groups-rings-fields

[next obsidian]

Unless you try and say there’s some topological ring thing and blah blah blah

[cobalt heath]

Yeah

[vivid tiger]

so every point at which being zero is possible (read: every point) ends up being detected via the ideals in the Spec

[cobalt heath]

Yep!

[vivid tiger]

What can you do with this?

[cobalt heath]

Note you can also consider topological space of Spec(M, R)

[vivid tiger]

yes

[cobalt heath]

(Well, Spem(M, R) because our 'points' are maximals)

[vivid tiger]

but the topology doesn't seem to tell you about that of M

[cobalt heath]

Can you identify the topology?

[vivid tiger]

the closed sets are just zero sets

[dull ginkgo]

disc eat

[vivid tiger]

assuming we think of points on M as being maximal ideals

[cobalt heath]

See if given a closed interval

You can find some function whose zero set is exactly the closed interval

[vivid tiger]

this is true by bump

[cobalt heath]

Ah wait, I guess we need compactness? Idk

[vivid tiger]

same for any closed set?

[next obsidian]

OKAY I THOUGHT YOU MIGHT NEED COMPACTNESS

[vivid tiger]

why would you need compactness

[next obsidian]

To show any ideal is contained in one of the nice ideals

For one

[vivid tiger]

suppose you have a closed non compact interval

[cobalt heath]

Ah, in case of M = R, you can find the desired function for closed intervals

[vivid tiger]

https://math.stackexchange.com/questions/1406035/spectrum-of-the-ring-of-continuous-functions-on-a-space

compactness be needed

Let X and Y be compact Hausdorff spaces. If C(X) and C(Y) are isomorphic as rings, then X and Y are homeomorphic.

[cobalt heath]

There are several strong correspondences between compact Hausdorff spaces (even locally compact) and their functions into C and R.

So locally compact Hausdorff can be enough

Annddd a manifold is..

[dull ginkgo]

For normal topological spaces, for every two disjoint closed sets A, B, there is a continuous function between the space and [0,1] where the fibers of 0 and 1 are A and B respectively by Urysohn's lemma

If we have a continuous f from normal space T to [0,1] such that there's an open set S that's 0 under f, then we can consider the (closed) singleton of any point in S, x_0, to be A and the complement of S to be B. Then there is a function g such that g^-1(1) = {x_0} and g^_1(0) = T\S. Then f is a zero divisor as f(x)g(x) is 0 everywhere

[vivid tiger]

https://ncatlab.org/nlab/show/Gelfand+duality

https://arxiv.org/abs/2009.09757

whole schting

[dull ginkgo]

huge fucking strengthening of the original problem lmao

[vivid tiger]

something about cosmos.

https://ncatlab.org/nlab/show/Isbell+duality

[dull ginkgo]

if f is a zero divisor, then the preimage of (0,1) under it's complementary zero divisor must be open,

[vivid tiger]

Let 𝒱 be a good enriching category (a cosmos, i.e. a complete and cocomplete closed symmetric monoidal category).

cosmos is another good word in mathematics

oh hey my keyboard has \script letters

𝒳

𝕏

and \mathbb letters

𝕊

that's pretty cool

[dull ginkgo]

Let T be a normal topological space, and consider the ring of continuous functions from T into R, C(T, R), then an element is a zero divisor iff there is an open set of image {0} under that function

that sounds like a god awful question in a book by Rudin

actually no Urysohn's is just straight topology lol

I have a friend who's good with topology, might give it to him

Might ask this in #point-set-topology

[cobalt heath]

Urysohn lemma seems to be too weak for this.

[dull ginkgo]

I don't think so

[barren sierra]

Dumb question here

where are we using that R is local?

Where does this break if m is maximal in R, but R is not necessarily local?

[dull ginkgo]

Being brain blasted by an abstract algebra problem:

Assume R is a noncommutative ring and uv = 1. u is a left zero divisor if u is not a unit @cobalt heath

vu = w neq 1
(vu - 1) neq 0
u(vu - 1) = (uv)u - u = u - u = 0
then u is a zero divisor

nevermind chat we're fucking back

[barren sierra]

ah I think I found where it breaks

[granite topaz]

Let M in GL_n(Z), and consider G = Z^n/(M*Z^n). This is an abelian group of order |det G|.

But is there any way to calculate the orders of generators?

[daring nova]

of order |det M| you mean ?

[last spoke]

All elements of GL_n(Z) have determinant +-1 🤔

[granite topaz]

It's an abuse of notation, I agree

M is an invertible matrix in R with coefficients in Z

[last spoke]

Ohhh

[granite topaz]

Maybe something with Smith normal form

Since a generating set for the quotient is easy to find, and a constructive version of the proof of theorem of classification of finite abelian groups should do the trick

[granite topaz]

Yeah, Smith normal form is computable in poly time

Took me too long to find

[daring nova]

if N is the normal form, then G / NG is simple, but how do you use that to know what G / MG is ? Do you argue that the other matrices have determinant +-1 so they must preserve the unit cube ? Then wouldn't that mean every matrix is just an homothety, as it's (unit square preserving) x (diagonal) x (unit square preserving) ?

[lilac mango]

Let $F$ be a field such that $\text{char}(F)=2$. Let $d_1, d_2\in F$ be both not squares in $F$. Prove that
$|F(\sqrt{d_1},\sqrt{d_2}):F|=4$ if and only if $d_1d_2$ is not a square in $F$.

[cloud walrus]

tirib00

[lilac mango]

So I did this but I'm a bit confused because I didn't use that char(F)=2 at any point I don't think

[south patrol]

rip

[lilac mango]

Yep

[south patrol]

Hm odd

Only thing is I've never used $$ on this bot but i assume it does work

[lilac mango]

Nope

[cloud walrus]

tirib00
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lilac mango]

The $\implies$ direction is the one that I think is the most likely to require $\text{char}(F)\neq 2$, so I'll explain what I did:

By contrareciprocal. Suppose $d_1d_2$ is a square in $F$ and take $\alpha \in F$ such that $\alpha^2=d_1d_2$. Thus,
$$\left(\frac{\alpha}{\sqrt{d_1}}\right)^2=d_2,$$

and since $\frac{\alpha}{\sqrt{d_1}} \in F(\sqrt{d_1})$, then $d_2$ is a square in $F(\sqrt{d_1})$. Take $\beta=\frac{\alpha}{\sqrt{d_1}}$. Then, $x^2-d_2=(x-\beta)(x+\beta)$, so $\beta=\pm \sqrt{d_2}$. Thus, $F(\sqrt{d_1},\sqrt{d_2})=F(\sqrt{d_1})$ and
$$\left|F(\sqrt{d_1},\sqrt{d_2}) : F\right|=2\neq 4.$$

Done

[cloud walrus]

tirib00

[south patrol]

I think this is a bit overcomplicated

In F(sqrt(d_1)), d_1 d_2 and d_1 are both squares

[lilac mango]

Ok yeah that's pretty much what I said, no?

