groups-rings-fields

[sacred wharf]

carney man 🥀

[white oxide]

Could I have a hint for showing that the powers of $(x_1, \dots, x_m)$ are primary? I am now fixing $k$ and letting $f$ be a zero divisor in $k[x_1, \dots, x_n]/(x_1, \dots, x_m)^k$. Then there is $g$ with $fg \in (x_1, \dots, x_m)^k$. Hence $f$ is not a polynomial in $x_j$ for $m < j \leq n$, and so $f \in (x_1, \dots, x_m)$. So $f^k \in (x_1 ,\dots, x_m)^k$. Does this work? I just don't see where we would use Exercise 7 but seems fine to me...

[cloud walrus]

okeyokay

[rocky cloak]

I feel like "hence f is not a polynomial in xj" is doing some lifting.

Are you thinking about writing down f explicitly or are you having some other argument in mind?

[white oxide]

Well I considered two approaches

The first approach was what I just posted but it's kind of hand wavy

The second approach was induction on the power

Or induction on the number of variables

And I kind of got stuck on that

[rocky cloak]

Anyway, you can prove this pretty quickly by using that powers of maximal ideals are primary and applying 7iii

[white oxide]

Yeah that's the hint I got 😭

[rocky cloak]

So you don't understand it or you want a different approach?

[white oxide]

Well I thought about it for an hour and how it was useful

I know that powers of (x_1, \dots, x_n) are useful but I wasn't sure how to apply that to powers of (x_1, \dots, x_m)

[rocky cloak]

Okay, so you can prove the result for m=n

Then you just apply 7iii and you get it for n=m+1, then induction it holds for all n>m

[white oxide]

Hold up I'm confused I thought only (x_1, \dots, x_n) was maximal

[rocky cloak]

Yes, so that covers the n=m case

After that you apply 7iii to get the rest

[white oxide]

You're my savior

I guess I forgot induction as well

I will think about how I can apply 7iii again 😂

[rocky cloak]

Like to take an example.

You know that (x1)^r is primary in k[x1]. Then the statement of 7iii is precisely that (x1)^r is also primary in k[x1, x2].

And then applying 7iii again it's primary in k[x1, x2, x3]

And so on, induction

[vagrant zinc]

Bourbaki ?

[white oxide]

Nah Atiyah Macdonald

[tardy hedge]

2.4 has always kinda bothered me. Just always felt like random linear algebra magic

[sacred wharf]

Kronecker delta

The goat

[tardy hedge]

Maybe learning why adj(A)A=det(A)I would make me feel better but idk

[karmic moat]

#groups-rings-fields message

[karmic moat]

This just follows from expanding the determinant

Anyway that identity is like the defining property of the adjugate matrix

[south patrol]

I will try to remember like there is a nice geometric way to think about it lol

Okay maybe not quite geometric but still very cute

[karmic moat]

hmm interesting

[south patrol]

I guess just like Cayley–Hamilton even in its usual form is kinda surprising (even if the proof is not too long or anything)

Hm

[tardy hedge]

I should review cayley hamilton too

[south patrol]

I mean like this is Cayley–Hamilton

Just in slightly more generality than normal

But like using the adjugate is used in some standard proofs of that in more or less this way I believe

(Though imo it is probably easiest to prove CH by reducing to triangular matrices)

[tardy hedge]

Yeah

Nice formula makes it work ig

[tough raven]

The key is that it's non-random linear algebra magic.

[tardy hedge]

Wdym non random

Lol

@tough raven

[tough raven]

The opposite of what you meant by random, i.e., that it's not arbitrary or unmotivated.

[tardy hedge]

Why? It doesnt seem motivated to me

Just seems like write this equation in this way and then uh multiply by adjugate

[elfin wraith]

My linear algebra class presented this as the complete proof but motivated that it’s massively unintuitive, but it also gave a kinda slightly bullshit intuitive proof, perhaps that’s of interest

[rocky cloak]

It might help to first think about the case where M is free.

Then all you're doing is showing that a matrix satisfies it's characteristic polynomial, which should feel more familiar.

Then the general picture is just lifting the endomorphism of M along a surjection R^n -> M

[elfin wraith]

But also with that being said I do think this is the easiest and clearest proof so maybe just dwell on it for a bit lol

[gusty thistle]

Oh shit

I never realised that wow

That is really nice

[swift tundra]

I am having trouble showing that spec(A) is disconnected implies A is the direct product of nonzero rings. Does anyone have a hint?

My trouble is that I can write spec(A)=V(I_1)\cup V(I_2), disjoint and I want to use the Chinese remainder theorem, but I_1 \cap I_2 is contained in the nilradical and might not be zero. I was able to show that I_1 and I_2 are comaximal, though.

[quiet pelican]

Ok so you know I + J = (1)
So take e \in I, f \in J such that e + f = 1
You also know ef is nilpotent
Can you modify your choice of e, f to some e’, f’ such that e’f’ = 0?

Hint: ||raise the equation e + f = 1 to a sufficiently high power, and divide the resulting terms on the lhs into two groups||
Hint 2: ||2n should work for your power, where (ef)^n = 0||

[swift tundra]

Ohhhh I see now! Thanks!

[swift tundra]

Though I don’t know if the second hint is correct. If (ef)^n=0, then we can take (e+f)^n. This will give us e^nf^n+ef* “other stuff”. We can subtract the “other stuff” to the other side and then 1-ef(…) is a unit say u. Then pick e’=u^-1e^n and f’=u^-1f^n.

Or maybe you had something else in mind with the hint.

In any case that was the little push I needed. Thank you!

[quiet pelican]

I was going to errr
(e + f)^2n = (e^2n + … + (binomial) e^nf^n) + (the rest)

[white oxide]

I believe an alternate proof is outlined in the exercises

This is what I used to see it

What theorem can this fact be attributed to: "If G is a finite cyclic subgroupof order n, then G has a unique subgroup of order m for each m dividing n."

[merry harness]

Lagrange's theorem, but you would also have to show uniqueness which is specific to cyclic groups here

i dont know if there is a name for that

[white oxide]

Oh I thought Lagrange is sort of the converse, that if H is a subgroup of G then the order of H divides G

[merry harness]

That is lagrange yes

[white oxide]

But it doesn't give existence of subgroups of given orders tho right

[karmic moat]

Idk if that theorem has a name but you do use lagrange in the proof

[candid patrol]

I would say that it is a corollary

[candid patrol]

Right, this is Cauchy for prime order

[tribal moss]

Claiming the subgroup of order m is unique goes beyond those statements, though. It needs to depend on the assumption that G itself is cyclic.

[proven stone]

I'm trying to do a group theory general result test: Let $g\in G, G$ a group with order $|g|=d$. Then for $s\in\mathbb{N}, \langle g^s\rangle=\langle g^{\gcd(s, d)}\rangle$.
What I tried: It is sufficient prove the two generators have the same elements. On the one hand, since $|g^s|=\frac{|g|}{\gcd(d,s)}=d/\gcd(d,s)$, we write $\langle g^s\rangle={e,g^s,g^{2s},...,g^{d/\gcd(d,s)-1}}$. From the other hand, $|g^{\gcd(d,s)}|=\frac{d}{\gcd(d,\gcd(d,s)}$. We can prove that $\gcd(d,\gcd(d,s))=\gcd(d,s)$, so $|g^{\gcd(d,s)}|=|\langle g^s\rangle|=d/\gcd(d,s)$. Then, $\langle {g^\gcd(d,s)}\rangle={e,g^{\gcd(d,s)},...,g^{\frac{d}{\gcd(d,s)}-1}$. Now?

Can anyone give me a hint or any helpful proof?

[cloud walrus]

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

[candid patrol]

🤨

[marble hinge]

I think you can prove that sets <g ^ s> and <g ^ gcd(s,d)> are the same. To this end we can prove that sets of powers of g are the same modulo d. For that we need to show that each element from the left set corresponds to an element from the right set and vice versa. We know that cyclic groups consist of powers of their generator. || So all powers on the left will be of form s * n where n is integer. Then you can show that for each of those you can find a corresponding element on the right, i.e. to find a matching power m such that gcd(s, d) * m = s * n. Note that s can be represented as gcd(s, d) * k for some integer k (because any common divisor of s obviously divides s), so we have gcd(s, d) * m = gcd(s, d) * k * n. We can cancel gcd now from both sides and get m = k * n, so we found our m and it is an integer. So for each element from the left set, there is a corresponding element in the right set. You'll also need to prove in another direction. How about that? ||

[marble hinge]

Mm, in another direction it's actually trickier 🙂 Maybe I need a better proof!

[proven stone]

It will help if you could write in latex

so you go about proving both inclusions

I'm tired now I'll try it tomorrow

[sacred wharf]

Paste into gpt and ask gpt to lateX it

[thorn jay]

have you no dignity

[sacred wharf]

Your dignity is as bad as ur music

[thorn jay]

well my music consists of only bangers so I'll take that as a compliment <3

[sacred wharf]

Indeed they bang my eardrums💯 fr

[thorn jay]

I make bass music

I'd be doing something wrong if that didn't happen

[twilit wraith]

i have to prove that a nonabelian group of order 21 has a unique composition series

but i honestly forget what it even means to have a unique comp series

no way i just got anki reacted by mq bru

i know that if we let T be that group

then $1 \unlhd Z_7 \unlhd T$

[cloud walrus]

hiidostuff

[sacred wharf]

What the heyl is that latex command

[twilit wraith]

the dumbass command for normal subgroups

i have it changed to \normal on overleaf

[sacred wharf]

\nmg

Normal group

So many choices

[thorn jay]

it means that every two compositions series are the same

[thorn jay]

well if Z3 and Z7 are normal then T = Z3 x Z7

[twilit wraith]

would it just suffice to show that T only has one maximal simple normal subgroup

[thorn jay]

maximal normal subgroup

in this case yes

because it has only two composition factors

(namely Z3 and Z7)

[twilit wraith]

yeah

so it just suffices to show that T has no normal subgroup of order 3

which is solved by the fact that |T| is only divisible by 3 once and so by properties of sylow subgroups, any subgroup of order 3 in T cant be normal

bc its conjugate to all the other subgroups of order 3 in T

and there must be more than one subgroup of order 3 as otherwise T would be abelian

[thorn jay]

yis

[twilit wraith]

man

idk why but for some reason i suddenly have like nilpotence and solvability and all that in the bag

or just like group classification that you see in an intro alg course

[thorn jay]

that's good no?

