groups-rings-fields

[wet zodiac]

yeah

[drifting plover]

And the Borel is always the p-Sylow subgroup for q = p?

[coral spindle]

No, the Borel is not the p-Sylow

[delicate orchid]

^

[coral spindle]

The unipotent radical of the borel is!

[delicate orchid]

it's index 2 for p odd

I think it is the sylow subgroup for p = 2

[coral spindle]

It is in the case p=2 yes

[drifting plover]

Right, close enough 😆

[coral spindle]

But for all others it is not a p-subgroup, let alone a maximal one

[delicate orchid]

I don't follow why the char poly being x^2-2x+1 implies M is GL-conjugate (funny fusion system on SL ;3) to an upper triangular matrix

[coral spindle]

Ah I should note, they are equal for *q = 2, not p=2.

[drifting plover]

So these are just the unitriangular matrices

[delicate orchid]

yeah sorry that's what I meant

[white oxide]

here, does a_(v) denote any product of powers a_i^v_i?

[coral spindle]

yeah just wanted to clarify

[delicate orchid]

cause in F_4 you can have all sorts of wacky things on the diagonal

[rocky cloak]

Similarity is determined by minimal polynomial. Or alternatively 1 is a root, so the matrix has an eigenvector with value 1

[coral spindle]

Yeah indeed it will, at the very least, be conjugate in the infinite group GL_2(\bar F_p)

[delicate orchid]

maybe one day I should learn linear algebra

[coral spindle]

which is sufficient

[karmic moat]

my guess is a_(v) denotes the coefficients a_i x_i^v_i

[coral spindle]

That's kinda a cool observation, that for the defining characteristic Sylow structure you can just look in GL_2. Nice.

[delicate orchid]

studying conjugation in different groups on a sylow subgroup.... someone should look into that...

[drifting plover]

I guess matrices with a certain trace pretty much have the same order 😮

[coral spindle]

For SL_2 this is true but in general nah bruv

[drifting plover]

Yeah, for SL_2

[tardy hedge]

Right now in class we started groups, subgroups etc that sort of stuff and for a lot of my hw questions im kinda just pushing symbols around without really a big sort of conceptual understanding of what it means

Is that par for the course at this stage ..?

I know what it “means” but not what it “means”

[white oxide]

i don't really see how the induction hypothesis applies here - are we assuming that f(b_1, \dots b_{n - 1}) = 0 for all a_i in S_i for i = 1 \dots n - 1? i would assume that because of induction.. but then we assume that f_j(b_1, \dots b_{n - 1}) \neq 0

[tardy hedge]

Eg: C(a) = { x in G st ax=xa}

Prove c(a) is subgroup

I just … did it

[white oxide]

are you in 113

[tardy hedge]

Like , uni class code?

I am taking a course at a school in halifax Nova Scotia canad

Canada

[white oxide]

oh nvm cuz i told a student named kian to join this mathcord and he was also in an abstract algebra course at my uni

[tardy hedge]

Oh interesting haha

Yeah this server is awesome

So great to have

[delicate orchid]

so this particular subgroup is called the centraliser and it's important that it's a subgroup but you probably won't know why for a while
it's a specific example of a stabiliser subgroup

[tardy hedge]

Yeah it did say it was called that

Ok

[drifting plover]

This is very cool

How shall I credit you lol

I think you helped me twice

[rocky cloak]

The definition of a group is quite minimal, so you can prove a lot of general things about them just by pushing symbols around. After working a bit with groups and seeing how they arise in different context, you'll hopefully build up a little more intuition about them.

[rocky cloak]

Like are you writing a paper or something? Otherwise I accept s

[drifting plover]

I'm happy to give a too but yeah it's a paper

It's nothing to do with group theory but there are some group theory calculations I just can't do

[rocky cloak]

Alright, I don't feel like I did anything worth mentioning, but if you want to you can credit my username or I can dm my real name.

[drifting plover]

Probably real name is better?

[rocky cloak]

Maybe, idk. Username is funnier

Depends what kind of paper it is I guess

[drifting plover]

It's an experimental number theory paper

[karmic moat]

with a million dollars

[drifting plover]

Actually I don't remember when you helped me the previous time, but your username is very familiar

[white oxide]

wait so does $\sum a_{(v)}M_{(v)}(x) = a_1x_1^{v_1} \dots x_n^{v_n} + a_2 x_1^{v_1} \dots x_n^{v_n} + \dots$?

[cloud walrus]

okeyokay

[night onyx]

yeah just an element of A indexed by an element of N^n

[white oxide]

ohh ok thanks

[karmic moat]

yeah ^

[rocky cloak]

Probably just a very common username

[drifting plover]

Nono, it's the question I had about the splitting of the group of size p|SL(2, p)|

Turned out I decided not to include this part because the group theory got too hard for me

[night onyx]

kindof common, like when working with tensor products instead of writing e_{i_1} \tensor e_{i_2} \tensor ... e_{i_n} you often see e_I where I = (i_1, ..., i_n)

simplifies a lot of formulas

[delicate orchid]

I've even just seen I on its own when the context is clear

crazy!

[white oxide]

where did this index d_n come from?

why didn't they just put m or something

[rocky cloak]

The number of summands depends on n

[coral spindle]

d_n........... deez............ nuts..............

[delicate orchid]

wtf

[coral spindle]

yeah also what jagr said

[white oxide]

ah okay i see

thx

[rocky cloak]

I mean they probably could have called it m, but they can also call it d_n, so why not

[white oxide]

dependence tings

[rocky cloak]

I don't remember that, what was the question?

[drifting plover]

This one, and above

I'll thank everyone here who contributed if I could

[delicate orchid]

I tried

[drifting plover]

Discord et al

[rocky cloak]

Ah right

[karmic moat]

ok so im trying to understand Ext and how it plays a role in considering the depth of modules over local rings:

let $(R, \mathfrak{m})$ be a local ring. then the depth of $M$ is defined as
[ \text{depth}(M) = \inf_i { \text{Ext}_R^i(R/\mathfrak{m}, M) \ne 0 } ]
what exactly does this have to do with the notion of depth? i get that Ext, in a sense, measures the exactness of the Hom functor, but i'm not sure what Ext = 0 would tell us about the structure of M

[cloud walrus]

ana(functor)mono(morphism)

[karmic moat]

for the record, im using the projective resolution for Ext

[white oxide]

would this proof work for showing that $\bigl(f(X)\bigl)$ maximal $\implies f(X)$ irreducible? Suppose we have $f(X) = g(X)h(X)$ where $g(X)$ and $h(X)$ are non-units and hence of degree greater than or equal to 1. In particular, $f(X) \in \bigl(g(X)\bigl)$, so $\bigl(f(X)\bigl) \subseteq \bigl(g(X)\bigl)$, contradicting the maximality of $\bigl(f(X)\bigl)$.

[cloud walrus]

okeyokay

[karmic moat]

i guess my current thought is that, considering the projective resolution
[ \cdots \to P_1 \to P_0 \to R/\mathfrak{m} \to 0 ]
then taking the cochain complex
[ 0 \to \text{Hom}(P_0, M) \to \text{Hom}(P_1, M) \to \cdots, ]
then $\text{Ext}^i = 0$ implies that, for every $\varphi \in \text{Hom}(P_{i-1}, M)$ and $\psi \in \text{Hom}(P_i, M)$, $\psi\varphi = 0$, but i'm not quite sure what this tells us about $M$

[cloud walrus]

ana(functor)mono(morphism)

[tender wharf]

you have to show its a proper subset

but it's basically that idea

yeah

just change subseteq to subset lol

[white oxide]

lol ok

th

x

[tender wharf]

or if you wanna do it this way

F[x]/(f(x)) is a field

then blah blah

[white oxide]

yea true but i'll pass!

