groups-rings-fields

[chilly ocean]

I'm not sure what it outputs I guess

[void cosmos]

it must output something in Hom(M,N)

[chilly ocean]

f(phi, n) -> something

[void cosmos]

right?

[chilly ocean]

what does an element in Hom(M,N) look like? Just some homomorphism right?

[void cosmos]

Hom(M,N) is defined to be the set of all R-module homomoprhisms from M to N

[chilly ocean]

right

so if you're taking f(phi, n) what would an output look like if it were in Hom(M,N)

[void cosmos]

u suggested nphi(y)

which is correct but the way you said it did not make snse

so what u probably meant was

[chilly ocean]

that's just what I thought of, but isn't that an element of N and not Hom(M,N)?

[void cosmos]

send (phi,n) to the map g_phi,n

such that g_phi,n(x) = nphi(x)

[chilly ocean]

what is g?

[void cosmos]

.

[chilly ocean]

ah

[void cosmos]

now this g is a map from M to N

and hence this makes sense

it takes elements from M

and spits out elements in N

[chilly ocean]

is x just a variable?

[void cosmos]

yes

x is the input of the function that is ssupposed to be spitted out from our homomoprhism that we are trying to find

cuz as we said

this f must send things to the set of all R-module homomprhisms from M to N

[chilly ocean]

I don't really get the meaning of g_\phi, n

like I don't get why you need it, what's the purpose of it

[void cosmos]

we are defining a map

this map must send this (phi,n) to a map

so g_phi,n is our map that we define to be g_phi,n(x) = nphi(x)

this g is our assignment

[chilly ocean]

so like this?

[void cosmos]

yea

[chilly ocean]

ok so now we have f, we need to show that it is bilinear right? Then we can invoke the universal property?

[void cosmos]

yeds

and now then find an inverse

and we are donzo

[chilly ocean]

how does finding an inverse work? Can we not just invert our opertaions?

or is it more involved

[void cosmos]

huh

ig

we can do some magic

with the Hom properties

cuz we now that they are free

know*

[chilly ocean]

so we want to find another map from Hom(M,N) to M*xN now?

[void cosmos]

no

[chilly ocean]

oh

what does finding an inverse mean then

[void cosmos]

u now have u

u want to find the inverse to this u

so its to M* tensor N

but i think this becomes a bit easier if we like

huh

1 sec

[chilly ocean]

ok

wait so does g send (M* oplus N) to N?

[void cosmos]

Hom(M,N)

this is your u

[chilly ocean]

so is f our g in the diagram you sent?

wait

what

g is our u?

[void cosmos]

wait

im confused now

[chilly ocean]

like in the universal property diagram

is g actually g

in the diagram

and is f f?

[void cosmos]

we found g : M* x N --> Hom(M,N) so this induces a map u:M tensor N --> Hom(M,N) such that u o f(x,y)) = g

f(x,y) is just x tensor y anyways

so u(x tensor y) = g now

we want an inverse to u

[chilly ocean]

induces a map u such that u o f(x,y)) = g
this is afforded by the universal property right?

[void cosmos]

yes

since g is bilinear which we didnt show

[chilly ocean]

what does this mean?

[void cosmos]

the map f in the diagram

is the "natural" (x,y) --> x tensor y

[chilly ocean]

ah ok

ok

what does an inverse mean in this case?

[void cosmos]

an inverse map

that is a homomprhism

just like when u show any map is an isomphirms

[chilly ocean]

so like u is a homomorphism from M* otimes N to Hom(M,N) and we want to find a map from Hom(M,N) to M* otimes N?

[void cosmos]

yes

such that they are inverses

[chilly ocean]

ok I guess I'm still a bit confused about g

So we now have
u(phi \otimes n) to g_phi, n right?

and we want to find g_phi, n to phi \otimes n?

[void cosmos]

yes

no

we want to find h : Hom(M,N) --> M tensor N

such that h o u(x tensor y) = x tensor y

the identity map

[chilly ocean]

hmm

thank you for your patient by the way

I'm not really sure how to find h

[void cosmos]

try it out

remember that M = R^n and N = R^k

[chilly ocean]

why does that matter?

[void cosmos]

R^n tensor R^m is something cool

but the thing is

we could have used that long ago

cuz

[chilly ocean]

oh, the fact taht it's isomorphic to R^nm?

is that what's important?

[void cosmos]

we could have went like Hom(M,R^k) is sum(Hom(M,R)) is Hom(M,R)^k which is M* tensor R^k

but i just got lost midway ig

anyways if M = R^n for some n

then R^n tensor N --> (R+R+R+...)tensor N = ( R tensor N ) ^k = N^k

[chilly ocean]

oh

so we want h to map to N^n instead?

[void cosmos]

we want to show that our map we defined is an isomporphism

[chilly ocean]

right

we want to show that u is an isomoprhism right?

can we just show injectivity and surjectivity instead of finding an inverse map?

[void cosmos]

yes but not of this map

[chilly ocean]

oh

I still don't really get what finding an inverse really means in this case

[void cosmos]

to recapa

we have our ingenius map from R^n tensor N to Hom(R^n,N) correct?

the "u" map..

[chilly ocean]

yea

[void cosmos]

now we know R^n tensor N is N^n

[chilly ocean]

right

[void cosmos]

and we know Hom(R^n,N) is N^n too

so now we have umm

two isomoprhisms on the left and right

and the bottom is the identity

so what does this tell us

[chilly ocean]

and the bottom is the identity
wait what?

[void cosmos]

these are all isomoprhism

isomorphisms

[void cosmos]

think of the square Hom(R^n,N) --> R^n tensor N

and then from R^n tensor N down to N^n

and from Hom(R^n,N) down to N^n

[chilly ocean]

right
So, N^n cong R^n tensor N --> Hom(R^n, N) cong N^n
So now we have a homomorphism from N^n to N^n?

[void cosmos]

then from N^n to N^n we have the identity

[void cosmos]

the identity

[chilly ocean]

oh

I see

and so that implies that u is isomorphic?

like by some chain of isomorphisms?

[void cosmos]

draw it out

it will look familiar

[chilly ocean]

like
can you argue:
R^n tensor N cong N^n cong N^n cong Hom(R^n, N)
So R^n tensor N cong Hom(R^n, N)?

[void cosmos]

cong

whats cong

[chilly ocean]

$\cong$

[cloud walrus]

Pancaker

[void cosmos]

wait

let me

draw it for u

using good old pen and paper

[chilly ocean]

I get the image I think

[void cosmos]

sorry bad handwriting

[chilly ocean]

all good

ok yeah that's the kinda image I had in mind, so we can use that path as the inverse map?

[void cosmos]

yea by inverting all of these kings

tbh idk if this is 100% correct tho

cuz we didnot use N being f.g

but it works the same if we assume N is f.g but M isnt

like the same map same same everything

[chilly ocean]

oh i see, if both weren't FG then there isn't an identity map between them right?

or wait

uh

how is Hom(M,N) isomorphic to N^n again?

