groups-rings-fields

[pastel cliff]

same argument could be made for sylow 3-subgroups

[robust pollen]

Let $R, S$ be rings, $M$, $N$ a left and a right $R$ module, $X$, $Y$ a left and a right $S$ module.
We can build the product ring $R \times S$. Then we get for example that $M \times X$ is a left $R \times S$-module by componentwise action, right?
So, let's now consider the abelian group $(M \times X) \otimes_{R \times S} (N \times Y)$. My question is whether this is isomorphic to $(M \otimes_R N) \times (X \otimes_S Y)$?

[cloud walrus]

expectTheUnexpected

[pastel cliff]

is this good enough? feels kinda handwavey but idk what else to write

[eager willow]

That's true, and in fact this is a R×S module isomorphism

[robust pollen]

How so? The tensor product is just an abelian group. (My rings are not assumed commutative)

[eager willow]

Oops it's not an isomorphism of abelian groups nevermind

Thought about it more. It respect the scalar multiplication structure but not additive

there's an obvious map defined on simple tensors

$(m , x)\otimes (n, y) \mapsto (m \

$(m , x)\otimes (n, y) \mapsto (m \otimes n, x \otimes y)$

[cloud walrus]

datorangeguy

[eager willow]

but this is not a homomorphism of abelian groups

oh nm it is hahaha I keep flipping

so yeah that's your isomorphism

well... shoot that's not injective is it...

no it is! if both entries on the right are 0 then one of m is 0 and one of y is 0. By tensor relations this means it is 0. And it's surjective. So that's an isomorphism as it's a homomorphism of modules

[eager willow]

tensor product of modules is between a right module and a left module, so that an R module structure is defined no matter if R is commutative. Without specifying if modules are given left or right action, there are multiple nonequivalent definitions of the tensor product as a group so it's assumed that M(x)N means M is a right module and N is a left module, with relations ar(x)b = a(x)rb but not ra(x)b = a(x)rb.

oh wait that's not right either ah

[robust pollen]

Your first sentence is incorrect. If M is right R, N is left R, then how do you define an R-module structure on (M ot N)?

[eager willow]

I'm just dumb ignore me. but those groups were isomorphic definitely

[robust pollen]

So basically this was your map in the one direction

[hallow kiln]

<= direction is wrong as you did not explain why |G| cannot have more than one prime factor.

[robust pollen]

Oh wait, that's the problem you had before @eager willow . The map I use to define f_1 is not additive. Is it?

[eager willow]

no it's bilinear like it needs to be. Add things in either the first two or the second two entries to check, not all 4 entries.

[robust pollen]

oh haha big dumb

[eager willow]

yeah that's what I went through on paper and got confused about

[robust pollen]

very thank

[chilly ocean]

I'm pretty sure that's because the "theory" in "wheel theory" doesn't even exist.

[coral spindle]

Wheels only exist so people on the internet can say "akshually you can divide by zero "

[rotund aurora]

imagine that for every object there were a "theory" of that object, xD

[eager willow]

just fyi there's a characterization that occured to me to make me guess it was true that I've thought more on and does work. If you can accept that tensor product over a fixed ring is an additive functor, then your situation is just a special case where the RxS action on A and C is just via R (S torsion), and similarly B and D, and you're looking at $$(A\oplus B)\otimes(C \oplus D).$$ Then if you distribute all the terms and cancel the tensor products that are torsion, you get
$$(A\otimes C)\oplus(B\otimes D)$$

[cloud walrus]

datorangeguy

[robust pollen]

I see, that is a lot easier and without whacky computations involving a million universal properties. Now I need only argue the simple fact that A \otimes_{R \times S} C \cong A \otimes_R C 👍

[sly storm]

You can consider the smallest subfield containing the set.

Hello. If $A$ is a finite subset of $\mathbb{C}$ and $r, s \in \mathbb{C}$ are algebraically independent over $A$, then is there an automorphism of $\sigma$ of $\mathbb{C}$ such that $\sigma |_A$ is the identity and $\sigma (r) = s$?

[cloud walrus]

MathPhysics

[karmic jacinth]

Probably

[tender wharf]

Honestly maybe if you showed some work

[sly storm]

I have no idea.

[tender wharf]

MathStackExchange?

[next obsidian]

The answer is yes

[tender wharf]

I'd like to prove that $|G|$ is prime. $G$ is a group with more than 1 element and G has no proper nontrivial subgroups. I have already proved that $G$ is finite. Will this argument work to show that $|G|$ is prime?

Let $|G| = p$. Suppose WPOC that $p$ is not prime. Then, $p = nm$. $1 < n, m < p$ Let $a \in G$ be arbitrary such that $a^n = e$. So $<a>$ is a proper subgroup of G which is a contradiction.

[cloud walrus]

DerpZ

[agile burrow]

The issue with your proof is that there doesn't necessarily exist an element of order n

[next obsidian]

Why does such an a exist

[tender wharf]

Right

Thanks for the help

[next obsidian]

This is the right train of thought

[tender wharf]

The contradiction is fine right?

[next obsidian]

But as Walter said this isn’t possible, take like Z_2^4, every element has order 2

But 16 = 4•4

You have to work a bit harder

[tender wharf]

Yup I'll go think about it more, thanks!

[sly storm]

Why?

[pastel cliff]

is this reasoning correct

[agile burrow]

that sounds fine

[pastel cliff]

i was starting an argument for each p=2,3,5 but then this occurred to me and it feels to simple ( ) to be true

[tender wharf]

Does this idea work: Pick some a in G and form <a>. <a> is a subgroup of G for sure. So |<a>| <= |G| if <a> = G then proper subgroups can be found which is a contradiction. If |<a>| < |G| but again that's a contradiction because <a> is now again a proper subgroup of |G|.

[agile burrow]

why does <a> = G imply proper subgroups can be found?

[pastel cliff]

im guilty of this too but usually when you have an idea, you can see for yourself if it works with like 5 minutes of work

[tender wharf]

we assumed |G| is not prime for contradiction

[agile burrow]

I'm not sure I follow.

I don't see how if <a> = G, then G has a proper subgroup

[tender wharf]

Wouldn't there be a proper subgroup of order |G|/k where k is a divisor of |G|

[agile burrow]

Yeah, that sounds better

You should try to elaborate on that step, can you be more explicit about how to construct the proper subgroup?

[tender wharf]

The subgroup is constructed by <a ^ n/k> where k is a positive divisor of |G| = n

Using the fundamental theorem of cyclic groups

Thank you very much for your help

[south patrol]

oh lol i was gonna say just cauchy it

but I suppose you want to avoid that / haven't been taught that

[next obsidian]

I mean if you have Cauchy’s theorem then the proof is immediate

Lol

[south patrol]

ye

lol

[tender wharf]

Yeah I wanted to avoid that since it's from a later chapter

I'm self learning so I definitely haven't been taught it

[pastel cliff]

can i get a hint for finishing this? mathoverflow soln uses group actions but im wondering if there's a direct argument i can use here, and if what i have so far is good

[coral spindle]

The typical way to do this is to show it has nontrivial center, then show that if G/Z(G) is cyclic then G is Abelian.

Neither of these are super hard, but the first part is harder than the second.

[pastel cliff]

yeah that's what i found online but id already started like this and i got stuck on the last bit

though idk if a group where every element has order p must be abelian

[coral spindle]

Well I think you should prove that first

[pastel cliff]

i'll do the one problem from yesterday soon i promise

[coral spindle]

, so <x, y> has order p^2
Why?

I don't think this is obvious

[pastel cliff]

i thought it would follow pretty easily from the fact that x is not in the cycle generated by y and also has order p

[coral spindle]

I don't think so

I want more justification

and if you can't justify it, then you shouldn't use that

There are groups, for example, where x and y are of order p, but <x,y> is infinite

[pastel cliff]

uhhhh

finite group is assumed

[coral spindle]

That still doesn't help me see why <x,y> has order p^2

[pastel cliff]

informally, can i say that since x is not in the cycle generated by y, it must generate its own cycle. by assumption they both have order p, so both cycles must be of order p

[coral spindle]

Yes that's totally normal

but that doesn't tell me why <x, y> has order p^2

at best it tells me that |<x, y>| >= 2p-1

Also this is a more fundamental error

but G = <x,y> does not mean that G is cyclic

and in fact, you conclude in every case that G is cyclic, and this is just wrong.

