groups-rings-fields

[coral steeple]

I know the order of the group too I suppose If I know the order of the group then I can find its elements

[vocal pebble]

Wait, what's your definition of Gal(f), f in F[X]? The most standard is that it's group of automorphisms of the splitting field of f over F that fix F

[coral steeple]

Yep

[vocal pebble]

Yeah, and that's all you need

Read this

[coral steeple]

Well, take a toy example like Q(sqrt2)/Q. We find its Galois group like this: the extension has degree 2 and is the splitting field of a separable polynomial, so the group has order 2; any automorphism permutes {\pm\sqrt 2} (and the action on this set determines the auto), and automorphisms of this form give at most two things in the group, so these facts mean that the group consists of the identity and the other thing. Is the takeaway that this is not the analysis I should be carrying out for this question?

[vocal pebble]

That's the correct analysis

And that does tell you what the galois group is in that case

It's C2

Because you have a map from the galois group to C2=S2, given as follows:
Let sqrt(2)=a1, -sqrt(2)=a2.
Then let sigma be an element of the galois group. Map sigma to the permutation that it induces on a1,a2.

This is injective and surjective. Injective because different automorphisms must carry a1 to different roots.

Its surjective because both a1 -> a1 and a1-> a2 are automorphisms fixing Q

[coral steeple]

Yes

[vocal pebble]

Now do the same for this case. Let the roots of x^2-3 be a0,a1 and that of x^2-7 be b0,b1, and the roots of 1+x+x^2+x^3+x^4 be c0,c1,c2,c3. Now let sigma be an element of the galois group.
Sigma carries a1 to ai for some i =0 or 1
Sigma carries b1 to bj for some j=0 or 1
Sigma carries c1 to ck for some k=0,1,2, or 3.

Map sigma into Z/2xZ/2xZ/4 as follows:
sigma -> (i,j,k)

Show that this is an isomorphism using almost identical logic as before

[coral steeple]

Yes, I agree. I guess the issue I was trying to point out is showing that your map is well-defined (which may just be pedantry on my part). How do we know that any of the sigma are automorphisms? I assume the fact to use is that if f1...fn are irreducible over K with no common roots in their splitting fields, then Gal(f1 f2 ... fn/K) is isomorphic to the product of the Gal(fi/K)?

For instance I don't know the degree of this extension. Who's to say the galois group doesn't only interchange the roots of the quadratic irreducible factors independently, and no automorphism moves any of the roots of the quintic factor? Hopefully it's to do with the statement I put forward in my last message

[vocal pebble]

Ah, okay, so for that it suffices to show that the minimal polynomial g of zeta_5 = e^{2pi i/5} over Q(sqrt(3),sqrt(7)) is the same as that over Q (the min poly over Q has degree 4) . Let f be the min poly over Q. We certainly have g| f. Now check that any subset of the factors of f containing zeta_5 don't give a polynomial in Q(sqrt(3),sqrt(7))[x]

So g=f infact, and that's what we wanted

[coral steeple]

So I'm guessing you'd check that you can't get an element of Q(sqrt3,sqrt7) by taking the product of a proper, nonempty subset of the factors of x^5-1

[vocal pebble]

Well you just need to take proper products of subsets factors of 1+x+..+x^4 but yes

[coral steeple]

Excuse me, yes

[vocal pebble]

And the nice thing is, for most of the subsets you should be able to argue that either the product or sum of the roots won't be in Q(sqrt(3),sqrt(7)), so by default those subsets are eliminated

[coral steeple]

Ah true

Yeah I think that answers things

[vocal pebble]

You will need to compute e^(2pi/5) (so that will involve computing sin(2pi/5)) for a full rigorous argument should you decide to prove it this way, or you can have the values memorised (maybe you can get away without knowing these values, not exactly sure)

[coral steeple]

Yeah

Thanks!

I fear this problem would take me more than the allotted 9 minutes to solve if I were to encounter it on my exam later today...

[rocky cloak]

Write
L = Q(sq3, sq7) and K = Q(zeta5).

As L and K are both Galois extensions you have [KL:L] = [K:L cap K].

So you just need to determine if they intersect. The degree 2 subextension of Q(zeta5) is Q(zeta5 + zeta5^-1) = Q(cos(2pi/5)) = Q(sq5). So should not intersect Q(sq3, sq7)

[coral steeple]

[KL:L] = [K:L cap K]
Never seen this before. Going out on a limb here but does that have a more well-known group-theoretic interpretation via the Galois correspondence?

[rocky cloak]

It's the second isomorphism theorem

[coral steeple]

Yeah I think so right?

Yes

[rocky cloak]

More specifically let
E/F be a Galois extensions, let K and L be intermediate extensions with K Galois.

Let G = Gal(E/F), N = Gal(E/K) and H = Gal(E/L).

Then HN = Gal(E/KnL) and HnN = Gal(E/KL), and G/N = Gal(K/F).

Then the second isomorphism theorem says HN/N = H/(HnN).

But HN/N = Gal(K/KnL) and H/HnN = Gal(KL/L)

[coral steeple]

E \subset K is Galois?

Excuse me, F subset K

This is very helpful, thanks

[copper tartan]

Is the inverse necessarily true? i.e., if f in R/<p> [x] is reducible, then f is reducible in R[x]?

For context, I'm trying to figure out if 9x^3 - 6x^2 - 7x + 3 is irreducible in Z[x], and figured I'd try to check it under F_2, which transforms it to x^3+x+1

and I'm fairly certain that x^3+x+1 is reducible, so can I say that because that's reducible, my original polynomial is also reducible?

[desert verge]

no

[desert verge]

the converse is not necessarily true

[copper tartan]

yeah I'm realizing I did my original modding wrong

and modded it into x^3 - x + 1, and not x^3 + x + 1

[desert verge]

same thing in F_2[x]

[copper tartan]

I've figured out its irreducible now

[copper tartan]

like I guess that makes sense

if only for the reason that the proposition would've stated otherwise

[desert verge]

you can see it's irredicuble in F_2[x] because it does not have a root in F_2

as a cubic, if it factorized, it would have a linear factor and hence a root

[copper tartan]

ok word, thank you

[torpid heath]

guys im ngl groups and rings is acc horror

i think ts has set my life span back about 5 years minimum

[lofty notch]

Should I get a real or lab grown diamond for my girlfriend’s wedding ring?

[fickle dirge]

this might be the wrong channel

[rocky cloak]

Wrong channel, but the answer is lab

[south patrol]

How can a labrador make a diamond

[rocky cloak]

Just hand them a pickaxe. They're pretty smart

[elfin wraith]

But what if you love slavery and paying ridiculous amounts of money?

[stone sky]

How trivial is it that a group that is torsion free cannot be isomorphic to a group with a torsion subgroup

I want to state it during my presentation but I don’t want to prove it because it’s not technically in a purely group theoretic context

[elfin wraith]

If you know what all those words mean I should think entirely

[stone sky]

Perfect

That makes my proof trivial lol

[swift root]

If your group is abelian then it is trivially equivalent too

also, if its isomorphic to a group with a torsion subgroup, then it is a group with a torsion subgroup

[stone sky]

I guess I won’t even be writing anything for that proof on the board then

I’ll just explain it verbally

[south patrol]

I guess I would say nontrivial torsion subgroup lol

[elfin wraith]

What are you trying to show?

[south patrol]

I would say yes it is barely worth mentioning from this

Also like any torsion element generates a torsion subgroup, so having a nontrivial subgroup which is torsion is just saying "not torsionfree"

[swift root]

oh right for some reason my mind stopped at the fact that the set of torsion elements is not necessarily a subgroup

[stone sky]

Im showing that m-fold sum of tori and the n-fold sum of projective planes are not homeomorphic. And it follows immediately from the homology groups

Which I described in terms of their ranks and torsion subgroups

In previous results

[south patrol]

I would just say that like two fg abelian things are same iff same ranks and torsion subgroups

If they don't know that then can take as a blackbox as it's very believable and not topology

[quiet pelican]

Classification of finitely generated abelian groups

[south patrol]

Indeed

[stone sky]

Precisely

[swift root]

also if you're doing homology then that should definitely be something you should be able to prove

[south patrol]

I am a sheaf.