[void cosmos]

the direct sum - product property of the Hom

Hom( directsum(P_i) , B ) = product Hom(P_i,B)

but since this is finite dimensional

these i's are finite

so the sum and direct product coincide

Hom(R,N) is N

so we get N^n

or whatever

[chilly ocean]

interesting

ok

I see

interesting

[void cosmos]

yea honestly once i saw this problem i kept looking if i could use the most cooll fact

in math

which is Hom(M tensor N,P) is Hom(M,Hom(N,P))

10x more cool but i didnt know how to use it here so meh

[chilly ocean]

rip haha

ok so

We can construct a bilinear map from M* x N to Hom(M,N) where (phi, n) gets sent to g_phi, n = nphi(x)
We also have our basic mapping of M*xN to M*otimes N which is also bilinear.
So then we can invoke our universal property, which means that there must exist a unique homomorphism u from M*otimes N to Hom(M,N). But we need to show that this homomorphism is invertible becaue then it is isomorphic. We can do this by showing that M*otimesN is isomorphic to N^n and that Hom(M,N) is also isomorphic to N^n. Since we have the identity map from N^n and N^n, we can use that as our inverse function between M*otimes N and Hom(M,N)

is this the end idea?

barring some specific things I need to prove in the middle, like bilinearity and isomorphism to N^n

[void cosmos]

that last part

its more like

wait let me write it out

first

do u see that R^n tensor N is N^n

and Hom(R^n,N) is also N^n

[chilly ocean]

yea, I think so

[void cosmos]

then yea

thats it

[chilly ocean]

ah ok, cool! Thanks so much, that actually made a lot of sense

[void cosmos]

yea cool

gl

isomoprhism to N^n is straight forward if u have R tensor N is N

u can do that just the same say as we did this

find a map R x N to N ( f(r,n)) = rn

and then find an inverse

just to cut things short the inverse would be g(n) = 1_R tensor n

oh assuming R has the identity ofc

if R doesnt have identity then everything i said is false

[chilly ocean]

we assume that R has the identity in this class

[void cosmos]

yea cool

[chilly ocean]

do you have any tips for getting better at algebra?

[void cosmos]

no

im not any good at algebra or anything in math really

u should ask people who are more serious

[chilly ocean]

wait, does this work even tho we have M* tensor N not M tensor N? is it still isomorphic to N^n?

[void cosmos]

those are vector spaces

M and M* are isomophric as vector spaces

[chilly ocean]

oh

[void cosmos]

finite dimensional ofc

[chilly ocean]

the dual space is isomorphic to M?

M* iso M?

[void cosmos]

yup

[chilly ocean]

😮

[void cosmos]

R is a field remember

and M is f.g

so u basically have two vec spaces that have the same dimension so meh

[chilly ocean]

ah

I see

[void cosmos]

just pick a basis first

its not natural but hwo cares

[chilly ocean]

that was question 1 I think kinda

[void cosmos]

let me see

[void cosmos]

this question is good

nice

also verify that counterexample

[chilly ocean]

yea

[void cosmos]

take M and N to be Z/2Z

[chilly ocean]

is it possible to show that this isn't an isomorphism by just counting the amount of elements on both sides?

[void cosmos]

What

queen im honestly watching breaking bad rn for the 10th time

so im not totally focused

sorry

[chilly ocean]

😭 sorry

[low wyvern]

Just out of interest I am learning group theory and using Dummit and Foote whilst supplementing it with Galilean. I noticed that whilst Dummit uses * (any binary operation) in proofs of, for example uniqueness of identity element, Galilean uses x, does this make any difference in the future or can these two approaches be interchanged?

[paper flint]

The symbol used (at least in this context) is irrelevant, so you can interchange them as long as you are consistent with it

[low wyvern]

Ok, which one would you do? Also which book would you recommend me paying more attention to? i.e. focusing on as the main text...

[paper flint]

That depends on you. I think Gallian is more beginner-friendly but it has a weird presentation (I remember it talks about homomorphisms or isomorphisms very late in Part A), and it absolutely avoids group actions which I think is pretty silly in retrospect

[low wyvern]

I think dummit does look more comprehense, hence why I planned on using it

Galilean looks a bit simplified tbh, but maybe it's not and I am just assuming wrong

[paper flint]

If you're learning on your own then comprehensiveness on a first pass is often not strictly necessary

[low wyvern]

I am a physicist and have taken this as an extra course, but I still value my pure maths ability, thus wish to do a good cover of it

[paper flint]

Well yeah, it does avoid very technical jargon but I see that as a plus than a minus with its target audience (presumably someone taking a first course in algebra with little or no exposure to writing rigorous arguments)

[low wyvern]

I have done analysis 1 and am self teaching metric spaces/topology atm

I'm not too worried about how rigoures it is

[paper flint]

I see. I think there are multiple textbooks on group theory with a physics audience in mind, but I'm not sure if one of their goals is to establish the results themselves

Ah okay, then DnF should be fine I guess

[south patrol]

(usually one wouldn't use any symol for multiplication tbh)

[low wyvern]

That's what I meant but to make it clear I used x

[south patrol]

lol i think that made it less clear aha

you can call it like "juxtaposition"

[low wyvern]

I see, well Ik in future 🙂

[paper flint]

Feel free to jump around. You can always refer back to relevant sections if something doesn't click.

[sleek flare]

E = mc2

[delicate orchid]

If only I E = mc cared

[cobalt heath]

So this is the only channel where I can say I really understand

[open sluice]

groumps, rimgs, and fielmds

[cobalt heath]

I dislike recalling what shape projective and injective modules give

As a diagram

Maybe I would always try deriving it from exactness of Hom Functor 💀

[delicate orchid]

injective is "in"
projective is "out" (It projects!)

[cobalt heath]

Well, that only explains one part

[ivory trail]

here's how i remember projective: a module P is projective iff for every "bundle" (||or "quotient bundle" from the AG point of view||, epimorphism N -> M) you can lift a map into the base space (P -> M) to a map into the total space (P -> N)

and bundles are like projections

injective: a module Q is injective iff for every subspace X of any space Y (||or subbundle from the AG point of view||, monomorphism X -> Y), you can extend any Q-valued quantity on X (map X -> Q) to a Q-valued quantity on Y (map Y -> Q)

here i am thinking about functions as generalized quantities, like how a scalar field is a number that depends on where you are in the space

[coral spindle]

cool perspective

[rocky cloak]

Which part is missing? In the diagram for projectives stuff points out of the projective, in the diagram for injectives stuff points into the injective.

Or are you not recalling which map is epi/mono?

[cobalt heath]

Yep, not recalling the arrows regarding the two modules other than the projective module.

[cobalt heath]

Ohh, cool perspective! (And I just discovered someone else already used this expression)

[rocky cloak]

Okay, but there the mnemonic is the same. Injective involves a mono (an injective map) and projectives involve an epi (a projection)

[cobalt heath]

And then you need to figure out one more

Namely, what is given and what is derived as "exists"?

[rocky cloak]

Well, that one should be clear. The one that is just the composition is not derived, because being able to compose morphisms is not a special property

Like you have
$\begin{tikzcd}
& P \ar[dl, swap, "f"] \ar[d, "g"]\
M \ar[r, "h"] & N
\end{tikzcd}$

[delicate orchid]

add the c in boss

[rocky cloak]

Then the fact that f induces g is true no matter what P is, so its that g induces f that is special

[cloud walrus]

jagr2808

[south patrol]

Lol using the word derived here is confusing my brain

[glossy crag]

I'm trying to recall why $x^p-x-t$ is irreducible over $\mathbb F_p(t)$. This is equivalent to the Galois group of the splitting field acting transitively on the roots, which I know are of the form ${\alpha,\alpha+1,\dots,\alpha+(p-1)}$ for a fixed root $\alpha$, however I can't think of a field automorphism that would move the roots like this.

[cloud walrus]

Ocean Man

[rocky cloak]

How about this:

f(x) is irreducible iff t^p f(x/t) = x^p - t^p-1 x - t is irreducible. F[t] is integrally closed, and x^p - t^p-1 x - t is irreducible over F[t] by Eisenstein.