[karmic moat]

(i guess to be specific if f : Hom(P_{i-1}, M) \to Hom(P_i, M), then psi f(phi) = 0 but it doesnt matter too much)

[white oxide]

and then for the converse: suppose $f(X)$ is irreducible and that we have $f(X) \subset I$ for some proper nontrivial ideal $I$ of $k[X]$. Since $k[X]$ is a PID, $I = \bigl(g(X)\bigl)$ for some $g(X) \in k[X]$. In particular, $f(X) \in \bigl(g(X)\bigl)$, so $f(X) = g(X)h(X)$ for some $h(X) \in k[X]$. now, $g(X)$ is not a unit and $h(X)$ is nonzero, so since $f$ is irreducible, we must have $h(X) = u$ for some unit $u$. But then [\bigl(f(X)u^{-1}\bigl) = \bigl(f(X)\bigl) = \bigl(g(X)\bigl)]

[cloud walrus]

okeyokay

[delicate orchid]

I think these work

[karmic moat]

wew can u help me

[delicate orchid]

tbh with you no probably not

[white oxide]

cool

[delicate orchid]

I don't know much about ext

[karmic moat]

god dammit

🤬

me neither

[unique valve]

Sylow

[delicate orchid]

alternatively, k[x] PID => (prime <=> irreducible) so k[x]/f not a field <=> k[x]/f not an integral domain <=> f(x) = g(x)h(x) for some other polynomials

[delicate orchid]

let us journey together

[karmic moat]

im trying to read this paper
Evans, E. Graham; Griffith, Phillip: The syzygy problem. Ann. of Math.
(2) 114 (1981), no. 2, 323–333.
that i found on a list of "accessible papers in commutative algebra"

[delicate orchid]

well I know that if Ext is the zero then Hom(A, -) is exact so P is projective which is nice

[karmic moat]

i knew that i would use these to like. learn algebra and learn as igo but like. damn i dont know this

[karmic moat]

aren't all P's projective by virtue of being a projective resolution?

[delicate orchid]

oh I just assumed P was a module and started ranting immediately

yeah see you're applying Hom(-, M) there, different functor

[karmic moat]

here's the start of this madness

[delicate orchid]

this conclusion is true by virtue of being a cochain complex no?

[karmic moat]

yea

im just trying to figure out like

how it relates to M

as opposed to the cochain of Hom

[delicate orchid]

hmm I see

[karmic moat]

here's the paper

for more context

i guess im just trying to figure out what depth really tells us about M

[delicate orchid]

a fair question, I don't really know myself. Thankfully "jagr2808" is typing so we will find out shortly

[karmic moat]

if Ext =/= 0 then you have that the cochain isnt exact at the i-th position and then something something

god bless jagr2808

[delicate orchid]

pass to... homoglogoy...

[karmic moat]

mmm yes

[void cosmos]

lmfao

this is literally how the last month went when i was studying for my algebra test

[rocky cloak]

The other definition of depth is the length of the longest regular sequence on M.

If M has a regular sequence of length 1, then multiplication by r_1 is injective on M. Then Hom(R/m, M) = 0, because r_1R/m = 0.

If M has a regular sequence of length 2, then consider the short exact sequence

0 -> M -> M -> M/r_1 -> 0

And apply the long exact sequence in Ext(R/m, - ). Since these all have depth >= 1, the Homs vanish. So you have that multiplication by r_1 is an injective map

Ext^1(R/m, M) -> Ext^1(R/m, M)

Hence these must be 0. Just continuing by induction, you see that if M has a regular see of length n, then Ext^<n(R/m, M) vanishes

[karmic moat]

oh yeah ok that kinda makes sense

i was aware of the definition just not the connection between the two

[terse crystal]

24 hours of local cohomology by Isengar proves a lot equalities of different but equivalent definitions like this

[karmic moat]

oh sick

i can't find a pdf so i'll look for it at my uni library

oh nvm i found an older version

[delicate orchid]

this is actually sick

[karmic moat]

gotta wonder how jagr knows all this stuff

im almost jealous

scratch that i am jealous

[delicate orchid]

never mind knowing it how do you even remember this junk

[karmic moat]

infathomably real

[delicate orchid]

like I read a paper this afternoon and I've already forgotten 90% of the details

[rocky cloak]

The trick is to just derive it on the spot

[delicate orchid]

that's what I do 90% of the time and I get it wrong 89% of the time

[wet zodiac]

#discussion message

i dont think this proof works for numbers where its prime decomposition has numerous identical primes (i.e 4, 9, 18, etc.)

[dim widget]

It doesn't but it works out anyway

[wet zodiac]

why

[dim widget]

(Z/p^k)^* \cong (Z/p)^* \times Z/p^{k-1} for p odd

[delicate orchid]

just fyi, numbers where this isn't the case are called "square free"

[dim widget]

so \phi(p)|\phi(p^k)

You can also prove by induction that \phi(p) divides \phi(p^k)

Or use that Z/p^k \to Z/p is a surjective homomorphism of rings

many ways

[wet zodiac]

ah i see

[graceful elm]

fast question, what does this symbol represent? it's a multiplicative group apparently?

[coral spindle]

Seems like that's $F_{11}^\times$, badly typeset. It's the group of invertible elements of the field with 11 elements.

[cloud walrus]

Thifford Cleory (Mantjie)

[coral spindle]

The field with 11 elements can be represented by the integers modulo 11

The point being, it's all nonzero elements of Z modulo 11.

[graceful elm]

oh ok

so if i were asked the order of 2 in that group, that would be the same as the order of 2 mod 11?

[coral spindle]

Yes that's right. N.b. that it's the multiplicative order, so it's 2 times 2 times 2, not 2 + 2 + 2 etc

[graceful elm]

right. we've only talked about multiplicative order afai aware so that's what i'm going with

[coral spindle]

Yes indeed the additive order would not be very interesting.

Maybe you can see why :)

[delicate orchid]

;3

[graceful elm]

the additive order of an element is its multiplicative inverse, no?

[delicate orchid]

the additive group is cyclic of prime order so every non-identity element is the same order

[rocky cloak]

Sort of the opposite. The additive order is how many times you have to add to itself to get 0, but the multiplicative inverse is how many times you must add to itself to get 1

[graceful elm]

oh to get 0, to get the additive identity. i was still aiming for 1, haha

prime order, so they're all additive order 11. ok i got it now

(except 0 which is order 1 bc of course it is)

[coral spindle]

:3c

[summer path]

[graceful elm]

How would one count conjugacy classes of group homomorphisms?

say, from Z/2Z to S_5?

it feels like a simple example but i cant quite wrap my head around it

Z/2Z has elements that are basically 1 and -1, so we need to map the identity to an identity and the element of order 2 to an element of order 2. and there are 2 kinds of elements of order 2 in S5, right? a single transposition, or two simultaneous transpositions. so those would be the 2 conjugacy classes of these homomorphisms

[delicate orchid]

a group homomorphism from Z/2Z to S_5 is the same thing as an element from S_5 of order 2

as we can just asign the generator of Z/2Z to that element

[delicate orchid]

this is correct yes

now the question is how big are those conjugacy classes

[graceful elm]

ok. i couldve sworn i put 2 on the test and it got marked wrong. but i might be misremembering

there are 10 possible transpositions, and each transposition leaves 3 untouched elements so you pick one of them and transpose the other two, so that's 30 pairs of disjoint transpositions. 40 homomorphisms total

[delicate orchid]

there's also the map sending everything to the identity

[graceful elm]

oh, that counts? for some reason i thought that distinction was part of the structure we were meant to preserve

[delicate orchid]

why wouldn't that count?

[graceful elm]

but i guess f(ab)=f(a)f(b) if f(a) and f(b) are both 1 so it preserves the operation just fine

[delicate orchid]

anyway, there are 10 unique 2-cycles (5 choose 2) and 15 (2,2)-cycles

so it should have been 26 in total

[delicate orchid]

the reason why it's 15 and not 30, is because (12)(34) = (34)(12)

[graceful elm]

wait is the map to the identity its own conjugacy class?

[delicate orchid]

ok clearly something is getting confused here, the conjugacy classes are just a nice way to enumerate the number of maps

there are no conjugacy classes in the set Hom(Z/2Z, S_5) because it's not a group

but

the identity is always in it's own conjugacy class

[north shore]

Find a number x such that x is not equal to x

Yes

[graceful elm]

i got the answers right and i feel decent about the reasoning, even if im not displaying it very well. thank you so much for your help

[long obsidian]

Can someone help me with a sanity check.. the reals R as a module over integers Z is not a finitely generated module right?

