groups-rings-fields

[chilly ocean]

well yeah, but we are trying to prove the converse

that it's irreducible in Z (or Q? idk anymore)

[hot lake]

No

You need the contrapositive

[chilly ocean]

yes, because we're trying to conclude the statement of Eisenstein lemma

[hot lake]

And that's equivalent

[wild solar]

may I ask how to prove this

[hot lake]

Take mod of everything

The leading coefficient is 1 so the degrees of the factors don't change

[chilly ocean]

3x is reducible over Z but it's not reducible mod 2

[hot lake]

So a non trivial factorisation in z gives a non trivial factorisation mod p^2

The leading coefficient of 3x is not 1

[chilly ocean]

the statement of Eisenstein criterion doesn't imply that your polynomial is irreducible over Z in general

[hot lake]

It does when the leading coefficient is 1

[chilly ocean]

But amelia didn't assume a_n = 1

[hot lake]

...

The coefficient of x^n here is 1

[chilly ocean]

that's a different criterion

different theorem

[hot lake]

......

[chilly ocean]

here the coeffiecient wasn't assumed to be 1, it only gets "inverted mod p" to 1

[hot lake]

What's Eisenstein criterion then

[wild solar]

I wrote that x_0*a_n =1 mod p(x_0 an appropraite integer) is this what you're referring to?

[hot lake]

Ah I see

[chilly ocean]

I was trying to ask amelia that and check if the leading coefficient has to be 1. But I got thrown off by what she wrote

she said before that yes, it has to be 1

[hot lake]

On Wikipedia it doesn't need to be 1

[chilly ocean]

yes, but they study irreducibility over Q

instead of saying that f'(x) is irreducible mod p^2 implies f(x) is irreducible mod p^2, say that
f'(x) is irreducible mod p^2 implies f'(x) is irreducible over Q, which implies f(x) is irreducible over Q

and that would be it

[wild solar]

I tried using the contrapositive, as @hot lake suggested. Does this look right to you guys?

[wild solar]

sorry but why? Is there any justification?

[chilly ocean]

Like I said before, you can't prove that f(x) is irreducible over Z from (assumptions of) Eisenstein criterion

so what you wrote must contain some kind of error

[wild solar]

oh

[chilly ocean]

yes, I see the error

if f(x) is reducible, it doesn't mean that it can be written as g(x)k(x) where deg g, deg k > 0

that works for fields (when deg f > 0), but not in general rings like Z

[wild solar]

：（

[chilly ocean]

example is that f(x) = 2x is reducible over Z, but can't be written as a product of two polynomials of positive degree

[wild solar]

I still don't understand how you got this implication relation

[chilly ocean]

well if f(x) is reducible over Q, then it can be written as f(x) = q(x)k(x) as you tried to write it, and we can assume q(x) and k(x) are elements of Z[x] (since f(x) is an element of Z[x])

[wild solar]

how is this related to Z mod p^2?

[chilly ocean]

well it's kind of how you wanted to prove this, but we replace Z with Q because in Z weird things can happen

[wild solar]

ohhhhh

I think I understand now

[chilly ocean]

and that q(x) and k(x) are elements of Z[x] is important, because then we can reduce them mod p^2

[wild solar]

sorry I'm kind of slow, we have only mentioned fields in one lecture, as this is an introduction to rings and fields course

kk makes sense

[chilly ocean]

In general f(x) = a*g(x) where g(x) is a primitive polynomial and a is an element of Z

and g(x) being irreducible does mean that we can write it as a product of two polynomials of degree > 0 over Z

but we can have this annoying coefficient on the front, that's why it doesn't work over Z

[wild solar]

jesus now I know what went wrong I shouldn't have said yes cuz in our textbook a_n doesnt have to be 1

now everything's starting to make sense

Thank you both so much for your patience

🌹

[chilly ocean]

You're welcome

[chilly ocean]

I was studying principal ideals and thought to do a practice problem about it. 2Z (or a subgroup of even Integers, of the additive group contained in a Ring of Integers) is surely a principal (two-sided) ideal according to my research, but I'm unable to properly use the definition of a two-sided ideal here to prove ideal I is indeed a principal one.

Generating element seems a=2 here because I is a subgroup of even Integers

[hidden haven]

You have to show that the ideal generated by 2 is the same as the ideal of even integers

So you have to show 2 inclusions

Pick an element in one, show that it is also in the other

Also, 2-sided is not important, since multiplication is commutative

Any left or right ideal is automatically 2 sided

[chilly ocean]

I see

[hidden haven]

Similarly in an abelian group, all subgroups are normal

[chilly ocean]

Yes

Thank you for the help o7

[chilly ocean]

I am still unable to properly understand this definition of a two-sided principal ideal

Is the definition RaR = {xay | x,y in R} not correct?

[hidden haven]

That won't be an ideal

Not necessarily closed under addition

The given definition is the closure of that set under addition

[chilly ocean]

I see , thanks 👍

[chilly ocean]

Ahh I get it now, every element xay is always a finite sum of such elements in the set 👍

[fiery berry]

ok, so suppose our chi is irreducible, then take the 1 dimensional trivial character which is also irreducible. Then <chi, chi_trivial> > 0 since chi(1) = dim V >=2 and chi_trivial(g) = 1. According to orthogonality relations we must now have that the representations of chi and chi_trivial are isomorphic which isn't the case since dim V >=2? Is this valid?

[stoic rose]

Yes that's the solution I had in mind.

For any nontrivial character the sum of the values is 0 because of orthogonality with the trivial character.

[fiery berry]

alright, thank you

[latent anvil]

It's still the case that the group is abelian in the infinite case, that bit of the argument doesn't require finiteness

Because as you say g must be in the center, but then if there was a non central element h then f(x) = hxh^-1 would be a non identity morphism fixing g

[wooden ember]

Oh yeah obviously you’re right

I should’ve probably actually read what you’d written in full

If anyone read that I apologize

[fiery berry]

https://math.stackexchange.com/questions/741191/composition-of-irreducible-representation-and-surjective-homomorphism why is the first equality of the chosen answer correct? I don't see how you get there so quickly

[latent anvil]

$\sum_{g\in G} \alpha(\varphi(g)) = \sum_{h\in H} \sum_{g\in \varphi^{-1}(h)} \alpha(\varphi(g)) = \sum_{h\in H} \sum_{g\in \varphi^{-1}(h)} \alpha(h) = \sum_{h\in H} |\varphi^{-1}(h)| \alpha(h)$

[cloud walrus]

Shamrock?

[latent anvil]

For an epimorphism the fibers all have size equal to the kernel

But |G|/|ker φ| = |H| by first iso, so all the fibers have size |G|/|H|

(α is all the stuff with characters, I just didn't want to write it out)

@fiery berry

Also I think this might be wrong, I wrote some stuff out about Z/p^aZ × Z/p^bZ on a whiteboard which I now realize is wrong. I think it might just be products of cyclic groups of prime power order (which would be even more restrictive!)

[fleet pasture]

Is there name for a Field-like structure which also allows multiplicative inverse for 0?

Riemann sphere comes close by adding ∞ to C and things like 1/0 become well-defined, but here 0 - ∞ remains undefined. Is there no way to close all numbers under addition and multiplication - without creating an exception for 0?

[chilly ocean]

"wheel"

[latent anvil]

Okay yes correction to this, you actually can't have multiple p_k group factors. It suffices to show there's an endomorphism of Z/p^aZ × Z/p^bZ for a < b which fixes (1,1) but not the entire group. Take φ(n, m) = (n, p^(b-a)n + (1 - p^(b-a)) m). This is a well defined endomorphism with φ(1, 1) = (1, 1) and φ(0, 1) = (0, 1) - (0, p^(b-a)) ≠ (0, 1), so φ ≠ id.

This means the final classification is that if G is a finite group and g in G is such that every endomorphism φ : G -> G fixing g is the identity, then G is actually cyclic generated by g

Which is a much nicer classification

[chilly ocean]

The Riemann sphere is an example of a wheel induced by the field of complex numbers

[fleet pasture]

@chilly ocean Yes, I will look into wheel theory. However, I think Riemann sphere does not have 0-infinity or negative infinity anywhere on it. That's the extension I am looking for.