[swift root]

(even if with a little effort)

[elfin wraith]

This is what i was gonna say lol

[stone sky]

I’m introducing it to the class. I don’t wanna prove it because I have little time

Even it’s easy

[swift root]

I was talking about the people listening

[stone sky]

Earlier results were harder and much longer

ah that’s true

[south patrol]

What is the best proof tbh

Can quickly reduce to classifying torsion things

For abelian groups ig can do an easier argument than general PID

[swift root]

classifying finite ab groups is pretty easy using chinese remainder theorem and induction

[south patrol]

Ye sure

Lol

I more.mean general PID

Besides smith normal form

[quiet pelican]

Just prove Smith Normal Form

[swift root]

[south patrol]

Lol

[quiet pelican]

As soon as I say “row reduce everything in sight” like
It’s obvious how you prove it

[wraith cargo]

We had a HW to prove it

But it's an easy corollary of fg modules over PID

[south patrol]

I guess can also like reduce to stuff killed by p^k for a prime element p by CRT and then probably some usual devissage ting

[south patrol]

What is lol

[wraith cargo]

Smith normal form

[south patrol]

Sure lol

[swift root]

the structure theorem of fg modules over PID

obv

[wraith cargo]

This is literally the only proof I've ever seen lmfao

[rocky cloak]

Fg torsion module has nontrivial annihilator. Modding out reduces to artinian PIR, its a product of local rings so reduces to local artinian PIR. Show that these are self injective with a quick zorn.

Break off free summand continue by induction

[south patrol]

Yeah ig what I said above reduces to local artinian PIR and then that is a nice way to finish that off

Though using Zorn feels unnecessary

Maybe it is though for this proof since you're proving a more general statement? (About local artinian PIR)

[rocky cloak]

Well, after reducing to the artinian PIR might as well just clarify all modules over them, not just the fg ones

[south patrol]

Ah

[elfin wraith]

This is what my group theory course did without ever mentioning it which I find funny, like just fully proved the smith normal form without putting it as an actual result

[wraith cargo]

There's a geometric way to prove PID but I forget it now

[south patrol]

Pedant hat is you don't prove the form but you prove its existence uniqueness smh

[south patrol]

Oh wait yeah I guess you can just avoid Zorn if you use fg modules, is that what you mean?

[rocky cloak]

Yes, then it's just normal induction

[south patrol]

Like ig Artinian local PIR are self-injective by Baer for example but you only actually need injectivity applied to a fg module which is an easier claim

Yeah okay this is a nice proof then thank

[wraith cargo]

There's a cool proof (that idk how standard it is) where you find a specific homomorphism in Hom(D^n,D) that decomposes D^n into ker(phi) (+) Dx_1 and then proceed by induction on the rank
And then the general theorem follows from this

[rocky cloak]

Can you elaborate a little more on this?

[quiet pelican]

/lh

[wraith cargo]

since D is a PID, every homomorphism D^n->D has image some ideal call it <a_phi>. Then by noetherianity you have a maximal such ideal <a_phi_1>
Now a_phi_1 is the image of some y_1 under phi_1, and you can show that there's also an x_1 such that y_1 = phi_1(y_1)x_1 and phi_1(x_1) = 1, and from this it follows that D^n = ker(phi_1) (+) Dx_1 and M = (ker(phi_1) cap M) (+) Dy_1
Then proceeding by induction on the rank of D^n, you have your decomposition and that every submodule M is free
From this you prove the general case by nothing that M = <m_1,...,m_k> and there's a surjective morphism D^k -> M, and since ker(this morphism) is a submodule of D^k, it's free and that way you get your decomposition of M

[rocky cloak]

Another possible proof:

Again R/Ann(M) artinian means M has finite length.

Let C = R/r be a cyclic submodule with largest possible length. Now for x in M not in C, let it have annihilator (s). If s does not divide r then it's easy to construct an element with annihilator lcm(r, s) contradicting maximality. So s divides r. Then if (x) intersects C you can shift x a little by something in C to make them not intersect, and a little induction should round it off.

[rocky cloak]

I'm not sure I follow. What are the morphisms D^n -> D you're considering?

[south patrol]

Hm I don't quite see the last bit

[wraith cargo]

any D-module hom between D^n and D

so it's all elements of Hom_D(D^n,D)

[rocky cloak]

Then I definitely don't follow

[wraith cargo]

hm well I was a bit vague because I didn't wanna write out all the details

but I can write everything out

I still have my notes from this class

but I don't remember who I was told originally came up with this proof

[rocky cloak]

So the first part is just proving that any homomorphism D^n -> D splits into ker (+) image

That's clear if that's what you're trying to say

So that's all you're doing I guess? And then at you end you say that since the kernel is free you're done? I'm not sure I see why

[rocky cloak]

Again pick x such that image in M/C has maximal length.

Then you get
R/r (+) R/s submodule of M. Pick next etc

[wraith cargo]

well it's a bit more specific than that because you aren't just showing it's an isomorphism this is a genuine equality

you show that D^n deconstructs in this very specific way that's compatible with any submodule

the actual statement we proved is

D^n decomposes (equality) as Dx_1 (+) ... (+) Dx_n

and for any submodule of D^n there exist a_1|a_2|...|a_n such that M = Da_1x_1 (+) ... (+) Da_nx_n

[rocky cloak]

Ahhh, so when you're talking about the image of a map D^n -> D you mean the image of a submodule of D^n

[wraith cargo]

oh yes sorry

[rocky cloak]

Then the first part makes more sense

[wraith cargo]

I knew I was being too vague 😔

[rocky cloak]

Cool, I see

[wraith cargo]

there's some geometric interpretation of this that I forget that my professor told us was how this proof was originally discovered

but I forget

I was taught this by Alireza Golsefidy at UCSD when I took his algebra class

[south patrol]

Ig I mostly meant how you get the nonintersecting but maybe I am being foolish

[rocky cloak]

Say tx in C, then r/t * tx = 0, so tx is a multiple of t (inside C = R/r)

Then (x - tx/t) doesn't intersect C

Horrible notation, but hopefully you get it

[south patrol]

Okay nice thanks

[tough raven]

Is it true that for k a field, k(x1, ..., xn)^⨯ / k[x1, ..., xn]_{(x1, ..., xn)}^⨯ is the free abelian group generated by x1, ..., xn?

What's the cleanest proof of this fact, if true?

IG Frac(A)^⨯/A^⨯ is free on the irreducible elements of A for any UFD A and S^{-1}A^⨯/A^⨯ is generated by precisely the irreducibles (dividing elements) in S, and the saturated multiplicative set k[x1, ..., xn]\(x1,...,xn) is generated by all irreducibles with constant term, so you get the free abelian group on all irreducible polynomials with no constant term... which is actually larger than {x1, ..., xn}.

[fossil spade]

having a hard time understanding what this theorem is saying, btw Integers are under addition here

[knotty badger]

so let $f : \mathbb{Z} \to G$ be a group homomorphism

[rocky cloak]

Any specific part that trips you up, or you just don't understand any of it?

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

then you can obtain an element of $G$ by evaluating $f$ at $1 \in \mathbb{Z}$, to obtain $f(1) \in G$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[fossil spade]

maybe understanding what the map epsilon does, maybe an example will help?

[knotty badger]

this defines a map $\epsilon : \text{Hom}(\mathbb{Z}, G) \to G$ by $\epsilon(f) := f(1)$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

so for example, you could set $G = D_4$, the group of rotations of a square

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

and then define $f(n)$ to be a clockwise rotation by $90n$ degrees

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

this is a group homomorphism, right?

[fossil spade]

im not sure to be honest...

[knotty badger]

well, let's check the definition

what is $f(n + m)$?

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[fossil spade]

wait sorry we have never define groups in terms of rotations of a square so thats what im confused about with this example

[knotty badger]

oh sorry

[fossil spade]

its cool

[knotty badger]

i meant $D_4$ as the group of symmetries of a square

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

that was a typo on my part

so this includes the rotations and the reflections

it's a type of group called a dihedral group

[fossil spade]

we havent done rotations either but im guessing the identity is no rotation. 1 rotation is by 90 degrees, 2 is by 180, and 3 is by 270, and 4 you are back to identity?