[rocky cloak]

They are not

[daring nova]

It's not even finitely generated as a Q-vector space
So even less as a Z module

[rocky cloak]

Think it should be isomorphic to Q^R, which is kinda cursed I guess

[long obsidian]

Thank you

[white oxide]

is there any non-heavy computational way to check closure conditions for matrix groups under multiplication or do you always have to multiply em out

[chilly ocean]

depends on the group

very broad question

[long obsidian]

Can someone remind me how I can go about showing $(Z^n)\otimes_{Z} R\cong R^n$ ? I think this is true

[white oxide]

i'm this dude's #1 opp fr

[cloud walrus]

HausdorffT1

[crystal turtle]

tensor product should distribute over direct sums

[delicate orchid]

I'll prove it explicitly anyway

[crystal turtle]

And then A otimes_A M = M for any ring A and A-modules M

[long obsidian]

Got you okay thank you that was straight forward actually

[crystal turtle]

Wew is still proving explicitly

[delicate orchid]

nah I'll stop now

[south patrol]

left adjoints preserve colimits

:)

[crystal turtle]

monoidal closed :3c

[delicate orchid]

just remember that (a_1, a_2) = (a_1, 0)+(0, a_2) and that's basically the whole problem solved lol

[long obsidian]

$Hom(Z^n,Z) =Hom (\oplus_n Z, n) \cong \oplus_n Hom(Z,Z) \cong \oplus_n Z$

Is this a correct argument that the dual module of $Z^n$ is itself?

[cloud walrus]

HausdorffT1

[delicate orchid]

I buy it

[summer path]

Hom(oplus Z, n)?

[delicate orchid]

?

[rotund aurora]

What

[summer path]

The second term in the equality

[rotund aurora]

Well ig you do use that n is finite

[delicate orchid]

probably a typo, the n turned into a Z

[summer path]

Should it not be Hom(oplus_n Z, Z) or an I missing something

Yeah ok

[delicate orchid]

I'm literally too dyslexic to be confused by typos because I don't even notice them

[long obsidian]

What was the issue? Isn't $Z^n\cong \oplus_n Z$?

[cloud walrus]

HausdorffT1

[delicate orchid]

look at what you actually wrote

[south patrol]

Also not sure this is the best notation anyway

Oplus_n Z to me looks like a countable direct sum indexed by naturals

[delicate orchid]

When the colimit preserves left adjoints

[south patrol]

Write Z^(oplus n) or smth

Or just Z^n here as no ambiguity

[delicate orchid]

I think the point is the moving of the inner product out

[south patrol]

Yes

Then write oplus 1 to n

[delicate orchid]

You want me to write Hom^n like it’s a trig function?!? Insane

[south patrol]

Idk oplus_n is odd

[delicate orchid]

I just replaced it with \oplus^n in my head

[crystal turtle]

Without the context that it's a finite sum I would've also interpreted it as a countable direct sum

[crystal turtle]

This is good

[vestal snow]

Let $R$ be an integral domain and $(f_1,\ldots,f_n)=(1)$. If $R_{f_i}$ is integrally closed for all $i$, then is $R$ also integrally closed?

[cloud walrus]

Finitely Many Bananas

[vestal snow]

Is it true that if $(f_1,\ldots,f_n)=1$, then $\cap_i R_{f_i}=R$?

[cloud walrus]

Finitely Many Bananas

[vestal snow]

Assuming R is an integral domain

This is true

Thanks for the help everyone

[next obsidian]

You’re welcome

[vestal snow]

I truly am a 21st Century Schizoid Man

[chilly ocean]

if anyone could help with this 🙏

[next obsidian]

Think of a suitable automorphism of Z[i] taking one ideal to the other

[chilly ocean]

conjugation mapping?

[stone depot]

is quotient group also a normal subgroup?

[coral spindle]

Typically it’s not even a subgroup

[rocky cloak]

Being able to realize a quotient group as a subgroup (in a natural way) is equivalent to being a semidirect product. Realizing it as a normal subgroup is equivalent to being a direct product

[stone depot]

actually, I was studying from herstein, ths first para in the image I attached its mentioned, a normal subgroup of G corresponds to normal subgroup of G*
So, in theorem 2.7.2 proof, I thought the same arguments goes, and
G|N is isomorphic to G* | N*

maybe its very stupid of me to think so

I have changed some notation here

[chilly ocean]

please y'all 😭

[rocky cloak]

You already solved that one, didn't you?

[chilly ocean]

nope ive been dying 🥹

[rocky cloak]

Yes

[chilly ocean]

he had mentioned a suitable automorphism of z[i]... afaik, the only ones that woudl be applicable to this siutation is the conjugation automorphism

but i dont see how that helps

[rocky cloak]

Conjugation maps a+bi to a-bi

[chilly ocean]

like do i just say $\phi(x + yi + \langle a + bi \rangle) = x + yi + \langle a - bi \rangle$

?

[cloud walrus]

okay bozo

[chilly ocean]

or do i also switch it to x - yi...

[rocky cloak]

You would also switch to x-yi

[chilly ocean]

oh, so just conjugate everything?

i wasn't sure for the longest time since i didnt know if i could just... switch the a + bi to a - bi in the principal ideal like that

[rocky cloak]

Like if you want to justify it a bit more consider the quotient map

Z[i] -> Z[i]/(a+bi)

Then compose with conjugation

Z[i] -conj-> Z[i] -> Z[i]/(a+bi)

Then show that the resulting map is surjective with kernel (a-bi)

[chilly ocean]

the z[i] conj is the conjugation automorphism?

[rocky cloak]

Yeah like conj(x + yi) = x - yi

[chilly ocean]

gotcha

okay never mind this was way simpler than i made it out to be

thank you sm

i think i just havent had good sleep

i appreciate the help a ton

[karmic moat]

just want to be clear on the definition of syzygys:

given a projective resolution
[ \cdots \to P_k \to P_{k-1} \to \cdots \to P_1 \to P_0 \to M \to 0, ]
the $k$th syzygy is $\ker(P_k \to P_{k-1})$?

[cloud walrus]

ana(functor)mono(morphism)

[karmic moat]

or am i totally misunderstanding syzygys in this context

[next obsidian]

Indeedy

Except the plural is syzygies

[karmic moat]

oh

dammit

[delicate orchid]

really cool word, I wonder why they chose it for this

[karmic moat]

i think it came from astronomy

[delicate orchid]

I only know it from astronomy where it means 3 objects in a line

but

[karmic moat]

something about like three astronomical bodies in a line

god dammit

[delicate orchid]

I ain't takin the projective resolution of Mimas ykwis!?!?!

[karmic moat]

why does this screenshot suck

[delicate orchid]

bros on that Commadore 64

[karmic moat]

https://en.wikipedia.org/wiki/Linear_relation#History

oh it's cos i got hdr turned on or something

stupid ass windows i bet a commadore 64 works better

[delicate orchid]

e 1 ' ... ' e n ...

[karmic moat]

wait is the k-th syzygy the kernel of P_k to P_{k-1} or ker P_{k-1} to P_{k-2} because wikipedia says the latter

[delicate orchid]

big if true

[next obsidian]

Oh yeah heehee

It should be the image of P_k

[karmic moat]

which is ker of P_{k-1} by exactness

ok thank u

[delicate orchid]

yeah nlab agrees with wikipedia here

[karmic moat]

wikipedia editors probably steal from nlab but dont cite it tbh

they add a "category theory" section to trivial articles and then copy-paste from nlab to look like geniuses

probably

[summer path]

there are some pages that are like word for word taken from nlab

iirc

[karmic moat]

lmfao

ok two more questions

1. i looked up "order ideal" and haven't been able to find anything on it outside of this paper, is there some more common notion/alternative name?
2. just making sure i understand the usage of Serre k-condition: we just use that instead of the "kth syzygy blah blah blah" so that we can do this over arbitrary rings instead of CM rings?

also i dont really get how it lets us drop the CM ring condition since the theorem from Auslander and Bridger assumes that it's over a CM ring

[tender linden]

(if this message doesn't belong here, tell me where to ask)
i am a noob at Galois theory but it showed up during my research. i know (or, rather, i believe) that if the Galois group of a field extension is cyclic then the defining polynomial of that field extension is solvable. does that also imply that the solution to that polynomoal is expressible in real/complex radicals? if so, is there a general approach to find such representation in radicals?

[delicate orchid]

not sure what it means for a polynomial to be solvable but all abelian groups are solvable so you would be able to express the polynomial in terms of radicals

[tender linden]

understood, is there some way to actually do it?

[delicate orchid]

not that I know of

maybe there's some nice way you can do it because it's cyclic

but even for C_2 the quadratic formula is pretty stinky!

