groups-rings-fields

[charred iris]

I'm like 99% certain this was the notation used in my algebra course

[thorn jay]

Q^+ and R^+ is diabolical

i saw that and was literally like "oh yeah the additive group o- WHAT DO YOU MEAN"

[south patrol]

yeah but that course was very idiosyncratic

[charred iris]

that's one way of putting it

[clear hatch]

Exam in ONE day

https://tenor.com/view/bro-is-finished-cooked-finished-dog-closing-eyes-dog-with-eyes-closed-gif-8933440936102921947

I just need 50% on the exam to pass

[proud vigil]

u got thiss

[rocky cloak]

Someone needs to make a :dogking:

[thorn jay]

whats it about?

[thorn jay]

that is a lot

i believe in you though!!

[clear hatch]

im only going to cover the first two (ring theory, polynomials)

thing is, i took the course last semester but taking the exam this semester cos i got sick last time. but last time when i took it we didn't do any of the applications stuff 💀

im just going to skip it and hope i can get 50% marks from the other two sections

[thorn jay]

should be doable, given the ratio of amount of topics

[quiet pelican]

Why do we care about Frobenius groups?
It kinda seems like an unmotivated definition
Like sure we can make the definition and prove things about it
But I just don’t see how it’s useful, except to prove things about itself

[crystal vale]

any hint?

so if [G:H] = m, then for all g in N, we have g^ms = g, for some fixed s in Z

[knotty pilot]

Have you learnt group actions yet

[crystal vale]

i know group action but author hasn't introduced it yet

[knotty pilot]

Hmm
The fastest proof i know is by group actions

[crystal vale]

how i can do it with group action?

[knotty pilot]

Sorry I have a mistake let me think about it

[quiet pelican]

Consider the index of the subgroup NH

||[G:NH][NH:H] = [G:H], and [NH:H] = [N: N \cap H] divides |N|, so [NH : H] = 1, ie N \subset H||

[knotty pilot]

Indeed I just drew the diagram lol

[crystal vale]

So N is normal implies NH is subgroup and [G: H] = [G:NH][NH : H] gives that index of NH is finite

[quiet pelican]

Yup
Now what else can you say about [NH:H]?

[crystal vale]

now i saw an hint, i got it

thank you

[vast verge]

I don't understand these two highlighted parts, can someone explain please?

[velvet hull]

for the first line, it's essentially just saying that the constant polynomials are units in R[x]

and if an ideal contains a unit then it is the entire ring

[vast verge]

If we know that the ideal A contains c, how do we know that 1/c is in A? Is it because A is a subring?

[vast verge]

How could the entire ring be generating from a unit?

[velvet hull]

recall that ideals need to be closed under external multiplication

if i is in I and a is in A, then ia has to be in I as well

[vast verge]

Oh I think I understand now

If $A$ is an ideal and $a \in A$ is a unit of $\mathbb{R}[x]$, then it will have a multiplicative inverse $a^{-1} \in \mathbb{R}[x]$.
So, for any $x \in \mathbb{R}[x]$, $a \cdot \left( a^{-1} x \right) = \left( a a^{-1} \right) x \in A$ since $A$ is an ideal

[cloud walrus]

Tropical Greens

[velvet hull]

if an ideal contains a unit then it contains 1

but if an ideal contains 1 then it contains everything

[vast verge]

Right, right

That is very helpful, thank you

[velvet hull]

for the second line, it's enough to show that every element in R[x] mod I is a unit

and that's what the proof is doing essentially

[glad osprey]

on the second highlighted line, I think it should say ax + b = r(x), not ax + b + r(x)

the second line uses the fact that f(x) and q(x)(x² + 1) are both in A, and A is closed under subtraction

[tulip otter]

let $G=\langle a\rangle$ and fix $z\in Z$. Define a map $f:G\to Z$ by $f(xy)=f(x)f(y)\ \forall x,y\in G$ and $f(a)=z$. Then this map is well defined and is the $\bf{unique}$ homomorphism that satisfy $f(a)=z$. \Indeed, given $x,y\in G$ such that $x=y,\exists k\in{1,2,\dots,n}$ such that $x=y=a^k$ so $f(x)=f(a^k)=f(y)$ and $f$ is well defined (is this really sufficient to prove that $f$ is well defined?). Now $f$ is a homomorphism by construction.\\Finally, let $g:G\to Z$ be a map defined just as $f$, ie $\forall x,y\in G, g(xy)=g(x)g(y)$ and $g(a)=z$, then $g=f$. Indeed, given $x\in G,\exists k\in{1,2,\dots,n}$ such that $x=a^k$ so that $g(x)=g(a^k)=g(a)^k=z^k=f(a)^k=f(a^k)=f(x)$ which completes the proof of the hint

[cloud walrus]

yassine

[tulip otter]

is this correct?

[velvet hull]

that's not how well-defined works

[quiet pelican]

That doesn’t prove it’s well defined
The same argument would go through to give you a map from Z/nZ to Z, but no such map exists

[velvet hull]

what you want to show is if a^i = a^j, then f(a^i) = f(a^j)

[tulip otter]

yea i figured that i am missing something there

[velvet hull]

you don't know a priori if i = j or not and indeed it may not be

[tulip otter]

ohhh i see

let me think about that

[tulip otter]

but couldnt i argue that a^i=a^j implies i=j since G=<a> is cyclic first and then this implication will follow?

[velvet hull]

prove that the map from Z/2Z to Z/3Z by sending 1 to 1 is not well-defined

using that example, see why your argument doesn't work

[tulip otter]

otherwise i am not sure how to proceed, f(a^i)=(f(a))^i=z^i and f(a^j)=z^j but then what. There is nothing more that i can do from the constructed map f unless i am missing something

[tulip otter]

is it a homomorphism too ?

[velvet hull]

it's not well-defined

[tulip otter]

well thats what i should reach, but before knowing that you could say that it is a homomorphism or not because you still dont know that it is not even well defined right?

[rocky cloak]

This is not true unless you have some bounds on i and j

[velvet hull]

well its not even well-defined

so it just can't be a homomorphism

when it's not even a set function

[rocky cloak]

In a cyclic group of order 2
a^1 = a^3

[tulip otter]

yes i see what you are pointing at, i always assume that the i in {1,2,...,|G|}, but that is not necessarily needed initially

[rocky cloak]

You could do it that way, but frankly it just makes the proof messier

[tulip otter]

ohhh

[rocky cloak]

Because then verifying the homomorphism condition is a mess

[tulip otter]

because when the sum of the exponents is >|G| it should be reduced mod |G|?

[rocky cloak]

Yeah, so then you would need to deal with that

[tulip otter]

yea i see

[tulip otter]

let f be this map, then 4f(1)=4=1 but 4f(1)=0f(1)=0 so that 1=0 and f is not well defined

is that a correct way to prove it ?

[velvet hull]

it's one way, I guess

[tulip otter]

alright now i see what you meant

[tulip otter]

ok so now WLOG assume that j>i and a^i=a^j, then j=i+kn (here n=|G|=|Z| and k in N). Now f(a^j)=f(a)^j=z^j=z^{i+kn}=z^iz^{kn}=z^i=f(a^i) so that f is well defined

does this now correctly prove that f is well defined?

[velvet hull]

looks good

[tulip otter]

what about the rest of the proof here, can you check it if you have the time

[velvet hull]

most of the work is in showing that it's well-defined

everything else is kind of obvious

[tulip otter]

yea i see

tysm both you and jagr

have a great day/night

i will probably be back very soon with more questions lol

[vast verge]

Prime and maximal ideals are pretty interesting

I'm looking forward to doing the exercises in this chapter

[fading acorn]

yes they are

we also do geometries on prime ideals, so called algebraic geometry

[vast verge]

Me who doesn't know what a geometry is:

The only group geometries I knoe are picturing the dihedral groups in my head

[velvet hull]

any graduate textbook lol

hell no i dont think youre ready for any of them

haha, you'll get there when you get there

[rocky cloak]

basic abstract algebra by bhattacharya etal maybe

Not to say there isn't occasionally a paragraph here or there saying something, but theorems and definitions are marked so they would be easy to skip if the goal is to avoid exposition

[elfin wraith]

There’s also books like jacobson Lang but you’ll be filling in half of the results yourself

Got my books confused

[karmic moat]

Bruns herzog but that’s not an intro to comm alg book

[ashen heron]

the dover book by allan clark

[karmic moat]

When I think of a book that just lists results I think of Bruns Herzog

[tardy hedge]

I have an R-mod hom f: M->N. If k->R is a ring hom then f can be viewed as a map between vector spaces.. and the kernel, image etc remains the same right

[elfin wraith]

You are that books biggest hater lmao

But yeah it is like 500 pages of just text, while still managing to say very little

[coral spindle]

The image depends only on the set structure, and the kernel depends only on the pointed set structure. They are of course preserved when you forget the additional info

[karmic moat]

It’s a good reference book but boring to read

[elfin wraith]

I was under the impression that that is how the books kinda supposed to be used anyway

[karmic moat]

Probably, makes sense

[elfin wraith]

It’s also possible I have that idea because my supervisor told me to look at it that way, presumably after he suffered through all of it himself

[tardy hedge]

that is true it def is like that

[elfin wraith]

I think it’s Herzogs style, Monomial Ideals is similar

[tulip otter]

let G/Z=<aZ> and let x in G, then there exists k in N such that xZ=a^kZ so that Z=x^{-1}a^kZ and x^{-1}a^k in Z. But then x^{-1}a^kx=xx^{-1}a^k=a^k which means that x^{-1} in Z and x in Z. Since x was any element of G, this shows that G=<Z but Z=<G so that G=Z and hence G is abelian (here =< means subgroup)

is this correct or is it missing something

[delicate orchid]

You’re missing the most important parts: love and kindness

[rocky cloak]

I'm not sure I quite see how you're concluding x^-1 being in Z...

Maybe it would make more sense to show that any pair of elements commute

[tulip otter]

ah wait, you are right. I didnt show that it commutes with all elements of G, i just showed that it does commute with a^k (?)