Maybe pushing more symbols but yeah

My point is I don't see where I could have used char(F)=2

[south patrol]

You don't need to

This is the easier direction which always holds

[lilac mango]

Hmm

[south patrol]

Yeah just shorter

[lilac mango]

Ok so for the other direction

[south patrol]

Tbh I thought for this problem you had to assume the characteristic was not 2

But maybe I am misremembering

[lilac mango]

Sorry yeah I miswrote it

You are right

Anyway

If d1d2 is not a square in F, then sqrt(d2)=a+bsqrt(d1)

For a,b in F

[south patrol]

Well i guess you mean like

If [F(sqrt(d_1), sqrt(d_2)), F] is not 4

so that F(sqrt(d_1),sqrt(d_2)) = F(sqrt(d_1))

[lilac mango]

Yep

So sqrt(d2) is in F(sqrt(d1))

[south patrol]

yes

and thenw hat you said

[velvet steeple]

Kepe Learns Abstract Algebra

[lilac mango]

So we consider sqrt(d1)sqrt(d2), which is bd1+asqrt(d1)

And supposing d1d2 is not a square that means bd1 is not 0

Sorry

a is not 0

And since d2 is not a square and sqrt(d2)=a+bsqrt(d1), b is not 0

Then we take d1d2=(sqrt(d1)sqrt(d2))²=a²+2absqrt(d1)+b²d1

And since ab is not 0, then we can express sqrt(d1) in terms of elements of F

Oh I see what I did

2=0 if char(F)=2

Yep that's it

lmao

[south patrol]

Yes exactly

Uh do you know any Galois theory

Actually tbf this is boring lol

[lilac mango]

No

[south patrol]

Dw then

[lilac mango]

Thnks 🙂

[south patrol]

Np

[chilly radish]

Just for posterity, it's because you are using nakayama, which only works when your ideal is contained in the jacobson radical

[barren sierra]

Yea

Thanks

[rotund aurora]

does anyone know if Lagrange actually proved his theorem?

because groups certainly appeared later to Lagrange I think, and Euler had already proven the corresponding statement for (Z/nZ)^times

or what did Lagrange actually do

[rocky cloak]

It appears they did, but in a very different language

https://www.jstor.org/stable/2690624#:~:text=Lagrange's Theorem first appeared in,theory a few years earlier.

[velvet steeple]

#1221844496967274546 message

[boreal inlet]

If n is an odd integer, How do we prove Q(zeta_2n) = Q(zeta_n) where zeta_r is the rth primitive root of unity over Q?

[rocky cloak]

I guess prove that zeta_2n = -zeta_n

[boreal inlet]

wait that works?

I thought the group is different

[rocky cloak]

What do you mean by "the group"?

[boreal inlet]

The group of roots of unity

[rocky cloak]

Well, it is different

[boreal inlet]

So if I say this : zeta_n is a root of the polynomial x^n - 1 over Q, and zeta_2n is a root of the polynomial x^(2n) - 1 over Q

[rocky cloak]

-zeta_n is not a root of x^n - 1

zeta_n is a root

[boreal inlet]

i wanted to just put a hyphen there, not a minus

[rocky cloak]

Alright

[boreal inlet]

We see that (zeta_2n)^2 is a nth root of unity

[rocky cloak]

Indeed

[boreal inlet]

As zeta_n is a generator, (zeta_2n)^2 = (zeta_n)^r for some r, less than n

...do we even conclude anything from this

[rocky cloak]

Not really

Or idk, maybe

[boreal inlet]

😭

[boreal inlet]

can we show one side inclusion?

(zeta_2n)^2 should be in Q(zeta_n)

[rocky cloak]

Yeah, so that zeta_n is in Q(zeta_2n) should be clear

[boreal inlet]

wait how

won't it be the opposite

[rocky cloak]

Because (zeta_2n)^2 is a primitive nth root of unity.

So Q(zeta_n) = Q((zeta_2n)^2)

[boreal inlet]

Q(zeta_n) is the splitting field of x^n - 1 over Q, hence, all nth roots of unity should be in Q(zeta_n)

[boreal inlet]

ohhhhhh right, order is n

Q((zeta_2n)^2) \subset Q(zeta_2n)

[rocky cloak]

That's right. So if you can prove that -zeta_n has order 2n, that would be the other direction.

[boreal inlet]

So one side is done

[boreal inlet]

that I think is fine enough

because zeta_n has order n, -zeta_n must have order 2n

so -zeta_n is a primitive 2nth root of unity

Hence Q(-zeta_n) = Q(zeta_2n)

oh wait actually isn't this sufficient

[rocky cloak]

Yeah, it is relevant that n is odd though. So probably it should require some argument

[rocky cloak]

This part I mean

[boreal inlet]

(-1)^n = -1 for n odd

[rocky cloak]

Yeah, that's the main point

[velvet steeple]

#1221844496967274546 message

[grizzled spindle]

Is this an infinite group?

I’m working with coxeter diagrams and I’m not generally sure how I can tell if a given diagram produces a finite or infinite group

[vivid tiger]

isn't the problem undecidable

or maybe it's decidable for coxeter groups

[dire siren]

aren't the elements str, (str)^2, (str)^3, ... all distinct?

[vivid tiger]

well, okay, suppose you have some word. it'll be composed of (st)^n, (sr)^n, (tr)^n, (ts)^n, (rt)^n

but for a coxeter group the relations are symmetric

so I assume you also mean (ts)^3=(rs)^3=(tr)^3=e

wait right that's implied if m_{ij} isn't infinite

(yx)^m = (yx)^m y y = y (xy)^m y = yey=y^2=e

[vivid tiger]

@grizzled spindle so first of all you can reduce any word to a word of these, and reduce these to have n<3

I believe the point of coxeter diagrams is that they classify the finite coxeter groups

[grizzled spindle]

I mean coxeter diagrams allows edges numbered with infinity

So st would have infinite order in such groups

Thx btw

I think one important theorems of coxeter groups is that the edges give the order

[vivid tiger]

okay some class of coxeter diagrams classifies them

[grizzled spindle]

Hmm actually I think this is right

[grizzled spindle]

This doesn’t cover every word, stuff like str for example

Maybe it was a misunderstanding from the notation I used

[dull ginkgo]

is this not S_4?

wait no

[vivid tiger]

no I just derped

[dull ginkgo]

It needs the (sr)^2 relations

[vivid tiger]

(str)^n are all unfortunately distinct

[grizzled spindle]

Welp

Yeah they Cayley graph I just did

Probably should have started with that

It gives a colored hexagonal tiling of the plane

This material is so weird to work with

[dull ginkgo]

I never know how to handle these damn type of problems

[grizzled spindle]

But I think if I had like one edge that was 2 instead of 3

It would be finite

Because then the structure would close up

[vivid tiger]

only if is easy

[dull ginkgo]

Even the part of uvu = u, vu^2v = 1 => uv = vu = 1 I can't even fucking do and keep going in circles no matter what route I am trying

[vivid tiger]

vuuv = 1

[dull ginkgo]

It's not true for monoids alone so I need to somehow use some ring nonsense

[vivid tiger]

uvuuv=u

uuv = u

[dull ginkgo]

that's not a valid jump

afaik

I tried that

[vivid tiger]

it's not valid because I can't do basic algebra

[dull ginkgo]

I think there needs to be some ring property usage in the mix

but idk where, might try applying a prior exercise?

