean]

No

I mean

C itself has the conjugate automorphism

But I can't think of any proper subfields of C that is not R

[rocky cloak]

Well that's not true because I'm sure you're aware of Q.

But just take any complex number and then take the subfield it generates

[little ocean]

I left out Q, since it is contained in R

[rocky cloak]

Subfields of C are allowed to be contained in R

[little ocean]

So for example Q[1 + i]

?

[rocky cloak]

Yeah or Q(sqrt(2)) or Q(pi) or whatever

[little ocean]

Okay

Why does this have an automorphism?

Or is our automorphism in the poset above allowed to be trivial?

[rocky cloak]

Well maybe you're able to come up with an automorphism of Q(sqrt(2))

[little ocean]

Maybe $f : a + b\sqrt{2} \mapsto a - b\sqrt{2}$?

[cloud walrus]

Nico

[rocky cloak]

That's right.

So following the Zorn argument that means C has an automorphism that maps sqrt(2) to -sqrt(2), so that's neither the identity or conjugation

[little ocean]

Is it unique?

[rocky cloak]

No

[little ocean]

No

[rocky cloak]

There's tons of them

[little ocean]

Yeah, ofc ofc

And why would this not be true if Zorn/choice doesn't hold?

[rocky cloak]

I mean, it's hard to give a good answer to why it might not hold.

Something you can say is that the identity and conjugation are the only continuous automorphisms.

But yeah, the proof requires choice so without choice you can't prove it.

[little ocean]

And this proof is the only proof that exists?

Or every proof ever requires choice?

[rocky cloak]

I mean you can prove it in tons of different ways depending on what you count as a different proof

[little ocean]

sqrt(2) is irrational has different proofs

'different' like that

I mean

[rocky cloak]

Well everything has different proofs

[little ocean]

And you can't give even an intuitive explanation of why we need choice?

[rocky cloak]

I mean I thought I did

[little ocean]

Yeah, so by this construction of posets, we need Zorn

[rocky cloak]

Discontinuous automorphisms are huge ugly beasts you can construct step by step, but requires a bunch of choices

[little ocean]

Right

And why are there 2^|C| automorphisms?

And with 2^|C|, we mean the power set of C?

[rocky cloak]

Well, if you use this proof instead. Then it's enough to count automorphisms of rational functions in uncountably many variables.

And there whether you send x to x+1 or x for example would be a binary choice for each variable

[little ocean]

Is this proof given somewhere in uni?

I'm currently only in my 2nd year, but will I see this proof?

Like, did you see it when you were in uni?

Ok but wait

Q[sqrt(2)] has a nontrivial automorphism

So why does R not have a nontrivial automorphism?

[rocky cloak]

You use that C is alg closed in the induction step of Zorn.

An automorphism that takes sqrt(2) to -sqrt(2) would need to take sqrt(sqrt(2)) to sqrt(-sqrt(2)) which is not a real number

[rocky cloak]

Might have seen it in a Galois theory class. I don't remember

[little ocean]

Gotcha

[little ocean]

I have a friend that took it last semester, I'll ask him

Thanks so much for your help, all of you!

[cloud walrus]

lexi

[tidal torrent]

That’s if the field F is a subfield of a field E and the operations of E are restricted over a field F

Something like that

[summer snow]

im having trouble getting a foothold on this question. Ig it makes sense to me that in the lower symmetry group, an irrep in a higher symmetry group reduces to a combination of irreps in a lower symmetry group, but im having trouble understanding how i can use orthogonality relations here to show this. Sorry if this is a stupid question, i feel very dumb right now as i havent been this stuck on a question in a long while

what i have done so far is recognize D6h can be obtained by a direct product of D6 with an inversion, which gives you 6 more conjugacy classes and 6 more irreps. My next thought process is to then see how the subgroups of D6 transform under the direct product which is what im trying to figure out now

[tulip otter]

isnt Hom_A(X',X) still a module even if A is not commutative?

