groups-rings-fields

[obsidian path]

why exactly? o-o''

just to be pedantic im a little mixed up with everything cause

[rustic crown]

since row rank = column rank... you get that rank <= 5 this would mean nullity >=1

[obsidian path]

i saw someone do this as a mapping from R^6 to R and idk why he did that

[rustic crown]

i'll write the equation for ya\
\
$\begin{bmatrix}
x_1^2 & x_1y_1 & y_1^2 & x_1 & y_1 & 1\
x_2^2 & x_2y_2 & y_2^2 & x_2 & y_2 & 1\
x_3^2 & x_3y_3 & y_3^2 & x_3 & y_3 & 1\
x_4^2 & x_4y_4 & y_4^2 & x_4 & y_4 & 1\
x_5^2 & x_5y_5 & y_5^2 & x_5 & y_5 & 1
\end{bmatrix}
\begin{bmatrix}
A\
B\
C\
D\
E\
F
\end{bmatrix}

\begin{bmatrix}
0\
0\
0\
0\
0
\end{bmatrix}
$

[obsidian path]

this makes sense yes

[cloud walrus]

det

[obsidian path]

could you more explicitly remind me of rank nullity theorem tho?

:pleading face or something like that yessir

[rustic crown]

like do you want me to prove it or something? cause it will just say rank + nullity = 6

[obsidian path]

no cause i don't see where you got the row rank = col rank

get that rank <= 5

wait

ok so its just null = 6 - rank(5) = 1

so there exists p /= 0 in p(x, y) such that T_p = 0
ie p(x_i, y_i) = 0 for all i = 1, .. , 5

something like that?

[rustic crown]

rank is generally defined at dimension of the image... that is dimension of the space spanned by the columns... but it turns out that this dimension is also the dimension of the space spanned by the rows

[rustic crown]

more like null = 6 - rank >= 6 - 5 = 1

[obsidian path]

"even if all our 5 points were same, in this situation, the rank reduces even further" can you remind me of the details for this?

this sounds like a basic property of rank that I forgot about

[rustic crown]

so if all points are the same... then all the 5 rows of our matrix are same which would mean that the row space is 1 dimensional so in this case null = 6 - rank = 5.

[obsidian path]

true..

[rustic crown]

this is just saying that there are lots and lots of conics passing through a single point

[obsidian path]

then you need to show that the union of kernels will be >= 1

even tho maybe thats trivial too

ok ok sorry, im probably being too anxious about this

thank you!!

[rustic crown]

you don't see the word "union" put together with "kernels" a lot.... maybe check this again?

[obsidian path]

ah yeah thats kinda weird

omg wait its makes so much sense

ok ty

nvm

what i said

have a wonderful day i appreciate ur patience

[rustic crown]

uwu

[severe osprey]

Hi how would I define a Ring and a Field

Not define

I don't really know the word

but how would i right a ring out

like you write a group as

$(\bZ,+)$

[cloud walrus]

Ilovehumaninteraction

[severe osprey]

so like the integers under addition

[rustic crown]

(R, +, .)

[severe osprey]

is there an equivilant for rings and fields?

ah thank you

and i assume that goes for fields as well

[rustic crown]

i haven't seen that a lot in books... they just usually say let k be a field...

[severe osprey]

Ah I see

i Ask because im looking at topological algebra

and it says a topological space that is also a field

is a topological field

[rustic crown]

i see, i don't know about it much. but (k, +, .) should suffice..

[severe osprey]

Alright thank you c:

[rustic crown]

uwu

[golden pasture]

you usually dont like specify it like that cuz it's usually pretty obvious

[severe osprey]

[golden pasture]

jus say

the ring Z

the field k

people assume + and • are the ring operations

the local field Qp

[scarlet estuary]

yeah the main reason you see this notation for groups at all is that its often ambiguous what your group operation is

consider Q for example

it could be a group under + or (as long as you exclude 0) under *

in rings, the operations are rarely "ambiguous"; theyre either the "obvious" operations (standard + and *) or some esoteric stuff that has to be explicitly defined

[real panther]

Hey, so I know that for a group ( G ) and an element ( a \in G ) and subgroup ( S \subseteq G ), we have the conjugacy set ( aSa^{-1} )

[cloud walrus]

Neil_P

[real panther]

I'm being asked to show that one set is conjugate to another set. I've looked online and I cannot figure out what that means

[carmine fossil]

A set T is conjugate to S,if T=aSa^-1

[real panther]

I'm guessing to show that subgroups ( V, W ) of ( G ) are conjugate, I need to show that there is an element ( x \in G ) such that [ xVx^{-1} = W ] ?

[cloud walrus]

Neil_P

[carmine fossil]

i.e.,for every element t in T, you can find a element s in S such that t=asa^-1

[real panther]

@carmine fossil So that seems right, cool

Thanks - I couldn't find that specific definition on proof wiki or my previous textbooks. Any clue why this is important?

[carmine fossil]

Well It's a way of defining a group action on subsets

a. S=a S a^-1

[real panther]

I know that my subgroups ( V ) and ( W ) are normal, so I'm guessing it has to the actions between cosets?

[cloud walrus]

Neil_P

[carmine fossil]

Yea

[real panther]

Interesting - the group is group action on a vertex transitive graph and V and W are stabilizers of the nodes v, w respectively

I don't expect them to be equal

Or, didn't

[mystic jungle]

a is an element in an abelian group G, e is an identity element

is this statement always true for an abelian group G, where a^2=e ?

yes

damn

[golden pasture]

any finite group that has that property is product of Z/2Z

[mystic jungle]

Lets say im given a set where all the a's are not equal to its own inverse, and im given this group G = {e,a1,a2,...,an} im asked to prove (a1,a2,...,an)^2=e

but i can't just multiply and commute because a != a inverse how should i approach this

[golden pasture]

if a!=a^-1 then a^2!=e

[strong valve]

do we have a book recomm. channel or can we ask on related server?

[cursive temple]

there is a book discussion channel which will probably suffice

#book-recommendations

[mystic jungle]

im doing 4

oh whoops

[viscid pewter]

what about the self-inverse ones

that's the only thing

i think we just reset the assumptions for 4-6

like i got tripped up by the diagram but i don't think it applies to 4

[mystic jungle]

really? so i can use a^2=e

[viscid pewter]

no

reset the assumptions

actually just where are you getting a^2 = e from

[mystic jungle]

a=a^-1

[viscid pewter]

why

[mystic jungle]

a^2=e iff a=a^-1

[viscid pewter]

yes

oh are you just using my thing that says only the self-inverse ones survive after a single thing

ok so yeah just say all the non-self-inverse ones get cancelled with one copy

and then the self-inverse ones get done by the square

if ykwim

[mystic jungle]

sorry if im being slow but why is this true again

[viscid pewter]

it's not

multiplying all the elements that aren't self-inverse together should give you e

[mystic jungle]

but why

not sure why i cant pick this up haha

[cyan marten]

Here's another way to see it:

Let x = product of all elements in a group that has no nonidentity elements that are their own inverses. You can check that x^-1 = x, so x must be the identity.

In general, we can only conclude that x has order 1 or 2.