[vivid tiger]

uv -1 = uv - vuuv = (1-vu) uv

uv ≠0

not helpful quite

[dull ginkgo]

Y E A H

[vivid tiger]

uvu-u = (uv-1)u

[dull ginkgo]

u(vu - 1)

W A I T

if (1 - ab) is invertible, so is (1 - ba)

[vivid tiger]

0 = uvu-u = (1-vu)uvu

[dull ginkgo]

might be useful somewhere

[vivid tiger]

= uvu - vuuvu

= uvu-u

wow how helpful

chat, 0=0

[dull ginkgo]

I defined l = vu, r = uv

lr = 1

ul = ru = u

which implies l^2 = l, r^2 = r

that's monoidal, so they're idempotent but not enough

[vivid tiger]

hmph

[dull ginkgo]

It's where exactly to use the ring part

Because we really can't say much about zero divisors

[vivid tiger]

u(l-1)=u(r-1)=0

l(l-1)=r(r-1)=0

=(l-1)l=(r-1)r

[dull ginkgo]

i tried hijacking left/right cancellativity of l and r respectively

(l - 1) = (r-1)r^2

1 = l - (r-1)r^2

[vivid tiger]

where you get cancellaticiry

[dull ginkgo]

lr = 1

so if xl = yl => x = xlr = ylr = y

same for rx = ry

[vivid tiger]

checks out

l is a unit

r is a unit

[dull ginkgo]

1 = l - (r-1)r^2 = l - (r-1)r = r - r^2 - r = r - r + r = r

[vivid tiger]

now show that l =1 and r = 1

[dull ginkgo]

wtf how did you get that

[vivid tiger]

lr=1

[dull ginkgo]

NAH WAIT

[vivid tiger]

oh wait

does rl=1?

[dull ginkgo]

didn't show that yet

[vivid tiger]

I don't remember what r and l are.

[dull ginkgo]

r = vu, l = uv

nah wait swapped em

r = uv, l = vu so lr = 1

[dire siren]

for (1)
from vu^2v=1 you get that v is left invertible and right invertible, so it is invertible
therefore u^2v=v^-1, and so u^2v^2=1, so u is right invertible
similarly, vu^2=v^-1, and so v^2u^2=1, so u is left invertible
it follows that u is invertible

now uvu=u implies both uv=1 and vu=1

[vivid tiger]

why

[vivid tiger]

why third equals? why fourth equals?

[dull ginkgo]

god I'm a fucking idiot

[vivid tiger]

ah, duh

wait, exists left inverse and exists right inverse implies exists inverse?

ab=1, ca= 1

[dire siren]

yes, multiply the first one to the left by c

[vivid tiger]

b=cab=c

sweet!

[dull ginkgo]

huh so it doesn't require ring properties

[dire siren]

indeed

[dull ginkgo]

i swear I always need help with these

[dire siren]

they are pretty tricky

[dull ginkgo]

that's no excuse

[dire siren]

for the second one I still don't have an idea

[dull ginkgo]

1 of 3 hell problems

[dire siren]

probably not the best introductory problems in ring theory

[dull ginkgo]

Trying to distill your proof into algebra so I don't have to write down as much in my workbook

nvm I barely understand it

wait

i miswrote it

Trying to do it somewhat symbolically

[dire siren]

basically what I did after I found that v is invertible was to multiply the relation by v^-1 to the left and then to the right by v
and then the other way around

[dull ginkgo]

vu^2v = 1
u^2v = vu^2(vu^2v) = vu^2 = w => wv = vw = 1

uvu = u => vu(uvu) = vu^2 = w => uv = 1

Making it easier to read with x and y

[delicate orchid]

every single one of these problems look disgusting

[dull ginkgo]

$1 = xy^2x \Rightarrow xy^2 = xy^2(xy^2x) = (xy^2x)y^2x = y^2x$ \
$yxy = y \Rightarrow xy^2 = xy(yxy) = (xy^2x)y = y \Rightarrow xy = yx = 1$

[cloud walrus]

Mizalign #1 simp

[dull ginkgo]

Now for part 2

assume uwu = u implies v = w

uvu = u => u(vuv)u = (uvu)vu = uvu = u => vuv = v

[summer path]

unexpected uwu

[dull ginkgo]

I didn't realize I posted that with "uwu" and I feel like a us soldier in vietnam after committing a war crime

anyway

because of that I am changing variables

I am trying to think of a way to hijack the quantifier for z there to show xy is 1

actually, gonna try an idea

[solemn dew]

does anyone know how a eigenvalue of a tensor product L x L : V X V -> V X V woud look like?
where L: V -> V

obviously thiis would be L(V) x L(V), but how would you write a eigenvector?

[dull ginkgo]

Assume uxu = u <=> v = u

Then uxu = 0 <=> x = 0
Proof: uxu = 0 => uxu + uvu = uvu = u = u(x + v)u => x + v = v => x = 0

u[v,u]u = u(vu - uv)u = (uvu - u^2v)u = (u - u^2v)u = u^2 - u(uvu) = u^2 - u^2 = 0 => [v,u] = 0 thus u and v commute

[dire siren]

@dull ginkgo part (2) is really tricky, I just figured how to solve it
||uvu=u => u(vu-1)=0 => u(vu-1)u=0||
||therefore u(vu-1+v)u=u, so vu-1+v=v, so vu=1, and similarly uv=1||

[rocky cloak]

If u is an eigenvector of f and v is an eigenvector of g, then u(x)v is an eigenvector of f(x)g. So the eigenvectors of L(x)L are exactly the products of any pair of eigenvectors of L.

[dull ginkgo]

@dire siren Assume uvu = u and v is the only x that satisfies uxu = u is v
Lemma 1: uxu = 0 <=> x = 0
uxu = 0 <=> uxu + uvu = uvu = u(x + v)u = uvu <=> x + v = v <=> x = 0

Lemma 2: ux = 0 <=> x = 0
ux = 0 => uxu = 0 <=> x = 0

Lemma 3: v u= 1.
0 =/= vu - 1 => u(vu - 1) = uvu - u = u - u = 0 => vu - 1 = 0 [CONTRADICTION]

i can probably condense the fuck out of this

[solemn dew]

okok but from the definition of an eigenvector, how is this true?

what I mean is that if you have eigen vector, say 111, and 123

then 111 x 123 would be an eigenvector of the tensor product

but how do you check this?

1 * (111 x 123) = (1* 111 x 123) or something?

[dire siren]

good job
basically I used the same ideas

[dull ginkgo]

actually wtf did I use contradiction

[dire siren]

yeah, no need of contradiction

[solemn dew]

by the way i asked long time ago why V (x) V / W is isomorphic to Λ^2V, where W is subspace to V (x) V

i will show you the proof tomorrow

if you remember it haha

[dull ginkgo]

u(vu - 1) = uvu - u = 0 => u(vu - 1)u = 0 => u(vu - 1)u + uvu = uvu = u(vu - 1 + v)u = u => vu - 1 + v = v <=> vu = 1

hypercondensed my proof

Let me jot down all the bullshit

This is the second part of the ring section

[rocky cloak]

Like W is the span of things of the form v(x)v you mean?

[solemn dew]

yes

[rocky cloak]

f(x)g (u(x)v) = f(u) (x) g(v) = ab u(x)v where a is the eigenvalue of u and b is the eigenvalue of v.

[rocky cloak]

Well isn't that just the definition of the exterior power?