Oops t^p f(x/t) = x^p - t^p-1 x - t^p+1

So I guess that doesn't work

[glossy crag]

oopsie

[rocky cloak]

Okay, attempt number 2:

It should be clear that alpha is not in F[t], so the minimal polynomial has some other root (alpha + i), so there is a Galois automorphism s mapping alpha to (alpha + i).

Then s^n(alpha) = alpha + ni, and since p is prime ni runs through every number mod p.

[dim widget]

One way to see is eisenstein’s criterion applied at t=0

[glossy crag]

What does "at t=0" mean though.

[dim widget]

And an easy change of coords

[delicate orchid]

I'm not seeing this change of coords

[dim widget]

So I guess I actually mean at t=infty

Since the change of coords was t \to t^-1

The point being that here you can apply eisenstein’s criterion since the Newton polygon is a straight line with slope 1/p

The intuition is that geometrically your cover defines an unbranched cover of A^1 but such a thing must be totally ramified at infinity because there are no unbranched covers of P^1

[delicate orchid]

I think I'm going to cry

learnt what the newton polygon is at least, cool construction

[dim widget]

The Newton polygon is p cool

[coral shale]

https://media.discordapp.net/attachments/418981682117345288/1159189663647535254/image.png?ex=651efb00&is=651da980&hm=2b674b6cf2bff2176259947c9ca85680aa6bc1ddb664c9f93fa0979d84ae0e8a&

^ what blumming group theory

from #elementary-number-theory

[dim widget]

It’s definitely easy using ring theory!

(Blumming)

[coral shale]

ok, so SE tells me thats Z2 x Z2^k-2

[dim widget]

Because 5 = 1 + 2^2 so we can expand 5^n to second lowest order in 2 as 1 + n*2^2 + ? which is only zero when n is 2^{k-2}

[coral shale]

and i presume 5 = (1, 1)

[delicate orchid]

wow ok I see how it is

[coral shale]

whattt noooo

weewwww

ur appreciated always

[delicate orchid]

basically what I was saying but way more concise why are you complaining

[dim widget]

We appreciate all wews here

[delicate orchid]

didn't help I was making the proof up as I went

I should probably really learn the pattern of orders of elements mod n but I can just generate them with python so who cares

[dim widget]

There isn’t a pattern it’s really hard

[coral shale]

oh ic

[delicate orchid]

even for n = p^k?

like going off of the proof that 5 generates the part of U_2^k that's not stupid

hmm no wait not so clear

[dim widget]

I think the thing is that even for just n = p it’s complex

I don’t know a smarter algorithm than multiplying something by itself at most p-1 times

[delicate orchid]

yeah I mean to me it's a total mess but I always just assumed there was some clever galois theory nonsense you could do

[cobalt heath]

Yea it is clear, but it needs some thoughts

I dislike that

[frank cosmos]

Given a set A, free groups are the group component of initial objects in the category where objects are functions f:A -> group G and morphisms are defined to be commutive diagrams in the obvious sense

But, suppose (f:A->G) and (k:A->H) are both initial

then are G and H isomorphic as groups?

[delicate orchid]

initial objects are unique up to isomorphism, and what would it mean for two objects to be isomorphic in this category

[frank cosmos]

right so these 2 objects are isomorphic

which means that there is a unique homorphism a:G->H such that k=af and a unique homormopshim b:H->G such that f=bk

my intuition is that b and k are inverses but im not sure how to prove this

[delicate orchid]

sorry just trying to figure out how to write this without diagrams

[frank cosmos]

oof

uhh

webwhiteboard then copy paste screenshot? idk how much effort that is tho

[delicate orchid]

so since these objects are initial we have a unique morphism from f : A -> G to itself, the identity triangle obviously
but the triangle H <- A -> G which induces our maps a, b gives us the map ab which also satisfies this identity triangle, so we have to have that ab = Id_G, likewise for ba = Id_H by an identical argument, so a is an isomorphism between G, H
does this work?

I'm using "the identity triangle" to mean like G <- A -> G with the identity map between the Gs

[frank cosmos]

1 sec im gonna draw this lol

[delicate orchid]

the only bit I'm not 100% on is ab satisfying the same triangle

[rocky cloak]

The two objects being isomorphic means that they are isomorphic. But yeah this together with what wew said is how you would show they are isomorphic

[frank cosmos]

the 2 objects being isomprhic means that the objects are isomorphic, not nessecarily the group components r

in this case yes

still checking the details

[delicate orchid]

yeah, they want that the codomains are isomorphic as groups

[frank cosmos]

wait free groups arent isomprhic?

[delicate orchid]

they are, but that's not what you're after

[frank cosmos]

yea

[delicate orchid]

wait no inital objects in this category are free groups sorry

[frank cosmos]

lol

[delicate orchid]

I'm getting mixed up

[frank cosmos]

i was confused

[rocky cloak]

A morphism is just a homomorphism of groups with some extra properties, so an isomorphism is just an isomorphism of groups with some extra properties

[delicate orchid]

yeah I wanted to make this argument but I wasn't sure!

cones as morphisms are scaryyy

[rocky cloak]

Anyway, k=af -> bk = baf -> f = baf -> ba = 1

[frank cosmos]

but this assumes range of f is G?

i wanted to make that arg

[rocky cloak]

The image of f is very much not G...

[frank cosmos]

ok @delicate orchid i checked your stuff its perfect!

[delicate orchid]

wow that's a once in a blue moon occurrence

[frank cosmos]

you have baf=ba and deduce ba=1

[rocky cloak]

Because f, G is initial

[frank cosmos]

how does that help here

[rocky cloak]

ba: G -> G is such that baf = f, hence is a morphism in your category. f, G is initial so such a morphism is unique hence equals 1

[frank cosmos]

also unrelated question: if (f,G) is initial then that does not imply (g,G) is also initial where g is a diff function A->G right?

[rocky cloak]

No, only if (f, G) and (g, G) are isomorphic, which need not be the case

[frank cosmos]

thx

[frank cosmos]

👌

idk why i needed that spelled out to me

i think thats what wew said though right?

just condensed

[rocky cloak]

Indeed it is

This argument also, basically immediately, generalizes to any category, i.e. initial objects are unique up to (unique) isomorphism.

[frank cosmos]

yes

nice argument, i just used it to also show that free groups on finite sets of the same size are isomorphic

[delicate orchid]

there's a much quicker argument for that

the map phi is a bijection between the two sets, which exists as they're of equal sizes, the diagonals are the obvious compositions which induce the dotted maps, which are unique and so their composition must be the identity on F(X) or F(Y) depending on the order you compose them

[frank cosmos]

thats the exact thing i did lol

just condensed

[dim widget]

getting condensed

[delicate orchid]

I prefer this argument because I don't have to think about coslice categories

[frank cosmos]

lol

[warm saffron]

Had another brain fart I need some confirmation on something....I'm thinking that <3> in U(8) is precisely <3>={1,3} right?

[delicate orchid]

3^2 = 1 so yeah

[warm saffron]

oh duh I'm so dumb lol

3 has order 2 slippd my mind

thanks for the confirmation

[mighty spade]

scholze alt spotted

[sly crescent]

What is that program called?

[chilly ocean]

https://q.uiver.app/

[crystal turtle]

quiver my beloved

[molten rivet]

Not sure how to approach this other than just listing the automorphisms of D_8 and just seeing that there are generators that satisfy the normal properties of D_8. Obviously there's a quicker way involving using the fact that D_8 is a normal subgroup of D_16, but I can't see it.