[chilly ocean]

it does, those are again infinity

depending on how you want to think about it

[fiery berry]

thats exactly what i needed, thank you

[long obsidian]

Is there some correspondence between algebraic varieties and ideals of some ring? I think this is Hilbert's null. But I'm not sure

[next obsidian]

Boy oh boy oh boy oh boy

[chilly ocean]

over algebraically closed fields, that's right. there's a one-to-one correspondence between algebraic varieties (specifically, algebraic subsets of A^n) and radical ideals in K[x_1, ..., x_n], and more

to say the very least

[next obsidian]

It’s way more than that, once you upgrade to schemes every single ideal gives a different closed subscheme so you don’t even need to restrict to radical ideals

Although to be fair when your ideal isn’t radical life kind of sucks

[latent anvil]

I think the statement for varieties is a bigger theorem

Like if we restrict to radical things, the affine scheme version is practically definitional

But for varieties this is the content of the nullstellensatz

And for non radical ideals it still feels more like a matter of definitions than a big theorem like the nullstellensatz

Like the proof that I(V(J)) = r(J) in Spec A is entirely formal, bc V(J) now has all these non closed points

[fallow plume]

https://youtu.be/g87CBjYqHWk

a bit before 8:00 Borcherds puts that K = M^G

where G is the Galois group, and M is a field extension of K

he says this is the fixed points, but is there something I'm missing that the notation should be reflecting?

using a group as an exponent seems interesting

does it have to deal with conjugation?

[next obsidian]

It’s just notation

Like it literally is notation for the fixed points of a group action

[chilly radish]

I am having trouble with this proof, specifically in (3). I've tried reasoning about this and I understand it when f is separable, but what about the inseparable case

D is the discriminant of the polynomial

when f is separable we have a very neat formula for the norm, but in the inseparable case that formula doesn't work

[runic hemlock]

if f is not seperable both sides are 0

[runic hemlock]

if f is not seperable, then obviously D(f)=0. For the other side, note that f'(alpha)=0 because it's a double root

[chilly radish]

Oh you're right

Idk how I missed that, thanks!

[runic hemlock]

I have a question. Let $G=S_p$ (permutation group), and $P$ a sylow p-subgroup, so we can assume $P=\left<\sigma\right>$ where $\sigma=(p12\dots p-1)$. Then I'm supposed to show that $|N_G(P)|=p(p-1)$. Any hints?

[cloud walrus]

Porphyrion

[cursive temple]

doesnt sylow 3 state that $[G:N_G(P)] = n_p$, where $n_p$ is the number of p-sylow subgroups

[cloud walrus]

Pappa

[cursive temple]

so it would suffice to show that there are (p-2)! p-sylow subgroups

[runic hemlock]

oh and that's easy (it's just the number of cycles divided by p-1)

Thanks!

[cursive temple]

np

[pearl saddle]

$S_n = {n : n^2 - 81n + 1681 = p, n \in \mathbb{Z}^+}$\
Where $p$ is a prime. Find the least $n_2 \notin S_n$.

[pearl saddle]

can anyone of you solve this problem using abstract algebra

I have found a solution using number theory

[chilly ocean]

this definition doesn't make sense, S_n can't be parametrized by n

[pearl saddle]

this is my solution

[pearl saddle]

sorry it should be n'

[cloud walrus]

ದೀಪನು

[chilly ocean]

how can S_n depend on n?

[pearl saddle]

it depends ig

[pearl saddle]

and forgive my set knowledge

[pearl saddle]

just assume that Sn is a collection of some n which satisfies the condition

[chilly ocean]

how can S_n be a set of n

[pearl saddle]

y not cud u elaborate

ooh god forgive my set theory knowledge

[pearl saddle]

cud u elaborate this plz

[chilly ocean]

$S_n = {n : ... }$ makes no sense

[cloud walrus]

Blitz

[pearl saddle]

aah, how shud I correct it

[chilly ocean]

I don't know

[pearl saddle]

I just want to say that $S_n$ is a set with elements n, such that $ n^2 -81n + 1681$ is a prime for $n\in\mathbb{Z}^+$

[cloud walrus]

ದೀಪನು

[chilly ocean]

you can't use the variable n to denote both the index of S_n and elements of S_n

this is simply ambiguous, who knows how the parameter n comes into the picture

[pearl saddle]

ok simply assume that Sn is a solution set for the problem

[chilly ocean]

what problem

[pearl saddle]

the one I put up

[chilly ocean]

it doesn't make sense

[pearl saddle]

I meant for all values of $n$ such that $n^2 -81n + 1681$ is a prime

[cloud walrus]

ದೀಪನು

[chilly ocean]

yeah, what about them

[pearl saddle]

S contains these values

[chilly ocean]

okay, we have some set S = {n : n^2-81n+1681 is prime}

[pearl saddle]

yep

[chilly ocean]

and you want to find its least element

[pearl saddle]

nah

[chilly ocean]

the least element of its complement*

[pearl saddle]

least x not in S_n and belongs to Z^+

[chilly ocean]

Note that (m-40)^2-81(40-m)+1681 = m^2+m+41

[pearl saddle]

yep euler primes

but can u use abs alge for this?

coz I have solved this using num theory

[pearl saddle]

that's what I meant in the first place!

[chilly ocean]

Is is always true that the characteristic of a ring is the generator of its principal ideal (if there exists one)?

"its principal ideal"?

Sorry let me reframe
Is is always true that the characteristic of a ring is the generator of the principal ideal contained in it (if there exists one)?

[tribal moss]

There's in general no such thing as "the principal ideal".

[chilly ocean]

Oh

[tribal moss]

And the characteristic of a ring is not even an element of it, so it cannot generate any ideal.

The characteristic of a ring is the generator of the particular ideal in Z that is the kernel of the unique homomorphism from Z to the ring.

That might be what you're remembering.

[chilly ocean]

Ohh yes that was what I was trying to remember about

Thank you o7

[wraith obsidian]

I have a meta-question about infinite galois extensions

So given L/K and on the other hand Aut(L/K)/1
We have the usual galois connection \varphi: H\to Fix H and \psi: E\to Stab E

Iirc we already know that \phi\circ\psi =id, so every subfield in the lattice is already closed in the sense of the closure operator \phi\psi

So it's natural to look at the closed subgroups in the sense of the closure operator \psi\varphi

So it's clear that the closure operator satisfies extensivity, monotonicity, and idempotence

But

1. why does it have to form a topology, ie why is the union of two closed sets closed again
2. Even further, why does this happen to be a group topology? That's not a pretty strong property for something that just drops out of a galois connection.

(not sure if #point-set-topology would be more suitable?)

[hidden haven]

Do the closed subgroups determine the entire topology on a topological group?

Because the galois connection just tells us those

Maybe you can prove that the union of 2 such closed subgroups is an intersection of closed subgroups from the connection

Then they'd form a basis of closed sets, and you could look at the topology they generate, which should be a group topology because nice basis

But you'd have to show that this coincides with the Krull topology? Unless this is how you define the Krull topology

[wraith obsidian]

What do you mean by topological _sub_group
We just have one topological group, we're not really considering all field extensions between L/K at the same time when we look at this one galois connection, right?

I need to look up the precise result but afaik these closed sets already form the entire topology of the group

[hidden haven]

ye I meant topological group

[wraith obsidian]

I will take a look at the details later. I think looking this result up will clarify things, perhaps it's just a base and I was wrong.

[hidden haven]

Usually you'd define the Krull topology in other ways, then show that the closed subgroups in that topology correspond to closed elements of the lattice

[hidden haven]

Union of 2 subgroups can not be a subgroup in any non trivial cases, so it's not even a basis

[wraith obsidian]

Lmao you're obviously correct

Thanks

[simple mulch]

Hey guys! I was doing some representation theory and noticed that my characters aren't irreducible... I wonder what would be my next move to decompose them?

I thought about finding invariant subspaces of V (the vector space) but I am kinda stuck

[fallow plume]

are \Q and F_p the only base fields?
or is it possible to have base fields that aren't countable/etc and if so what's their name?

[next obsidian]

Base field is just any field when you’re considering a field extension

Q and F_p are the prime fields / prime sub fields because every field is an extension of exactly one of those fields

But eg if you’re thinking about C as an extension of R then R is the base field

[fallow plume]

oh weird, idk if I've been taught that base field = prime field or just assumed it myself, but I always thought they were the same

good to know

because every field is an extension of exactly one of those fields
but yea that answers my question

weird to know you can't construct different prime fields, but neat

[next obsidian]

Uniqueness follows because what your prime subfield is determines the characteristic

F_p iff char p

And Q iff char 0

[fallow plume]

ah yea

[waxen hedge]

As chmonkey says there are very natural.
If you have a field $K$, the canonical morphism $\mathbb{Z}\to K$ factors as $\mathbb{Z}/p\mathbb{Z}\to K$ with $p=0$ or prime. Using the field of fractions property you get a morphism $\mathbb{Q}\to K$ or $\mathbb{F}_p\to K$ induced by $1\mapsto 1$

[cloud walrus]

Adrien

[fallow plume]

any tricks for the galois of a somewhat arbitrary polynomial?

i can prove it has 3 real roots and 2 complex

so it's a direct product of \Z_2 and some subgroup of S_3 right?

or is it more complicated than that

[cold frigate]

You'll need to be more careful. if your polynomial is irreducible, for example, you can show that it has galois group S_5

[hot tinsel]

Suppose $M$ is an $R$-module, where $R=\mathbb{C}[G]$ is a group ring. For any $x\in M$ can we have an isomorphism of the form $$Rx\cong Ry\oplus Rz\oplus T$$ for some $y,z\in M$ and $T$ some torsion module?