[knotty badger]

yep

[fossil spade]

i see, wait can we use integers mod 4, under addition

[knotty badger]

mhm, sure

that's $C_4$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

then, we can define a group homomorphism $f : \mathbb{Z} \to C_4$ by $f(n) = n \pmod 4$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

right?

[fossil spade]

mhm

[knotty badger]

in this case, $\epsilon(f) = f(1) = 1 \pmod 4$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

which is an element of $C_4$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[fossil spade]

yup

[knotty badger]

in general, what $\epsilon$ does is take a homomorphism $f : \mathbb{Z} \to G$, and evaluates it at $1$ to obtain $f(1) \in G$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

thus, it's a map $\epsilon : \text{Hom}(\mathbb{Z}, G) \to G$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

the domain is the set of homomorphisms from Z to G

the codomain is G itself, viewed as a set

in particular, you should just think of epsilon as a function between sets, not a group homomorphism

[fossil spade]

I see

[knotty badger]

then, what your screenshot is saying is that $\epsilon$ is a bijection

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

for any group G

[wraith cargo]

the best way to interpret this theorem is that it says "any homomorphism from a cyclic group is uniquely determined by where it sends the generator"

so you don't need to specify where every element is sent

just where 1 is sent

and this determines the homomorphism in it's entirety

[knotty badger]

-# Z represents the forgetful functor

[fossil spade]

ahhh i see

i understand now

if two homomorphisms $f_a $ and $f_b$ have it so that $f_a(1) = 1 = f_b(1)$ then they agree on the generator and that forces them to agree on all other elements so $f_a(n) = f_b(n)$ $\forall $n \in \mathbb{Z}$

so that makes it 1-1

[wraith cargo]

yes

[mighty echo]

Hihi! I'm doing some reading of the basics of group theory, and I have need some help understanding something.

It says that given a group G and subgroup H, then G/H has a natural operation (xG)(yG) = (xy)G, but warns that in general that this is not well defined. What I don't understand is how this can not be well defined? And is it ever the case that G/H is not a group?

I have attached the actual passage

[elfin wraith]

This happens if the subgroup H is not normal, and in that case G/H cannot be a group because the action isn’t well defined.

As an example to see this, take S_3, and think of a non normal subgroup, write out what happens

[mighty echo]

Ah, the text defined normal right after it states this and I got stuck here and have not moved on yet

I'll have a look

Thanks!

[elfin wraith]

Yeah normal essentially just means that the multiplication on G/H is well defined, but yeah you should write out what the multiplication looks like in S_3/H where H isn’t a normal subgroup (there’s 3 to choose from)

[mighty echo]

Ohhhhhhhh because coset representatives are not unique

Got it! Thanks!!

[knotty badger]

this is why i prefer a different def for the quotient group operation

given any two subsets $A, B \subset G$, you can form their product $AB = {ab : a \in A, b \in B}$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

you can then show that, if $H$ is normal, the product of two cosets is always another coset

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

and this is what's happening in the quotient group, under the hood

[mighty echo]

ooooh

icic

[mighty echo]

how hard is that to prove? would it be a good exercise?

[knotty badger]

it would be a good exercise

[fossil spade]

another good exercise relating to this would be: Suppose $G$ is a group, $H \leq G$ and $N \trianglelefteq G$.
\begin{enumerate}
\item Show that $HN = NH$, and conclude that $HN \leq G$.
\item If $H \trianglelefteq G$, show that $HN \trianglelefteq G$.
\end{enumerate}

[cloud walrus]

BigBalla

[mighty echo]

I'm still at the definition of normal subgroups lol. I am still before isomorphisms theorums and quotient groups so it looks a bit out of my depth. Maybe later. But thanks!

[fossil spade]

you shouldnt need the isomorphism theorems for that proof

[elfin wraith]

That’s basically just applying the definition of a normal subgroup, it’s a good thing to do to make sure you’ve absorbed it

[mighty echo]

oh okay. the notation was new

[fossil spade]

yeah the triangle less than equal is

N is a normal subgroup of G

try it, if you get stuck go and reread the definition for normal subgroups, it should help with internalizing what you just read

[mighty echo]

I haven't read it a first time yet 😭😭

[fossil spade]

lol whenever you feel ready or wanna test yourself

[barren heath]

1+2 = 4 right

[kind crypt]

yes

[small yacht]

How do i do this problem? I thought something among the lines of that every elements square should be unique, therefore by the pigeonhole principle all elements should have a square.
But i'm pretty sure there's something wrong with this reasoning

[frail shoal]

i think this is valid reasoning

essentially you're reasoning that squaring is injective and injective self-maps on finite sets are bijective

the finite assumption is important, cuz the group (Z, +) for instance

yay pigeonhole

[small yacht]

I don't know how to prove that the squares should be unique, though...
This isn't true for (R^x, *) for example, where x^2 = y^2 doesn't necessarily imply x=y

[frail shoal]

yea

this is where the odd size assumption is critical

cuz the reason it fails for that group is cuz of the {-1, +1} subgroup (isomorphic to Z/2)

[small yacht]

how does |G| being odd ensure that squares are unique?

[frail shoal]

ah ok this is the key to the exercise

i assumed you figured it out cuz of how you worded your original statement

but yea you're on the right track

i can hint if u want

[small yacht]

Oh, hm, okay, i'll think about it, then

[small yacht]

Yeah a hint would be good

[frail shoal]

Lagrange's theorem is huge here

wait i might have quoted the wrong theorem

i should check

[small yacht]

For every subgroup H of G, cardinality of H divides cardinality G?

[frail shoal]

yea that one

ok this certainly works although idk if it's overkill

have you learned it yet

[small yacht]

No it makes sense, it was introduced in this chapter

[frail shoal]

ah in that case it was probably the intention

good luck

[frail shoal]

ok the thing is that you can use the theorem to prove the theorem im actually thinking of but idk the name

||the order of an element divides the size of the group||

[small yacht]

Hm, thinking of your {-1 , 1} example, do x and y with same squares form a subgroup?

[frail shoal]

interestingly squares form a subgroup of any abelian group but not in general

yea sorry i feel like i misled you a little

[frail shoal]

this is what i should have hinted with

[small yacht]

I thought of this, but couldn't remember if this is the case with all groups or some special case

[frail shoal]

yea it's an all groups thing

i forget the optimal proofs so it sorta got lumped with lagrange in my head cuz it's sorta a strict generalization

cuz the order of an element is the size of the subgroup generated by it

[small yacht]

oh buh hat was one of the examples in the tetbook

[frail shoal]

that makes sense

maybe that is how you prove that theorem

i wasn't sure if i was being circular

[small yacht]

What theorem?

sorry

[frail shoal]

idk what to call it

the order of an element divides the order of the group theorem

[small yacht]

Oh yeah you use lagranges theorem for it

if |g| is finite then <g> is a subgroup and therefore |g| divides |G|

[frail shoal]

hell yea

[small yacht]

hm ok let me think

[frail shoal]

yea there's still a few nontrivial steps here

[small yacht]

Yeah uhm i still have no idea...

[frail shoal]

so, what we want to show is that x^2 = y^2 implies x = y. we know that x^m = 1 and y^n = 1 for m, n | |G|, and that |G| is odd

because |G| is odd, what can we say about m, n?

