groups-rings-fields

[thorn jay]

We call a group centerless if its center Z is trivial

This is equivalent to the conjugation group action being faithful

And is in particular interesting because G is centerless implies Aut G is centerless, but i believe it is not the case that Aut G = Inn G => Aut Aut G = Inn Aut G

Every automorphism tower of a finite centerless group terminates finitely

🔥

[hidden wind]

tfw S6

[glad osprey]

I see, interesting

[glad osprey]

coincidentally I recently saw this being linked somewhere on this server: https://math.ucr.edu/home/baez/six.html

[hidden wind]

you’re the second person to link me to this today

[glad osprey]

I think I got the link from the person who linked it to you the first time

[acoustic igloo]

Let $R$ be an ordered integral domain. Show that if $a^{2n+1}=b^{2n+1}$ where $a,b\in R$ and $n$ is a positive integer, then $a=b$.

[cloud walrus]

Axe

[acoustic igloo]

am i meant to do this by factoring a^{2n+1}-b^{2n+1} ?

$(a-b)\sum_{i=0}^{2n} a^{2n-i} b^i = 0$

[cloud walrus]

Axe

[tardy hedge]

Yeah i need to learn and review group actions too

And just group theory in general

[crystal vale]

I think what if we extend this into its ring of fractions

Ordered integral domain?

[acoustic igloo]

an order is given by a nonempty subset P with
closure: a,b in P => a+b in P and ab in P
trichotomy: exactly one of (a in P) (a=0) (-a in P) holds

P represents the positive elements

a < b iff b-a in P

[patent harbor]

that makes sense to me.
So if a-b =0 then done.
suppose the other part is 0.
Then in the case n=1,
we have a^2+b^2=-ab.

So we have 9 cases left by trichotemy.
In all cases we may conclude a=b.
(and the arugment generalizes)

[acoustic igloo]

i think we can narrow down to 3 cases
a=b=0
a,b < 0
a,b > 0

[patent harbor]

yea

[acoustic igloo]

if the sign of a and b differ, then the sign of a^{2n+1} and b^{2n+1} differ

so it's not possible

[patent harbor]

elegantly put

[round portal]

good morning guys

I am trying to prove the following assertion

Let G be a group and |G| = pq where p and q are prime numbers

Then for all proper subgroup H of G, we have that H is cyclic

what I've done so far:

by Lagrange's theorem: |G| = [G:H] * |H| => pq = [H:G]*|H| => |H| = 1 or |H| = p or |H| = q
if |H| = 1 then H = {e} => cyclic
if |H| = p then H isomorphic to Z_p => cyclic
same for |H| = q

am I on the right track or have I missed something?

[velvet hull]

there are only 3 proper subgroups of G, there isn't much to it

[round portal]

thanks buddy!

[finite turtle]

this from dummit and foote, how does one deduce K_i subgroup of N? and using this fact how does one conclude N is of the form G_I = product of K_i and identity?

[rocky cloak]

Ki \cap N is a normal subgroup of Ki so is either Ki or trivial.

If it's Ki, that means Ki < N

[woeful sage]

Chat is there such a thing as a "minimal subgroup"?

maybe something like "Given a group G, a non-trivial subgroup K is called a minimal subgroup if K is a subgroup of H for every non-trivial subgroup H of G"

[delicate orchid]

well you seem to have defined it so I guess there is!

non-trivial example is <-1> in Q_8

[velvet hull]

something something zorns lemma actually, does it work? I think that would only guarantee the existence of the trivial subgroup, sooo... not exactly useful

[thorn jay]

Not sure how common it is for subgroups, but for normal subgroups/ideals/subspaces/congruences you have basically the same idea, which is called the monolith

[woeful sage]

Intereting

I actually came up with this based on a problem that me and my friend were trying to do

[woeful sage]

nani

[thorn jay]

It's existence is equivalent to not being able to nontrivially subdirectly embed your structure

[woeful sage]

wait you are the UA guy

[thorn jay]

I am

[woeful sage]

is this some crank UA stuff?

jk jk

[thorn jay]

Lmaoo

[woeful sage]

I see

[woeful sage]

What does zorns lemonade say again?

[thorn jay]

Kind of? Algebras with such a monolith, called subdirectly irreducible algebras, essentially tell you the structure of your desired algebraic structures

[thorn jay]

In reply to this

[woeful sage]

I only know the well ordering property which is a consequence of Zorn's lemma

[glad osprey]

Hmm, this definition doesn't seem to mirror the definition of a maximal subgroup. I'm thinking a minimal subgroup would be a non-trivial subgroup H such that there is no subgroup K where {0} < K < H

[thorn jay]

That guaranteed a maximal subgroup/congruence im pretty sure

[thorn jay]

Yeah this is a totally minimal subgroup, in a sense

What you have is that the trivial subgroup is meet-irreducible in the subgroup lattice)

?

I believe?

[woeful sage]

yea this seems more inline with your classic "maximal ideal" definitions

[glad osprey]

Is there a lattice-theoretic term for these definitions? I'm guessing Neam's definition is a minimum in the poset of subgroups, while my definition is minimal in the poset?

[woeful sage]

this is another sign that I need to learn some basic order theory

I saw that the first chapter of Munkres had some nice stuff on this

[thorn jay]

I think Neam's definition is genuinely called a monolith

Haha

Lemma look it up

[woeful sage]

Also tomorrow I will Re:Start learning Algebra

I will absolutely grind group theory and ring theory

[thorn jay]

No it seems that it is only standard for congruences

Shame

I think its only useful there anyways

[thorn jay]

Lets goooo

[woeful sage]

what exactly are congruences?

I am only familiar with the number theoretic term

[thorn jay]

Its the same there actually!

A congruence on an algebraic structure is one that preserves the fundamental operations

So, if a_i ~ b_i for all i = 1, ..., n, and f is an n-ary fundamental operation:
f(a_1, ..., a_n) ~ f(b_1, ..., b_n)

The congruences on the ring of integers align with congruence relations on numbers

Because the set of numbers congruent to 0 is always an ideal of Z, and vice versa

For groups, congruences are the same as normal subgroups, modules have submodules etc

[woeful sage]

Ohh I've seen this for groups! lol I thought the congruence for integers was just a special case of that, but it seems congruence for groups themselves is a special case of UA congruence

[thorn jay]

Yessir

Those congruences form a lattice, and the structure and properties of that lattice is one of the main problems in UA

For example, if for sufficiently many algebras the congruences permute (meaning that R • S = S • R for all congruences S, R, • denoting the relational product), then they must have a term p(x, y, z) with
p(x, x, y) = y = p(y, x, x)

For groups this term is
p(x, y, z) = xy^-1 z

[delicate orchid]

been thinking about this some more. It's clear that if such a subgroup exists and isn't trivial, your original group must be a p-group (otherwise there will exist C_p, C_q subgroups for two different primes, which intersect trivially). It is also clear that any such subgroup must be cyclic prime order, so you're essentially asking for a classification of p-groups with a unique C_p subgroup, which there aren't a lot of

I legitimately cannot think of any other than C_{p^k} and Q_{2^k}

[thorn jay]

Ill add to this that the group must also be subdirectly irreducible

[delicate orchid]

sure whatever that means

[thorn jay]

Nevermind, thats probably not useful here

It just means that the group has a unique minimal nontrivial normal subgroup

[delicate orchid]

the humble intersection of all normal subgroups:

[thorn jay]

Minimal in the sense that a maximal subgroup is maximal

[delicate orchid]

https://tenor.com/view/biden-joe-biden-microphone-interview-walk-gif-2478024445377708296

[amber burrow]

where can i find some interesting group theory problems?

like something that uses a lot of different tools

[elfin wraith]

Interesting group theory problems
See that’s your issue right there

[thorn jay]

[elfin wraith]

Slightly more seriously, dummit and Foote has an obscene number of problems across its like 250 pages of group theory, if you want a slightly more curated list, googling “group theory qualifying exam” brought up loads of interesting (modulo your definition of the word) looking problem sets from various universities

[glad osprey]

I thought a group cannot be a union of subgroups at all, unless the subgroups contain each other?

[elfin wraith]

|| I haven’t seen this before or tried it, so I could be wrong, but is this some orbit stabiliser argument?||

[thorn jay]

Any algebraic structure is the union of its finitely generated subalgebras

Simply because any element generates a finitely generated subalgebra

[rocky cloak]

It cannot be the union of two subgroups

[elfin wraith]

Yeah I think I see an argument along those lines, a classic group theory apply some theorem, count, things go wrong because of the identity type argument I think

[rocky cloak]

But for example (Z/2)x(Z/2) is the union of 3

[elfin wraith]

I really need to start studying group theory, the exams in a month and it’s been about 2 years since I thought about groups

[glad osprey]

Ah, I see 👍 I was thinking the proof for 2 subgroups generalized

[elfin wraith]

I essentially just ignored the course last semester since “it’s group theory, how hard could it be” which is fine, other than the fact I also did that for 2 more classes this year

[south patrol]

Another fun one is that a vector space over an infinite field can't be the union of finitely many proper subspaces

[delicate orchid]

oh yeah? watch THIS

[thorn jay]

I wonder if thats a Mal'cev condition..

[south patrol]

I still want a better proof of this tbh lol

[elfin wraith]

Oh that sounds cool

Im unsure if ive ever actually seen a proof of wedderburns little theorem now, i cant recall one

[rocky cloak]

How does the proof go, if I may ask

[elfin wraith]

Apparently kcyznski gave a different group theoretic proof

That’s, something I suppose

[thorn jay]

Better known for other work

[elfin wraith]

This is why I started including the question I was solving above my solution to all my homework’s

I remeber getting really frustrated by that exact thing once and I’ve just kinda ran with it since

[amber burrow]

is this real 😢

[amber burrow]

idk i just tired of more exercise oriented problems, where you can kinda immediately you can kinda run down a small check list of things and one of them will work

actually one problem i saw recently on the usa-primes test was counting the number of subgroups of index 2,3,4,5,6 of PSL(2,Z)

not sure if discussion is allowed or not, even though they've already sent out acceptances

i found like a verbatim solution online, but most of it was way beyond the scope of my knowledge

[elfin wraith]

No im very much joking, group theory is very interesting and an incredibly important subject

[rocky cloak]

I mean if you want a list of accessible problems, some of them will trend to be similar in which case things start to boil down to a trick you can put on a list.