[mystic jungle]

Ok, im putting it together now thanks folks

[chilly ocean]

I am trying to understand primary ideals

an ideal q is primary if xy in q implies either x is in q or y^n is in q for some n

this is fine

the radical of a primary ideal is always prime and is the smallest prime ideal contain q

if q is a primary ideal and its radical is the prime ideal p, we say that q is p-primary

opps i see now

[sullen bloom]

woooooo all modules are group rings

[cyan marten]

I find it easier to interpret this as a statement about A/q

[chilly ocean]

yeah this is nicer

starting learning algebra i was always tricked into thinking its better to think about the elements themeselves

but it is nicer thinking about them in terms of the quoitent

saying A/p is an integral domain is much more succint than

p is an ideal such that if any product of elements from the ring is in theideal then at least one of those elements must be in the ideal

obviously this information is still being said when you have to define an itnegral domain

but assuming we all know what integral domains are by now it is nice

can someone explain "minimal primary decomposition" to me, I'll type in what i understand first then then where i get stuck

An ideal is not gaurrented to have a primary decomposition. A primary decomposition is simply a way of writting an ideal as a intersection of finitely map primary ideals.

A primary decomposition $a=\bigcap^n q_i$ is said to be minimal if:
(1) each $q_i$ is $p_i$-primary with each $p_i$ distinct
(2) for all $j$, $q_j$ does not contain the intersection over the other $q_i$

[cloud walrus]

lime_soup

[chilly ocean]

we can always reduce a given primary decomposition down to one that obeys (2) because if there was some j where qj cotained the the interesection of the others then removing this qj does not change the itnersection

ah

we can do 1 because if q_1,...q_j are primary withthe same p then we can replace this by their intersection

okay nevermind i have no question

[next obsidian]

rubber ducky

[chilly ocean]

Let $\mathfrak a$ have a primary decomposition into the primary ideals $q_1,...,q_n$.
There are things called associated primes, minimal/isolated primes, and embedded primes.
First question, I assume these definitions are only in terms of a minimal primary decomposition of $\mathfrak a$?

Second question: The minimal primes are the minimal elements in the set all the $p_i=\sqrt(q_i)$. I assume we mean minimal wrt inlcusion? If none of the primes include in each other then they are all the mimimal primes and there are no embeeded primes? So lets say $(2),(3),(7)$ were the associated primes of some primary decomposition in the integers. Then they are also minimal primes

[cloud walrus]

lime_soup

[next obsidian]

associated primes are defined without respect to a primary decomposition

As a result, the minimal primes of a ring are the same with respect to any primary decomposition

even though the primary decomposition can vary

I forget the nuances of some of these things though

[chilly ocean]

hmm? I thought the associated primes are independant of the choice of minimal primary decomposition, but we could choose a stupid non minimal decomp that would gives us extra associated primes

and because of this we choose to define them in terms of a minimal decomp?

[next obsidian]

associated primes are defined for an A-module right?

they're primes of the form ann_A(m) for an m in M

[chilly ocean]

hmm I haven't seen that

oh sorry you are right

[uncut girder]

ASS(M)

[vestal snow]

If R is an integral domain and M is a square matrix, prove that M and M^T have the same characteristic polynomial and the same annihilator ideals

I proves that they have the same characteristic polynomial

How do I prove that they have the same annihilator ideals?

[uncut girder]

For char poly remember that the matrix satisfies its char poly

=0

What happens when you take transpose of both sides?

What would happen if the transpose matrix had a smaller degree char poly?

Idk what annihilator ideal means

I have an unrelated question about representation theory

Say you have an irreducible representation V of G. My book says
"The center of the image G -> GL(V) has split rank one"

What does this mean?

Why does it follow from Schur's lemma?

Oh its literally just schurs lemma

[obsidian path]

can someone help me prove this?

[chilly ocean]

[obsidian path]

cross product yikes lmao

two subspaces

[chilly ocean]

you prove this the same way you prove most statements about dimensions: by furnishing a basis

find a basis of W_1 x W_2

ideally in terms of bases of W_1 and W_2

[carmine fossil]

Shouldn't it be less than or equal to?

[obsidian path]

I don't think so but im not sure

[chilly ocean]

it is an equality

[carmine fossil]

Is this the cross product on R^3? Generalised cross product otherwise takes more than 2 arguments

[latent anvil]

it's a direct product of two vector spaces

[latent anvil]

$\dim(W_1 \times W_2)$

[cloud walrus]

Shamrock emoji

[chilly ocean]

what the fuck does it mean to cross product two vector spaces

[carmine fossil]

Makes sense

[carmine fossil]

Ask this guy

[obsidian path]

.-.

[latent anvil]

sorry deku

I feel like I've heard it called that

[obsidian path]

no its okay

I'm confused so im confusing people

idk

think its cartesian product

[chilly ocean]

$W_1 \times W_2$ is indeed the Cartesian product.

[cloud walrus]

(T*(Terra), -dτ)

[snow flint]

shamrock's new pfp looks even more like the old pfp than the old one

[latent anvil]

lmao

thank @chilly ocean

He thought this was what my previous pfp looked like

I think he said he thought it was a "gremlin wizard"

[snow flint]

no fr thats also what i pictured whenever i saw it

[uncut girder]

Wow

Place the images side by side

[carmine fossil]

[cyan marten]

Is there a slick proof of Chevalley-Warning theorem employing group actions?

Suppose f(x1, x2, ..., x^n) has degree d < n, and is homogeneous with constant term zero. The theorem states that the number of solutions is divisible by p.

If G is a p-group acting on X, then |X| = |S| (mod p) where S is the subset of fixed points. In particular, if X is nonempty and has pk elements, then there is a fixed point.

Maybe we should look for a group that acts in a way that fixes roots

[tight otter]

scratching my head a bit over this one

showing $u|a_0$ is easy, because the constant term $a_0$ is equal to the product of the constant terms of the factors

[cloud walrus]

bela

[tight otter]

and because $u$ and $v$ are coprime then $u/v\mid a_0$ implies $u\mid a_0$

[cloud walrus]

bela

[tight otter]

but the v part is tricker

[thorn delta]

hm, plug in u/v into f.

multiply both sides by v^n

[tight otter]

$a_nu^n+a_{n-1}u^{n-1}v+\cdots+a_1uv^{n-1}+a_0v^n=0$

hm

[thorn delta]

ur missing v^n next to a_0

[tight otter]

oop yeah

[cloud walrus]

bela

[tight otter]

it's like the binomial theorem without binomials lol

ohhhhh

oh

i get it

i see

you factor out v

and then because u and v are relatively prime if you put a_n on the RHS it must be true that v divides a_n

thank you good friend

[thorn delta]

nice. np

[tight otter]

in my prof's notes for a proof he just states that $F[x]/\langle x\rangle\cong F$ where $F$ is an integral domain. what's the logic and reasoning behind this? i don't understand where this assumption comes from, is it just one of the three isomorphic theorems?

[cloud walrus]

bela

[rustic crown]

We have the identity map F --> F. Extend this by the universal property to the map F[x] --> F by sending x to 0. is this surjective? what's the kernel? what does the first isomorphism theorem say?

[tight otter]

ohhh yeah!!

the kernel would be all polynomials with no constant term

i.e. the principal ideal generated by x

and the first isomorphic theorem is just F/K is congruent to G

thank you, im a bit rusty on isomorphic stuff, gonna need to brush up on some of those rules

[rustic crown]

[vestal snow]

Let T be a matrix with entries in an integral domain R and f(x) in R[x]. If f(T) = 0, is it true that the characteristic polynomial divides f(x)?

[rustic crown]

nope

[vestal snow]

dang

Is a matrix in an ID similar to it its transpose?