[delicate orchid]

this is odd, the orbits of elements of D_8 under the action of Inn(D_16) is just the regular conjugacy classes, perhaps there's another automorphism of D_16 which fixes the copy of D_8?

ah no that's not true, {r, rt^2} and {rt, rt^3} are fused

yes yes that's exactly the action of the outer automorphism of D_8

[molten rivet]

oh wait

[delicate orchid]

the only other outer automorphism of D_16 fuses conjugacy classes of elements of order 8 I think so it's irrelevant here

but why do we have Aut_{D_16}(D_8) \cong Aut(D_8)

Aut_{D_16}(D_8) \cong N_{D_16}(D_8)/C_{D_16}(D_8) \cong D_16/Z(D_16) which is iso to D_8 so that's good

[molten rivet]

wait ok

what does the notation Aut_{D_16}(D_8) mean?

like

the action of the automorphism of D_16 on the subgroup D_8?

[delicate orchid]

yeah

[molten rivet]

ok

[delicate orchid]

all of the maps in Aut(D_8) which arise as conjugation in D_16

hmm

[molten rivet]

right

[delicate orchid]

this question seems REALLY stupid!

[molten rivet]

lol

[delicate orchid]

why would you do it like this?! nobody does it like this!

[delicate orchid]

we still don't know if these are even all the automorphisms!

[molten rivet]

a previous exercise had us prove that |Aut(D_8)|<=8 so that's fine actually

[delicate orchid]

oh then we're done

[molten rivet]

so wait

[delicate orchid]

cause we have it's iso to D_8

can you envoke the NC-theorem though that seems a bit

cringe...

[molten rivet]

I don't remember us proving that

but I'll see if I missed something

[delicate orchid]

it's the one that tells you that the normaliser quotiented by the centraliser is isomorphic to the subgroup of Aut(whatever) that's given by conjugation of elements in the larger group

[delicate orchid]

first \cong in this

[molten rivet]

I think we've shown that the centraliser is a normal subgroup of the normaliser, but we haven't shown the second part iirc

We only pretty recently started doing things with the automorphism group

[delicate orchid]

ah ok

hmm

maybe then just directly show that Inn(D_16) is D_8 and then D_8 is fixed under these maps?

so they restrict to automorphisms of D_8

and then we know that Aut(D_8) is size 8 or smaller so we're done

that's probably what they wanted you to do in hindsight

sorry

[molten rivet]

it's all good

[delicate orchid]

anyway see if you can make a rigorous argument out of that

[molten rivet]

ok lol

[cobalt heath]

How do i do math without thinking

[open sluice]

join another server

[cobalt heath]

To be precise

I want to chase diagram without much thoughts

And memorize algebraic structures such that I do not need to think

[delicate orchid]

some things just aren't diagram chases

[cobalt heath]

Yea, I mean I want to do diagram chasing mindlessly

Not all math, some deserves a thought

[arctic fog]

but why this channel specifically

[cobalt heath]

Because I face diagram chasing a lot with module theory. (And groups, rings as well ofc)

[crystal turtle]

Diagram chases are just "monkey at a typewriter"

Try your different options (you only have like 2-10 reasonable ones lol) until it works

no thought needed (usually)

[cobalt heath]

Huh, so I've been doing it wrong

Hmm, yea. So my intuition is right that I've been doing it wrong

Should I memorize rules like e.g how monomorphisms act on composition?

Like If f is monomprphism and f . g = 0, then g = 0

[topaz solar]

Monic is basically injections

Epic is basically surjection

Done gg

[crystal turtle]

mono = injection = left cancellable
epi = surjection = right cancellable

[cobalt heath]

Ahh

[crystal turtle]

(not literally = of course, but with how they work with composition and w/e)

[topaz solar]

They’re not identical, consider Z -> Q as rings, but it’s still essentially surjective because it’s all determined by the image after all

[delicate orchid]

my mummy says I'm epic

[coral spindle]

Why can you speak to a pharaoh

[topaz solar]

Who said they weren’t a Buddhist monk

[crystal turtle]

Idk "Thifford Cleory", why don't you tell me?

[broken quartz]

how would i prove that k^2/ p = 1 and that (1 + k^-1 / p) = (1+k)/p? cuz in my proof i think i just assume it, but now that im thinking about it, idk how to prove either lol.

[cobalt heath]

Huh, was that epi

[topaz solar]

Yes

Consider right cancellation

[crystal turtle]

#discrete-math maybe

[delicate orchid]

https://cdn.discordapp.com/attachments/482958702211497994/1153062592508592128/pharaoh_posting_15fiw27.mov

anyway where can you find a a group theorist with no legs

[cobalt heath]

I laughed too hard than I was supposed to, this shouldn't be this funny

[broken quartz]

oh damn i thought this was the discrete math channel lol

mb

[delicate orchid]

right where you left them

[topaz solar]

Just pretend things are sets

[cobalt heath]

I have hard time diagram chasing sets

[crystal turtle]

If you're working with modules you can literally do that

[topaz solar]

Literally just apply functions & preimages

I don’t think modules are too screwy with mono/epi either

[cobalt heath]

Yea, they aren't

[topaz solar]

Or if it’s abelian & small enough

[cobalt heath]

Tho I did not know ring homo is.. well

[topaz solar]

Because it’s essentially modules

[cobalt heath]

Problematic

[crystal turtle]

Hellfire.

[topaz solar]

It’s not problematic

[topaz solar]

Why, embedding theorem good

[cobalt heath]

Z-> Q tho

[topaz solar]

Yeah why is that problematic?

Every element in Q is determined by Z in Q

It’s basically surjective

[cobalt heath]

Yea but

One more thing to think abt.

[crystal turtle]

Maps with dense image in Haus are also epix but not necessarily surjective

[topaz solar]

Yep

Epic = essentially surjective

[cobalt heath]

Haus?

[crystal turtle]

category of hausdorff spaces

[cobalt heath]

Is it hausdirff

Ahh

[delicate orchid]

category of hausdorff mfs

[topaz solar]

It determines the whole space, not genuine surjective

[delicate orchid]

this is a little bit insane chat but makes sense I think

[cobalt heath]

I see

[cobalt heath]

Wdym

[topaz solar]

cHaus is algebraic

[broken quartz]

since i accidentally posted it here, do any of u mind going to discrete-math and answering my question? if not, ill just wait.

[delicate orchid]

continous functions are defined by their values on dense sets right

that's like a thing

[topaz solar]

Yep

[crystal turtle]

What I said also holds on cHaus

[delicate orchid]

#discrete-math

[crystal turtle]

Just not interesting there lol

[cobalt heath]

I guess cHaus is not compact hausdorff

[topaz solar]

The further cHaus comment is that the forgetful to Set reflects isos or wtv

[crystal turtle]

Yeah

[topaz solar]

Which is what I mean by algebraic

[crystal turtle]

Yes I know

[cobalt heath]

Wtv, now thats a new word

[topaz solar]

Cursed

[topaz solar]

It is

[crystal turtle]

But maps with dense image are still epic in cHaus since they're surjective

[cobalt heath]

What does wtv mean?

Is it a new math concept

[topaz solar]

But yeah, epic just means it’s basically surjective

[crystal turtle]

monadic categories

[delicate orchid]

reflects isos??? this crazy talk... what Coexter System we got on this thang?? What's the WEYL GROUP NERD

[topaz solar]

Not necessarily in the sense of forgetful functor, but you get some internal logic stuff (which is what diagram chasing seems like to me)

[topaz solar]

I don’t care

[delicate orchid]

what do you actually mean by reflects isos though

I've heard this thrown about

maps isos to isos?