[twilit wraith]

yes very

its just nice to feel confident on stuff that was seemingly advanced

[thorn jay]

sometimes you forget what you were even struggling with in the first place

(why I'd probably make a horrible tutor)

[twilit wraith]

true

hopefully one day i will see that with topology

today is not that day

[thorn jay]

topology, like category theory, is unreasonably effective

but to that end hopelessly abstract without good intuition

[twilit wraith]

its beautiful but unfortunately it feels as if my prof is making it harder to understand it

i dont say that lightly because i dont like blaming professors but its bad

[thorn jay]

oh damb

how so?

[twilit wraith]

we barely went over quotient spaces

as in i have a comfortable understanding with pretty must every topic in pointset except for quotient spaces

which is not good considering that many have told me theyre one of the most important concepts

my prof also doesnt really argue anything rigorously which makes understanding the nuts and bolts of topology take much longer than it does for any of my other courses

i also think our weekly quizzes are unreasonable in my opinion

for example were expected to answer a question like "the klein bottle minus one point is homotopic to the figure 8 space" in about 30 seconds to a minute after only having known about homotopy for 3 days

im assuming the answer can be reached intuitively by someone who understands homotopy but i dont feel as if i had enough time to settle with it in order to handle these problems

[thorn jay]

that sounds reasonable tbh

as in its reasonable that its true

not the question

30 seconds to a minute for that is crazy

[twilit wraith]

i see

the answer is true btw

[thorn jay]

yeah okay

[twilit wraith]

its just so overwhelming

[thorn jay]

my intuition didn't fail me

[thorn jay]

I can imagine

I guess I'd imagine removing a points and blowing up the gap, but the klein bottle is alr hard to visualise

[twilit wraith]

i mean i think i know how to construct it via quotients but i kinda forget how you have to orient the boundary of the unit square

i think its both factors having sides directed in different directions

[thorn jay]

could be it

if you do it wrong you might end up with projective space iirc

[twilit wraith]

smth like this

[thorn jay]

you spin me right round baby right round like a record baby round round-

[twilit wraith]

lol exactly

i pretty much just took the mobius strip orientation but did it both ways

damn it seems i wasnt right

one of the sides has to be going the same way

[thorn jay]

hm

[twilit wraith]

its been even more stressful as i have an exam on pointset on monday

and im just so unsure as it how ill do on it

[marble hinge]

To follow up on that, here is the proof in another direction, @proven stone:

Note that this part is where we actually use the fact that d = |G|, we didn't need it for another inclusion

[winter niche]

Can smn give a light hint on showing that GL2(R) is isomorphic to R* x SL2(R)

[vocal pebble]

Construct an isomorphism using the determinant

[candid patrol]

You need to find a subgroup of GL2(R) which is isomorphic to R* with only one element with det = 1

[tribal moss]

I don't think the "obvious" way of using the determinant here gives rise to a direct product when the dimension is even.

(Some cheek we have, trying to lecture Niels Henrik Abel on group theory!)

[winter niche]

Yes i had the idea of sending a matrix A to (detA, A/(detA)^(1/n)) but it has some problems

[rocky cloak]

In some sense it's much easier.

Like what is an obvious matrix with determinant r for a given r?

[winter niche]

diag(r,1,1,1,..)

[tribal moss]

That's what I would answer too ... but that doesn't commute with SL(2,R).

[rocky cloak]

Oh, I thought we were talking semidirect product

[vocal pebble]

Is this even possible?

[tribal moss]

One idea might be to split R* up further as (0,infty) × {-1,1} and map one part to the squared magnitude of the determinant and the other to the sign.

[vocal pebble]

That you have a surjective homomorphism from GL2(R) to SL2(R)

[mental lake]

How do you find all the ideals of Z12?

[rocky cloak]

So what are you sending -1 to in that description?

Because diag(-1, -1) is in SL2

[tribal moss]

It's a bit fuzzy for me yet.
My rough idea is that I can make the subgroup of matrices with positive determinant as a direct product of SL(2,R) and the subgroup of positive multiples of I -- and then wave my hands and say something like the subgroup has index 2, so it must be a direct factor. But I'm not sure of the exact argument for the second step.

[rocky cloak]

But subgroups of index 2 aren't in general direct factors

[tribal moss]

Well, darn.

[vocal pebble]

Hmm this answer says there is no surjective map so you can't have an isomorphism from GL2(R) to R*×SL2(R) because otherwise you should be able to restrict to the second component and get a surjective map from GL2(R) to SL2(R)
https://share.google/pDNXNiICN0is5zVb2

[tribal moss]

As a quotient of a principal ideal ring, Z/12Z is itself a principal ideal ring, so there are only 12 candidates to check.

[mental lake]

im confused

[rocky cloak]

You know what the elements of Z12 are, yes? Are you able to determine which ideals they generate?

[tribal moss]

By one of the isomorphism theorems, every ideal of Z/12Z is the projection of an ideal of Z, and all ideals of Z are principal, so every ideal in Z/12Z will be generated by a single element too.

[rocky cloak]

In general, a good method for finding all the ideals of a ring starts by determining the principal ideals first anyway.

Then considering sums of them gives you everything

[vocal pebble]

But how practical is the set of generators it gives you?

[median seal]

"The center of a group $G$ (denoted by $Z$) is defined to get the elements ${z_1, z_2, ...}$ that commute with all elements of $G$, that is $z_i g = g z_i$ for all $g$. Show that $Z$ is an abelian subgroup of $G$"
I have no idea how to proceed with the group ${z_1, z_2, ...}$

[cloud walrus]

Absolute Chips

[rocky cloak]

So... do you know the definition of subgroup and being abelian?

[median seal]

Have to satisfy the four axioms of the group, and abelian means commutative relationship holds

[rocky cloak]

Alright, so are you able to show that the commutative relationship holds?

[median seal]

I'm just a bit confused by the element ${z_1, ...}$

[cloud walrus]

Absolute Chips

[rocky cloak]

They are just giving names to the elements

[median seal]

Because it doesn't seems like it's being finite or closed

Not sure how to show the set being closure

[vocal pebble]

it need not be finite (or even countable), but it is closed. it is true that the numbering is a bit misleading

[median seal]

I'm still really new to this

[rocky cloak]

So say x and y are such that
xg = gx and yg = gy for all g.

Then your goal is to show that
xyg = gxy

[median seal]

Such that $g_i g_j = g_k$?

[cloud walrus]

Absolute Chips

[vocal pebble]

if you are calling the elements of the center as g_x then yes you need to show g_i * g_j = g_k for some k. that is also the "goal" in the above message

[median seal]

I think I should say $z_i z_j = z_k$ instead

[cloud walrus]

Absolute Chips

[rocky cloak]

So exactly what i, j or k is not important. All you need is to show that
z_i z_j
satisfies the property of being in Z

Namely that it commutes with everything

[median seal]

First by proving Z is subgroup of G?

[rocky cloak]

I'm not sure I understand what you're asking.

Showing that Z is closed is the first step in showing it is a subgroup yes

[median seal]

Hmm let me try it first

[median seal]

So is this it?

[rapid cave]

I thin its fine.

[thorn jay]

i thin so too

[tardy hedge]

Lol

[thorn jay]

#alg-top-geo-top
shaking my head

kids these days

[rocky cloak]

Not sure what the point of bringing in banana and pizza here is, but the answer is just
🍔[🌮]/(🌮^n+1)

[tall igloo]

🍔[🌮]/(🌮🔼🇳➕1️⃣)

[rocky cloak]

And the bottom index for 🍌 and 🍕 should be 🥪, not🍔.

The whole line is messed up actually

[fading acorn]

technically uses group structure on elliptic curves so its indeed #groups-rings-fields stuff /hj

[thorn jay]

this just in! Homotopy theory is group theory!

[rocky cloak]

https://youtu.be/Ct3lCfgJV_A?si=S0P3vU89bqhit4bv

[winter niche]

So did u get anything about the isomorphism between GL2(R) and R* x SL2(R)

[tribal moss]

Asteroid found an MSE answer that argues it is not possible.
What makes you expect such an isomorphism ought to exist?

[winter niche]

I had to prove it

As an exercise

[tribal moss]

Hmm, this is the point where we generally ask to see a picture of the exercise, just to make sure you haven't misunderstood it.

[winter niche]

Yep

[thorn jay]

what is R here?

arbitrary ring?

[winter niche]

Real numbers without 0

This is the exercise

[tribal moss]

There isn't any instructions to go with it that could have been "prove or disprove" or "is this true"?

[thorn jay]

that isn't an exercise, that's a statement

[winter niche]

There is a Show that out of this image

[tribal moss]

Hmm, so there must be a mistake either in the exercise or in the argument Asteroid linked to.

[winter niche]

Ok thank u

[coral spindle]

It seems to me that they meant to write C instead of R

[tribal moss]

Or 3 instead of 2. Who knows?

[white oxide]

But with induction we have to show first that the powers of (x_1) in k[x_1] are maximal right, we can't just start from the last case with all the variables

In any case I will just show that (x_1)^r is primary in k[x_1] like you said and then use induction like you said

I am still lost as to why we are first considering the powers of (x_1, \dots, x_n) (given that there are at most n variables) but everything else makes sense

Maybe I just got confused about the indexing

Yeah I think that was it

[rocky cloak]

The powers of (x1) are not maximal... I don't understand what you mean.

Like the point is that
(x1, ..., xn) is maximal in k[x1, ..., xn], therefore its powers are primary.