[rustic crown]

its false when R is a field... just take f to be the minimal polynomial

[vestal snow]

I know they are similar in an alg closed field

and thus any field

[rustic crown]

ah, then just go to the fraction field

[vestal snow]

Yeah but is being similar in frac(R) => being similar in R?

[rustic crown]

need to think about that.... i feel that since we have both A and A^-1 in the field we can probably cancel the denominators

but it also feels that things could easily go wrong 😛

[vestal snow]

I can't wait until we get to field theory lol

1+ month of working with matrices has broken me

[rustic crown]

oof

[tight otter]

we just started field theory this week in my course

[vestal snow]

Nice

What book are you using?

[tight otter]

Ash

Basic Abstract Algebra

not sure what the normal rate to work through it is but i think we're working through it a bit slow, but i'm glad we are if that's the case

[vestal snow]

Haven't heard of that book before

Seems pretty cool

[tight otter]

between this, graph theory, cryptography, and the satellite project im working on + my procrastination im too overstretched to commit to a fast pace lol

yeah it's a neat book

pretty brief in its explanations and proofs

the first course in my abstract sequence last term had me thinking that i was much dumber than i thought i was

but doing graph theory this term has made me feel like im a fuckin genius with algebra

[vestal snow]

I don't need algebra to do that for me lol

[tight otter]

hehe

im liking the class though

looking forward to starting my masters next year bc im no longer gonna be bogged down by CS courses

[vestal snow]

What are you doing your masters in?

[tight otter]

math!

[vestal snow]

Oh nice

Applied or pure?

[tight otter]

Pure, probably

im studying pure mathematics that is

but i'm interested in real-world applications

[vestal snow]

Seems like a fun time

[rustic crown]

i really liked aluffi's field theory

but chapter 8 is too scary for me at the moment 🥲

[vestal snow]

That's the one I'm waiting for lol

Because the stuff in chapter 8 allows us to state everything in chapter 6 without bases

[ivory cosmos]

I'm re-reading Aluffi rn

I stopped before chapter 7, now solidifying my understanding of commutative ring theory, since that's a bit flimsy

[rustic crown]

lol i was waiting for that chapter to get scared 😂

[vestal snow]

Have you done Tensor product stuff before?

[rustic crown]

nope, i just read them today morning

[vestal snow]

It usually shows up in commutative ring theory

A-M has a terse, but nice discussion about them

[rustic crown]

Tor functor scared me for the day, before that it was adjoint functor

[ivory cosmos]

have you done much category theory before?

[rustic crown]

nope

i'm in my second year undergraduate

category theory was one of the reasons to pick up aluffi

[vestal snow]

Same

[ivory cosmos]

you've actually seen quite a few examples of adjunctions, funnily enough

Yeah, adjoints are tricky to get a hang of, but very useful

For example, all of the free ring / group etc. constructions are adjunctions

or, arise from an adjunction, rather

[rustic crown]

yea, free-forgetful is nice

btw you usually don't think about the naturality square while talking about free-forgetful adjunction... right?

[ivory cosmos]

I think about the triangle identity

[vestal snow]

guys I did. I solved racism the problem about annihilator ideals

[vestal snow]

All I had to use was the property that (A+B)^T = A^T + B^T

[rustic crown]

its just that right now, when i look at the naturality square its has data about 2 functors, and you combine it with the Hom thing to get two bifunctors and then we're talking about natrual transformation between these. These many layers make it hard to get a nice intuition about it

[ivory cosmos]

oh, you're using that definition of an adjunction

[rustic crown]

how else is it defined?

[ivory cosmos]

Hom_C(L -, -) = Hom_D(_, R) or whatever

[rustic crown]

yea these 2 are required go be naturally isomorphic

[rustic crown]

i really don't understand this word "natural"

[tight otter]

i always feel super dumb when people mention studying stuff outside of classes i always want to learn more but i dont have the energy

[ivory cosmos]

A transformation is natural if it respects the structure of the category, essentially

It's kind of like how, I want F (f g) to be (F f) (F g)

[rustic crown]

i have heard this saying that "Categories and Functors are there just to define Natural Transformations, and these are actually what are important"

[ivory cosmos]

sort of

that's historically kind of true in algebraic topology iirc

But it's kind of like, there's no point really in talking about functions between groups that aren't homomorphisms, because you can't really say anything about the structure

Similarly, a good notion of morphism between functors is a natural transformation, and without that naturality assumption, you don't respect the structure

[rustic crown]

I'm at the stage where i have to verify that every functor is actually a functor and same for every natural transformation... and because of that its kinda hard to get an intuition...

idk what i'm saying.. sorry if it doesn't make sense

[latent anvil]

I think it should be true that if R is a subring of S and you have matrices A, B over R which are similar over S then they are also similar over S. It's certainly true if R, S are fields due to rational canonical form

[ivory cosmos]

another way of thinking about natural transformations is that it operates purely on the outer part of something

like, if you have F A, an I transform that to G A

this transformation can only observe the "shell", i.e. the F part

it can't do different things based on A without violating naturality

[latent anvil]

Any numerical invariants like determinant or trace will have to be the same

[ivory cosmos]

So, I'm transforming different "shells" without observing the contents

[vestal snow]

I figured out another way to do the problem without relying on this. Thanks.

[latent anvil]

And like, the characteristic polynomials

Nice!

I'm still curious about it though

[rustic crown]

that's a nice way to think about it

[latent anvil]

hmm

[ivory cosmos]

But, with adjunctions, if you have L : C -> D, and R : D -> C, then for any A in C, you have the slice category with objects morphisms A -> R B, with B in D, and morphisms between these commuting diagrams:

Then, RL A is initial in this category

so like, if you have a free forgetful situation

e.g. Set and Group

then this category would be set morphisms from a set to a group, and commuting triangles as above, with the morphism on the right being a group homomorphism

and the free group is initial here

This is the usual characterization of the free group

You also have a dual property, with a category

and LR A being terminal here

[rustic crown]

i need to verify that this is same as the earlier definition... but this seems a bit more comfortable way to put it

[ivory cosmos]

(this is equivalent to being an adjunction, but more concrete)

The proof is ass though

[rustic crown]

Thanks!

[latent anvil]

det, have you done any homological algebra? Like, do you know what a projective object is?

[rustic crown]

no

[latent anvil]

oh okay, nevermind then

[rustic crown]

I saw some exact sequence in R-Mod

[latent anvil]

I was going to talk about one reason we care about adjoints/how they show up in algebra

[rustic crown]

but i don't know what projective object is

[ivory cosmos]

Discrete Topology ⊢ Forgetful ⊢ Indiscrete Topology

[latent anvil]

I don't really either

[ivory cosmos]

is another fun and trivial example if you know some topology

[latent anvil]

I mean I know the definition and how to work with them

but I still don't feel like I have any intuition

Something something vector bundles

[ivory cosmos]

homological algebra feels weird to me, because you're like, synthesizing a field that comes from spatial intution

but you can study the objects in a purely synthetic way

but all the terms come from the geometric intuition

like "flat rings" or whatever

[latent anvil]

I sort of agree, but there's two different kinds of geometric intuition at play

like the formalism of homological algebra rose out of algebraic topology initially ("cycles") but was then expanded on by algebraic geometry

[ivory cosmos]

a lot of things are like that 😛

[latent anvil]

Eg general abelian categories and derived functors were defined by grothendieck

[ivory cosmos]

based

[latent anvil]

Sure, I'm just saying it's a common synthesis of two different things

Well, at least two

MacLane was also doing group cohomology stuff

[ivory cosmos]

Yeah like, the idea of working in a general abelian category is smart af

but an obvious generalization of like, modules

[unique juniper]

Prove that G cannot have a subgroup H with $|H|$ = n - 1, where n = |G| $>$ 2.