[pastel cliff]

dihedral group should counter example this right

[coral spindle]

Well yes, but I should just say that that's not the definition!

[pastel cliff]

nah i see why this is wrong anyways

i'll do it the Z(G) way

t minus 20 hours until my midterm

[chilly ocean]

Hey. I'm tired and v sleep-deprived.
Can I get a sanity check?

Let either one or both of G and H be NON-cyclic groups.
Then the direct product of G x H is necessarily also NON-cyclic.

In other words, if G is a NON-cyclic group.
Let H be any group.
Then G x H is necessarily NON-cyclic.

This is true, right?

[chilly ocean]

Yes because any subgroup of a cyclic group is cyclic, and G can be identified with a subgroup of G x H

you could sanity check it after getting a good 8 hours you know

[chilly ocean]

Thank you! X__X

[chilly ocean]

Wow, why didn't I think of that?

[south patrol]

Or I guess you can use the projection G x H -> G to see that any generator of G x H would give you a generator of G, but that amounts to the same thing more or less

[white nymph]

is there a nice way to simplify $\prod_{i=0}^{n-1} (q^n-q^i)$ ? im looking to compare that quantity with $q^{n^2}$. i didn't really know where to put this so i put it here since it comes from a problem where im figuring out the probability a random nxn mtx over a finite field (with $q$ elts) is invertible.

[cloud walrus]

EricW

[delicate bloom]

compare in what way? since $q^n-q^i<q^n$ you can get the inequality that the product is less than $q^{n^2}$ if that's what you're after

[cloud walrus]

Merosity

[glossy crag]

You can write it as $q^{\frac{n(n-1)}{2}}\prod_{j=1}^{n}(q^j-1)$, so it's better comparable with $q^{n^2}$, but I don't see any other simplifications.

[cloud walrus]

Ocean Man

[white nymph]

i was looking to divide that product by q^(n^2)

[delicate bloom]

you could write it as a product of q^n with the same indices, then reindex to k=1 to n of (1-1/q^k)

[glossy crag]

You can divide the expression I wrote by it and get q^(-n/2)m, where m is P_j(q^j-1). If q happens to be prime, m will not be divisible by q.

Does this automatically prove that every symmetric polynomial is a polynomial in the s_k? Every such polynomial is an element of L^{S_n}, so there must be a rational function in the s_k equal to it. Does it also follow that there must also be a polynomial in the s_k equal to it?

[peak hornet]

what is the constant trivial map?

[shell brook]

The zero morphism, no?

[chilly ocean]

[shell brook]

I guess in Grp that is a homomorphism A -> B that sends everything to 1_B

[chilly ocean]

this shouldn't be an "i guess"

[shell brook]

I mean I'm right

[peak hornet]

so my idea for this question

is just $\phi(h) = e_g$, the identity element in the group $G$

[cloud walrus]

failingphysics

[peak hornet]

i don't want to know the solution if this is wrong

but if it is wrong can someone help me understand why it's not the right one?

I know it's well defined because if h_1 = h_2 then $\varphi(h_1) = \varphi(h_2) = e_G$

[cloud walrus]

failingphysics

I know it's well defined because if h_1 = h_2 then $\varphi(h_1) = \varphi(h_2) = e_G$
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                                           1 = h_2 then $\varphi(h_1) = \var...
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.

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[peak hornet]

so it is a function

and furthermore it is a group morphism because $\varphi(h_1h_2) = e_G$ and $\varphi(h_1) \cdot \varphi(h_2) = e_G \cdot e_G = e_G$

[cloud walrus]

failingphysics

[tribal moss]

That won't work -- it's a homomorphism all right, but it doesn't make the diagram commute.

[peak hornet]

why not? we have $\alpha(h) = e_G$ and $i \circ \varphi(h) = i(e_G) = e_G$

[cloud walrus]

failingphysics

[shell brook]

(Check the LHS triangle, that's what you're interested in making commute)

[tribal moss]

You don't necessarily have alpha(h) = e_G, only f(alpha(h)) = e_G'.

[peak hornet]

ahhhhhh

yes i misread the question

ok let me go back and try to figure it out then

thanks

[tribal moss]

(The diagram is missing a null map from H to G', really).

[peak hornet]

null map?

also so

is $i$ here basically the inclusion morphism?

[cloud walrus]

failingphysics

[shell brook]

Yes

[tribal moss]

"trivially constant" map or whatever your exercise called it

[shell brook]

Tropo is saying that f \circ \alpha being constant trivial is the same as saying that there is a constant trivial map H -> G' and the diagram commutes

[pastel cliff]

is there a proper non-abelian subgroup of S_n which acts transitively on {1, ... , n}

assuming n \geq 4

actually

ive asked this before but i'll reask

what does it mean for the "natural action" of a subgroup of S_n on {1, ... , n} to be transitive

im assuming that the natural action simply refers to the permutation described by each element of S_n

so that being transitive means that there is only one orbit of this action

[tribal moss]

A_n ought to qualify.

[tribal moss]

Yes.

[pastel cliff]

will try showing it

[tribal moss]

Or D_2n.

[pastel cliff]

can i get a small nudge pls - mechanically, it seems like showing that forall x,y \in {1, ... ,n}, there exists a g \in A_n s.t. g(x) = y would be easier than showing that it has only one orbit but idk how to start that

i know that those two things are basically the same, i moreso mean the actual method of the proof

[tribal moss]

It's a "just do it" thing. After you set g(x)=y, fill in the rest of the values of g randomly to get a bijection. If that ends up producing an odd permutation, swap g(a) and g(b) for some a and b that are both different from x.

[pastel cliff]

im not sure i follow, wdym fill in the rest of the values randomly

[tribal moss]

Any which way you like.

Or if you want to be more concrete than that, pick a z distinct from both x and y and consider the 3-cycle (x y z).

[pastel cliff]

i understood it but may have over simplified?

mildly related but how often to short exact sequences come up

im studying for an exam and my understanding of semi direct products is pretty lacking atm, but group extensions are directly (ha) related to that right?

[tribal moss]

(x y) is not in A_n.

[pastel cliff]

oh wow im

[pastel cliff]

that's why you said this lol

[tribal moss]

Pretty much all the time, inasmuch as "short exact sequence" is the category-theoretic ways to talk about quotient objects, and quotients are everywhere.

[pastel cliff]

another quickie but does this look sound?

my main concern is the handwavey argument that that is in fact the klein 4 group in A_4

[tribal moss]

No, that sounds convincing.

[pastel cliff]

what is the motivation for considering semi direct products

that's probably an incredibly broad question so if there's something more refined that i should be asking about that would also be useful and col

[next obsidian]

Within group theory they help@you classify groups

They also are exactly the split exact sequences

And show up in places groups show up in nature, like in NT

[chilly ocean]

I only saw them ever used as "yeah that group is just a semidirect product of these two groups"

Which honestly doesn't tell you that much

[pastel cliff]

the way im understanding it is that it’s moreso a property of a group

[chilly ocean]

It's more like a construction

We can just think about it internally

For other structures we do similiar things, like direct sum of vector spaces, or subdirect product of rings

I'm actually not sure if internal subdirect product is a thing

[pastel cliff]

this smells important?