Then you 7iii tells you that (x1, ..., xn)^k is also primary in k[x1, ..., xn, x_{n+1}]

[white oxide]

Yeah I understand I just got confused about the indexing

I interpreted it as this: suppose n = 7 say. Then I was confused about why we were starting with the powers of (x_1, \dots, x_7) in k[x_1, \dots, x_7] when we should be starting with the powers of (x_1)

Anyways I understand now

Ty

[white oxide]

I guess this also implies that (x_1, ..., x_m)^r [x_{m + 1}] = (x_1, ..., x_m)^r

within k[x_1, ..., x_{m + 1}]

Something like this: Fix $1 \leq i \leq n$. Then $k[x_1, \dots, x_n]/\mathfrak{p}i \cong k[x{i + 1}, \dots, x_n]$ is an integral domain. Hence, $\mathfrak{p}i$ is prime. Let any $r > 0$ and suppose that $m < n$. Then $(x_1, \dots, x_m)^r$ is primary in $k[x_1, \dots, x_m]$, for $\mathfrak{p}m = (x_1, \dots, x_m)$ is maximal in $k[x_1, \dots, x_m]$. If $A \coloneqq k[x_1, \dots, x_m]$, $k[x_1, \dots, x_m, x{m + 1}] = A[x{m + 1}]$. By 7iii, $\mathfrak{p}m^r[x{m + 1}] = \mathfrak{p}m^r \subset k[x_1, \dots, x{m + 1}] = A[x_{m + 1}]$ is primary. By induction, this completes the proof.

[cloud walrus]

okeyokay

[vale crypt]

Hello perhaps a more general question: I am (re)learning basic group theory for a representation theory class.

I'm not used to the arguments because I am a mathematical physicist that does more analysis, hence I'm used to arguing with pictures where you can see that a property leads to a bound somewhere or sth similar.

How should I heuristically think of proofs where being coprime is an important property? Should I begin instinctively by contradiction and look for a divisor?

[knotty badger]

Probably the most important property of coprime integers is that they give you a resolution of the identity

This is one way I like to think about them, at least

The formal statement of what I mean is Bezout’s identity

[vale crypt]

Is there a proof you like that demonstrates this idea?

[knotty badger]

Proving Euclid’s lemma requires this

If d | ab and gcd(d, a) = 1, then d | b

[vale crypt]

i mean in algebra

[knotty badger]

Using Bezout you get integers x and y such that dx + ay = 1

So then bdx + ab y = b

d divides the LHS, so d divides the RHS

[tribal moss]

Perhaps it would be more targeted if you could point to an argument where "coprime" appears in a way that's surprising or unclear to you, and we'll see if we can find an intuition that works for you?

[vale crypt]

To be honest the reason I'm asking is because whenever I see it in the proofs in the lecture I can understand where it was used and how we got to the end result using it, but when I go to the exercises and it's a condition I blank out on how I'm supposed to use it

which is why I'm just looking for general ideas

where I can re-read the lecture notes with these ideas in mind

to try to build an intuition

I just don't really know how to deal with integers tbh

[tribal moss]

Can you show some exercises where you "blank out" then? I think we need something to hold on to; otherwise all we can do is blindly recite favorite facts about coprime numbers, which is not terribly likely to be helpful.

[vale crypt]

I think as a concrete example of the exercise I'm blanking out on now (I don't want a solution)

I am to show that if U,V are subgroups of G and their indices in G are coprime, G = UV.

I start with the examples:

The dihedral group is the product of the reflection and the rotations.

The cyclic group cannot be generated by a product of some subgroups (like, choose the subgroup with index 2 and index 4 in C_8 or sth)

But I don't see what's going on

constructing the examples didn't really teach me anything beyond "yeah that happened", which, from my experience elsewhere, isn't meant to be the case

I'm meant to learn something from it that will help me with the problem, but I can't figure it out

[tribal moss]

Hmm, what I would immediately think (no guarantee that it leads to a solution, though) is "it doesn't seem to be easy to use the coprime condition directly here, so I'll first try to prove it in contraposed form: if UV is not all of G, can I then show that the indices must have a common factor?"

And in order to connect UV to indices, my first idea would be that UV is a union of right cosets of V, but is also a union of left cosets of U.

Hmm, is G known to be finite such that it makes sense to calculuate with the sizes of cosets?

[vale crypt]

oh right sorry it says finite

lemme see are there any other words I've missed

nope, finite was the only one

[glad osprey]

My first idea was to show that the index of UV in G is 1, since it's a common divisor of both [G:U] and [G:V], but this only works if we know UV is a subgroup. Anyways, I think it's worth pointing out that you can have intuition for why a result is true without knowing how to prove it. Sometimes a proof is just a trick or some specific technique that you'd need to have seen before

[tribal moss]

So to get back to "coprime", I wouldn't think of that as a freestanding concept here. Rather, my experience is that pretty much every integer that shows up in the context of a finite group is a divisor of the size of the group, so I'd be thinking of the lattice of divisors ordered by divisibility. "Coprime" means that the two numbers don't meet downwards before we get to 1. But going from "order" to "index" corresponds to turning the lattice upside down, so the assumption also means the orders of U and V don't have a common multiple other than |G|.

In order to make use of this way of thinking, my first priority would be to see if I can prove that |UV| is a divisor of |G|. It feels like something that ought to be right, even though I don't currently know it is.

Hmm ... but then again I've just argued that |UV| is a multiple of |U| (since it's a union of right cosets) and also a multiple of |V| (since it's a union of left cosets). So |UV| is a common multiple of |U| and |V|, and the turn-the-lattice-upside-down intuition implies that the assumption that [G:U] and [G:V] is coprime means that |G| is the least common multiple of |U| and |V|. And then we're done and I don't actually need to argue that |UV| divides |G| in general.

[vale crypt]

i vaguely understand the individual steps but I will be converting this into rigorous writing to better understand it, thanks for the help

...oh

i see it

[tribal moss]

My goal here was to demonstrate the (necessarily vague) intuition and exploration I'd use to solve the problem. [This was my actual thought process -- I genuinely didn't know how I'd get through before I started writing the last post].

[vale crypt]

yeah that helped a lot

[rocky cloak]

Presumably the notation UV here means the subgroup generated by U and V

[glad osprey]

Hmm okay, I thought it was just the product of subgroups

[vale crypt]

it's written like that in my exercise sheet, but I thought it meant the product of subgroups too

now that you mention it i don't actually know what the convention for the former is in this class though

[rocky cloak]

Conventions vary, but the set
{uv : u in U, v in V}
is not particularly interesting. So usually UV means the subgroup generated by this set

[vale crypt]

okay I just checked, apparently we \langle and \rangle it if we want the cyclic group

[rocky cloak]

And also if someone says "the product of subgroups" I would assume they meant the subgroup generated by them.

If they actually meant "pointwise product of elements" I would expect them to say something more along those lines.

[vale crypt]

Ah right this makes some other counterexamples more clear, like if I choose a dihedral D_12 or sth, C_3\subset C_6, and the reflection this doesn't work because it's "too small"

because those 2 meet at 6, which is half of D_12

[tribal moss]

In this case the conclusion is true about that set too, though.

[rocky cloak]

Indeed, so the author could have meant either

And I guess might aswell do the most general one

[tough raven]

I think this set is rarely interesting when it's not a subgroup, but it is quite useful to know when the subgroup generated by U and V is "UV" rather than "...UVUVU...". So (I believe) the notation UV for {uv : u in U, v in V} has survived, with <U, V> being more common for the subgroup generated by U and V.

[thorn jay]

Id rather people call it the join of two subgroups ngl

and write U ∨ V

[rocky cloak]

I'm just glad everybody agrees on the product of ideals at least

[noble nexus]

yeah I would write <U,V> instead of UV

and use UV for the set of products of pairs

[thorn jay]

whats wrong with U ∨ V

[elfin wraith]

If only we could get people to agree on the definition of a ring

[rapid cave]

Ring is this thing on people's finger

[elfin wraith]

Only if you like it

https://tenor.com/view/pieceofmani-renaissance-world-tour-beyoncé-nodding-gif-15406475241457126069

[quiet pelican]

Rings have identity but aren’t necessarily commutative
Rings may also be non-associative if they are algebras

[crystal vale]

C[x,y] -> C[t], by x-> t^2 and y -> t^3, I have to prove that kernel of this ring morphism is an ideal generated by x^3-y^2, how do I show it any hint?

Do I need to write f \in C[x,y] in terms of x^3-y^2 ?

[rocky cloak]

So it might be helpful to frame it as the map
C[x, y]/(x^3 - y^2) -> C[t] being injective.

Not that in the first ring since you can replace y^2 by x^3 every element can be written as
f(x, y) = g(x) + yh(x)

[crystal vale]

Ah, and image of f(x,y) will be g(t^2) + t^3(h(t^2) ), and if we simplify it we have, g(t^2) all terms in this polynomial have even degree and in t^3(h(t^2)) all terms have odd degree so therefore no terms of g(t^3) and t^3h(t^3) cancel out each other, therefore all such coefficient zero if the image of f(x,y) = 0, right?

[crystal vale]

And if I have to find the explicit form of the image of this morphism, isn't the subring generated by {t^2, t^3} ?

[tribal moss]

Not quite, you also need some constants as generators.

[rocky cloak]

The subalgebra generated by t^2 and t^3 at least

[tribal moss]

You get all the integers for free, of course, but the image contains for example pi, which is _not_in the subring generated by t² and t³.

Even more explicitly, I would describe the image as consisting of all the polynomials whose t^1 coefficient is 0.

[crystal vale]

I see

[crystal vale]

What is subalgebra?

[crystal vale]

I got it, thank you

[thorn jay]

a subset containing all constants and closed under the operations

[vapid vale]

i was going to say if you know sub-(anything else) the answer is immediate

[thorn jay]

its an object thats sub

[vapid vale]

so are quotient objects bus then

[thorn jay]

no they're cosub

[vapid vale]

cos sub is like something i failed to do on my GRE

[thorn jay]

gonna have to do cos sub on friday 💔

[crystal vale]

I see

I have to show if the ring has characteristic p, then if a is nilpotent then some power of (1+a) is equal to 1.