[barren sierra]

I guess technically it'd be an (A, A)-bimodule?

[quiet pelican]

af as defined isn’t necessarily an A-homomorphism if A is not commutative

[barren sierra]

it isn't?

[quiet pelican]

af(rx) = arf(x) is not necessarily equal to raf(x)

[barren sierra]

oh yea

[tulip otter]

ah right

tysm

[tulip otter]

a bimodule is a left and a right module? If yes then shouldnt right multiplication be defined to begin with in order to talk about that?

[barren sierra]

(fa)(x) = f(x)a but same issue

[tulip otter]

oh yea right lol

tysm

[rocky cloak]

In general if X' is an R-S-bimodule and X is an R-T-bimodule, then
Hom_R(X', X)
is an S-T-bimodule.

Is R is commutative then any R-module can be made into a bimodule in a canonical way by just defining
xr = rx

[tulip otter]

ohhh i see, tysm jagr

[rocky cloak]

I think I would just write down the two representations and observe which representations of D3h, C3 and C3v they correspond to

Like D6h is generated by one element of order 6h and one of order 2, which matrices do they correspond to in these representations? What are the generators for D3h, C3v,

[summer snow]

yes i think i am getting it now. My main issue was thinking the conjugacy classes should be related, but i should be comparing traces of elements instead since which elements are conjugate in one group arent going to necessarily be conjugate in another.

the other major hurdle here for me here is getting used to all of the notations for all of the types of symmetry operations 🫠

[white oxide]

Can I please have a hint for this question? If $\tau = 2 \omega + 1$, $\tau^p \equiv (-3/p) \tau \text{ mod } p$. But I'm stuck from here

[cloud walrus]

okeyokay

[white oxide]

Well I guess 2^p is always congruent to 2 mod p...

[wraith cargo]

for p = 2,3 the situation is trivial, so assume p>=5

you should show that 2w + 1 is always gonna be relatively prime to p in Z[w]

now the reason why this is important is basically, you can assume p>=5

so you have that p is odd, i.e. you can write it as 2k+1 for some k

so t^p = (2w+1)^2k (2w+1) = (-3)^k (2w+1)

[white oxide]

I felt like I went wrong somewhere here so far: Let $\tau = 2 \omega + 1$. If $p$ is an odd prime, then
[\tau^{p - 1} \equiv (-3/p) ; (p)] so that $\tau^p \equiv (-3/p) \tau ; (p)$. By Proposition 6.1.6, $\tau^p = (2\omega + 1)^p \equiv 2^p \omega^p + 1 ; (p)$. Since $p \geq 3$ we have $\tau^p \equiv 2^p + 1 ; (p)$. Now, $2^p = 2^{p - 1} 2 \equiv 2 \text{ mod } p$, so that $\tau^p \equiv 3 \text{ mod } p$.

[cloud walrus]

okeyokay

[wraith cargo]

why is w^p = 1 mod p?

[white oxide]

I just assumed that w^3 = 1, and so when we project into Z_p it is also 1

Am I high

[wraith cargo]

this would imply all primes are divisible by 3 lol

[white oxide]

Because p >= 3

Oh well anyways 2^pw^p + 1 is congruent to 2w^p + 1 mod p

I would like w^p to be congruent to w mod p lol

[wraith cargo]

for dat you gotta show that w is relatively prime to p

which isn't hard to do

Z[w] is an euclidean domain so you just look at their norms

and you know that w is prime in Z[w] anyways

[white oxide]

Oh right because w^{p - 1} \cong w mod p if and only if w^{p - 1} \cong 1 mod p if and only if w is nonzero

[white oxide]

I'm not really too familiar with norms lol they haven't been introduced

I remember studying them earlier in an abstract algebra class but I forgot everything

Uhhh

I feel like this is easy to show without that but nothing's coming up 😂

[copper kestrel]

im guessing Z x {0} is a prime ideal since {0} is a prime ideal, and im trying to show that it is a prime ideal by computing Z x Z / Z x {0} and showing that its isomorphic to an integral domain but the computation is confusing

i know we have the class of (a,0) for all a in Z, then maybe like 0,1 + (a,0), etc etc

which makes me think its isomorphic to Z?