[chilly ocean]

It's like Chmonkey already said. You describe groups using semidirect products, it's just kinda an useful way

Like for exemple the dihedral groups, you havy x, y as generators and you can write yx = x^a y^b

[pastel cliff]

semi direct product of C_2 and C_n right

[chilly ocean]

And this gives you a semidirect product

I think so

[pastel cliff]

im reviewing my prof's lecture notes from a week where i was terribly sick, we apparently spent three whole lectures on semidirect products lol

[chilly ocean]

Dihedral group is what I think about for semidirect products

Oh. Maybe they have more applicatioms than I thought. Spending 3 whole lectures on them

[pastel cliff]

something something finitely generated abelian groups

i would post the notes but i dont wanna doxx myself lol

[chilly ocean]

If it's abelian then it's a direct product though

I'm pretty sure

[pastel cliff]

yeah that's like the ultimate thing it proves

[chilly ocean]

Oh

[chilly ocean]

I guess you could state it as

Any finitely generated abelian group is isomorphic to a product of quotients of Z

[pastel cliff]

yeah i see that

there's no way i learn all this tonight lmao, t minus 9 hours until exam

[chilly ocean]

It seems a little bit more like something that has potential generalizations

Then

Any finitely generated R-module is isomorphic to a product of quotients of R perhaps?

I think that's a theorem?

[pastel cliff]

idk what a module is

[chilly ocean]

A vector space but over ring

[pastel cliff]

the only ring i fuck w is elden ring

[lethal dune]

over PID yes

ftofgmopid

[chilly ocean]

Is there a homological proof?

[lethal dune]

meaning?

proves I've seen uses SNF

[pastel cliff]

this should just be an application of orbit stabilizer right

[lethal dune]

do you need orbit stab for this?

[pastel cliff]

the equality looks very similar to |G| / |Stab(g)| = |Orb(g)|

[chilly ocean]

SNF?

[lethal dune]

smith normal form

[chilly ocean]

Ah. It's a known acronym to me

[lethal dune]

what do you mean by a "homologica proof"?

using homological algebra?

[chilly ocean]

Yes

[chilly ocean]

It's not a known acronym by me*

[robust pollen]

In hindsight, a motivation could be: linearizing the concept of "semidirect product of groups" leads to "smash product of Hopf algebras", and these have some very nice categorical interpretations.

If I have a ring R, an element x in R, what does the ideal R x R look like? Is it elements of the form rxs with r, s in R, or is it elements of the form r_1 x s_1 + r_2 x s_2 + ... + r_n x s_n ?

[chilly ocean]

The latter

[robust pollen]

Ya I'm dumb... otherwise it wouldn't even be a subgroup of the abelian group lmao

[chilly ocean]

Because it contains rxs and you make it an ideal so throw all additions too

And then it satisfies the definition so we're done

If R were just a semigroup then it would be the former though

[robust pollen]

I was a bit confused by, for example, how Rx looks like.

[chilly ocean]

Yeah, it's confusing

But even more confusing is remembering which of Rx, xR is left and which one is right ideal

Commutative rings ftw

[robust pollen]

hm, I don't think so. Rx captures multiplication on the left, so it's a left ideal

[robust pollen]

Nah 😄 tbh I've been doing abstract stuff so much that I don't even want to consider the category of vector spaces anymore, I

I'd rather consider the category of bimodules over a field. But that really is confusing.

[chilly ocean]

Lol

If that's something you like

I think it's okay to consider left/right ideals as long as you stick to one. Probably better to remember too if you do that

[coral spindle]

Have you seen that irreps of Abelian groups over an ACF are dim 1?

This is that + Mashke

You just consider it on the subgroup <g>.

This will have been discussed earlier in these notes so I suggest you go back and check

<g> is Abelian.

[coral spindle]

Like I said here…

[sly storm]

Hello. If $A$ is a finite subset of $\mathbb{C}$ and $r, s \in \mathbb{C}$ are algebraically independent over $A$, then is there an automorphism of $\sigma$ of $\mathbb{C}$ such that $\sigma |_A$ is the identity and $\sigma (r) = s$?

[cloud walrus]

MathPhysics

[trail stump]

so I guess an orbit of a permutation of a set A can only have one element?

[hidden haven]

Yes. Let k denote the algebraic closure of ℚ(A) inside ℂ, which doesn't contain r or s. Then you want an extension of the identity automorphism of k to a map with r ↦ s. Extend both r and s to transcendence bases B and C of ℂ over k, and B and C must have the same cardinality. Take a bijection between the 2 that maps r to s. Extend this to a map of the purely transcendental part of the extension, then extend to a map of the algebraic closures (which is ℂ in both cases)

Commuting operators have the same eigenvalues

In the sense that any eigenvalue of one is an eigenvalue of the other

[sharp dirge]

how to show 3 is irreducible in $Z[\sqrt{-5}]$

[cloud walrus]

pewdssssssss

[sharp dirge]

just straight from defintion ?

[chilly ocean]

N(x+sqrt(-5)y) = x^2+5y^2, then N(3) = 9
But there is no a with N(a) = 3

[coral spindle]

We consider the representation restricted to <g>. It is diagoalisable by the theorems I mentioned (i.e. it has a basis of eigenvectors, which are just 1d reps) so we are done

[sharp dirge]

wym

[chilly ocean]

N(u) = 1 iff u is a unit, so the only way that 3 could be reducible if 3 = ab where N(a) = N(b) = 3

because N is multiplicative

[sharp dirge]

why cant N(a) = 9

[chilly ocean]

because then N(b) = 1

and that means b is a unit

[sharp dirge]

ye so 3 is irreducible

so you are saying that N(a) or N(b) has to be 9

@chilly ocean

[chilly ocean]

Rather, I'm saying N(a) or N(b) has to be 1

so one of them is a unit

I'm not an algebraic number theorist in disguise

[sharp dirge]

kk in turn one of them has to be a 9

and norm = pm 1 iff unit

[chilly ocean]

But I don't care if they are 9

[sharp dirge]

yea but it implies that the other hast o be 1

whats a good way to show that (3) is not prime in Z[\sqrt{-5}] ?

[lethal dune]

look at (1+sqrt(-5))*(1-sqrt(-5))

[sharp dirge]

thnx

[cloud walrus]

koaaa

[weak flower]

Also I wonder if $V^* \otimes_{kG} W$ is the vector space of the module morphisms. It is naturally a quotient of $V^* \otimes_k W$, which reminds me of some projection formula. However the projection formula requires the finiteness of $G$ but this construction only of $V,W$, so I guess I’m missing something here.

[cloud walrus]

koaaa

[weak flower]

Wait no, it’s some sort of biggest quotient of Hom on which G acts trivially… I don’t know how to describe that.

[pastel cliff]

how is the normalizer condition weaker?

[sly storm]

Thank you.

[patent ocean]

If shown that it is well defined. Do I need to use sequences to show that 1/f is continuous on this open ball

[chilly ocean]

algebra?

[chilly ocean]

like TTerra is trying to say, it's not algebra, but this just follows from continuity of 1/x

[lethal dune]

talgebra

[chilly ocean]

"trying to say"?

[patent ocean]

So do I assume the distance fund used in the reals is the modulus

Oh my bad which channel is metric spaces

[chilly ocean]

#real-complex-analysis

#point-set-topology

[pastel cliff]

#cats

[lapis trail]

How do I solve this?

[chilly ocean]

assuming if f:Z/2Z to S_5, by conjugate of f by g in S_5 you mean x -> gf(x)g^-1, note that f(1) must be trivial or 2-cycle

also, g(i, j)g^-1 = (g(i), g(j))

[lapis trail]

Oh, I think I see what I did wrong

[chilly ocean]

I don't like how they say "the number" though

seems like they want number of conjugacy classes for a non-trivial homomorphism

[chilly radish]

It can also be a product of 2 disjoint transpositions

[chilly ocean]

Oh right.