[tender linden]

good point
thank you!

[brittle ether]

Question: is it true that for every possible finite group G, there exists some finite permutation group such that G is isomorphic to one of its subgroups?

Basically, can permutation groups always form a "supergroup" for any finite group?

[delicate orchid]

this is cayley's theorem and yes it's always possible
in fact, it's true for infinite groups as well you just need to take infinite permutation groups

see if you can prove it

[brittle ether]

🤯

Thanks, I'll look into this

[alpine island]

Can an ideal generated by an infinite number of elements contain infinite combinations of those elements?

[rocky cloak]

There is no concept of infinite linear combinations in general

[alpine island]

So no?

This came up because I was thinking about the definition of real numbers as equivalence classes of Cauchy sequences of rational numbers. The set of Cauchy sequences can be a ring, and I was wondering if I could define the set of zero sequences as an ideal.

There might not really be a good reason to do this, it just occurred to me that this could be possible

[rocky cloak]

The set of sequences converging to 0 is indeed a (maximal) ideal in the ring of Cauchy sequences

[alpine island]

Yeah so I was trying to think of a set of elements that generate it

which lead me to that question of "infinite linear combinations"

[rocky cloak]

Yeah if you allow "infinite" linear combinations here, then sequences converging to 0 would generate the whole ring

In general you need some form of topology/form of convergence to talk about infinite linear combinations. But if you are studying topological rings, you would be interested in closed ideals

[restive birch]

what does G/pG mean for an abelian group G

[delicate orchid]

you can consider G as a Z-module

[restive birch]

a what now

doing algtop and it came up but i havent done algebra yet

[delicate orchid]

so pG is like {pg : g \in G}

[restive birch]

ah i see

[delicate orchid]

how are you doing algtop without alg

[restive birch]

and then quotient

cool

thats what i thought but i wanted to check

[restive birch]

you know, im not quite sure

itll be fine

[deft pumice]

Hey guys quick question; This is false right

if f,g,h,... are polynoms of K[T1,...,Tn] and none of them are associated and all of them are prime then
K[T1,...,Tn]/ <f,g,h...> does not have any nilpotent elements

I thought so too but I cant come up with an example. Im just blanking

[alpine island]

No, that is right. if f is prime, then it has no roots or is linear. if it's linear this does nothing, and if it has no roots then this will not have zero divisors

[deft pumice]

But when I thought about it I came up with the following: I have a primeideal and if its maximal then I would have a Field which then would mean it has no nilpotent elements

[lime badge]

Where does the 3rd isomorphism here come from?

The first is just by F being free, the second comes from how the Hom functor commutes with limits, but I was confused by the third step

[void cosmos]

which one exactly

can u write out which statement

that ur having trouble with

im assuming M^v means the dual of M

[coral shale]

nth step means nth === sign, presumably

[void cosmos]

too hard for me

[lime badge]

Hom(M, R)^n = Hom(M,R) \otimes_R R^n

This is the step I was confused by

[void cosmos]

(f1 +f2 + f3 +f4 +...) --> f1 tensor (f1 +f2+f3....)i think

[void cosmos]

yeah i dont think its immediate like the other ones for me

or nvm just

do it the usual way ig

[abstract rock]

whats corollary 5.3 it referennces?

[void cosmos]

like find a bilinear map to Hom(M,R) x R^n

and then find an inverse for the map that comes out of the tensor product by the universla property

[lime badge]

Oh wait, I see. R^n = (R (+) R (+) ... (+) R) n times. Then since Hom_R(M,R) is an R module, Hom_R(M,R) \otimes R = Hom_R(M,R)

So you just distribute the tensor product across the direct sum

[void cosmos]

yeah

[lime badge]

Perfect, thanks!

[void cosmos]

yeah this is trash its just as u said

nice :d

[rocky cloak]

Consider k[x, y] and f=y-x^2 and g = y.

Then k[x, y]/(f, g) = k[x]/x^2

[deft pumice]

Im sorry im lost. What would this mean?

Wouldnt I have a nilpotent element then with polynom x

Or did I misunderstand

[rocky cloak]

Yes exactly

Where did you get this statement from? Seems very strange

[deft pumice]

It was a true or false question from a practice exam

So I was wondering if it was true or false

Would this be true? I think so:
R is Integrityring; then R[T] is integrityring

[rocky cloak]

Yes, R is an integral domain iff R[T] is

Assuming R[T] means polynomial ring anyway

[deft pumice]

Yes R[T] means polynomial ring sorry

[deft pumice]

Sorry one more question:

Is it correct to say: Char(R) = Char(Q(R))?

I know that Char(Q(R)) is smaller or equal to Char(R)

But is it possible to show that its equal?

Its not equal right? Since Z/4Z has char: 4; But the quotientof the ring would be a field and there the Charaktersitik would have to be a prime number

Ah nvm I cant do that example

Since Z/4Z has zero divisors

[rocky cloak]

Any subring of a ring will have the same characteristic.

[deft pumice]

Ah okay and then since I can see R as subring of Q(R) it holds as true

Thank you!

[teal vessel]

I'm really dumb and I know that this is true but I can't find a good way to say it. Because h^n=1, |h| divides n. where you assign X is basically "step 1" in the cycle, and suppose you assign X to some bigger value h^d+1 where d=m|h| for some integer m, then you still have an equivalence class modulo |h| in the target subgroup to deal with, which should bring... oh wait.

x to h ~ x to h^(d+1). In the former case, x^2 goes to h^2, but in the latter case x^2=xx, so f(xx)=f(x)f(x)=(h^[d+1])^2, but because d+1 is congruent to 1 mod |h|, it must be the case that x^2 always goes to h^2 (and the argument may be continued for any arbitrary exponent).

I feel like I'm missing something important.

[plush quartz]

The integers form a group under the shift map, right?

[coral spindle]

What is the "shift map"

Do you mean addition

Like, +

[plush quartz]

s: N->N defined by s(n)=n+1

I have to show that s has no right inverse

[coral spindle]

But what is the group structure

[plush quartz]

but I don't know if s is a group or not

[coral spindle]

A group involves a map G x G → G, not just a map G → G

So what's the map G x G → G. What's the group operation.

[plush quartz]

I am given the set of natural numbers

and the shift map as defined

[coral spindle]

So the group consists of natural numbers?

So what's the group operation

[plush quartz]

s

it's a function

[coral spindle]

But that's a map N → N

The group operation must be a map N x N → N

Are you clear on what a group is?

[plush quartz]

Yeah, it's a set that is associative, has an identity, and an inverse

but I'm not given N x N

[coral spindle]

In particular you will note it has an operation!

Anyway, it's clear to me now that you are just not actually working with a group on N.

[plush quartz]

I thought you have to be given an operation in a problem?

[coral spindle]

You are working with the group of functions N → N. We'll call it Fun(N).

[plush quartz]

lemme just paste the problem. I'm very confused on group operations hold on

[coral spindle]

Well, your job is to prove it's not a group.

No I am going to tell you your confusion.

You are looking at the set Fun(N) of functions from N to N

Now s is an element of Fun(N), and Fun(N) has a natural operation Fun(N) x Fun(N) → Fun(N) on it.

This natural operation is composition: doing one function after another.

Now I ask you this: despite this not being a group, Fun(N) does have an identity element. What is it?

Remember, this is going to be a map f : N → N.

Do not worry about your particular question yet; you need to understand this

[plush quartz]

okay, gimme like a minute I need to think about this.

Sym(N) is a symmetric group of all the naturals

[coral spindle]

I did define this

[plush quartz]

Is this a definition?

[plush quartz]

oh woops

[coral spindle]

Actually, I made a very bad mistake with naming it Sym(N).

Because in fact we typically define Sym(N) differently. I'm going to choose a different name quickly.

Sorry about that confusion.

[coral spindle]

A definition of the operation, yes.

Please note I'm calling it Fun(N) instead of Sym(N), because Sym(N) refers to something else typically.

[plush quartz]

okay that looks way better. Fun(N) makes more sense

To answer your question, the identity would be the identity function.

[coral spindle]

Indeed it would be the function defined by f(n) = n

OK

Now let's look at your question

You have defined s(n) = n+1 and you are asked to show that s in Fun(N) has no inverse in Fun(N)

So let's use contradiction to prove this. Let's say s had an inverse, call it r maybe

What equations would this mean that r and s satisfy?