[tulip otter]

sure but i tried some roundabout way because i wasnt sure how to proceed if i start with something like "let x,y in G, then xy=.... to then finally reach =yx"

[rocky cloak]

Well you know x = a^k z for some z in Z

[delicate orchid]

You’re halfway there. From here you can conclude it commute with everything

[tulip otter]

ah wait, i honestly dont know why i didnt think about writing any element of G in that way

so given x,y in G, there exist k,i in N and z,w in Z such that x=a^k z and y=a^i w. So xy=a^k za^i w=a^{k+i}zw=a^ia^k wz=a^i wa^k z=yx

[tulip otter]

i see, it was straightforward after writing the elements of G in this form

tysm

isnt this done by the exact same way ?

first H is in the center of G so it is normal in G. now (G:H)=p where p is a prime so G/H is cyclic

and then given x,y in G. there exist k,i in N and z,w in H (and thus in Z(G)) such that x=a^k z and y= a^i w, so xy=...=yx

[tulip otter]

then this can be generalized a bit more by replacing the (G:H)=p for p prime part by G/H is cyclic?

ohhh i see thats nice

[crystal vale]

Book?

[tulip otter]

lang's undergraduate algebra

[crystal vale]

Okay

Good choice

[tulip otter]

tysm

i am enjoying the exercies tbh, they are nice

[crystal vale]

Yes G is abelian iff G/Z(G) is cyclic

[tulip otter]

is p here meant to be prime? because otherwise the dihedral group of order 36 for example would be a counterexample

I see

[tall igloo]

im a bit confused by this - isn't abelian G = Z(G)? so G/Z(G) = 1? I guess one could call this cyclic of order 1, is G/Z(G) = Z/n just never possible otherwise?

[crystal vale]

I have to complete this, anyone wants to join?

[rocky cloak]

Yeah, that's right.

The only way for G/Z(G) to be cyclic is if it's trivial

[tall igloo]

huh neat

is there some nice way to see this

[proud vigil]

i remember reading this in aluffi! the proof is fun

[glad osprey]

I've also been confused by this result, but I think its usefulness is that it's sometimes easier to prove that G/Z(G) is cyclic than proving it's trivial

[proud vigil]

i may or may not have the book open at all times on my laptop

[crystal vale]

Yes, you can use this to show every group of order p^2 is abelian

[tall igloo]

huh very cool

thanks for sharing!

[tall igloo]

may i ask what book this is? i need to study for my prelims and its been a while since i took a group theory class 🥲

oh wait you said the name

lol

[proud vigil]

yaa it's "algebra, chapter 0" by paolo aluffi

[tulip otter]

so now for this exercise, i have an idea. (Z:1)=p,p^2 or 1. Suppose that (Z:1)=p, then (G:Z)=p so that G/Z is cyclic and hence G is abelian which means (Z:1)=p^2, a contradiction. So now (Z:1)=p^2 or 1.

Now i am thinking about why (Z:1) neq 1

[quiet pelican]

||Consider the possible conjugacy class sizes||

[tulip otter]

tysm for spoiling it like that so i can think about it instead of just checking the answer

[quiet pelican]

(And that’s also more a nudge in the right direction than an answer)

[rocky cloak]

(7) was tricky. ||Managed to do it using (6)||

[crystal vale]

Okay, after a few hours I will start this assignment

[crystal vale]

okay let's start

[tulip otter]

i tried not to look at this but i didnt reach anywhere

so i looked and then didnt reach anywhere too

[tulip otter]

so the answer to this question that i came up with is that each conjugay class can either have size p^2-p+1 or size 1

the conjugacy class has size 1 only for e where e is the identity element in G since Z(G)={e}

otherwise, let a neq e and assume that G is not cyclic (otherwise G would be abelian and (Z:1)=p^2 neq 1). Then the subgroup H=<a> generated by a has order p since (H:1)|(G:1) and (H:1) neq 1

[quiet pelican]

The size of the group is the sum of the sizes of the ccls

[quiet pelican]

No
Hint: ||orbit-stabiliser||

[tulip otter]

i am not familiar with this

but let me look it up

actually the problem has a hint in the textbook but i didnt look at it yet for the same reason that i didnt look at your hint at first

[keen canyon]

is it fine if I ask a question as well or should I wait?

[tulip otter]

of course you can, the channel is for everyone

[keen canyon]

i am not 100% sure this belongs to group theory more than algebraic geometry but here it is

I just cant solve this exercise:

By considering the Cayley graph for the group G generated by the side pairing transformations of F, or otherwise, show that G is defined by the relations $(T_{i_k}^{\epsilonk} ... T{i_1}^{\epsilon_1})^p = 1.$

[cloud walrus]

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

[keen canyon]

For context:
\begin{theorem}\label{conditions}
If a compact, finite-sided polygon $F \subset \mathbb{H}$ is the fundamental region of a Fuchsian group $\Gamma$, then the following
statements hold:
\begin{itemize}
\item For each side $s$ of $F$ there is exactly one other side $s'$ of $F$ of the form $s' = T(s)$ with $T \in \Gamma$.
The elements $T$ are called \textbf{\emph{side pairing transformations}} of $F$.
\item If each side $s$ is identified with the corresponding side $s'$, then each set of vertices identified as a result
corresponds to a set of corners of $F$ with angle sum $2\pi/ p$ for some $p \in \Z$. These sets of vertices are called
\textbf{\emph{vertex cycles}}.
\end{itemize}
\end{theorem}

[cloud walrus]

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

[keen canyon]

More context

Consider a vertex cycle $v_1, \dots, vk$ of a fundamental polygon $F$. This means we have a sequence of side-pairing transformations
$T{i1},..., T{i_k}$ which relate the vertices in our cycle in the fashion of Figure \ref{fig:4.3.a}. The second statement of theorem
\ref{conditions} tells us that the angle sum $\theta_1 + \dots + \thetak = 2\pi / p$ for some $p \in \N$.

Let $r{\theta_j}(v_j)$ denote the rotation about $v_j$ through $\thetaj$, i.e. $s{i{j-1}}'$ is mapped to $s{ij}$. We have
[ T{ik}r{\theta_k}(vk) \dots T{i2}r_{\theta_2}(v2)T{i1}r{\theta_1}(v1) = 1 ]
because the isometry on the left hand side is the identity on $s{i_k}'$ (He means the entire left hand side!) and is orientation-preserving
(since the $Ti$ are, by hypothesis).

Also
[ r{\theta_2}(v2)T{i1} = g{i1}r{\theta_2}(v1) ]
because $T{i_1}$ sends $v_1$ to $v2$ and therefore

[ r{\theta_2}(v2) = T{i1}r{\theta_2}(v1)T{i1}^{-1}. ]

Similarly
[ r{\theta_3}(v3)T{i2}T{i1} = T{i2}T{i1}r{\theta_3}(v1), ]
and so on, eventally giving
\begin{equation}\label{itsarotation}
T{ik}\dots T{i1}r{\theta_k}(v1)\dots r{\theta_1}(v1) = 1
\end{equation}
after all rotations have been brought to the right.

But $r{\theta_k}(v1) + ... + r{\theta_1}(v1) = r{\theta_k + ... + \theta_1}(v1)$. Therefore \ref{itsarotation} is a rotation through
$2\pi/p$.

Hence $(T{ik} \dots T{i1})^p$ is the least power of $T{ik} \dots T{i_1}$ which equals $1$.

[inner owl]

Is question 21 (i) wrong? I am talking about a = bc?

[cloud walrus]

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

[keen canyon]

my exercise is kind of giving word-problem vibes

My idea was to assume a word T in my group is minimal in length with respect to the given relations and also assumed to be a cycle in the cayley graph, i.e. T = 1. I want to show T is the empty word.

But if anyone has other ideas Im happy for any help!

[velvet hull]

it looks right to me, what might you think is wrong?

[inner owl]

If b = a^s and c = a^r then bc = a^(r+s) = a then a^(r+s-1) = 1 which is a contradiction to the order being rs unless it is 1

@velvet hull

[velvet hull]

well, why does b have to be equal to a^s?

all you've proved is that b=a^s and c=a^r cant be true

a^s is certainly an element in G with order r, but how are you sure there can't be any other element of order r?

for instance take n=6 = 2*3.
Then 2+3 = 5, which is not 1, but 4+3 = 1, and 4 also has order 3

[inner owl]

Then isn't the uniqueness part of the problem wrong?

[velvet hull]

no, you haven't even proven existence yet

[rocky cloak]

That's what makes the uniqueness true

[velvet hull]

don't worry about the uniqueness when you haven't done existence

[inner owl]

Ok I think I read the problem wrong 😭😭😭

Sorry everyone

[tulip otter]

from what i saw after searching about orbit stabilizer, the size of the ccls for this group G would be either p or p^2 if Z(G)={e}. is that right?

[quiet pelican]

Yup (or, of course, the ccl {e})

[crystal vale]

If G has a normal subgroup of each order dividing |G|, then Sylow subgroup will be normal and then G will be nilpotent.

Now if G is nilpotent we have to show it has a normal subgroup of each order dividing |G|.

So it is enough to show every p-group has this property, because G is nilpotent implies G can be written as direct product of its Sylow p- subgroups.

And for p-group, it comes from induction, and because for group of order p, it is trivial and for the group of order p^m, take Z(G) which is non-trivial and if Z(G) = G, we are done, if Z(G) ≠ G, then take quotient G/Z(G), and apply induction hypothesis and then pull out that subgroup, right?

[south patrol]

Sounds good

[crystal vale]

I am thinking if G is nilpotent then is that necessary it's quotient group has to be the same?,

[velvet hull]

do you mean to say that quotients of nilpotent groups are nilpotent?

[crystal vale]

Yes

I am thinking, is it true or not?

Finding counterexample

[velvet hull]

yeah, it's true because the derived central series of a nilpotent group terminates

[crystal vale]

Hint?

[velvet hull]

and so the derived central series of its quotient also has to terminate

[south patrol]

Ig derived series should be replaved by a central series here

[velvet hull]

the same

[south patrol]

Pretty sure they arent

Derived series is to do with solvability rather than nilpotence

[crystal vale]

Yes, Dummit used the central series for nilpotent

[velvet hull]

oh yeah it's the central series

okay I got it mixed up

[south patrol]

But the principle is the same

[velvet hull]

the lower central series is defined slightly differently from the derived series

huh, I thought they were the same

[coral spindle]

Z(G/N) is at least Z(G)N/N right

That's the core of the argument surely

You just show the central series gets bigger mod N

[crystal vale]

I have to show if G is nilpotent then maximal subgroup H of G has prime index.

So H is the maximal subgroup and G is nilpotent implies H is normal, G/H will be simple and nilpotent so G/H has prime order.

[crystal vale]

Here at least means?