[chilly radish]

Seems to me they want the number of conjugacy classes of Hom(Z/2,S_5), that is "the number"

[chilly ocean]

Ah. I was thinking of amount of elements in each conjugacy class. Makes sense

[chilly radish]

This isn't a group I don't think, but I don't think that should matter either way

[lapis trail]

The next question is the amount in each

[pastel cliff]

just finished my midterm, if i knew like 3 more facts about semi direct products i would've done sm better

[sharp dirge]

I want to show $(2, 1 + \sqrt{-5}) \neq (a + b\sqrt{-5})$ in $\mathbb{Z}[\sqrt{-5}]$. i.e not a principle ideal in the quadrratic ring. Suppose that $(2, 1 + \sqrt{-5}) = (a + b\sqrt{-5})$. Then $2 = \alpha (a + b\sqrt{-5}), Applying norm. 4 = N(\alpha)(a^2 + 5b^2)$ . Which implies that $a^2 + 5b^2 = 1,2,4$. It cant be 2 cuz no solutions. If $a^2 + 5b^2 = 4$, Then $a = \pm 2$ which implies $1 + \sqrt{-5} = \pm 2c \pm 2d\sqrt{-5}$ which cant happen. So if $a^2 + 5b^2 = 1$ then $a = \pm 1$. Which implies that $2x + (1 + \sqrt{-5})y = 1 $. after multiply by $1 - \sqrt{-5}$ it implies that 2 divides $1 - \sqrt{-5}$ which cant happen. so $(2, 1 + \sqrt{-5})$ is not principle

[cloud walrus]

pewdssssssss

[sharp dirge]

anyone can say if this looks ok?

[lethal dune]

what did you do in the last part? 2x+(...) shows 2 divides 1-sqrt(-5)

[sharp dirge]

if a = plus minus 1

then (2, 1 + \sqrt{-5}) = (1)

so 2x + (1 + \sqrt{-5})y = 1

x,y in Z[\sqrt{-5}]

then multiply by 1 - \sqrt{-5}

for contradiction

does that work?

[lethal dune]

you are getting 2x{1,-1}+6y = 2{1,-1} => 2(x-1){1, -1} = -6y => (x-1){1, -1} = -3y

what after that? {1, -1} = 1-sqrt(-5)

[sharp dirge]

so $2x + (1 + \sqrt{-5})y = 1$ then $2x(1 - \sqrt{-5}) + (1 - \sqrt{-5})(1 +\sqrt{-5})y = 1 - \sqrt{-5}$

[cloud walrus]

pewdssssssss

[sharp dirge]

then $2x(1 - \sqrt{-5}) + 6y = 1- \sqrt{-5}$

[cloud walrus]

pewdssssssss

[sharp dirge]

then $2 (x(1-\sqrt{-5}) + 3y) = 1 - \sqrt{-5}$

[cloud walrus]

pewdssssssss

[sharp dirge]

so 2 divides 1 - \sqrt{-5}

does that work?

@lethal dune

[lethal dune]

I see

looks fine now

[sharp dirge]

awesome

[lapis trail]

Is using a shit ton of symbols the norm in writing for abstract algebra, and I just need to get gud; or is it poor, unclear writing? I borrowed a book and I have to keep searching to know what their notation means.

It might be easier to understand if I completed the book cover to cover but I don't have time for that

[lethal dune]

you have time to complain

[lapis trail]

I take that as a yes?

[lethal dune]

take that as 'idk'

[coral spindle]

Without an example of what you're referring to it's hard to tell.

[lapis trail]

"Group Theory" by W.R. Scott is the book if anyone knows it

[sharp dirge]

taking algebra course was mistake for me

too many defintions

[coral spindle]

You're right, we should only talk about things that we already know about. No new things.

[pastel cliff]

i only know that i know nothing

[chilly ocean]

what about the new things that we already know about

[coral spindle]

Define what you mean by—WAIT NO

[formal ermine]

why is $(ab)\inv = b\inv a\inv$

[cloud walrus]

illuminator3 (#eric4honorable)

[coral spindle]

Remember that the inverse is unique?

[formal ermine]

yes

ahhhh

yes

I get it

thanks

[coral spindle]

Well notice that $(b^{-1}a^{-1})(ab) = b^{-1}(a^{-1}a)b = e$

[cloud walrus]

Boytjie

[formal ermine]

yeah

[coral spindle]

Well you noticed it before I could finish

[formal ermine]

lmao

thanks

what does "structure preserving" mean?

[chilly ocean]

it's informal speech

if your structure has addition for example, we preserve addition

if it has order, we preserve the order

if it has multiplication, it preserves multiplication

[formal ermine]

what does 'preserve' mean here?

[chilly ocean]

commute with operations/if relation holds then it holds under the map as well

it depends on if we are talking about preserving a particular n-ary operation or an n-ary relation

moreover we can talk about preserving something else

so it's more complicated than that

[formal ermine]

no idea what that means. for context, my prof said "a homomorphism is a structure preserving map" so I was wondering what it means in that context

[chilly ocean]

it means that it commutes with operations

so what the definition of homomorphism says

[formal ermine]

what does to commute mean here

[chilly ocean]

xy = yx

but for maps

and well

not exactly

[formal ermine]

wdym?

[chilly ocean]

did you read the definition of homomorphism

I already said: it's informal

[formal ermine]

yes

I'm just having a hard time right now because my lecture is in german and all of the terms are different in english

[chilly ocean]

symbols are the same

just stare at the definition of homomorphism and you'll see it

[formal ermine]

I still don't understand what "it commutes with operations" means

[chilly ocean]

the order in which you apply the map and the group operation doesn't matter. you can do them in either order and get the same result. they commute

[formal ermine]

but that's the definition of a homomorphism, no?

[chilly ocean]

yes, and?

[chilly ocean]

Can Fraleigh be a enough prerequisite for Eisenbud?

[coral spindle]

I would say so, yes

[chilly ocean]

But I've read fraleigh and then read eisenbud but couldn't understand...
Is it cause of that I didn't read chapters with a asterisk? ..

[coral spindle]

So you're saying you only read parts of fraleigh? I think that explains it.

[soft skiff]

can someone check help 17

its serious

[coral spindle]

Don't spam requests for your help channel in unrelated channels.

[ruby sundial]

Funny question

Is there an algebraic way to talk about integration?

[sage spruce]

I guess in C*-algebras

[zenith quiver]

if H is a subset of G, it's automatically associative right?

[sage spruce]

Yes

But the product does not necessarily remain in H

So it's not an internal law anymore

[upper pivot]

Yeah I mean integrals are basically just linear operators on e.g. compactly supported functions on X

Riesz representation is maybe what you are looking for

If you look into some of Pedersons work, he hates measure theory and wants to replace it with a more linear algebraic formulation

Maybe you will find some of this useful

(His book analysis now has a chapter that does measure theory 101 in this framework)

[zenith quiver]

how do we get the identity in C(a) {gag^-1=a} I'm assuming we can set a as g to get g(gg^-1)=g, can we say the identity is in C(a) because of this?

[zenith quiver]

or wait if g and g^-1 exists in C(a) can't i just say gg^-1=e thus e is in C(a)?

[drowsy summit]

I'm taking my first formal abstract algebra course (mostly group theory stuff), and something that has confused me consistently is why, if the definition of "operation" suggests that we may only call a rule an operation if it maps an ordered pair in a set to one element of the set. However, in computer science and philosophy, we often refer to unary operators and ternary operators. If something like logical not isn't an operation, what is it?

[sage spruce]

It's not an operation it's an internal law

Which is bery specific

You can define all kinds of operations in fact

[drowsy summit]

Interesting okay. I'll look internal law up

Thank you

I imagine then that it's just lazy language to call something like logical not an operator?

[sage spruce]

Just gag^-1=agg^-1=a in C(a)

[sage spruce]

It's a boolean operator defined on specific sets

[zenith quiver]

we didnt assume its abelian

[sage spruce]

Yes

But you're in the commutant of a

So every element commutes with a

[drowsy summit]

So I can refer to n-ary rules with arbitrary n as operations?