[coral spindle]

OK fine that's OK

Let's assume r is a right inverse specifically. What would that mean

[plush quartz]

pick some random natural number.

Then that number divided by iteself is 1

[coral spindle]

?

[plush quartz]

zz^-1=1

[coral spindle]

Why is this relevant

[plush quartz]

z^-1 is the right inverse of z

and left is vice versa

[coral spindle]

This is totally irrelevant.

I'm not talking about right inverses of numbers.

The question says prove that s has no right inverse.

What does it mean to be a right inverse of s? Remember, s is a function.

[plush quartz]

but s is not a set, it's a function

[coral spindle]

:|

[plush quartz]

uhhhh

[coral spindle]

Do you know what it means to be a right inverse?

[plush quartz]

lemme get back to you on this in like a day or something. I just started reading this chapter

[coral spindle]

Just tell me.

:|

OK fine, just get back to me in a day then.

[plush quartz]

Trust me, I'm just as frustrated as you are. It's not quite clicking

I have a rough idea of what a group is

but this all to me right now is just set theory stuff it feels like

[coral spindle]

I think rather than not clicking, you need to read the material preceding this exercise. These words have specific meanings which will have been given.

Whether or not it clicks is secondary to actually being able to reference the meanings.

[teal vessel]

S is a set, because all functions are sets

but that's beside the point

[coral spindle]

This is an unhelpful thing to mention

[teal vessel]

I don't think there's much helpful to mention here tbh. There's too much information asymmetry

[coral shale]

🤨

[teal vessel]

as in they're missing a lot of information that's necessary to even engage in the exercise

[coral shale]

Functions having (left or right) inverses can be considered to be a disjoint matter to the elements of groups having inverses really

until you see more of abstract algebra.

[coral spindle]

I was going to lead by example and just not say anything, but I think this must be said: if you don't have anything helpful to say you can just say nothing

[coral shale]

Also - all functions are sets... in ZF.

This need not necessarily be true more generally in terms of encapsulating what a function is.

I view ZF as one way of "implementing" math. Everything is a set.

But the idea of a function is... not necessarily tied to this implementation.

eg. in lambda calc, functions aren't sets, no.

[teal vessel]

meh, tomato, vegetable

I find it most useful in many cases to see functions as just a subset of a cartesian product because it gets a lot of the higher assumptions about functions out of the way that get baked in inductively (not math-induction). It's certainly not the only way to look at them, but I find the incredibly deflationary effect of "it's just a set" to be quite helpful in many cases.

there are other definitions that have similar deflationary effects that work just as well. I just have a preferred flavor

[coral shale]

Well of course, everything is a set is a very handy way of doing math. Your proofs can often quite easily become definition and symbolic bashing

expand into set definition, do some set manipulation, and you have what you wanna show.

[teal vessel]

a common one I use for my students who I don't think would like the idea of making sets is "a black box that takes in [usually numbers] and spits out a single [usually number] reliably." I find that there's a lot of students that get hung up on functions being either A) continuous or B) algebraic constructions and get really confused by things like logarithms or trig without realizing that we can just assign outputs as we get inputs arbitrarily. I like using the isint (maps integers to 1, non-integers to 0) or the d6 (randomly assigns a number 1-6 to every real number) functions as examples to help break those assumptions.

[white oxide]

silly question but i'm a little bit confused about the phrasing here, do we say a monomial occurs if it has the maximum degree in a polynomial f or if it's just simply exists in the expansion of f lol

[coral spindle]

a monomial a_(v)X_q^v_1 etc occurs in f if a_(v) =/= 0.

So it occurs in f if you see it in f when you write it out :)

[white oxide]

lol ok thanks, i always get lost in the weeds

is this a typo? we don't necessarily have X_i^{\mu + 1} equal to X_i^{v_i} right? and if that's the case, we'll get a different polynomial, but what do they mean by the same function?

[coral spindle]

Seems so.

[white oxide]

actually nvm i guess

[coral spindle]

Oh, we're in a finite field. Indeed.

x^q = x in every finite field (as functions, not as polynomials).

N.b. I mean a finite field with q elements

[white oxide]

right... need to recall stuff from field theory

[coral spindle]

OK this is actually very simple

F_q has q elements. The group F_q \ {0} has q-1 elements

So if x is not zero, what does this tell us about x^(q-1)?

Finish this off by incorporating the x=0 case.

[white oxide]

it's the identity and then x^{q - 1} = 1, whence x^q = x

[coral spindle]

Exactly.

Simple, right?

[white oxide]

ah yes i remember that

yeah lol

oh yea cuz nonzero elements of field form a group

wait i thinkkkk the group is cyclic

yeah

wait

yea

[coral spindle]

Indeed, any finite subgroup of a group of units (of any field) is cyclic.

So in particular F_q \ {0} is cyclic.

If you'd like you can prove this via a minimality argument. It doesn't really matter, I'd just keep it in mind.

[white oxide]

yeah, i'll come back to it some time

why do we say that we have a "new polynomial" then if the function is the same?

just appearance tings?

[coral spindle]

The polynomial x^q is not the polynomial x

do you agree with that?

If you do then that's the end of the story.

[white oxide]

i don't agree with that, or rather i don't see how that's the case

because they're equal as functions

or do we define them being equal separately

or something

[coral spindle]

But we're not considering them as functions.

We are considering them as polynomials, which are formal sums.

[white oxide]

ohhh

[coral spindle]

If you want to know some practical reason why we'd want to distinguish it, notice that there are only four functions F_2 → F_2, but there are infinitely many polynomials F_2[x].

Supposing that we denote by, idk, PF(R) the set of polynomial functions R → R on a ring R

There's clearly a homomorphism R[x] → PF(R)

And it is surjective by definition (since we are looking at polynomial functions)

Fact: in the case that R is an infinite domain, this homomorphism is an injection.

I had to remind myself of how the division algorithm works for a sec there lol

[white oxide]

ah yeah, i guess that makes sense

[white oxide]

wait sorry i'm dense, what is the point you're making here?

[coral spindle]

.

I'm explaining, in essence, why you have likely not thought about the distinction before.

Since you are so used to working with infinite domains (Z, R, C) where there is no distinction

You can safely not distinguish polynomials and the corresponding functions in these cases bc of the fact I pointed out

[white oxide]

ohhh okay i see

[coral spindle]

Conversely, in finite fields it's obviously false, since you can just take the product of all X - k for every k in the field.

That's constantly zero, and yet not the zero polynomial.

[coral spindle]

I'm feeling lazy so I won't check this carefully, but come to think of it I think this holds for any infinite ring.

[low wyvern]

guys I hate to interupt with a simple question but I can't do the last part 🤣

[coral spindle]

Set up the equations you need to and solve

[low wyvern]

Oh ofc yes lol.... Pretend I never asked

[coral spindle]

Sidenote: x . y in this case is equal to (x + 1)(y + 1) - 1 :)

[low wyvern]

I don't see how that helps

[white oxide]

right

thanks

[coral spindle]

Take it as a hint then.

[low wyvern]

I have the answer its ok. I just for some reason, was trying to do it in my head

and whilst u may be able to do that... I can not yet

[coral spindle]

's not about me mate

[white oxide]

i don't understand the purpose of assuming this for contradiction/where the inductive hypothesis comes in. we're assuming "if f(a_1, ...., a_n)"so why are negating it?

[coral spindle]

Oh lol this has as a corollary a weaker version of what I was just saying

Let's see

[white oxide]

or are we maybe doing contrapositive or some shit and then contradictoin with that 😭

[delicate orchid]

Yeah

[white oxide]

wait rlly or were you saying yeah to something else

cuz you responded in like 1 millisecond

[delicate orchid]

Yeah

[coral spindle]

Yeah we are doing the contrapositive

[white oxide]

bro literally got it in 1 millisecond

that's impressive

[delicate orchid]

I didn’t read anything and just said yeah

I have no idea what we’re talking about

[white oxide]

LMFAOOO

honestly i usually ask trivial verifications so that's fair

[delicate orchid]

Yeah

[plush quartz]

Are we trying to show that an element in s has a right inverse (contradiction) or the function s has a right inverse?