[cloud walrus]

Yes

[rustic crown]

do you know lagrange's theorem?

[unique juniper]

yes, am aware lagranges theorem beats this

just a question about alternate solution

not completely sure how they knew x^-1 y was in H

[rustic crown]

because x^-1 y is not y

[unique juniper]

oh yh

[rustic crown]

since we assumed that x is not the identity.

[unique juniper]

:D ty

[vestal snow]

Det did you do this exercise in Aluffi?

How would one prove that adj(tI-A) equals a sum of that form?

This seems trivial from the definition but my hw set has a comment that if this problem seems easy I'm doing it wrong

I think that since we "delete" one column and one row when calculating each entry of the matrix, we get rid of exactly one t-a_ii in the determinant

[rustic crown]

yep i did it

[vestal snow]

And so we get a polynomial of degree n-1 in each entry

And then we can just split it

Does this seem correct?

[rustic crown]

yea... that's pretty much the reasoning... each cofactor is a determinant of size n-1 x n-1 so the adj(tI-A) is a matrix of polynomials each has degree atmost n-1

[rustic crown]

yea

[rustic crown]

they probably meant that "plugging A in f(x) = det(xI-A)" to get "f(A) = det(AI-A) = det(0) = 0"

[vestal snow]

Ah gotcha

Arigatou

[rustic crown]

hehe doitashimashite

[vestal snow]

@rustic crown did you do this exercise?

[rustic crown]

ye

[vestal snow]

My proof is almost complete

I only need to prove the dim 1 case

that is, I have to prove $e^{\sum_{r=1}^{\infty}\frac{(zt)^r}{r}}=\frac{1}{1-zt}$

yeah

[cloud walrus]

Have a Banana, Bitch

[rustic crown]

yep

[vestal snow]

How did you do it?

[rustic crown]

either you can use the series for log(1-x) = -(x+x^2/2 + ....), so basically you're computing e^(-log(1-zt)) = 1/(1-zt)

or you can do it formally without referring to convergence in C

[vestal snow]

Let's do the latter

[rustic crown]

the other way i thought used formal derivatives

so you need to prove the chain rule for them first...?

its enough to show this

call that weird sum lol(t). and take f(t) = exp(lol(t))

now f'(t) = exp'(lol(t)) * lol'(t) = exp(lol(t)) * (1 + t + t^2 + ...)

(1-t)f'(t) = f(t)

(f(t)*(1-t))' = 0

f(t) = a/(1-t)

t=0 tells a = 1

proving the chain rule is a bit weird but not hard

[vestal snow]

where did the a come from?

[rustic crown]

derivative = 0 => thing = constant

[vestal snow]

Oh gotcha

Thanks

[rustic crown]

[carmine fossil]

Do you really need exact sequences for this?

[vestal snow]

I think so

You need the splitness of short exact sequences of vector spaces

[carmine fossil]

Won't tr(alpha^r) just be sum of eigenvalues of alpha raised to r th power?

Because find a matrix in which alpha is upper triangular,and the diagonal entries will be just be the eigenvalues

[ivory cosmos]

,tex $rM := {rm \ |\ m \in M}$

[cloud walrus]

CronoKirby

[ivory cosmos]

Maybe I'm missing something, but Aluffi presents this as a submodule, provided M is an R-module

but doesn't this only work if R is commutative?

because if I take some element rm, then s(rm) isn't necessarily contained in here, unless I can swap the two elements of R?

[rustic crown]

we generally used two-sided ideals... iirc

(r) is 2 sided...

[ivory cosmos]

I don't think this is the ideal (r)

rm denotes the action of r in R, on an element m in M

[rustic crown]

but i don't remember him writing rM... i remember seeing IM

[ivory cosmos]

let me check the errata

[rustic crown]

oh yes, this was an error. I was using the other edition

[ivory cosmos]

ah well then yeah, it follows immediately

[chilly ocean]

Say we have a ring morphism from A to B

If B is an integral domain then the kernel of the morphism is a prime ideal?

[ivory cosmos]

I think the argument you're trying to make would hold if the morphism is surjective

[chilly ocean]

f(xy)=0 implies f(xy)=f(x)f(y)=0 so one of x or y is in the kernel

[ivory cosmos]

Err, yeah no it works regardless

[chilly ocean]

I am just worried about zero divisors in A

[ivory cosmos]

1. a subring of an integral domain is an integral domain

[chilly ocean]

Oh that doesn’t matter

[ivory cosmos]

2. A / ker f = f(A)

3. By definition, an ideal is prime iff the quotient is an integral domain

qed

[chilly ocean]

Ah that is slick

Ty king

[ivory cosmos]

also, for maximal ideals, replace integral domain with field

quite elegant definitions imo

[rustic crown]

but subring of a field may not be a field...

[ivory cosmos]

Indeed, just thought I'd mention the alternate definition of maximal ideals, while we're at it

[rustic crown]

oh... my bad

[ivory cosmos]

nah, was good to point that out

[rustic crown]

[ivory cosmos]

Fun corollary:

If f : R -> S, R is a PID, and S an integral domain, then f(R) is a field

Well, you'd have to exclude trivial kernel here I guess

[rustic crown]

yep

[next obsidian]

Isn’t that true for any dim 1 integral domain really

[rustic crown]

are they not just pids?

[latent anvil]

No

,w plot y^2 = x^3

[ivory cosmos]

if a is in J, then (a) is a subset of J

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by shamrock#0598

[ivory cosmos]

so you'd have dimension at least 2 provided an ideal isn't principal, no?

[latent anvil]

(a) might not be prime

It's true that minimal nonzero primes in a UFD are principal

By essentially your argument

[ivory cosmos]

ah, right

an ideal where no element in that ideal is prime huh

🤔

despite the ideal itself being prime

[latent anvil]

The ring $\C[x, y]/(y^2 - x^3)$ is a 1 dim integral domain which is not a UFD

[cloud walrus]

Shamrock emoji

[latent anvil]

Because x^3 = y^2

The ideal (x, y) is probably not principal

Since that's where the curve is singular

[rustic crown]

oh right

this was something liek this ring C[x^2, x^3] i remember now

[latent anvil]

It's isomorphic to this ring

[rustic crown]

Thanks! uwu

[obsidian path]

det wholesome

[subtle current]

Does anyone have errata for Jacobson Basic Algebra 1 and 2?

I couldn't find it anywhere on internet

[smoky cypress]

Let $A$ be a ring. Can a $A$-algebra (defined as a ring $B$ with a homomorphism $A\to B$) be defined just as a ring $B$ that is an $A$ module such that $(ax)(by)=(ab)(xy)$ for all $a,b\in A$ and $x,y\in B$

[cloud walrus]

Whoever

[smoky cypress]

All commutative rings

[oblique river]

that's kind of how I think of algebras tbh

"an A-module but it's also a ring"

[smoky cypress]

Well atiyah mcdonalds defines it to be a ring $B$ with a homomorphism $A\to B$ which isn’t very intuitive

[cloud walrus]

Whoever

[smoky cypress]

So I’ll take it that I have the correct definition

[chilly ocean]

atiyah mcdonalds

[smoky cypress]

[median tree]

Do the integers mod m define all possible subgroups of the integers under addition?