[crystal turtle]

If the underlying function is a bijection, then it was an iso

[topaz solar]

Uhh F(f) bijection -> f is an iso

[cobalt heath]

Ahhh

[topaz solar]

So you can pull em back

[crystal turtle]

Or generally, a functor F: C --> D reflects (whatever) if, whenever Ff is (whatever), then f is also (whatever)

[cobalt heath]

So forgetful reflects iso into the simpler category

[topaz solar]

Absta what are you even saying

[cobalt heath]

(Meanwhile idk forgetful somehow)

[delicate orchid]

ah ok you go backways

[crystal turtle]

whatever can be iso, mono, epi, weak equivalence, ....

[topaz solar]

Limits

[delicate orchid]

cool cool

swag, even

[crystal turtle]

oh yeah those are things

swag

[topaz solar]

Anyhow

[cobalt heath]

Well I was just rambling

[crystal turtle]

I have tried to forget that

Ever since product of locales

[cobalt heath]

Because you did not show me whatever wtv means

[topaz solar]

The internal logic of Ring should not look like applying the forgetful functor, because it doesn’t preserve monic, epic or wtv?

But internal for modules should

I think

[crystal turtle]

wtv = whatever

[cobalt heath]

Welp.

[crystal turtle]

[cobalt heath]

Now I feel like I learned entirety of math (just by learning what that abbreviation means)

[crystal turtle]

what

[topaz solar]

Common ring category L anyway

[cobalt heath]

So bijective morphism is sometimes not isomorphisms. Maybe ring is an example?

[crystal turtle]

All bijective morphisms are isomorphism

For pretty much anything in algebra

[cobalt heath]

Ahh

I also read it backwards.

[topaz solar]

Not sure how that’s backwards

[cobalt heath]

Rather, isomorphism might not be bijective

..well, ring isomorphism seems bijective to me

[delicate orchid]

what

[topaz solar]

They are, if you can map them to sets

???

[crystal turtle]

I have no clue what the fuck you're talking about by this point.

[cobalt heath]

Huh, right.

[delicate orchid]

an isomorphism is an invertible morphism, so if we're working in a concrete category.... I'll let u fill in the rest

[cobalt heath]

I'm trying to grok this

[crystal turtle]

Functors preserve isos moment

[topaz solar]

This is unrelated?

[cobalt heath]

Was looking for counterexample to: forgetful functor to Set reflects isomorphism.

[crystal turtle]

Topological spaces

[cobalt heath]

Aha!

[crystal turtle]

Again, not gonna find an example in algebra. So topology is your next best bet

[topaz solar]

[crystal turtle]

just don't consider compact Hausdorff spaces since they are also algebra

[topaz solar]

Categories which do this are called algebraic for a reason

You gotta have some relational structure

Or be something preimage-y like topology

I don’t think topological spaces nicely fall into some model theory language stuff

[cobalt heath]

Huh, interesting.

[warm saffron]

Having some trouble proving a theorem without using Cauchy's theorem....here's the theorem.

If G is a group with order p^n, p is prime, show that G must have a proper subgroup of order p. If n>=3 is it true that G will have a proper subgroup of order p^2?

I'm thinking bout proving by contradiction. FSOC assume not. Let g in G-{e}. Then g has order p^k for some k in {2,3,...,n}. Not really sure where to go from here.

[topaz solar]

If p isn’t 2 then uh

Having a subgroup of order 2 is not happening

[warm saffron]

sorry lol typo

For the first part, I realize I need to essentially find an element k of G such that <k>={e,k,k^2,...,k^(p-1)}

i.e., some k with order p

[terse crystal]

k^2=k^2 mod p so Legendre symbol (k^2/p)=1
Not those two terms equal, but taking sum are equal since both are taking sum for units in Z/pZ plus (0/p)

[topaz solar]

I think some group action approach is best?

[crystal turtle]

In Z/(p^k), what elements will have order p?

Or rather, there is an easiest one

[warm saffron]

I'm unfamiliar with that notation

is that Z_{p^k}?

[topaz solar]

Z/p^k Z yes

[warm saffron]

oh um let me think

[topaz solar]

<g> is gonna have order |p^k| and be cyclic for some k

[warm saffron]

right

I need to show k=1 pretty much

[topaz solar]

No group action needed since only prime divisor is p but I was thinking just doing Cauchy

[topaz solar]

Nope

[topaz solar]

See here

[warm saffron]

nontrivial elements

[topaz solar]

No

No need for contradiction either

[warm saffron]

I'm trying to unpack the modulus p^k

my brain is getting lost in the NT

[topaz solar]

No NT needed here

What’s an order p subgroup for Z/p^2Z

[terse crystal]

Although I think your picture has errors. It should be sum from k=1 to p-1 in the beginning, so k=2 to p in the end

[warm saffron]

um an order p subgroup for Z_{p^2}

[topaz solar]

yes that’s what I said

[warm saffron]

I feel like I'm missing something so trivial

[topaz solar]

what’s the order of p

[warm saffron]

p must have order 1 or p or p^2. Can't be 1, might be p?

[topaz solar]

There’s no need to split it up like that, literally just add p, 2p, … till you get p^2

[warm saffron]

I know the only possible positive divisors of a p^k are nonnegative powers of p up to and including k.

wait

ohh

[topaz solar]

Don’t overcomplicate this one

[terse crystal]

If you are allowed to use Sylow then use Sylow directly, if not forget this

[topaz solar]

This is Cauchy not Sylow anyway

[warm saffron]

okay this makes sense so <p>={p,2p,3p,...,pp}

[terse crystal]

I mean to prove his statement

[topaz solar]

Order p not a maximal p-Subgroup

[terse crystal]

Sylow doesn’t require maximal

Just any r such that p^r | |G|

Number of subgroups of order p^r=1 mod p

[topaz solar]

This is obviously overpowered for this

[warm saffron]

wait so how does this work for a general group G of size p^n when G isn't Z_{p^n}?

[topaz solar]

What’s the order of g in G

[terse crystal]

So I said if he doesn’t know Sylow then forget this

[warm saffron]

I'm supposed to prove it's p, but I have no idea why that's even true

<g>={e,g,gg,...,g^{p-1}}

[topaz solar]

Don’t overcomplicate it

What possible orders could there be for g in G

[warm saffron]

For a nontrivial g in G it must have order p^k for some k in {2,3,...,n}

[crystal turtle]

The subgroup generated by an element is cyclic.

So once we have an element of order p^k for some k, we can reduce it to the cycli case.

[warm saffron]

right

[topaz solar]

No need to say it’s n>1, no contradiction needed

[crystal turtle]

Okay

[warm saffron]

i'm thinking for nontrivial tho so k can't be 1

[topaz solar]

Pick g that isn’t e, and look at <g>, this is Z/p^k Z

[warm saffron]

because <e> doesn't have order p, so I'm assuming g is not e

[topaz solar]

There’s no need to assume g doesn’t exist with order p, also, just take any g with g=/=e

[crystal turtle]

So essentially, you want ot think about it this way:
In the integers modulo p^k, what number, when added to itself p times, will give you zero. Or in other words, what number in {0, 1, ..., p^k-1}, when you multiply it by p, will be divisible by p^k

(aside from 0 of course)

[warm saffron]

p^{k-1}

[topaz solar]

And once you have this, that’s your element of order p

[crystal turtle]

[warm saffron]

So nontrivial g has order p^k for some k in 2,3,...,n. So then <p^{k-1}> is a subgroup of order p

for such k

[topaz solar]

Why do you keep saying k isn’t 1

[warm saffron]

because if g isn't k can't be one right?