[small yacht]

they are both odd

ohhh wait

[frail shoal]

correct

[small yacht]

x^2k * x = 1
y^2j * y = 1

y^2k = x^-1
x^2j = y^-1

y^2k+1 = x^-1 y = 1
x = y

wait

no

that doesn't wrok

ugh

wait it does

because y^2=x^2

y^2k+1 = x^2k+1=1

[next obsidian]

Maya the Maya

Since x^2 = y^2

[small yacht]

Chmonkey Monkey

[next obsidian]

x^4 = y^4

x^6 = y^6

So on so forth

[small yacht]

so true

[next obsidian]

Can u make it so that at some point the thing on the left is actually = x

And the thing on the right is actually = y

[small yacht]

Uhm

can i

[small yacht]

also doesn't this do it already

[next obsidian]

Oh

Yeh cool

I was gonna say you can take the orders of x and y, call them m and n

And multiply them

And you get x^mn = 1 = y^mn

But x^mn is odd so it’s a 2k + 1 and then you eliminate

Similar for y^mn

But yeh

[small yacht]

x^2k+1=y^2k+1
x^2k * x = y^2k * y
x = y
WOWIES you are so smart

[next obsidian]

I mean what u did is functionally the same

[small yacht]

I took the long way

[next obsidian]

Yes but sometimes the long way is good

Here’s an exampl

https://youtu.be/gJxObMyVHoM?si=QFJT9X1Zc5axQ3-T

[small yacht]

SO TRUE

[frail shoal]

this feels like vine

[next obsidian]

The crazy thing is I think it isn’t

[frail shoal]

yea ikr

i wish vine were still real that was the shit

tiktok doesn't count, you actually needa limit people to 6 seconds

[next obsidian]

Bro how old r u

Maybe it wa the monkey pfp but I clocked you as like… 20

And that’s too young for vine lol

[knotty frigate]

wait fr?

huhhh

[next obsidian]

P sure that’s a TikTok

Vine died in like 2017 or something

It has vine vibes

[knotty frigate]

i wasnt even on vine for the most part but yeah it totally does. vine had a pretty different culture or at least thats what it felt like as someone who was predominately on yt

[next obsidian]

That awkward time when it was music.ly 😂😂😂😂

Kind of the dead zone btw vine and TikTok

[knotty frigate]

completely forgot that music.ly was a thing

[next obsidian]

I struggled to remember the name for a while

[woeful sage]

me when I enter #vines-tiktoks-reels

[maiden crater]

really silly question

isn

isn't F already over a field ans this is its field of fractions

[chilly radish]

F is a field, F[x,y] isn't

[maiden crater]

right

got it

[woeful sage]

How can a ring (which is not a field) F[x, y] be the same as a field (the fraction field of R)?

I don't understand how it follows that \alpha^m \in F (in the line before last line of the screenshot)

[rocky cloak]

If g divides f then
g(x) = (x - alpha)^m
so
g(0) = ± alpha^m is in F

[frail shoal]

22

[frail shoal]

vine was popular when i was in elementary school so like, even if i were 20, id be able to remember it

[next obsidian]

I guess I was kinda the last little generation that elementary schoolers didn’t all have phones

Even if you go 5 years later it started kinda being the norm huh

[rocky cloak]

You'll be amazed to learn that there are university students younger than the iphone

[woeful sage]

How can I be so oblivious

thanks jagr

I spent close to an hour trying to understand this

it was this simple

[next obsidian]

He is younger than the iPhone for an amazing reason

[next obsidian]

Cuz ur Neam

Idk what that is supposed to mean

[woeful sage]

It's modern art, it means whatever you want it to mean

neam = mean

[woeful sage]

I'm never gonna be able to learn AG if I'm like this

[frail shoal]

oh yea i didnt have a phone, my classmates just quoted it all the time

[next obsidian]

Yeah but some of ur friends probably did

[frail shoal]

i only ever learned the sources from looking them up on youtube at home on my pc

[next obsidian]

When I was in elementary school nobody had a phone

[frail shoal]

i forget tbh

[next obsidian]

I didn’t know what YouTube was until 3rd or 4th grade I think

[frail shoal]

i dont remember phones being a thing when i was in grade school but they became things when i was in upper grade school

[next obsidian]

Maybe 5th

I think someone told me about raywilliamjohnson

And that’s how I found out

[woeful sage]

The telephone hadn't been invented when you were in elementary school 🥀

[sudden condor]

groups rings and fields exam just completed c:

[sudden condor]

im about 4 months older

and in 3rd year

oh wait 11 months, im 4 months older than the announcement of the iphone

[fossil spade]

I think I have an idea of what this theorem is saying but maybe not fully?

So I understand that here the main idea is that the only way $f$ can factor through $\pi$ is if $f$ is constant on each coset of $\ker \pi$. If $f = \psi \circ \pi$, then it must be that $\ker \pi \leq \ker f$, since $\pi$ collapses each coset of $\ker \pi$ into a single point in $Q$, and then $\psi$ maps that point to an element of $K$. (Recall that the cosets of $\ker \pi$ partition $G$.) So we need $\ker \pi \leq \ker f$ because that makes $f$ ``blind'' to the elements of $\ker \pi$ that differ between elements of the same coset: if $g_1$ and $g_2$ lie in the same coset, then $g_1 = g_2 \cdot n$ for some $n \in \ker \pi \subseteq \ker f$, so $f(g_1) = f(g_2) \cdot f(n) = f(g_2)$. Thus $f$ sends the entire coset to a single point in $K$, which is exactly what's needed for $\psi$ on $Q$ to be well-defined.

[cloud walrus]

BigBalla

[knotty badger]

categorically, this says that surjective group homomorphisms are epimorphisms

[small yacht]

Hm can I apply FIT backwards

I.e

If G/H ~= M, there is a surjective homomorphism from G to M

[wraith cargo]

Yes just compose your isomorphism with the canonical surjection G->G/H

[small yacht]

Sorry, I don’t think I get your explanation very well

[frail shoal]

wdym by lagrange backwards

[small yacht]

Converse of it

( I think that’s what converse is?)

[frail shoal]

yea

if A -> B is the original statement then B -> A is the converse

[small yacht]

Yes that’s it

[frail shoal]

lagrance says that if H ⊆ G is a subgroup, then |H| divides |G|

[small yacht]

Wait I’m so stupid

[frail shoal]

so you're asking, if |H| divides |G|, then is H ⊆ G?

[small yacht]

I meant FIT

[frail shoal]

should be false

[frail shoal]

ohhhhh

[small yacht]

Yes it is

[frail shoal]

that makes way more sense w what you posted

[small yacht]

Second time posting something stupid in here!!

[frail shoal]

FIT says, if φ: G -> M, then G/ker φ ≅ im φ, and you're taking the case when im φ = M, i assume

[small yacht]

Yes

[frail shoal]

yea then what irony said is right

essentially, if H ⊆ G is a normal subgroup, we have an induced map G -> G/H

we send an element g ∈ G to its coset gH ∈ G/H

[small yacht]

Yes I know that much

Their explanation was just uhh

Very concise

I know the terminology

And etc

[frail shoal]

yea true

oh yea i forgor to add on the isomorphism

but yeah a surjective homomorphism is in one-to-one correspondence with a choice of normal subgroup

it's pretty beautiful

[small yacht]

It is indeed

[frail shoal]

[up to isomorphism of the image]

im kinda brainrotted i forget to say that a lot

actually this is kinda subtle

cuz sometimes the image can be isomorphic but you have distinct kernels

i can explain the subtlety if you're curious

[small yacht]

Sure actually but maybe after figuring out the converse of FIT thing…?

[frail shoal]

oh yea that makes sense

what are you stuck on understanding rn

[small yacht]

Wellll
They said take the isomorphism and compose with projection, right?
Yuh uhh I have no idea what they meant

Do I take the bijective function between them?

[frail shoal]

yee

i guess let's refresh the setup

[frail shoal]

let's name the map i: G/H -> M

also, let's name the map p: G -> G/H

so we use i ○ p

[small yacht]

Ohhh

Omg how did I not see it buh

Thanks

[frail shoal]

np

sometimes it takes a lot of different explanations to make it click

ur doing great

[small yacht]

Thxx

[small yacht]

Do you still wanna explain the subtlety?