[solemn dew]

i mean probably

but it is not how my professor defined it

[solemn dew]

alright thanks

trying to find some eigenspaces 😄

[rocky cloak]

Well, there's probably many equivalent definitions, so it just comes down to which one you're working with

[dull ginkgo]

Here's the whole fucking proof:
u is a unit with u^-1 = v iff either:

1. uvu = u and vu^2v = 1
   2)uxu = u <=> x = v

Necessary:

1. vu^2v = 1 => vu^2 = vu^2(vu^2v) = (vu^2v)u^2v = u^2v
   uvu = u => vu^2 = vu(uvu) = (vu^2v)u = u => uv = vu = 1

2. u(vu - 1) = uvu - u = 0 => u(vu - 1)u = 0 => u(vu - 1)u + uvu = uvu = u(vu - 1 + v)u = u => vu - 1 + v = v <=> vu = 1
   u(vu - uv)u = (uvu - u^2v)u = (u - u^2v)u = u^2 - u^2 = 0 => u(vu - uv)u + uvu = uvu = u(vu - uv + v)u = u <=> vu - uv + v = v => uv = vu = 1

Sufficient:
uvu = u(u^-1)u = u
vu^2v = 1 satisfying one side of the OR

holy fuckdoodle

[solemn dew]

i solved my problem thanks brotha

[celest furnace]

I feel like your use of swear words are greatly exaggerated. Surely you didn’t feel so adverse to this problem to use such strong language.

[solemn dew]

would had taken me 1 more hour if you didnt help

!!

very grateful

[dull ginkgo]

this one sucks

rather inconvenient

by sucks it's mainly just a lot of steps

[celest furnace]

Sir to me it sounds like you just don’t enjoy the problem

[dull ginkgo]

true!

Not bad nvm

[summer path]

are you sure you arent having fun

it looks like you're having a good time

[dull ginkgo]

I committed to myself that I would get this section done today

but the problems take

a while

[dire siren]

the direction should be "<=" or "<=>" , not "=>", but it seems like all the steps are indeed reversible

the problem also asks to prove that those expressions are units

[dull ginkgo]

I did that in my actual notebook

oh dear god

[velvet steeple]

#1221844496967274546 message

[tribal niche]

what does it mean to add e.g. m1 and m1' ?

is the + there just a formal symbol

[rustic crown]

nup. each M_i is already an R-module

there + in M_i is well-defined

[tribal niche]

but + in M1 could be different to + in

oh nvm, i'm dumb. thanks 🙂

[hollow mica]

• with k+1 different meanings

[dull ginkgo]

I've been on this problem for like an hour and it's really stumping me

I've proven thus far that e = e^-1:

f(e^-1) = ef(e)e^-1 = f(eee^-1) = f(e) => f(f(e)) = e = f(f(e^-1)) = e^-1

But I seek to show that e commutes with every element

[cobalt heath]

Wait, does (iv) hold once we assume theta(x) = 1 - x

It seems something is off to me

[dull ginkgo]

worth a shot to try to compute

just let y =/= x^-1

[dull ginkgo]

I don't think so actually?

f(x)f(y) = (1 - x)(1 - y) = 1 - x - y + xy
f(f(x)f(y)) = x + y - xy = y + x(1 - y)
f(f(x)f(y))(1 - y) = (y + x(1 - y))(1 - y) = y(1-y) + x(1-y)^2

i think I'm gonna skip it and the problem's flat out typed wrong

-1 (1 - x) x^-1 = 1 - 1/x so that part's true at least

I'll go fuck myself

I spent so long on a problem not even stated correctly

[cobalt heath]

Ahh, it happens

[dull ginkgo]

I wonder how it's supposed to be stated

[dull ginkgo]

I think I'm burnt out

I can't even get myself to prove that if AB = 1 then BA = 1 for matricies

$(AB){i,j} = \delta{i,j}$

[cloud walrus]

Mizalign #1 simp

[dull ginkgo]

so $(BA){i,j} = \sum{n = 1}^{N}{A_{i,n}B_{n,j}}$

[cloud walrus]

Mizalign #1 simp

[pearl wasp]

Enab. NJ.

{ENB.. ANI

[dull ginkgo]

waht

[cobalt heath]

Ya know, this is not at all easy

[dull ginkgo]

winging it now

[dull ginkgo]

It's the first exercise for this section

so it has to be

nope I am an idiot idk what to do

I've tried 3 ways where non yielded any results

[coral steeple]

Have I shown that maximal ideals are prime? Suppose for contradiction ab is in a maximal ideal M contained in R, but neither a nor b are in M. Then M+Ra=R, so Rb=(M+Ra)b=Mb+Rab. However, Rab and Mb are both contained in M, so Rb is contained in M. Thus M+Rb is strictly contained in R, contradicting maximality of M.

I don't quite trust the formal ideal operations here.

[dull ginkgo]

I can't even use the fucking determinant

I'm just going to hijack a lemma used in the text

Assume AB = 1, then det(A)det(B) = det(B)det(A) = 1. Let A* = det(B)adj(A). Then A*A = det(B)adj(A)A = det(B)det(A) = 1
A*ABA = A*(AB)A = A*A = 1 = (A*A)(BA) = BA thus BA = 1

fuck that

it was asserted combinatorically that adj(A) and A commute so

I just used that

[chilly radish]

If A and B are square (otherwise you couldn't use determinant either) this is just the assertion that a linear endomorphism is injective iff it is surjective iff it is an isomorphism

[dull ginkgo]

It wasn't introduced via that formality yet

just through the ring operation

Though I understand that and would use it, I am trying to keep with the pace Jacobson is setting

[vivid tiger]

AB=1. Then BAB=B.

det(A)det(B)=1 shows that B is invertible. QED.

[dull ginkgo]

he did prove that A and adj(A) commute and their product is det(A)I so

[vivid tiger]

isn't this it?

[dull ginkgo]

Jacobson uses the adjugate to get a formal definition of the inverse

[vivid tiger]

yes A^{-1}=adj(A)/det(A)

[dull ginkgo]

that's how he defined it lol

[vivid tiger]

just by linear algebra

[dull ginkgo]

so my A* is B

[vivid tiger]

even without adjugates you see that det(B) ≠ 0 and so B is invertible

simply because det(B) ≠ 0 so rank is the whole dimension

and so it's surjective

[dull ginkgo]

yeah I guess that works, no need for stronger statements

[vivid tiger]

by rank nullity it's injective.

[dull ginkgo]

no need for stronger statements

[vivid tiger]

and so this bijective linear map has an inverse

that you can show is a linear map

and then you are done

[dull ginkgo]

This is the ring theory section fam

[vivid tiger]

fam "matrices"

[dull ginkgo]

I haven't gotten to modules yet

[vivid tiger]

fam "matrices". you know what a matrix is

[dull ginkgo]

It's an endomorphism of a finite-dim module

[vivid tiger]

an example about linear algebra will assume you know linear algebra

idk if the corresponding statements i made are true for endomorphism rings of modules.