You can aim to find problems where those tricks are not on your list though I guess. Competition problems might be a good source...

A problem I remember thinking was hard was showing that there's no simple group of order 96. But I don't know if it's particularly interesting necessarily

IMC usually has group theory questions.

https://www.imc-math.org.uk/?year=2024&section=problems&item=day1

Problem 4, day 1, 2024 for example

[green fox]

Is $F_{a,b}$ abelian, where $F$ is the free group generated by $<a,b>$?

[cloud walrus]

Luke

[void cosmos]

no

[green fox]

So can you explain this to me. I am struggling to understand this notation, and the map $f$.

Firstly, $f$ is a map from $F_{a,b}$, which is not an abelian group (is it even a group)?. So, I don't see how the "obvious map" extends to words of the form $a^2b^{-1}a^{-12}b^{13}$ or other words of this form.

My guess is that $f(a^2b^{-1}a^{-12}b^{13}) = (-10, 12)$, i.e. it sums the exponents for $a$ and $b$ respectively. It is clear this a homomorphism.

Also, $ker(f) = {a^{x_1}b^{y_1}\ldots a^{x_n}b^{y_n} : \sum_{i=1}^n x_i = 0 = \sum_{i}^n y_i, n \in \mathbb{N}}$.

Is this correct? This means $<aba^{-1}b^{-1}>_n = ker(f)$, but I haven't tried showing this yet as I am not sure if this is correct.

[cloud walrus]

Luke

[void cosmos]

yes

<a,b> is a group

ur guess is right

the thing is

[amber burrow]

Why no langle and rangle 😭

[void cosmos]

when u deal with free groups u wanna use the universal property. the universal property is basically saying that this is the object where defining a map on the generators is enough. sort of like when ue xtend by linearty for vector spaces.

if u wanna define a function on a vector space, it is enough to define it on the basis

its basically the same thing with free groups. and so since u know where a and b go u should be able to know how <a,b> goes

but ur guess is right

[green fox]

How would you show that $<aba^{-1}b^{-1}>n = {a^{x_1}b^{y_1}\ldots a^{x_n}b^{y_n} : \sum{i=1}^n x_i = 0 = \sum_{i}^n y_i, n \in \mathbb{N}}$

[void cosmos]

i would simplify things first by looking at just a and b and how f acts on them

[cloud walrus]

Luke

[void cosmos]

and seeing what sort of combination gives me (0,0)

u should see that the combination is exactly the subgroup generated by a and b under the relation ab=ba

[elfin wraith]

Just as an FYI, for the group generated by a you should write \langle a\rangle rather than <a>

[green fox]

but isn't this just $\langle a^n b^m : (n,m) \in \mathbb{Z} \times \mathbb{Z}\rangle$

[cloud walrus]

Luke

[green fox]

How can you get $a^2b^2$ just from multiplies and inverse multiples of $aba^{-1}b^{-1}$

[cloud walrus]

Luke

[void cosmos]

no

again, look at how f acts

and see how can u get (0,0) , the identity in Z x Z

a--> (1,0) and b --> (0,1)

firstly where does the word aba^-1b^-1 go

[green fox]

into kernal

[void cosmos]

and if so where does (aba^-1b^-1)^n go for any n

so now u have one inclusion

[green fox]

oh, 😅

[void cosmos]

i meant under f. but yes u mean that aba^-1b^-1 goes to (0,0)

a cool way to look at it is look at the figure eight

and try to trace all loops that u can form around a basepoint in the middle

or nvm about that ig

[green fox]

How to show $ker(f) \subseteq \langle aba^{-1}b^{-1} \rangle_n$

[cloud walrus]

Luke

[vapid loom]

Reading up on lattices, where does the term "covolume" come from?

ah kk, reasonable enough

Is volume of a quotient group (or any group for that matter), a common thing?

[hot goblet]

Can someone help me think about what happens to the identity element (of a ring) in a direct product of rings?

eg in Z clearly 1 is the multiplicative identity, what happens in Z*Z, is it (1, 1)?

and how does this relate to polynomial rings, R[X]/<X^2+1> isomorphic to C isomorphic to R*R but surely it should be that 1 in R[X]/<X^2+1> gets mapped (by the isomorphism map) to (1,0) in RxR or C?

[muted fog]

$\mathbb{C}$ is not isomorphic to $\mathbb{R}\times\mathbb{R}$ as a ring. But they are isomorphic as real vector space. In fact $\mathbb{R}\times\mathbb{R}$ is not even a field. For example it has zero divisors e.g $(0,1)\cdot(1,0) = (0,0)$

[cloud walrus]

Sascha

[hot goblet]

true, I am getting my complex analysis course and my algebra course confused

I also think im getting additive and multiplicative identity properties mixed around in my head

hmm

[thorn jay]

Noooooooo i thought for a second it was an order theoretic lattice and got happy qwq

[vapid loom]

what are order theoretic lattices? I've seen the definition but have no idea what they're used for

[thorn jay]

Theyre partial orders where every pair of elements has a lowest upper bound and a greatest lower bound

These are in correspondence with particular algebraic structures

You also have complete lattices, and they are partial orders where every set of elements has a LUB and GLB

Those pop up everywhere, anytime you have a Galois connection / closure operator the closed sets form a complete lattice

Complete lattices are to Galois connections / closure operators what frames are to topological spaces

In particular, you could see them as one of the languages that universal algebra is written in, but they pop up in other places too

[woeful sage]

well it's not a field with that multiplication structure but say... you defined multiplication such that (0, 1) \cdot (1, 0) = (0, 1) then it'd be a field

well nevermind, you guys were talking about direct product of rings, so yea you'd have component-wise operation

[dim wagon]

im not sure whether this is the right place to ask but one project im working on involve taking modulo arithmetic on algebraic integers so for example 3+3alpha is considered to congruent to 0 mod 3, anyone what is the name for this kind of thing?

[rocky cloak]

I mean it's just a quotient ring. A/(3) where A is the ring of algebraic integers

[dim wagon]

oh okay thank you

[languid trellis]

The reverse direction of this exercise is clear, if it has a root in F, then we can divide by the linear factor, so x^p-a is reducible. For the direction irreducible implies no roots, I'm quite stuck. Can I please get a pointer of where to look?

Actually one observation is that x^p-a is seperable, because it's derivative is px^p-1, and p and char F are coprime

if a isn't 0

[tardy hedge]

I really want to solve that question lol. It is kind of weird you're right. It seems so doable tho i just gotta write stuff down

[languid trellis]

I've written a lot of stuff down and have kinda gotten nowhere lol

[tardy hedge]

Wow dang

Ok so thats saying x^p-a reducible equivalent to x^p-a has a root in F right

[languid trellis]

I think so

Which is reconciling the fact that this equivalence doesn't hold if char F =p

[tardy hedge]

can we do something with field degrees

multiplicative thing

[glad osprey]

Aren't you supposed to prove that it's either irreducible or has a root? By proving irreducible implies no roots you're proving that it cannot both be irreducible and have roots. I think you mixed up P v Q being equivalent to not P => Q, as opposed to P => not Q

[languid trellis]

Well I think looking at field extensions is probably a good idea, because this chapter introduces the normal closure and stuff

[next obsidian]

You haven’t interpreted the problem correctly

[languid trellis]

Hm, maybe

[next obsidian]

In mathematics or means inclusive or

You need to show not irreducible => has a root, or that no root => irreducible

[languid trellis]

Okay I think it's time for a rethink on my part

[spare isle]

I always see the condition $gNg^{-1} \subseteq N$ and $gN = Ng$ for normality, but I never see the condition $gN \subseteq Ng$; however, the latter is still a valid condition right? Since it immediately follows from it that $gNg^{-1} \subseteq N$.

[cloud walrus]

JJCUBER

[rocky cloak]

It is in fact also true in characteristic p. But you would need a different proof

I'd actually argue it's easier if char(F) = p

[languid trellis]

Well I've made some progress, suppose x^p-a is reducible. Then x^p-a = g(x)h(x), for some g,f in F[x]. Let E/F be a splitting field, b a root of x^p-a in that splitting field. Then g(b)h(b) = 0 in E/F, and because x^p-a is seperable, we know that either g(b) =0 xor h(b) = 0. So, suppose g(b) = 0. g = x^k + a_{k-1}x^k-a + ... + a_1x + a_0.

Now from here, i feel that I should use the fact that any two splitting fields are isomorphic to say that E \cong F(b*zeta_p)

But I need to end up with b in F somehow

[rocky cloak]

I think it might be smart to think about which pth roots of unity E contains

[languid trellis]

I'd imagine it has all of them

If F doesn't contain a primitive p'th root of unity, then [E:F] = [E:F(zeta_p)][F(zeta_p) : F] \geq p-1? If [E:F] =p-1, then divide x^p-a = (min poly)(x-b), so x^p-a is reducible and has a root?

[rocky cloak]

So if it has all pth roots of unity, how many roots of x^p - a are in E?

[languid trellis]

Surely p

[rocky cloak]

So then E is the splitting field of x^p - a.

[languid trellis]

I agree, but I said it was already

[rocky cloak]

Oh you did? I'm sorry

[rocky cloak]

The fact that F(zeta_p)/F has degree p-1 is true in characteristic 0, but not in general.

For example F2(zeta_7)/F2 has degree 3, not 6.
x^7 - 1 = (x+1)(x^3 + x + 1)(x^3 + x^2 + 1) over F2

[languid trellis]

I didn't even think about this

Hmmm

I am now even more helplessly stuck

[rocky cloak]

I mean reducing it to easier cases is always a good strat.