[topaz solar]

No?

obviously k can be 1

Consider 1 in Z mod p

[warm saffron]

if g isn't e, and g=g^1=e, then isn't this just a contradiction because g wasn't supposed to be e

[topaz solar]

p^1 is not 1

That’s k=0

[warm saffron]

I'm saying g=g^1, right?

and then so if g has order 1 (and if g is not e), then we get g=g^1=e

[topaz solar]

homie the exponent on p is not the exponent on g

p^1 = p

[warm saffron]

oh god

[topaz solar]

g^p = e

[warm saffron]

my brain is fried

ur right

[topaz solar]

Now go off and write down the answer to this homework

[warm saffron]

okay yes to using LaTeX it would be

\begin{proof}Let $g\in G\setminus{e}$. Then $g$ has order $p^k$ for some $k\in{1,2,3,\dots,n}$. For such $k$, $\langle g^{p^{k-1}}\rangle={g^{p^{k-1}},g^{2p^{k-1}},\dots,g^{pp^{k-1}}}$, which has order $p$. \end{proof}

[crystal turtle]

(again, use \langle g \rangle)

Also, it should only go up to g^{ (p-1) * p^{k-1} } since g^{pp^{k-1}} = e

[topaz solar]

k is 1 to p
k is at most the exponent of p in |G| = p^n

[cloud walrus]

logician

[warm saffron]

ur right oh god I keep making silly mistakes

[topaz solar]

So just proofread a bit

[warm saffron]

guys I'm lost on something about this

How do we know that all the elements of <g^{p^(k-1)}> are all distinct? and that g^{p^(k-1)} has order p. I know based on how we wrote <g^{p^(k-1)}> it's a subgroup of size p

[thorny oasis]

Hi guys, I have the polynomial f(x) = 3x^5 + x^4 + 15x^3 + 12x + 4
and I need to represent it as a product of linear factors, can someone tell me if this question is related to this topic or no?
https://math.stackexchange.com/questions/4781062/representing-a-quintic-polynomial-as-the-product-of-its-linear-factors

[topaz solar]

Because the order of g?

[crystal turtle]

If not, try to see how that would contradict the order of g being p^k

[crystal turtle]

If there aren't rational roots, good luck lol. Quintics and above can get really wacky

Famously, you can't solve all quintics with radicals/roots. So it's not just a "put it into a formula" question.

And I don't see anything obvious that would let you factor this one?

[warm saffron]

oh okay so since g^(p^k), there are p^k elements that are all distinct. So all natural number powers i up to p^k yield g^i being something not already in <g^(p^k)>. That makes sense

[topaz solar]

There’s specific quintics you can’t write via roots even

Not just “can’t do em all with one formula”

[crystal turtle]

Well okay it does have a nice real root it seems (unless that rounding error by desmos) so after that you can apply the quartic formula for the other (complex) roots (probably rounding error I think)

But uhhh have fun with quartic formula? Also nasty

[warm saffron]

This makes sense now too since p^k doesn't divide any of those integer powers 1p^k, 2p^k,...,(p-1)p^k since they're all natural numbers smaller than p^k so it would contradict the minimality of p^k if one of those gave e back

[cobalt heath]

That reminds me

Is there a method to find an algebraic integer root of a (non-monic) polynomial?

[warm saffron]

For this second part I'm thinking it's false because g may have order p^k with k>1 and k<= n, but g^{p^(k-2)} may not have order p^2 since k-2 could be negative.

[cobalt heath]

Well..

A group of order p^2 need not be cyclic.

[warm saffron]

right if p=2 I see that

[cobalt heath]

Zp * Zp is of order p^2

(Well, Zp being group of order p)

[warm saffron]

I'm confused how does this help me anser if n>=3, and |G|=p^n, will G have a proper subgroup p^2?

p is prime

[cobalt heath]

Well, consider the case where you found 2 elements of order p

What happens then?

[warm saffron]

I'm confused why is that a case tho, couldn't we just have only one element of order p which I proved in the first part that there is at least one, and another element having order p^2.

[thorny oasis]

hi guys

can someone help me find the galois group for this poly?

f(x) = 3x^5 + x^4 + 15x^3 + 12x + 4

[cobalt heath]

Well, if you have an element of order p^2

Then you are done.

[rocky cloak]

I assume this is over the rational numbers.
It has only one real root, so the galois group contains a double trasnposition (complex conjugation). If the polynomial is irreducible (which it appears to be), then the galois group contains a 5-cycle.

The subgroups of S5 generated by a 5-cycle and double-transposition is D5 and A5, so at the very least the galois group will contain D5.

Now, you say earlier that what you want is represent this as a product of linear factors. I doubt computing the galois group would be very helpful in that regard. I don't know what kind of expression for the roots you're after. There is a general formula for the quintic involving radicals and the bring radical. But the resulting root will just be a useless mess. So I guess you first need to ask why you want to express this as a product of linear factors.

[somber sleet]

does somebody know how I can prove the existance of the inverse in $\mathbb{F}_p$
(it should not be too hurd, since I need to explain this to one first year student)

[cloud walrus]

damn_guuurl

[rocky cloak]

Do they know about bezouts theorem or Euclids algorithm or fermats little theorem?

[somber sleet]

Nope, that's the problem

[rocky cloak]

What do they know then?

[somber sleet]

They only know the definition of a field and finite field and didn't have abstract algebra yet

Like literally

And the proof in the skript is left ad an exercise 😎

[coral shale]

little isnt too hard from first principles right

but zzz

[rocky cloak]

okay, so a nonzero mod p, means a and p relatively prime, hence a^n and p is relatively prime. By pigeon hole a^n = a^m -> a^n(1-a^m-n) = 0. Since p doesnt divide a^n it must be that a^m-n = 1 (mod p)

so the inverse is a^m-n-1

this is just the proof of FlT

[somber sleet]

Okay, I think this is quite understandable for them

Otherwise I don't know

[rocky cloak]

similar proof but without FlT is to consider ab for all b nonzero mod p. If this is never 1, then by pigeon hole ab = ac, but then a(b-c) = 0, which is a contradiction since p is prime

[thorny oasis]

does this mean the poly is radical solvable?

[rocky cloak]

If the Galois group contains A5 it is not. If it doesnt then it will be

I would guess it is not

[thorny oasis]

you said there is a general formula for quintic, is this true? and if so can you send it here (?

[rocky cloak]

The formula is given by transforming the quintic into normal form, the applying the bring radical. A description is given here https://math.stackexchange.com/a/542228/306319

[thorny oasis]

using a quadratic Tschirnhausen transformation, it will take me a bit processing that out

[thorny oasis]

thanks for linking me up , jagger

[rocky cloak]

btw
`x = polygen(ZZ, 'x');
L.<v> = NumberField(3x^5 + x^4 + 15x^3 + 12*x + 4); G = L.galois_group(names='y')
G.order()

120`

sagemath says the galois group is S5

[thorny oasis]

so strange that it is not radical solvable, since
my prof, told us to represent the poly as a product of linear factors

[rocky cloak]

what kind of a class is this?

[thorny oasis]

Real Analysis

[rocky cloak]

then Im even more confused

[thorny oasis]

I will send an email, maybe he did a mistake

[rocky cloak]

and thats the whole exercise? just write this random polynomial as a product of linear factors?

even if the polynomial had all rational roots, that would be kind of a weird exercise, like what would be the point?

[thorny oasis]

I mean there is a filter that I have to pass in order to enter university, here the university is free to enter like there is no requisites beforehand but there is a whole year where you need to solve multiple puzzles in real analysis and algebra

[thorny oasis]

[rocky cloak]

and this is supposed to be a real analysis puzzle?