[zenith quiver]

im too early to know what commutant is

[sage spruce]

C(a) is defined as the commutant of a isn't it? (Like the set of elements that commute with a)

[sage spruce]

Yes

[drowsy summit]

Thank you that clears up a lot actually

My professor was faily adamant about operations needing to be binary, which seemed off

[zenith quiver]

to me they just defined it as C(a)={gag^-1=a} and asked to prove its a subgroup of G when a is in G

[sage spruce]

Which means that ga=ag

[zenith quiver]

how

[sage spruce]

Multiply by g on the right

[tender wharf]

from gallian

It's the same thing as that definition yeah although this one is a bit clearer imo

[chilly ocean]

based

[zenith quiver]

SL2(R) is a group under addition right?

[chilly ocean]

no

I + (-I) isn't even in there

it is a group under multiplication, though

[hollow mica]

I'm trying to see why every abelian group admits a unique Z-module structure

I know how to find the module: just define scalar multiplication as n * g := (g + g + ... + g) where there are n g's

I'm not sure how to prove that this module is unique

[rotund aurora]

I think its just because 1*v should be v, and Z is generated by addition from 1

[hollow mica]

ohh

that makes sense

this is an alternative proof that I'm trying to understand

it's the same for the most part except it uses the "endomorphism ring of a group"

I'm not sure what they mean by "By applying this to End(G)...", could anyone explain?

[zenith quiver]

can anyone find the wiki for this? im dying to get information on it i know its all the 2x2 matrices with det not equal to 0 but i cant find visual confirmation of it anywhere

[hollow mica]

usually M_n(F) means all n by n matrices with entries in a field F

sometimes M_{x, y}(F) means all the x by y matrices with entries in a field F

maybe here it is the latter and they forgot the comma

if you don't want matrices with det equal to 0 you usually notate this as GL(R, n), the "general linear group"

[zenith quiver]

so in this case, theyre talking about GL(2,R)

[hollow mica]

no where do they say the det has to equal 0

[hollow mica]

so this

[hollow mica]

like H very clearly has elements with det equal to 0

[zenith quiver]

if H very clearly has elements with det equal to 0 then how is it a subgroup since it's inverse would be 1/det

[hollow mica]

the group operation in M_{2, 2}(R) is addition

that's why it also has people with det equal to 0

whereas GL(R, n) doesn't (group operation is multiplication)

[zenith quiver]

so is the identity for this group 0? or (0,0,0,0)

[hollow mica]

the identity is the 0 matrix, the matrix with entries all equal to 0, but it is very common to denote the identity in groups with an additive structure as just the number 0

[hollow mica]

I'm confused

isn't R the set of two variables polynomials in x and y

so like some sum a_{i,j} x^i y^j

and then the ideal generated by x, y is I = {ax + by | a, b \in k}

for a module to be free over R, it can't be isomorphic to any power of R (definition)

why is I not isomorphic to any power of R?

[agile burrow]

do you understand why a Z-module structure on G is the same as a ring homomorphism Z -> End(G)?

[hollow mica]

yes, it's like currying

[pastel cliff]

walter that graduate+ role looks mighty fine on you

[hollow mica]

z -> (g -> g) is equivalent to (z, g) -> g

[agile burrow]

Do you understand why there's a unique ring homomorphism Z -> End(G)?

[pastel cliff]

could i possibly bother you about some grad-related questions? i have an important decision to make soon and ive been told talking to other students is a good idea

[agile burrow]

i'm not a grad student lol

[wicked zephyr]

is a module over itself, so that ideal is {ax+by with a,b in k[x,y]}

[hollow mica]

oh ah I see it now

[hollow mica]

oh yeah

that

[pastel cliff]

nvm lmao

i also realize im doing the literal "hey can i ask u dis help me pls urgent" but talking to literally anyone would be helpful, dont kill me mods

[hollow mica]

but isn't that just k[x, y] without elements with constant term = 0

[agile burrow]

if (x, y) were free, what do you think the rank would be?

[wicked zephyr]

in deed

[hollow mica]

(x, y) is the submodule generated by x, y? (the same one as above)

[wicked zephyr]

yeah

[hollow mica]

i would think the rank is 2

in fact i don't know why it isn't

isn't it by definition generated by a two element set

[agile burrow]

The issue is that x and y don't form a basis for a free module

Can you see why?

[wicked zephyr]

it is not difficult to prove that in general, a free ideal over domains are generated by just one element

[hollow mica]

well

I have two definitions of free module rn

[wicked zephyr]

so, if a,b are distinct elements, then a(b)+b(-a) is linear depende, and therefore these cannot form a basis.

But then it reduces the problem to prove that (x,y) is not a principal ideal in k[x,y]

[hollow mica]

"isomorphic (as modules) to some power of the ring it is over"
and
"has some finite basis"

is {x, y} not a basis for the submodule we just generated by definition?

[wicked zephyr]

which you can do using coprimality of the variables I think

[hollow mica]

oh

I can multiply by polys

that makes sense

[wicked zephyr]

You have to be careful, we are considering as a module over k[x,y]

[hollow mica]

ok let me write up a proof

[wicked zephyr]

Did you get the idea?

[hollow mica]

I think

[zenith quiver]

this is honestly a bit confusing, a lot of variables and I'm not sure what i should work with

[hollow mica]

consider the module k[x, y], and the submodule {a*x + b*y | a, b \in k[x, y]}.

wait shit

we only showed {a, b} isn't a basis

we didn't show there exists no basis

[agile burrow]

Potitov06 explained why if I is an ideal which is a free module, then I must be principal

[hollow mica]

ah ok I see

another side question

R is clearly free over R for any ring R

by the definition "must be isomorphic to some power of the base ring"

but how do I see this from the definition "has a finite basis"

what's the basis of the R-module R

[agile burrow]

{1}

[hollow mica]

oh true

[lapis trail]

Is there a way to construct a homomorphism based on what properties must be conserved?

[delicate bloom]

maybe?

[lapis trail]

How about an homomorphism from a space onto itself?

[delicate bloom]

whos gonna stop me if i do?

[next obsidian]

Me

[delicate bloom]

[zenith quiver]

if i have an element lets say A with order n, and an element B with order m can AB be infinite?

[agile burrow]

Yes. Consider the group G with presentation <A, B | A^n = B^m = e>

Alternatively, you could probably construct some examples with small matrices

[zenith quiver]

yeah ive got a matrices problem where the order of A is 4 and order of B is 3 but it feels like it doesnt make sense because when i take AB then find the order it turns to be infinite but shouldnt AB just be 12? 4*3?

[agile burrow]

No, you can't expect that to be true in general when your elements don't commute

[zenith quiver]

so A^12B^12 =/= (AB)^12?

[agile burrow]

Right

[zenith quiver]

if my elements dont commute?

[agile burrow]

Exactly

And you can see this very explicitly by just writing out what (AB)^12 is; it's just (AB)(AB)...(AB) 12 times

You can't necessarily rearrange this to be A^12 B^12 when A and B don't commute

[zenith quiver]

makes sense

thank you boss

i have no idea how to even begin with this

[agile burrow]

What are the properties a subset of G needs to satisfy to be a subgroup?

[zenith quiver]

non empty, has the identity, a,b belongs to the subset then ab^-1 must belong there too

[zenith quiver]

but here, do i prove its a subgroup of H or G, since if its a subgroup of H then its a subgroup of G but here they're talking about x in G and h in G

[agile burrow]

You prove it's a subgroup of G. In fact, K is not necessarily a subgroup of H. As it's defined, K is the set of elements x in G such that x = aha^-1 for some h in H. However, there's no reason aha^-1 should be in H if a is not in H.

[zenith quiver]

if we take the inverse of x=aha^-1 it'll be a^-1h^-1a right?

how is this possible?

[agile burrow]

The inverse they compute is correct. Try multiplying it out to see why and note that inverses in a group are unique

[zenith quiver]

wdym multiplying it out?