[thorn delta]

the integers mod m aren't subgroups of Z

u might be thinking of nZ for natural numbers n

Every subgroup of Z has the form nZ for some natural number n

[median tree]

yes, that's what I mean

thank you

[thorn delta]

np

[median tree]

ah, so just to make sure I have this all straight in my head, the integers mod is actually the set of indices of some set of equivalence classes, is that correct?

[latent anvil]

To answer the original question, every subgroup of Z is of the form nZ for a unique nonnegative integer n

[median tree]

okay, am I incorrect in believing there's an explicit connection to equivalence classes? Those different subgroups are each equivalence classes?

[thorn delta]

the integers mod is actually the set of indices of some set of equivalence classes, is that correct?
yea, basically. If you know about quotient groups, the integers mod n are just Z/nZ.

[prisma ibex]

right the subgroups are nZ, the quotients are Z/nZ

and this exhausts all of them

[median tree]

Oh, I see, the subgroups are nZ! So the quotients of some subgroup like 2Z would be [1] [2]?

[prisma ibex]

Z/2Z isn't a subgroup

Z/2Z itself has two subgroups: the trivial one and the whole thing

so it has two quotients: the trivial one and the whole thing, the association is just reversed

larger subgroup <=> smaller quotient

and yes the quotient of Z by a subgroup nZ is the set of mod n congruence classes

so {[1],...,[n]}

[thorn delta]

it might make more sense to write [0], [1] as the elements of Z/2Z though

[prisma ibex]

yea that is perhaps a better choice

[thorn delta]

but [2] = [0] so its fine as well

[prisma ibex]

mhm

[median tree]

Ah, so Z/2Z is not a subgroup of Z, though?

[prisma ibex]

nope!

2Z is

[median tree]

what's the technical term for what Z/nZ is? I think I need to review the definition

[latent anvil]

It's a quotient

[median tree]

oh, okay

[prisma ibex]

to be more precise, you're defining an equivalence relation on Z

where two elements are considered equivalent if they differ by a multiple of n

and then you're taking the quotient of Z by that equivalence relation

i.e. you're collapsing everything in the same equivalence class into a single element in the quotient

so e.g. 7 and 10 become the same element, namely [1] in Z/3Z

does that make sense?

[median tree]

yeah, that makes sense. I guess my confusion then

is that in a particular problem I'm dealing with I'm working with a cyclic group C_m that's isomorphic to (Z/mZ, +)

but that doesn't seem to make sense if Z/mZ isn't a group?

[prisma ibex]

Z/mZ is a group!

[median tree]

oh, it's just not a subgroup?

[prisma ibex]

right

if you have a subgroup H of a group G

then the quotient G/H need not be a group

but it is when H is what is called a normal subgroup

these subgroups nZ of Z are all normal subgroups

so the quotients Z/nZ are groups

if H is not normal then the quotient G/H is only a set, passing the group operation from G to the quotient G/H isn't well defined; it's only well defined when H is normal

[median tree]

I see, I haven't deal with normal subgroups yet, but that sounds nifty

Ah, okay, so other question

This process of defining an equivalence relation on Z so that the elements of Z are collapsed into classes

Is this a function of some sort?

[prisma ibex]

I mean the "collapsing" is exactly the projection homomorphism Z->Z/nZ

that sends an element of Z to its equivalence class

this turns out to be a group homomorphism

I mean one way to think of normal subgroups is they are precisely the kernels of group homomorphisms

(that is, if you have a group homomorphism f:G->G' its kernel is the subgroup of all elements in G that map to the identity in G')

you can define Z/nZ by itself, it's the group of integers mod n

and this projection p:Z->Z/nZ is defined by p(N)=N mod n

nZ is its kernel

so in particular nZ is a normal subgroup of Z

[median tree]

Hm, I think I'm going to have to spend a bit more time with this particular idea when I have a chance later. But this cleared up my initial confusion, so thank you so much! (first day here)

[prisma ibex]

No problem! Group theory is wonderful, good luck! 🙂

[untold sapphire]

Anyone who knows about Combinatorial commutative algebra know if it's worth picking up Stanleys book? I've already got CCA by Miller and Sturmfels

[ivory cosmos]

Given some R modules, is and an automorphism, is this equality ever not true?

There's an exercise to show that if phi^2 = phi, then this holds, but this seems like it works in general?

[rustic crown]

idts

i can see why it would be true with phi^2=phi

[ivory cosmos]

So is this identity not true in general?

[rustic crown]

yea... consider M a pid?

it is definitely true for vector spaces

but do you think Z is isomorphic to 2Z oplus Z/2Z

[ivory cosmos]

is it not?

[rustic crown]

the right has this element (0, 1) which has finite order

but only finite order element in Z is 0

[ivory cosmos]

hmm

I guess so

hmm, what's the easiest way to show the first identity, assuming phi^2 = phi

right now I'm leaning towards showing that M satisfies the properties of the coproduct

🤔

[rustic crown]

given m in M, stare at the triviality... m = (m - phi(m)) + phi(m)

this shows M = ker phi + im phi... showing their intersection is 0 is even easier.

[ivory cosmos]

right

I actually managed to get the coproduct argument to work

except for uniquness

For the unique function, you can take

but I wonder how you show that this is unique?

[rustic crown]

because image of m-p(m) is forced... and same with p(m)

[ivory cosmos]

ah right, like, any other alpha would be equal at each point of m

because you're either in the kernel, or in the image

or 0

[rustic crown]

i don't quite get this

you can have some component in kernel and some in the image

[ivory cosmos]

I don't get this argument then

[rustic crown]

so sigma(m) = sigma(m-p(m)) + sigma(p(m))

[ivory cosmos]

oh yeah, by that decomposition right right

[rustic crown]

lol i feel like you basically showed that internal direct product is a coproduct

[ivory cosmos]

no subsets where I live buddy

[rustic crown]

oh

[ivory cosmos]

nah I just wanted to see if this was possible

the direct argument is definitely easier

[rustic crown]

I was wondering this.... in the category of k-vector spaces, "direct-sum" works as a finite product. So is there some category where the categorical product is actually the tensor product..?

ig i found the answer.... tensor product is coproduct in the category k-Alg

[ivory cosmos]

the trivial category

8^)

[ivory cosmos]

the category of finitely generated R modules works

[rustic crown]

but there the product is direct sum... not the tensor product.

i think what I wanted was something like (R-Alg)^op

[ivory cosmos]

Ah woops, misread

[next obsidian]

Ah you mean Sch / Spec R?

[molten silo]

Can anybody give me an example of a ring with no multiplicative identity

will R[1] work?

[sturdy marsh]

any ideal

(except the whole ring depending on your convention)

[molten silo]

(3) = {3a : a is a ring element]

of the integers

[sturdy marsh]

yes

[molten silo]

k

[sturdy marsh]

3Z is a ring

w/o mult identity

[molten silo]

thanks bro

[sturdy marsh]

some authors like to call those 'rngs'

instead of rings

[ivory cosmos]

that may not be a ring depending on your definition of ring

[north widget]

is it true that if you have a ring of unity R then it can be generated by its unity element?

my guess is no but i dont get why

[mild laurel]

Uh yes?