[coral spindle]

Z(G/N) >= Z(G)N/N.

[tulip otter]

let a in G with a neq e, if the ccl(a) has size p^2, its centralizer has size 1, but a commutes with itself and with e. so the centralizer cant have size 1. thus the only possible size of ccl(a) is p, in this case the subgroup H=<a> of G would have order p

and there would be another element b in G such that G=<a,b> and such that <b> has order p too (?)

[crystal vale]

So G is nilpotent, and |a|, |b| are relative prime numbers.

G is nilpotent so we can write as a direct product of its Sylow p-subgroups.

Say f: P1×P2×..×P_n -> G be isomorphism given by (a_1,..,a_n ) -> a1a2..a_n.

Now take the inverse image of a and b, we can observe that if a = a_1..a_n and b = b1...b_n.

Then, if a_i ≠ 1 then b_i = 1 and if b_i ≠ then a_i = 1.

So its inverse image commute implies they will commute

[quiet pelican]

Ok
So there’s one ccl of size 1, and some amount of size p
And every element is in exactly one ccl, so we can count the number of elements this way

[tulip otter]

the number of elements of G?

[quiet pelican]

Yes

[crystal vale]

Is it correct?

[tulip otter]

so there are p^2 ccls, p^2-1 of them have p elements so that G has p(p^2-1)+1 elements?

[quiet pelican]

We don’t know how many ccls there are
But we can count that G has ap + 1 elements, where a is the number of ccls minus 1

[tulip otter]

ah i see, but now ap+1 doesnt divide p^2 so it is definitely neq p^2

[quiet pelican]

Yup

[tulip otter]

which is a contradiction

woah that was very nice

meanwhile lang's hint

[tulip otter]

tysm for helping me all the way till the end even though i was stupid there. have a great day/night

[rocky cloak]

What's exercise 30? I'm curious

[crystal vale]

So if n=2^k, then it will 2- group so it is nilpotent.

Can I say if n = 2k, then D_2n/Z(G) is isomorphic to D_2k? If yes then
And if n is odd then Z(D_2n) = {1}, therefore n can't be odd, but if n = 2×m, where m is odd then Z(D_2n/Z(D_2n) ) = 1, so D_2n can't nilpotent, same reasoning if n = 2^k m, if m is odd.

[quiet pelican]

It’s not gonna be fun
I now have to trawl through a bunch of half translated sites to find PhD info
Helping here’s the only thing keeping me mildly sane

[minor gazelle]

Is it true the group action that has been consistently torturing me has always been a functor mapping from a category C such that |ob(C)|= 1 and C(A,A) consists all isomorphism to set?

And the category is locally small

[coral spindle]

A group action is indeed a functor G -> Set where G is a group viewed as a groupoid with one object.

N.b. not all categories with one object are groups.

Now yes as you say, you can look at the subcategory of Set whose only morphisms are isomorphisms

[knotty badger]

yeah in general a category with one object carries the same data as a monoid

[tulip otter]

ohhh thats sad to hear, i hear that it is usually difficult when one wants to go for a PhD (eventually i will experience this too since i am planning to go for a PhD). I hope that you find great results as soon as possible

[thorn jay]

I love categorists always talking about "data"

"X consists of the data of so and so"

[knotty badger]

Well I’m a physicist technically

[thorn jay]

I label you under "category theory person"

you're really shooting yourself in the foot there with your name 💀

[knotty badger]

Well being a fan of something is different from being that thing right

I am a big fan of cat theory

But at the end of the day I am a physicist

[rocky cloak]

What would you call it?

Things consists of data and axioms.

Or thingymagings and rules.

Or stuff and relations.

[thorn jay]

it's not bad, but i dont of it being used in a non-categorical / AG setting

[rocky cloak]

You don't have a name for it in UA?

[thorn jay]

we just say "an algebra is a tuple (A; F) where F is an interpretation of the ..."

so rather than say "consists of the data of" we say "is a tuple"/"is defined as"

or in this case "a set equipped with operations"

[tribal moss]

I think it's mostly a matter of category theory having a lot of differently-shaped concepts to keep track of and has a tradition of not writing some of the data down explicitly in formulas and diagrams when it's "obvious" what they are (for example, it is common to name functors just by what they do to objects). So it relatively often becomes relevant to say "wait a moment, I might be confused, does a thingamajig remember such-and-such?" to the extent it's relevant to have a word for it.

[thorn jay]

that does make sense

in UA it's very easy to just alter your signature, still an algebra

[clear hatch]

am i tripping or the proof only showed $p(x) = cf_1(x)g_1(x)$ while the lemma statement said we can find $p(x) = f_1(x)g_1(x)$?

[cloud walrus]

ffflick

[coral spindle]

since c is in D and f_1, g_1 are in D[x], wlog we can replace f_1 with cf_1 (or equally g_1 with cg_1) and get an appropriate factorisation. This is what the proof is stating implicitly.

[south patrol]

Lemma statement is a little dodgy lol

like I would state it as "p = f1 g1 for f1, g1 in D[x] with deg f = deg f1 [...]" but yeah

[clear hatch]

i dont really understand this example. does the part in yellow follow because $f(x) - r(x) = p(x)q(x) \implies f(x) + \langle p(x) \rangle = r(x) + \langle p(x) \rangle$? I think i kind of understand why r(x) must be one of the four possibilities, but then how can we conclude that E is the given factor ring?

[cloud walrus]

ffflick

[clear hatch]

(that is, how do we know that the factor ring Z2[x]/<p> is the extension field E that we are looking for?)

[clear hatch]

[next obsidian]

It’s a polynomial over Z_2 of degree < 2

There’s only 4 polynomials

So you found a ring with 4 elements

It’s a field because it’s a quotient of a ring by a maximal ideal

And the class of x is a root of p(x) because when you plug it into p(x) you get x^2 + x + 1 + <x^2 + x + 1> which is 0

[clear hatch]

compris

[crystal vale]

Any hint for converse? How can I show G is nilpotent

It is enough to show every Sylow subgroup of G is normal

[crystal vale]

Okay thank you

[clear hatch]

Dawg why is the section on extension fields so long 💔 ive veen studying it the whole day

[crystal vale]

Yes every q Sylow subgroup of G normalise P, where p≠q

And yes we can write G = P1...P_n

Where P_i is p_i- Sylow subgroup

And N_G(P) is a subgroup therefore N_G(P) will be G

[clear hatch]

how are these basis elements obtained? i know the first few are just the original basis elements, but what about the rest

[velvet hull]

if we have field extensions E/F/K

such that ei is a basis for E over F, and fj is a basis for F over K

then the set {eifj} is a basis for E over K

[clear hatch]

im so dum 😞

[minor gazelle]

i got so addicted to this cat theory and how they are used to prove very basic group theory concepts like learning a language not because of that I can do more with it still need to build up fundamentals but the mental clarity is really profound to phrase such way

[minor gazelle]

I must say what you have suggested really led me to a fever dream the arrows collection

[knotty badger]

That’s great!

You may be interested in this paper:

https://webhomes.maths.ed.ac.uk/~tl/elements.pdf

It shows how to do categorical arguments at an “element” level without needing to draw a bunch of commutative diagrams

[minor gazelle]

❤️ ❤️ ❤️

[minor gazelle]

I will finish reading this before the summer breaks 🥰🥰🥰

I’m gonna show this thing off to my first year undergrad peers🥰🥰🥰🥰🎊🎊🎊

[knotty badger]

[minor gazelle]

I’m gonna be a legend in term of how cool it is 🥰🥰

[knotty badger]

-# what monster have i created...

[fading acorn]

[dusty raft]

Can I get some help with 37(b)

I think that matrix looks like this

[minor gazelle]

Not yet maybe monster in training I’m still too vague hopefully in a year or two I can be one monster is defo better than boring 🫠

[dusty raft]

Okay nvm I had a brain fart, it's obviously this i think

[clear hatch]

https://tenor.com/view/sticker-yung-joc-crying-meme-man-crying-meme-thebeyonderrrr-gif-828328460466032722

[distant summit]

From Halmos.

For (f), is the following correct?

(-1)(-1)=1 iff (-1)(-1)-1=0 iff (-1)(-1)+(-1)=0 iff (-1)[(-1)+1]=0 and since the final statement is true, it follows that (-1)(-1)=1 is true

[crystal vale]

because we know that |HK| = |H||K|/ |H and K|, and we can extend this by induction

and intersection are trivial in case of sylow subgroups

[crystal vale]

I see

Thank you for pointing out my mistake

[barren sierra]

How do I prove (iii) not using the fact that for p the smallest prime dividing the order of G, any subgroup of index p is normal

using that fact seems overkill

(although of course useful to keep in mind regardless)

[rocky cloak]

So if H contains ZG, then it's normal by (ii).

If H doesn't contain ZG, then they intersect trivial which means G = ZG x H, contradiction

[barren sierra]

ah nice

thanks

[barren sierra]

wait how does the proof of this first sentence not use the fact I don't want to use 💀

if H properly contains Z(G) then H is index p and so H normal

[elfin wraith]

I think I’m just being dumb but what’s the contradiction we get if H doesn’t contain the centre?

[barren sierra]

H has order p^2 so it's abelian, Z(G) is abelian, direct product of abelian groups is abelian so G is abelian contradiction

although the fact that they form a direct product is not clear to me cause Z(G) and H form a direct product to G if and only if they intersect trivially, their orders multiply to the order of the whole group, and both Z(G) and H are normal in G

but apriori we just have Z(G) normal in G

[proud vigil]

wait i forgot the normality condition ugh

[rocky cloak]

H is always indeed p... I don't understand what you mean.

I'm saying instead of using this fact you just use exercise (ii) from your picture.

[elfin wraith]

Ahh yeah I knew it would’ve been something obvious

I was forgetting groups of order p^2 are always abelian

[barren sierra]

sorry how does (ii) show that if H contains Z(G) then H is normal

[rocky cloak]

Every subgroup of Z/p x Z/p is normal

[barren sierra]

but you don't know G/H is a group in the first place

cause you don't know if it's normal or not

[rocky cloak]

I'm taking about G/ZG

H/ZG is a subgroup of G/ZG

[barren sierra]

Oh oh I see

oh it's the fucking correspondence theorem 😵‍💫

[elfin wraith]

Many such cases

[barren sierra]

group theory is nonsense

gimme my rings and fields

[elfin wraith]

So many times I’ve been so stuck on a problem only to realise it’s second iso and correspondence spam

[barren sierra]

which one is second iso

[elfin wraith]

The intersection one

[barren sierra]

gun to my head I only know the first one is the first one

and then the others are just others

I could not tell you which is second or third or fourth lol

[elfin wraith]

There’s a fourth?