[agile burrow]

Just multiply x and x^-1 to see that the inverse they computed is actually correct

[zenith quiver]

sound good? @agile burrow

[scarlet estuary]

please use LaTeX

[chilly ocean]

Using desmos to write math is the funniest thing I ever seen

If you want something external then you can write in markdown or on a whiteboard... or a piece of paper

In #resources there's a cheat sheet about how you can type in LaTeX, here directly using a bot

[lethal dune]

omg

[trail stump]

for 39,d isn't A4 a counter example?

oh wait nvm

[woeful sage]

${\bZ}_{1009}$ means a ring of integers with addition mod 1009 right?

[cloud walrus]

Neamesis

[tribal moss]

Yes, sometimes.

It may also mean the 1009-adic integers, which is a quite different beast.

[chilly radish]

If you have to ask it's probably that

(integers mod 1009)

[woeful sage]

so i saw a problem "find the multiplicative inverse of [17] in ${\bZ}_{1009}$"

[cloud walrus]

Neamesis

[south patrol]

Lol

[tribal moss]

Definitely integers mod 1009 then.

[woeful sage]

so how would multiplication work in that? also mod 1009?

[trail stump]

i thought it really depends on the context

[tribal moss]

Yes.

[woeful sage]

okay thanks

[woeful sage]

hey u became mod

[chilly radish]

Indeed

[tribal moss]

If there's a "[17]" in the context, then p-adics looks unlikely.

[woeful sage]

what do the brackets signify?

[chilly radish]

Residue class

[tribal moss]

The congruence class of 17, i.e. the set {..., -992, 17, 1026, 2035, ...} of numbers congruent to 17 modulo 1009.

[trail stump]

that seems right

[woeful sage]

hm i see

[chilly ocean]

And I'll say equivalence class of 17 just to be different

[south patrol]

And I'd just abuse notation and say 17

[tribal moss]

Yeah, the brackets are only really used in introductory texts that need to emphasize how things work formally.

[chilly ocean]

If alpha is the real sixth root of 2 and zeta = exp(i * pi/3) is the primitive 6th root of unity, then I know that Q(alpha, zeta) is the splitting field of the polynomial x^6 - 2. We can define the automorphism sigma : alpha -> alpha * zeta; using this, how can I show that the fixed field Q(alpha, zeta)^{<sigma^2>} = Q(sqrt(2), zeta)?

The generator <sigma^2> gives the set of automorphisms: {id, sigma^2, sigma^4} I think

[chilly ocean]

no, take everything to e

i get the feeling you meant to write more than "homomorphism"

what is the "category theory definition"?

[formal ermine]

f : G -> G is an endomorphism though

[chilly ocean]

to be distinguished from "automorphism"

[formal ermine]

yes

homomorphism: G -> H
endomorphism: G -> G
automorphism: G -> G, bijective
isomorphism: G -> H, bijective
monomorphism, G -> H, injective

[chilly ocean]

i feel like you're not really addressing the point illuminator

[lament dawn]

Nobody likes ur boy epimorphism 😔

Oops I thought this was discussion myb

[chilly ocean]

the morphisms in this definition are not going to be the morphisms of the corresponding group, but the elements of the corresponding group instead

... something like that

so you might say that the morphisms of the corresponding group are "morphisms of the morphisms of the groupoid" or something

there's probably some category theory jargon for this but i don't care

there's a reason people typically learn about groups before they do categories

i didn't say you haven't

[formal ermine]

$n\bZ + k = { nz + k | z \in \bZ }$?

[cloud walrus]

illuminator3 (#eric4honorable)

[coral spindle]

Yes

[formal ermine]

thanks

the square u operator here is union?

[coral spindle]

Disjoint union

[formal ermine]

what's that

[coral spindle]

A union of disjoint sets

[formal ermine]

what's a disjoint set

[coral spindle]

Two sets are disjoint if they have no common elements

[formal ermine]

ah

[formal ermine]

why do we include H itself in G/H where H is a subgroup of G

[coral spindle]

I don't know what you mean by that. H is annihilated under the natural map G -> G/H

Oh are you saying why we consider H as a coset of H?

Because H = e.H 🤓

and in fact for any h in H, H = h.H

It is natural and turns out to be important as a coset.

[chilly ocean]

I mean, there's no particular reason for that

[formal ermine]

ah

[chilly ocean]

We just use congruence classes as elements of G/H, but we could use something else

like representatives of such classes

[formal ermine]

it just feels weird cuz left coset is linksnebenklassen in german which translates as like left classes besides

if $|G/H| = 1$ then $G = H$?

[cloud walrus]

illuminator3 (#eric4honorable)

[chilly ocean]

I mean it's beneficial to represent them as whole congruence classes since then it's clear how can we act with G on elements of G/H

[coral spindle]

Yes

[formal ermine]

ahhhhhh wait I think I finally understand what G/H is

G/H are H and the different disjoint sets needed to represent G?

[coral spindle]

I don't like that interpretation particularly

G/H consists of equivalence classes of G. These classes are elements of G that differ by something in H.

[formal ermine]

equivalence classes?

[coral spindle]

Equivalence classes.

[chilly ocean]

it's because you're quotient by a normal subgroup, you really are quotienting by a relation generated by it

[coral spindle]

Slow down there, we haven't seen normal quotients yet

This is just an arbitrary subgroup H

[chilly ocean]

oh wait. My bad lol

still though, what you really are doing is quotiening by a relation, which says that x ~ y iff xH = yH

[formal ermine]

what's an equivalence class

[coral spindle]

A set of things that we consider "the same" up to some definition of what we mean as "the same"

You likely know about equivalence relations

an equivalence class is just a set of all things which are equivalent to each other under an equivalence relation

[formal ermine]

ah

and what is meant by "equivalence classes of G"

[coral spindle]

These classes are elements of G that differ by something in H.

Also see Blitz' messages

[chilly ocean]

Because this relation is on G. By itself it doesn't mean anything

maybe they should teach group actions before introducing quotients

[coral spindle]

lol maybe

certainly they should introduce equivalence relations...

[formal ermine]

I don't understand that sentence

like uh

how can a class be an element of a set

isn't a class a set itself

or am I misunderstanding what a class is

[chilly ocean]

because then I could just say that G/H for some group H is what you obtain when you want to act with G on some quotient of G by an equivalence relation

which kind of justifies the need to quotient by stuff

[chilly ocean]

everything is built from sets

[formal ermine]

but how can a class be an element of a set that e.g. only contains numbers

[chilly ocean]

numbers are sets

[formal ermine]

yes von neumann ordinals

but I still don't understand boytjie's sentence

[chilly ocean]

It means that elements of G/H consist of subsets A of G such that any two elements of A differ by something in H

[formal ermine]

but then the subsets aren't disjoint anymore if they only differ in one element?

wait

[chilly ocean]

you misunderstood me. I'm not sure what you understood, but it's not what I'm saying

[formal ermine]

yeah

so for any two elements a_1 in A_1, and a_2 in A_2, a_1 = h \circ a_2 for some element h in H?

[chilly ocean]

no, a_1 in A and a_2 in A, a_1 = a_2 h for some element h in H

[coral spindle]

That is quite the opposite of the definition

if A_1 and A_2 are distinct classes, there should not exist an h for which that works

[chilly ocean]

Yeah. I'd give you something non-commutative but I'm not sure of a good early example, but do try this on G = Z and H = 2Z

[formal ermine]

that's the example they also used in the lecture

[chilly ocean]

try to maybe, plug this in in what we are saying

and see if it makes sense

[formal ermine]

I'm sorry but I have no idea what you guys are saying :c

I still don't know what "equivalence classes of G" means

[chilly ocean]

.

it means we are thinking of some equivalence relation, and under this equivalence relation, they are equivalence classes of G

[formal ermine]

what's 'they' here?

[chilly ocean]

elements of G/H

[coral spindle]

Illuminator, do you know what an equivalence relation is?