am i on the right track at all

or i guess i could do 1 + (a,0) maybe?

i dont know

oh wait no i think i got it

its the equivalence classes of Z x {0}, Z x 1 + Z x {0}, dot dot dot for all z in Z, so it should be isomorphic to Z

[noble nexus]

yeah

it might just be easier here to use the definition of a prime ideal

[next obsidian]

make a surjective map to Z with the correct kernel

and apply 1st isomorphism theorem

[copper kestrel]

i figured the cosets out :]

[vapid vale]

that is the definition :)

[copper kestrel]

i dont like the definition they gave me

its like if ab \in I then either a in I or b in I

for a,b in R

[vapid vale]

the contrapositive of that statement is a little easier to parse imo: that if both a and b are not in I, then ab is not in I

and that aligns with the idea that the quotient is an integral domain; in a domain, a and b not equal to 0 means ab doesn't equal 0

[maiden crater]

here I can't assume R has no zero divisors right

nvm, that doesn't matter here

[hexed aurora]

is there any fast way / trick to calculate multiplicative inverse of of a group Z[n] ?

[delicate orchid]

there are many cryptographic algorithms that are based on that precise problem being very difficult in general

[hexed aurora]

i remember i found a few tricks online few years ago but i cant remember any of them now

[rocky cloak]

Specifically multiplicative inverses in Z/n ?

That you can do with Euclids algorithm. Which is very efficient

[delicate orchid]

oh is that what Z[n] meant

[hexed aurora]

Ring of Z n

[rocky cloak]

I guess one can think of the discrete logarithm problem (as it appears in e.g. RSA) as finding multiplicative inverses in Z/n, but the main issue is that you don't know what n is in that case

[hexed aurora]

Yeah

[twilit wraith]

i dont understand where Delta being a simple system comes into play

i mean computing the matrices for the reflections is just a matter of computing the images of the basis vectors under each reflection

regardless of whether or not the vectors generate a simple system

maybe theres some quick way of computing the image using the simple system i really dont know

[maiden crater]

I have solved this before , but just wanted to calrify

here S bar is the image of the projection onto R/I?

[tardy hedge]

The dimension of R/I is the supremum of the coheights of the minimal primes over I?

[astral ivy]

Hello, I struggle with some tasks that are usually proofs around normal groups or something involving ideals/principal ideals. Is anybody willing to help me privately? I asked here before but got overwhelmed with a bunch of answers and couldn't follow the pace.

[elfin wraith]

You’ll have more luck just posting your problem here than looking for someone to privately DM you.

Hopefully with this being said one person will respond and let some time for that. You also shouldn’t feel any pressure to respond so quickly, this is very low commitment!

[wraith cargo]

yeah this sounds true

[astral ivy]

Okay I'll ask here. (this isn't really related to normal subgroups just something I found along the way)

Let $G$ be a cyclic group of order $90$.

1. Find all subgroups of $G$ that don't contain elements of order $5$.
2. Find $|\text{Aut}G|$.
3. How many homomorphisms is there in $f: G \rightarrow \mathbb{D}_4$?

I think the 1. is the simplest but I still don't quite understand it. I can rewrite $90 = 2\cdot 3^2 \cdot 5$, this is also the elementary form. So I just 'kick out' $5$ and I'm left with $2\cdot 3^2$ which is I presume the number of subgroups that don't contain elements of order 5? I'm not sure how to find those specific subgroups though.

2. I genuinely have no idea what 'Aut' means? Is it automorphism? I don't understand the context behind it at all and why there are absolute brackets.