[thorny oasis]

yea

[rocky cloak]

Maybe they just want you to prove it has only one real root

that would at least be analysis

[cobalt heath]

That would make much more sense

[crystal turtle]

Why the hell would you need to factor a polynomial in real analysis???

Much less a not-so-nice 5th degree poly

[summer path]

Just graph it

[open sluice]

fake analysis

[peak rain]

Yeah this seems more like an algebra exercise with Eisenstein criterion

[mighty spade]

Polynomials over the ring of regular functions, maybe?

[rotund aurora]

how can you make this metatheorem rigorous

[delicate orchid]

characteristic subgroups are unique under the action of automorphisms of G - that is to say that they are the only orbit representitive, so they are "the something" in G

I guess

[rotund aurora]

so you have some formula describing some subgroup of an unspecified group G. Then if H satisfies the formula sigma H should satisfy the formula too

[delicate orchid]

well if it's characterisitic sigma H is H

I'm not talking up to isomorphism here I mean as sets

[rotund aurora]

ik

[delicate orchid]

oh right we're going the other way

yeah I agree with you then

[rotund aurora]

but the point is, you prove that every group G has a subgroup H satisfying that formula and that it is unique, therefore it must be characteristic. This is the metatheorem

[delicate orchid]

nice example is Sylow groups, even though they're all isomorphic you wouldn't say the Sylow group unless there's only one of them - and when a Sylow p-subgroup is normal it's characteristic

[rotund aurora]

yeah

[delicate orchid]

I like this intuition

[rotund aurora]

like

the intersection of all sylow subgroups is characteristic

because if S is Sylow, then sigma S is also Sylow

but what confuses me a little is that the definition of Sylow uses the order of the group and the subgroup, which is not purely group theoretic. Although ig this is fine because automorphisms preserve size

[delicate orchid]

you can do it entirely by looking at p-elements

[rotund aurora]

what is a p-element?

[delicate orchid]

an element of p-power order

these all live inside sylow subgroups obviously and generate them

[rotund aurora]

ahh ok

nice

[delicate orchid]

yeah and I hope it's fairly clear you can't make a group bigger than a Sylow subgroup with them

although let me actually think about that

[rotund aurora]

you can?

[delicate orchid]

no nvm that's definitely not true

[rotund aurora]

like if it has more than one sylow, what they generate will be larger

[delicate orchid]

take 2 different embeddings of D_8 inside S_4

you can make 3-cycles here if I'm visualising it right

shame!

[rotund aurora]

uhm well

a Sylow is just a maximal p-subgroup

[delicate orchid]

yeah?

[rocky cloak]

Maybe some model theory stuff. Like any subgroup that is defined by formula in the language of groups is characteristic.

[rotund aurora]

yeah, but what is the language of groups?

[delicate orchid]

I think this is what they were getting at

[rotund aurora]

because we are also allowed to invoke |H|

[rocky cloak]

Right, you want some second order stuff

[rocky cloak]

You should be able to talk about the order of elements with just first order stuff though

[rotund aurora]

yes, elements is fine

[rocky cloak]

Okay, but so, first order group theory + quantifying over subsets is not enough?

Wait, your meta theorem is not stated as an iff

So it's not saying that every characteristic subgroup is a "the something"

[rotund aurora]

yeah

I just wanted to try to make precise "the something" part

[rocky cloak]

Alright, and you want something stronger than first order group theory?

[rotund aurora]

isnt every characteristic subgroup a special instance of a "the something"? like pick generators and relations, then H can be described in that way inside G

[rocky cloak]

I mean, I guess H is always "the group in the same orbit as H"

[rotund aurora]

Im just asking how to make that precise, not claiming anything

[rocky cloak]

Make what precise?

[delicate orchid]

https://cdn.discordapp.com/attachments/496784958430380033/1159536763702878218/image.png?ex=653161c3&is=651eecc3&hm=2a85f99c4d723eacc165b5e6c022608def219a0f5eb3f4787e1c55838a72eccd&

[rotund aurora]

this might not be true

but idk

If G can have two distinct characteristic subgroups H and H' that are isomorphic this is false

is this possible? because idk

[rocky cloak]

Possible
https://math.stackexchange.com/a/4285987/306319

[coral spindle]

I think the point they may be trying to make here is that sets defined by formulae, without parameters, are characteristic? Like they might be saying "the something" instead of "the something of something else"

But yeah also the uniqueness thing idk

[rocky cloak]

Also, what does the next sentence say. Since it does look like it's clarifying what "the something" means

[coral spindle]

Context....................... CONTEXT!!!!!!!!

[rotund aurora]

zzz

[coral spindle]

Yeah indeed

Their immediate example of something non-characteristic is something that depends on some other object

So it seems to me what I am saying is correct.

Model-theoretically, they are saying something about definable sets I reckon.

[delicate orchid]

O_p(G) is characteristic? neat

[rocky cloak]

Alright, so at the very least the definition of "the something" should include Z(G') and O_p(G).

So then first order theory is not enough. But if you can quantify over subsets it should be fine.

[restive birch]

im reading something and it mentioned "the direct sum of the infinite cyclic group and the cyclic group of order 3", but online i could only find a definition of the direct sum in terms of subgroups. what is it likely they mean?

[coral spindle]

Z x C_3.

[restive birch]

lol thats so confusing

why didnt they just say the cartesian product

[delicate orchid]

because it's also the direct sum

cartesian product is usually just for sets

[coral spindle]

For Abelian groups, like all modules, we usually call it the direct sum.

[coral shale]

i get confused when we talk about direct products for Z_n things

[abstract rock]

just think bout them partitions

[cyan schooner]

Could someone help me in a thing about rings?

[coral spindle]

https://dontasktoask.com/

[cyan schooner]

Let us consider the ring of remainder classes modulus m, and take one of its elements [a] (an equivalence class), [a] is not [0]. Why if GCD(a,m) > 1 then [a] is a divisor of zero? That is, there exists another class [b] different by [0] such that [a][b] = [ab] = [0]?

English is not my mother language, I had to translate, I hope it is understandable

[ivory trail]

m/gcd(a,m) * a = 0 mod m

[delicate orchid]

^

and obviously if gcd(a,m) = 1 this doesn't mean a is a zero divisor cause you're multiplying it by 0 in that case

[cyan schooner]

It does actually

[coral spindle]

No, wew is correct.

[delicate orchid]

then every element of every ring is a zero divisor.

which is plainly nonsense

[rotund aurora]

welcome again, I remember seeing you once or twice some time ago, and never again 😆

[cyan schooner]

In the definition a is a != 0 is zero divisor if there is b != 0 such that ab = 0. You don't take 0 by definition

[coral spindle]

Oh dang I didn't recognise nasty either

welcome back

[coral spindle]

That is exactly what wew is saying

Clearly there's some miscommunication here, but I think in any case the question has been answered.

[cyan schooner]

Oh yeah, my bad

[abstract rock]

how long ago? i've come n gone a few times now

[rotund aurora]

Several months

[abstract rock]

yeah that woulda been the last time i posted, back in the grinder

[thorny oasis]

at the end, the prof did a mistake with the poly, it was incomplete 🤣

[topaz solar]

Model theory-y?

[topaz solar]

Multiplication, identity, & inverses

Basically, there’s not a whole lot you can do with them, but if you have extra stuff like, say

Totally transcendental, or if there’s any extra structure, then there’s a fair bit you can do

The intersection of all definable finite index subgroups is a definable finite index subgroup, if you have a totally transcendental group, for example

Since, if you can define a subgroup H, you can do cosets because a ~ b iff ab^-1 in H is definable

Also, there’s a way to make Sylow 2-subgroup stuff work?