[frail shoal]

sure

ok so, for example, there's many different surjective group homs Z^2 -> Z, and they have many different kernels

one, for instance, sends (m, n) to m

another sends (m, n) to n

another sends (m, n) to m+n

what gives?

the idea is that, for the statement to fully work, we need to consider the entire map, up to post-composing by an isomorphism

for example, we consider the map that sends (m, n) to m and the map that sends (m, n) to -m "the same" in a sense

this is because we can get one from the other by post-composing by the map Z -> Z that sends n to -n

[small yacht]

Ahh

[frail shoal]

honestly i think im confusing myself

cuz i said it right the first time

by saying surjective map

ah, no, i didn't specify what i meant by isomorphism of surjective map

that's why i felt the need to say this

cuz it's not as simple as an isomorphism of the images

you need to specifically post-compose by an isomorphism

anyway hopefully im not being too confusing rn

ur free to ignore this yap but, yea

[small yacht]

I think I see what you meant

[small yacht]

OHH wait this ties in so nicely with this problem

Since the order of elements divides the order of the group, there aren't any elements of order k, so G/H where H is all eleemnts of order k is isomorphic to G

so g -> g^k is surjective

Well

hm

it's actually elements whose order divides k

so it's just the identity

[velvet hull]

is it?

you are only showing that there is one element mapped to the identity

[small yacht]

I'm using First iso

[velvet hull]

that does not imply that the map is surjective, because it is not necessarily a group homomorphism

[small yacht]

the converse of it

[velvet hull]

the function is not necessary a group homomorphism

the theorem does not apply

[small yacht]

nooooo
if G/H ~= M, i:G/H -> M is a homomorphism, and pi:G -> G/H is a homomorphism. so i * p: G -> M is a homomorphism

furthermore becase pi isn't necessarily injective, i * p is surjective

[remote anvil]

I have the answer. Would it be ok if I posted it?

[small yacht]

Answer to what?

[frail shoal]

yea sadly the power map is only guaranteed to be a homomorphism if our group is abelian

[remote anvil]

to 8.14

[frail shoal]

essentially we want (gh)^k = g^k h^k

[small yacht]

sure

[small yacht]

How is my proof invalid, then?

[frail shoal]

ah i haven't read ur proof with good eyes yet

lemme see

[small yacht]

I can write it out more clearly

[frail shoal]

yea this is valid

wait

nvm i gotta read more carefully

ah the issue is that elements of order k [edit: i meant to say k-th powers, oops] aren't a subgroup in general

but tbh it's close to the right idea

we do want to use divisibility stuff

[small yacht]

in this case it is, isn't it?

since it's only the identity

they are also the elements g -> g^k maps to the identity

so they are forced to be the kernel and therefore the subgroup

so it may be something unique to groups where k divides |G|

[frail shoal]

wait im like

i said the wrong thing

oops

[frail shoal]

i mean k-th powers

just fyi stupid mistakes never go away

[small yacht]

lol

well still

if k-th powers are mapped to the identity by a group homomorphism

then they have to be a subgroup

so if they aren't usually

maybe not all g -> g^k are homomorphisms

but in this case we knwo that it is

so they are forced to be a subgroup

[frail shoal]

not necessarily

[small yacht]

actually

yeah you're right

[frail shoal]

here's an example: consider g -> g^5 on S_3

(1 2)^5 (2 3)^5 ≠ [(1 2)(2 3)]^5

[small yacht]

-# well in this case we know that g^k = 1 is only true for g = e
-# so it definitely is a subgroup

[chilly radish]

Here's the intuition:

If k is coprime to n, it is invertible mod n. Note that if G is of order n, powers of an element g can be reduced mod n, i.e. g^n=1, so g^(a+nb)=g^a

[frail shoal]

ok actually in this case kth powers do form a subgroup but

that's cuz it's G itself

but that's a theorem, not the way you prove the theorem

ok im kinda tired and disoriented imma have to tap out

[small yacht]

well
it's easy to prove

ig g^n=1 then n | |g|

[chilly radish]

No

Opposite

[small yacht]

oh

you're right

im stupid

wait

hmmm

of

we are given that GCD(k, |G|) = 1
then if g^k=1, |g| divides k and |g| divides |G|, so |g|=1, g=e

if H are elements such that g^k=1, trivially
G/H ~= G

so the function g -> g^k is surjective

i just noticed that the problem doesn't require it being a homomorphism

Does it even apply then

ughhhh

[small yacht]

You were right al along

[small yacht]

i think with some manipulation of FIT you can show that it does

[tidal torrent]

what exactly is a group extension

do they have similar properties of a field extension (is this basically how supersets work?)

idk how this has anything to do with the schur multiplier

[rocky cloak]

Given a normal subgroup N < G we say that G is an extension of
N by G/N.

One is often interested in the situation where you start with N and G/N and see what different extensions can be built from them.

For example if N = G/N = C2 the cyclic group of order 2, then G could be either C2xC2 or C4

[tight brook]

I've recently thought about a problem im not sure how to solve:

What is the chromatic number of a Cayley graph of S_n generated by elements of order 2?

In other words if I make a graph of all permutations of order n, with each two connected iff they differ by 2 placements. What would be the minimum number of colors needed to color the nodes with no neighbors of the same color

Oh wait that's a dumb question

It's a bipartite graph

But what about a general set of generators of order k?

[tight brook]

Well apparently this also is not that interesting

I checked some cases

It's mostly χ=k

Except for like special cases and stuff

But it does seem like an interesting graph

So like I wanna ask questions about it

Automorphism group, diameter, hamiltonicity and stuff like this

[copper kestrel]

does anyone happen to know why the group averaging operator produces a G invariant vector?

[noble nexus]

I assume you mean you have a finite group acting on a vector space and you average the action?

try applying an element of the group to the average and seeing what happens

[copper kestrel]

yep yep

[true bolt]

The map x->gx is a bijection G->G

[copper kestrel]

hm

that makes sense but like

idk it's just weird

i got malware from arxiv or an arxiv copy so i dont want to look for any papers anymore ☹️

[vocal pebble]

Its just really that addition is commutative and the action of an element induces a bijection

[vocal pebble]

Are you sure? Afaik you have to submit a .tex file and the pdf which is served is compiled by arxiv

[copper kestrel]

well something happened when i clicked on an arxiv link, my computer got really slow, then my google turned to
yahoo

[south patrol]

What about in characteristic p

Joking

Premium ragebait

$g. (\sum_{h \in G} h.v) = \sum_{h \in G} (gh).v$

[cloud walrus]

Prismatic Potato

[south patrol]

and this is a sum over all group elements

[noble nexus]

Importantly the action must be linear for this to work

[copper kestrel]

is this it? i thought it was dividing by the order of G too?

[swift root]

scaling an invariant vector produces another invariant vector

[copper kestrel]

bc i'm doing serre's linear representations of finite groups

[copper kestrel]

ah

[swift root]

i mean, the set of invariant vectors is a submodule

[noble nexus]

The advantage of dividing by the order of the group is that it gives you a projection operator

i.e. if you average an invariant vector you just get the same thing back

[copper kestrel]

ohhhhhh so thats why he did it

i thought the averaging operator included that

[noble nexus]

It does yeah, dividing by the order of the group is the averaging part

[copper kestrel]

yeye

[copper kestrel]

okay im confused a bit bc then what is this

like i understand this is what happens by acting g on something

[noble nexus]

what do you by what is this

[vocal pebble]

You're just moving terms around in the sum

That doesn't change the value of the sum

[glad osprey]

The sum is the averaging operation except the division. Then you have g acting on the sum, and as h ranges over G so does gh, so g * sum(hv) = sum(hv)

[south patrol]

Lol ye I just omitted that 1/n as it's linear

[copper kestrel]

ahhhhh

[chilly radish]

The idea here is that, by taking g.v for all g, you have exhausted all of the possible actions by elements of G. Multiplying by an element of h then only permutes the elements of g, so the resulting vector is the same since you only permuted the sum (so you have the same factors)

[knotty badger]

another way i like to think of it is, given your vector v and your group action, the most general possible vector you can make is $\sum_{g \in G} f(g) g \cdot v$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

then you can check what f you need for this to be invariant, and it turns out a constant f works

this is kind of the perspective of the group ring, i suppose

[broken tulip]

Any hints as to how to approach this? I was able to show that the chain (x₁) ⊆(x₂) ⊆(x₃) ⊆ ... has the property that if xₙ₊₁ ∈ (xₙ) then x₂ ∈ (x₁). I tried to go for a contradiction in this case, by considering a homomorphism τ from R into ℝ that sends x₁ to 2 and each further indeterminate to the square root of the previous indeterminate, and the image of that homomorphism is a subring S of ℝ that contains all integers and all numbers of the form 2^(1/2ⁿ), but the issue is that I don't know how to show that this subring doesn't contain √2/2. It seems like there should be some field theory lemma for this, but I don't know any field theory.