3. Clueless about this part.

[cloud walrus]

dan

[proud vigil]

finite cyclic groups are convenient because if the group order is |G|, then there is a bijection between factors of |G| and subgroups of G

there wouldn't be 2 * 3^2 subgroups, but 2 * 3^2 is an important number that will help you solve 1

[proud vigil]

this is related to the fact that every subgroup of a cyclic group is cyclic

Aut is automorphism, it wants to know how many automorphisms of G there are (brackets refer to cardinality)

a good method to find automorphisms is to map generators to generators in a way that preserves relations

[astral ivy]

yeah figured that out. Forgot I should've used the Euler's totient function

so $\phi(90) = 12$

[cloud walrus]

dan

[astral ivy]

those are all the subgroups, now I pick those divisible by 5 and eliminate them?

[wraith cargo]

this just tells you how many generators of G there are

[astral ivy]

I know what an automorphism is (bijective endomorphism) but like I don't really understand it past the definition. It just maps its entirety to itself I just don't understand the point of it

[wraith cargo]

yeah but it shifts the points around that's the point

this is like saying permutations aren't important because they're bijections of finite sets which makes them all "basically the smae"

[astral ivy]

yeah because G is cyclic

[astral ivy]

I didn't know that

what exactly does it shift?

[wraith cargo]

take Z/3Z and take the automorphism defined by f(0) = 0, f(1) = 2 and f(2) = 1

so it mixes up 1 and 2

which isn't a trivial thing

[astral ivy]

I can't comprehend that

I can't visualize it

[wraith cargo]

how do you usually visualize groups?

[astral ivy]

as a set with the ability to perform operation with its elements to get another element thats also the part of the set. Like Z

[wraith cargo]

Here's how you visualize it

[astral ivy]

Is $\mathbb{Z}/3\mathbb{Z}$ equivalent to $\mathbb{Z}_3$? I haven't used the first notation thus far, not sure if that's a quotient group?

[cloud walrus]

dan

[somber goblet]

they are typically equivalent

[astral ivy]

so basically something like $\mathbb{S}_2$? 2 elements switch places? (but here there's more elements ofc)

[cloud walrus]

dan

[astral ivy]

or am I mixing it all up

[somber goblet]

in the case of $\mathbb{Z}/3\mathbb{Z}$, the automorphism group is indeed isomorphic to $S_2$

[cloud walrus]

lexi

[somber goblet]

(which is just $\mathbb{Z}/2\mathbb{Z}$)

[cloud walrus]

lexi

[astral ivy]

wait what?

[somber goblet]

$S_2$ is isomorphic to $\mathbb{Z}_2$

[cloud walrus]

lexi

[astral ivy]

but that would mean $Z_2$ is isomorphic to $Z_3$ and I don't think that's the case

[cloud walrus]

dan

[somber goblet]

no, the automorphism group of $\mathbb{Z}_3$ is isomorphic to $\mathbb{Z}_2$

[cloud walrus]

lexi

[astral ivy]

ahh got it

Okay so how do I find |AutG|?

like what am I supposed to do?

G is a huge group of order 90

[somber goblet]

in general it is difficult to find automorphism groups

however for the cyclic case it is quite simple

cyclic groups are generated by a single element

[proud vigil]

(btw yes it is a quotient group)

[somber goblet]

yes i forgot to mention that

in any case cyclic groups are generated by a single element

but a cyclic group can have many elements that can act as a generator for the entire group

in particular, you can show homomorphisms (and thus automorphisms) preserve powers: $\phi(a^n) = \phi(a)^n$

[cloud walrus]

lexi

[somber goblet]

and from this you can say that an automorphism preserves the orders of elements in the group

if you select a fixed generator element of a cyclic group, all that is needed to produce a (unique!) automorphism is to pick another generating element to map it to

that should make the question a lot simpler

[astral ivy]

but 90 has 12 generators so 12 different elements that can single handedly generate it?

[somber goblet]

yes

the cyclic subgroup generated by any of those elements is the entire group

[astral ivy]

how do I do this? This looks like rewritting it and that's it

[somber goblet]

wdym?