So if a^k = 0, then I am thinking of taking (1+a)^kp, but my calculations are not so good, any good candidate for power of (1+a)?

[thorn jay]

use binomial expansion

[tribal moss]

I think it would go better with (1+a)^(p^k).

[crystal vale]

I think so

How do I show < x^2, y > is not the principal ideal in F[x,y].

If it is then there is f(x,y) which divides both x^2 and y but how do I get a contradiction from here?

[rapid cave]

y is prime/irreducible

so f must be an associate of one of 1,y

[vocal pebble]

I mean, gcd x^2,y = 1 and <x^2,y> isn't the unit ideal

[dim wagon]

Can i check if its dangerous to try to define Group homomorphisms/R-module homomorphisms from tensor products to something directly on the simple tensors(and then extend it) instead of the universal property? I notice that in alot of the examples in my course, we seem to go back to the universal property construction even if the map on the simple tensors seem obvious.

For example, lets say i want to define the group homomorphism $\phi: D \otimes (P\oplus Q) \to D\otimes (P)$ thats act in the expected way of projecting the second second part. Could we just define $\phi$ on the simple tensors by. $\phi(d\otimes (p,q))=d\otimes p$ and then claim that we extend linearly and call it a day?

[cloud walrus]

somethingwrong

[vocal pebble]

if you can show its well defined then i dont think you should have any problem. the universal property tells you directly that a well defined map exists

[thorn jay]

the universal property tells you on the level of elements that this is exactly the induced map

the universal property basically tells you "hey this map is well defined no need to worry abt it 👍"

[dim wagon]

ah okay makes sense thanks both. for example my map may not be well-defined if two simple tensors are actually the same right? whereas I will never encounter this if i just consider the direct product D \times (P\oplus Q)

[tardy hedge]

In what sense are you talking about gcd here? Since F[x,y] is not a pid

The only common divisor between x^2 and y is 1, and 1 is not in (x^2,y). Saying it like that makes sense to me

[vocal pebble]

yes. (i used it as a short hand if you want)

[tardy hedge]

GCD’s always exist in ufd?

Im just thinking in a pid i know that we can define gcd of a and b as being the generator of (a,b) and we argue “greatest” bc of ideal inclusions and stuff

I guess im talking about the use of “greatest” here specifically

[wraith cargo]

yes

in very large generality you have gcd domains

which are a larger class than UFDs

[tardy hedge]

Oh yeah we just define greatest as the one where everything else divides into

U dont need it to be a pid to talk about that i guess

[wraith cargo]

yeah so in a UFD you decompose the elements into primes and take the ones they both share

and that's the gcd

[tardy hedge]

Oh ok yea makes sense haha

[wraith cargo]

you have something similar going on in dedekind domains where every ideal can be decomposed uniquely into a product of prime ideals

tho okay in hindsight this isn't really related it's just another version of unique factorization

[tardy hedge]

Yeah

I did learn about that a bit at some point but i dont really have too much intuition for it

[wraith cargo]

oh you were learning about dedekind domains?

weren't you on your local cohomology arc hahahahaha

[tardy hedge]

Yeah lol, this was early this year i took some sort of number theory course but it was very introductory, so i think the term dedekind domain was only mentioned in passing

[balmy python]

whatre sylow theorems?

[rapid cave]

they are three theorems concerning so called p-sylow subgroups of a finite group G

[balmy python]

are they stuff ive probably done without being told the name?

[rapid cave]

probably not

[elfin wraith]

https://en.wikipedia.org/wiki/Sylow_theorems

I would doubt youve come across them without being told their name unless you learned maths in another language

[balmy python]

😭

Okay im trying to prove that G/G_tors has no non-trivial elements of finite order

[velvet hull]

the sylow theorems should not have anything to do with that question

[elfin wraith]

Isnt that pretty much by definition

[balmy python]

so if i assume there is a non-trivial element of finite order which we can call gG-tors for g \in G, then g^nG-tors = G-tors which implies that g \in G-tors = e which is trivial since g G-tors = G-tors

[balmy python]

oh nah this is a separate q sorry 😭

[balmy python]

can someone verify this?

[rapid cave]

g^nG-tors = G-tors
this implies that g^n has finite order, so g has finite order

[balmy python]

how does this help with my argument

[rapid cave]

you said g^nG-tors = G-tors implies g is a torsion element

[thorn jay]

this about abelian groups?

[balmy python]

yeah

[balmy python]

yeah

is that wrong?

[rapid cave]

no, but I added a bit more detail since I think its not immediate

[balmy python]

ohhh okay

i get it

thanks

[white oxide]

I want to make sure that I'm understanding this proof right. Here, $x$ is algebraic over $K$, since $x$ is a zero of a polynomial with coefficients in $\mathfrak{a} \subseteq A$, and we can just project that polynomial into the field of fractions? Also, what does the sentence ``The coefficients of the minimal polynomial of $x$ over $K$ are polynomias lin the $x_i$" mean? The minimal polynomial is listed in the proposition, and its coefficients are the $a_i$; how can these be interpreted as polynomials in $x_i$? Just as constants? Moreover, wouldn't it be easier to say, the coefficients lie in $\mathfrak{a} \subseteq A$, hence are clearly integral over $A$?

[velvet hull]

the minimal polynomial of x in L, f(t), is a product of the galois conjugate t-x_i's

[cloud walrus]

okeyokay

[white oxide]

Sorry I don't understand. I don't get why we can't just say, each $x_i$ satisfies $x_i^n + a_1 x_i^{n - 1} + \dots + a_n = 0$, and since $a_i \in \mathfrak{a}$, this shows that each $x_i$ is integral over $\mathfrak{a}$. Also, don't we have $a_i \in \mathfrak{a} \subseteq r(\mathfrak{a})$?

[cloud walrus]

okeyokay

[white oxide]

Maybe I need to review Galois theory or I'll just blackbox it for now, I'm prob wrong

[velvet hull]

well yeah they are

that is true

but we're not trying to show that the x_i's are integral over a

we're trying to say something about the a_i's

[white oxide]

It's displayed in the Proposition right

I'm confused but the coefficients of this minimal polynomial displayed in the proposition are the a_is

[velvet hull]

yes

so we have t^n + a1t^n-1 + ... + an = (t - x1)^m1 (t-x2)^m2 ... (t-xn)^mn

and when we expand out the coefficients you find that each ai is a polynomial in the xi's

[white oxide]

Ahhh okay I see

[velvet hull]

so thats what it means by the coefficients being a polynomial in the xis

[white oxide]

Ahh got it thanks

[balmy python]

so a p-primary torsion subgroup is a subgroup of a torsion group with elements of p^n order?

[wraith cargo]

Uh I would guess so but the terminology is a bit ambigous

I'm not sure if this implies it's a subgroup of the torsion subgroup that's the p-primary component or that the whole torsion subgroup is the p-primary component

tho I'd bet on the former

[balmy python]

ah okay

what is an example of a easy non-faithful action?

[noble nexus]

the trivial action

most actions are not faithful really

for example the action of a group on itself by conjugation is faithful only if the center is trivial

[balmy python]

ooh that's an action?

G x G -> G

woah

g(x) = gxg^-1

[noble nexus]

more than the factorial of the acting set*

[balmy python]

yeah okay

😭

[noble nexus]

since for example Sn acting on n points is faithful

[balmy python]

i guess with infinite order groups it's different

[noble nexus]

You should think of a group action as being a manifestation of the symmetries of the group on an actual object

And it's faithful if the symmetries on the object actually "faithfully" represent all of the symmetries of G

[balmy python]

matrix multiplication by invertible matrices on a vector is an action right?

[noble nexus]

yep

[balmy python]

can i have an example?

[noble nexus]

yeah sure so for example the cyclic group C_n you can think of being the abstract ideal of rotational symmetry

[balmy python]

abstract ideal?

[noble nexus]

Then an action of C_n is basically picking out some object with a rotational symmetry

[balmy python]

is that just like

[noble nexus]

not a precise term

[balmy python]

isomorphic to the behaviour of rotational symmetry

[noble nexus]

I mean it in the sense that like, the number 3 is an abstract concept

but 3 apples is something concrete

group actions are kinda like that

[balmy python]

ahhh it's like we're actually using our group now

[noble nexus]

yeah it's really the main reason why groups are useful is because they encode symmetries of objects

[balmy python]

haven't actually seen that use though 😭

todays lecture was on orbits and stabilisers

but also before that commutators and torsion subgroups i guess

[noble nexus]

So an example of a non-faithful action is for example the symmetric group S_n acts on a two point set {x,y} where even permutations do nothing and odd permutations swap x and y

[balmy python]

okay yeah it's non-injective

[noble nexus]

so in this case most of the symmetries of S_n are kinda being forgotten by this action since you're only using the fact that it's even or odd

[balmy python]

yeah

[noble nexus]

and yeah an action of G on X is faithful if the action identifies G with an actual subgroup of the group of symmetries of X

(For actions on sets the group of symmetries is just the permutation group on X)

[balmy python]

yeah

[noble nexus]

but you can actually have groups act in many interesting ways, so for example an action on a vector space V (usually called a linear representation but it's really just an action) is a homomorphism from G into the general linear group GL(V)

which are the symmetries of the vector space V (that respect the vector space structure

[balmy python]

symmetries just mean respect the structure of the group?

[noble nexus]

probably the #1 most useful theorem in group theory

[balmy python]

no wayyy

[noble nexus]

yeah typically, it's not really precisely defined

[balmy python]

what's G?

[noble nexus]

but typically the symmetries of a mathematical object X means the isomorphisms from X to X

just a group

[noble nexus]

where "isomorphism" depends what you are studying

[balmy python]

ah so you can describe the action as G x V -> V?