[quiet pelican]

The image of the homomorphism you defined lies in the ring of algebraic integers

[frail shoal]

boo

[frail shoal]

oh yea that's clever

[frail shoal]

i guess one strategy you could do is show that if you let x_1 = 0 (i.e. add x_1 to the ideal), then x_2 ≠ 0 in the resulting quotient ring

ohh i get ur proof idea tho

you're tryna embed R into R

i think your intuition is right, that every 2^(1/2^n) is linearly independent

wait i take it back im not sure

but yea it seems kinda tricky

[frail shoal]

so you might wanna go w/ something more like this

[broken tulip]

Yeah, the issue is that algebraic integers aren't covered in the book just yet so I'd have to include a bunch of lemmas about them into the proof. I guess this shouldn't be too bad

[broken tulip]

Oh, that's clever! I'll see if this works

[frail shoal]

pona o tawa sina

[broken tulip]

pona a!

[swift root]

new pfp

[frail shoal]

yes! although it's also probably temporary

for the record i didn't change it just to scare that dude

[swift root]

[frail shoal]

it's cuz it kinda looks like monk key color scheme wise

[swift root]

i like it!

[frail shoal]

yay

[swift root]

oh true

[frail shoal]

btw i posted a question to MO relatively recently that's kinda UAG pilled

it was worded like shit originally but it got slowly fixed over time

do u wanna see it (i suspect the answer is yes)

[swift root]

share

:0

[frail shoal]

https://mathoverflow.net/q/510133/533379

[frail shoal]

one consequence of this is the title is long now 🥀

[swift root]

these correspond to what i call completions

so let L ⊆ L' be an extension of algebraic languages, and V, V' be varieties over L resp L'.

V' is a completion of V if V' restricted to L generates V, and restricting does not lose any information (i.e. the restriction functor reflects isomorphisms)

[frail shoal]

is this a coinage

i also had a coinage but i think the language often used is tha L' is implicitly definable from L

or something like that

[frail shoal]

how does this deal with the ring case btw

[swift root]

i briefly looked into this a looong time ago ¯_(ツ)_/¯

[frail shoal]

fair

[swift root]

oh right this is specifically for totally defineable operations

so it doesnt really correspond to your case

just a nice subcase of it

[frail shoal]

yea it's weird how rings specifically just like

fuck with you

im looking for that in particular tbh

i wanna be fucked

but yea it's also cool how monoids complete to groups and bdl's complete to ba's

[swift root]

units too

the magma to quasigroup is also a completion im p sure

every edge in the groupoid lattice is a completion if you remove the condition that V' generates V

[frail shoal]

holy canoli and spicy guacamole

wait, how do you define this

[swift root]

well the idea is that being a quasigroup is a property of the magma, and not extra structure

this can be formalized by asking that the functor that takes the magma reduct (i.e. forgetting the division operations) reflects isomorphisms

[frail shoal]

ok tbh i wouldn't be surprised if like

your notion of completing

is entirely distinct from my notion

[swift root]

any completion induces an instance of your notion onto the variety generated by the reduct

[frail shoal]

for a simple example, consider N-actions vs Z-actions, how do we get my notion

i.e. arbitrary unary f, vs. f, g such that f ○ g = g ○ f = id

oh i guess maybe it fails the generation condition

well that tells me what to interrogate at least

quasigroups generate magmas?

context: i wrote monoids instead of magmas, oops

[swift root]

quasigroups arent associative

[frail shoal]

mental typo

[swift root]

yeah i just realized
your notion is about specific elements

[frail shoal]

fixed

[swift root]

mine is about operations

[frail shoal]

yea perhaps

[swift root]

in a sense my notion is your notion applied to the "variety of equational theories"

[frail shoal]

that sounds plausible

wait what's a variety of theories

is this vibes or real

[swift root]

there are several ways to define it

you can have a single associative infinitary operation, or you can use the formalism of heterogeneous algebras

i.e. you define the variety of clones and use the correspondence of clones <-> varieties

[frail shoal]

what do we close clones under

[swift root]

composition

[frail shoal]

wait, clones are like, families of maps A^n -> A

how do we compose two different clones

oh wait maybe you thought i was asking how to define clones

no i mean how do we define a variety of clones

[swift root]

products of clones, subclones, and quotient clones :P

[frail shoal]

wtfrick

you can do that ?!

[swift root]

i mean why not right

subclones correspond to taking term reducts, quotient clones correspond to subvarieties

and products of clones are the tensor product of varieties

[frail shoal]

well my best guesses seem reasonable

[swift root]

A nice treatment of this is given in W. Taylor's paper on classifying Malcev conditions

[frail shoal]

like given a map A^n -> A and a map B^n -> B we have a map (A × B)^n -> A × B

just by functorality

[swift root]

those are concrete clones

talking about abstract clones here

[frail shoal]

and i assume subclones are just subfamilies closed under composition and projections

[swift root]

but yes this is the idea

[frail shoal]

oh! and a quotient is like

how quotients work on algebras

probably

[swift root]

right

[frail shoal]

like is it on the underlying set as opposed to the family

[swift root]

no

[frail shoal]

and it has to be compatible

oh

[swift root]

we work with abstract clones

[frail shoal]

hm

what's an abstract clone

is that like, a monad

tbh i never really thought about quotients of functors much

but when i think through this lense it does become apparent we want to identify terms together

ah yea subvarieties

as you said

makes sense

so we close under reducts, subvarieties, and tensor product

[swift root]

N-graded set { A_i }_i in N and operations
C^n_m : A_n × (A_m)^n → A_m
π^n_i ∈ A_n for 1 ≤ i ≤ n
satisfying the necessary axioms

[frail shoal]

reduct in the general sense of, we can pick some terms to choose that weren't part of our basis

why sadness

[swift root]

only thumbs up i could find

[frail shoal]

yea fair

[frail shoal]

wait we generalizing the idea of composition?

bruh these people smoking some real shit

[swift root]

if we have two variety V and W with signatures Σ and Σ', then their tensor product V ⊗ W consists of algebras A ⊗ B, where A ∈ V and B ∈ W. These have underlying set A × B, and operations f × g for f ∈ Σ and g ∈ Σ' exactly like you described

[frail shoal]

i guess traditionally we can represent this by band shenanigans

or something fucky like that

[swift root]

band?

[frail shoal]

like those things equivalent to A × B

structurally

and in terms of maps it's literally equivalent to Set^2

rectangular band or something

[swift root]

oh, factor bands or smt?

i think an old book about sheaves and UA talked about them

[frail shoal]

makes sense

[swift root]

old as in "not yet accepted functorial definition of sheaf" old

[frail shoal]

nobody has standard names for these things

[frail shoal]

the cave times

[swift root]

yes..

anywho, variable sets for clones have to be graded, of course

a set of variables for each arity

If we let V be the variety of clones, and Σ(x, y) = { C^2_1(x, π^1_1, ι^0_1(y)) = π^1_1, C^2_1(x, ι^0_1(y), π^1_1) = π^1_1 } (x is arity 2 and y is arity 0), then we indeed have an implicitly defined nonexplicit partial operation

[frail shoal]

im debating whether i wanna actually parse that

[swift root]

the fact that this is an instance of your notion parses to (let K be a class of algebras with binary operation m and nullary operation 1, 1'):
K ⊨ m(x, 1) ≈ x ≈ m(1, x) and K ⊨ m(x, 1') ≈ x ≈ m(1', x) implies 1 = 1'

[frail shoal]

yay

yea identity is only unique if you allow substitution in your deduction

[swift root]

yee

and completions need that

[swift root]

and this corresponds to the completion of magmas to unital magmas

this also has a lot to do with Malcev conditions

a strong Malcev condition is nothing more than taking a clone M and considering the heterogeneous variety where you add an operation for every term in M

because these correspond to clone morphisms M -> A

[frail shoal]

yay

[swift root]

if one has a completion V ⊢ W, and some clone morphism W -> A, this always lifts to a completion A* ⊢ A, because of the fact that induces your notion