[tulip otter]

I think that you should use the first notation when you are talking about integers modulo 3

because Z_3 is used to denote something else

so its not a good notation if you use it for Z/3Z

[somber goblet]

this is true but i don't think we're worrying too much about p-adic integers rn

[tulip otter]

if Z/3Z is annoying, then maybe write Z/3 or something like that if you want

[somber goblet]

i prefer $C_3$ when talking about groups

[cloud walrus]

lexi

[tulip otter]

sure but its better to avoid using that notation early on to avoid picking up the habit of using it later

[somber goblet]

very true

[tulip otter]

also Z/3Z emphasizes that this is indeed a quotient group

which raised dan's question in the first place

[tulip otter]

by dan's question i meant this

[somber goblet]

yeah

[astral ivy]

well to show something is a homomorphism (from $(G, \cdot)$ to $(H, \ast)$) I'll need:

1. $\phi(g_1 \cdot g_2) = \phi(g_1) \ast \phi(g_2)$
2. $\phi(e_G) = e_H$
3. $\phi(g^{-1}) = \phi(g)^{-1}$

And I don't know how I'd show that on the above automorphism example.

[cloud walrus]

dan

[astral ivy]

fair, we were taught to use it but I understand.

[tulip otter]

yea i am not blaming you, because Z_n was the notation used in my elementary nt course too

[somber goblet]

a homomorphism is completely determined by what it does to the generating set of a group

[tulip otter]

sorry to interrupt in the middle of your work, but just a small note: You dont really need 3 because it follows from 1 and 2 (try proving it later if you want)

[somber goblet]

in the special case of a cyclic group, which is generated by a single element, you can characterize any endomorphism (homomorphism into itself) by where it sends 1, since the group is entirely generated by that element

[somber goblet]

if you have inverses, you can determine 2 from 1 as well (though this doesn't work for monoids in general)

[tulip otter]

right

[astral ivy]

well that's the 'abstract' part of algebra and I didn't understand it

so Automorphism in this case is just the number of generators?

'shifting' the group

and by shifting i mean permutations and not coset stuff

[proud vigil]

you only need to show 1, that's the definition, everything else follows from that

oh someone else pointed this out but yes

[tulip otter]

you also need 2 no?

[cloud walrus]

lexi

[proud vigil]

let g2 = e_G, then

phi (g_1) = phi(g_1) * phi(e_G) -> e_H = phi(e_G)

[tulip otter]

oh right lol

[somber goblet]

this doesnt work for rings

[tulip otter]

right, it works for 0 but not for 1

[somber goblet]

yeah

that’s probably why 1 and 2 are usually paired up

[proud vigil]

1 and 2 are the requirements for a monoid homomorphism, 1 is the minimal requirement for a group homomorphism

[somber goblet]

i forgot to answer this but yes

(if your group is cyclic)

[astral ivy]

thanks!

How many homomorphisms is there in $f: G \rightarrow \mathbb{D}_4$
What about this part? How do I connect the dots with D_4

[cloud walrus]

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

[somber goblet]

i’m not quite sure

maybe use the fact that D4 has a subgroup of order 4 and 2 of order 2?

[velvet hull]

No, for this part you need to use what you know about G

[astral ivy]

Hmmm order 90, subgroups are: 1,2,3,5,6,9,10,15,18,30,45,90.

D4 has the order 2*4 = 8.

I don't know how to connect it

[velvet hull]

Not the subgroups

It is the exact same philosophy as what lexi was telling you

If you want to understand the homomorphisms coming out of a group, it is enough to know what it does to any generating set of that group

[somber goblet]

oh yes

[velvet hull]

So, can you find a good generating set for G?

[astral ivy]

I think it's the Z/2Z right?