Every element r is just r * 1

[north widget]

i mean like

if u add 1 n times

or if you keep adding 1

will you get entire group?

i think you should because if you keep adding 1 and you are missing out on that element then there would be a problem.

[latent anvil]

it may not generate it as an abelian group

consider the ring $R = \Z/2\Z \times \Z/2\Z$

[cloud walrus]

Shamrock emoji ☘

[latent anvil]

the unit is $(1,1)$

[cloud walrus]

Shamrock emoji ☘

[north widget]

yea

oh ok

yeah i meant does it generate it with additive group

thats really clear example ty

[molten silo]

Could The reals work here because the unit are everything but 0. So cant be a subring since there is 0 in the subset

[thorn delta]

So cant be a subring since there is 0 in the subset
do you mean 0 isn't in the subset?

[molten silo]

yes

[thorn delta]

that works. 0 is not in the set of units of any ring either.

[molten silo]

true

lol

[latent anvil]

Not true

There's exactly one ring of which 0 is a unit

[molten silo]

what

[latent anvil]

The ring {0}

[molten silo]

no

According to this r is a unit if and only if

wait is that the empty set

[thorn delta]

in {0}, "0 = 1"

[molten silo]

i see

[vestal snow]

Given an Artin-Schreier $E$ extension of the function field $k(x)$ given by the equation $y^p-y=f(x)$ where $f$ is a polynomial with coefficients in $k$, can we say anything about the ramifications at the places of $E/k(x)$?

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

Assuming that $k$ is algebraically closed

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

It would be really nice if I could somehow get that all places are unramified (or at least, all places not over the infinity place are unramified)

[prisma ibex]

Artin-Schreier extensions will be everywhere unramified

@vestal snow

I mean somehow the way I remember this is that A^1_k is not etale simply connected, and all the nontrivial finite etale covers come from Artin-Schreier extensions of the function field k(t), with no restriction on f(t)

but such an extension being finite etale means it is everywhere unramified; the only ramification lives at the point at infinity

[vestal snow]

Gotcha

Wait so if it ramified at infinity, it is not everywhere unramified then?

Is everywhere unramified defined to be unramified at every place except infinity?

Will the ramification index at infinity be equal to the degree of f?

If so, will the uniformizer at infinity be x^-deg(f)?

[prisma ibex]

well okay it's ramified at the place at infinity sure

I just mean that it's unramified at all the finite places

coming from actual points on A^1_k

yea I should have been more specific sorry

[vestal snow]

No it's okay!

I figured that's what you meant

[prisma ibex]

and yea that sounds correct regarding the normalizer and the ramification index I think

would have to think a little more to be totally sure

one amusing corollary: this means that the etale fundamental group of A^1_k is a free pro-p group of rank equal to the cardinality of k

which is utterly gigantic, even in the case where k is as small as possible (i.e. k=\bar{F}_p)

[vestal snow]

It's nice to have someone who knows AG verify things

Thanks!

[prisma ibex]

🙂

[vestal snow]

Also one last question

Will there be any holomorphic differentials in E?

[prisma ibex]

what does holomorphic differentials mean in the positive characteristic setting?

are you just asking about what the genus of the Artin-Schreier covering X of A^1_k can be?

the answer is yes, in general these can have genus g>0, hence they can have some nontrivial H^1(X,\Omega^1_X) (which is maybe the appropriate replacement for holomorphic differentials in this setting)

[vestal snow]

w is a holomorphic differential iff ord_P(w) is greater than or equal to 0 for all places P in E

[prisma ibex]

the way you can compute the genus is the version of Riemann-Hurwitz that holds in positive characteristic

[vestal snow]

Is my definition of holomorphic the same as yours?

[prisma ibex]

yes I believe so

you just shouldn't use the term "holomorphic" in the positive characteristic setting

[vestal snow]

Okay. That's an unexpected result then

Because of this theorem

Since f has denominator 1, $t^{(\mu)} = 0$

[cloud walrus]

Have a Banana, Bitch

[prisma ibex]

yea I think what it means by holomorphic differentials are classes in H^0(X,\Omega^1_X) as opposed to classes in H^1(X,O_X) which might appropriately be called anti-holomorphic differentials

these make sense algebraically in char p despite the naming

(over C these are holomorphic and anti-holomorphic differentials respectively in the usual sense)

[vestal snow]

So this theorem suggests that the basis is empty right

[prisma ibex]

no I don't think so

[vestal snow]

in the case when f is a polynomial

since $t^{(\mu)}=0$, the $x^v$ in the basis can't range over anything

[cloud walrus]

Have a Banana, Bitch

[prisma ibex]

e.g. you might try the hyperelliptic curve in char p=2 given by y^2-y=f(x) which will have genus g=(deg(f)+1)/2-1

e.g. if deg(f)=3 you'll have genus g=1 and you should be able to find precisely one "holomorphic differential" up to scaling

[vestal snow]

Yeah I guess that makes sense

So what will the basis, according to the theorem, be in the case f(x) is a polynomial?

[prisma ibex]

ugh let's see

idk I would have to think about this for a bit and unpack this theorem, need to get back to work sorry 😛

something to think about though, sorry I can't finish the example right now

[vestal snow]

No problem! Thanks for the help!

[prisma ibex]

[vestal snow]

In an Artin-Schreier extension E of k(x), what is the uniformizer at the place of E over infinity?

[chilly ocean]

I'm trying to do this exercise: Let M be an artin module, let u be an injective endomorphism of M. Then u is also surjective.

So the idea is that we want to assume u is not surjective, and then show this breaks the descending chain condition

A good candidate for a descending chain is going to be im(u) contains im(u^2) ...

Now it is clear that each im(u^n) contains im(u^n+1)

What is not so clear to me is how do we know this sequence is not constant?

For the similar exercise for noetherian modules where we show that surjective implies injective, its clear that if the kernel is non zero then we have a non constant asscending chain

where the inclusions are strict

its not clear to me why the inclusions are strict in the case with the images

[golden pasture]

suppose u isnt surjective

im(u) is proper subset of M

im im u is proper subset of im u is proper subset if M

im im im u is proper subset of im im u is proper subset of im u is proper subset if M

suppose somehow

i just remembered a very nice reason

injective means f(x)=f(y) implies x=y

so if f^n=f^{n+1} can just cancel away all the f

hence im f=the whole thing

[chilly ocean]

gg thanks very much

[chilly ocean]

I’m a little confused with this

A sub ring of a noetherian ring need not be noetherian

But we do have if 0,→A→B→ C→0 is an exact sequence

B noetherian implies A and C are noetherian

Does the second not imply the first

[rustic crown]

consider a non-noetherian integral domain R and take its field of fractions...

[rich ravine]

Here you are talking about modules over a fixed ring

[chilly ocean]

Yeah

[rich ravine]

A ring is noetherian if it is noetherian as a module over itself

[rustic crown]

fields are definitely noetherian

[chilly ocean]

Sorry

A sub ring of a noetherian module I am working with

I am tying to do this exercise in atiyah Macdonald

[rich ravine]

Oh sorry I misread your question

Are you asking, if A and C are noetherian then does that imply that B is noetherian ?