[noble nexus]

in a first course in group theory you usually stick to actions on sets

[balmy python]

but obviously it's mapping toa subset of V

[noble nexus]

Yeah

[balmy python]

GL(V)? I thought they were square, invertible matrices over a field?

[noble nexus]

so I assume you've seen the fact that an action G x X-> X for sets is the same information as a homomorphism G -> Sym(X)

[noble nexus]

yeah if you pick a basis then that's what it is

[balmy python]

My first course of group theory was way more simpler than this 😭

[balmy python]

how is that a subset of V?

isn't that a subset of L(V,V)

[noble nexus]

GL(V) is just the invertible linear maps from V to V

which are the same thing as matrices at least in finite dimensions

anyway that's not super important it's just good to know that groups can encode the symmetries of many different types of objects in math

[balmy python]

yeah okay

i understand that now

i really sucked at groups last year 😭

they barely did it tbh

think it was like 10 lectures of groups

[elfin wraith]

My favourite theorem to just create random really slick proofs that I never spot

[noble nexus]

I had to use it in an assignment for an operator algebras course the other day so it just keeps showin up

[thorn jay]

its a silly trick mathematicians like to pull out of thin air

[delicate orchid]

algebra is the study of group actions. Maybe monoid actions if we're feeling fresh

[merry harness]

wait you are kind of cooking

[knotty badger]

Orbit stabiliser is quite helpful

[thorn jay]

expanding on the idea to clone actions and youre technically correct lol

[balmy python]

how do conjugacy classes relate to normal subgroups

[delicate orchid]

a normal subgroup is a union of conjugacy classes

[balmy python]

always?

[delicate orchid]

yup. but a union of conjugacy classes isn't always a subgroup

[balmy python]

ah

does that make it quicker to work out normal subgroups or smth?

[delicate orchid]

proving a subgroup is a union of conjugacy classes is equivalent to the question of it being closed under conjugation

if you're given complete knowledge of the conjugacy classes by some oracle I guess it could be faster

[balmy python]

ah

so a conjugacy class is essentially the set of conjugates of an element and all those conjugates

[delicate orchid]

what other definition of conjugacy class is there

[south patrol]

A conjugacy class is a maximal set of elements whose values under every complex character coincide

[delicate orchid]

you win this round, chucklefucks

[delicate orchid]

Let Z[G] be the integral group algebra of a finite group G with natural basis B of elements of G. Let C be the centre of Z[G] and C^+ be the positive cone of C. Then a conjugacy class of G is precisely a subset of B such that the sum over B is an element in the minimal (under the obvious ordering) hilbert basis for C^+

the worst part is this is true

[elfin wraith]

I’m not sure I’ve seen a more cursed definition of anything before, well played

[errant maple]

nvm I'm just sleep deprived (ignore the messages that i deleted)

[crystal vale]

I see y is prime because F[x,y]/(y) is an integral domain, and in integral domain every prime element is irreducible element

Thank you

[crystal vale]

I have one question, how many automorphisms of C there? Don't give the whole solution just give me a direction, I know that automorphism gives identity on Q and i maps to ± i

[sly crescent]

A lot

[restive idol]

Depends on what you mean by automorphism!

As a field, there should only be 1 (the identity) and as a group, there are countably many at least

[tough raven]

Unless it has index 2 🥸

[restive idol]

I think that $\text{Aut}_{\mathbf{Ab}}(\mathbb{Q}) \cong \mathbb{Q}^\times$

[cloud walrus]

Lucas

[sly crescent]

His point still stands

[tough raven]

It was about ℂ, not ℚ.

[tough raven]

Correct BTW.

[restive idol]

OH, I missed it.. Then there is only two (as a field)

[tough raven]

Nope.

[tough raven]

As they said.

[crystal vale]

What about conjugation?

[restive idol]

Nono, ignore me. I misread your question (twice)

When I said that, I thought you meant the (field) automorphism group of Q. In which case, you get the trivial group

[rocky cloak]

A hint could be that it's consistent with ZF that there are only 2.

[candid patrol]

wrong channel

[wraith cargo]

#real-complex-analysis

[low quiver]

sorry

[mental lake]

I have the quotient ring R=Z[t]/(1-t)^3 and you have to show that (t^3-2)bar=(6t^2-6t)bar but idk what to do

[rapid cave]

They are equal in R iff their difference is in ((1-t)^3) which is the ideal you quotient with

[mental lake]

am i wrong or are they not equal?

it did say show that so im confused on why it would be false

@rapid cave

[rapid cave]

Yeah they are not equal

It should be 2t^3 maybe instead of t^3

[mental lake]

Ok

[vocal pebble]

Can you send the original question?

[mental lake]

The a question also says to show that the polynomial is equivalent to a degree lower or equal to 2

it’s in Dutch tho

[rapid cave]

Yeah so its probably a typo

[mental lake]

ok

is the second one true tho

the one w 1-4t^3

[rapid cave]

Yep

Its true

The difference is (1-t)^4 which is 0 in R

[violet spade]

nederlands!:o

[thorn jay]

nederlands!!

[topaz solar]

with choice

[thorn jay]

implicitly assumed

[merry harness]

All my homies use choice

[crystal vale]

can you tell me more about it?

[rocky cloak]

Like you want me to elaborate on why this is a hint, or just tell you why it's true?

[crystal vale]

both

i don't see how ZF comes into the game?

and here how aoc related to it

[rocky cloak]

So my point about the hint is just that, since it's consistent with ZF that conjugation and the identity are the only automorphisms you will need aoc to construct the remaining ones.

[crystal vale]

so you mean there are more than 2?

but how do i show it?

and sorry but i don't get fully understand what does it mean by it's consistent with ZF

[rocky cloak]

It means you cannot disprove it.

So without using aoc you cannot prove that there exists other automorphisms.

[crystal vale]

and how i can construct other by using aoc?

[rocky cloak]

Are you familiar with transcendence bases?

[crystal vale]

is it something like C is vector space over Q which has infinitely uncountable dimensional?

[rocky cloak]

That's part of it yeah

[crystal vale]

ah, no

[rocky cloak]

Alright, that makes it trickier. Let me see if I can come up with a good hint

[crystal vale]

but how Q-linear morphism related to ring morphism?

[rocky cloak]

Well a ring homomorphism C -> C is in particular a Q-linear map

[crystal vale]

but not converse, right?

[rocky cloak]

Correct

[rocky cloak]

Alright, so maybe the best hint is just to tell you what they are.

A subset S of C is a transcendence basis for C if it is algebraically independent, meaning there is no polynomial
f(x1, ..., xn)
with rational coefficients that is 0 when evaluated on n (different) elements, and
C/Q(S)
is an algebraic extension.

Can you prove transcendence basis exists?

And notice that Q(S) is just isomorphic to the field of rational functions in |S| many variables

[crystal vale]

what does it mean by algebraic extension?

[rocky cloak]

E/F is algebraic if every element of E is a root of a polynomial with coefficients in F

[crystal vale]

okay

[crystal vale]

so if i take S be the basis of C over Q, then S will work

right?

so transcendence basis exists

should i mention how i did it?

[rocky cloak]

A vector space basis?

That won't work as S won't be algebraically independent

[crystal vale]

yes vector basis, i see

[crystal vale]

i am not sure but my guess what if i take all transcedental number in C ?

[rocky cloak]

Well pi and 2pi are both transcendental.

Also notice that you're not using aoc

[crystal vale]

sorry, i have no idea

do i need to use Zorn's lemma?

And say i have a transcendence basis then how do I use it in my problem?

Say S is transcendence basis, is S basis or do I need to work more ?

[rocky cloak]

Well, are you able to construct automorphisms of the field of rational functions Q(S)?

[crystal vale]

If R is ring then R(x) = R[x, x^-1 ] ?

R(x) = R[x,y] with xy = 1

[tribal moss]

No, it's larger than that. R(x) contains all rational functions, so for example also 1/(x+1), which is not in R[x, x^-1].

[crystal vale]

I see

[crystal vale]

Q-linear automorphism or ring morphism?

I think you mean ring automorphism

What if there exists two distinct s1 and s2 exists in S, and i map s1 to s2 and s2 to s1 and remaining s_i to s_i?

If I have to describe Q(S), how do I do it?

[rocky cloak]

Well the elements are of the form
f(s1, ..., sn)/g(t1, ..., tm)
with f and g polynomials in Q and s1, ..., sn, t1, ..., tm from S

[crystal vale]

I see

[crystal vale]

Yes

So using Q(S) automorphism, do I have to induce automorphism C to C?

[rocky cloak]

Yup, that would be the next step

And again Zorn will be helpful

[tribal moss]

After you have that automorphism, you need to appeal to the fact that an isomorphism between fields can be extended to an isomorphism between their algebraic closures -- which also depends on choice.

[crystal vale]

Okay thanks @rocky cloak @tribal moss

[crystal vale]

Why for maximal subset B, k(B) \ K, is an algebraic extension?

I think because if there exists a in K such that there is no polynomial with coefficient in k(B), which has root a.

Then a not in B, so B \subset B u {a}, now I have to show B u {a} is algebrically independent.

If it is not then there is polynomial f(x1,..,x_n) such that f(a,x_2,..,x_n) = 0, i choose x1 as a, because otherwise it will contradict the algebrically independent of B.

Now I am not sure how to proceed further?

Maybe f(a,x2,..,x_n) = 0 implies there is polynomial in k(B)[x] which has root a.

[rocky cloak]

Indeed, f(x, x2, ..., xn) is that polynomial

[crystal vale]

Yes exactly i was confused

So my approach is correct?

[tardy hedge]

R(x) is the same as R[x,y] with xy=1?

[tribal moss]

Still no.

[tardy hedge]

Notknow said R(x) = R[x,y] with xy=1 so i was a lil confused

Yea i didnt think so

[crystal vale]

I am stupid

Okay, does every automorphism of k(B) induce the automorphism of K?

[crystal vale]

So the algebraic closure of F means extension field K, such that every polynomial in F has a root in K?

[tardy hedge]

That stuff is interesting

[crystal vale]

Have you learned it ?