[velvet hull]

The what

[astral ivy]

💀

90 has a subgroup of order 2 and D4 has a subgroup of order 2 as well

that was the logic behind my answer

[tulip otter]

he probably meant Z/2Z if that was the problem

[astral ivy]

ahhh yes I merged the 2 different notations 😭

[velvet hull]

I’m asking you to give a generating set for G

[somber goblet]

hint: we already know the generators are those coprime to 90

[astral ivy]

I'm not sure how. I don't know what are the elements of G and what is the operation, maybe {1} could be a generating set idk

[velvet hull]

It’s a cyclic group

What do you know about cyclic groups

[astral ivy]

generated by a single element

[velvet hull]

Right, so that’s your generating set

[astral ivy]

my generating set is a single element?

or alternatively any element that generates the entire G

[velvet hull]

Okay, good

So now where can it go under a homomorphism

[astral ivy]

I don't understand this question

[cursive spindle]

if g is a generator where do you map it?

[astral ivy]

$f(g^n) = (f(g))^n$?

[cloud walrus]

dan

[somber goblet]

yes

you can show this inductively

[tulip otter]

if g is a generator, can f(g) be anything? or does it have to have a certain property

[somber goblet]

and keep in mind we’re just looking for homomorphisms, they don’t have to be surjective or injective

[astral ivy]

ahh well yeah I think it needs a certain property

whatever the operation is in dihedral group?

rotation/reflection

[somber goblet]

does it?

[astral ivy]

hmm well g is from G so maybe not

g will use the operation from G so we don't really care about the operations in D?

[somber goblet]

try to construct a homomorphism by picking an arbitrary element of D4 to map the g to

see what you can conclude about that map

verify if it is or isn’t a homomorphism

[astral ivy]

I'm confused because D4 is completely different.

You need two elements for a homomorphism tho?

$\phi(\sigma) \ast \phi(\rho) = \phi(\sigma\cdot \rho)$

[cloud walrus]

dan

[astral ivy]

idk man im confused

[somber goblet]

if you know $\phi(g)$, can you find $\phi(g^2) = \phi(g \cdot g)$?

[cloud walrus]

lexi

[somber goblet]

what about $\phi(g^3)$?

[cloud walrus]

lexi

[somber goblet]

or arbitrary $\phi(g^n)$? what does this tell you about how $\phi$ acts on $G$ as a whole?

[cloud walrus]

lexi

[somber goblet]

(given that g is a generator for G)

[mortal copper]

$P_u = u(u^Tu)^{-1} u^T\ P_{col(A)} = A(A^TA)^{-1}A^T$

[cloud walrus]

alimohammed5

[mortal copper]

I love Projection matrices

[somber goblet]

assume $\phi$ is a homomorphism such that $\phi(g) = h$ for some generator $g \in G$

[cloud walrus]

lexi

[somber goblet]

and work from there

[mortal copper]

this equation is genuinely bueatiful

[astral ivy]

huh?

$\phi(g^n) = \phi(g)^n$

[cloud walrus]

dan

[astral ivy]

like i'm stuck in a loop im not sure what youre asking me

[somber goblet]

if you know this, and that any $g_1 \in G$ has $g_1 = g^n$ for some $n \in \mathbb Z$

[cloud walrus]

lexi

lexi

[somber goblet]

i’m sorry if i’m not explaining this well

[copper kestrel]

i know that this is a way to do it, i'm just having trouble justifying why this m exists, im pretty sure its because no zero divisors, but i dont see why no zero divisors implies that d doesnt bounce around

also please ignore the last sentence, i deleted it but im too lazy to take a new picture

also please ignore the number of times i say "consider" LMAO

[rocky cloak]

By pigeonhole there exists m and n with d^m = d^n (and m not equal to n)

See if you can work from there

[copper kestrel]

well i dont know if we can cancel

oh wait yes we can

bc its an integral domain

so d^{m-n} = 1

[unkempt comet]

This is also true if you remove the commutative part of this

[copper kestrel]

me when its taken me 5+ days to figure out how to show every ideal in Z/12 is principal and i still havent figured it out

i'll be back tmr to scream about it bc my homework is due tmr

[copper kestrel]

same for Z/2 x Z/4

[velvet hull]

Z/12Z is straightforward: ||the quotient of principal ideals are always also principal ideals||

[copper kestrel]

we never proved that

so i have to prove that

and i dont know how to prove it

there's nothing in my book about quotients of ideals

[velvet hull]

let I be an arbitrary ideal in the quotient. what can you say about it?