[formal ermine]

uh I don't think I know exactly what it means

[coral spindle]

So the answer is no

[formal ermine]

yeah

[coral spindle]

You are likely to have seen it before if you are doing a math degree, but iirc you're not are you

[chilly ocean]

[coral spindle]

This is an OK introduction. https://byjus.com/maths/equivalence-relation/

[chilly ocean]

you should probably go and read some set theory introduction

[coral spindle]

Woah even better: http://www-math.ucdenver.edu/~wcherowi/courses/m3000/lecture9.pdf

[formal ermine]

this one works

[coral spindle]

OK then just use the PDF I linked :)

[formal ermine]

lemme read it

[coral spindle]

If you have questions about equivalence relations and classes, ask them in #discrete-math as here is not the right place.

[formal ermine]

alr

[formal ermine]

equivalence classes in(?) what equivalence relation?

[coral spindle]

.

[formal ermine]

mhm alright

thanks

another question

what's this operator here

the disjoint union with the dot on top

[chilly ocean]

people, usually in probability, denote this to mean the sets are disjoint

it's just a normal union

[formal ermine]

and what about this one

[formal ermine]

so this one is disjoint union

[formal ermine]

and this one is regular union?

[chilly ocean]

both are regular union

[formal ermine]

wait I think this one is regular union

[formal ermine]

and this one is disjoint

[formal ermine]

because after this line

[chilly ocean]

both are regular unions

the dot is only an additional information than those sets are disjoint

same with square unions

[formal ermine]

than or that?

[chilly ocean]

that

[formal ermine]

ahhhhh

yeah

that's what I meant

probably used the wrong terminology

alright

thanks

what's $\bZ_k$? like how is it defined in set builder notation

[cloud walrus]

illuminator3 (#eric4honorable)

[celest cairn]

Do you mean like $\bZ_n$?
$\bZ_n$ = /{0,1,…,n-1}

[chilly ocean]

[celest cairn]

Idk how to have the brackets.

[chilly ocean]

backslash them

[celest cairn]

oh thx lol

[formal ermine]

but then (Z_n, +) isn't a group

or 1 + n-1 = 0?

[chilly ocean]

addition is modulo n

[formal ermine]

ah

ok thank you

[celest cairn]

What is this?

Idk what this means.

[chilly ocean]

go to #latex-testing

[celest cairn]

Ok anyways what’s the picture above?

[next obsidian]

That’s just Q(-1) lol

[celest cairn]

ok how about like
Q(zeta 5 + (zeta 7)^2)?

[chilly ocean]

If I have G a group and x->x^3 a surjective morphism. Is G abelian ?

[eternal schooner]

what does this notation mean? K is a field

[lavish gull]

i think <x1,x2> is the ideal generated by x1 and x2

[eternal schooner]

we denote ideals with ( )

here they re defining a quotient K<x_1,x_2>/ (x_1x_2-c)

[gloomy night]

Might just be the field of rational expressions in x_1 and x_2 over K, but I usually see that as K(x_1,x_2).

Or might even the polynomial ring K[x_1, x_2].

[coral spindle]

Typically K<a,b> is the ring of non-commuting polynomials

so it's one in which ab is not ba

[gloomy night]

So K<a,b>/(ab - ba) = K[a,b] (isomorphic)

[coral spindle]

Yeah, if indeed this is the notation intended. It is a bit hard to tell without more context though

[south patrol]

free associative (not necessarily commutative) K-algebra on {x1,x2} ig

[celest cairn]

What is this and how would I find the minimal polynomial?

[gloomy night]

Two very different questions in one. Are these squiggly symbols roots of unity or smth?

[celest cairn]

Roots of unity

[chilly ocean]

Can someone please help me with this question

[lavish gull]

algebraic numbers, i would think

[chilly ocean]

I'm not sure but note that it implies that ab and ba are conjugates. Maybe that's important

[lavish gull]

not clear what the subscript is to me though

[delicate bloom]

looks like zeta with subscript, this notation is commonly used to denote roots of unity of order equal to the subscript

[chilly ocean]

How ?

[chilly ocean]

ab = x^3 y^3 = y^-1 (yx)^3 y = x (yx)^3 x^-1 = y^-1 ba y = x ba x^-1

[lethal dune]

look at GL_n(ℂ) maybe

or some subgroup

GL(n, ℂ) works as every matrix here have a cuberoot, so x ↦x³ is surjective

[thorn delta]

i don't think x^3y^3 = (xy)^3 is true in GL(n, C) though

[lethal dune]

do they require it to be a homomorphism? only said morphism

ig they do

[thorn delta]

morphism of groups surely

[lethal dune]

forget it then

[delicate bloom]

well can't hurt to look for a counter example maybe to prove it wrong I guess lol

[chilly ocean]

I think the result is right

[celest cairn]

Anyone know

I know it’s zeta and the roots of unity, I just don’t know how to find the minimal polynomial of Zeta.

[delicate bloom]

one way to try to do it might be take $\alpha-\zeta_5$ and plug it into the cyclotomic polynomial $\Phi_7(\alpha-\zeta_5)=0$ and then try to work further on playing with the $\zeta_5$ terms to simplify more until it's just $\alpha$ in terms of rational coefficients, then you can gauge to the degree of $\bQ(\zeta_5,\zeta_7)$ to see how good you are to try to see if you can simplify it some

[cloud walrus]

Merosity

[delicate bloom]

maybe there's a better way, but this is something to get started

[celest cairn]

Thanks! 🙂

[sharp dirge]

how showing (x,y) is not principal in Q[x,y]

i was thinking ..

[willow mason]

Suppose it was principal and use the fact that x and y are irreducible

[ember field]

if i have 1->A->G->H->1 a group extension and A abelian, why is the action h*a=h'ah'^(-1) an action on A by H with h' s.t. pi(h')=h ?

i am trying to check the identity requirement but im not sure why e*a = e'ae'^(-1)= a ?

[next obsidian]

pi(e) = e

So you can take e’ = e

[hollow mica]

Let P be the submodule generated by x, y in k[x, y] (as a module over itself)

how do I show that P is not generated by a single element

in fact it isn't free, and I was able to show the rank can't be greater than 1, but I am stuck now

in other words, I want to show P is not a principle ideal

[hollow mica]

basically trying to understand this @wicked zephyr

Over the past 24 hours I learned what ideals and principle ideals were

[wicked zephyr]

As soon as you prove that it cannot be generated by a single element, it is non-free

suppose (x,y)=(f)

[hollow mica]

I was trying to show f + 1 isn't in (x, y)

[wicked zephyr]

then $x\in (f)$ and $y\in (f)$

[cloud walrus]

Potitov06

[hollow mica]

wait actually that isn't true (my comment)

[wicked zephyr]

so $f\mid x$ and $f\mid y$

[cloud walrus]

Potitov06

[wicked zephyr]

so...

[hollow mica]

wait

x = af
y = bf

x + y = (a + b)f

what's the problem

oh but

we have

bx = ay

so a = b = 0

yay!

@wicked zephyr

thanks

[wicked zephyr]

hmm

that is not the point

$f$ is common divisor of $x$ and $y$... What are those?

[cloud walrus]

Potitov06

[hollow mica]

x and y are multiples of f

that's why I wrote

x = af
y = bf

and then from this you get bx = ay (for some nonzero a, b) because we're in a commutative ring

and then that's clearly impossible

right?

[wicked zephyr]

no

b and a are in k[x,y]

[hollow mica]

oh shit

[wicked zephyr]

so, certainly xy=yx

but that is not the point

f divides x and f divides y

and the only common divisors are constants

sorry, units

so then (f)=(x,y)=k[x,y], which is not ture

[hollow mica]

the only common divisors between x and y are units?

[wicked zephyr]

i mean, just think about itttt

what are the divisors of x??