[barren sierra]

https://en.wikipedia.org/wiki/Isomorphism_theorems

supposedly

[elfin wraith]

Oh 4th is just the correspondence theorem

[barren sierra]

lattice theorem good

I would recommend unskipping

[thorn jay]

it's because its an isomorphism of lattices

[elfin wraith]

Isn’t the lattice theorem just another name for the correspondence theorem?

[barren sierra]

ya

but lattice is a cooler word

[elfin wraith]

I’m the same in fairness, I’m aware that lattices are a thing, but not much more than that and it seems to have done me well enough

I think enpeace shills for them because they’re useful in UA, but im yet to see them in “normal” algebra

[barren sierra]

UA?

[elfin wraith]

Universal algebra, enpeaces area of expertise

[clear hatch]

https://tenor.com/view/drug-anime-girl-gif-26426197

[coral spindle]

I hope you'll be contributing to this channel more in the future rather than just posting gifs!

[rocky cloak]

I mean the lattice of ideals or the lattice of subgroups appears all the time.

[thorn jay]

i believe that you guys do use them, but still just in the language of the specific fields

[south patrol]

Lattices are cool

[karmic moat]

I remember looking at the subgroup lattice of Sn in my rep theory class

For some n

Or maybe it was the Bruhat order on Sn

[thorn jay]

a good example is the Jordan Hölder theorem which has a nice analogue in modular lattices using, instead of isomorphic factor groups, so-called projective intervals

[south patrol]

Main thm of Galois theory is an isomorphism of lattices

Gotem.

[thorn jay]

for example, yes

closure operators have close links to complete lattices, and closure operators pop up everywhere

[elfin wraith]

Maybe I have secretly learned about them at some point without really knowing then

[rocky cloak]

It's just a poset with supremums and infimums

[thorn jay]

finite supremums and infimums

[elfin wraith]

Yeah reading up on them I’ve definitely used them I’ve just never explicitly studied them

[thorn jay]

arbitrary ones make it a complete lattice

[elfin wraith]

The best line on the Wikipedia page may be “a bounded lattice can be thought of as a commutative rig without distributivity”

I love an X can be realised as an adjective soup Y statement

[thorn jay]

that is a very useless statement

[elfin wraith]

I know, it’s literally just the definition lol

It brings exactly 0 additional information, but I find it amusing none the less

It would possibly be slightly more meaningful if people cared about rigs

[thorn jay]

i think what sucks is that, in the algebraic structures people actually care about, most if not all tools that UA gives you are either trivial or have been known

people just dont study new algebraic structures anymore 😔

[barren sierra]

Bruhat order mentioned

[elfin wraith]

I don’t think that’s strictly true, I know there’s a prof at my UG uni who studies a relatively new algebraic structure, like 2008 or something they were defined, which still has a lot of “low” hanging fruit and tough problems

[thorn jay]

oh whats it called?

[elfin wraith]

And as much as I joke about people not caring about rigs, there is some interest in rig categories in quantum computing circles

[thorn jay]

probably something from alg logic, I know BCK algebras are of interest

[elfin wraith]

Braces, they’re a weird object, I don’t know much about them but my good friend did his UG thesis on something to do with them and got a slightly novel result

[thorn jay]

oh yeah, theyre related to the Yang Baxter equation, somehow

[elfin wraith]

Yeah the prof at my old uni was interested in set theoretic solutions to Y-B

[thorn jay]

id like to look into it but all the stuff with hopf algebras and coalgebras is scary

[elfin wraith]

I still don’t really know what Yang Baxter is, we did some stuff with it in my quantum computing course, my friend mentioned them in his presentation/defence for his UG presentation

[elfin wraith]

I know a little about Hopf algebras and it’s not too bad

At least what I know, the basics are approachable at least

[thorn jay]

maybe i just get scared easily

ill put it off for later though already got enough to do

[elfin wraith]

The book Categories for Quantum theory introduces them in a really love way

[thorn jay]

oo

[thorn jay]

going to read through an AG book with a couple friends

[elfin wraith]

It’s a bit of a different perspective of course but they’re built up quite naturally, and after discussing those is where the book starts to get more into quantum rather than just monoidal categories, but it builds up to them nicely

[thorn jay]

thank you for the recommendation!

[elfin wraith]

Your welcome! I’m always happy to shill for that book I really enjoyed it

[thorn jay]

🔥

[rocky cloak]

A hopf algebra structure is exactly what you need to make Mod R monoidal abelian.

[thorn jay]

you mean if A is a hopf algebra then Mod A can be given a canonical abelian monoidal category structure and vice versa?

[south patrol]

Does monoidal abelian = symmetric monoidal here

[clear hatch]

Me too 🥹

[rocky cloak]

Nope, but a Hopf algebra is (quasi-)triangular if it's category of modules is (braided) symmetric monoidal.

[elfin wraith]

Category theory is such silly adjective soup

[knotty badger]

And also really fun :D

[elfin wraith]

That course I took on categorical QM spent a long time talking about symmetric monoidal braided dagger categories before we introduced compactness and got to chop that down

Just a lot of words

[thorn jay]

does monoidal abelian mean that the functors X ⊗ - and - ⊗ Y (where ⊗ is the monoid bifunctor) are additive?

[rocky cloak]

I would say they need to be exact, but I guess definitions vary

[thorn jay]

right, so exactness guarantees that equivalence you mentioned

[dull ginkgo]

Hey chat, general question and yall have unbelievable intuition

Why are irreducible representations (i.e G mapping into Aut(K) for some object K with no invariant subobject) so important?

Are there decomposition theorems or the like that make them important besides just being the “simple” of their type

[quiet pelican]

In some nice cases, yeah
If G is finite and K is a vector space over a field of characteristic not dividing the order of G, then this is the content of Maschke’s theorem

[dull ginkgo]

Interesting

That makes sense for vector spaces because you can “diagonalize” matricies into invariant blocks for finite dim vec spaces

[dull ginkgo]

Is this finite dim or arbitrary

[quiet pelican]

I believe arbitrary

[dull ginkgo]

I know there is some epic C* algebra stuff about involutions but that remains to be my next rabbit hole

I am doing functional analysis rn but my algebra roots call to me

[quiet pelican]

(When I say arbitrary, I mean “arbitrary non-topological”)

[dull ginkgo]

Epic, thank you

[rocky cloak]

(with the assumptions mico said) every G-module is a (possibly infinite) direct sum of irreducibles in a unique way

[dull ginkgo]

Interesting

[thorn jay]

In general I believe they aren't much important, unless subobjects of K correspond to quotient objects. If you have a notion of "normal" subobjects, then they are of some importance because then you can have a thing similar to a Jordan-Hölder theorem, although the definition of decomposable would then be no invariant normal subobject

this is the case in universal algebra, at least

since Jordan-Hölder is just lattice theory I have trouble imagining it not holding in general categories with a notion of normal and quotient objects

[barren sierra]

been bashing my head at this one for a while

only thought that comes to my head with the non-normality of H is that the conjugation action by G on H is non-trivial

but then from there idk what other things I can deduce / idk any other paths of attack

[south patrol]

Oh lol by monoidal abelian you mean like monoidal (abelian category)

sure

[rocky cloak]

Instead of the conjugation action, think about the action of H on G/H.

(by left multiplication)

Then p>n is pretty straight forward.

Not sure if p=n needs to be a separate thing or you can do it in one woosh

[barren sierra]

I'm looking at the contrapositive, so that if H is simple and p divides |H| then p < n. I was able to show p <= n and now I gotta show p < n, hmmmm.

I feel a common trick for this stuff is to restrict the action slightly, so instead of looking at G/H look at G/H minus 1 element

so maybe that'll work

[rocky cloak]

Yeah, H fixes H, so really you can consider the action on G/H - {H}

[barren sierra]

yup

so then p divides (n - 1)! and that gets me everything I need

thanks

[muted lichen]

Can I have a little help with this. I know the theorem but I dont understant how to apply it

[coral spindle]

I also don't know what they expect you to do here

Oh wait, this is part (b) of a larger question. Post the full question.

[rapid cave]

The theorem tells you D5 can be embedded in S10. Do you want to construct such and embedding?

[muted lichen]

Yes

[vivid kestrel]

let $k$ be a field and choose $a, b$ from the algebraic closure of $k$, then if $a$ and $b$ have the same minimal polynomial, the fields $k[a]$ and $k[b]$ are isomorphic. does the converse also hold, that is, can we deduce that $a$ and $b$ have the same minimalpolynomial if those fields are isomorphic?

[cloud walrus]

eggman

[tulip otter]

why does the author choose 5 distinct integers instead of 6? doesnt that exclude the latter case?

[quiet pelican]

Wdym?

[tulip otter]

i mean why take the two 3-cycles to be in the form of σ=[ijk] and τ=[krs] instead of something like σ=[ijk] and τ=[trs] where all of these integers are distinct? is it because if this is done then στσ^{-1}τ^{-1} would be the identity?

[quiet pelican]

It’s done because having one overlap gives us another 3-cycle
If you have 6 distinct yeah it will be trivial

[tulip otter]

i see, is there a general way to figure out the length of a cycle that is the result of a product of cycles (for example a product of cycles with the same length)

[velvet hull]

sounds kind of hard, especially since every permutation is a product of 2-cycles

and every alternating permutation is a product of 3-cycles

[tulip otter]

ohhh i see

tysm mico and HChan

[rocky cloak]

No, consider for example a=0, b=1

[coral spindle]

For a nontrivial example, take k[sqrt(2)] = k[sqrt(2)+1]

[rocky cloak]

I guess it's also worth noting that if n=5, there are not 6 distinct integers 1 to n

[tulip otter]

true

[vivid kestrel]

oh of course, thanks

[velvet hull]

still an interesting question though, because it seems like it's a statement that should be true for "most" simple field extensions in any isomorphism class

[rocky cloak]

Maybe it seems like it at first glance to someone.

But a polynomial has very few roots, while a field typically has quite a few elements, so it would be strange if almost all of those elements had the same minimal polynomial

[velvet hull]

ah yes, its the same minimal polynomial up to automorphism of the polynomial ring

actually Im not sure if that even covers everything

[rocky cloak]

I'm not sure if that's true, but that does strike me as an interesting question

So cbrt(2) and cbrt(4) generate the same field extension over Q. And they have minimal polynomials
x^3 - 2 and x^3 - 4

An automorphism of the polynomial ring takes x to ax + b.