[tardy hedge]

A little

[tribal moss]

Yes, in particular a minimal such extension.

[crystal vale]

I see

[tribal moss]

Whoops, not quite. You need every polynomial on F[x] to split into linear factors in K[x], not merely have one root ther.

[crystal vale]

Why is it not equivalent to that one?

[tribal moss]

(This turns out to be equivalent to wanting a minimal K such that every polynomial in K[x] splits into linear factors.)

[crystal vale]

I got it

[tribal moss]

I can't rattle off a complete counterexample, but consider that for R over Q, the polynomial x³-2 does have a root in the extension field yet doesn't split.

[tardy hedge]

Isnt it equivalent

[tardy hedge]

Oh right

I thought u could factor it and the process continues

I forget the detail im missing

[tribal moss]

Yes, so it is equivalent to requiring that every nonconstant polynomial in K[x] has a root in K.
But it's (as far as I know) not enough to require it for polynomials in F[x].

[tardy hedge]

What r the relationships between the fields in question rn

[tribal moss]

We're defining what it means for K to be an algebraic closure of F.

So together with the condition about polynomials splitting or having roots, there needs to be an additional condition saying that K is algebraic over F, or is minimal.

[tardy hedge]

Like K is alg closure of F?

[tribal moss]

We're talking about the definition of that.

[crystal vale]

Where can I learn more about transcendence basis? In field theory?

[tardy hedge]

Yeah every polynomial over F needs to split completely in K, and K is the minimal field to do that

[crystal vale]

.

[tribal moss]

Not necessarily. At least in the screenshot, K doesn't have to be algebraically closed, so for example it might be that some of the B have square roots in K and others don't. An automorphism of k(B) that interchanges them wouldn't extend to an automorphism of K.

[rapid cave]

I think milne has something in his field theory notes

[tardy hedge]

I honestly have no clue how one would come up with examples of this stuff

[rapid cave]

Of what

[tardy hedge]

Examples of this ig

K is algebraically independent over F or whatever it was ig

Examples illustrating some of this stuff basically

[crystal vale]

if K is alg closed, then?

[tribal moss]

Then (assuming choice) an automorphism of k(B) always extends to an automorphism of K.

[tardy hedge]

How does that work?

Oh i think i know how

[crystal vale]

how?

[tardy hedge]

You’re allowed to choose any x in K to some conjugate

And since its alg closed it contains all conjugates

So thats why it can be bijective i think ..?

[crystal vale]

what is conjugate in arbitrary field K?

[tardy hedge]

I think its like the image of it under some automorphism

That fixes the ground field

( that part necessary im pre sure)

[crystal vale]

but then identity can be one of them

[tardy hedge]

Cuz it needs to fix the polynomial equation

[tardy hedge]

Ya

[tribal moss]

I think you can get through by using Zorn's lemma on extensions of your automorphism whose domain and range are subfields of K, ordered by set inclusion.

[tardy hedge]

Oh u cant finitely choose right

To construct the map ig

[elfin wraith]

Question about automorphisms of field extensions (rather on topic). Ive got 2 similar problems on this homework and im thinking about how theyre different. The first question says, let w be non-real root of X^p-1 for an odd prime p, show that Aut(Q(w)) is non trivial. The second say let a be a root of X^p-2 for p an odd prime, show that Aut(Q(a)) is trivial.

Im a little stuck with the second and but I think im getting there, I just need a nudge. The first difference I spot is that in the first problem, the polynomial is reducible, it has 1 as a root and that is in Q. We know the automorphisms are non-trivial because we can just take like complex conjugation. Apparently we cant do this with the second one, and I am admitadley not sure why. Like, we can work over the splitting field of f, Q(a,w) and then the automorphisms are determined by where we send all the p roots, but im not sure how to bring this all together tbh

I guess I dont know where to go because im not actually conviced that Aut(Q(a)) is trivial, like Im sure it is true, but I dont have a feeling for why this is the case yet

(I can tex that up if its hard to read btw lol)

[crystal vale]

how do i show transcendence basis S of C over Q is infinite?

[tribal moss]

If it wasn't, C would have only countably many elements.

[tardy hedge]

You can always do complex conjugation?

Like its not complex conjugation isnt it like just replace + with -

I may be off or confused its been a while

[elfin wraith]

I think the issue is that in the second case the root could be real

So there conjugation doesnt do anything

[tribal moss]

The complex conjugate of a is not necessarily in Q(a).

[tardy hedge]

But an extension Q(w) is not like C

Tbh idk why the odd prime part matters rn

[elfin wraith]

Ok, so I know this is true for the pth roots of unity, because they all give the same field extension, but im not sure how to make anything happen for the second case

[tribal moss]

"odd prime" is just so -1 won't be root in x^p-1.

[crystal vale]

is there any relation with transcendence basis and vector basis?

[tardy hedge]

Ohh primitive root of unity but p is prime so any root of unity generates da shi

Thats why odd prime matters i think

[elfin wraith]

Yeah im happy enough with that one

[tardy hedge]

Any basis for field extension stuff is formulating it as vector space basis

Im pretty sure anyway

[elfin wraith]

But im stuck with showing the second group is trivial

As I said I dont really have a feeling for why it should be trivial, so im not sure where to start looking

[tribal moss]

A transcendence basis generates the field uniquely by plugging the base elements into polynomials.
A vector space basis generates generates the field uniquely by linear combinations.

[tardy hedge]

Yeah true

I got kinda lost there too

I thought u could get all pth roots by multiplying by roots of unity

And since its odd prime adjoining any root of unity u get all of them

[elfin wraith]

So like $f = x^p-2$ and then $\alpha = \sqrt[p]{2}$ and $\omega$ is a primitive root of unity. We then get that $f$ splits as $f = (x-\alpha)(x-\alpha\omega)\cdots(x-\alpha\omega^{p-1})$ and we can call these $p$ roots $\alpha_i$ in the sensible order

[cloud walrus]

Nope

[tribal moss]

Whoops, that's not quite true. It's more involved than that for transcendence bases.

[crystal vale]

which are base elements and what are they polynomial?

[elfin wraith]

then any automorphism is determined by where we send each of the $\alpha_i$

[cloud walrus]

Nope

[tardy hedge]

So we’re restricted in choices somehow ig

[elfin wraith]

Yeah Im wondering if we can say one of them has to be fixed for some reason and then it follows from that, that the rest must be too

I think the important thing here, thats different to the first case, is that 1 isnt a root I guess?

Ive not worked out exactly how this helps yet, but 1 is a root, and ring homomorphisms preserve 1 so idk this likely means something

[tardy hedge]

Doesnt that just mean something like the extension should have nontrivial automorphisms

I didnt think that statement through but

Nah the root u adjoin isnt assumened to be in base field anyway

[elfin wraith]

Yeah its just a root (and in this case none of them are in Q anyway)

[tardy hedge]

Oh i didnt notice the nonreal vs real prt

[elfin wraith]

For $i\neq 1$, do all of the $\alpha_i$ not just generate the same field extenstion?

[cloud walrus]

Nope

[tardy hedge]

Or just nonreal

[elfin wraith]

No I dont think thats true actually, I think thats key, because thatll be why the conjugate doesnt work

[tardy hedge]

What do u mean exactly by conjugate

[elfin wraith]

For clarity this is the actual question

[elfin wraith]

The map sending a+xb to a-xb where x is whatever youve stuck onto your field

[tardy hedge]

Gtg doing tutoring

[tribal moss]

I think the right way for (b) is to argue that you get an isomorphic subfield of C no matter which root of x^p-2 you adjoin, namely one isomorphic to Q[x]/<x^p-2>. Since at least one of those subfields -- namely the one where you adjoin the real root -- doesn't contain any other roots, none of the subfields contain a root other than the one you adjoin.

[elfin wraith]

Yeah this is what I was working out there after realising that alpha_i dont actually generate the same subfields, then I think I should be ok from there?

[tribal moss]

That's not even well-defined unless the extension has degree 2.
(And it's only a homomorphism if x happens to be a square root of something in the base field).

[elfin wraith]

Yeah of course, I was just kind thinking out loud and working out why that didnt always work

[tulip otter]

i have a question: given an abelian additive $p$-divisible group $A$, \let $p_A:A\to A, x\mapsto px$. why is the tate group given by $T_p(A)=\varprojlim A[p^{n+1}]$ instead of $\varprojlim A[p^n]$ where $A[p^n]=\ker p_A^n$. I mean if each element (sequence) $(x_0, x_1,\dots)$ of the tate group starts with $x_0=0$ then $p^nx_n=0$ and $x_n\in A[p^n]$ for all $n$ no?

[cloud walrus]

ali yassine

[tulip otter]

here is the context

[tribal moss]

Does the difference matter? One of those expressions leaves out A[p^0] but everything it contributes is encoded in A[p^1] anyway -- you can leave out any finite number of steps at that end of the inverse system without making a difference in the limit (up to isomorphism anyway).

[tulip otter]

is it because here if x_k\in A[p^k] then x_k\in A[p^m] for any m>k?

[tribal moss]

No, it's because whenever you have (x1,x2,x3,...) you can reconstruct what x0 must be be by applying p_A to x1, so leaving it out of the limit doesn't lose any information.

In general, whenever we have an inverse limit of a system of the form
... -> A4 -> A3 -> A2 -> A1 -> A0
we can omit A0 without changing what the limit is, because the A0 component of a limit can always be recovered from the A1 component by applying the A1->A0 map.

And so forth -- in fact you can omit just about any combination of the objects as long as there are still infinitely many of them left. For example, omit all the even-numbered A's.

[chilly ocean]

Hello everyone. I am new to the server. I would like to discuss something about cayley's table D4 and some observations I made.

I am still a novice and only know till like cyclic groups and generators and a first year undergrad.

So the idea is mostly to compute cayley's table of D4 without exhaustively computing each elements of the table individually.