[cloud walrus]

K零ꓘ

[hot tinsel]

I dont think this should be true in this case

but i am unable to show this

[hidden haven]

Take y = x, z = 0, T = 0 🤡

If things aren't allowed to be 0 you should be able to get a counterexample by taking G to be the trivial group/taking action of G on M to be trivial and then the given decomposition can't happen for dimension reasons

@hot tinsel

[hot tinsel]

i phrased it poorly, I am given a group $G$ acting on $M$, so like my beginning choices are fixed. Now i want to show that direct sum of cyclic modules is not cyclic.

[cloud walrus]

K零ꓘ

[hidden haven]

well if y and z are given then the internal direct sum of Ry and Rz may not exist

if they are already linearly dependent

Is it that you want an example where this happens?

[hot tinsel]

I apologize for the vagueness, I am like trying to do something and I thought to see the problems in a module setting. So can a cyclic module be decomposed as internal direct sum of cyclic submodules? In the case when $M$ is a complex vector space as well?

[cloud walrus]

K零ꓘ

[hidden haven]

Yes it should be. Take G = C_3 = {1, x, x²}. Look at R = ℂC_3 as an R module. This is cyclic, but decomposes as the direct sum
R(1+x+x²) ⊕ R(x-x²)

I hope this is right, do check the final decomposition

[hot tinsel]

Thanks a lot i will try checking the details

[chilly ocean]

How do I prove that if a module is flat then it disappears on Tor_n(•, B) for all n > 0?

[next obsidian]

Surject onto it with a free module (or something you know has no Tors), and stare at the resulting LES very hard

[chilly ocean]

Thanks, got it. That was pretty cool

[next obsidian]

(b) < (ai) < (b) so (b) = (ai) for all the latter ai

[chilly ocean]

Why is (r) proper

[sharp sonnet]

primes are non units

[chilly ocean]

@sharp sonnet wym

[sharp sonnet]

wait i misread lmao

replace the word prime with irred

irreducible elements are not units

so 1 = rx has no solution in x

so its not in the ideal generated by r

[chilly ocean]

(r) cannot contain 1

[sharp sonnet]

yes

[chilly ocean]

Why is P principal

@chilly ocean what ring is R

Commutative ring with unity

Then it doesn't have to be principal iirc

In k[x, y] where k is a field the ideal (x, y) is maximal but not principal

[chilly ocean]

What does (x, y) mean

Ideal generated by x and y

We need some assumptions on R

[chilly ocean]

What does that mean

[south patrol]

I assume PID or smth given what was said about ascending chain conditions?

[chilly ocean]

The smallest ideal containing both x and y

[iron vessel]

So my professor said that $\left(\mathbb{Z}/9\mathbb{Z}\right)^{\times} \cong <2>$. This is not correct right? Because $2^3 \ne 1$, and $2^2 \ne 1$, and also $2^9 \ne 1$.

[cloud walrus]

Évariste Galois

[delicate orchid]

Consider 2^6

[iron vessel]

Yeah 2^6 is 1 mod 9

So $<2> \cong \mathbb{Z}/6\mathbb{Z} \cong \left(\mathbb{Z}/9\mathbb{Z}\right)^{\times}$. Right?

[cloud walrus]

Évariste Galois

[delicate orchid]

Seems like it

[iron vessel]

Ok, i was a bit confused because 2^6 = 1 mod 9 not mod 6

[delicate orchid]

I’d just work in (Z/9Z)* directly ngl

Going through Z/6Z seems like a hassle

Especially since I’m assuming you’re taking Z/6Z as an additive group which confuses my brain when it’s mapped to a multiplicative group

[iron vessel]

Yeah that is probably better; I have another question tho, more related to galois theory. I need to prove or disprove wether the 36-gon can be constructed with ruler and compas starting from the subset ${0,1,\zeta_9 + \zeta_9^{-1}} \subset \mathbb{C}$. I know that the 36-gon is constructible as 36 is the product of a power of two with the product of distinct fermat primes; more precisely, $36 = 2^2 \times (2^{2^0} + 1)(2^{2^0} + 1)$. But how do I approach it with the set above?

[cloud walrus]

Évariste Galois

[runic hemlock]

36 is not a product of a power of 2 and distinct fermat primes

Because 3 and 3 are not distinct

So it can't be constructed just from {0,1}

[iron vessel]

Lol how could i not see that. So how do I know if it can be constructed from {0,1,\zeta_9 + \zeta_9^{-1}}? @runic hemlock

[runic hemlock]

it can

because from Re(zeta_9) you can get zeta_9 (just build a perpendicular and intersect with the unit circle), and now just bisect the angle (zeta_9, 0, 1) twice

[iron vessel]

So let me see if I understand what you mean. So, zeta_9 = cos(2pi/9) + isin(2pi/9), and zeta_9 + zeta_9^{-1} = 2cos(2pi/9), so since you can construct cos(2pi/9) by dividing 2cos(2pi/9) by 2, you obtain that you can construct the real part of zeta_9. Then, if you build a line perpendicular to the real axis going through Re(zeta_9) you intersect teh unit circle exactly at zeta_9. Then, a line going through the origin and through zeta_9 will make an angle, say alpha, with respect to the real axis. Finally, if we bisect alpha, which is possible, we get zeta_18; bisecting it again, we obtain zeta_36. So, we may construct the 36-gon.

Is this a good argument @runic hemlock ?

[runic hemlock]

yes

[iron vessel]

Nice, this really helped. Thanks!

[simple mulch]

is bourbaki still readable for algebra?

[chilly ocean]

It's most likely not

[simple mulch]

yeah I guess so xd

[waxen hedge]

It's very readable but it depends of your goals

It's not a standard textbook

[simple mulch]

Well, it would be mostly for references

[sly storm]

Hello. Is there a simple and intuitive proof of the fact that the countable direct product of Z-module Z is not free?

[chilly ocean]

What if you just prove that it's not projective

[sly storm]

It will be good if it has a simple and intuitive proof.

[chilly ocean]

Yeah, exactly

Well if you just take product of Z and then a map into the free sum on the generetors, how would you extend it to the whole product?

By generators I mean that 1 is only on one coordinate and the rest is 0

On the free sum it has to be the identity, right

But say, what does something like (1, 1, 1, ...) gets mapped to

Let's think about it

Hmm... I don't think this approach really works

https://mathoverflow.net/questions/46475/infinite-direct-product-of-the-integers-not-a-free-module-over-the-integers

If it's not here then it probably doesn't exist

[simple mulch]

orbits of H in X means {hX : h in H} or {xH : x in X} ?

[hidden haven]

Oh god I read this construction last month it's so damn awful

I think it means the first one

[simple mulch]

a bit xd

thanks

[hidden haven]

I would also recommend looking at Serre instead of Lang

It also has this construction

[simple mulch]

this is Serre

[stark sigil]

Wow this is indeed impossible to read

What is A…

Also X/H suggests orbits means the second option?

[hidden haven]

Oh damn lol

[stark sigil]

At least no undefined variables in this one but this is way too beginner unfriendly still

[hidden haven]

oh my bad

[hidden haven]

Ye the set of orbits is {Hx : x in X}

[simple mulch]

What would be more beginner friendly? I think its just the notation that kills it

Thanks!

[stark sigil]

Haven’t looked at abstract algebra textbooks in a while, someone else should answer that

[hidden haven]

My rep theory course used lang and it was horrible Serre was a million times better

But this section is still horrible lol

Lang gives this construction as an exercise

[simple mulch]

Yeah I kinda like the book too

ok brb

[stark sigil]

Ikr, I have to concentrate to figure out what’s going on in that screenshot

And I’m not a beginner

[simple mulch]

Ok I kinda need help understanding what's going on in there. I don't understand the orbits

Like Hf = {hf : h in H} and f in Hom(A,C)

take f1, according with that construction we have (h f1) (a) = f1(h a h^-1) for all h in H which is again a function in Hom(A,C) since A is a normal group

What I can't see is how does h would "change" something in f1

Because class functions are constant on conjugacy class this wouldn't change anything? But then how many orbits would we have?

Can we describe such orbits?

[hidden haven]

Conjugacy classes of A, but now you are conjugating by something in H

Maybe take a small example of a semidirect product and see what the construction gives

I haven't tried that myself but I assume that would give some idea

[simple mulch]

Good point

[simple mulch]

uh what?