(ax + b)^3 - 2 = x^3 - 4
forces a=1 and b=0 right way, so not possible

[velvet hull]

it doesn't have to take it to x^3-4 exactly, because that doesn't work, but it does take to that up to a unit

[rocky cloak]

It still doesn't do that though.

Like it can take
x^3 - 2 to (ax)^3 - 2 which you can rescale to
x^3 - 2/a^3

But there's no rational a that makes 2/a^3 = 4

[velvet hull]

hmm

[rocky cloak]

It makes sense though.

Like an automorphism of the polynomial ring is just taking something to ax + b. So you would only really expect this to work if every element that generated the extension was of this form (a quadratic extension for example)

[velvet hull]

yes, and the issue is that you're allowed to multiply units and add units to the simple element itself, at the very least

and the automorphisms of the poly ring cant account for that

there's probably some more wacky things you can do to the simple element to get isomorphic extensions

[tulip otter]

now that i am solving an exercise where i have to find the sign of the given permutations, i am decomposing the permutations into products of transpositions (2-cycles). I noticed that the number of 2-cycles in the product is less than the length of the original cycle by 1. That seems to be true in general, is it?

I think an induction argument can be made to show this

[velvet hull]

it doesn't have to be

the 2-cycle product is not unique

but if you assume that no cancellations can happen then it sounds true

but you don't need that for your exercise though, once you've written out the permuation as a product of 2-cycles you are done

just count the number of swaps

[tulip otter]

by no cancellations you mean that by assuming that there is no even number of the same 2-cycle multiplies with each other in the decomposition?

[velvet hull]

there is no string of 2cycles in the middle of the product that cancels to 0

[tulip otter]

yea just count the number of 2-cycles, but i thought about that because i remembered my question earlier

[velvet hull]

or can be written out as a shorter product of 2 cycles

[tulip otter]

the only way for this to happen is to have 2 transpositions which are inverse to each other multiplied together right?

[velvet hull]

I'm not sure

and I don't think so

[tulip otter]

ohhh i see, tysm.

[south patrol]

I wouldn't say it is entirely standard nptation but it sometimes means that

[rocky cloak]

I would think no mostly because why would you ever need to talk about Z without 0

[velvet hull]

what if you hate monoids

[vocal pebble]

It is a multiplicatively closed set

[rocky cloak]

But 1 and -1 are the ones with multiplicative inverse

But yeah, if I just saw Z* I would probably think it meant {±1}

[cloud walrus]

mq in his algebra/stats arc

[rocky cloak]

For a field F, F* definitely means F without 0

[vocal pebble]

For a ring R, R* is sometimes used to denote the group of units

[velvet hull]

but yeah as asteroid said, the real reason is to want to localize Z at {0}

[crystal vale]

is there any better proof of centre of Gl(n,R) is scalar matrix, R is real number here, i know we have to compute it with elementary matrices, but it goes computational?

yes

and it has to be invertible

i got it, mse , it is helpful if i find AB = BA, where i assumed A is in centre and B is elementary matrix first row is multiple of some non-zero number

[rocky cloak]

You can cut down on a lot of computation by also thinking about non-invertible matrices.

Like say E_ij is the matrix with 1 in position i,j and 0 elsewhere.

Then E_ii A kills everything but row i and A E_ii kills everything but column i. Then notice A commutes with E_ii iff A commutes with I+E_ii which is invertible.

This then gives you that A is diagonal. Commuting with E_ij says that element i and j on the diagonal are equal

[crystal vale]

actually yes you told me about this but here group is Gl(n,R) so how can i take non invertible matrices

[rocky cloak]

I mean matrices dont stop existing just because your considering the invertible ones

[crystal vale]

oh A commutes with Eii iff A commutes I + Eii

[untold torrent]

I didn't understand the 4 symbols mean?

i know what are even permutation

Which is formed with odd cycle

And identity too

[crystal vale]

asking in you in S4

[rocky cloak]

probably permuations of the symbols 1, 2, 3, 4

"begin with 1" is less clear. Meaning in cycle notation maybe...

[untold torrent]

So there even will be

I, cycles made of (12)(34)

And (123) types

(12)(34) is even?

But how can I sure it starts with 1?

[rocky cloak]

I think you need to allow things like

(1)(234)

for one of the oprions to be correct

[untold torrent]

6 is not?

[rocky cloak]

what about (12)(34)?

[untold torrent]

Yeah that is what i am confused

[rocky cloak]

Whats the confusion?

[crystal vale]

i got it, thank you

[untold torrent]

Thanks

(12)(34) will be in or out?

[rocky cloak]

well it has a 1 on the left

[elfin wraith]

Depends on your convention for cycle notation I guess lol

[rocky cloak]

I think it mostly depends what "begins with 1" is supposed to mean

[elfin wraith]

It’s a vaguely worded question but yeah I would assume (12)(34) to be in

[untold torrent]

Then i will go with 12

Okay guys guys will you help me how do we make Cycles?

I meant i know the partition

S4

1-1-1-1 (identity)

(13)

(2)(2)

(4)

[elfin wraith]

I guess it’s 12 if you allow the identity cycle on (1), that feels weird to be though but this is the issue with poorly worded questions

[untold torrent]

Okay so i want to make even permutation of length 3 cycle

(1)(3) where one element map to itself

How many of it?

[rapid cave]

choose which element to fix, then there are 3 3-cycles you can add

but this way you overcount the identity so you need to remove 3

total 9

[untold torrent]

How do we make it by hand?

Any pattern?

(123)(4)
(132)(4)

(234)(1)
(243)(1)

(134)(2)
(143)(2)

(124)(3)
(142)(3)

6

[rapid cave]

8

yeah identity doesn't count mb

[distant summit]

Maybe better to ask in linear algebra

I would like some clarification on the terminology in C.

Let a and b be scalars and x and y be vectors.

Associating the pair (a,x+y) to a(x+y) is a map (-,-): FxV -> V
Associating the pair [a+b,x] to (a+b)x is a map [-,-]: FxV-> V

So are we calling (-,-) "multiplication by scalars" and [-,-] "multiplication by vectors"?

The terminology doesn't seem great because in both cases an element of FxV is sent to an element of V, so it seems like you could plausibly call both of those multiplication by scalars and both of those multiplication by vectors

[thorn jay]

it's about what object is fixed

multiplication by a vector is a function from F to V, and multiplication by a scalr is a function from V to V

[white oxide]

What am I missing here? If f: A -> B is injective (since the sequence is short exact), isn't the first condition satisfied since injective maps have left inverses?

[velvet hull]

they have left inverses that are set functions

[white oxide]

Oh, is it because the inverse is not always a homomorphism

[velvet hull]

the lemma is saying that it's not just a left inverse, but a left inverse that is also a group homomorphism

[white oxide]

Ah okay, thanks

Is the correct approach here to use the splitting lemma then? If I can construct a suitable linear left inverse for the first map

Wait nvm that doesn't seem like it'll be linear

Bc I know that M' \cong Im f and M/Im f \cong M''

And intuitively, M = Im f \oplus M/ Im f

Or idk

[velvet hull]

you don't have to use the splitting lemma, here, you just need the fact that M'' is isomorphic to M/M'

[white oxide]

I will ponder this

[thorn jay]

i think I know of a way you can reduce this to an instance of the 9 lemma

if you've heard of that that

[white oxide]

Nope I only know of the 5 lemma

[thorn jay]

nvm

[white oxide]

I would think the argument is similar to this one?

Or maybe not

[rocky cloak]

You don't need a splitting homomorphism. But picking preimages of thing (set theoretically) can be useful.

Particularly you want to turn a generating set for M' and M'' into one for M

[white oxide]

Since $M' \cong \text{Im } f$ and $M/\text{Im } f \cong M''$, $M/M' \cong M''$. Let ${x_i}_{i = 1}^n$ be elements of $M$ whose images generate $M/M'$, and let $N$ be the submodule of $M$ generated by the $x_i$.

This feels sus bc it doesn't use the fact that M' is fg

[cloud walrus]

okeyokay

[rocky cloak]

I mean you're not done yet

[white oxide]

Wdym

Yeah I'm just showing what I have so far

[rocky cloak]

Right, so just use the fact that M' is fg now

[white oxide]

Oh maybe we can append the generators of M' to this generating set

[rocky cloak]

Ding ding ding

[white oxide]

Good man for that

[crystal vale]

we know that

1. if H char K and K normal in G then H will be normal in G.
2. [G:G] is normal in G.
3. if H char K and K char G then H char G.

so if i show G^{i+1} is char in G^i then we are done.
so we have to show for any group H, [H:H] char H.

if f:H ->H, any automorphism, so image of all commutator element is also commutator, so im([H:H)) contained in [H:H].

Now pick any commutator element xyx^-1y^-1, since f is automorphism we get xyx^-1y^-1 = f(aba^-1b^-1), where f(a) = x and f(b) = y, so [H:H] contained in im([H:H]). Hence [H:H] char H.

is that correct proof?

[velvet hull]

lmao I forgot about the H char K lemma, I need to write that down

[abstract spear]

How can you use fermat’s little theorem in groups?

[elfin wraith]

I’m not sure how useful it is in group theory specifically, though i can definitely see it being used in some proofs about finite groups, I just can’t think of any off the top of my head. It’s really fundamental in number theory though and in cryptography, it’s how one of the early public key systems works

RSA iirc, but it’s been a while since I did anything with that so I could be misremembering

I do remember having to implement whatever one it was in sage for a class though

[abstract spear]

I just walked through the proof for it and I was wondering how the number a is interpreted with respect to finite groups.

[elfin wraith]

I mean modular arithmetic is just working over the groups Z_n so I guess there’s an “application” to groups

I’ll try to rack my brain for a non trivial example where it comes up in some proof about finite groups because I’d bet there’s something

[abstract spear]

If I am interpreting it correctly, a is just some integer while we consider the multiplicative group Z_p.

Or is a in the group?

[dim wind]

This I think

[velvet hull]

you can prove fermat's little theorem very easily with groups, if that's what you mean

[rocky cloak]

Fermat's little theorem is essentially a special case of Lagrange's theorem which says that the order of a subgroup divides the order of the group.

This comes up quite frequently in group theory.