In a meta way Im sure i could prove that the malcev condition of W is a completion of the malcev condition of V

anwaysyysy

[frail shoal]

i just realized something kinda cursed

sets with binary operations definitely have the ES property (epics are surjective)

given two sets with binary operations (magmas) with A ⊆ B, we can define a magma on (B ⊔_A B) ⊔ {0} by sending every "undefined" operation to 0

⊔ here being over Set

[swift root]

oh cool

so it isnt a malcev condition

[frail shoal]

ES?

i actually have very little intuition for when ES will hold. it definitely holds for some minimal varieties, but not others. it's not closed in either direction of passing to subvarieties either, that is, it isn't preserved by passing to subvarieties, nor is it preserved by extending your variety

part of my aim with my MO question is to learn more "prototypical" examples of where ES fails

[tribal moss]

You can also say that if you have a magma morphism f: A -> B that is not surjective, then you can define two different morphisms B -> ({0,1}, ×) that each compose with f to the same, namely either map everything to 1, or map only the image of A to 1.

[frail shoal]

oh you're hella right

it's kinda like a subalgebra indicator

ok it can't literally be one cuz that would break math

i needa think on this

[swift root]

would it?

[frail shoal]

most varities shouldn't have them

[frail shoal]

nvm

[frail shoal]

the preimage of both {0} and {1} must be subalgebras because they themselves are

we have no guarantee that B \ A is also a subalgebra

[tribal moss]

Hmm, that sounds convincing, let me think some more.

[frail shoal]

im slowly starting to see the "pushout with itself is itself" view of epics as intuitive

more than the actual definition of epic

[rocky cloak]

I mean, the two views are essentially the same.

One is saying there arent two maps B -> C that make a commutative square
A -> B
v v
B -> C

and the other is saying there arent two maps B -> C where the composition with A->B is the same. Which is just the definition of the square being commutative.

[frail shoal]

oh i just realized you meant, magma completion stuff

yea

[swift root]

no i meant ES

[frail shoal]

ah

[swift root]

i could've seen that from Set anyways im stupid

[frail shoal]

yea, you do need to prove that B = B is the initial object of B / our cat, but, that's not particularly hard

[vestal jay]

So I'm doing an exercise on free abelian groups, is this the right channel to ask something related to it?

[rocky cloak]

Yes

[vestal jay]

I made a channel for my question, I'd really appreciate it if you or someone looked into it

[small yacht]

You are encouraged to use these channels to ask about more advanced math

[lime magnet]

anyone want to chat about modules?

brb

[desert verge]

sure

[swift root]

mfs ask a question then disappear for over an hour

[desert verge]

happens all the time

[candid patrol]

Don’t ask to ask 🤓

[rocky cloak]

They never said they wanted to chat about modules 🤓

[balmy python]

can you only have a torsion subgroup for abelian groups?

[elfin wraith]

Yeah, it wont be closed in general otherwise

https://en.wikipedia.org/wiki/Torsion_subgroup#Examples_and_further_results

[balmy python]

ah yes i remember that group

[elfin wraith]

I suggest maybe going through the proof that it forms a subgroup again and work out where you use the fact that your group is abelian in a crucial way

[balmy python]

if x^m = y^n = e and you want a k s.t. (xy)^k = e then the group being abelian is a sufficient condition for k to exist

what's useful about this?

[knotty badger]

Oh this is the correspondence theorem

Gives you information about the lattice of subgroups of G/N

[balmy python]

lattice?

could i have an example 😭

[knotty badger]

Basically the subgroups of any group G form a partially ordered set by inclusion

This poset has some extra properties which warrant calling it a lattice

[balmy python]

ohhh

[balmy python]

so i guess like the grid of even points in Z^2 is a lattice?

poset is such a funny name

https://tenor.com/view/fighting-stance-po-kung-fu-panda-4-try-me-you-wanna-fight-gif-4596534491769696639

[knotty badger]

Well there’s two different senses of lattice here

[balmy python]

ah

[knotty badger]

The lattice I mean is the order-theoretic version

This is different from “regularly spaced collection of points”

[balmy python]

ah so youre more so comparing lattices to lattices with the order?

[knotty badger]

The poset of subgroups forms an order-theoretic lattice

[balmy python]

ahhh

[elfin wraith]

This is massivley useful, youll use it all the time doing algebra. It tells you that normal subgroups of G/N are the same as normal subgroups of G contining N which gives you a lot of control over things. This also applies to modules etc and you can typically throw in whatever adjectives would be reasonable and its still true

[knotty badger]

ok to be fair i don't remember ever actually using this

but i'm not an algebraist

[balmy python]

oh damn

[knotty badger]

sounds like it might be useful for galois theory

[balmy python]

maybe i can try something with the groups right now

[elfin wraith]

To be fair I dont know how much ive used it for groups

[balmy python]

nvm 😭

[elfin wraith]

But in like ring theoretic contexts it shows up all the time, and I probably have often used it for groups, I just dont actually think about it because its such a fundamental result in my mind

[quiet pelican]

It’s useful but it’s one of those things you kinda just use subconsciously

[knotty badger]

it might be useful to prove the following set-theoretic facts

[balmy python]

is this more of a post grad thing?

[elfin wraith]

No not really

[balmy python]

oh

i haven't seen anything to do with it at all in second year

[elfin wraith]

I think you just tend to argue with inclusions etc more in the context of rings. But like me and micoi said, its honestly so fundamental I probably do use it a lot just without even noticing

[knotty badger]

let $f : A \to B$ be a function. Then $f(f^{-1}(S)) = S \cap \text{Im}(f)$ for any $S \subset B$, and $f^{-1}(f(T))$ is the closure of $T$ under the equivalence relation $a_1 \sim a_2 \iff f(a_1) = f(a_2)$ on $A$, for any $T \subset A$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[balmy python]

oooh

[knotty badger]

so $f(f^{-1}(S))$ is a kind of "set-theoretic interior" operation, while $f^{-1}(f(T))$ is a kind of "set-theoretic closure" operation

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

https://pseudonium.github.io/2026/01/21/Subset_Images_Categorically.html

i go through this in a bit more detail in this article of mine

[tribal moss]

Hmm, my attempts to salvage my flawed construction ended up with reinventing essentially this.

[balmy python]

wait this is lowkey insane

like if you have Z and 4Z as a normal subgroup you can check 2Z is normal by taking its cosets mod 4Z and checking if it's normal in Z/4Z

quite convoluted but it works

[elfin wraith]

Yeah, this kind of argument is often useful. Of course thats maybe a difficult way to do that problem but its often a nice way to approach it in other cases

[balmy python]

after going through group theory again i'm starting to appreciate the theorems we rushed over in lectures a lot more

i would want to do some algebra next year but there's so many choices that i have very little background to make a judgement on

[frail shoal]

i agree, it's sad that a chain / total order / linear order isn't called a "toset"

[swift root]

everything

it's for example super useful when you want want to move subnormal series between quotients

esp in combination with the third isomorphism theorem (G/N)/(M/N) = G/M

also, like Nope said, in ring theory this pops up

[knotty badger]

i'm curious, where does it pop up in ring theory?

i don't remember using it

[balmy python]

series?

does anyone have tips on how to answer questions like these? they seem to always use groups like D8 and i feel like checking with groups like that is really long especially because you get like 2 mins in an exam to answer it

[swift root]

for example the fact that surjections induce closed immersions rely on this fact

[knotty badger]

@.@

[swift root]

also stuff with proving that the quotient of a Noetherian/artinian ring is Noetherian/artinian

[rocky cloak]

Say you want to compute something like
Z[i]/(p).

Then you would probably do
Z[i] = Z[x]/(x^2 + 1)
Z[i]/(p) = Z[x]/(x^2 + 1, p) = Z/pZ[x]/(x^2 + 1)

[knotty badger]

right

[swift root]

that implicitly uses the correspondence theorem

but it's something that's essentially so intuitive you don't even notice it

[knotty badger]

oh, that makes sense!