[dull ginkgo]

that works but once again, I am provided the operation and I am using that

I don't want to overcomplicate it yet

[vivid tiger]

you are overcomplicating it

adjugate stuff is annoying algebra

[dull ginkgo]

that's how jacobson defines it

[vivid tiger]

all you need is the statement that det(B) ≠ 0 => B is invertible

[dull ginkgo]

I am following the goddamn textbook

[vivid tiger]

you can use the adjugate stuff to see that statement

but otherwise no need to introduce it

[vivid tiger]

and i am following god

[dull ginkgo]

fair enough I reintroduced a proven statement

[vivid tiger]

unlike this unholiness

:studyingmanmadehorrorssothattheyarenolongerbeyondmycomprehension:

[fading field]

lol

[dull ginkgo]

AB = 1, det(B) is a unit, thus B^-1 exists (not constructing it like I did)
so BAB = B => BABB^-1 = BA = (BA)(BB^-1) = 1

[fading field]

[dull ginkgo]

jacobson is a textbook

very masochistic textbook

[fading field]

did you row-reduce the 4x4 matrix yet

[dull ginkgo]

what

[fading field]

nothing

[dull ginkgo]

I had to row reduce a non-sparse 4x4 in Z/7Z for a fuckin assignment before

god

[vivid tiger]

why is that so bad

it's 4x4

and in Z/7Z

[dull ginkgo]

pain in the balls

takes forever

[south patrol]

Lol

That sounds like a waste of time as an exercise

[dull ginkgo]

yes

[south patrol]

In a sense maybe you are the masochist if you chose to do that problem

[south patrol]

Here's an interesting q for tou

If A is a matrix over a commutative ring R, then it is injective iff det A is nilpotent

I only learnt this like this year lmao

[dull ginkgo]

oh jesus

actually no

[south patrol]

I think one direction is easy but yeah

Wait this sounds false as stated

I mean a non-zero divisor

[dull ginkgo]

[south patrol]

I think

[dull ginkgo]

if det(A) isn't a zero divisor

[south patrol]

Hmmmmm

[fading field]

uh

check identity

your statement is false

[fading field]

this is better

oh y’all said that

i have selective sight

[south patrol]

Yes that is correct

[south patrol]

What lol

Identity is not nilpotent

[fading field]

oh i misread your statement anyways

LMAO

oops

shhhh

[south patrol]

Lbut zero divisor is what I meant

[fading field]

wait i am right

[hollow mica]

wait in ufds aren’t primes and irreducibles the same, so the unique factorization up to associates can be primes or irreducibles?

[fading field]

det of identity matrix is not nilpotent and also identity is injective i meant!!

[south patrol]

Lol

Sorry I may have said injective rather than not injective

Me eepy

[fading field]

lol

[south patrol]

Hopefully obvious that that was a mistake

Lol

[dull ginkgo]

wait a fuckin second

isn't A injective iff det(A) is 0

[fading field]

no what

[south patrol]

Do you mean non injective

[dull ginkgo]

nvm that's for field

[south patrol]

Anyway

[fading field]

the identity is injective

[south patrol]

Yes that is for a field

And yes non injective

Note that the n=1 case is very clear

multiplication by a is injective iff a is not a zero divisor

Lol

But yeah n > 1 is interesting

I've never sat down to prove this I should say aha

[dull ginkgo]

I thought A is injective iff det(A) neq 0

[south patrol]

that's over fields

[dull ginkgo]

not that it's a zero divisor

[south patrol]

Non zero divisor

[dull ginkgo]

oh

[south patrol]

Ut yeah

That's over fields or even domains

[dull ginkgo]

idk fam, i want to finish these problems and head to bed

it's 11:36

[south patrol]

Since domains embed in fields lpl

Sure dw

It's 3:36 Here

😭

Hmm to be honest neither direction of this is clear to me lol

[dull ginkgo]

I still at some point want to try to do some schenanigans to prove Hilbert Basis Theorem

[south patrol]

I think like there's an easy way for one of them, like

[fading field]

(Z/4Z)^2-> (Z/4Z)^2, (2 0; 0 1) determinant is 2 and it’s not injective since (2,0) and (0,0) map to the same thing

[cobalt heath]

Ah, Hilbert basis theorem

[dull ginkgo]

I don't like the standard leading-coefficient proof

[cobalt heath]

Are there alternative proofs?

[fading field]

hilbert smaysis theorem

[dull ginkgo]

I came up with one but I didn't like its use of choice to show every ideal is in a prime

[cobalt heath]

I guess some categorical shenanigans

[south patrol]

Well comm alg needs choice for most interesting things ig lol

But hm

Idk i remember looking for alternative proofs and found the one in e.gm atoyah macsinald is probably optimal

[fading field]

atoyah macsinald

[south patrol]

Lol

[vivid tiger]

rank nullity

[cobalt heath]

Somehow I never thought of looking into alternative proofs

[south patrol]

For a field

[fading field]

modules

[south patrol]

Lol

[vivid tiger]

noooooo

(rank nullity is false for modules, then?)

[fading field]

@vivid tiger counterexample here

[vivid tiger]

ah

[cobalt heath]

Module shenanigans

[vivid tiger]

so indeed the det thing is also false

maybe you can say some stuff given info about the torsion

[fading field]

i mean rank nullity might not be false actually

[dull ginkgo]

Assume R is noetherian.
Assume we have a strictly increasing chain of ideals K_n in R[X].

Now, ideals in K_n at some point has to intersect both R and (X), as otherwise it would stabilize in R, or stabilize in R[X]/(X) ~= R
Now, the intersections A_n = R \cap K_n is an ideal of R, and B_n = (X) \cap K_n is an ideal of R[X]/(X) and thus has an isomorphic ideal in R

I wonder if it's impossible for K_n neq K_m but A_n = A_m, B_n = B_m

[cobalt heath]

Also submodules can have larger rank, right

[south patrol]

Well yes that is the point of my question

[dull ginkgo]

NO WAIT

[fading field]

because it’s a statement about the matrix itself

[dull ginkgo]

R[X}\R = (X)\{0}

[vivid tiger]

right, forgot that bit of weirdness

[dull ginkgo]

wait does that actually fucking prove hilbert basis

[fading field]

no it celibate proves it

idk i didn’t read it i just wanted to say that

[vivid tiger]

abstinence is what you mean

hilbert's basis theorem has a proof

[south patrol]

Anyway I found smth nice online lol. ||if A is not injective then so is its nth tensor product, e.g. because if v1...,v_n are linearly independent then r v_1 ^ ... ^ v_n = 0 for some r||

[vivid tiger]

im used to it just being like a counting segumentt

err, a diagonal argument

[cobalt heath]

How is it possible to not intersect R

[dull ginkgo]

I mean R*

R{0}

not TRIVIALLY intersect R

[south patrol]

What about the ideal (x+1) of R[x]

[vivid tiger]

i remember the proof in lang being boring

[south patrol]

It is, I don't think it is worth worrying too much about tbh

Hmm

[cobalt heath]

Why is it stabilized once it is in this one?

[dull ginkgo]

Let me kind of work it out a bit nicer

[south patrol]

I mean this is just false lol

X+ 1 for example

[cobalt heath]

0 still is in this ideal

[dull ginkgo]

is (x + 1) \cap R not an ideal of R?

[south patrol]

0 is in every ideal

[cobalt heath]

Ye

[south patrol]

Lol

[south patrol]

Yes this is always the case

[cobalt heath]

Ah, so it means nontrivially intersect

[south patrol]

The intersection is 0

[dull ginkgo]

what about if we have some zero divisor bullshit

nvm not for x + 1

[south patrol]

Yeah it's not a zero divisor over any ring

[dull ginkgo]

is (0) not an ideal. I'm not restricting myself to nontrivial ideals, just proper ones

[south patrol]

It is an ideal

I said it was

Maybe it was unclear sorry

[dull ginkgo]

oh just my wording

My point is, if we have two ideals A and B of R[X], where A \cap R = B \cap R, and A \cap (X) = B \cap (X), then does that mean A = B

[cobalt heath]

I still don’t see how this means something

[dull ginkgo]

Because R \cup (X) = R, and their intersection is {0}

[cobalt heath]

Hmm, I have a doubt about this for general ring

IIRC this likely hold over UFD (faint memory tho)

[dull ginkgo]

NO WAIT IT DOES

It's just set theory

[cobalt heath]

Huh

[hollow mica]

I’m trying to show these two definitions of Noetherian rings are equivalent:

1. no strictly ascending sequence of ideals
2. all ideals are finitely generated

for 1 -> 2 contrapositive is easy
but for 2 -> 1, I’m trying to do contrapositive but it’s harder. if I_1, I_2, … is a strictly ascending sequence of ideals, then I can take r1 in I_2 \ I_1, r2 in I_3 \ I_2, and so on, and I want to show that the ideal I that is generated by r1, r2, …, is not finitely generated.