I have tried explaining my observation in this reddit post https://www.reddit.com/r/learnmath/comments/1o8cakr/studied_basics_of_group_theory_and_cayley_table/?utm_source=share&utm_medium=mweb3x&utm_name=mweb3xcss&utm_term=1&utm_content=share_button

I am open to criticism and want to know more well read individual's opinions on the method.

[tribal moss]

That's a pretty long text, and I haven't the energy to look at every detail, but if you're getting the right results, then most likely what you're doing is valid.
It's definitely good for your learning and understanding to engage with the material in this way, but I doubt you have discovered something radically new. On the other hand, if it was new to you, then it has served its main purpose, so well done.

For what it's worth, the comment by Wofster to the Reddit post describes the usual way to calculate with elements in the dihedral group.

(Actually writing down Cayley tables is something nobody really ever does, except during the first few weeks of a course in group theory in order to get familiar with how the definitions work. Pretty much every group that matters in practice has better ways to calculate with it when you need to calculate -- which is less often than you'd think at first ...)

[rocky cloak]

It is in fact enough, but it is not completely immediate to show.

[tulip otter]

I see, can one also omit uncountably many objects too? (of course when the indexing set of the collection of groups is uncountable itself so that it can contain uncountable subsets)

And also when it meets the conditions necessary for the definition of an inverse limit to make sense

As in the indexing set I being a directed poset and the family of groups being directed

[tribal moss]

As long as everything you omit was reachable from something you keep, and you don't delete so much that the diagram stops being directed, the inverse limit should be the same.
Hmm, actually that's probably not true in the general case where the diagram is not linear. I can't immediately rattle off the right condition in that case.

[tulip otter]

Ohhh makes sense

[chilly ocean]

Thanks for the review! Btw, I have also observed that It also worked for like cayley table for D2(with some tweaking here and there)

And yea it's a bit too long because I thought I wanted to display that the method works instead of writing some examples.

And thanks again, I appreciate your effort to read the post.

[chilly ocean]

Just curious, is that grothendieck in yr pfp?

[tulip otter]

No it's a wizard yes

[thorn jay]

is there really any difference?

[rocky cloak]

So one approach is to show that alpha is the only root of X^p - 2 in Q(alpha).

Another is to calculate the Galois group of X^p - 2 and find the subgroup corresponding to Q(alpha). Then show that it's equal to its own normalizer.

[tulip otter]

No, but I had to clarify

[tribal moss]

That surprises me. On the other hand it's not a question that lends itself to simple counterexamples ...
Hmmm, I actually know it to be true when the base field is finite. Perhaps we can lift it to arbitrary fields by some kind of model-theoretic shenanigans.

[elfin wraith]

No Galois groups introduced yet, but I went for the first thing. I just kinda had to think out loud for a minute first lol

Thanks though

[rocky cloak]

So for perfect fields it's not so bad.

For any element polynomial over F, the splitting field is seperable, so by the primitive element theorem it is generated by a single element. Then any conjugate of this element generates the same extension, so just having one root of the minimal polynomial is enough.

So this covers characteristic 0 and finite fields, among others.

[tribal moss]

by the primitive element theorem
Ah, if I ever knew that, I had forgotten it again. Yes, I see that would do it.

[azure lichen]

Hey, I'm considering a few books on ring theory to get, and I was wondering if I could get some recommendations

The ones popping up for me right now are:

A Course in Ring Theory by Donald S. Passman
Lectures on Modules and Rings by T. Y. Lam
Ring Theory (Volume by Louis H. Rowen)

And there was a broad suggestion to read Serge Lang

[rocky cloak]

Here's an argument for the non-prefect case
https://www.jstor.org/stable/2315743?seq=1#metadata_info_tab_contents

The idea is once you adjoin all pth roots to F you get a perfect field, then you can apply the perfect case argument to this to show that the extension is algebraically closed and you're done

[tribal moss]

Makes sense.

<shakes fist at paywall>

[tardy hedge]

Bruh just get dummit and foote or atiyah macdonald

Ok like im biased cuz thats what i use lol but those ones u mentioned i never heard of (besides lang, which i dont think is good for beginner text)

[elfin wraith]

Lam is a very common book for non com rings, I don’t know how beginner friendly it is though. I’ve had a look at passman before too but it’s quite dated iirc, I used another one of his books a lot of a short paper I wrote though and it was well written

But yeah basically these are just noncom texts

[thorn jay]

some people want to do noncomm rings

[tardy hedge]

wowie

[tardy hedge]

Makes sense why i wouldnt know

I mean i wouldnt be opposed to learning about noncomm stuff

But i dont know where they’re important

[elfin wraith]

From a mathematical point of view, I just think they’re more interesting, and it makes sense to study the more general object

As for more tangible reasons, there’s plenty of things that don’t commute, differential operators being the clear example, there’s a whole field of differential algebra and that’s all noncom stuff

[tardy hedge]

Yeah i guess for me personally im only just feeling like i have some sort of handle on ordinary commutative rings

So noncomm is like eh

Might be nice for putting some stuff into context ig

[azure lichen]

Thanks. I’m planning on going into a PhD program for data science so I want to understand the concepts deeply. But I also have not yet looked at ring theory related topics. Currently am a research assistant for a professor’s idea to improve Gaussian Processes

[azure lichen]

Would you say that these authors are generally superior?

[tardy hedge]

Theyre more common for commutative algebra purposes which is what most ppl usually study at first

Atiyah macdonald in particular is a well liked book

[azure lichen]

Okay. For context, the concept of ring theory came up when I was looking up the relationship between the different forms that the Bellman Ford equation takes, if you’re familiar

[tardy hedge]

Im not , but if its commutative stuff then ur thinking commutative algebra lol

[azure lichen]

Thanks I’ll look into these books

[twilit wraith]

ive been asked on my hw if the set of all functions with finitely many zeroes along with the zero function is a subring of {f: [0,1] to R}

the answer is yes as elements of rings can only be expressed as finite sums and products of other elements, right?

[noble nexus]

I think you have issues with sums

what if you take some function thats equal to 1 on [0,0.5] and add it to a function equal to -1 on [0,0.5]

[twilit wraith]

oh duh

ive been conditioned by analysis to only think about the positive real numbers

[noble nexus]

unless im misunderstanding the question, because it seems kinda strange

are you sure it doesn't mean the functions with finite support

(as in, finitely many nonzero values)

[twilit wraith]

its finitely many zeroes

[noble nexus]

hm weird question

[twilit wraith]

yeah

[vapid vale]

i guess its just wanting you to think about closure properties somewhat abstractly

[twilit wraith]

yeah

[twilit wraith]

is there a not incredibly tedious way to do this

the first relation is obvious

but the others just seem like such a bore

[velvet hull]

oh yeah, there's another way to derive this automorphism

hold on i can dm

[crystal vale]

In Atiyah, they said a ring R is called Noetherian if R is Noetherian as R-module

But say M is a finite set, which is an A-module then it is always Artinian and Noetherian, right?

[thorn jay]

yes

what is the problem here?

[crystal vale]

So when I am saying some A- module is Noetherian, it is very important that what is A, if I change A to some B then maybe that's not true, right?

So now what if one of them is subring the other? [ I think it depends on the scalar multiplication too ]

[thorn jay]

well if youve got a morphism f : B → A and you pull back an A-module to a B-module, you usually lose a lot of structure (as f might fail to be surjective), so there will appear more submodules

for example, the underlying group of a Noetherian R-module usually fails to be Noetherian

and this is obtained by consider R as a Z-algebra and performing a change of basis

[crystal vale]

If R is a ring, and if I show any maximal ideal is finitely generated then R will be Noetherian, right? No.

Can you give me a counterexample?

I know why it cannot be true

[boreal inlet]

There are nice counterexamples if you look into ring of smooth functions of certain manifolds

[velvet hull]

perhaps if you localize a non noetherian ring you can get something

[boreal inlet]

Also you can check for local rings with their maximal ideal being principal, though I have no idea if Non-Noetherian examples of this exists

[thorn jay]

Take R = ∏_i∈I F_i, where F_i ≈ F is some fixed field, and I = {0, 1, 2, ... }. This is certainly not Noetherian; consider the chain of ideals
0 < F×0... < F×F×0... < ...
however, each maximal ideal is of the form F×...×F×0×F..., which is generated by the element (1, ..., 1, 0, 1, ...)

[velvet hull]

ah, great example!

[thorn jay]

products always make for great non-Noetherian rings which are still relatively well-behaved

[rocky cloak]

There are actually other maximal ideals given by non-principal ultrafilters on I. And I have a feeling these won't be finitely generated...

[thorn jay]

agh of course

bamboozled again by infinity

take the ideal generated by all (0, ..., 0, 1, ...)

what would the resulting quotient be?

the ring of "almost equal" sequences

or ig equivalently "eventually equal" sequences

[rapid cave]

Enpeacian sequences

[thorn jay]

reduced products are weird

given an algebraic modular lattice L such that the top element is compact, is there always a ring such that the ideal lattice is isomorphic to L?

i cant imagine that

[rocky cloak]

Do those lattices necessarily have maximal elements (under the top)?

[thorn jay]

whoops forgot completeness

[thorn jay]

oh yeah right should add that

top element is compact

https://math.stackexchange.com/questions/4621500/is-every-modular-lattice-the-ideals-of-a-commutative-ring

[crystal vale]

N_(i-1)/N_i and M_(i-1)/M_i as a set are different right? I mean both have empty intersection as a set, so how do I say N_(i-1)/N_i \subset M_(i-1)/M_i?

[tough raven]

The inclusion N_i into M_i, composed with the quotient map from M_i to M_i/M_{i-1}, has kernel N_{i-1} and hence its image is a submodule of M_i/M_{i-1} isomorphic to N_i/N_{i-1} by the first isomorphism theorem.

(This is an argument applicable to any algebraic structure and not specific to modules.)

[crystal vale]

I see

Thank you

[gentle wolf]

what are irreducible integers?