[coral spindle]

https://media.discordapp.net/attachments/496784958430380033/1162486646684524544/image.png?ex=653c1d0e&is=6529a80e&hm=372b242bf99fd1ac80a02935b2a15e5d7657841ce92594b9b95a5215c5c4cb7c&=&width=1186&height=113

Prove that s has no right inverse,

[plush quartz]

I've never heard of a function having an inverse, only elements

like numbers

a function can be invertible though

[coral spindle]

Did you take my advice and read what the definition of an inverse is, earlier in your book?

[white oxide]

to be invertible means that it has an inverse

[delicate orchid]

It does not sound like you’re prepared for AA then

[coral spindle]

The answer to this can be as simple as 'yes' or 'no.' I'm not invested in your grade or your future or whatever – I'm not your teacher or your dad and there's no consequence to not knowing something. It just takes more time if you don't answer this outright, and that's just annoying.

[delicate orchid]

I am your father

[coral spindle]

Wew is actually ur dad tho and will publicly embarrass you.

[white oxide]

ur a father figure to me tho

[coral spindle]

I hope I'm not just a father figure, but an enemy to you all 😍 🥰

[plush quartz]

I will just try to continue through the proof. I have to find a function that, when composed with s, gives me the identity function.

[delicate orchid]

You will be defeated as I am puppet for divine will

[coral spindle]

Bunker. Sincerely answer me this: if you literally do not know what a term means, how are you supposed to complete the proof involving that term?

Imagine trying to derive the quadratic formula without knowing what the symbol sqrt(x) means

You need to learn the terminology!

I'm glad you've at least got a correct idea, but you're missing a crucial detail here which is unclear if you've realised or not.

[plush quartz]

I will show you the section.

[coral spindle]

Great.

[plush quartz]

The "1" in that section when applied to this problem is the identity function

[coral spindle]

Write down, in mathematical symbols rather than words, what a right inverse to s would do. Call such a hypothetical inverse r, and write down the equation it would satisfy.

[white oxide]

boys i'm not going insane right, by theorem 1.8 lang means corollary 1.8

cuz 1) i can't find any theorem 1.8 and 2) it seems like he would be referencing the result he just showed

nvm yea that's the case

[white oxide]

so here is there any reason why he's not defining the content as the gcd of the polynomial? (we're considering A to be any UFD) because surely there's a generalization of a gcd in the UFD

i know these definitions are equivalent obviously but wouldn't it just be much easier to define it as the gcd

oh nvm

the coefficients are in the quotient field

if i ever write a textbook i'm writing like this mf

[summer path]

Lang often writes like that

so in some sense, either you learn to teach yourself a lot or you don't learn anything at all, and use another text instead

[white oxide]

Nah I know I rlly like his style of writing

Imo you learn a lot better that way since you’re always verifying stuff

And usually the proclaimed obvious and trivial things are indeed obvious and trivial

I just find it funny how often he uses it

[lament dawn]

This is so cap

[white oxide]

Ok straightforward, given you’ve seen these things before xD

[summer path]

lang will say things are obvious/trivial/exercise to the reader for things that are in fact not obvious nor trivial

lang is probably better as a "review book" than a studying book tbh

or a problem book i guess

there are just much better texts to use for actually trying to learn the material it hink

[white oxide]

yea, i guess it's only obvious/trivial if it's a review

which, if you're taking a second course or even a grad course in algebra and this is your assigned textbook, most likely is

[summer path]

you do realize that if the entire class is just a review, you wouldn't be taking the class right?

surely at some point you're going to be seeing material that is new to you

[white oxide]

well you know what i mean

just the material that is a review

[summer path]

and at that point, if lang writes like how he does, it may not be so trivial

but yes it is nice to work things out on your own; though i think that there are definitely points where it's just like maybe another resource would be nice to supplement with

[plush quartz]

s /circle r

r: N->N r(n)=n-1

that should give us the identity.

compose r with s

[shy python]

Can someone give me some pointers for this problem? It was given in my algebra class but I can't figure out a way to do it, which might be because my linear algebra knowledge is very lacking.

all I have deduced is that B preserves the eigenvectors of A and vice versa

[ebon pine]

For questions like these, there's no way I am supposed to show closure by trying so many products right?

I basically have to show that this is a nonabelain group with matrix mul as binary operation

[terse crystal]

Choose non zero y such that Ay=y, (an eigenvector y of A of eigenvalue 1), let x=((B-I)^(p-1))y, you see x is a common eigenvector of A and B with respect to eigenvalue 1. The rest follows by induction process.

When ((B-I)^(p-1))y=0 in the first place I need to think

Oh never mind, choose minimal r such that ((B-I)^r)y=0, (B-I)^p=0 so this r exists and 1<=r (<=p), then let x=((B-I)^(r-1))y

Now correct

[stuck fiber]

So I'm trying this problem specifically proving it is a ring homomorphism and it seems like I have a ton of cases to go through is there a more efficient way to do this than saying something like
Case 1: k+l=0
Case 2: k+l>0
Case 3: k+l<0
then breaking these down into subcases

[delicate bloom]

maybe just look at where -1 is sent

[stuck fiber]

Ah so use the fact that Z is cyclic so everything is determined by a generator type argument?

[delicate bloom]

well I'm thinking f(-n)=f(-1)f(n)

[stuck fiber]

Wouldn't you have to say that f(-n)=-nf(1) since it's not necessarily a ring homomorphism

[dim widget]

f(1) is 1 by definition

and n*f(1) = n = f(n)

also by definition

[alpine island]

What's the name of the theorem that if prime p divides ord(G), there is an element of G with order p?

[delicate bloom]

cauchy's thm

kinda like baby sylow thm

[deft pumice]

Is there a epimorphism from S3 to Z3 ? What about Z2

S is the symmetric group

Z is the cyclic group

I thought to Z3 yes due to caley

[delicate orchid]

There’s definitely one to Z_2, <(123)> is normal so just take the canonical surjection S_3 -> S_3/<(123)>

Alternatively the sign function maps into <-1,1>

[deft pumice]

And Z3

[deft pumice]

Yeah this is what i thought off

[delicate orchid]

Nah

The kernel would be order 2 and there are no normal subgroups of order 2

[deft pumice]

Okay. One last question, what about S4 to S3

[deft pumice]

I thought due to caley

[delicate orchid]

What does Cayley have to do with anything

[delicate orchid]

There’s a normal subgroup of order 4, the quotient must be isomorphic to S_3 because there are no elements of order 6 in S_4

[deft pumice]

Caley says that for each group of order n there is a subgroup in Sn such that those are isomorph

[rocky cloak]

A fun way to see this is that S4 has 3 copies of C4 as subgroups. And act on them by conjugation.

[delicate orchid]

Yes, you’re going the other way around

You’re asking about Hom(S_n, -) not Hom(-, S_n)

[deft pumice]

Allright I understand

Maybe one more question: is there a finite field with char 0?

[delicate bloom]

think about it

[deft pumice]

I thought since its a finite field i have a injective automorphism from K to K

[delicate bloom]

what does characteristic 0 mean

[deft pumice]

It means that 1+1+1+1+ is never 0

If I recall correctly

[delicate orchid]

Yup

[delicate bloom]

are any of 1, 1+1, 1+1+1, ... the same

or are they all distinct

[deft pumice]

In a finitenfield some should be the same

So then i would subtract those right

[dim widget]

Yes but if the characteristic is 0

What does that mean about the homomorphism from Z to your field

[delicate bloom]

if some are the same, what does that mean

[deft pumice]

That if k1 is same as m1 , k and m being my amount of adding

That k-m is my characteristik

Since (k-m) *1 equals 0

Or other way

[dim widget]

No all you know is that your characteristic divides k-m

[deft pumice]

Ah

Yes

[dim widget]

But anyway since it’s 0 what does that mean about k, m?

[deft pumice]

But in a field i cant have zero divisors right?

[dim widget]

Definitely not

[deft pumice]

Couldt i say that since char divides k-m

But char 0

There exists a zero divisor

[dim widget]

Well or k=m is more clean/principled

It’s good to avoid unnecessary proofs by contradiction

[deft pumice]

Ah okay

So k must equal m cause otherwise i have zero divisors

But this means that my assumption k and m are not same has been nulled

And so my assumption of F finite is hurt

[dim widget]

Yes so F cannot be finite

[deft pumice]

Ah and then in total if char is 0 F cannot be finite

[dim widget]

Another way to say this is that every commutative ring including a field has a natural unique ring map from the integers

And the characteristic of a field is the positive generator of the kernel of this ring map

[deft pumice]

But arent mappings from field to commutative rings always injective

So then kernel is 0?