[cobalt heath]

Hm, how about (X + 1) vs. (X(X+1))

[dull ginkgo]

$R[X] = R \cup (X)$ and $R[X] \cap (X) = {0}$
\
Let $A$ and $B$ be ideals,thus both must contain $0$. Assume $A \cap R = B \cap R$ and $A \cap (X) = B \cap (X)$, then
$A = A \cap (R \cup (X)) = (A \cap R) \cup (A \cap (X)) = (B \cap R) \cup (B \cap (X)) = B \cap (R \cup (X)) = B$

[cloud walrus]

Mizalign #1 simp

[cobalt heath]

X + 1 is neither in R nor in (X)

[dull ginkgo]

so?

that's not my point

It intersects both

wait

[cobalt heath]

R[X] /= R \cup (X), so you cannot simply use union

[dull ginkgo]

it's the sum

I am wrong

you're right

[vivid tiger]

is the problem that you were given Hilbert's basis theorem as an exercise?

[hollow mica]

ok I see now: if I is generated by a finite number of the r_i then eventually one of the ideals in the sequence would contain all of the r_i, contradicting the definition of some r_i for sufficiently large i

[vivid tiger]

how does T Stepped differ from T Teppa?

[dull ginkgo]

no, it's something I've wanted to do for a long time

[vivid tiger]

as an exercise

[hollow mica]

They copied the name from me

[dull ginkgo]

yes

[vivid tiger]

because you are the ultimate one?

is this why you don't have me blocked?

[dull ginkgo]

But every ideal K of R[X] must intersect either (X) or R I assume

not lie strictly within

nvm

(X + 1) still doesn't

[hollow mica]

Lmao what did you do

[cobalt heath]

Seems like this one require choice axiom

[hollow mica]

countable choice

[cobalt heath]

Yeah

[hollow mica]

Not so bad

[cobalt heath]

IIRC there was a proof using Zorn’s lemma

Maybe transfinite induction can be used succinctly as well?

[hollow mica]

big scary words in chat

[cobalt heath]

Ah you already did the choice step. Sorry

So basically, assume it is finitely generated

[dull ginkgo]

I frankly don't like the standard proof of hilbert basis because it uses the leading coefficient map which isn't a homomorphism of rings, and I was wondering if there was a way that did

[cobalt heath]

Then, once assuming union of ideals is f.g. , you can notice something is wrong

[vivid tiger]

suppose you have a strictly ascending infinite sequence of ideals. union them all. if this was finitely generated, then

after finitely many terms in your sequence you'd get every generator

but then you'd get the union in finitely many steps

QED

[vivid tiger]

sorry what

okay I'll get my copy of Lang

[cobalt heath]

I think R[X] ~= Z[X] \otimes R, so one can maybe use tensor product

[vivid tiger]

take an ideal in the poly ring

take the ideal \fraktur(a)_iof elements that appear as a leading coefficient of a poly of degree X^i in the ideal

[dull ginkgo]

I came up with another proof a while back but it requires every ideal being in a prime ideal.

[dull ginkgo]

yepp

it's an ideal of R

[vivid tiger]

this is an increasing sequence

so it must stop

[dull ginkgo]

and you basically just use a diagonal arg

or something amongst those lines

What I really want to do is try to understand how the ideal structure of R[X] depends on R

a lot of it EXPLICITLY has to do with primes

[vivid tiger]

let a{ij} be so that fixing i and varying j gets you generators for \fraktur{a}_i

take poly's f_{ij} with leading coeffs a_{ij}

f_{ij} in fact generated our ideal

[cobalt heath]

Ah, turns out this is not easier.

[vivid tiger]

let f be of degree d in the ideal

if d>r where r was how long it took for the sequence of ideals earlier to stop

[dull ginkgo]

Like I don't want to have to invoke zorn's lemma for the primes

[vivid tiger]

then we can do X^{d-r} f_{rj} to generate \fraktur(a)_d

just because the sequence stopped and so later things are equal

thus we can match the leading term of f

likewise if <d we can match the leading term of f

QED

[cobalt heath]

Turns out, it was just Grobner basis

https://pi.math.cornell.edu/~dmehrle/notes/old/alggeo/10HilbertBasisTheorem.pdf

[dull ginkgo]

My proof that required choice used this

Because i needed the existence of a prime

[cobalt heath]

Nvm, it still uses the leading term. Hmm

[vivid tiger]

isn't this problem in some sense much of algebraic geometry

[dull ginkgo]

Yeah

[cobalt heath]

Hmmm

[dull ginkgo]

well AG i know focuses a lot on the prime ideals via the spectrum and Zariski topology

[vivid tiger]

just learn all of algebraic geometry

what's the problem?

[next obsidian]

ummackshually the existence of a prime ideal of a ring is equivalent to the ultrafilter lemma which is strictly weaker than the axiom of choice

[dull ginkgo]

[dense raven]

trying to understand group rings rn. the formal definition is a bit weird, but if im correct, group rings basically have the same vibes as polynomial rings right

[dull ginkgo]

Poly ring is a monoid ring

So yeah

[unique anchor]

Okay, it's for a field theory qn. Can one say lcm($\varphi{(n)}$,$\varphi{(m)}$) = $\varphi{lcm(n,m)}$ ???

[cloud walrus]

piss_master49

[dense raven]

if im correct, every poly ring can just be embedded in a group ring where the group is Z, right

in the same way that N can be embedded in Z

[dull ginkgo]

no,

[dense raven]

oh?

[dull ginkgo]

Wait yeah

Embedded, not iso to

[cobalt heath]

R[Z]?

[dense raven]

yeah

[dull ginkgo]

Iso to R[N] embed into R[Z]

[cobalt heath]

Yeah, that sounds like R[X, X^(-1)]

[dense raven]

i guess thats another way to view it

ok got it thanks fellas

needed an intuition check

[cobalt heath]

@dull ginkgo welp it seems like Hilbert basis theorem inherently requires choice

https://mathoverflow.net/questions/222923/alternate-proofs-of-hilberts-basis-theorem

First comment of first answer imply that

It requires dependent choice to prove Hilbert basis under standard definition.

[dull ginkgo]

There’s a few whacky things I’ve wanted to try out

Like if we have a field F, and a finite automorphism subgroup G, we can consider the fixed point subfield F^G

I wonder how F^G[G] behaves

F^G[G] is an algebra with an underlying vector space of dimension |G|… same as F

[cobalt heath]

Good to think of examples

For zeta a root of unity, F = R(zeta), then zeta -> zeta^k gives Aut(F).

Consider subgroup G generated by zeta -> zeta^(-1), then
F^G = R[zeta + zeta^(-1)].

Then basically, F^G[G] = F^G \oplus F^G w, where w^2 = 1

Hm, this is quite different from F itself.

[dull ginkgo]

I was going to try to prove something but ended up accidentally proving independence of characters lol

[long obsidian]

Hey I'm confused by some language on basic rep theory.