[vocal pebble]

Prime numbers probably

[charred iris]

Presumably defined in the actual chapter? But yeah they're the prime numbers here ('irreducible' and 'prime' technically refer to different properties, but for integers they're equivalent)

[gentle wolf]

yeah probly
i just tried to attempt the exercises first

[harsh gale]

Okay, it is written in the book, that tensor product is a function $ \omega \otimes \eta (v_1,v_2) = \omega(v_1) \eta(v_2) $ from $V \times W$ to $\mathbb{R}$, where $V,W$ are vector spaces and $\mathbb{R}$ are their Field.

Am I correct that It's exactly the linear morphism I would get with universal property of tensor product if I factor map to just the ring of tensor product $M \otimes_R N$?

[cloud walrus]

Pyramid

[wraith cargo]

The tensor product isn't a function

The tensor product of two functions is the function you get when you factorize your map V x W -> R through V (x) W

So it's a map like η(x)ω(v_1 (x) v_2) = η(v_1)ω(v_2)

[wraith cargo]

But yeah your other paragraph is correct

[harsh gale]

I believe it's elementary way to introduce tensors in analysis book.
And yeah, that's good. Thanks for verifying

[delicate orchid]

you might want to look into the Kronecker product of matrices if you want more of an intuition for this

[knotty badger]

wait kronecker product?

ohhh i mixed that up with hadamard product srry

kronecker is the tensor one

[elfin wraith]

Does the hadamard product have any uses? I feel like I have come across it before outside of the context of “why doesn’t this naive idea work” but I couldn’t tell you where

[harsh gale]

$\operatorname{Hom}(M \otimes N, K) \cong \operatorname{Hom}(M, \operatorname{Hom}(N,K))$
I thought it could be somehow related to this, but type-checking said no.
Idk If I can reverse tensor product from it though. Although it's unique, so I can reconstruct it knowing r-modules

[cloud walrus]

Pyramid

[knotty badger]

yeah this is the tensor-hom adjunction

[glad osprey]

I'm not sure I understand what you mean by "the linear morphism I would get with universal property of tensor product", are you asking about the relation between ⊗ defined as (w⊗p)(u, v) = w(u)p(v) compared to ⊗ as the canonical multilinear map from VxW to V⊗W? Or are you asking whether the former map factors through V⊗W?

[harsh gale]

It's about category-theoretic definition of tensor product

[knotty badger]

essentially the thing is that

multiplication on the real numbers is a bilinear map $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

so if you have linear functionals $\alpha : V \to \mathbb{R}, \beta : W \to \mathbb{R}$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

then you get a linear map $\alpha \times \beta : V \times W \to \mathbb{R} \times \mathbb{R}$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

composing gets you a bilinear map $V \times W \to \mathbb{R}$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

and then using the universal property gets you a linear map $V \otimes W \to \mathbb{R}$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

which is defined on simple tensors $\vec v \otimes \vec w$ as $\alpha(\vec v) \beta(\vec w)$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[glad osprey]

yeah, and this is how you identify V*⊗W* with (V⊗W)*, right?

[knotty badger]

oh uh hm

is that true

$(V \otimes W)^* \cong \text{Bil}(V \times W, K)$ by the universal property

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

now there's certainly a map $V^* \times W^* \to \text{Bil}(V \times W, K)$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

by sending $\alpha, \beta$ to $(\vec v, \vec w) \mapsto \alpha(\vec v) \beta(\vec w)$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

so that gets you a linear map $V^* \otimes W^* \to (V \otimes W)^*$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

i can believe this map is injective

[harsh gale]

Oh really

Yeah, it just cleared my intuition a little bit

[knotty badger]

i'd have to check but i think it is

i don't see why it's necessarily surjective though

[glad osprey]

I think this is only a bijection for finite dimensional vector spaces (not sure if both or only one of V, W need to be fin. dim)

[knotty badger]

i think if one of them is finite-dim you'll get a bijection

[harsh gale]

I'm not sure if it's true but I have possible false memory, that in Aluffi it was proved through showing func and it's inverse

[knotty badger]

and thus an isomorphism

[harsh gale]

Proving it's section and a retraction categorically (intentionally rephrasing for cat fan)

[knotty badger]

but if both are then you won't

[knotty badger]

yes yes i'm aware

[glad osprey]

wait, you mean neither, right? Or are you saying it's not a bijection if both are fin dim?

[knotty badger]

oh sorry i mean neither lol

the issue is i don't see a natural map $(V \otimes W)^* \to V^* \otimes W^*$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

this would be a decomposition of a bilinear form into products of pairs of 1-forms

i don't expect such a decomposition exists generically

[knotty badger]

yeah so

$(V \otimes W)^* \cong \text{Hom}(V, W^*)$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

and it's a standard result that if either $V$ or $Z$ are finite-dimensional, then $\text{Hom}(V, Z) \cong V^* \otimes Z$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

in fact, in general, $V^* \otimes Z \cong \text{Hom}_\text{Finite Rank}(V, Z)$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

and so $V^* \otimes W^* \cong \text{Hom}_{\text{Finite Rank} }(V, W^*)$ for example

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[glad osprey]

yeah, that makes sense

[knotty badger]

Yay

[harsh gale]

Oh my god, haha, I was 1 page away from the answer and more general approach to tensors

Okay, it make sense to review chapter before going through every detail

[knotty badger]

oh nice, what’s he general approach

[harsh gale]

Define the tensor product of the spaces $V_1, \ldots, V_k$, denoted by $V_1 \otimes \cdots \otimes V_k$, to be the following quotient vector space: $$V_1 \otimes \cdots \otimes V_k = \mathcal{F}(V_1 \times \cdots \times V_k) / \mathcal{R},$$ and let $\Pi: \mathcal{F}(V_1 \times \cdots \times V_k) \to V_1 \otimes \cdots \otimes V_k$ be the natural projection. The equiv- alence class of an element $(v_1, \ldots, v_k)$ in $V_1 \otimes \cdots \otimes V_k$ is denoted by

[cloud walrus]

Pyramid

[harsh gale]

Just common tensor product in algebra

[knotty badger]

Ah yeah, this is the standard way to construct it

There is an alternative construction I prefer but it uses a little cat theory

[harsh gale]

You mean concrete construction in r-mod? What is it?

As left-adjoint to multilinear map?

[knotty badger]

Well that’s just a universal property, it’s not a construction

Are you familiar with natural transformations

[harsh gale]

Yeah, even with abelian and monoid categories

[knotty badger]

Awesome awesome

So here goes

Given these spaces V_1, …, V_n

You can define a functor from Rmod to Set, via Z -> n-Linear(V_1 x V_2 x … x V_n, Z)

So it sends a R-module Z to the set of n-linear maps from that product to Z

Do you agree this is a functor?

[harsh gale]

I believe yeah

[knotty badger]

Cool

Then, you consider the set of natural transformations from this functor to the forgetful functor U

Which just sends an R-module to its underlying set

[knotty badger]

This actually forms an R-module, via pointwise operations on your natural transformations

This is in fact the tensor product

[harsh gale]

Interesting

By the way starting as construction of free object

[knotty badger]

In other words, the tensor product is an R-module of natural transformations

In fact you can view any free object this way

As a kind of “algebra of natural transformations”

The trick is

You look at what functor the free object is supposed to represent

In this case, it’s n-linear maps

Then you consider natural transformations from that functor to the forgetful functor

These obtain a natural algebraic structure via pointwise operations

And that correspond to your free object

[harsh gale]

It's pretty interesting. A pretty universal way and uncover some properties of categories and linkage between free objects and tensors

[knotty badger]

Yep yep

This works essentially because of the yoneda lemma

[knotty badger]

You do have to check this of course

The set of natural transformations isn’t guaranteed to form an algebraic structure if the type you want

And while this is a “construction” it doesn’t tell you too much about what the elements of your free object are

Unless you have a good idea of what such natural transformations look like

[thorn jay]

I prefer the view of the free object over X being the set of all terms with variables in X over your signature quotiented out by all the equations your class of algebraic structures satisfy

it is, at least, the interpretation which is much more useful in UA

[gentle wolf]

how the hell did someone work out this addition definition so that the new point is always on the elliptic curve??

[thorn jay]

well that's the end product

the begin product iirc is you take two points, and the line through them, and the third is the other point intersecting the curve and your line

then you do some algebra do derive this form

[gentle wolf]

ah, so it has a geometric interpretation
ic

[thorn jay]

ofc, mathematicians would never pluck stuff out of thin air!

( they absolutely do)

[karmic moat]

Some plucker embedding joke

[copper kestrel]

im a bit confused on finding a counter example for this, bc i know its not true and that its not closed under composition but like i just dont know what to do from here

[thorn jay]

well, this is true only if B is the orbit of b under some subgroup of S_A

🙏

[copper kestrel]

whats an orbit

[thorn jay]

Take A = { 1, 2, 3 }, and B = { 1, 2 }, and b = 1. Then (321) maps 1 to 2, but (321)(321) = (231) which doesnt send b to an element in B

[heady oar]

i cant send gifs

?

its math gifs

[thorn jay]

you're sending gifs for no reason, and adding nothing to the conversation

[heady oar]

is that illegal?

[true bolt]

handled

[next fog]

Not here, in #chill if you want to

[copper kestrel]

hmm, doesnt 1 get mapped to 3 though?

[thorn jay]

thank you mods!

[true bolt]

@thorn jay please don't delete mod pings in future

[copper kestrel]

ty mods :3

[thorn jay]

oh, I'm sorry! I thought it was already deleted by the mods lol that's why, but I won't in the future

[true bolt]

not to worry, we'll react to the mod ping in future, and its not really an issue for mod pings to be in chat

[thorn jay]

exactly, so the subset { s in S_A | s(b) in B } cannot be closed under composition

[copper kestrel]

ohhhh

[thorn jay]

because the cycle (321) maps 1 to 2

[copper kestrel]

oh right i get it

sick

[thorn jay]

but then, if it was a subgroup, then (231) = (321)(321) must also be in the subgroup, and that sends 1 to 3

tip with finding counterexamples is try the simplest possible examples