[chilly ocean]

Sorry I’m not being very clear

I believe the following two statements to be true

1. Let M be a noetherian module over a ring A. Let N be an A-submodule of M. It does not follow that N is noetherian.

[rustic crown]

it does follow that N is noetherian... i think

[chilly ocean]

2. Let 0 to M'' to M to M' to 0 be an exact sequence of A-modules. Then M is notherian iff M'' and M'' are noetherian

[rich ravine]

Yes it does ; a chain of submodules of N is also a chain of submodules of M

Point 2) is ok

[chilly ocean]

oh

oooh

its quoitents that it does not work for

big ty

[rich ravine]

No it also works for quotients

[rustic crown]

M is noetherian if and only if N and M/N are noetherian

[chilly ocean]

damn

[rich ravine]

The inverse image by the canonical projections of your chain stabilizes, so the image of the inverse image = the original chain stabilizes

[chilly ocean]

is there some "nice" property where noetherian doesn't carry over

I thought I remember hearing some warning that M noetherian and N being some nice relation to M does not imply notherian

maybe something like

If notherian things hold for N, it doesn't imply it holds for M unless M is also finiteyl generated?

I

I'm really not sure what I am trying to think of

[rich ravine]

You have to be careful when talking about noetherian rings maybe, but modules over a fixed noetherian ring behave really well

[chilly ocean]

hmm thats strange

since a noetherian ring is just a ring that viewed as a module is noetherian

is there a classic example of what can go wrong with rings

[rich ravine]

See the answer of det earlier

[chilly ocean]

thats very strange

ooh

so let R be our non noetherian integral domain

let k be its field of fractions

k viewed as a k-module is noetherian

so we say k is a noetherian ring

R is a submodule of k, so it is noetherian as a k-submodule

but it need not be noetherian as an R-module

[rich ravine]

It is not a k-submodule

[chilly ocean]

oh

[rich ravine]

Otherwise it would contain k

[chilly ocean]

oh yes, k is an R-module

but R certainly R will not be a k-module

[rich ravine]

For 1) look at where a given vertex is sent under r^i

Wait ; I assume you define D_2n as the group of isometies of the regular n-gon ?

If my memory is right, the regular n-gon is the convex enveloppe of its n extremal points (say by definition) and an isometry preserves convex combinations, so an isometry of the regular n-gon is uniquely determined by its actions on the n extremal points

So you can indeed just look at what happens on vertices

It is the definition of 2pi that rotating once by 2pi goes back to where you started

Well maybe do a first proof assuming that you can just look at what happens on vertices

And then use your built up intuition to do a full rigorous proof

Have you actually shown that |D_2n|=2n ?

Maybe distinguish odd n and even n

It should work out better

For 3) you can just say that the determinant of a symmetry is -1 and the determinant of a rotation is 1

This is something really useful to remember

It can tell you whether a given element is actually a symmetry or a rotation

A rotation matrix of angle theta (or maybe -theta, I'm not sure) is (cos(theta) sin(theta))(-sin(theta) cos(theta))

so the determinant is 1

All those things are talked about when classifying isometries in euclidean space

and specifically in R^2

A non-trivial symmetry satisfies s^2=id and s!=id

Then you can show that there is a direct sum decomposition R^n= ker(s-id)+ker(s+id)

Pick a basis of each, this represents s by a matrix with -1s and 1s on the diagonal

How much linear algebra have you done ?

Ok let's do it another way. Your reflection has a fixed axis, and an orthogonal axis. Any vector on the fixed axis is fixed, and any vector on the orthogonal axis is sent to its opposite

(we're in R^2)

So you pick a vector on the fixed axis and a vector on the orthogonal axis

say x_1 and x_2

Those form a basis of R^2 since the axis were orthogonal

Actually take them to have norm 1

So they form an "orthonormal" basis of R^2

Then in that basis your reflection has matrix (1 0)(0 -1)

Well about the new x axis

[obsidian path]

can anyone explain part b a little more to me?

[rich ravine]

But yes

[obsidian path]

oh sorry for interrupting go on

this should've went in lin alg anyway

perhaps

[rich ravine]

Yes, by definition the fixed axis in given by the axis origin-first vertex

There's a neat trick now

Yes

The trick is to notice that the map {rotations}->{symmetries} given by composing with a fixed symmetry is a bijection

by symmetries I mean reflections sorry

It's not exactly that

Actually composition of isometries are isometries

is more like what we want

There are as many reflections as there are rotations

Also a rotation is determined uniquely by what it does to the first vertex

so you should be able to use this somehow for 5)

Np, good night !

@obsidian path Would you care to explain a bit more what you don't understand ?

[obsidian path]

yeah so

is it asking to show that the rank of T is n?

[rich ravine]

No ; you can see the columns of A as column vectors, that is elements of F^n (F is your ground field)

Then the span of those vectors is usually what's called the image of A, noted im(A)

Then it's asked to show that im(A) and im(T) have same dimension

[obsidian path]

I see, it's weird because I can't differentiate between A and T

[rich ravine]

A is a matrix, T is a linear transformation

Those are really different

A is just a tabular of number

[obsidian path]

I thought matrices and linear transformations were isomorphic?

[rich ravine]

Yes exactly

[obsidian path]

oh that's what they want shown? lmao

[rich ravine]

The point is to show that this isomorphism preserves the notion of rank

[obsidian path]

I see.. makes much more sense now thanks

but now the question is like

how do you relate the coefficient matrix to F^n perhaps

[rich ravine]

Note that the isomorphism between matrices and linear transformation is really non-canonical, you have to chooses bases

[obsidian path]

makes sense

[rich ravine]

But there is one vector space that has a "canonical" basis, that's F^n. So you can view a matrix as the associated linear transformation F^m -> F^n in the canonical bases

You then have to see that under this identification, the span in F^n of the columns is actually the image of the associated linear transformation F^m -> F^n

Then you've basically reduced to point a)

Actually not basically maybe, but you have to show then that the situation reduces to point a)

[obsidian path]

so I want to express the linear maps in terms of matrices, show its rank and show that its preserved in the linear maps as well? sort of?

what is F^m I'm not sure I understand that point

sorry last question

[rich ravine]

I don't understand what you've said

m is the dimension of X, sorry should have said that

[obsidian path]

ah its okay, I'll just think more about what you said

ah I see

hm

oh makes sense

[rich ravine]

The point is that you have to be careful about wether you're handling matrices or linear transformations

[obsidian path]

true, I see that

[rich ravine]

Because there is a subtle distinction between the two

[obsidian path]

alright thanks! appreciate your thorough explanation (:

think I can do this now

[rich ravine]

Ok, have fun !

[fringe lagoon]

https://cdn.discordapp.com/attachments/767576686324875265/810672358452756480/485820733537386526.png

[chilly ocean]

can you biject Z with Q and be done?

[fringe lagoon]

i know there is a bijection between the 2, but how does that help?

[chilly ocean]

let $\phi: \bZ\rightarrow \bQ$ be the bijection, then define $a \oplus b = \phi^{-1}(\phi(a)+\phi(b)), a\otimes b = \phi^{-1}(\phi(a) \times \phi(b))$

er, sorry, i made a small error (forgot to pull back to Z)

[cloud walrus]

8da

[fringe lagoon]

ok that blew my mind, idk if this makes sense but why exactly is this allowed? since Z lacks fractions, wouldn't getting Q into the mix be sort of cheating?

i guess the edit answered that mb

[chilly ocean]

but what do we even mean "fractions"? if we are interested in giving Z a field structure, then at this point Z is still just a set, so there is even no notion of a fraction

[fringe lagoon]

maybe scalar multiples of the "standard" multiplicative inverses? then it doesn't matter since we are not talking about that at all?