If p:G->GL(V) is a rep we might call V a G-module according to Fulton/Harris' rep theory. But V is only a genuine module over the field yet V is not a module over G right cause G isn't a ring right?

[fading field]

i mean it’s close enough that the terminology makes sense - for example the only additional structure that you need for an abelian group A to be an R-module is an action of R on A

G is not a ring of course

[coral steeple]

bump

[mighty kiln]

Yes

[coral steeple]

Yay, thanks

[fading field]

real

[coral steeple]

Rings are lowkey more interesting than groups

[mighty kiln]

Modules be nice indeed

[old hollow]

I'm suspicious of this proof

why is he allowed to assume the map phi respects the derivative?

[fading field]

we are allowed to use properties of maps from R^n to R like the derivative

like the output of a polynomial will be a smooth map

and we know what the derivative of a polynomial function is without algebra

(you can verify that differentiating the polynomial formally corresponds to the derivative map via calculus, is what i mean)

[old hollow]

If i understand correctly, his proof is basically:

Suppose the nonzero polynomial f maps to the zero function f~. Taking derivatives of f, eventually we will hit some f'''' which is a nonzero constant polynomial. Since the map phi sends constant polynomials to constant functions, we have f'''''~ is a nonzero constant function. But the problem is that f'''''~ = f~'''''', and we assumed f~ = 0, so f~''''' = 0, contradiction.

is this a valid summary

oh I'm an idiot.... he says "suitable derivative" to mean ∂/∂xi....

I thought he meant a suitably high order derivative

[fading field]

also suitably high order

[old hollow]

oh shit

[fading field]

but it’s a suitably high d/dx^i

no mixed partials needed

[old hollow]

hm

I really don't understand this proof then

[fading field]

do it for one variable

[old hollow]

I mean, is it following this logic at least?

(for one variable)

[fading field]

the proof for one variable goes: suppose p is not the zero polynomial. it has a nonzero term, ax^n. then the associated map of the nth derivative of p doesn’t vanish at 0 (why?)

if p itself corresponded to the 0 function, then the nth derivative of the corresponding map would have to vanish everywhere

[old hollow]

yes okay

but aren't you implicitly assuming the correspondence respects taking the derivative?

[fading field]

yes, but that is not a problem

because we know the power rule from elsewhere

here is the fact:

suppose f_p is the map associated to p. then f’p(x) = f(p’)(x)

[old hollow]

if p --> p~, and p' --> (p')~, then the claim is (p~)' = (p')~

ah yes, exactly

[fading field]

this is a calculus fact

it’s the power rule

[old hollow]

ohhhhh shit

wait that makes sense

[fading field]

i find it neat that the power rule has an actual use other than to help calculus students memorize derivatives

personally

[old hollow]

okay wait maybe this is totally ridiculous question, but I just learned about 'commutative diagrams', is this a relevant idea

[fading field]

yes

[old hollow]

the claim is that this diagram commutes......

is that right

[fading field]

yes the claim is that that diagram commutes

[old hollow]

cool! let's go

i've doen a diagram now

wow

okay wait this makes so much more sense

these diagrams look scary usually but here it boils down to the idea that this map agrees with the derivative

maybe it's more complicated for other diagrams idk

[fading field]

yeah that’s right, in fact eventually the diagrams become more intuitive than writing this stuff out horizontally

because you can just see elements of the objects in the diagram

and where they travel

and claims about the diagram commuting correspond nicely to rules about how the maps interact with each other on an element level

[old hollow]

yeah

[fading field]

(this of course is if the objects in your diagram have elements, which you can go your entire life without caring about element less objects if you want)

[old hollow]

I read that there's this tendency to move away from talking about individual elements

[fading field]

yep

[old hollow]

and that talking about arrows is usually more elegant and deeper than talking about elements

[fading field]

lol

sniped

[fading field]

yes it’s true (to me)

[old hollow]

lol

reminded me of my professor

basically im in an intro abstract algebra course

he defined a module, as a generalization of a vector space

but he said, you can rewrite the definition as just an abelian group M with a ring R, with a ring homomorphism R --> End(M), and this homomorphism records all the data of the rules of associativity and whatnot

[fading field]

that’s because the second def is more natural

hats how i think about modules

[old hollow]

well, it's definitely not immediate, if you've learned about vector spaces as abelian groups with a group action, and a module is an attempt at generalization toward a ring

[fading field]

have you ever read about group actions

yeah

[old hollow]

ah we didn't do anything with group actions

[fading field]

okay great you have

ah

[old hollow]

is this more intuitive when you learn about group actions

[fading field]

yes

[old hollow]

hmmm

let me guess

[fading field]

it’s because a module is an abelian group with a ring action

and if the ring happens to be a field you get a vector space

[old hollow]

if R is an action on S, then there is some sort of map R --> A(S), where A(S) is the set of permutatinos of S

oh shit

that's wild

okay maybe I should look into group actions...

they seem to be pretty crucial

[fading field]

yeah it’s the most natural setting for group theory

[hollow mica]

I guess scalar multiplication defines a group action into the endomorphism ring, but really this is just a ring homomorphism no? @fading field

[fading field]

since using group actions you come to the perspective that all groups are groups of symmetries of things

[old hollow]

damn i rly thought i could skip the chapter on group actions

hmm ok

[vivid tiger]

point is a module is an abelian group with a ring action

[fading field]

#groups-rings-fields message start here i have said some things on this

[vivid tiger]

you mean with a field action

note also that if you have a representation G->GL(V,k) with V a vector space over k

then if you extend by linearity to the group ring k[G]

[fading field]

too much machinery

[vivid tiger]

you get a map k[G] -> End(V,k)

but then

V is a k[G] module!

[fading field]

you don’t need this much machinery you’re showing off

[vivid tiger]

:kapow:

this is what made me care more about modules

[fading field]

hmm

i think i cared about modules not in terms of representations actually

i like homology

[cobalt heath]

V is R-module if you have a R-linear map R -> End(V)

[vivid tiger]

all the extension by linearity does is ringify

[cobalt heath]

But yeah, it is also a generalization of vector spaces

[old hollow]

okay this is

[vivid tiger]

End here is matrices

[old hollow]

idk what a group ring is lol

[vivid tiger]

k[G] is like k-linear combinations of G

[fading field]

yeah, but ring actions on X are ring homs into End(X) for an object X

[vivid tiger]

so like ℂ[G] is \sum z_i g_i

and we map this sum to a matrix

by mapping g_i to a matrix via the map G->GL

[hollow mica]

ohh right

[vivid tiger]

and then multiplying by z_i

and then adding

so that we get a map k[G] -> End

[old hollow]

ahh, yeah my professor mentioned cayley's theorem in class, and I was looking at the multiplication table, and thinking x --> gx is maybe the permutation which the element g could correspond to. I couldn't get farther than this, and he showed me the rest

but that was all my exposure to actions

anyway this chapter looks fun

lots of pictures

[vivid tiger]

example:

Z/2ℤ acting on ℝ^2 by swapping axes

[mighty kiln]

X has to be an Abelian group for end(X) to be a ring

[vivid tiger]

so 0 goes to the identity matrix and 1 goes to
[ [ 0 1]
[ 1 0 ] ]

[mighty kiln]

With addition being pointwise addition and multiplication being composition

[fading field]

well that’s good because i only care about modules

[vivid tiger]

let's call 0 and 1 in Z/2Z by the letters g and h to avoid confusion

[mighty kiln]

Hence modules

[vivid tiger]

so k[G] here would look like x g + y h

[hollow mica]

I have no idea what an abelian category is but I have a feeling what you said is true for objects in abelian categories

[fading field]

arki to the rescue, let X be an abelian group instead, or just any category enriched over Ring

[vivid tiger]

so this gets mapped to
x 0
0 x
plus
0 y
y 0

which becomes
x y
y x

[fading field]

abelian group

not abelian category

oh

idk if it’s true for abelian cats (ignore me listen to jagr)

[old hollow]

okay I'm retiring ty for your help smay

[hollow mica]

ok it was just a guess

[cobalt heath]

Not module over groups?