[dim widget]

No the map goes the other way

From Z to the field F

[deft pumice]

Ah I see

Okay thank you for your help!

[hot goblet]

can someone give me like a 2 sentence summary of group actions? and maybe another 2 sentences on why we care about them? Orbit Stabiliser formula is pretty cool

[void cosmos]

sylow theorems

cayleys theorem

ig

like men groups will be known by their actions

[abstract rock]

wreath products usually are nicer to write down if you make a choice of group action

[granite bay]

I'm too noob for stuff here but this is how I understand it: You know the "equilateral triangle with labeled vertices, rotations and reflections stuff" when groups are introduced in Interwebs? What really happens is that D_3 (defined in abstract "some set with binary operation blablabla") acts on the set of possible positions of vertices in that triangle.

[hot goblet]

ah yeah cool thats a good way of thinking about it, groups are simply concepts related by structure that act on sets

[granite bay]

The set can be bigger, like you can also think of D_3 acting on hexagon (or 3n-gon generally) with labelled vertices. Now in this case, starting from a labeling L of vertices, some other labeling can be reached and some others are not reachable. That subset of reachable labeling from L is the "orbit" of L, I think.

[tardy hedge]

Sorry bad pic but its Q12

Im stuck on c

Basically the question is show that every proper subgroup of Z x Z that contains C1 is Cn for some n

C1 is the set of all (a,b) with a=b

Cn is set of all (a,b) with a = b mod n

Really wanted to solve it on my own but cant spend too much time or else i wont finish the homework

Seems tricky to me

[abstract rock]

you can just do this as a chain of homomorphisms, try to find a set of h_n: Z times Z to Cn

[tardy hedge]

Without that

We havent covered that yet

[abstract rock]

you havent covered homomorphisms?

[tardy hedge]

Correct

My break is over, but basically only done intro to groups

SubgroupS, cyclic group definitions

No isomorphism or homomorphism yet

[abstract rock]

you can still consider the set of maps, then show that the only maps with a preimage that looks like a proper subgroup of Z times Z is closed under the Cn's group operation

[tardy hedge]

Not sure what u mean

I was trying to use just properties of groups and stuff like that i guess

Like using the basica

Basics

[abstract rock]

then you were done lol

what you outlined eas fine enough

[tardy hedge]

Oh yea i mean i was trying to do that but didnt succeed lol

I guess the problem rn is that i don’t intuitively see why this is true

The only way to have C1 as part of the subgroup is if the subgroup in given by some modular equivalence thing? Not sure why

Like it seems reasonable but idk why it must be true

Hmm

Is it cause if u have C1 and then two other elements that cant be related by some mod thing, then it makes it so the subgroup is really all of the group?

Maybe i can show that

[shy python]

Im sorry but can you explain a few parts of this to me? How do know there is a eigenvector with eigenvalue 1? And further how do you know that (B-I)^p=0 unless p is not 2? And what exactly is the induction process here, since I imagine you can't just pull out indefinitely many eigenvectors of eigenvalue 1?

actually it doesn't matter, thank you for the help anyways!

[grizzled spindle]

What about this subring of 2x2 matrices then

(We take matrice addition and multiplication as usual)

[prime sundial]

what is the subring

[delicate orchid]

The bottom bit

[prime sundial]

like are you asking if the matrices of the form
a 0
0 0
form a subring of the set of 2x2 matrices

[prime sundial]

right

[grizzled spindle]

Wouldn’t S also be a subring with a different 1

In that case

[prime sundial]

no

is the 1 from R in S?

is
1 0
0 1
in S?

[grizzled spindle]

Well no but I wasn’t sure if that definition implied that it was the 1 in R

[prime sundial]

it does

0 and 1 refer to the 0 and 1 in R

[grizzled spindle]

I see

[prime sundial]

the additive and multiplicative identities

showing that they also behave like additive and multiplicative identities for S is a quick lemma

[grizzled spindle]

So would there be a name for that S if it’s not called a subring

[prime sundial]

subset i guess lol

[grizzled spindle]

I see

[tribal moss]

For maximal confusion, some authors will sometimes use different definitions where this is a "subring".

[prime sundial]

since 0 in S?

[grizzled spindle]

So some authors will not ask for the ring and subring to have the same identity elements?

[tribal moss]

It depends on whether you consider the identity element something that's part of the definition of which ring you're looking at, or simply require rings to satisfy a "there exists an element that behaves like an identity" axiom.

The latter is a bit confusing, but it appears to be sometimes useful when working in the borderlands between rings and rngs.

[prime sundial]

fridge where did you get your definition from

[grizzled spindle]

My notes for algebra

[prime sundial]

i assume you are dealing exclusively with commutative rings with identity

[grizzled spindle]

We don’t have that multiplication is necessarily commutative in a ring for this class

But if it was then the subring would have the same identity element for multiplication I imagine

We are often just given that something is a ring and not necessarily that it is commutative

[prime sundial]

i guess the 2x2 matrices should have made that clear for me lol

[tribal moss]

Commutativity is not enough to force identities to match. Consider the subset {0,3} of the integers modulo 6, for example.

[grizzled spindle]

I see

[grizzled spindle]

Is multiplicative identity not required for rings in some textbooks?

[prime sundial]

yes

[grizzled spindle]

I thought the definitions of rings was the same everywhere given how important the concept seems to be

[prime sundial]

most i've seen only require a ring to have an additive identity

and a lot of the time you just assume it has a multiplicative identity, if the math you're doing benefits from it

so a lot of books will just say "let R be a ring," but will specify at the start that every ring is assumed to be with multiplicative identity

[grizzled spindle]

Fair enough, if it’s specified it probably isn’t too bad to deal with the differences

[rocky cloak]

I think the fact that it's an important concept is why the definition varies.

Different rings come up in different contexts, and if all the rings you care about are commutative you might use the word ring and "commutative ring" interchangeably. Similarly for rings without 1.

[snow condor]

what is a group presentation? how does it tell you how the group is defined?

i’ve had to do exercises with groups defined by a presentation in the past which i could get through but i realize i didn’t really understand what was going on

i would like some intuition for it

[tardy hedge]

Yeah i think this solves it, but im curious about other proofs

[snow condor]

wait actually smth on wikipedia just gave me the intuition i was looking for nvm lol

[terse crystal]

1. because A has an (actually all) eigenvalue 1, since (A-I)^p=0, so det(A-I)^p=0, det(A-I)=0
2. (B-I)^p=B^p-I^p=I-I=0
3. there exists invertible matrix S having x as the first column, then S^-1AS=(1, something; 0, A1), S^-1BS=(1, something ; 0, B1). A1, B1 are of size n-1 times n-1 and they still satisfy the conditions of the problem.

[shy python]

is Fp isnt algebraically closed are we sure A has only eigenvalue 1

otherwise yeah i got it

but thank oyu it was super helpful

[terse crystal]

By induction process then all its eigenvalue is 1

det (A-I)=0 give us one eigenvalue 1, then we still have det(A1-I)=0 for that n-1 times n-1 matrix A1, so in the end A and B are simultaneously conjugate to upper triangular matrices, whose elements on their diagonal are 1, meaning A,B have n many 1 as eigenvalues

[shy python]

how come det(A-I) has a 1 eigenvalue in it though im still a little confused on that

[delicate orchid]

Definition of an eigenvalue

[terse crystal]

I said det(A-I)=0, that means A has an eigenvalue 1

[shy python]

oh okay sure

thank you!

[coral shale]

Do we consider algebraic structures with an underlying object that isnt a set. Like a class.

Im thinking for example:
left/right identities/inverses of functions in general. And to do this you need to consider the class of all functions under composition, right?

[crystal turtle]

Congratulations. You just discovered the category of sets.

[coral shale]

monke

[crystal turtle]

(not a joke btw, category theory would be the closest thing to handling something like that)

[coral shale]

yh but i dont see why necessary in a sense?

can we not build algebra on top of classes

[crystal turtle]

I mean, proper classes are pretty unwieldly, you need a pretty strong choice of foundations to properly work with them, as opposed to just saying "yeah, there's technically no set of sets but we're gonna pretend there is to define the category of sets and just kinda gloss over some shit"

[next obsidian]

You can’t eg define the center

You can’t use specification for one

[coral shale]

ive also thought about a similar issue with matrices under composition in general.