[hidden haven]

When you say constant on conjugacy classes, you mean f1(a) = f1(b^-1 a b) for all b in A

But now this b is in H

[hidden haven]

Because you started with a character of A

[simple mulch]

oh I see

yeah

indeed

I think I see how it changes

Thanks

Will do some example

[wispy pivot]

I have a basic question about working with modulo integers. There's a question from dummit here:

Would I be allowed to do the following in a proof?

Let $b = \dfrac{n}{gcd(a,n)}$ (which satisfies $1 \leq b < n$ since $gcd(a,n) \leq n$ and $1 < gcd(a,n) $), then
\begin{center}
$ab = a\dfrac{n}{gcd(a,n)} = \dfrac{a}{gcd(a,n)}n \equiv \bar{0}\cdot\bar{1} = \bar{0}$.
\end{center}

Suppose by contradiction there exists $c \in \mathbb{Z}$ such that $ac \equiv 1$ (mod $n$). In other words $ac = nx + 1$ for some $x \in \mathbb{Z}$, then
\begin{center}
$\dfrac{ac}{gcd(a,n)} = \dfrac{nx+1}{gcd(a,n)}$
\end{center}
\begin{center}
$\Rightarrow \dfrac{a}{gcd(a,n)} \cdot c = \dfrac{n}{gcd(a,n)} \cdot x + \dfrac{1}{gcd(a,n)}$
\end{center}
\begin{center}
$\Rightarrow \bar{0} \cdot c \equiv \bar{0} \cdot x + \dfrac{1}{gcd(a,n)}$
\end{center}
\begin{center}
$\Rightarrow \bar{0} \equiv 1$
\end{center}
resulting in a contradiction.

[cloud walrus]

HimmyHow

[west violet]

you should use judson

[simple mulch]

What's judson?

[west violet]

oh its a textbook

for abstract algebra

[simple mulch]

Oh okok, thank you for the recommendation!

[west violet]

yeah it's really nice for beginners.

[spice whale]

what's the difference between product amd intersection of ideals

wikipedia says it's encoded by the Tor functor but i don't really get it

[chilly ocean]

Well, product is contained in the intersection

But converse doesn't have to hold

For example, take (x)

In k[x]

Then (x)^2 = (x^2) is a proper subset of (x)

[hidden haven]

That wikipedia line seems interesting can you send the full quote

Or is that all it says

I guess Tor(R/I, J) is the kernel of the multiplication map I ⊗ J → J

Follows from tensoring the exact sequence 0 → I → R → R/I → 0 with J

And using the fact that multiplication R ⊗ J → J is an isomorphism

But if every ring is a domain anyway then the kernel should be 0 🥴 so what does this even do

[hidden haven]

No this is wrong not everything is a pure tensor lol

[iron vessel]

You probably heard this one already but I learned Algebra from groups to fields using Dummitt and Foote. With it, since it is so popular, you are able to get solutions for proofs which is absolutely amazing and not something you have for less known I guess.

[hidden haven]

I see

Ty

[chilly ocean]

0->a->R->R/a->0 is exact
It leads to an exact sequence
0 -> Tor_1(R/b, R/a) -> R/b tensor a -> R/b -> R/b tensor R/a -> 0

I guess you need to manipulate this sequence now

[hidden haven]

Ye

R/a ⊗ M = M/aM

✓

[chilly ocean]

Tor_1(R/b, R/a) is isomorphic to kernel of the map R/b tensor a -> R/b tensor R

Given by Id tensor inclusion

R/b tensor a = a/ab -> R/b as Moldi noticed so this is just (a \cap b)/ab

[chilly ocean]

Consider a ring homomorphism f from Z×Z to Z, which in terms of Binary Relations is defined as:
f = {((a,b),c) | (a,b) in Z×Z and c in Z, c = b, b=a}

It's kernel is {(0,0)}. From what I have researched is that the cokernel of this homomorphism is Z/im(f).

Since Z is an abelian group under addition, it's subgroup {(0,0)} under addition is a trivial normal Subgroup of Z, which is the smallest amongst all other normal subgroups of Z (although any normal subgroup of Z might work here). The underlying set of the subgroup is also the image of the kernel of f. This subgroup is also an ideal in Z.
Hence, the cokernel of f will be:
cokernel(f) = Z/im(f) = {a + im(f) | a in Z}

Infinite cosets will be formed here, each singleton containing a distinct integer. Hence the quotient ring which is the cokernel will be {...{-2}, {-1}, {0}, {1}, {2}...} or just {...-2,-1,0,1,2...}

Is my notion of understanding cokernel correct? What does it really indicate here?

[chilly ocean]

This is not a function from Z x Z to Z!

[delicate orchid]

That is the wackiest way I’ve ever seen someone “define” a “function”

[chilly ocean]

I'm confused now lol

Like for every (a,b) in Z × Z, there exists c in Z

If ((a,b),c) = ((a,b),d), c=d

[delicate orchid]

Oh my god did you fall for the set theory meme

What’s wrong with f(a,b)=c

[chilly ocean]

We cannot treat (a,b) as a single entity?

[hidden haven]

The problem is that your function is only defined on pairs of the form (a,a)

[delicate orchid]

Yes, we can, it’s an element of ZxZ

[hidden haven]

Because you imposed that condition in the set definition

[chilly ocean]

So considering a ring Z × Z with only pairs (a,a), will it be a ring homomorphism then?

[hidden haven]

Then it won't be called Z x Z, but yes

[chilly ocean]

Although by definition Z×Z is a set of all ordered-pairs

Yes

[hidden haven]

The ring of pairs of integers of the form (a,a) is isomorphic to the ring of integers

(a,a) ↦ a
a ↦ (a,a)
are homomorphisms that are inverses of each other, so are isomorphisms

[chilly ocean]

Ohh I see

Thank you very much 👍

Wait so is my cokernel correct?

I guess both the kernel and cokernel are trivial and the homomorphism is also bijective, so it could be correct

[odd flame]

Hi guys

This is my working for question 1 im not sure if im doing this right at all

Can someone tell me if it is

[pastel cliff]

what kind of question is this lol

"for which sets does it have an identity" for the ones that have indentity?

[chilly ocean]

This seems alright to me.

[lethal dune]

[chilly ocean]

it's asking you to find which sets S are such that this operation has an identity element

Yeah @pastel cliff I think you misunderstood the question.

[delicate bloom]

my take is you're not going back to the definition of what an identity element is and writing that out to see what happens

[lethal dune]

rational root theorem might be helpful here

for e]

[chilly ocean]

Oh oops yeah I didn’t check E. Just use RRT.

[odd flame]

I just used the fact that since it’s irreducible in F_3 it would be irreducible over Z

[chilly ocean]

Or you can use some other irreducibility lemmas like Eisenstein.

[odd flame]

Since f is monic

[pastel cliff]

an identity for a law of composition is an element e of the set S s.t. for all elements s of S, es = s

[lethal dune]

okay

[odd flame]

And 3 is prime

[chilly ocean]

Yeah that works too great job bruh

[odd flame]

Thanks

[pastel cliff]

the law will have an identity if there's an element that satisfies this

what i mean is like

the sets for which that "law" (im just thinking of it as an operation) has an identity are just the ones that have such an element that satisfies es = s for all s in S

or maybe i need to be more specific to the ab = a operation?

im definitely overthinking this lol

[delicate bloom]

yeah keep going

[chilly ocean]

I mean isn’t the identity commutative too? We need ea=ae=a

[delicate bloom]

true, but what they wrote was sufficient for making an important discovery so I didn't want to side track with that until after

[chilly ocean]

Oh okee

i'd be rich if you gave me a dollar every time someone posted "i mean (solution to the problem)" in this server

Oops

[pastel cliff]

i get it tho

this is what happens when i get cocky when starting a book on something i kinda know lol

[odd flame]

Any hints for exercise 2

[delicate bloom]

focus on combining this @pastel cliff

[odd flame]

Can i use eisensteins criterion here

[chilly ocean]

all you need to show is that it is not reducible

[odd flame]

Thanks

[pastel cliff]

it should be no sets right

bc es will always be e

and we'd need it to be s

[delicate bloom]

so close

[pastel cliff]

ok well i guess any set of order 1

but that's it

[delicate bloom]

sure, not the terminology I'd use but sounds good

[pastel cliff]

set with one element

appreciate the guidance

i took an algebra class this past semester but then did 0 math for like 3 weeks so now im trying to do most of the problems in Artin

will be in here frequently

[delicate bloom]

I like artin, it's a good book for review, and if it's ever too terse or whatever you can always look at a different book

[pastel cliff]

it seems to focus a lot on LA

at least early on

idk if i like that or not

kinda embarrassing that the ch1 problems are taking some time too kek

[delicate bloom]

well it's good you're attacking them head on and not running away from them lol

[pastel cliff]