[delicate orchid]

using a = a^-1?

like, how would you make that work

[topaz solar]

Ah no I mean a lot of the Sylow subgroup theorems, the subgroup needn’t be definable

[delicate orchid]

I think I might cry

[topaz solar]

But you can do like a connected component of it I think

Which is definable?

Not quite sure how to guarantee it’s a nonempty collection, but intersect ur finite index definable subgroups of the Sylow subgroup

[wraith cargo]

Sharp what are you talking about lol
Why is the group not defineable

Or sorry subgeoup

[delicate orchid]

wait I only care about finite groups

[topaz solar]

Because you might not have a first order formula for it

Since G is infinite & we guarantee it exists by zorn

[wraith cargo]

Formula for a subgroup?

[delicate orchid]

right so the centre should be definable, for example

[topaz solar]

Definable = there’s a first order formula s.t. the formula exactly carves it out

[wraith cargo]

Ah okay

I believe there is a theory of infinite sylow subgroups

[topaz solar]

Yeah, $\phi(x)\equiv \forall y. xy=yx$ right?

[wraith cargo]

At least in the profinite case

[cloud walrus]

Dragonslayer Sharp

[delicate orchid]

yeah

[topaz solar]

A subgroup carved out by phi is normal iff forall g phi(x) iff phi(g’xg) ye?

Anyhow, not sure how to guarantee we have a definable finite index subgroup of the Sylow 2-subgroup

They work out nicely for finite RM groups because it’s a sort of finiteness condition

No infinitely descending definable subgroup chain, for example

Which guarantees those connected components I mentioned exist after you have that definable finite index subgroups exist. Since G is definable and finite index gg

Also, because of how these formulas are built, for any automorphism f, we know phi(x) iff phi(fx)

Because f preserves inverses, multiplication, identity, equality

So definable subgroups can’t be shuffled around, though you can’t necessarily say it’s pointwise the identity there though

Boytjie mentions definable without parameters, but if you have parameters from any set that’s fixed under all automorphisms I think you can get away with more

If that set is itself definable we can kinda toss it out, but you can get undefinable stuff fixed under every automorphism. For example: try defining Q without parameters in R or C

Well, the case of R is also because no automorphisms but still (as fields, in this case ofc)

Pretty sure C is like, omega-(super)stable, funny trolled finite RM or smth?

[delicate orchid]

the descent into madness is complete

[topaz solar]

Same

[crystal turtle]

Insanity

[topaz solar]

Well, R is not totally transcendental and is very much so unstable, so it has a lot of nastiness despite any sort of o-minimality stuff you can get?

[cloud walrus]

elgato

[mossy lintel]

R is a commutative ring with unit

[cobalt heath]

Depends on the definition

[mossy lintel]

ehh, basically R[x_1, .... x_n] is a polynomial ring of x_n where the coefficients are taken from R[x_1, ..., x_{n-1}] and you can commutate x_i, x_j when multiplying

[cobalt heath]

Well, isn't this obvious in that case?

Maybe there's some more details involved

[topaz solar]

consider f:left -> right, where f(r) = r for r in R, and f(x_i) = x_sigma i

try showing that turns into a bijective ring homomorphism pretty nicely?

[delicate orchid]

Invoke da universal ploperty

[mossy lintel]

So two rings are equal up to isomorphism, I guess that would work.

[coral shale]

I thought you were supposed to prove equality

ie. there is a ring iso between them which is the identity map

[woven obsidian]

They're not equal as sets

Just like Dedekind cut reals and Cauchy sequence reals are not equals as sets.

[peak root]

Wait, they are not the same set? I think ultimately you get the real numbers, right

[coral shale]

im confused what

[coral shale]

if you consider the symbol "x" used in both to be equal

then they are equal as sets

x1 + 2x2 in one is 2x2 + x1 in the other, for example

Am I missing something?

[open sluice]

I think the issue is when you consider multiplication

is x_1 x_2 the same element as x_2 x_1

[coral shale]

sure, but n is finite

and this is a ring?

comm*

oh ok, thats defo worth asking sure

[coral shale]

^

[crystal turtle]

Yes

[delicate orchid]

if you wanted non-com you'd write something like R<x_1, ..., x_n>

[coral shale]

ok yeah, i didnt think of that detail myself

no idea how poly rings with non-comm

if R is non comm, R[x, y, z] means ???

[crystal turtle]

:thevisceralsoundofthegatesofhellopening:

[coral shale]

i feel like u have R[x, y, z] and [x, y, z]R or something

you make the variables commute with everything ig

[crystal turtle]

R[x_1, ..., x_n] is the monoid ring on the free monoid generated by n elements. Which should make sense without R commutative.

[cloud walrus]

grothendieckfan1 (Ryx)

[crystal turtle]

wait fuck s in S not S in S

[delicate orchid]

0 in N
le troll face

[peak root]

Just a minute. Let's get back to the definition of R[x]. It is a R-algebra, which means it is a commutative ring with a ring homomorphism R-->R[x]. Iterate this definition, we have a comm ring homo R[x]-->R[x,y].

[delicate orchid]

oh yeah R need not be commutative, R<x_1, ..., x_n> is for when you don't want x_ix_j = x_jx_i

[crystal turtle]

Jail.

[rocky cloak]

Do you define a polynomial ring as some weird equivalence class of strings of symbols or how do you make this true?

[coral shale]

i uh... no follow for now

ah N is which power you want for the variables

eh wait (nope me no get)

[delicate orchid]

slam |S| copies of N together than put a ring under it. I don't like this description of the polynomial ring. It SUCKS!!

[topaz solar]

That “free monoid” is commutative

[rocky cloak]

I guess you're defining it as the monoid ring, but then R[x, y] is not literally equal to R[x][y] I guess

[crystal turtle]

It should be finitely supported functions from R --> N^n (or more generally N^S). Which sounds like a polynomial to me

[topaz solar]

Really you’d want a free (non commutative) monoid if you don’t want the x_i commuting?

[crystal turtle]

Well yeah. Asking for isomorphism is the correct method anyways, if you dig deep enough, asking for equality if nonsensical

[delicate orchid]

just take the free object in R-alg nerds...

[topaz solar]

Since ya know (1,0)+(0,1) is kinda

[delicate orchid]

the singular free object

[crystal turtle]

I never claimed that what I said what noncummutative. I just said poly ring

[dim widget]

You need to fix coherence data for the monoidal structure coming from sums of sets

[rocky cloak]

Then I guess the answer is "yes, it's a stupid question"

[crystal turtle]

Oh wait no I see now I forgor noncummutative monoid exist. So it should be free cummutative monoid on S, my bad.

[coral shale]

can someone elif. wtfs the map

thats a map from S -> N right

[delicate orchid]

it's not a map it's a monoid

[coral shale]

free commutative monoid?

its too early for me

[crystal turtle]

S-tuples of natural numbers, only finitely many of which are nonzero

(Or if you want, finitely supported functions S --> N)

[coral shale]

yh so like like like, this is the free commutative monoid right?

otherwise im not making sense still

if it was free monoid, u just have strings, abbbababaababbbab

[crystal turtle]

Yeah my bad it's commie

[coral shale]

😅

free comm group, free ab, is the same except Z

which makes sense yh yh ...

[tardy hedge]

Is the maximum order of a permutation on 16 elements 140?

Consisting of 3 disjoint cycles of length 7 5 and 4

[lusty marlin]

To verify answers for questions like these, I suggest you look up the relevant sequence on the OEIS, in this case, Landau's function.