Now you've covered the case of F characteristic 0 without containing pth roots of unity. Then the next reasonable case might be if F contains all pth roots of unity

[thorn jay]

Bro is assuming the axiom of excluded middle

[delicate orchid]

it's because there aren't any lol. I just realised that we're talking about 2-groups of 2-rank 1, which were classified by Gorestien in 1967, and these are the only ones out of that classification with a unique involution

[tardy hedge]

woohoo

[next obsidian]

Indeed

[glad osprey]

@languid trellis I found these couple of exercises in Galois Theory through Exercises. 5.12a is what you're working on now, but I think 5.11 is also relevant. There are hints and solutions in the back, let me know if I should send it (I'm working through these myself now btw)

[languid trellis]

I'm completely exhausted after thinking today, I'll take a look tomorrow, thank you 🙂

[tardy hedge]

thats funny u said that^

i was just about to say "jeez doing math feels like a workout sometimes"

[glad osprey]

math is like working out, except there's no limit to how strong you can get 💪

[tardy hedge]

woahh cool

thats true

i dunno abt u guys but its like the final grind for me

exams comin up and everythin

i be grinding

[glad osprey]

everybody wants to be a mathematician, but nobody wants to lift this heavy ass weight

[tardy hedge]

Ong im exhausted

[elfin wraith]

Coleman’s theorem, a classic

[tardy hedge]

math people also liking working out

is this true or like

[thorn jay]

I dance

But not necessarily work out

[tardy hedge]

I love da gym

I am waiting Magma man

[keen badge]

I am trying to find a group $G$, such that $C(G):={[x,y]=x y x^{-1}y^{-1}: x,y \in G}$ is not a subgroup of $G$.(By definition, the derived subgroup is generated by $C$, and not $C$ itself.
The group $F_2$(free group with two generators) may help here, but I am unsure how to compute $C(F_2)$.
(if y'all have any other ideas instead of $F_2$ I don't mind.

[tardy hedge]

Group theory 😢

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[thorn jay]

.>

[tardy hedge]

I barely know it

Teach me group theory boss

[thorn jay]

[x, y] • [x, y] = xyx^-1y^-1xyx^-1y^-1
This looks like it wont evaluate to a commutator

Maybe you can make a reduction argument hm

[keen badge]

If I could prove things with looks like I would get a million bucks for solving the Riemann Hypothesis LOL

[thorn jay]

Im just saying with free groups if it looks like it wont work then 99.9% of the time it wont

But perhaps pick generators of S_n and start computing

[arctic trail]

isn't it easier to pick all distinct elements

[x,y][z,w]

[rocky cloak]

Here's the smallest finite group with said property
https://groupprops.subwiki.org/wiki/SmallGroup(96,3)

[arctic trail]

oh wow, 5 generators

[thorn jay]

Oh yeah probably hahehs

[tardy hedge]

What the baloney is this

Ong^

[rocky cloak]

If you want simpler computation you could instead look at the free product of C2 with itself.

Then every element just looks like some alternating word of x and y, so you can very explicitly compute the commutators.

(I haven't verified that this example works, but it's at least easier to check then F2)
Edit: learning towards thinking this example might not work...

[glad osprey]

Probably with a computer, 96 is too big a number for humans to wrap their heads around

Here is someone who did it with GAP: https://math.stackexchange.com/a/7813

[thorn jay]

x = x^-1

So
[x, y] = xyxy

Wait no

It should

[x, y] • [w, v] = xyxywvwv
Hows that gonna be a commutator

[rocky cloak]

Yeah if you go to 4 generators there might be more hope

But then there's also more cases to check

[thorn jay]

cries

[rocky cloak]

[x (yx)^n, (yx)^n] = (xy)^2n

So everything should be an actual commutator :/

[thorn jay]

Yeah

Qwq

[tardy hedge]

bruh im falling apart

ive been working for like 5 hours straigt

[thorn jay]

pat pat

make it 6

[rocky cloak]

all work and no play makes jack a dull boy

[glad osprey]

kiand after working for 12 hours straight

[thorn jay]

ohh that's where I heard it

[tardy hedge]

Btw my name is kian

[hidden wind]

tfw simply checking all finite groups in GAP until you find an example

[elfin wraith]

Unit conjecture moment

Well I guess that was more, it’s probably this group, get GAP to randomly pick elements

[thorn jay]

d123

[spare isle]

I always see the condition $gNg^{-1} \subseteq N$ and $gN = Ng$ for normality, but I never see the condition $gN \subseteq Ng$; however, the latter is still a valid condition right? (Since it immediately follows from it that $gNg^{-1} \subseteq N$.)

[cloud walrus]

JJCUBER

[thorn jay]

Sure, but gNg^-1 \subseteq N is often a little easier to prove

[elfin wraith]

My intro to group theory class defininetly mentioned the last codition, but im not sure ive ever run into a situation where that condition was the easier one to use

[spare isle]

Yeah it doesn't seem the helpful since equality is a much stronger statement (though I guess one could argue something similar with the fact that $gNg^{-1} = N$ is also equivalent to $gNg^{-1} \subseteq N$).

[cloud walrus]

JJCUBER

[spare isle]

I think the difference is how $gNg^{-1} \subseteq N$ is typically used to show normality (which makes working with subsets preferable), while $gN = Ng$ is often used as a starting point/stepping stone in proving something (which makes working with equality preferable).

[cloud walrus]

JJCUBER

[tropic obsidian]

bit confused here

Z isnt local, but is looks like a valuation ring to me

edit: lol nevermind

[elfin wraith]

It took me like 5 minutes to work out why every ring isn’t a valuation ring there, Christ

[tardy hedge]

Ive never seen valuation rings before. Thats kind of interesting

[tardy hedge]

x is in K?

[tardy hedge]

Its not, right?

If im understanding the defn of valution ring correctly

[elfin wraith]

Yeah, they really should explicitly say that but it must be since it talks about x^-1

I also hadn’t seen the definition before, it took me a while to realise you get stuff like 2/5 but neither 2/5 or 5/2 are in Z but are perfectly good elements of Q

[tardy hedge]

Yeah

Kind of weird when Z is a nonexample

For ring stuff

[elfin wraith]

Which just like yeah is obvious now but took me so long to work out, don’t do ring theory at 2 am

[elfin wraith]

Yeah I was trying to think of more esoteric non examples because obviously Z must be an example

[tardy hedge]

Hahaha yeah

[grizzled crow]

Hey, can anyone tell me what this symbol is?

the thing before the (x)

[sharp ice]

$\mathcal{O}$?

[cloud walrus]

conan

[grizzled crow]

mathcal eh

do people also write this as 0(x)?

I guess mathcal is to emphasize it's a function and not a scalar?

[kind temple]

these are script fonts. i don't know which tho...
the F is close to \mathscr{F} using the mathrsfs package

the O in that same font is not tho...

[velvet hull]

Well, finite fields of the same size are isomorphic

[trail cave]

My professor said my proof of r1(x) + <f(x)> =/= r2(x) + <f(x)> here was wrong; I originally did it by assuming they were equal then showing r1(x) = r2(x). He said instead assume r1(x) =/= r2(x) and show when the generator is added to both sides they're not equal. How do I do that?

[boreal inlet]

Let q be a prime and let a and b be two primitive (q-1)th roots of unity.

When would be the sum

1 + (ab) + (ab)^2 + .... + (ab)^(q - 1) be zero?

This question came across to me as I was trying to find when two induced representations of SL(2, q) from one dimensional characters of F_q* are non-equivalent, and all of it reduced to this question.

Wait nvm

We don't need this condition

Anyways

My bad

[wraith cargo]

The sum is 0 when they're q-th roots of unity

Not (q-1)-st

[boreal inlet]

Okay. I was taking elements from the multiplicative group of the field F_q, hence it's isomorphic to Z/(q-1)Z

So I precisely needed (q-1)th roots

[cloud walrus]

Riku

[boreal inlet]

I was trying to find when this sum is zero, and mistakenly assumed w(1) is always a primitive root, but that need not be the case.

In any case, w(x)w'(x) also defines another representation, and then the sum just becomes something like this:

[cloud walrus]

Riku

[boreal inlet]

So as long as that product isn't a trivial representation, we're fine

[thorn jay]

Universal algebra rots the brain

[arctic trail]

is there any notation for like

well I'll try to explain what I have

I have a space X whose elements are things like
Aa + ABA^(-1)Bb + ... - A^(-1)a.

And the point is that this is kind of a free construction, where X is mapped to a vector space, a,b are mapped to vectors and A,B correspond to invertible operators on that vector space

I have no idea how to denote this space X

note the coefficients are always integers, but I want to map to vector spaces

Maybe Z[F_2] ox_Z Z^{a,b}?

[delicate orchid]

Z<a,b, A, B, A^(-1), B^(-1)>/(aA, bA, aB, bB, ... , AA^(-1)-1, BB^(-1)-1) might be what you want

free non-commutative algebra on those symbols but we just delete any composition that doesn't make sense

[arctic trail]

I don't really need an algebra

[delicate orchid]

well you have one

[arctic trail]

what do you mean?

[delicate orchid]

you have the sum of products of things? That's a ring - which are Z-algebras

[arctic trail]

no, it's clearly a module

the coefficients are elements in F_2 with an integer

[delicate orchid]

_ _

elements in F_2 with an integer what?

I regret saying anything

[arctic trail]

wdym?

like take the free group on A,B call it F.
Is what I want not exactly Z[F] tensored_Z Z^{a,b}?

it gives coefficients in Z[F] to a and b

[delicate orchid]

why are A and B commutative

[arctic trail]

they're not

[delicate orchid]

I'll presume you meant Z<F> then

[arctic trail]

Z[F] is the group ring

if I take elements u,v in F. Their product in Z[F]is uv and is still different to vu

this is very common in representation theory

[delicate orchid]

ok don't speak down to me

[arctic trail]

???