[chilly ocean]

What book is this

[pastel cliff]

trying to stick to the idea that there are no bad problems, only ones i can do and ones i can learn from

Artin

[chilly ocean]

Oh that’s a good one, you’ll have fun for sure

Good luck in your study of AA broskie

[pastel cliff]

merci

[pastel cliff]

this is more of a meta question but is this a valid proof by contradiction? like should i be more explicit about it being contradiction?

im trying to make my proofs shorter (i tend to be overly verbose)

[chilly ocean]

"suppose that a has two unique inverses"

you are to show the inverse is unique

[pastel cliff]

yeah i changed that to distinct

[agile burrow]

really you don't even need it to be a proof by contradiction, you've just shown that if there are two inverses then they are necessarily equal

[chilly ocean]

you can make it shorter 😉

[agile burrow]

proof by "because I said so"

[pastel cliff]

ommitted

the "there exists" part is probably unnecessary right

[chilly ocean]

proof: "see p. 87 of (ancient book lost to library burnings)"

[pastel cliff]

i love these

actually this is unintentionally a good example of what i mean

should i have statements like "we can see that this is a contradiction since..." or "this contradicts our original assumption and thus..."

are these usually redundant or is it just author's preference

[chilly ocean]

if you're submitting it as homework then you might want to include them

[pastel cliff]

i'll take that to mean it's preference

but nah im just reviewing easy stuff and typing up for practice

[untold basin]

Hello

1. Is to show that A is a subring of Q and I had to determine A* which I did
   For the second point, I have to show that m is a the unique maximal ideal of A but I struggle a bit

Can someone give me a hint pls ?

[dreamy fiber]

You only have to show that the sum of 2 non-units is also a non-unit (why?)

[red imp]

Does the first isomorphism theorem hold for Lie groups? As in like, does it respect the topology of the manifolds?

[hidden haven]

Whenever you have a quotient by an equivalence relation ~ defined for any kind of object X, the following holds
There is a canonical "quotient" map
X → X/~
which respects ~ (equivalent things have the same image)
such that whenever you have any map
X → Y
that respects ~, there is a unique map
X/~ → Y
making the relevant triangle commute

If this is what you mean by the first isomorphism theorem, then yes. But the induced map X/ker(X → Y) → im(X → Y) will not be an isomorphism in general. If X → Y was surjective to begin with, this new map will also be surjective. If ~ was saturated with respect to X → Y (meaning 2 elements with the same image must be equivalent) then this map is injective. So if both hold then you get a bijective homomorphism. So if your first isomorphism theorem is about the isomorphism im f = X/ker f, then it reduces to the question of whether there exist bijective homomorphisms of Lie groups that aren't isomorphisms

So if you take G to be discrete top on ℝ so 0 dimensional Lie group and H to be ℝ with usual manifold structure, then the identity map works as a counterexample

[chilly ocean]

lee's introduction to smooth manifolds

[red imp]

ahhh yes so the generalisation of the first isomorphism theorem is something akin to the path lifting lemma

[hidden haven]

Oh I was talking about maps of Lie groups of different dimensions in my counterexample 🤦

[red imp]

Thanks Terra and Moldilocks!

[chilly ocean]

Hi. I've been solving this problem for a while. Any hints or guides? Appreciate any help.

A,B,C and D are all block upper triangular (it's still defined for non-square matrices)

a_ij = 0 for i > j

[untold basin]

My bad but I don't see links

I've been trying for 1h30 now

I got a lot of things but I can't conclude

[dreamy fiber]

Ahh you fine

So the point right is you wanna show the set of non-units form a unique maximal ideal

[untold basin]

the set A - A* precisely

[dreamy fiber]

Can you see why it’s maximal (if it is an ideal)

[untold basin]

Not really

[dreamy fiber]

Well suppose it is not maximal

[untold basin]

I can't use anything but the definition

[dreamy fiber]

Then there exist some ideal J \neq the whole ring such that m is a strict subset of J

But since m is the set of non-units, what does J necessarily have to contain?

[untold basin]

Wait a minute pls

I would say 0 but it isn't helpful

1 would be interesting but it is in A*

[dreamy fiber]

Well every ideal contains 0

Well m is the set of non-units, J is a strict superset of m

So what does J have to contain?

[untold basin]

I hate when it's simple

wait

[dreamy fiber]

Got it?

[untold basin]

no

[untold basin]

1 ?

[dreamy fiber]

Yeah exactly

J has to contain a unit

So J has to contain 1

So J is the entire ring

Contradiction! Thus the set of non-units is a maximal ideal

[untold basin]

contradiction because J =/= A

[dreamy fiber]

Yep

Now can you see why this maximal ideal is unique

[untold basin]

Wait pls

[untold basin]

How did you see that

Because I developped what does "sub set strict" means

And I got it

But how do you see it

[dreamy fiber]

Well m is the set of non-units, J is a strict super set of m

Well so there’s a difference between subset and strict subset right

[untold basin]

There is an element of J which is not in A-A*

[dreamy fiber]

If A is a subset of B, then A can still equal to B

Right exactly

[untold basin]

Ok I understand

[dreamy fiber]

Now can you see why this is unique?

[untold basin]

There is a detail I want to understand first

[dreamy fiber]

Yeah what’s up

[untold basin]

Why this element has to be 1 ?

I just know there is an invertible

[dreamy fiber]

Well it doesn’t have to be, but it is a unit

So call this unit r

[untold basin]

ok

By def of ideal

1 is in J

[dreamy fiber]

Then you know ideal is closed under scalar multiplication, so r^(-1) r = 1 is an element of the ideal

Yes exactly

[untold basin]

I feel bad

[untold basin]

Anyways I'll try this

[dreamy fiber]

Oh you are fine

[untold basin]

It's some trivial stuff I have to know this

[dreamy fiber]

Keep in mind that maximal ideals do not contain units

[untold basin]

For now I have that if A\A* is a subset of I_0 or I_0 is a subset of A\A* I have a contradiction

I_0 is the maximal ideal I obtained by supposing A\A* is not unique

[dreamy fiber]

Yeah exactly

[untold basin]

But now I have to treat the other case

[dreamy fiber]

Alright so now you know that if A\A* is an ideal, then it is uniquely maximal

[untold basin]

I'm not exhaustive, am I ?

[dreamy fiber]

What other case are you worried about?

[untold basin]

This case

[dreamy fiber]

The uniqueness of the maximal ideal more less comes from the fact that every non-unit ideal is necessarily a subset of m

[untold basin]

wait

I think

Ok I see

[dreamy fiber]

Alright

Do you know why A\A* is an ideal in the first place

[untold basin]

I want to clarify first why there are not other cases

I have :

[dreamy fiber]

Okay?

[untold basin]

A\A* not a subset of I_0 AND I_0 not a subset of A\A*
But second gives me an element of I_0 a which is in A* so 1 is in I_0 so I_0 = A which is a contradiction because I_0 is maximal

[dreamy fiber]

Right so I_0 has to be a subset of A\A*

If it’s a strict subset then I_0 is not maximal

If it’s equal to A\A* then well A\A* is unique

[untold basin]

I didn't get that

ok I got it

since A\A* is maximal

then A\A* = I_0 or I_0 = A

first is false since it's strict

so I_0 = A

I_0 not maxi

ok

[dreamy fiber]

Alrighty

Okay so

Do you know why A\A* is an ideal in the first place

[untold basin]

I can show it with the definition but it seems to be too long

So if you ask that then there is a shorter way

[dreamy fiber]

I mean that sounds like a fine proof

So in general the only thing you actually really need to show is that the sum of 2 non-units is a non-unit

The product of a non-unit with any other element in the ring is always a non-unit

[untold basin]

When you told me that

[dreamy fiber]

This fact is true in any commutative ring with 1

[untold basin]

I thought about irreducibility

[dreamy fiber]

Huh interesting

[untold basin]

I have to prove that A\A* < A first no ?

[dreamy fiber]

If you want to know more I guess

[untold basin]

subgroup

[dreamy fiber]

Oh dear

[untold basin]

c:

[dreamy fiber]

So to show something is an ideal

You only have to show that it’s closed under addition

And that it’s closed under multiplication by any element in the ring

Now I mean an equivalent defintion 😂 is that an ideal is a subgroup of the additive group and closed under scalar multiplication

But in practice you only have to verify 2 things

[red imp]

Hi, I'm trying to show that the mapping cone of the Hopf map is homeomorphic to CP^2. Any hints?

[dreamy fiber]

This sounds more like a question for top?😂

[red imp]

oh whoops wrong channel lol

[dreamy fiber]

Lol you fine

[untold basin]

Why did I receive a definition like that then

[dreamy fiber]

French math be weird?