[chilly ocean]

yeah, the field structure i just gave on Z has nothing to do with the usual addition and multiplication on Z

it is just sort of a fake Q, or more like Q in disguise

[fringe lagoon]

ok, that makes a lot of sense

btw that was really clever lol

[fringe lagoon]

since in a "set" setting, Q and Z are isomorphic?

[chilly ocean]

yeah

[fringe lagoon]

oki, ty very much

[chilly ocean]

EPSILON

[carmine fossil]

You should be considering the position where vertex 1 was

Initially

Not where it is now

s(1) should be 1 ,s(2) should be n

and s(n)=2

[latent anvil]

nice!

Sort of nice

well you wouldn't be talking about a single element of a group

[carmine fossil]

Find a point fixed by (rs)

And show a point to left side of it ,after applying (rs) goes to the right side

Yes

I think you still have to show a fixed point exists

Or like atleast find the axis of symmetry

A reflection is defined by an axis of symmetry

The axis of symmetry for (rs)

That is for s

[carmine fossil]

For odd cases, It's just find an invariant point and the axis of symmetry will be the line through that point and origin

For even cases,it is finding a line passing through midpoints of 2 lines and origin

[thorn delta]

Let $A$ be a nonzero submodule of a rank 1 free module $F$ over a PID $R$. Suppose ${x}$ is a basis for $F$. Define the set $I = {r \in R : r.x \in A}$. It's clear that $I$ is a nonzero left ideal of $R$, so $I = (a)$ for some nonzero $a \in R$. I am trying to show that ${a.x}$ is linearly independent. Suppose that $r.(a.x) = 0$ for some $r \in R$.

[cloud walrus]

kxrider

[thorn delta]

i think i should be able to somehow conclude r = 0 using that R is an integral domain, but im tired and can't see how atm. Any ideas

(for context, this is supposed to be the base case for the proof that a submodule of a free module over a PID is free)

[chilly ocean]

i think r(ax)=(ra)x=0 implies ra=0 right? since {x} is a basis

[thorn delta]

oh.... yea i think that's correct lol. im losing my mind

[rich ravine]

Is rs an isometry, knowing that r and s are ? What is its determinant ?

[carmine fossil]

What's a determinant here?

Like what does det(a)=-1 mean here?

Like What's a

Ok,makes sense

[rich ravine]

Yes it is exactly that. We even discussed it yesterday. You have to know the classification if isometries in 2d space though, which says that there are only rotations and reflections (I'm only talking about those fixing the origin)

Wait, you wrote "a reflection is a reflection iff..." but you meant "an isometry is a reflexion iff"...

right ?

Anyway, you know that for the isometries of the regular n-gon are either rotations or reflections

So if you pick an arbitrary isometry of the regular n-gon, just look at the determinant to know what type it is

[mild laurel]

What are you confused about?

Another way to say the first part is that if you remove relations, your group either gets bigger or stays the same size. So since the group of symmetries of an n-gon has those relations (but maybe it has more) and this group of symmetries has 2n elements, any group with that presentation in (1.1) must have at least 2n elements.

I'm not sure what their argument for X_2n is, but if I had to guess, it probably shows that you can write each element of D_{2n} in the form s^a r^b with a being either 0 or 1 and b from 0 to n-1 and this representation is unique so since there are only 2n things of the form s^a r^b, the group can only have 2n elements at most

@chilly ocean

I'm saying that the group of symmetries of the n gon has at least as many relations than D_2n which means D_2n must be at least as big as the group of symmetries of the n-gon

Maybe, depends what they showed. If they showed that the representation is actually unique, then this is sufficient already

The second shows it can't be more than 2n since every element can be written in that form, but maybe it can be less than 2n because its possible that two of the things in that form are actually equal

No

The identity can be written as both 1 and r^n and s^2 for example

Right

Right

Do you see how if we add another relation to a presentation, the group can only stay the same size or get smaller

There's no way an additional relation can make your group bigger

That's all that's happening

The group of symmetries of an n gon has the same presentation as D_2n, but it could have more relations

But since the group of symmetries of the n gon has 2n elements, by maybe removing relations, the group D_2n must have at least 2n elements

Other way around

D_2n is exactly defined as having those relations

It's the group of symmetries of an n gon that might have more relations

np

[chilly ocean]

I think they just mean if the group had more relations it could have fewer than 2n elements

Fwiw D and F makes this sort of thing more precise in something like ch 6

[carmine fossil]

Atmost 2n

You can map r to a rotation in a ngon and s to a reflection along some axis

Note that this map is well defined(because of those relations)

D and F is trying to prove any group defined by that presentation should be D_2n

How did you prove order of D_2n is 2n?

[scarlet estuary]

you could also use a presentation of the group and do an inductive argument

if you prefer

this is less elegant though

mirza's demonstrates an isomorphism between D_2n and a subgroup of a symmetric group

which is imo more meaningful

[carmine fossil]

The proof is about showing that presentation actually describes D_{2n}

[chilly ocean]

I might be doing bullshit math, but I think you could say something like: the presentation of D_2n consists of r, s|r^n=s^2=1,rs=sr^-1,.. where ... Consists of some relations, a priori could be empty or not. But this can be obtained as a quotient of r, s|r^n=s^2=1,rs=sr^-1, this the latter must be at least 2n

[carmine fossil]

You can map r to the rotation and s to the reflection actions

[chilly ocean]

That's basically the explanation in ch 6 I think

[carmine fossil]

And since there are 2n such actions,no of elements in group described by the presentation is atleast 2n

[hot lake]

because if it was any less then it means there would be some more relations

[carmine fossil]

r->rotate 360/n,r^2->rotate 2*360/n... r^{n-1}-> rotate (n-1) 360/n, r^n-> rotate n 360/n
s->reflect,(rs)->reflect and rotate 360/n ....
(2n different possible actions, implying there should be atleast 2n different elements)

[hot lake]

(new relations that are not implied by the ones you already have)

[chilly ocean]

Can I check if I under stand quotient ideals correctly

Let $A$ be a ring and let $\mathfrak{a},\mathfrak{b}$ be two ideals. Then $(\mathfrak{a}:\mathfrak{b})={y\in A: \quad y\cdot \mathfrak{b}\subset \mathfrak{a}}$.
If $\mathfrak{b}=(x)$ is a principal ideal we write $(\mathfrak{a}:\mathfrak{b})=(\mathfrak{a}:x)$.

[cloud walrus]

lime_soup

[chilly ocean]

My question is then. Are the following two sets equal. ${y\in A \mid y\cdot x\in \mathfrak{a}}$ and ${y\in A \mid y\cdot (x)\subset \mathfrak{a}}$

[cloud walrus]

lime_soup

[subtle current]

If $yx \in \mathfrak{a}$, then $y(rx) = r(yx) \in \mathfrak{a}$ for any $r \in A$ as $\mathfrak{a}$ is an ideal. So, $y(x) \subset \mathfrak{a}$. On the other hand, if $y(x) \subset \mathfrak{a}$, then clearly $yx \in \mathfrak{a}$.

[cloud walrus]

evilbuggy

[chilly ocean]

ty

this is what i had in mind but i felt suss about zero divisors

[rustic crown]

I hope your ring is commutative and has 1

[chilly ocean]

yessum

Definition: a ring is a commutative ring with a unit

good.. good

[next obsidian]

Good moment

[carmine fossil]

I will make it better:A ring is an integral domain

[chilly ocean]

a ring is a field