I don't understand why you're tripping up

I wasn't speaking down to you

[delicate orchid]

lets think about linear maps out of this tensor product, Hom(Z[F] (x) Z^{a,b}, M) \cong Hom(Z[F], Hom(Z^{a,b}, M)), so any linear map is indeed associating M-valued operators on Z^{a,b} to elements of Z[F]

you said you wanted a free construction which is why I gave you my original answer, which if you examine - does indeed contain the group ring Z[F]

[delicate orchid]

I also read this as the field with two elements, because why wouldn't I?

but we've got it all sorted now at least

[arctic trail]

no, I really just said that it was kinda free

as in resembled a free construction in some way

anyway thanks

[delicate orchid]

I wonder if my first thingy worked

[arctic trail]

the thing is that your space has elements like A

so it really has too much structure

[delicate orchid]

https://tenor.com/view/morgan-freeman-true-morgan-freeman-true-nodding-gif-13973817878387504960

a fair assessment

[rocky cloak]

I'm not sure I understand exactly what you're asking here. You just want the free Z[F2]-module on two generators?

[arctic trail]

yeah that was it essentially. I wasn't quite sure what I wanted there

[rocky cloak]

Well there is notation for that, i.e. Z[F2]^2

[arctic trail]

Touché

[rocky cloak]

It will also be ZF2Z^2 sure

[arctic trail]

yeah I will put Z[F2]^A

so it's more clear

thanks

[rocky cloak]

I'm curious what context you got this from though

[arctic trail]

some density stuff in free groups

[tardy hedge]

Omfg love it omgggg

[jovial snow]

Hi everyone! I'm currently in high school and I'm trying to get into higher math. I've found the topics I most enjoy are things in the calculus route (multi, odes, pdes, tensor calculus), but the farther I go in these topics, the more I realize the importance of algebra and analysis, I'll probably ask a similar question in the analysis section after this, but I'd figure I'd start here. I currently have books on abstract algebra, galois theory, lie algebra, and clifford algebras, but I'm far from understanding it all. A big part of that is that I'm struggling to get into the subject as a whole. With other topics I've studied, it's easy to find applications, but most of what I've done so far just feels like random definitions and proofs. My question is this: what are the applications of advanced algebra, set theory, group theory, etc. The only actual application I've heard of so far is the unsolvability of the guintic, which does sonlund really interesting, but what else is there? I don't need "super duper applied" reasons, just interesting truths that algebra covers. Anyways, I know this is super long, but thanks for reading and hopefully responding!

[velvet hull]

You’re going to hard pressed to find actual direct applications of abstract algebra and real analysis that are easy to state, because they’re the foundation of all higher level math, and the ideas and techniques you see in them are going to serve as the blueprint for everything that follows

In a sense, it’s kind of like asking “what’s the applications of addition” - I don’t know how to answer that question, it’s everywhere

[jovial snow]

@velvet hull I get that, but even in an abstract sense, what's the point? Where do these things show up in higher math? Are these ideas ever used outside of their own bubble? I see things like linear algebra used a lot in differential equations, but just in general, why should I learn this?

[chilly ocean]

Real analysis techniques show up outside of it all the time

[velvet hull]

It’s a good question, and the answer is if you’re not interested in higher level math (and also the sciences and engineering to a lesser extent) then they’re probably not going to be useful or needed at all

[chilly ocean]

Point-set topology for instance

and Topology shows up in a lot of other places

[jovial snow]

@velvet hull I'm definitely interested in higher math, I just don't see how they would show up there

[languid trellis]

One thing I can think of is that a lot of elementary number theory can be rewritten in terms of abstract algebra, such as euler's theorem, wilson's theorem, divisibility being able to be extended to PID's in general are just a few off the top of my head

[velvet hull]

It’s less so the theorems and more so the philosophy, if that makes sense

[jovial snow]

@velvet hull could you eleborate?

[velvet hull]

Abstract algebra teaches you how to regconize and formalise symmetry, and real analysis teaches you how to give technical arguments in a rigorous and concrete way

[velvet hull]

Also this, when you learn abstract algebra it comes with an crashcourse in elementary number theory for free

[delicate orchid]

I disagree that algebra doesn't require technical arguments in a rigorous and concrete fashion

[velvet hull]

Of course I’m not implying that

[delicate orchid]

however, yes

[velvet hull]

I’m just saying that analysis definitely leans more towards that

[delicate orchid]

sure

I agree with you there

especially at the under grad level

[jovial snow]

@velvet hull I keep hearing about symmetry in things, but why do we care if things are symmetric or not?

[delicate orchid]

at the most general level, symmetry in a problem allows us to reduce it

if you integrate an odd function on an interval centred at 0, you automatically know that it is 0 by exploiting a symmetry - for instance

[velvet hull]

Almost any meaningful argument or statement you can make in mathematics uses some kind of symmetry, even if you are aware of it or not

In a very precise sense, you can even say mathematics is the study of symmetry

[languid trellis]

yappa yappa yoneda lemma

[velvet hull]

Yoneda lemmaaaaaaaa (wait why are we talking about yoneda)

[delicate orchid]

in the other channel, jagr mentioned Noether's theorem - which I was going to bring up

that's a very real direct application of symmetries to physics

[languid trellis]

Reciprocity laws are another example

[wooden rain]

When this statement pops up, you know you're talking to an algebraist

[languid trellis]

perhaps

[delicate orchid]

do u mean like... adjoint functors or what

we talkin QUADRATIC?

[velvet hull]

:caught:

[languid trellis]

yes...

[delicate orchid]

Shapiro's lemma...

[thorn jay]

I'd argue that abstract algebra is the language that a lot of statements in higher level math are written in

Besides literally being written in set theory

[languid trellis]

I think I agree. I think this is quite clear from the historical development of algebra too. Kronecker had already investigated "ideal numbers" before we had any algebraic definitions (I think noether was first with the definition of a module at some point in the late 1800s)

[delicate orchid]

written in commutative diagrams more like

[velvet hull]

Still irks me that ideals arent called normal subrings

[thorn jay]

Thats just a partial monoid

smh

[delicate orchid]

yes categories are mononoids but you can express pretty much any algebraic structure as "internal" in some category

[thorn jay]

Its not a subring :(

[delicate orchid]

they ARE subrings if you stop being woke (requiring unity)

[thorn jay]

Just any, plain and simple, Lawvere's algebraic theories in Set

[delicate orchid]

I'm actually so anti-untiy pilled now it's crazy

[languid trellis]

i think we had klein & co. do the erlangen program before a group was defined in full generality

before group = automorphism group

[velvet hull]

Sorry normal subrngs, fixed :))))

[delicate orchid]

not entirely convinced this is related. I'm just talking about the ability to make group/algebra/module objects in a category

[languid trellis]

die die die die die

[delicate orchid]

it could be though

[thorn jay]

Its the same

A group object C is an instance of the lawvere theory of groups in C

[delicate orchid]

that would be good to know if I knew more than zero about categorical logic

[thorn jay]

Or algebraic theory, im forget the difference

[delicate orchid]

point is u can put them in diagrams

[thorn jay]

But
But
What if i dont want to :(

[delicate orchid]

therefore. All of algebra is about gluing spheres together

[velvet hull]

Back to the psych ward you go

[delicate orchid]

or squares. Or perhaps even triangles

[languid trellis]

he never left

[thorn jay]

Holy homotopy?

[delicate orchid]

or whatever the fuck a "pasting diagram" is

[thorn jay]

Ong wanna get into that sometime

Would complement universal algebra nicely

[twilit creek]

hey, can someone explain me the link between primitivity and irreduciblity in this context

[rocky cloak]

You can consider Z[X] -> Q[X] and the polynomial 2x. This is irreducible in Q[X], but not in Z[X] where it factors as 2*X. The problem being that its not primitive. You can factor out the common factor 2, which because it becomes a unit in Q can be ignored

[twilit creek]

I understand but I cannot understand how to make a proof of the statement using the fact that its primitive

[rocky cloak]

Well assume f = gh and show that one of g or h is a unit

start by looking at sigma(f) = sigma(g) sigma(h)

[twilit creek]

But I am not sure its rigourous

[tardy hedge]

if alpha is separable over F, then its also separable over any extension of F right?

[languid trellis]

alpha seperable means that the minimum poly factors into linear stuff in F[x], so if we look at the minimum poly of E[x], E an extension field of F, then can we not just use the same factorisation as in F[x]? Lmk if someonething seems wrong about this

I feel that i'm tripping balls

[tardy hedge]

oh yeah youre right, alpha separable means the min poly factors into distinct linears in the splitting field

[languid trellis]

yh i fucked up slightly but yeah

[tardy hedge]

I think also we need to mention the minimum polynomial over any extension of F divides the minimum polynomial over F, so if the min poly of F has no distinct factors then you cant have something dividing it with distinct factors

[languid trellis]

something like that

[tardy hedge]

wow

[elfin wraith]

Simplical complexes are a lie by big combinatorics to trick you into thinking combinatorics isn’t just painful

[delicate orchid]

the trick is you're not actually doing combinatorics. You're just gluing triangles

and then telling the computer to do smith normal forms

[elfin wraith]

This is true, you just make computer science and software people do the painful stuff for you

Every day I’m thankful Perseus already exists 🙏

I actually went to a talk about persistent homology today and the speaker said they wrote their own software in Haskell for fun

Some people should be observed

[trail cave]

How do I go about this problem?
R is Z adjoint sqrt(2) with norm |a^2 - 2b^2|. For pi = a+bsqrt(2), show that a rational prime p either factors as p = u(pi)(pi_conjugate) where u is 1 or -1 and pi, pi_conjugate are prime in R, or p remains prime in R

[rocky cloak]

Do you know if R is a UFD?