[untold basin]

xD

Even on english wiki I see that def

[dreamy fiber]

They call compact “quasi-compact” 😂

Well it’s just an equivalent defintion

The longer defintion might be better from a structural viewpoint 😂

But yeah, this type of ring the question asked you is what’s called a “local ring”, which are rings where it has 1 unique maximal ideal

Maybe it’d help to look into that?

[untold basin]

I'm currently trying to convince myself about that

[dreamy fiber]

Local rings play a really important role in ring theory bc it turns out you can actually turn a general commutative ring with 1 into a local ring in a process called “localization”. And it turns out many of your favorite properties of a ring (ex. Zero, injectivity, surjectivity) is true if and only if it’s true in every localization of that ring

But that’s really beyond the scope of your course

[untold basin]

Btw I'm Belgian x)

[terse crystal]

Yeah his ring is actually Z_p, which clearly is a local ring with the unique maximal ideal pZ_p. Z_p/pZ_p is isomorphic to Z/p since p=5Z is maximal

[dreamy fiber]

Sorry 😂 xD

[dreamy fiber]

Yeah exactly 😂 maybe that way would have been shorter honestly

It also conveniently answers whether the unit group is cyclic

Finite fields!

[terse crystal]

Yeah… it will be so nice if it is from commutative algebra course… so many steps can be omitted…

[dreamy fiber]

Yeah my initial suggestion was 😂 to just show sum of 2 non-units is a non-unit

I guess in these later classes you can get very hand wavy

[terse crystal]

That’s equivalent to the ring being a local ring right

[dreamy fiber]

Yep

If you assume axiom of choice

[terse crystal]

Yeah

[dreamy fiber]

I wonder what kind of intro to abstract algebra class toys with a local ring 😂 like that

[untold basin]

Z_p = Z/5Z ?

[terse crystal]

No localization

S^-1Z where S=Z\5Z

[untold basin]

Oh okay that's the limit of my course like Scarlet said xd

[dreamy fiber]

Yeah don’t worry about it 😂

[untold basin]

Every year I learn that last year was hard because we were not using handy methods to solve exs

[dreamy fiber]

Though since your course is teaching you about quotients

Localization is not too far off the tree

[dreamy fiber]

It comes with experience. I am sure you will be just fine

[untold basin]

It's not about understanding the objects but if I use a property of smth that's isn't in my course I will have to prove it during my exam

[dreamy fiber]

Well I mean who doesn’t like an extra lemma written on the exam 😂 Jkjk

[untold basin]

With pleasure I'd like to understand localization

[untold basin]

xD

[dreamy fiber]

Well if you are interested 😂 I think you should look into it

[glacial prism]

Can we find groups $G$ and $H$ such that $G\times H$ has a normal subgroup $N$ that contains neither $G\times 1$, $1\times H$ or any other normal subgroup of $G\times H$? Basically I'm searching for an example of a product group where the lattice of normal subgroups has an isolated chain (with minimal non trivial element $N$).

[cloud walrus]

MrMonday

[untold basin]

Thanks for your help Scarlet

[dreamy fiber]

No problem! Glad I was helpful

[odd flame]

Hey guys can someone give me a hint for exercise 3?

[dreamy fiber]

I might be wrong but C_4 times C_4 has normal subgroup C_1 times C_2, it doesn’t contain any other normal subgroup except for the trivial group

[dreamy fiber]

So every element of your ideal is essentially

f(x)(x - 2) + 5g(x)

f(x)(x -2) + 5(integral polynomial) + 5k

Now given some integer polynomials h(x) can you show that h(x) - one of 1, …, 4 can always be decomposed to the expression above?

[chilly ocean]

Hi everyone. I've been solving this for a while. Kindly give a hint on how to approach this problem.

[dreamy fiber]

I am planning on taking a course in Lie Algebra next semester would you recommend learning this subject or no 😂

[chilly ocean]

My professor sucks. Moreover it's difficult that you have to master differential geometry and algebra to understand it.

[dreamy fiber]

It do be like that sometimes :/

Sorry about your Professor

[chilly ocean]

It also depends on whether the class will start with matrix Lie group or you just go straight ahead to general Lie groups

[dreamy fiber]

I took a class on analysis of manifolds this semester and my only exposure to “lie things” have been lie brackets, levi-civita connections, and Riemann curvature tensors

You think that’s sufficient enough to start?

[odd flame]

Thanks i will work with this

[glacial prism]

Oh you are right, but that's unfortunately not what I'm looking for. I misphrased what I actually want :/ We get an embedding of the normal subgroups of G times the normal subgroups of H into GxH by sending (N, N') to NxN'. I wanted to exclude all of these as well. Unfortunately C1xC2 is precisely one of those. Ideally I'm searching for a normal subgroup of GxH that is both maximal and minimal in the lattice of normal subgroups of GxH (so nothing bigger except GxH and nothing smaller except 1x1) and not of the form NxN'.

[dreamy fiber]

I think you might have better luck trying this on sage 😂

Good luck

[glacial prism]

:') very likely

[untold sapphire]

100%

It's one of the most beautiful areas of mathematics imo

Although not a fan of the courses / books that just look at different matrix groups

[untold sapphire]

I know less DG than you and I've been fine

Humphreys is great

[dreamy fiber]

Thank you so much!

Representation Theory and Lie Algebra are very foreign 😂 subjects to me but I think it will be great

[terse crystal]

f(x)=f(2) mod I, and f(2)=r mod 5 for some 0<=r<=4, so f(x)=r mod I

[terse crystal]

Sure, G=H=S_3,N={e,(12)(45)}

I mean G=S({1,2,3}),H=S({4,5,6})

Oh or simply <(1,1)> in Z/2Z times Z/2Z

[glacial prism]

Right, makes sense

Thanks! :D

[untold basin]

is ℚ x ℂ x ℚ x ℂ is isomorphic to ℚ² x ℂ² ?

I just have to define the morphism which moves the components ?

[glacial prism]

yes

[untold basin]

Hello,
1 : Show that [..] is isomorphic to the product of 4 fields which are
Q[X]/(X-1) times Q[X]/(X²+X+1) times Q[X]/(X+1) times Q[X]/(X²-X+1) (I did that question)
I struggle a bit for 2nd one which says :
Determine all P(X) in Q[X] s.t X^6 - 1 divides P(X)²
I found that
X^6-1 divides P(X)² <=> P(X)² mod (X^6-1) = 0 mod (X^6-1)
But from here I don't really know what to say

[hidden haven]

You have already factored X^6 - 1

Use that factorization

Q[X] is a UFD

[shell agate]

You're almost there: X^6-1 divides P(X)² iff P(X)² = 0 mod (X^6-1). But Q[X]/(X^6-1) is isomorphic to a product of 4 fields: by this isomorphism, P(X)² = 0 mod (X^6-1) iff P(X)² is sent to 0 in all these fields

So, P(X)² = 0 mod (X+1), P(X)² = 0 mod (X-1), etc.

[untold basin]

yes exactly

I found that

But it's useless

because I have that P(X)² is in (X+1) ... P(X)² is in (X²-X+1)

So P(X)² is in the intersection

But the intersection is (X^6-1) 🤣

In french we say "I bite my tail"

[sly nexus]

use the fact theyre all irreducible

[untold basin]

Since they all are fields, there are domains so P(X)mod(X^6-1) * P(X)mod(X^6-1) = 0 mod(X^6-1) => P(X)mod(X^6-1) = 0 mod(X^6-1)
I conclude that P(X) is in (X^6-1)

[shell agate]

Don't undo what you did

P(X)² = 0 mod X^6-1 means it 0 mod the 4 other bits

Now, let's see. When is P(X)² 0 mod X+1 ?

Well, it means P(X)² = (X+1)Q(X), for some Q.

[untold basin]

When P(X)² is in (X+1)

Yes

[shell agate]

Ok, now because P(X)² is a square, in fact, (X+1) must also divide Q(X), so we can say P(X)²=(X+1)²R(X), for some R(X)

And in fact, we now know R itself is a square.

So, P(X)² is 0 mod (X+1) iff P(X) is 0 mod (X+1) iff P(X)=(X+1)Q(X), for some Q.

The same holds for any irreducible polynomial.