[copper kestrel]

i have no idea

other than the definition

[velvet hull]

what is the definition

[copper kestrel]

aN subset N and Na subset N

for an ideal N and for all a in R

[velvet hull]

thats just the definition of an ideal

[copper kestrel]

yep

[velvet hull]

I am also telling you that the ideal is inside the quotient

[copper kestrel]

i dont know whats special about an ideal in a quotient, i know you can only form a quotient using a two sided ideal

[velvet hull]

well what are special about quotients in general?

[copper kestrel]

they make new rings

[velvet hull]

I can make new rings plenty of other ways

the point is that the quotient ring can be completely understood by looking at the original ring and the ideal you are quotienting out by

so whenever you have a question about a quotient, you almost always first pass that question into a question about the original ring

[zealous sandal]

So idk if this is the right place to ask this, but I am looking for advice on learning algebra. I am currently taking the first term of graduate group theory, and while I feel like I understand the proofs and concepts presented, I struggle a lot with problems. I know that this is a symptom of lack of practice, but I feel like I'm missing the "bigger picture." When I'm stuck, I end up throwing things at the wall hoping something sticks without being able to spot a clear line forward. As a result, I've found the homework to be a massive time sink where I just end up staring at the paper without feeling much ability to unstick myself. This is the first time where I've really struggled like this in a math class, so I am looking for advice on what to do.

[tidal torrent]

i read all the theorems at least 10 times

rewrite the proofs for each until i understand them

use the theorems and definitions through many examples

(i still struggle with abstract algebra to this day)

most of abstract algebra that stumps me is probably the number theory stuff (if not the proofs themselves)

[tidal torrent]

even thought abstract algebra doesnt require much prerequisities

its not a bad idea to relearn everything you once knew in linear algebra (going back and learning stuff in stuff past basic set theory)

once you know most theorems/definitions in AA by heart and understand most of linear algebra you should be fine

the examples are what helps you most in intermediate/elementary group theory (proving that Aut(V) is isomorphic to S3 for example)

[craggy oar]

Sometimes try checking the first line of the solution if nothing clicks for a long time, try to take hints. But you need to be careful not to see the whole solution. I personally prefer having physical pages for solutions because hiding them is easy

Also you can't do every problem at once. Take time to figure out if you can solve it, if you can't then leave
I believe after a few days of extreme struggle (unable to solve even a bit) you'll gradually be able to solve 1-2 problems then more. It comes naturally once you start practicing everyday

[kind temple]

let G be a group. if S is a finite subset of G such that element of S has finite order, then the subgroup generated by S is finite.

i think this is true because we can syntactically exhaust all possible finite strings of powers of elements of S.

[somber goblet]

if the group is abelian than i think so

[velvet hull]

it seems like your claim amounts to the question if the product of torsion element is torsion

[kind temple]

ah, right. in a non-abelian group that might not be true

[somber goblet]

if not then not necessarily? because you can have $ab$ with infinite order

[cloud walrus]

lexi

[kind temple]

yea

[velvet hull]

Aut(Z[x]), f(x) = -x and g(x) = 1-x

<f,g> should be infinite

[kind temple]

classic

[somber goblet]

take the group generated by ${a,b}$ under $a^2 = b^2 = 0$

[cloud walrus]

lexi

[somber goblet]

yeah

plenty of counterexamples

[kind temple]

i knew something was off

thanks

[somber goblet]

it is true in the abelian case tho

[kind temple]

yea, the argument i have goes through in that case

this means Z with xor is not finitely generated

[somber goblet]

Z with xor?

[kind temple]

the integers with xor as the group operation

[somber goblet]

how is xor defined on the integers

like bitwise

?