[rocky cloak]

You think of a as an element of the multiplicative group of Z/p yeah

[elfin wraith]

Oh yeah lol that’s a much better example

For some reason in my head the generalisation of the little theorem was just the totient theorem

[void cosmos]

just confirmign here:

L = ad(L) , given L is semisimple

(this is abuse of notation, this is an isomorphism under x -->ad(x))

every endmorphism of L (over an algebrically closed field) can be decomposed into a semisimple part and a nilpotent part

these parts are derivations and hence they are in Der(L)

and so every element in L corresponds to a semisimple part (semisimple part of its image in ad(L) and a nilpotent part

is this the "abstract jordan decomposition" of an element? just using that x can be thought of as ad(x) and then decomposing it as a matrix in End(L)?

if L is a linear lie algebra, then every element can be decomposed directly (jordan)

and it is a theorem that these parts coincide ? is that what this paragraph is trying to say?

[tulip otter]

what is meant by translation on cosets here, is it the homomorphism T_y:{xH}->{xH} defined by T_y(xH)=yxH?

[coral spindle]

I don't know what you mean when you say "homomorphism" here since they aren't algebraic structures (traditionally)

Translation on the cosets of H (a subgroup of G) is the action of G on the cosets of H where g sends some coset aH to gaH

We can call it translation because, if you like, if you consider a vector space V and a subspace W then the cosets of W are translations -- literally translations, geometrically -- of W

[tulip otter]

is that because H is not necessarily normal ?

[tulip otter]

actions arent yet inroduced in the textbook, but from the definition that i saw this action is a homomorphism G->Perm(xH), but in that case how would it apply to A_4 and S_3? I mean S_3 doesnt contain Perm(xH) does it?

[coral spindle]

A coset is not an algebraic structure (traditionally). The set of all cosets of a subgroup may be a group, indeed if the subgroup is normal

[coral spindle]

This action is a homomorphism G -> Perm({gH | g in G})

[elfin wraith]

Traditionally?

[coral spindle]

maybe you call the set {gH | g in G} by the name G/H

[tulip otter]

well a coset of H isnt but the set of cosets of H is if H is normal no?

[coral spindle]

You can see cosets as heaps. I anticipated enpeace um acktually-ing me

[coral spindle]

...you read the full message right

[tulip otter]

I read it, but your first message seemed weird after you sent this

because i was referring to {xH} the set of cosets of H

and not one coset

[coral spindle]

Well that's not correct notation

But in any case, you still cannot guarantee that the set of all cosets is a group, so you cannot talk about a homomorphism between these things unless you mean a homomorphism of G-sets

[elfin wraith]

Ah, I see, another random ass algebraic structure

Would be amusing to hit your professor with a heap homomorphism in intro group theory

[tulip otter]

i mean i probably should've written it as {xH: x in S_3} but i wouldnt call this notation incorrect tbh (i have actually seen it in a textbook before so i assumed that it might be understood like this)

[coral spindle]

Heaps are just unpointed groups. Pointed heaps are groups in a precise way

I wrote an imaginary exercise based on this once... someday...

[tulip otter]

i see yea i get your point (not the G-set part because idk what these are yet)

[elfin wraith]

I actually have heard of heaps but I can’t think of how, some sort of alggeo context maybe?

[coral spindle]

perhaps you're thinking of stacks? I don't think heaps matter at all lmao, except perhaps as torsors

[coral spindle]

Exercise: we say that a subset X of a group G is 'good' if
(1) it is nonempty
(2) if a, b, c are elements of X, then so is ab^-1c.
Show that X is a coset of a subgroup of G.

[elfin wraith]

No I’ve heard of torsors too so maybe I’ve just seen someone here ramble about them

[coral spindle]

Torsors are heaps in a nice way

[elfin wraith]

This seems fun, Ill maybe do this tomorrow

[tulip otter]

so in the case of the exerise that i sent it would be A_4->Perm({gh| g in A_4})? but in that case how is this a map A_4->S_3 although Perm({gh| g in A_4}) isnt a subset of S_3?

[coral spindle]

I'm going to write G/H instead of {gH | g in G} since this is standard anyway

[tulip otter]

yes i am aware of this notation

[coral spindle]

but in that case how is this a map A_4->S_3
Perm() of any set with 3 elements is isomorphic to S_3.

[tulip otter]

i thought you were avoiding it because H may not be normal

[coral spindle]

I was avoiding it since it can confuse people who might think that we can only write G/H if H is normal

[tulip otter]

so maybe me

[coral spindle]

I do try not to confuse people!

[tulip otter]

ah sure

i see tysm for your help

[coral spindle]

Best of luck with the book!

[tulip otter]

tysm, have a great day/night

[barren sierra]

pretty stuck in this one 😵‍💫

2010 = 2 * 3 * 5 * 67 and there's a unique Sylow-67 subgroup

A group of order 67 is supersolvable clearly as 67 is prime

And so I think it suffices to show that a group of order 2 * 3 * 5 = 30 is supersolvable

not sure how to do that

[quiet pelican]

How many sylow 5s do we have?
||Either 1 or 6||
||If we have 1, then we’re done (prove it)||
||If we have 6, then we’ve accounted for 24 non-trivial elements||
||So we have one sylow 2 and one sylow 3, so we’re done||

[barren sierra]

ah

[elfin wraith]

That’s a nice problem I like that, the usual Sylow procedure but not quite as obviously so

[south patrol]

ye from a strong form of the correspondence theorem

[proud vigil]

in general, groups of size pqr, for primes p < q < r , have a normal subgroup of size qr, i think that makes this super speedy

[barren sierra]

yea that got me to the end

[barren sierra]

wdym strong form?

[south patrol]

Oh I just mean like you need that normal subgroups correspond to one another

[barren sierra]

oh yea

[south patrol]

r being a prime feels bad aha

But this is a nice fact!

any hints for how to prove it?

[barren sierra]

I think you just Sylow bash

[proud vigil]

yaa

[barren sierra]

you need distinctness also?

[proud vigil]

yeah i think so

the inequalities are important here, u force a massive number of sylows

[barren sierra]

yea

[south patrol]

ah i have seen that before then

i think

[barren sierra]

can't wait for this prelim exam to be over so that I can forget finite group theory again

[south patrol]

was in my first year course i think for some reason

iirc

i think there is a weird elementary way w/o sylow or smth

[barren sierra]

I'd be curious how to show that without Sylow stuff

[south patrol]

by counting carefully

[barren sierra]

sounds like Sylow with more steps

[south patrol]

actually maybe i am misremembering and we did have sylow lol

lol

[south patrol]

lmao

it is giving me weird flashbacks cause at my uni our first year (undergrad) exams are called prelims

[dull ginkgo]

Yeah the supersolvable part probably boils down here to the fact that the order is a product of distinct primes

[barren sierra]

yea

that was basically it

[barren sierra]

Is there a way to finish this last part other than showing that the groups of order 12 that aren't A4 have exactly 1 subgroup of order 3?

[proud vigil]

I think you can instead constructively show that having 4 subgroups of order 3 implies A4

in that setting, we can define a conjugation action on the 3-sylow subgroups, and we can just think of these as permutations, so if the action is injective (Im pretty sure it is but I’m struggling to prove this lol) then our group is isomorphic to some subgroup of S4 of size 12.

We can just compare orders of elements after that, we know there’s 8 elements of order 3 from the 3-sylow groups and 3 unknown orders from the 2-sylow group.

there’s only 8 elements of order 3 in S4 (4 choose 3 * 2, there aren’t enough elements to permute on to get order 3 otherwise). After that, combinations of these get us the 3 order 2 permutations, which fills in the entire permutation group and this is just A4

this is… more involved than i thought 😭

I also still don’t actually know if that action is injective or how to prove that

[barren sierra]

Yea hm I guess the solution is to just check the cases for the other possible groups

Reason I asked is because I forgot about one of the groups of order 12 💀

[proud vigil]

Okay I just totally forgot about that last part of the theorem

But if 4 = G : N(P) where P is any of the 3-sylow subgroups then N(P) = P, but if the kernel is a subgroup of every N(P), then the kernel must be trivial

Therefore, the conjugation action of G on the 3-sylow subgroups, or the homomorphism from G to S4, is injective

And then the 8 order 3 elements specify the entire group structure

And it has to be A4

And now you don’t have to check any other order 12 groups

@barren sierra does this check out I have very little confidence in this

I also don’t know what this motivates for other problems beyond “conjugation actions are randomly useful”

[barren sierra]

Hmmm I think it checks out

[barren sierra]

Yea 😵‍💫

[crystal vale]

So in p-group G, we can always find a normal subgroup of each divisor of |G|.

We can do by induction on order of G. If Z(G) is G then we are done, if not then apply induction hypothesis to G/Z(G) because it has less order than |G|. And the pull back the subgroup of that order.

I think here I have to be careful, if Z(G) has order p^k, then for any p^m, 1≤m≤k, we can get normal subgroup, for p^n, n>k, we need to do G/Z(G) stuff

[south patrol]

Yeah last paragraph is great

I guess you could also pick any nontrivial element g in the centre with order p and consider G/<g>

[crystal vale]

yes that would also be great

[crystal vale]

upper central series looks good, why do we need lower central series?

if i have to prove G is nilpotent then every subgroup of G is nilpotent.
Say first, G is finite group so G can be written as direct product of its sylow subgroups because G is nilpotent.

say G = P1 \times \ldots......\times P_n, then any subgroup of G will be of the form H_1\times \ldots ...\times H_n, where H_i is a subgroup of P_i.

Since P_i is p group so H_i will be p group so H_i will be nilpotent, hence H will be nilpotent because product of nilpotent groups is nilpotent.

but i want to prove the same when G is infinite group, any hint?

[rocky cloak]

The more the merrier

[rocky cloak]

Maybe compare one of the central series of G with one for H

[crystal vale]

but what is the relation between Z(H) and Z(G) ?

[south patrol]

Z(G) cap H is contained Z(H) ye

[crystal vale]

yes Z(G) intersection H contained in Z(H)

[crystal vale]

is there any good resource where i can get equivalent between upper central series and lower central series?

[crystal vale]

from lower central series it is easily comes that every subgroup of nilpotent is nilpotent

[rocky cloak]

The argument is essentially the same for the upper central series, no?