If so since the norm of p is p^2 anything dividing p will have norm ±p^2 or ±p, which you can show corresponds to your two cases

[trail cave]

I showed that R is a Euclidean Ring, and I think it's an integral domain since a, b are integers so it should be right?

ty

[steep cliff]

i’ve been doing some group theory questions and i had to use the answer sheet for most of this.. is anybody able to talk me through the solution? it’s probably way less complicated than i’m worrying but cycles and equivalence classes aren’t my strong suit haha

[trail cave]

which points are you most confused on? Roughly these proofs go like: we want to show G/R = G' for some groups G, G', R (for you, G is Zm, G' is Zn, R is the cycle). first define a function from G to G'. Check that it's well defined (I think you have a small error here, it should say that [z]n = [w]n, not [z]n = [w]m), check that it's surjective, check that it's a homomorphism. Then you have a surjective homomorphism and may apply the FTH to find G/Kerf = G' (f here is theta). But Kerf = R (by showing both inclusions), so G/R = G'.
So overall you're just defining a function that's a surjective homomorphism, showing that its kernel is the cycle, and applying the fundamental theorem

[steep cliff]

Honestly, i’m most confused by all the strange notation. i’m not the best at understanding things in lectures, and most of the content i’ve found on groups isn’t the most helpful.
i’m showing that the integers mod m / (etc etc) is isomorphic to integers mod n, but i’m not entirely sure what that actually means.
i absolutely love abstract algebra, and i know i’m supposed to be finding some preservation in structure here, but i struggle to believe what i’m calculating if i don’t see it from the very start if that makes sense?

thanks for spotting my error too!!

[trail cave]

I'm an undergrad so maybe with more experience that kind of intuition comes (and maybe someone else can help more), but typically when I'm told to prove an isomorphism from the isomorphism theorems I don't really have an intuition for it either

[tardy hedge]

I think the first isomorphism theorem you should understand how / why it works

the other ones kind of just reduce to appying the first iso theorem

[trail cave]

I roughly see how it works but is there an intuition you grow or a way to see that a quotient group and another group are isomorphic?

[steep cliff]

i understand that it works but i find geometric interpretations dumbed down almost into real world examples the most useful. for example, to remember some theorems for vector calculus i memorised diagrams of bubble mixture on a closed wire frame. what a funny brain i have lmao

[tardy hedge]

Hm I'm not an expert either but it seems like in practice when you're dealing with writing homomorphisms between structures there is usually a "natural" map to consider and that one usually turns out to work

[steep cliff]

so the most obvious map is usually the right one?

[tardy hedge]

If you have a quotient group H/K and another group G and you want to see if theyre isomorphic, that would be the same as writing a surjective homomorphism H->G with kernel K

[tardy hedge]

Yeah

[steep cliff]

wait this is so helpful thanks! and is the kernel is just the set of elements in H that are mapped to the identity of G?

[tardy hedge]

Yeah, and you should understand the connection between the kernel of a map and if the map is injective

[steep cliff]

what’s the connection?

[tardy hedge]

A homomorphism is injective if and only if ker(f) = 0

i.e f only sends the identity to the identity

[steep cliff]

is that literally 0 or are we just using 0 as a placeholder for whatever the identity could be? like an n x n identity matrix for example

[tardy hedge]

yes the latter

when you see symbols like 0 in abstract algebra it always means the abstract "0 element"

ker(f) = {identity element}

[steep cliff]

ah okay we use the letter e in lectures

[tardy hedge]

ok ker(f) = {e}

[trail cave]

How do I show the second part here? I think it's probably easy but I'm blanking

[hidden wind]

fundamental homomorphism theorem my beloved

[tardy hedge]

Im not actually sure how to conclude that the separability degree is multiplicative

I was trying to make some field tower but i got confusee

Confused

[tardy hedge]

I dont think it should be hard no?

[tardy hedge]

Poop

[thorn jay]

It was hard..

Wasnt it

[trim wind]

it shouldn't be too hard if you use the definition using automorphisms

[steady field]

I think I found a way to represent connect 4 positions as a semiring, is there anything useful I can do with this?

like are there any lemmas/theorems for semirings that I might be able to apply.

[tardy hedge]

What is that

[tardy hedge]

I didnt do it yet lol but tbf i didnt sit down and really work it through much

Just so tired

Bro said Ok

[trim wind]

Embeddings that fix the ground field

Or what do you mean?

[tardy hedge]

For my question about separability degrees?

I never seen a definition with automorphisms

Im just using min poly is separable for each a

Thats what i hve to use

[trim wind]

Whats your definition of the separable degree?

I guess as the degree of the separable closure then?

[tardy hedge]

Yea

[trim wind]

And you never saw any automorphism stuff?

[tardy hedge]

about separability? no

We just defined E/F is separable if every a in E is separable over F

[trim wind]

That's not what I mean. If you know how automorphisms of separable field extensions behave, then it is easy to see that the degree of the separable closure of k in E is equal to the number of automorphisms E->K which fixes k, where K is an algebraically closed field

And then it's easy

[tardy hedge]

Oh, ok I will study that then

It talks about that sort of thing in the next question in my hw actually

which is maybe a bit weird?

[tropic obsidian]

is there a local ring whose maximal ideal is principal, but the ring is not notherian? edit: not sure how much harder finding one that is also a domain would be

[tardy hedge]

that made me think of what kind of ideals can be subset of principal ideals

[tropic obsidian]

if your domain isnt gcd then the intersection of principal ideals may not be principal

[tardy hedge]

Gcd?

[tropic obsidian]

from wikipedia page of GCD domains

[tough raven]

Yes, you can construct valuation rings with principal maximal ideal that are not DVRs.

Explicitly

Let k be a field. Let K be the field of series ∑_{m, n ∈ ℤ} a_{m,n} X^m Y^n with coefficients in k such that (i) there is a lower bound on the n such that a_{m,n} ≠ 0 for some m (ii) for each n, there is a lower bound on the m such that a_{m,n} ≠ 0.

Within this, let R be the subring of series with no negative powers of Y and no negative powers of X with Y^0, i.e., R = {∑_{n ≥ 0} ∑_{m ≥ M_n} a_{m,n} X^m Y^n : M_n ∈ ℤ for each n ≥ 0 with M_0 = 0, all a_{m,n}'s in k}.

The maximal ideal m of R is those series for which a_{0, 0} = 0.

The intersection of all powers of m is the ideal of those series with no X^m Y^0 terms, i.e., the ideal I = (Y, X^{-1} Y, X^{-2} Y, ...). (This would contradict the Krull intersection theorem if R is Noetherian. You can also verify by hand that I is not finitely generated.)

[tough raven]

Another way of describing this condition on the series which may be easier to understand: order ℤ ⨯ ℤ lexicographically (comparing the second factor first); then K consists of those series for which {(m,n) : a_{m,n} ≠ 0} ⊆ ℤ ⨯ ℤ is well-ordered.

(This generalises to what are called Hahn series.)

[tropic obsidian]

isn't this generated by (x, y) though?

[tough raven]

No, just X.

X^{-1} Y ∈ R, so X divides Y anyway.

[tropic obsidian]

oh I see, the ring is just the polynoimals with a lowest coefficient

not nec. possitive coefficients

or maybe Im confused

I thought R is the subring of the series with no negative powers

so how would mutiplying be X^{-1} be allowed?

[tough raven]

It's in the definition of R. 1 + X + X^2 + ... + X^{-3}Y + X^{-15} Y^2 + X^{-1} Y^3 + ... ∈ R for example.

[tropic obsidian]

ah I see, I didnt fully read your definition of R, makes sense now

so the valuation here put on K to get R is the minimal power of the x^i's and y^i's terms?

[tough raven]

v(X^i Y^j) = (i, j) ∈ ℤ ⨯ ℤ with the lexicographical ordering I mentioned before.

(You can't use a valuation valued in ℝ for this one.)

[tough raven]

And v of a sum is the minimum (by the lexicographical order) of the v's of the monomials with non-zero coefficients.

[tropic obsidian]

oh, I just realized you can have valuations that arent into Z

[tropic obsidian]

is there a name for this field?

[tough raven]

There must be, but I don't know it.

Hahn series are a generalisation (replacing ℤ ⨯ ℤ with any totally ordered abelian group) but I don't think looking that up would turn up much about this specific case, which ought to exist.

[green fox]

How would you show that $<aba^{-1}b^{-1}>n = {a^{x_1}b^{y_1}\ldots a^{x_n}b^{yn} : \sum{i=1}^n x_i = 0 = \sum_{i}^n y_i, n \in \mathbb{N}}$

[cloud walrus]

Luke

[green fox]

$<aba^{-1}b^{-1}>_n$ is the smallest normal subgroup containing $aba^{-1}b^{-1}$

[cloud walrus]

Luke

[hollow imp]

is this in the free group with two generators or something?

[green fox]

yes

[crystal vale]

First implication, Since n = p^k for some k in N, so |H| = p^r and |K| = p^s for some r and s in N. If r = s then H = K ( because in the cyclic group there is a unique subgroup of each order).

Without loss of generality, we can take r < s, but K is a cyclic subgroup of order p^s, so it has subgroup T in K which has order p^r, but then T also subgroup in G so T = H (same argument, uniqueness).

Second implication comes from Sylow, or Cauchy theorem

Right?

[lethal hornet]

first is slicker if you say r <= s or s <= r, in which case H <= K or K <= H respectively. I think the converse is easier framed as considering a cyclic group C with two distinct prime divisors of its order, then apply Cauchy's

[lethal hornet]

this doesn't answer your question, but you can use $\langle$ and $\rangle$ for your angle brackets

[cloud walrus]

Flip

[lethal hornet]

induction maybe? on the number of "aba^-1b^-1"s and its inverse you have in your word?

[crystal vale]

yes

[tough raven]

For a general finite group, yes. For a cyclic group, you can also be 100% explicit, specifically using that the lattice of subgroups of C_n is the same as the poset of factors of n.

[tough raven]

Which inclusion are you stuck on?