So, doing the same argument, P(X)² is 0 mod (X^6-1) iff P(X)=(X+1)(X-1)(X+1)(X²+X+1)(X²-X+1)Q(X)=(X^6-1)Q(X) iff P(X) is 0 mod (X^6-1)

[shell agate]

Ah sorry, I didn't read this

This is correct

[broken stirrup]

I've done exercise 6 and now trying to do 7. I'm trying to use previous exercise but can't proceed and maybe I'm not in the right path to follow

[next obsidian]

Maybe try to show that if u,v are transcendental over F? Then K(u) ≈ K(v)

Hint: a transcendental element is like a formal variable

[patent ocean]

for 7c) does this make sense? The cardinality of G being less than or equal to 2

[untold basin]

@hidden haven @shell agate thanks btw

[untold basin]

thx too

[patent ocean]

but there are normal subgroups that have cardinality greater than 2 so I don't understand

[sly nexus]

np bud

[patent ocean]

oh it says any two elements

ignore me

[sly nexus]

normal subgroups dont have that property

oops

kk

[junior harbor]

I'm reading this book on abstract algebra, and they mention that a cyclic group is a group that has an element a such that it is equal to [a], then mention [a] consists of all $a^n, n > 0$. Problem is, then they define the positive integers as a cyclic group of 1 or -1... how? Wouldn't that just be the set {1}?

[cloud walrus]

Duhon

[chilly ocean]

the positive integers isn't a group, do you mean integers?

[junior harbor]

Oh yeah right

Misread it

[chilly ocean]

a^n means applying the group operation to a, n times. the group operation in the integers is addition

[junior harbor]

Integers relative to addition

Ohhh

Gotcha. They exemplified earlier with multiplication

So I thought we were talking about that still

Thanks

[broken stirrup]

Ok so I showed that K(v) =~ K(f(u)/g(u)) but i don't see how result follows 😦

[untold basin]

When i have that morphism

And it isn't precised that it is a morphism of what structure

It is ambiguous right ?

[south patrol]

I mean it's gonna be either groups or sets right

[chilly ocean]

"morphism of sets"

[untold basin]

Why not rings ?

[chilly ocean]

are these rings?

[untold basin]

I don't know

The first question asks me to prove that 1+p mod p²Z is invertible in Z/p²Z

And then I have to prove that the order of that element is p in (Z/p²Z)*

So I guess it is the additive order and then multiplicative ?

[chilly ocean]

rings have an additive identity ("0") and these don't, so all you can say looking at them is that they are groups, under multiplication. so f is a homomorphism of groups

[untold basin]

So by Z/p²Z they mean (Z/p²Z,.) ?

[chilly ocean]

$\bZ/p^2\bZ$ is a ring, and $(\bZ/p^2\bZ)^\times$ means its group (under the ring's multiplication) of units

[cloud walrus]

TTerra (ping spam = block)

[untold basin]

^ so here it is the additive order ?

[chilly ocean]

group (under the ring's multiplication)

[untold basin]

Oh okay

[chilly ocean]

it says "order in (Z/p^2Z)*"

[untold basin]

"Show that x is invertible in Z/p²Z and that its order is p in (Z/p²Z)*"

[junior harbor]

Consider a set of mappings S and a mapping of those into S itself (say integration). I guess S can't be a group because integration isn't a binary composition, and also because some functions can't be properly integrated. What would S be then?

[chilly ocean]

Integration is a linear operator

[junior harbor]

I guess, but it does map a function (a mapping) into another one doesn't it?

[chilly ocean]

Yes

[junior harbor]

Well, then I guess it does count as a mapping

But it isn't a mapping that takes 2 values and mixes them together with some rule into a single thing, so what would S even be?

Not a group/semi-group/subgroup for sure

[chilly ocean]

S together with integration?

[junior harbor]

Yeah

[chilly ocean]

some unary algebra

[junior harbor]

Well, haven't gotten to that part yet lol

But I'm guessing a similar thing to generators and cyclic groups exist in there?

[chilly ocean]

It means a set A together with operation u:A to A

[junior harbor]

Huh?

Hm

[chilly ocean]

cyclic groups in something that isn't a group... right

[junior harbor]

I guess, but is there some definition/notation for it?

[chilly ocean]

for an unary algebra?

[junior harbor]

Yeah ig

[chilly ocean]

it's just denoted as a pair (A, u)

[junior harbor]

Gotcha, thanks

[chilly ocean]

There's a general notion of algebra, which is a set together with a family of n-ary operations (where n can be different for each operation)

we can demand an algebra to satisfy some equations, and then this class is called either an equational class or a variety

we can talk about subalgebras of a unary algebra, of their congruences, quotients, products

things such as first isomorphism theorem etc. hold for those as well

[junior harbor]

Interesting. Well, thanks for the help

[chilly ocean]

Oh. I thought you'd inquire about this more. Sad

but yes, you can talk about generators and "cyclic algebras" in an unary algebra too

by itself this is not a very interesting concept, so to study it algebraically you'd like it to satisfy some equation, like semigroups satisfy associativity for example

[untold basin]

hello

I want to describe explicitly what are the elements of order 2 in this set

So what I did is :
I'm searching all the P in Q[X] such that
P²mod(Phi_5) = 1 mod (Phi_5) <=> there exist T in Q[X] : P²-1 = T * Phi_5

[tribal moss]

What is Phi_5?

[untold basin]

Phi_5(X) = X^4 + X^3 + X^2 + X + 1

I noticed that everytime I want to explicit elements of a quotient like this one I often get a horrible equation

[cursive temple]

i think $\bQ[X]/(\Phi_5) \cong \bQ[e^{\frac{2\pi}{5}}]$

[cloud walrus]

Pappa

[cursive temple]

isnt the most useful thing ever

[tribal moss]

Yeah, so in particular the quotient ring is a field.

[untold basin]

In my course

[tribal moss]

And so x²=1 can have at most two solutions.

Which must be 1 and -1.

[untold basin]

I have a property which is "Phi_5 is irreducible in Q[X]"

[tribal moss]

Of which only -1 has order 2.

[untold basin]

Is it useful to know that it is a field ?

[tribal moss]

Yes, because we know that in a field (or more generally an integral domain) a polynomial cannot have more roots than its degree.

In particular there cannot be more solutions to t²=1 than the two that Q already has.

[untold basin]

How did you obtain that equation ?

I have to process all the informations

[tribal moss]

x²=1 is what make an element have order 2 (if it doesn't have a smaller order).

[untold basin]

and here x is an element of Q[e^2pi/5] ?

[tribal moss]

Yes.

It should be e^(2pi·i/5), by the way.

[untold basin]

oh okay

[tribal moss]

I.e. a primitive fifth root of unity.

[untold basin]

I couldn't have found that by myself

So everytime there is a quotient we try to find an isomorphism to see the elements more clearly ?

[tribal moss]

With a bit of experience you will recognize x^4+x^3+x^2+x^1+1 as (x^5-1)/(x-1) -- so it's the polynomial whose complex roots are exactly the fifth roots of unity, except for 1.

[untold basin]

I started from X^5 - 1 lmao

Okay I see

Thanks both of you @tribal moss @cursive temple

[chilly ocean]

Troposphere's solution doesn't use any explicit characterization of (the roots of) this polynomial though, other than that it's irreducible to conclude that the quotient is a field

[tribal moss]

(But, for full disclosure, the way I convinced myself that this particular polynomial is irreducible was by noticing that its roots are the primitive fifth roots of 1, such that if one extends Q with one of them, the all the others automatically follow. Christophe's class might have followed a different path to that fact, though).

[chilly ocean]

Oh, alright. How I think about this is using Eisenstein criterion

[untold basin]

That's how my prof did prove

We used the fact that p divides the binomial coefficient p
k

[cursive temple]

A nice fact is that every cyclotomic polynomial is irreducible

[tribal moss]

That looks like the argument Blitz was describing.

[chilly ocean]

Yeah, it's the same thing that I was thinking about

[untold basin]

Can I ask what were the intuitions for this ?
I don't have smth similar in my course I think I'm quite curious

[delicate bloom]

well once you know $\Phi_p(X) = \frac{X^p-1}{X-1}$ modding out by it means $X^p-1=0$ and so $X^p=1$ so $X$ is a $p$th root of unity.

[cloud walrus]

Merosity

[tribal moss]

The general intuition is that Q[x]/(P) where p is an irreducible polynomial, gives Q[alpha] = Q(alpha) where alpha is a complex root of P.

[untold basin]

Yes I understood that

[untold basin]

Okay I think I will check some proofs for more details

thanks guys

[chilly ocean]

I think you need more. Check out Loring Tu Introduction to Manifolds. He does a great job introducing the topic.

[chilly radish]

The standard proof of this fact, for n not prime is very cool

[chilly ocean]

Over Q the rational numbers. This is not true for any arbitrary field.

[chilly radish]

Yes

Sure

, you can always just adjoin roots of unity to your field

[sage wagon]

Hi, so I'm confused about the definition of an FG-module. In Dummit and Foote it seems to mean a vector space V with an action FG x V -> V, but in other places it says it's a vector space V with a multiplication G x V -> V with some properties. Why is there a discrepancy?

Or is it actually the same thing by linearity?