Like
H^i < G^i
is one and
Zi(H) > Zi(G) \cap H
The other

[crystal vale]

yes

i proved first part already, now second part is if gcd( |a|, |b| ) = 1 then ab = ba. So it implies if P1 and P2 are two distinct p sylow subgroup and q sylow subgroup respectively then P1P2 is subgroup because P1 commute with P2. Similarly for P1P2 and P3 commute each other then P1P2P3 subgroup,..., so G = P1P2P3...P_n.

therefore N_G(P_i) will be G, right?

[velvet hull]

well this just follows from the fact that a finite group is nilpotent iff it is the direct product of its (unique) sylow subgroups

[crystal vale]

yes

[crystal vale]

if i have to show G^i is characteristic in G, so induction on i, works

[crystal vale]

So G/G" acts on G", how it helps me here?

[crystal vale]

How do I show any normal subgroup H of nilpotent group has non trivial intersection with centre ?

Any hint?

[void cosmos]

[G,H] is a subset of H

[G,H]^n is a subset of H^n

[crystal vale]

yes

so it gives [G:H] is nilpotent

i got it, thank you

[barren sierra]

Is there a definition of length l(G) that doesn't assume that all composition series have the same length?

[karmic moat]

If a group admits more than one composition series, then all of the composition series are the same length by Jordan-Holder

[barren sierra]

right

[karmic moat]

And thus definition of length is well defined

There's no assumption that they're all the same length

That's given to you by Jordan-Holder

[barren sierra]

so I guess like something that doesn't use Jordan-Holder if that makes any sense

I only ask because this is from a past qual

and one of the instructions is "don't use theorems that trivialize the problem"

and that feels like it trivializes the problem 😵‍💫

[karmic moat]

Hmm I don't think Jordan-Holder trivializes this problem

[elfin wraith]

Can you even do it without Jordan-Holder? Dont you need to know theyre the same length for it to be well defined anyway

[karmic moat]

I guess you could define the length of a group to be the min length of a composition series of it

Then Jordan-Holder gives you that actually every composition series has the same length

[barren sierra]

well that was my question, is there an equivalent definition that works that doesn't use length to be well defined

[barren sierra]

Hm I guess

[karmic moat]

Iirc that's how atiyah-macdonald defines length

Idk I think some ppl will define Jordan-Holder then define length as a consequence of it

[barren sierra]

Aityah Macdonald talks about groups?

[karmic moat]

And some define length first and then use Jordan-Holder

[elfin wraith]

That might actually have been how my group theory course did it iirc

[karmic moat]

Atiyah-Macdonald talks about modules

[barren sierra]

Oh wait nvm JH is a thing for modules also

yea

[karmic moat]

Yeah

[rapid cave]

thats how they did in my class

[elfin wraith]

It also applies to artinian modules

[rocky cloak]

The infimum of lengths of all composition series is a natural one

[barren sierra]

Yea I wrote out a proof using that one just now

[rocky cloak]

This is the definition used by my professor when I first learned about length

[regal mango]

A little help would he good for Q2

[rapid cave]

if some $g \in G$ where of infinite order, then $<g> \cong \mathbb{Z}$. In that case the intersection in the question is contained in the intersection of all nontrivial subgroups of $<g>$ which is trivial.

[cloud walrus]

ExpertSqueeSQUEE

[tardy hedge]

HomR(R,S) is isomorphic to S as S-modules?

if S is a ring

[elfin wraith]

oops I was thinking of R endomorphisms ignore what i said before, Im not sure if it also holds for any rings R and S, I wouldnt think so but Im not sure I could prove it

[elfin wraith]

I found this though, maybe of use: https://math.stackexchange.com/questions/3706239/when-does-rm-hom-sr-s-cong-r

[barren sierra]

My solution was use the fact that p groups are nilpotent

[dull ginkgo]

Is this Sisyphean study session for abs alg for quals lmfao

[barren sierra]

and then prove the more generally true statement that for all subgroups H of a nilpotent group G that N(H) is strictly larger than H

yes

[dull ginkgo]

Rip bozo

[barren sierra]

anyways I can't figure out a proof that doesn't use this fact

[barren sierra]

can't wait till I'm done with quals to forget group theory

unless I fail qual and have to take the algebra I course

[dull ginkgo]

Seems in similar vain to the proof of Sylow in The first place

[barren sierra]

?

[dull ginkgo]

The whole gist of Sylow is a group acting on it’s subsets of a given prime power size

[barren sierra]

ah right

[dull ginkgo]

Yeah

[elfin wraith]

I think I may have done this or a similar problem at some point in my group theory class, IIRC you assume that N(Q) = Q and you see that Z(G)<Q and then do some more waffeling

I think you then quotiented out by the centre and do some tedious argument with results in a contradiction somwhere down the line, but I only vaugley remeber this, its been a miniute haha

[dull ginkgo]

Why are elementary group theory proofs so fucking ass

[barren sierra]

I really hate them

[thorn jay]

lotta computation I assume

the real fun part imo starts with subgroup series and stuff

I'm a sucker for filtrations

[dull ginkgo]

Can you do do this in an orbit stabilizer way

[rocky cloak]

Consider the action of Q on P/Q by left multiplication.

The fixed points of this action is N(Q)/Q.

Since Q is a p-group the number of fixed points is equal to |P/Q| mod p. So 0 mod p.

Since Q is a fixed point there is at least 1, so [N(Q):Q] is at least p.

This is also how I would prove the sylow theorems

[dull ginkgo]

Like P acts on its power set by conjugation, and this action preserves order

DAMN

[dull ginkgo]

SOULREAD

[dull ginkgo]

A lot of finite group theory is a group acting on its power set via multiplication or conjugation, huh

[rocky cloak]

Well, no power set in sight here, but groups acting on stuff is a big deal yeah

[dull ginkgo]

Or it’s quotient by a subgroup

I interpret the normalizer as the stabilizer for conjugation on a group’s power set

Never thought of it like that, pretty peak

[white oxide]

Here, is S^{-1}A/S^{-1}a a S^{-1}A module, and Ann(S^{-1}A/S^{-1}a) = Ann(S^{-1}M)?

This is my attempt at filling in the details, can somebody let me know if it looks alright so far? Let $a/s \in \text{Ann}(S^{-1}M)$, so that $am/st = 0$ for all $m/t \in S^{-1} M$. Now, $S^{-1}A/S^{-1} \mathfrak{a} \cong S^{-1}(A/\mathfrak{a}) \cong S^{-1}M$, via $\frac{a}{s} + S^{-1} \mathfrak{a} \mapsto \frac{a + \mathfrak{a}}{s} \mapsto \frac{am}{s}$ (where $m$ generates $M$). Therefore, if $\frac{b}{t} + S^{-1} \mathfrak{a} \in S^{-1} A/S^{-1} \mathfrak{a}$, $\frac{ab}{st} + S^{-1} \mathfrak{a} \mapsto \frac{ab + \mathfrak{a}}{st} \mapsto \frac{abm}{st} = 0$, since $\frac{a}{s} \in \text{Ann}(S^{-1}M)$. Since isomorphisms are injective, $\frac{a}{s} \in \text{Ann}(S^{-1}A/S^{-1} \mathfrak{a})$, so that $\text{Ann}(S^{-1}M) \subseteq \text{Ann}(S^{-1}A/S^{-1} \mathfrak{a})$. Conversely, let $\frac{a}{s} \in \text{Ann}(S^{-1} A /S^{-1} \mathfrak{a})$. If $\frac{m'}{t} \in S^{-1} M$, then $m' = b m$ for some $b \in A$. Then $\frac{am'}{st} = \frac{a bm}{st} \mapsto \frac{ab + \mathfrak{a}}{st} \mapsto \frac{ab}{st} + S^{-1} \mathfrak{a} = S^{-1} \mathfrak{a}$ since $\frac{a}{s} \in \text{Ann}(S^{-1} A/ S^{-1} \mathfrak{a})$. Since isomorphisms are injective, $\frac{am'}{st} = 0$, whence $\frac{a}{s} \in \text{Ann}(S^{-1}M)$. It follows that $\text{Ann}(S^{-1}M) = \text{Ann}(S^{-1}A/S^{-1} \mathfrak{a})$. It remains to show that $\text{Ann}(S^{-1}A/S^{-1} \mathfrak{a}) = S^{-1} \mathfrak{a}$.

[cloud walrus]

okeyokay

[rapid cave]

But you know $S^{-1}M \cong S^{-1}A/S^{-1}\mathfrak{a}$ so of course they have the same annihilator

[cloud walrus]

ExpertSqueeSQUEE

[white oxide]

Well I mean set theoretically

Of course they're isomorphic but as sets they could be different a priori

[rapid cave]

If some $x \in A$ acts on $S^{-1}M$ by 0 then it also acts on $S^{-1}A/S^{-1}\mathfrak{a}$ by 0 and vice versa.

[cloud walrus]

ExpertSqueeSQUEE

[white oxide]

I see

Why is it the case that the annihilator of S^{-1}A/S^{-1}a is S^{-1}a? I can see the \supset inclusion but not the \subset

Never mind

We can just multiply it by 1/s

[rapid cave]

its the same as Ann(A/a) = a

[white oxide]

Wait

[rapid cave]

btw I answered earlier your question on #elementary-number-theory

[white oxide]

Yeah I saw, thanks

Switched over to comm alg tho

[void cosmos]

sl(3,F)/Z(sl(3,F)) , F char 3

why is this a non-semisimple lie algebra?

i can't think of any solvable ideal

id say any solvable ideal must stablizie some flag with x.V_i /subset V_(i-1) .. as in that elements of I can be put in upper triangular form ... the diagonals must sum to 0 tho

the diagonal can't be (1,1,1) and so i thought maybe the strictly upper triangular matrices.. but this isn't an ideal of sl(3,F)/Z(sl(3,F))

[dull ginkgo]

ORE LOCALIZATION

[white oxide]

What

[south patrol]

Sorry i didnt see a reply to ur stuff

And repeated stuff that has been said

[white oxide]

oh lol

What's stopping f(A)/q being zero?

[velvet hull]

An ideal q in a ring A is primary if q != A (...)

[white oxide]

Sorry my brain is not working right now

I thought q is a primary ideal of B

I'm looking at the sentence "Also the contraction of a primary ideal is primary"

And the logic which follows it

[velvet hull]

I can look at the sentence too

I can look in general in fact

[white oxide]

I still don't understand why A/q^c is not zero since we don't have q^c primary immediately lol

[velvet hull]

oh yeah I don't think it's true actually

it's an omitted assumption, I think

it's not something you should worry about, most of the time you can be pretty confident your ideal is a proper ideal

[white oxide]

So it's only true under the condition that q^c =/= A?