[green fox]

Taking an element of the the right set and showing it is in left

[tough raven]

So the left consists of products of conjugates of aba^{-1}b^{-1}. Such a conjugate can be used to "turn" a ba into an ab, and its inverse does the opposite. Assume for simplicity that x1, y1 > 0 and try to left-multiply the expression on the right by the inverse of f (aba^{-1}b^{-1}) for a suitably chosen f, such that the first occurrence of ab is replaced by ba, i.e., to get a^{x1-1}bab^{y1-1}....

[crystal vale]

Yes

[crystal vale]

your idea is better because yeah it looks like poset of factors of n

[merry summit]

what books would you recommend to an absolute beginner in abstract algebra?

[merry summit]

is joseph gallian good?

[crystal vale]

yes it is good

[amber burrow]

How would you do this? The best I could think of was for the prime index ones, the subgroup must be normal, so kernel of a homomorphism, so it suffices to consider all possible homomorphic images, and I can find the group presentation online.

By the way this is from the USA-primes problem set, but since they have already sent out admission decisions I think it is safe to ask

[languid trellis]

I had to look at the solution eventually but I made so many right steps haha, including identifying the splitting field explicitly as K(epsilon, a). This is a really good resource though, and I'm really glad you've introduced it. Thanks 🙏

galois theory is hard asf

lord end me

[glad osprey]

Awesome I love that book

[rocky cloak]

It's not correct that prime index implies normal in general.

Index 2 implies normal, but other than that any index could appear for a non-normal group.

As for how to approach this, and index m subgroup will correspond to a transitive group action on an m-element set.

So it could be worth examining homomorphisms to Sn.

If you can find all of them you can just go over and see which the subgroups you want. But it might be a lot of work as m grows bigger...

[crystal vale]

Given $G = GL(2, \mathbb{Q})$ and $H$ be its subgroup of diagonal matrices i.e, $H = { A \in G | \text{A is a diagonal matrix} }$.
\vspace{0.5cm}
\Now we have to determine $H_G = \bigcap_{g\in G} gHg^{-1}$
\vspace{0.5cm}

Take $K$ be the set of all scalar matrices in $G$, then $K\subset H$ and since every element of $K$ is scalar matrix so $K\subset gHg^{-1}$, for all $g\in G$ implies $K\subset H_G$
\vspace{0.5cm}

Also, $H_G \subset H$ so element in $H_G$ is form of $\begin{pmatrix} a & 0 \ 0 & b \end{pmatrix}$, where $a, b\in \mathbb{Q^{\times}}$
\vspace{0.5cm}

Claim: $H_G = K$ i.e, $ H_G = \begin{pmatrix} a & 0 \ 0 & a \end{pmatrix}$, where $a\in \mathbb{Q^{\times}}$
\vspace{0.5cm}

\Let there exist $A = \begin{pmatrix} a & 0 \ 0 & b \end{pmatrix} \in H_G$, where $a\neq b$
\vspace{0.5cm}
\By definition of $H_G$, $A \in gHg^{-1}$, for all $g\in G$, take $g = \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix}$
\vspace{0.5cm}

If $A\in gHg^{-1}$ then there exist $B = \begin{pmatrix} c & 0 \ 0 & d \end{pmatrix} \in H$ such that $ A = gBg^{-1}$
\vspace{0.5cm}
\Compute, $gBg^{-1} = \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix} \begin{pmatrix} c & 0 \ 0 & d \end{pmatrix} \begin{pmatrix} 1 & -1 \ 0 & 1 \end{pmatrix} = \begin{pmatrix} c & d \ 0 & d \end{pmatrix} \begin{pmatrix} 1 & -1 \ 0 & 1 \end{pmatrix} = \begin{pmatrix} c & d-c \ 0 & d \end{pmatrix}$
\vspace{0.5cm}
\Now if $A = gBg^{-1}$ then by comparison the entries we have, $ c = a $, $ d = b $ and $ c = d$, implies $ a = b $, but we assumed that $a \neq b$, so that's a contradiction. Therefore, $H_G = \begin{pmatrix} a & 0 \ 0 & a \end{pmatrix}$, where $a\in \mathbb{Q^{\times}}$

[cloud walrus]

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

[crystal vale]

is it correct?

[velvet hull]

you can shorten the proof to straightforwardly show that any diagonal matrix in H_G is necessarily a scalar identity matrix, but the idea looks good

oh actually it seems like you've only shown that H_G is in K, and not K is in H_G

so you'll need to do that as well to finish the proof, it's pretty easy

[crystal vale]

Actually I also showed that K in H_G , first three line

[velvet hull]

oh okay

[crystal vale]

I skipped that part, because that is trivial

[crystal vale]

How do I do that?

[velvet hull]

like, there's no point in assuming a!=b

since you're gonna show it anyway

so just start with [a 0; 0 b] in H
then [1 1; 0 1] [a 0; 0 b] [1 1; 0 1]^-1 = [a b-a; 0 b] is also in H

and therefore b-a=0 and b=a and we're done

[crystal vale]

Right

[languid trellis]

Okay it turns out the exercises are much easier if you use the results in the chapter. I am so silly i spent two days when the answer was in front of me

[elfin wraith]

Who’d have thunk it

Not universally true though I suppose, there’s always that exercise in Lang

[thorn jay]

so me tbh

[untold pendant]

what’s a set with addition and subtraction defined and closed, but not multiplication and division?

[glad osprey]

The imaginary line { ix | x in R} in C for example

[thorn jay]

Do you mean a subset of a ring not closed under multiplication but closed under and subtraction?

[rocky cloak]

the set of half integers for example

[velvet hull]

the set of all (real) polynomials of a fixed degree

[rocky cloak]

They're not really closed under addition, but I guess homogeneous polynomials would work

Or polynomials less than a given degree

[velvet hull]

oh, right lmao they're not closed under addition

maybe I don't know what a ring is after all

[tulip otter]

How is the induction done here

[elfin wraith]

Am I being dumb? Is this not an abelian group

[thorn jay]

I interpreted it as a subgroup of a ring that is not a subring

[elfin wraith]

Fair, I was just confused by the answer because I took it to be a set where we just can’t multiply (I kinda assumed the person wasn’t asking since they mentioned division rather than like a multiplicative inverse)

And confused because I felt they were strange examples of abelian groups to give lol

[thorn jay]

Lmao yeah

[amber burrow]

I’m trying this but I think I’m already messing up

Say for index 2

Google tells me the group presentation of PSL(2,Z) is a^2,b^3

So either a(1) =2, a(2) = 1 or a(1) = 1, a(2) = 2

This gives two distinct actions

But google tells me there’s only one index 2 subgroup

[rocky cloak]

a(1) = 1 is just the trivial action

You need the action to be transitive. So for m=2 there's only one nontrivial map to S2, so there it's pretty easy

[amber burrow]

Oh yeah forgot about the transitive part

[rocky cloak]

For m=3 it already gets a little trickier. You have the normal one, which is just the normal subgroup generated by a.

Then what remains are those corresponding to surjective homomorphisms onto S3. But just all of them will have some overcounting. The subgroup corresponds to the preimage of the stabilizer of 1 (or some fixed choice of element in {1, ..., n})

[amber burrow]

For m=3 shouldn’t it be 3?

Any non-trivial way you pick the action of b gives a transitive action

You have two options

Either 1,2,3 or 1,3,2

Then you can choose a however you want

That actually is 6

[rocky cloak]

There are 6 surjective homomorphisms onto S3, but it seems to me they pairwise correspond to the same subgroup.

And then there's the normal subgroup, so that's 4 in total.

[amber burrow]

yeah i checked ur right

im so bad 😢

i like have a very rudimentary understanding of group actions

The best i can do is like applying Burnside's lemma or figuring out all possible actions of a group on a set

wait how do you know which homomorphisms correspond to transitive actions?

so i follow that every group action of G on {1,..,m} can be viewed as a homorphism G -> Sym({1,...,m}) = S_m

[rocky cloak]

If the image is a transitive subgroup

[amber burrow]

ok your approach makes sense

ty

where would you recommend studying group actions from?

artin doesnt really do much

idk most textbooks stop at orbit stabilizer

[royal shadow]

are you thinking of groups?

[acoustic igloo]

the only ideals of a field are {0} and the field itself right?

[barren sierra]

yes

[acoustic igloo]

thanks 🙏

[velvet steeple]

Z(A) is here? Hi!

[thorn jay]

Haiii

I annoy people with UA and make false statements which I hastily correct from time to time

[velvet steeple]

When did you get into UA?

[thorn jay]

About a year ago i discovered it because I wanted a toolbox to study some algebraic structure i derived from some Latin squares stuff

But i wasnt good enough at math to properly do it yet, so i dropped it for a couple months

After a while went back to properly learned it because I didnt suck balls anymore and now it's been my main interest

[velvet steeple]

that's cool, sounds like those Ramanujan magic square problems

[velvet steeple]

So you picked up all pre-reqs in a couple months?