[vivid tiger]

so now we have ℝ^2 as a ℝ[ℤ/2ℤ]-module

which is neateroo

[fading field]

eh idk not really

[vivid tiger]

in general
R->End(V) aka a module is kinda like a "ring representation"

[rocky cloak]

It is indeed true for abelian categories. Also true for preadditive categories.

[fading field]

hm

why is it called an abelian category anyways

[rocky cloak]

Because it looks like the category of abelian groups

Like you just take the main properties of Ab were interested in, axiomatize them and boom, that's what abelian categories are

[hollow mica]

yay

[cobalt heath]

Can you embed abelian category into category of abelian groups

..what am I saying, it of course does

[rocky cloak]

At least with small abelian categories you can

[prisma ibex]

you can embed Abelian categories into categories of modules over a ring

[rocky cloak]

That's Freyd-Mitchell embedding theorem

[prisma ibex]

you need to consider RMod for any ring R not just R=Z but yeah

[rocky cloak]

Only if you need it to be a full embedding

Otherwise ModR embeds into Ab anyway

[prisma ibex]

yes, this is the statement of the Freyd-Mitchell embedding theorem...

[rocky cloak]

Indeed it is

[prisma ibex]

indeed it is, but you're quoting it in a misleading way, but whatever

[cobalt heath]

Yeah

Hmm, what was full subcategory again?

[rocky cloak]

Were it has all the morphisms between it's objects

[cobalt heath]

Ahh, I see

[hollow mica]

Mfs had an indeed it is off

[rocky cloak]

For an example of an abelian category that can't be embedded like that. Take a semisimple category were the simple objects are indexed by sets (all sets).

No ring can have that many simple modules

[still dew]

One side implication is clear

I have trouble coming up with the map for the other side

[rocky cloak]

I guess you can inductively construct an injective map G -> Q.

Starting with mapping one element x to something, then inductively extending the map by one element at a time. Use that G is countable to set up nice induction

[still dew]

Okay hmmm

So I start with say I can enumerate elements of G as {g_1,g_2,...}

then for g_1 is contained in cyclic group generated by g_1 so I send it to somewhere in Q and map every element of the cyclic group accordingly?

Then uhh

Take subgroup generated by {g_1,g_2}?

As it's contained in a cyclic group

I take it's generator

Map it so that everything falls in place

But how do I know that I can map it so that everything falls in place?

[cobalt heath]

Maybe write down the explicit conditions for when/how "it falls in place"?

That could make this clearer for you

[still dew]

Uhhhh okay...

It falls in place when there's containment in the respective images?

Like one is a subset of the other?

And it's like a chain

[rocky cloak]

Like you have the subgroups (g1) < (g1, g2), and you have a map defined on the first one

Then you want to extend it to a map from the second one, such that restricting to the first gives the original map

Think about what constraints that puts on where you map the generator of (g1, g2)

[still dew]

[still dew]

The constraint is literally this

Is there something more?

[cobalt heath]

Can you elaborate this concretely?

[rocky cloak]

Say g1 maps to 1.
And g generates (g1, g2). That means g1 = ng for some n. What would that imply about the image of g?

[still dew]

1/n?

[rocky cloak]

Yes, and in the rationals we can divide by n

Hence it works!

[still dew]

Okay!

So we divide by the map of the previous setting

Something like that?

I still have some trouble overcoming the inductive step

Okay nvm

I can start by mapping the first set to 1

And then everything follows isn't it

[rocky cloak]

Pretty much

[still dew]

Thank you

And apologies for so much trouble

I am dumb 😭

[still dew]

I think this automorphisms are just GL(Z_p)

And that every automorphism is actually just a bijective linear transformation

So I was trying to see if there was any p^2 element in GL group

I tried looking at the char and the minimal poly

But couldn't find anything

[rocky cloak]

If A^p^2 = I, then every eigenvalue of A satisfies x^p^2 = 1, so 1 is the only eigenvalue of A. Then A = I + N, where N is nilpotent.

Then use that n <= p

[still dew]

Z_p is not algebraically closed how do we know that A^p^2 has eigenvalues in the first place?

[rocky cloak]

It has eigenvalues in the algebraic closure

[still dew]

So we show for algebraic closure in general?

[rocky cloak]

I'm not sure what you mean. But if A satisfies some polynomial, then the eigenvalues will be roots of that polynomial.

So you don't even need to go to the algebraic closure when all the roots are already in Z/p

[still dew]

Like rotation matrices satisfy A^4=I but have no eigenvalues in R?

[vivid tiger]

but if there are eigenvalues, they will satisfy it

[still dew]

I agree

[vivid tiger]

thus this goes through

[rocky cloak]

Yes, but x^4 - 1 has roots not in R

[still dew]

So as char of Z_p is p x^p^2-1 factors as (x-1)^p^2..?

[vivid tiger]

why can't -1 also be an eigenvalue?

[still dew]

Hmm

I wasn't relying on eigenvalues cause I thought that I didn't have enough to say that they exist in the first place

[still dew]

Oh but it's Z_p right

And if p is 2

Then 1 and -1 are same ?

[vivid tiger]

oh right okay

indeed (x-1)^{p^2}

[still dew]

Okay but what's the definition of eigenvalue?

[vivid tiger]

anyways point is

an order p^2 element A would be such that

A=I + N

but then (I+N)^{p^2} = I

so since N^{m} for m > n is already 0

this just means that
\sum_{i=1}^{n} N^{i} = 0

but then

A^n = I + that sum = I

[rocky cloak]

||A^p = I^p + N^p = I ||

[still dew]

Yeah as we have char p right?

And I and N commute

[rocky cloak]

Yes, the characteristic p is the key through all of this

[vivid tiger]

oh, right, that too

I mean shouldn't it also be true in any char that if A = I + N and A^{m}=I and the dimension n < m then we have by my reasoning A^{n}=I?

No, we don't

because I somehow entirely forgot to put the binomial coefficients.

[still dew]

Ok so far everything makes sense to me just that why must the eigenvalue exist

[vivid tiger]

who cares if they do

[still dew]

Huh

[vivid tiger]

the point is that any that do are 1

so that in the jordan decomposition the semisimple part is I

[rocky cloak]

Notice (A - I)^p^2 = 0, so A-I is not invertible, hence there is a vector such that (A-I)v = 0, i.e. Av = v

[still dew]

Cool

[vivid tiger]

this is also true, interestint

[rocky cloak]

In particular the eigenvalues are the roots of the characteristic polynomial

[still dew]

But we never used the characteristic polynomial

We just took the anhilate poly we had

[vivid tiger]

pretty sure Jordan uses it

well, Jordan I think only uses it for existence of eigenvalues

[rocky cloak]

Well, theres nothing special about the characteristic polynomial in that regard. Just any polynomial in which the Matrix vanishes