[kind temple]

yea

[somber goblet]

or like $\bigoplus_{\alpha\in\mathbb{Z} }C_2$ type deal

[cloud walrus]

lexi

[somber goblet]

actually i think these might be isomorphic?

wait how do you define negative integers

[kind temple]

ah right

i think you can work with non-negative integers and its fine?

[somber goblet]

yeah

in that case i believe what you have is isomorphic to this

$\bigoplus_{\alpha \in \mathbb N} C_2$

[cloud walrus]

lexi

[somber goblet]

infinitely long sequences of {0, 1} for which finitely many are nonzero

[kind temple]

yea

neat

[somber goblet]

also probably isomorphic to the additive group of $\mathbb F_2 [x]$

[cloud walrus]

lexi

[kind temple]

yea, i see that

working with negatives requires that you only work with strings of length n

[celest rover]

hi!!!!
i have dumb question
take A as the weyl algebra over 1 variable
if you have A/Ax^2 as a left A-module, isn't it non-isomorphic to A/Ax + A/Ax?
the action by x^2 is trivial in the second but not in the first

[quiet pelican]

Yes that’s right

[celest rover]

huh

[quiet pelican]

Why?

[celest rover]

this is in the pdf i'm reading

[celest rover]

yay someone typing

[rocky cloak]

I don't what they were going for / if it's just a mistake. But A/Ax (+) A/Ax is cyclic, which is notable I guess

[celest rover]

oh

imma assume there's a typo somewhere then

[rocky cloak]

Also, I don't think the action of x^2 is trivial on the second. Unless I'm stupid

x^2 d^2 = d^2 x^2 - 4dx + 2
which is 2 modulo Ax.

So maybe they actually are isomorphic by some clever choice of generator

[celest rover]

huh
imma check again

[rocky cloak]

So (A/Ax)^2 is generated by (1, d) and this one is annihilated by x^2

I don't know if Ax^2 is it's annihilator, but if the exercise is true I would guess it is

[celest rover]

i am getting x^2 d^2 = 2dx = 0 (mod Ax)

ok imma check again cause i'm not sure if i'm doing something very dumb now

-[d^2, x^2] = x^2 d^2 (mod Ax)
[d^2, x^2] = 2 + 2xd = 2dx
so x^2 d^2 = -2dx = 0

oh wait

i forgot something

oh yeah i see it now @rocky cloak
i got the commutator for [d2,x2] wrong

sorry

[rocky cloak]

That's my calculation at least

[celest rover]

yeah it's right

i knew my question was going to be dumb....

[rocky cloak]

Alright, so want
||A/Ax^2 -> (A/Ax)^2 by
1 |-> (1, d)
to be an isomorphism.
It's well defined and surjective, so just need an inverse. ||

||(f, g) |-> f + fdx - gx
is a good candidate. Is it will defined? Clearly the g part is. For the f part
x + xdx = x + (dx - 1)x = dx^2
well defined!||

So they are in fact isomorphic

[copper kestrel]

i think i figured out how to show that every ideal of Z_n is principal, but now i need to show that every ideal of Z/n x Z/m is principal and can be of the form I x Z/m and Z/n x J

[rocky cloak]

It might be beneficial to describe what all ideals of a product R x S looks like in general, and prove the more general thing that if R and S are principal ideal rings, then so is RxS

[copper kestrel]

:')

i dont even know how i'd go about doing that

like i guess i can show I x J is an ideal for some R x S

[rocky cloak]

If your stuck I guess it's helpful to consider an example.

Like what are the ideals of Z/4 x Z/6? Any pattern to them?

[copper kestrel]

well its all combinations of ideals I subset Z/4 and J subset Z/6 in the form I x J, and also allowing for I x Z/6 and Z/4 x J

but idk how to prove it

i did it with Z/2 x Z/4 for my homework, and i found the prime and maximal ideals

i just cant justify the ideals are of that form

[elfin wraith]

How did you do it for that case?