[thorn jay]

I had a strong base in group theory, and in a couple months i was basically doing module and ring theory

I basically figured out the rest of the prereqs during the learning of UA though lol

Especially the logic / lattices part took me a while to fully "get"

[velvet steeple]

ah, so you basically picked up algebra

Alright, yeah that's cool

[thorn jay]

My learning curve is scuffed

Heavily

Just recently got into closure operators and systems, and from that also learning topology to apply it to universal algebraic geometry

[velvet steeple]

my learning is usually rather linear, I'm afraid that I miss out on some topic otherwise and a solid base

like I mean going cover-to-cover on stuff

[thorn jay]

Yeah, that is very smart lol

[velvet steeple]

well in the end you get to the 'interesting stuff' much slower

[thorn jay]

The only reason i got this far was because of the fact i had a project to apply everything to

Because somehow I devolved into module theory studying Latin squares which is just

Lmao

[velvet steeple]

yeah this is the big advantage

you right away have an application

[rich granite]

Oh btw if you take the right kan extension of the inclusion of Fields into commutative rings along itself then you get a codensity monad $\text{Ran}{\iota} \iota (A) \coloneq T(A) = \prod{\mathfrak{p} \in \text{Spec}(A) } \text{Frac}(A/ \mathfrak{p})$ essentially getting the spectrum functor "for free". What I mean by this is that if you can somehow globally define a topology on T(A) then maybe this works more generally for any inclusion functor and its codensity monad giving you generalised affine schemes

[cloud walrus]

Chrizzl_

[rich granite]

Learning everything in the wrong order is such a vibe

[thorn jay]

The connection is that certain pairs of latin squares (MOLS) correspond 1-1 to modules over Z[t, t^-1] / (1 + t) I think

Its been a while

[thorn jay]

This sounds a lot like coherent conditions

Lmao

[rich granite]

Maybe they're connected idk

But like you only need the codomain to be complete for kan extensions to exist

And that's true for any algebraic category (Eilenberg Moore)

[thorn jay]

Ill have to look into it

Thx btw

[rich granite]

I'm trying to figure out open subsets without reference to the prime ideals in order to generalise this. I mean closed ones are easy just take the image of the quotient A → A/I under T (just that the direction of the arrow is wrong but ig turn the codomain into the opposite category idk)

But then you'd somehow need to get their complement

[thorn jay]

You can define a poset category using the closed sets

Then take the opposite category

And all you need tbh for schemes is a locale

(Read: frame)

[rich granite]

For S={1,f,f^2,...} The map SpecS^-1R → SpecR is a homeorphism onto its image and I can produce other open sets by using limits and colimits. Now if I have a notion of localisation in other categories I can apply this procedure to them too

[thorn jay]

You should look into noncommutative geometry

And duality theory

[rich granite]

I should

[thorn jay]

https://www.amazon.com/Dualities-Algebraist-Cambridge-Advanced-Mathematics/dp/0521454158

This isnt necessarily NC

But this should be interesting for you

[rich granite]

ZAMN

[thorn jay]

I KNOW

[rich granite]

Hmmm 300 pages for almost 200€? 🤨

Eh

Worth it

[thorn jay]

👉 👈

[rich granite]

And no I'm not pirating because I prefer irl books

[thorn jay]

I'm sure it'll magically appear as a pdf on my phone

[rich granite]

You're a magician

[thorn jay]

Im harry potter if he was even remotely cool

[pearl fog]

Is it true that Gal(F_p^n/F_p)=Aut(Fp^n)?

algebraic ext arises from polynomial, count how many polynomials you have with finite field coeffs

[cloud walrus]

wudu991

[pearl fog]

well you can always exclude the identity

[cloud walrus]

wudu991

[pearl fog]

hes wrong then

i recognize the book

[rocky cloak]

The algebraic closure is not for example

The one you responded to in that comment? No

The two statements aren't really related

[pearl fog]

and mine pls im stuck

[rocky cloak]

It's just a typo. It's not fixed by the 3 nontrivial automorphisms

I mean it's such a small mistake I probably wouldn't even put it in the errata if I were them

[pearl fog]

this arises from answer 3 of this https://math.stackexchange.com/questions/774814/prove-that-n-divides-phian-1-where-phi-is-eulers-phi-function, DF 14.3.10

one must say Gal is Aut before claiming #Gal=phi(p^n-1)

[rocky cloak]

Yes, any field automorphism will necessarily fix Fp

You can try to prove this yourself

[pearl fog]

but why? all i can think of is that fpnx cyc so fp must be subgroup

am i on the right track

[rocky cloak]

Well a hint could be that a homomorphism satisfies f(1) = 1

[pearl fog]

oh i see characteristic

dumb me

can someone help me decipher this solution? i forgot what my TA said

those equals look like quotient by x^p^n-x+1 but first i cant work out exactly how quotients look like that and idk why this implies x^p^n-x+1 irred over Fp

I can think of another one namely x^p^n-x are all the irreds in Fp of degree d where d runs through divisors of n but that doesn't rule out the non divisor irreds

[tardy hedge]

Hey dudez i dont wanna do math no mo

I shuld go and make money

Haha

[rocky cloak]

What's the point of money if you can't spend it on doing math?

[tardy hedge]

Lol

Making the decision to just stay in academia is a big one

Im doing a masters degree rn, but yeah i was thinking im probably gonna only want to do a phd if i actually manage to get accepted to a good school of something

Like UofT in toronto

Otherwise …

[rocky cloak]

Can always spend all your free time doing math on discord

[tardy hedge]

Very true! I was doing that last year and enjoyed it …

[crystal vale]

Same condition, I am confused I have now to decide where I have to go, either placement course or research course

[tardy hedge]

I see. Are you in undergrad though?

[crystal vale]

Yes

And I got a better placement college this year and also a little bit better college for a master in pure mathematics

Mainly I am focusing on research but my family is concerned about my job

And I have no idea how the life of a PhD student

[tardy hedge]

Ask jagr he been a phd student. So far in my program ive been going to class kind of just like undergrad

So im notnsure

In my program i have one class left to do and write a thesis

So this summer i will mainly focus on the thesis . No idea what that would be like

My supervisor also studies combinatorial comm alg which i dont like

Im hoping i can write stuff more about algebra instead of the simplicial complex combinatorics stiff

Stuff

[thorn jay]

You know, universal algebra always can benefit from new researchers...

[tardy hedge]

Lol imagine i do that

Reframe the problem my supervisor is doing into universal algebra

Lol

I mean mayb

Could that be a thing?

[south patrol]

What is this

[tardy hedge]

Stanley reisner theory

https://en.m.wikipedia.org/wiki/Stanley–Reisner_ring

Idk im just not that interested in thinking about simplicial complexes tbh

[thorn jay]

Throw some commutator theory at that thang

[tardy hedge]

What is that lol

Yeah if i could write something about this stuff that isnt directly going down and dirty with simplicial complexes then that would be cool

I will talk to my supervisor about it

She already knows i dont like combinatorics stuff too much lol

She prolly annoyed ngl

[tardy hedge]

Like, my supervisors papers are really a lot of combinatorics stuff

Proof by induction

[thorn jay]

Hmm, basically generalises the notion of a commutator to arbitrary algebras

Also gives a natural generalisation of abelian groups and stuff, basically "nice" algebras

[tardy hedge]

Honestly I wouldn’t mind learning about that stuff if i could apply it here. I think its just like the combinatorics stuff is a niche that isnt that related to other parts of math too much

[thorn jay]

Also using commutator theory you can analyse and say things about the structure of your algebras but I'd need to properly read the book

[tardy hedge]

So id rather think about stuff that gives me more bang for my buck type sho

Shi

[thorn jay]

Makes sense

[tardy hedge]

From what ive seen its an area of math that is very like … ripe with papers to write or smth

Ppl just study random ass shit ngl

[thorn jay]

Combinatorics?

[tardy hedge]

Yeah this algebraic combinatorics stuff

[thorn jay]

Makes sense

[tardy hedge]

Yeah

[thorn jay]

I couldnt

Combinatorics ❌

[tardy hedge]

Yea its boring man …

I found it kind of weird how its a relatively new field

Stanley reisner ring only introduced in 1970s

Bizarre

[thorn jay]

It is weird

Formally i suppose

Before that combinatorics was done

[tardy hedge]

Yeah

[tough raven]

Sounds fun

[tough raven]

Fair

I think I've just realised that the proof of the PBW theorem secretly involves the Coxeter presentation of S_n.

[cloud walrus]

wudu991

[rocky cloak]

what is the field of absolute numbers?

the algebraic closure of the prime field?

[thorn jay]

often you use symbols and notation others don't, so a quick explanation never hurts

[rocky cloak]

so what dont you understand about it?

[thorn jay]

this looks like the largest algebraic field extension of F_0 included in F

[rocky cloak]

where are you getting this name though? my google fu hits nothing on this name

[tough raven]

It is, but I also don't see why that deserves a name, much less this specific one.

[thorn jay]

everything has interesting properties if you look hard enough ¯_(ツ)_/¯

[rocky cloak]

They’re preserved under any homomorphism at least

And just from this paper I guess its usefull to prove some decidability stuff.

[rocky cloak]

Yes that's true

There is only one field of a given order up to isomorphism

I was assuming d was an integer, are you meaning for d to be able to equal infinity?

So then the extension is finite

Integers are finite

If q is finite and d is finite then q^d is finite

Yeah, that's the definition

[cloud walrus]

wudu991

wudu991

wudu991

wudu991

[thorn jay]

You can also see it from the perspective that every finite dimensional vector space is free, so is isomorphic to a finite direct sum = finite direct product of copies of K, with the number being the dimension

Although I suppose it amounts to the same argument

[toxic sapphire]

I don't think I'll need help with this problem but what exactly is it asking for?

what does it mean for a function to "induce a natural homomorphism"?

[velvet hull]

If B is a basis for a vector space V, such that f: B->W is some fixed arbitrary function from B into another vector space W, then there is a unique linear function f*:V->W that agrees with f on B. In which case we say that f* is induced from f.

The idea is very much the same here

[toxic sapphire]

got it 👍

thanks guys

[velvet cave]

should I use Silverman's Abstract Algebra or D&F to study abstract algebra? I've got basically no experience with abstract algebra and I really like Silverman's book so far, but I've heard that D&F is more advanced since it's a grad level book. Is D&F self contained enough for me to learn from it directly, or should I just continue with Silverman's book since it seems far more introductory

[sturdy spear]

sounds like for graduate level

[velvet cave]

thanks for the advice, will def continue with it then!

[sturdy spear]

silvermans abstract algebra?

Abstract Algebra: An Integrated Approach
oh yeah a nice book

ah yes. I saw contents he introduces cat theory as well

so Borcherds’ lectures for group theory are suitable for this book

he has 32 vidoes

yeah i think so

[crystal vale]

I did the first part but for the second part I have a problem.