Well its the same issue tbh

oh nvm no, those are a set. finite dim matrices.

[chilly ocean]

I'm dumb but what is a class?

[coral shale]

proper class

"class of all sets"

[chilly ocean]

how is it different from sets

[coral shale]

just a collection too big to be a set

[ivory trail]

you can't have a set of all sets

[chilly ocean]

you just give it a different name so its not a set?

[coral shale]

too big in the sense that it cannot be under ZF

[ivory trail]

because of russell's paradox

[coral shale]

all sets, all groups, all rings, etc etc

these are proper classes

[chilly ocean]

what would the collection of all proper classes be named?

[crystal turtle]

[coral shale]

theres some theory about that zzz

[ivory trail]

well "class" is usually used informally

https://en.wikipedia.org/wiki/Von_Neumann–Bernays–Gödel_set_theory but there's this if you must be formal

[coral shale]

reminds me. The set of all ordinals is not an ordinal (not a set)

[alpine island]

it's just a notational helper as I understand it

[coral shale]

so, the first place you see classes being relevant is probably category, right?

idk where else

[alpine island]

if you define Set to be the class of sets, then "X in Set" is just defined to mean "X is a set"

[crystal turtle]

Yeah iirc, "class" doesn't make too much formal sense in ZF(C).

But I am not that well-versed on foundations tbh

[coral shale]

no it definitely cant in ZF

u need a containing theory or something

[ivory trail]

there's an area of concern when doing category theory that people call "size issues"

sometimes you bring in the big guns (inaccessible cardinals, possibly one after the other)

[alpine island]

Is that what my girlfriend was talking to me about?

[ivory trail]

so i guess classes are relevant to CT if you're thinking about size issues

but i try not to

[crystal turtle]

Size issues don't matter

(except for when they do)

So yeah I think if you want to treat them more properly you can work with proper classes with NBG or Morse-Kelley set theory. But otherwise, if you want to be careful even with categories which is often where you'll see the term "class" used, you just take inaccessible cardinals like Rho said

[ivory trail]

for example, afaik wiles's proof of FLT may or may not depend on inaccessible cardinals in some way, just because it goes through a bunch of algebraic geometry abstract nonsense

but if it does, it could probably be excised out, idk how much (nontrivial) work that would take though

[coral shale]

wait wut lol

[ivory trail]

https://mathoverflow.net/questions/35746/inaccessible-cardinals-and-andrew-wiless-proof

[alpine island]

Actually I'm curious, I know classes don't behave as sets, but I don't actually know how. what statements can you make about sets that you can't make about classes?

[coral shale]

i feel like it should be otherwise itd surely be pointed out right? his proof wouldnt work in zf?

[ivory trail]

all the axioms of ZFC only apply to sets

notably powerset and the filtering one (specification/separation/whatever)

[alpine island]

Okay yeah powerset makes sense

[alpine island]

Is that cardinal larger or smaller than this one? http://bp3.blogger.com/_eDsYe_W-c-E/R1RGpBaBGBI/AAAAAAAAAdI/6IxcWGAuaVU/s1600-R/Cardinal-Norther,+male+IMG_0048b.jpg

[ivory trail]

this cardinal looks kind of accessible to me

so probably larger

[alpine island]

I don't know where this cardinal is, or where the photo was taken. For me, this is not an accessible cardinal

[coral shale]

ah ok, its probably good according to the 2nd answer https://mathoverflow.net/a/35772/478637

proof by faith

[crystal turtle]

I mean my understanding is that assuming the existence of an uncountable grothendieck universe is basically just a formality to make some things easier to state and work with. You should essentialy be able to prove anything you can using it, without actually using the assumption, but it makes some "formal" constructions technically work better

[ivory trail]

yeah the answers in that MO post seem to suggest so

that is, it can probably be removed pretty easily

i'm mostly referring to the last answer

[white oxide]

ok i take back what i said yesterday

how does this hold? isn't cont(bf) going to be a product of powers of primes

and b times cont(f) is not going to be a product of powers of primes

that is, under the assumption that $(bf)(x) = ba_0 + ba_1x + ba_2x^2 + \dots + ba_nx^n$ where $f(x) = a_0 + a_1x + \dots + a_nx^n$

[cloud walrus]

okeyokay

[white oxide]

oh

it's because it's we can multiply it by a unit

wait no

not by a unit in K

ok now i'm confused

[ivory trail]

why not?

[white oxide]

never mind i forgot that the definition included negative powers as well

[white oxide]

are there any polynomials with content not equal to one which are irreducible in Q[x]? i'm trying to think of some but nothing is coming to mind

[south patrol]

Just multiply through a constant

[white oxide]

oh right

so 2x^2 + 2 is irreducible and has content 2

if i'm not going insane

[white oxide]

damn you're a mod now

that's crazy

i don't really understand how this "gives us existence of the factorization" - doesn't the first display provide the existence of the factorization? what good does knowing that c = cont(f) have towards the "existence of the factorization"? i know that in a UFD the existence of the factorization is given by a decomposition into a product of prime powers up to multiplication by a unit, but how exactly does c = cont(f) show this?

also how does it follow that each p_i(X) is irreducible in A[X]? how do we know that p(X) has content 1 => p(X) is irreducible? i wrote p(X) = h(X)g(X) and assumed that p(X) had content 1, but what's stopping h(X) and g(X) from having content 1 as well?

[south patrol]

Yes, or just do 2x

[white oxide]

like is this part in the proof just something i'm missing that follows immediately?

[chilly radish]

having a brainfart, if $v_1,\ldots,v_n$ is a basis of $L/K$ a finite extension, why does $v_{n}\notin L(v_1,\ldots,v_{n-1})/K$

[cloud walrus]

ShiN

[warm ember]

its a basis

[chilly radish]

obviously it's not a linear combination but what stops it from being a linear combination of powers of v_1,\ldots,v_n-1

[warm ember]

so they are linearly independent

oh

no

its not a basis then

[chilly radish]

what

[warm ember]

you dont take the power of basis

[chilly radish]

yea it needn't be a basis

[warm ember]

if there is a power for example x and x^2

then the basis needs to be x,x^2

the basis just does the same as in linear algebra

[chilly radish]

I know how field extensions work, i'm asking why $v_n$ is still not inside this extension

[cloud walrus]

ShiN

[warm ember]

that extension is not the same as the ring generated by the basis

i mean it is if v1,...,vn is a basis of the extension

but that means powers of v1 are among the vi

[chilly radish]

sure it is, all the elements are algebraic

[warm ember]

by definition you cant square a basis and say its in the field

[chilly radish]

what?

[warm ember]

its the same as linear algebra

[chilly radish]

if $x$ is algebraic over $K$ then $K[x]=K(x)$

[cloud walrus]

ShiN

[warm ember]

the basis dont interact with each other

[chilly radish]

that doesn't matter though since you're taking power...

[warm ember]

you cant

the basis dont interact with itself either

[chilly radish]

these are field extensions

of course you can take powers

[warm ember]

you need to seperately add the interactions

[chilly radish]

what are you talking about

[warm ember]

what are you talking about

[chilly radish]

field extensions

[warm ember]

I know

idt you understand what a basis means here

[chilly radish]

I don't think you do

[warm ember]

ok

for example

[chilly radish]

$v_1,\ldots,v_{n-1}$ needn't be a basis for $K(v_1,\ldots,v_{n-1})$ if $v_1,\ldots,v_n$ is a basis of $L=K(v_1,\ldots,v_n)$

[cloud walrus]

ShiN

[warm ember]

consider the field extension Q(i,w) where w is the third root of unity

what is the degree of this extension

[chilly radish]

i'm not sure how that's relevant to what i'm asking. The statement is false anyways as stated, e.g. if I take C/R with basis 1,i then 1\in R[i]

[warm ember]

no

[chilly radish]

uhhh yes

[warm ember]

this has degree 4 because (x^2-1)(x^2+x+1)

the basis isnt i,w

it is 1,i,w,w^2

hence degree 4 extension

[chilly radish]

I never said it was?

i'm taking v_1,\ldots,v_n a priori to be a basis

[warm ember]

yes

but thats just because i^2=-1 which is in R

[chilly radish]

yes....

[warm ember]

yes

[chilly radish]

its still a counterexample to my question