[hollow mica]

i've never talked about divisors of elements in a general ring

but intuitively it's 1 and x

[wicked zephyr]

yeah, or 1/2 and 2x

[hollow mica]

kx

and k

for nonzero k

[wicked zephyr]

inverse of k

ok

and what are divisors of y

[hollow mica]

ky and k for all nonzero k

[hollow mica]

wdym

[gritty skiff]

im a little confused with this problem

the given solution is (x-1)^2(x-2) but i dont see how that could be possible given that 2 isnt a zero of this polynomial

[hollow mica]

the units of k[x, y]

[wicked zephyr]

yeas

so $f$ is a unit

[cloud walrus]

Potitov06

[wicked zephyr]

and a unit generated the entire ring

but $(x,y)\neq k[x,y]$

[cloud walrus]

Potitov06

[hollow mica]

hmm

so we showed that the element generating the entire ring must be a unit

[wicked zephyr]

no, we showed that f is a unit

and you assumed $(f)=(x,y)$

[cloud walrus]

Potitov06

[wicked zephyr]

contradiction

[hollow mica]

oh

how do we know units generate the entire ring

[wicked zephyr]

an ideal is closed by multiplication

Let $I$ be an ideal that contains a unit $u$. If $a$ is any element of $R$, then $a=u\cdot (u^{-1}a)\in I$, so every element is in the ideal.

[cloud walrus]

Potitov06

[hollow mica]

oh hmm

i think i understand

wait i'm going to try to give an overview of the proof from the beginning

can you see if my ideas are right

question: prove that (x, y) = {fx + gy | f, g \in k[x, y]} is not a free submodule of k[x, y] (as a module over itself)

proof:

assume (x, y) is free

if it has rank > 1, then we can easily produce (b)a + (-a)b = 0 for some a, b in a basis

if (x, y) is generated by a single element, then that element divides both x and y

but the only common divisors of x and y are units

but a unit clearly generates the entire ring, done

[hollow mica]

this is the only shaky part for me

i'm trying to rigorously classify all divisors of x and y and am struggling

well actually since deg(fg) = deg(f) + deg(g), I know the divisors must be of the form ax + by + c, for some a, b, c \in k

but if fg = x, and f = ax + by + c, by the same degree thing above, b = 0

so the divisors of x are of the form ax + c

if fg = x, and f = ax + c, then g = d for some nonzero d \in k

but fg = (ad)x + (cd) = x, so we must have either a = 0 and c \neq 0, or a \neq 0 and c = 0

this gives what I had earlier

I was also going to ask why why the units in k[x, y] are only the constants, but that again is true because of the degree thing

nice

[wicked zephyr]

it is simpler, if fg=x then f or g is a constant (ny taking degrees). suppose f is the one of degree 1 and write f=ax+by+c, and g=d. Then x=adx+bdy+cd, so b=c=0 and ad=1, so a is a unit

thus if f divides x, then f is a unit or a unit times x

[hollow mica]

got it

[hollow mica]

Can someone explain to me the connection between the transpose of a linear map and the transpose of the matrix corresponding to the linear map (wrt some basis) if any

reminder: the transpose of a linear map A from vector space V to vector space W is the natural map from W* to V* that takes an element f in W* to f ○ A

[chilly ocean]

@hollow mica the transpose of the matrix representation of a map is the matrix representation of the transpose in the dual bases

in symbols: if $T\colon V \to W$ is a map and $\beta,\gamma$ are bases of $V, W$ with dual bases $\beta^,\gamma^$, then $$\left([T]\beta^\gamma\right)^T = [T^*]{\gamma^}^{\beta^},$$ where $T^*\colon f \mapsto f \circ T$ is the transpose

[cloud walrus]

TTerra

[chilly ocean]

this is quite straightforward to prove

i encourage you to do it

[hollow mica]

ohhhh

wow you have a talent of stating things in the most concise way possible lol

[old hollow]

If H, K are subgroups of G with H intersect K = {1}, then is the product map p: H x K --> G, (h,k)-->hk bijective?

[lethal dune]

Bijective to G? No you need further assumptions

[old hollow]

well the book proves it's injective

[lethal dune]

it's injective yes

[old hollow]

but the surjective part

nvm I'm kinda convincing myself it could be the case that

[lethal dune]

Don't you need normality for this also?

[old hollow]

there are some elements of g which can't be written as a product of h and k

for injective no

[lethal dune]

ig you don't NVM

[old hollow]

I've never seen that before

pretty cool

[analog pumice]

If k is an algebraically closed field such that char(k) != 2, then x^2 - 1 has two distinct roots, right?

[latent anvil]

yes

[analog pumice]

thanks

I think I have a proof

[latent anvil]

Sure, do you want me to check it?

[analog pumice]

sure!

if $\alpha^2=1$, then $\alpha\neq0$, and $-\alpha$ is another root of $x^2-1$. Suppose for contradiction that $\alpha=-\alpha$, then $\alpha+\alpha=2\alpha=0$, and $\alpha\neq0$, $k$ is a domain, so necessarily this would imply $2=0$, which is not true if $\mathrm{char}(k)\neq2$

[cloud walrus]

cucaracha

[latent anvil]

Yep!

[analog pumice]

thx

[latent anvil]

This works for x^2 - r for any nonzero r, yeah?

But actually in the specific case of x^2 - 1 we can be more direct

the roots are always 1 and -1

[analog pumice]

I think so, yea

[coral spindle]

You can also use the old derivative trick

[latent anvil]

yeah, also that

[analog pumice]

In general, what conditions do I need to put on an algebraically closed field k so that k has n distinct n-th roots of unity?

Does it suffice to take char(k) > 2?

[latent anvil]

no

[coral spindle]

that char(k) does not divide n

[analog pumice]

I didn't think so

[coral spindle]

Like for example

[analog pumice]

this makes sense

[coral spindle]

field of char p

[latent anvil]

Do you know what a separable polynomial is?

[analog pumice]

yea

[coral spindle]

x^p - 1 = (x - 1)^p due to freshman's dream, right

[latent anvil]

Okay, so you're exactly asking whether x^n - 1 is separable

Right?

[analog pumice]

yes

[latent anvil]

Do you know how to check if a polynomial is separable via the derivative?

[analog pumice]

I vaguely remember learning about that

sorry, it's been a while since I took algebra

[latent anvil]

Sure, so p(x) is separable iff p is coprime to p'(x)

[analog pumice]

oh right!

[latent anvil]

this is because of the product rule

[analog pumice]

yes I remember that now

[latent anvil]

If p has a double factor x-a, so p(x) = (x-a)^2 q(x), then p'(x) = 2(x-a) q(x) + (x-a)^2 q(x). So p'(a) = 0

If you's seen complex analysis, this is like the order of a zero thing

Okay so

This tell us a field has distinct nth roots of unity iff x^n - 1 is coprime to n x^(n-1)

Yeah?

when is that possible?

[analog pumice]

right, when char(k) does not divide n

[latent anvil]

Yep

[analog pumice]

thanks! that was enlightening

[latent anvil]

We know the irreducible divisors of x^(n-1)

[dreamy fiber]

Does anyone have an example of a function between ABELIAN groups $f: G \to H$ such that $f$ sends identity to identity, for any $g \in G$, $f(g^{-1}) = f(g)^{-1}$, but $f$ is not a group homomorphism?

[cloud walrus]

ScarletScorch

[dreamy fiber]

I don't think these conditions are sufficient to guarantee f is a group homomorphism

so i couldn't quite come up an example yet

[hollow mica]

Let G = Z_5, and define f:G -> G by f(0) = 0, f(1) = 2, f(2) = 1, f(3) = 4, f(4) = 3. Then f(2 + 4) = f(1) = 2 and f(2) + f(4) = 1 + 3 = 4.

[chilly ocean]

You could put f(n) = n for integers not -1, 1 and switch 1, -1 around. In Z

[neat valley]

f : Z --> Z sends x to sign(x) * x^2

[dreamy fiber]

oh great

thanks everyone

[neat valley]

or even easier take x --> x^3 from Z --> Z

[chilly ocean]

hello

i don't quite understand this part: https://i.imgur.com/CCEa1Ca.png

why direct sum has that requirement ?

"all but finitely many are identity" ?

does "direct product" not necessarily mean "commutative", and does "direct sum" conventionally mean "commutative" ?