So let G is a non-abelian group of order 8, yes Z(G) has order 2, and let h in G, | h | = 4, then I divide the question into cases.

Case I, When G - < h > has all elements of order 2, then it is isomorphic to D_8.

Case II, When G- < h > has all elements of order 4, then it is isomorphic to Q_8.

Case III, When G - < h> has some element of order 2 and order 4. Here it can be only 2 elements of order 4 and 2 elements of order 2

In case III, Group has total 4 elements of order 4 and 3 elements of order 2

Yes

There are 5 elements of order 2 in D_8

[fierce breach]

I don't understand why we're looking for a non-zero element divisible by every prime.

Oh wait, nvm I get it but I'd still like someone to verify my reasoning.

[fierce breach]

||If such an element exists in a subgroup, this element would be the identity but we know the identity must be unique||

[amber burrow]

Wait so if I take a purely group action approach

Every index m subgroup is the stabilizer of an action of G on {1,…,m}

*transitive action

And every transitive action of G on {1,…,m} has a stabilizer of index m

[rocky cloak]

G acts transitively on G/H which is a set with m elements

[rocky cloak]

That's the orbit stabilizer theorem yes

[amber burrow]

Yes, ok

So my problem is equivalent to counting distinct stabilizers of 1 under transitive actions

So for m=3

I have to have either b(x)=x for all x, or 1->2->3 or 1->3->2

And then I can choose a however I want

But the different choices of b don’t affect the stabilizer of 1

[crystal vale]

Which book is this?

[rocky cloak]

Well it does. For example if
a is the cycle (1 2) and b is (1 2 3) then ab is in the stabilizer, but if b = (1 3 2) then ab = (1 3)

[amber burrow]

Hmm ok yeah

I guess I’m getting confused because when I’m counting it I’m getting 6, not 4

Cuz two options for b, and huh can choose a however you want

So either a is one of 3 transpositions or the identity

Actually that’s 8 total

[rocky cloak]

If you want it to be surjective a must map to one of the 3 two cycles, so that's 2*3 = 6

Then I'm saying they should pairwise correspond to the same stabilizer, so that gives you 3. And then you have the one normal subgroup so 4 in total

And how to see they pairwise match up:

if you have a surjective map G -> S3 and you compose it with an automorphism of S3 you get another map. There are 6 automorphisms, so you get 6 such maps (i.e. all of them).

So then the question is which automorphisms preserve the stabilizer of 1, which is just conjugation by (2 3) and the identity. So we're overcounting by a factor of 2

[amber burrow]

Why do u have to consider tbe normal subgroup separately?

[rocky cloak]

It's just easier

[glad osprey]

What does N_f(\alpha) mean here?

[thorn jay]

I think they're called absolute because these are the elements which are algebraic over every subfield of F

Though I'd rather call them "absolutely algebraic" or something of the like

[tough raven]

Just restating the definition here...

I could argue "absolute" = "fixed by every homomorphism" in which case you get the prime subfield

[tough raven]

Since it has a multiplicity, I'm guessing it just means that alpha is a zero of f.

[thorn jay]

Naming is hard

[tough raven]

Which is why you shouldn't do it unless it's really necessary.

(Permanent) names have weight.

[thorn jay]

What would you consider really necessary?

If it isnt just a simple combination of already established definitions?

[tough raven]

Well, it's very much "(occasionally) know it when I see it", but

[glad osprey]

That makes sense, thanks I'm guessing it must have a typo though, since there's some weird recursion with the statement a \in N_f(a). Maybe it should be just a \in N(f)

[tough raven]

no, it's not about how simple or complicated it is, it's about how much it actually occurs

if there's a combination of 2 conditions that occurs as a hypothesis in 20 theorems you should name it

If there's a combination of 15 conditions for a single theorem I would probably say name it "locally" i.e. the name is only supposed to take up that meaning in this theorem/section/paper/book (e.g. you could give it a name like "good" or "special" or a more descriptive name).

If it turns out to be more useful it can be agreed later to recognise the name out-of-context

[thorn jay]

For "sufficiently nice" objects
Something like that

[tough raven]

Essentially naming a definition implicitly asserts that that combination of data/conditions is worth paying extra attention to, so you shouldn't name it if it isn't.

[thorn jay]

Gotcha

Thank you :3

[tough raven]

I mean I was just throwing my opinion at Discord for no reason 🤡 but sure, you're welcome.

[empty perch]

Hey folks, just a quick reminder that the Abstract Algebra Lecture series is meeting regularly again ^_^
We will meet to discuss Sylvester's Law for symmetric forms in about 10 minutes over in the ⁠⁠#1055201711679082516. More information can be found in our thread: ⁠⁠#1317307081535000606

[thorn jay]

I like hearing people's opinions on things

[wintry sluice]

Why do you need ideals for quotient rings? Why can’t you just use sub rings?

[tough raven]

Try taking the quotient of ℚ by ℤ.

You can equip it with the additive structure of an abelian group (even a ℤ-module), but multiplication won't be well-defined.

If you investigate what it would take for multiplication to be well-defined, you will rediscover the concept of (two-sided) ideal.

[thorn jay]

(I will take commutative rings here for simplicity)
In short: ideals act like 0, and essentially the defining property of 0 is that r • 0 = 0 for all r in your ring. As ideals act like 0, multiplying and element from your ideal with an arbitrary element from your ring must also be in your ideal

Subrings cant have this property without being the whole ring itself, as subrings contain 1, and r • 1 = r

[tardy hedge]

So im trying to compute Galois groups over Q. I factored
x^5+x-1 = (x^2-x+1)(x^3+x^2-1) into irreducibles. So, at this point, I’m not really sure how it goes. I suppose the Galois group acts on the roots with two orbits..?

Like, can I just say its C2 x A3 or C2 x S3? depending on discriminant of the cubic or smth

[rocky cloak]

Well, you would need to establish some kind of independence for that.

For example the Galois group of
(x^2 + x + 1)(x^3 - 2)
is simply S3, not C2xS3

[tardy hedge]

Yeah, im not really sure how it works then

the roots need to be not related somehow right, in what sense? there is no algebraic equation relating them?

and so if they are independent then the choice of where you send each root of each factor in the automorphisms does not depend on each other

[rocky cloak]

Well does one polynomial have roots in the splitting field of the other? That's something you should be able to determine from discriminant in this case

[acoustic igloo]

angle brackets <a> denote a subgroup generated by an element. can they be used to denote a subring?

[tardy hedge]

in the ring context <a> means the ideal generated by a

[acoustic igloo]

an exercise asked me to find ideals of Z_12

is it ok to use angle brackets?

is it generated by just addition, or mulitplication too?

i think i'll just avoid the brackets for now since they weren't introduced in the book

[acoustic igloo]

What is the relationship of the ideal $\sqrt{N}$ to the nilradical of $R/N$? Word your answer carefully.\
The nilradical of $R/N$ is ${a+N\mid a\in\sqrt{N}}$.

[cloud walrus]

Axe

[acoustic igloo]

right?

[acoustic igloo]

thanks 🙏

[vestal sapphire]

Hey, I doubt this is 'advanced' but I couldnt find other channels to ask this. How do I approach equations like these:
We want to solve in S_7

[velvet hull]

Cycle decomposition, oh hey it’s already in that form lol

[vestal sapphire]

Could you elaborate? You can ignore (b) but it would be great if you could show me first few steps of solving for a)

[velvet hull]

Do you know about cycle type decomposition

[vestal sapphire]

I dont think so. I only know that we can decompose any permutation, and there is exactly one unique decomposition

[velvet hull]

Well there isn’t a unique decomposition, because S_n is a group, so if p, q are any two permutations than pq^-1 * q is technically a valid decomposition of p

It’s specifically that the decomposition of a permutation into disjoint cycles that is unique

And that’s the cycle type decomposition

[vestal sapphire]

oh right thats what I meant sorry, just woke up

[velvet hull]

So let’s say p =abc is its cycle type decomposition. Then the powers of p are exactly the powers of a, b, c

[vestal sapphire]

So in this case i need to find a s.t a^2 = (123) and b st b^2 = (456)? The answer would be a•b? Is there an efficient way of finding these roots? Because in the next one I see we have 4th power so guessing the cycles might be difficult

[velvet hull]

well, there's one technical detail that you're overlooking here

the cycle type isn't 100% preserved - consider the 4-cycle (1324)

(1234)^2 = (12)(34), so a 4-cycle becomes a 2,2-cycle

[vestal sapphire]

So I also have to consider ‚merging’ the cycles right? What about (b), would I have to consider splitting this cycle into several shorter ones?

[velvet hull]

the powers of an n-cycle can only be either another n cycle or a bunch of d-cycles, where d is some divisor of n

now apply this to the entire cycle decomposition

of course I'm telling you the most general result here, you don't need this result for your specific question since the cycle type of the permutations in your questions all happen to be composed of primes (why does this simplify things?)

[vestal sapphire]

oh okay that clears this up

[distant sleet]

What does "homogeneous elements" of an algebra stands for?

[rocky cloak]

In a graded ring homogeneous means only the sum of elements of one degree. So for example in the polynomial ring k[x] with the usual grading things if the form ax^n would be homogenous.

I'm assuming that's the meaning here, though you've provided no context.

[distant sleet]

The context

[boreal inlet]

Yeah it should be the same idea

[rocky cloak]

When they say "the elements of A^i are the homogeneous elements of degree i." That's actually the definition

[distant sleet]

Ohhh I see

Yeah Im not used to such terms

Thanks

[distant sleet]

What does "if it is of order two" mean for a linear transformation? Im probably misunderstanding it or not getting it

[thorn jay]

probably an invertible transformation such that T^2 = I

It means that it is of order 2 in the general linear group

[distant sleet]

That doesnt mean V must be of dimension 2 too?

[thorn jay]

Mo

No*

For example, take R^n and the linear transformation switching two basis vectors

[distant sleet]

Ohhh I see

[distant sleet]

I wrote it down and yeah all sense now