[velvet hull]

yes

[karmic moat]

contraction of primary ideal is primary, so if q is a primary ideal of B, then q^c is a primary ideal of A. you can verify this by just looking at the definition which uses elements

[thorn jay]

it's a general fact that there is an injective homomorphism from A/I^c to B/I. And as an ideal is primary iff it's quotient has no non-nilpotent zero divisors, a fact which then also holds for all subrings, I^c is primary if I is

in particular, I^c won't be A if I isnt B, because there is no homorphism from the zero ring to any other ring

[velvet hull]

I had to pull up Atiyah to confirm this, and the issue is that the author is seemingly implying that this is true for arbitrary ring homomorphisms

[thorn jay]

wdym?

[velvet hull]

there is no injective assumption

but to be fair for integral extensions and all that jazz this is not an issue

so whatever

[thorn jay]

you mean that there is an embedding of A/I^c in B/I?

that's a general fact, hell it's technically universal algebraic

[velvet hull]

no, the issue is that if you drop the injective assumption the contraction might be the whole ring

I'm not 100% convinced it goes away even if you assume injectiveness

[thorn jay]

how can there be an embedding of the zero ring into a nonzero ring

or a homomorphism at all

[velvet hull]

I'm getting confused, this is not a detail I really care that much about lmao have a nice day

[thorn jay]

I^c can't be A, because else there would be a homomorphism from 0 = A/I^c into B/I =/= 0

that's impossible, ergo, I^c is never A when I isn't B

using elements, I guess you can see that, if I^c = A, then f(1) = 1 in I, so I = B

[karmic moat]

if I is a proper ideal and f : A -> B a ring hom, then I^c = {a in A | f(a) \in I}. If I^c = A then, in particular, 1 in I^c, so f(1) = 1 in I, contradiction

sniped

[thorn jay]

lmao

it's really just the same argument in disguise

[karmic moat]

yeah

i don't see any missing assumption here

[thorn jay]

man, rings are so nice

[karmic moat]

real shit

[karmic moat]

See above why this assumption isnt needed

[velvet hull]

oh, wait, lmao my prof insisted on dropping f(1) = 1 assumption when he taught us ring homos

yeah that makes sense

[thorn jay]

😭

they are teaching you LIES

and algebras

[karmic moat]

doesnt that just make it a rng homomorphism and not a ring homomorphism

[thorn jay]

it does but that's too based for some people

[karmic moat]

what a weird prof

[elfin wraith]

My linear algebra/intro ring theory had course notes written by someone who said homomorphisms don’t need to map 1 to 1 and lectured by someone who did.

That did get confusing on many many occasions

[thorn jay]

if there's no 1, then you are able to have

short exact sequences of rings

the horror

[thorn jay]

what

we're talking about mapping the multiplicative identity to multiplicative identity

[proud vigil]

are you saying "if you require ring homomorphisms to map multiplicative identities to multiplicative identities, then the difference between a homomorphism and an isomorphism is only whether or not the map is surjective"? i just want to make sure i understand what you're saying

oh wait is 1 to 1 supposed to mean like an injective map

okay that explains the discrepancy then i think LOL

[terse umbra]

Found another error in Pinter 😔 "two errors in any codeword can always be detected". What if 000000 is changed into 001001?

You'd need a minimum distance of 3 not 2

[languid trellis]

So far, I've been able to figure out the p | a case by using the fact that Z/pZ is an integral domain, so either x+y or x-y=0. I'm somewhat lost with the other case however, I was thinking that I could treat the congruence uv = a (p) 'one unknown at a time', essentially by treating only v as unknown and applying previous theory but clearly this won't work. Does anyone have any advice on where to start?

[south patrol]

So if uv = a mod p and p does not divide a, can you show that u is determined by v?

[languid trellis]

as in, do you want me to write an expression for u in terms of v?

[south patrol]

Ye

[languid trellis]

right ic

[south patrol]

Dw this is quick like

Just i wanted to point out u and v determine one anothsr

[languid trellis]

Are we assuming that uv = a (p) too

well then u = a/v (p) i guess lol

This only shows that u is determined by v when p doesnt divide a because if p |a then u = 0 or v= 0, and they are thus not related

The other thought I had was to say that (u,p) = 1 so there exists x',y' with ux' + py' = 1, then multiply by a to get uax' + pay' = a, reducing mod p, we get that ax' = v, but this is too general, as we're not using that u=x+y and v=x-y

Wait I can assume that uv=a (p) because i'm just counting solutions for variable x,y

[south patrol]

Yes but you already did the other case right

[south patrol]

Ye

So how many choices of u, v are there

I guess also of course you should deal with the case p = 2 separately lol, but that is easy enough

[languid trellis]

well then there can be at most p-1 choices for u and v because there are at most p-1 choices for u.

[south patrol]

How about the exactly p-1

Can you show that any nonzero u works

[languid trellis]

Oh can we linear algebra this actually lol

Like

[south patrol]

You are essentially there lol

If i pick any nonzero u, what should you set v to be

(You have already written it down)

[languid trellis]

Well I should set v = a/u, but the point I'm making is that if we put the system of equations x+y = u, x-y = a/u, then by linear algebra we always have a unique solution (x,y). Like the determinant is -2.

So we have exactly p-1 solutions

[south patrol]

Oh sorry sorry i didnt realise that was where you were stuck and not here

[languid trellis]

as we have p-1 choices for u

[south patrol]

Again need p not 2 but yes

Gg

[languid trellis]

Why p not 2? what goes wrong

Or maybe i should figure that out myself

Well everything goes wrong for p=2 i guess

oh we have 2 solutions for each case .. whereas this theory tells us that there should be 1 for p not dividing a

hm

[south patrol]

-2 is not invertible in Z/2

[languid trellis]

thank you broski

[south patrol]

But then you can just try like all 4 possibilities for x and y

[languid trellis]

ya

[south patrol]

Wait I am confused lol. Does this problem actually break when p = 2

Like x^2 = y^2 implies x = y in Z/2 so there should be 2 instead of 3 = 2p-1 solutions for 2nd bit

Im goated...

[thorn jay]

if only..

we could drop the "char ≠ 2" assumption in soooo many places

[languid trellis]

Well I have just tried solving x^2 - y^2 =0 (2) and x^2 - y^2 = 1 (2), and there are 2 solutions for each case. but this theory tells us the first equation should have 2p-1 = 3 solutions and p-1 =1 solutions in the second case

[languid trellis]

ya exactly

[south patrol]

I am sad cause I have smth in what I am working on and one case needs p >= 5

😭

[languid trellis]

oh how hideous

[elfin wraith]

I feel like that’s quite common, 2 and 3 are the weird freaky primes

They’re primes by default rather than by merit

[south patrol]

2 is often the most interesting prime tho smh

[languid trellis]

2 is a quad residue mod p iff p is congruent to 1 or 7 mod 8. awesome

[south patrol]

Yes very fun fact

Usually stated as 1 or -1 though aha

I forget like what is the nicest way to prove this fact

Ah yeah one is using Gauss' lemma and just explicitly computing lol

[languid trellis]

Yeah the way I know uses Gauss' lemma. Very down to earth. big fan

[rocky cloak]

Maybe this:
Consider Z[i]/p.

Then
(1+i)^p = 1 + i^p
= 1 + i * (-1)^((p-1)/2)
so is 1+i or 1-i = (1+i)*i^3 depending on whether (p-1)/2 is even.

But also
(1 + i)^p = (1+i) * (1+i)^(p-1)
= (1+i) * (2i)^((p-1)/2)

So if s = (p-1)/2 is even
2^s i^s = 1
2^s = (-1)^(s/2)
So 2 is quadratic residue iff s/2 is even iff p-1 is divisible by 8.

If s is odd you have
2^s i^s = i^3
2^s = (-1)^((3-s)/2)
So quadratic residue iff 3/2 - (p-1)/4 = (7-p)/4 is even iff 7-p is divisible by 8.

[south patrol]

Ah very nice. I guess this is basically a Gauss sums argument

this is very nice

I was wondering if you could study Z[sqrt(2)]/p directly to do this

[rocky cloak]

It would be cool, but I don't quite see how.

Like you would have to show whether it's a field only using the value of p module 8...

[south patrol]

Yeah lol

[somber sleet]

I will now change chat: If I have a ring R and an ideal P in R, what are the conditions that R and P need such that the localization of R at the ideal P, R_P, is an integral domain?

[south patrol]

The first clear criterion is that is an integral domain, then the localisation at any multiplicative set is a domain (unless it contains 0 lol), and be identified with a subring of the field of fractions of R

Beyond that I think it is a little subtle. For example e.g. Z x Z has two prime ideals corresponding to each factor of Z, and after localising at one of those prime ideals you get back Z. So you go from a non-domain to a domain

[somber sleet]

Is it thus enough for the ideal to be prime? It is clear to me that it works in case P is maximal, but what if it's not? Of course the ring must be id

[south patrol]

Yes, exactly

S = R \ I being a multiplicative set is equivalent to I being prime i guess

[rocky cloak]

I guess writing it out completely explicitly:

If there is an s not in P such that
sxy = 0
then there should be a t not in P such that either
tx = 0 or ty = 0.

I'm not sure you can reformulate this into anything nicer, but maybe

[somber sleet]

I mean 1 must be in S and ab for each a,b in S

[rocky cloak]

Which is exactly the definition of prime ideal

[somber sleet]

exac

thanks guys!

[rocky cloak]

I feel like there should be some geometric argument, at least for reduced rings.

Like the product of two rings is just the union of their spectra, so doesn't change this local condition.

And then something like C[x, y]/(xy), the spectrum looks like a cross. So locally appears irreducible everywhere except the origin. And indeed localizing somewhere other than the origin gives some localization of C[x] or C[y].

[barren sierra]

Let $G$ have a composition series and let $K$ normal in $G$. I would like to show that there exists a composition series of $G$ which contains $K$ as a term. If we consider the normal series ${e} \trianglelefteq K \trianglelefteq G$, then by Schreier refinement we can take this normal series and the one for $G$ that we know exists and find a common refinement. Thus $G$ has a composition series that includes $K$. Is there a simple proof that doesn't use Schreier refinement?

[cloud walrus]

Spamakin🎷

[velvet hull]

K has a composition series, and so does G/K