[swift root]

same with most isomorphism theorems tbh

[foggy tartan]

Consider classes of groups you already know.

Say, D_n.

[balmy python]

would i just have to memorise facts about D_n?

can't really investigate things about it in 2 minutes

[foggy tartan]

You can use a geometrical argument. Try the 180deg rotation.

That commutes when it exists.

You usually just want to have the intuition for it I guess. Not that that's great advice.

[balmy python]

hm yeah

that's still kinda frustrating to think about very quickly

[delicate orchid]

you should memorise facts about all common families of groups tbh

[balmy python]

they should make a pdf with general ones for group theory

[foggy tartan]

Knowing important properties gives you reason to care.

[balmy python]

the analysis ones are fine because they are intuitive

group theory is stupid

[delicate orchid]

analysis is notorious for having strange edge cases idk where that opinion is coming from

[balmy python]

really? 😭

[delicate orchid]

yes?

[foggy tartan]

That's like, all of real analysis.

[delicate orchid]

https://en.wikipedia.org/wiki/Weierstrass_function first one that comes to mind

[foggy tartan]

You usually memorize the edge cases specifically to stop yourself from conjecturing things that seem true.

[balmy python]

for me something that comes into mind is like the convergence in measure implies convergence almost everywhere on a finite measure space

[knotty badger]

a first course in real analysis consists of two parts:

• proving lots of statements that seem intuitively true
• disproving lots of statements that seem intuitively true

[balmy python]

but it seems possible to come up with a counterexample without being too specific with what you're working with

whereas with groups i feel as though i have to look at specific groups and then do it and i can't really picture something in my head

[delicate orchid]

the statement about D_n is obvious if you just view it's presentation <r, t | r^2 = t^n, rtr = t^-1> there's only one element that's fixed by the conjugation action if n is even and it's t^(n/2)

it's centreless if n is odd

[balmy python]

oh yeah you told me about this presentation

[delicate orchid]

did I lol

[foggy tartan]

It's also not too far-fetched of an idea. Rotate a square in your head.

It's nice to keep the geometric intuition behind small groups for occasions like this.

[delicate orchid]

you could also work with Q_8. It only has one C_2 subgroup and thus that subgroup must be central. Then because i, j, k don't commute that must be the entire centre

[balmy python]

my adhd brain would rather make a type writer function over rotating and reflecting a bunch in my head

[balmy python]

true

[delicate orchid]

you could also just take any centreless group

like A_n n > 4

and say G = A_n x C_2

[balmy python]

i'm thinking about how to do this

would you have to represent every group element as a combination of r and t?

[foggy tartan]

Rip a square off of a piece of paper. I did this.

[delicate orchid]

ok true I suppose you need the foresight that every element of D_n is writeable as r^it^j for some i, j but that's clear if you think geometrically

[balmy python]

ah yeah

[foggy tartan]

For small n that's not so bad.

[delicate orchid]

D_{2^n} is a maximal class 2-group and therefore the centre must be C_2

le troll face

[delicate orchid]

I do actually think this is the best method for generating examples of a group of some arbitrary minimal size and arbitrary centre

[sly crescent]

n=1

[delicate orchid]

https://cdn.discordapp.com/attachments/1431035017135525961/1491188420431843478/391.gif

[sly crescent]

What does that mean?

[inner owl]

Look up cantor distribution function/devils stair case

[swift root]

brainrot

[rocky cloak]

No need to do it in your head, cut out a square piece of paper and label the vertecies

[balmy python]

true but an examioner might think im hungry

hungry for groups

very right

[fossil spade]

stuck on this problem not really sure what to do here...maybe a hint?

[rocky cloak]

So I guess the key idea here is what can you say about h^n?

[fossil spade]

hmmm oh is it h^n = phi(g)^n = phi(g^n) = phi(e)? or is that wrong

[rocky cloak]

That's right

[fossil spade]

oh cool

lol i was overthinking it

thanks

[rocky cloak]

So then h^n = e and maybe you've already seen that means m divides n

[fossil spade]

yeah thats a theorem

i believe

[rocky cloak]

Well probably good for you to try to prove it if you don't recall the proof

[fossil spade]

yeah maybe i should, need it for the final anyway

thanks!

[fossil spade]

so for this problem I was thinking that maybe show that one of the maps acts as a right only identity or left only identity

is that enough?

to show that M is not isomrphic to the opposite monoid M^{op}

[velvet hull]

well, you would have to show that it doesn't pair up with any other element when you take the opposite

i.e. if there is a right only identity then you will have to show that there isn't a left only identity

(but that's not too hard)

[fossil spade]

ahh got you , thats just computing the composition of the right only identity as a left only in the opposite and showing it doesnt work

[brisk matrix]

How can we construct a continuous-time evolution operator that is strictly covariant under the action of the rotation group $SO(3)$?

[cloud walrus]

Canvas123

[lethal mortar]

How do i check isomorphism here?

What is m there?

[lusty marlin]

A group isomorphism is a group homomorphism whose inverse is also a homomorphism

So to prove that f is an isomorphism, you want to check if it has an inverse and if that inverse is also a homomorphism

[velvet hull]

You should also check that the f given is itself a homomorphism to being with

[foggy tartan]

m is a free variable here. You can also check that f is an isomorphism if it is injective and surjective.

[lusty marlin]

While it is true that group homomorphisms are isomorphisms iff they are bijective, I would argue that it is slightly bad practice to make it a habit of only checking for bijectivity, since that doesn't generalise to morphisms in other categories.

[knotty badger]

It works for “algebraic” categories

[foggy tartan]

Of course. This is ameliorated by keeping the structure-preserving part in mind.

[lusty marlin]

Could you elaborate a bit on this? I'm not familiar with a classification of categories in which this holds.

[foggy tartan]

No homomorphism proof is complete without proving the map is actually a homomorphism, but that should be by default.

[knotty badger]

Read: monadic over Set

[lusty marlin]

Oh ok

Thanks

[knotty badger]

This is because every monadic functor is conservative

[lethal mortar]

F(x)=x^m=y

x=y^(1/m)

f^-(x)=x^(1/m)

[lusty marlin]

y^(1/m) isn't necessarily well-defined

[lethal mortar]

But this can be inverse function no?@lusty marlin

[long nebula]

wdym covariant

[lusty marlin]

By definition, if the map x ↦x^m is invertible, then its inverse sends y to an (the unique, since we have assumed invertibility) element whose m-th power is y.

[lethal mortar]

Thanks

[brisk matrix]

if $\mathbf{x}$ is rotated such that it becomes $Rx$, then applying my function to both gives essentially $f(R\text{v}) = R f(\dots)$

[cloud walrus]

Canvas123

[brisk matrix]

it means if you rotate the input, a truly equivariant operator ensures your output rotates by that same amount R

[fast kelp]

Bit of a random question:

It is possible to define the complex numbers as a ring on R^2 with a particular multiplication rule. It is also possible to create a projective plane from R^2. Typically, we talk about the Riemann Sphere as the completion of the complex numbers with a single point at infinity. Is possible to use RP^2 instead, with its line at infinity and still have a (mostly) coherent system of arithmetic? (I am aware of wheel theory, and I assume I'll need to also join with a bottom element to make sure everything is defined.)

Motivation: recently a paper was released that describes how to generate most of the "useful" functions in mathematics with a single function eml(x,y) = exp(x) - ln(y). But they needed to cheat a little to generate a function that simply negates a value through taking ln(0) = -inf. This triggered a bit of a discussion on a different discord, trying to see if there was any formalism of the complex numbers that could allow this. I had thought of RP^2, but I will admit to not knowing enough to be confident in any inferences I might make on this subject, and I thought surely this must have been considered before. But unfortunately in the year of our Linux desktop 2026 search engines are awful, so I couldn't find anything.

[long nebula]

oh so you're using covariant to mean an equivariant action—I think this is impossible for time evolutions defined by a vector field because S^2 doesn't form a Lie group

(hairy ball theorem)

haven't thought about it deeply enough though

[finite dust]

he means a u sub