[cosmic oriole]

It's not the same thing, but by linearity, you can extend it to FG x V -> V.

[sage wagon]

Ah ok that's what I meant, but essentially the two definitions are equivalent right?

[cosmic oriole]

Basically, yes.

[sage wagon]

Amazing, thank you! 😄

Oh one more question, in this case do they technically mean unital left FG-module?

[cosmic oriole]

Yeah, it will usually be a left module. If by unital you mean there is some element e that acts on V as an identity, then yeah.

[sage wagon]

I see, thanks a lot!👍

[grand cliff]

Hi, I've been prepping for my final exams and there are a few exercises I've been struggling with, could anyone provide the solutions:

1. Show that there are no group homomorphisms $$\varphi: S_9 \to \mathbb{Z}_7 \times \mathbb{Z}_3 $$ whose image is $$\mathbb{Z}_7 \times \mathbb{Z}_3$$

2. Let $$g = (12)(345)(678)$$ and $$h=(3456)$$ (elements of S8) Show that $$ \langle g, h \rangle $$ has an index two subgroup.

[cloud walrus]

Cursor

[agile burrow]

For 1, the first isomorphism theorem implies S9 / ker(\varphi) = Z3 x Z7. In particular, |ker(\varphi)| = |S9| / |Z3 x Z7| = 17280, hence ker(\varphi) is non-trivial. The kernel of \varphi is a normal subgroup of S9, but the only non-trivial normal subgroup of S9 is A9, which has order 181440.

[grand cliff]

oooh okay, I forgot that S9 only had A9

I got everything up to there

[agile burrow]

yeah, that's where most of the work is hidden anyways

For 2, the not smart but completely valid way to approach it is to compute the order of the <g, h> by writing out elements and then probably using Sylow or something

[grand cliff]

Yeah, the question says not to do that actually

or at least

it says you can do it without doing that

[agile burrow]

understandable, i'll think about it a bit more

[chilly ocean]

Hi everyone. Does anyone know what is the centralizer of the lie algebra of the set of upper triangular matrices?

[patent ocean]

For part a) of the proof can someone explain to me why for the associativity argument. The proof has used the brackets in this way? Like why can't I immediately go from right to left since composition of permutations is associative

[delicate orchid]

We’re proving that composition of permutations is associative. You cannot assume what you’re proving

[lethal dune]

or can you

[patent ocean]

Ok why does the proof use the brackets in this order

like what is the reasoning behind it

[shell agate]

I'm assuming the action you have from Sym(S) to S is a function written x -> xf, where x in S and f in Sym(S)

x((fg)h) is the image of x under the element (fg)h of Sym(S)

This is the same as first taking the image under (fg), then the image under h. So, it is the same as (x(fg))h.

x(fg) is now the same as (xf)g

so (x(fg))h is the same as ((xf)g)h

I take x, send it to its image under f, send that to its image under g, and then to its image under h.

Same as taking x, sending it to its image under f, then to its image under (gh)

That is, x(f(gh))

And there we go

(fg)h and f(gh) agree on x

But there's nothing special about x

So they agree everywhere on S

That is, f(gh)=(fg)h

[subtle ivy]

does anyone happen to know what the coproduct in the category of inverse semigroups is?

or if we have one lol

[chilly ocean]

Inverse semigroups can be treated as triples (S, *, ^-1) where (S, *) is a semigroup, and ^-1 is an unary operation satisfying the equations x = xx^-1x and x^-1 = x^-1xx^-1

Since they're an equational class, I think that necessarily means they have coproducts

No wait, this doesn't tell us about the uniqueness of x^-1 as an inverse I think... so it's a different structure

https://encyclopediaofmath.org/wiki/Inversion_semi-group

this article defines them as a variety i. e. equational class

they just need more equations than what I wrote

https://encyclopediaofmath.org/wiki/Free_product

check this for existence of coproducts - they are also called free products in the context of universal algebra

[subtle ivy]

ok thx. im more interested in an explicit coproduct so ill look at this some other time maybe. exploring some other ideas first

[chilly ocean]

You'd probably want to take your inverse semigroups S and T, take their free product as a semigroups with involution, and quotient by some equations to make it an inverse semigroup

The construction of free inverse semigroup on wikipedia might be helpful
https://en.wikipedia.org/wiki/Inverse_semigroup#Free_inverse_semigroups

[patent ocean]

nicely explained, thanks

[wraith obsidian]

IIRC in one half of Birkhoffs HSP theorem you essentially introduce the „Free algebra modulo equations“ (ie congruence + substitution closure), which is a particularly important construction.

Why does the coproduct exist? The coproduct of A and B in an equational theory T is just the free product A\star B modulo T.
Why does the free product beteeen A and B exist? Well, it's just the Term algebra over A \sqcup B modulo the equational theories Eq(A) and Eq(B).

tl;dr you can always force equations to hold and that makes everything work

(Modulo mistakes because ||I'm tired, drunk, and took this course 4 years ago||)

[devout crow]

I never got the reason for notation 53

if it's not standard why use it

[wild solar]

can anyone tell me why is one allowed to write f(x) like this , as (x-a)g(x) where g(x) has coefficients from F(a)? Pls help I'm going crzay with this...many thanks, algebra is hard.

[next obsidian]

a is a root of f

So you can just rip away a factor of (x - a)

You can use polynomial long division to prove this

[wild solar]

okay, but why deos g(x) has coefficients from F(a)? And how does this help the proof?

[next obsidian]

Because a exists in F(a)

I’m sure you showed that if K is a field then K[x] is a Euclidean domain

Applying this with K = F(a) let’s you conclude that you can factor out (x - a) inside of F(a)[x]

It helps the proof because g(x) is now of lower degree than f so you can apply induction

[wild solar]

actually that does not sound familiar...I'm gonna try to find that in the textbook, but sounds like it makes sense.

[next obsidian]

I mean

[wild solar]

many many thanks

[next obsidian]

This is just saying

Take p,q

Then you can write p(x) = h(x)q(x) + r(x) where r has degree < q(x)

This is dividing p(x) by q(x) with remainder r(x)

If you assume that p(a) = 0 and q = (x-a)

Then

1. r(x) is a constant since it has deg < deg q = 1

And then if you plug in a you see
p(a) = h(a)q(a) + r

But then you get 0 = r

[since r is just a constant now]

So you showed that if a is a root of p, then you can write p(x) = h(x)(x-a)

You might’ve seen this proof in high school or something, or maybe it was just stated without proof but

You probably learned like, if f(a) = 0 then you can write f as (x-a)g(x) for polynomials over the real numbers

[wild solar]

oh wait I remembered

found it

[wild solar]

yeah

[next obsidian]

That’s not the theorem

You need it to be a Euclidean domain

In order to do division with remainder

[wild solar]

okay

[next obsidian]

I can try to produce the proof right now

[wild solar]

There are so many theorems on herstein

[next obsidian]

I think the key is that every polynomial (after multiplying by a unit) is monic

[wild solar]

a lot of the times I do't remember what theorems I learnt

[next obsidian]

Probably by induction… 🤔

Yeah I mean there’s a lot to remember

And if you don’t use a theorem you tend to forget it

Okay so let’s try to prove this by induction if you want

[wild solar]

I found something that says if R is a UFD then R[x] is UFD

[next obsidian]

Or you can try to find a proof

Yeah that one is still not what you want

Although that’s super useful

What you want is that if F is a field then F[x] is a Euclidean domain

Basically you just want to be able to do polynomial long division

[wild solar]

it'd be great if I could understand it

[next obsidian]

Okay so

[wild solar]

[next obsidian]

If we induct on the degree this seems good

So the statement is that if you have p(x), q(x) nonzero then you can write
p(x) = h(x)q(x) + r(x)

Where r is lower degree than q

Or 0

So we’ll induct on the degree of p

The thing we want to divide by q, with remainder

So if it’s degree 0 then this is easy

Just let h(x) = 1, q(x) = p(x) and r(x) = 0

The degree 1 case is also pretty easy, and this is kind of how we’ll do the general one but it’s worth writing out

So first of all, if deg q(x) > deg p(x), we can let h(x) = 0, and r(x) = p(x)

Maybe take a second to verify that this works

Also I don’t think we’ll even need induction lol

But whatever

[wild solar]

wdym

[next obsidian]

Like I don’t think we need to ever use the inductive hypothesis

But it’s whatever

[wild solar]

oh okay

[next obsidian]

Does the proof make sense so far though?

[wild solar]

yes

[next obsidian]

Okay so we need to handle the case when q(x) is degree 1 or 0

So let’s write p(x) = ax + b

And assume q(x) is degree 1 for now

Write q(x) = cx + d

So okay, first note that p(x) - aq(x)/c is a constant

Because we fudged q(x) so it has the same leading term as p(x) now