May someone bring me some motivation for why in algebra course we initially considered a group G as "operations" but later as elements being manipulated by the group elements of its symmetric group SymG? Now the focus even shifts to AutG

I mean, what's the idea of looking for the same structure in "higher" levels and exploring it

I'm getting stuck on a basic thing. If $\nu$ is a signed measure, $f$ extended $\mu$-integrable, it is claimed $\nu(E)=\int_Ef\,d\mu$ is finite iff $f\in L^1(\mu)$. A signed measure $\nu$ is finite if $|\nu|$ is finite. For $\impliedby$ direction, if $f\in L^1(\mu)$, I see how $$|\nu(E)|\leq\int_E |f|\,d\mu<\infty,$$but $|\nu(E)|$ being finite is not the same as $|\nu|(E)$ being finite, is it?

In fact, $|\nu(E)|\leq |\nu|(E)$, so how can I conclude that $\nu$ is finite? :(

@X4J You have some mathematical structure $X$ you want to study. One way to do this is to look at its "symmetries" (whatever that is), this will give you a group $G$. These may give you quite a bit of information, so $G$ becomes an object of study in its own right. How do you study $G$? Well, you could look at its "symmetries"...

You should establish for yourself that groups are something worth studying, and then ask "how do I study groups?" and you will find that these things arise in the same way as groups in the first place.

@BenSteffan Hrm... I got stuck on step 1.

That makes sense until the part of considering the automorphism group

Sure it would be nice if the operations preserve the structure of the group they're acting on

and yeah I guess it's natural to look for the specific subgroup AutG, once considering SymG. But is that it

Sym G doesn't know anything about the group structure on G

in what sense

if there are two group operations G_1 and G_2 on the set X, then Sym G_1 and Sym G_2 are isomorphic. Sym G only depends on the underlying set of G

@XanderHenderson maybe you should invent fractal groups

oh yeah but there's a subgroup of SymG that is isomorphic to G so it "includes" the structure of G in some sense, isnt it?

yes, but Sym G only "includes" all other group structures on the underlying set of G in the same way

Aut G is the part of Sym G that takes the group operation in G into account

You mean that the only relevant subgroup of SymG to G is AutG?

I wouldn't say that necessarily

oh yeah I get it now

that's all about the operation

then when you consider Inn(G) it's even more specifically "linked" to G's structure in the sense that you can derive properties about G by considering Inn(G)?

@LukasHeger gross.

@psie the Hahn decomposition gives $P,N$ that are positive, negative respectively. Then you can get an "explicit" representation for $|\nu|$.

just wondering if we could tap Hannibal Lecter for some government position.

OMG OMG I just solved the problem

Funniest shit is i solved it in my dream

I have to take a minimal normal subgroup M of G and show that it's abelian. Then apply induction on G/M

I would answer my own question on mse but unfortunately I'm on my phone :(

that's a problem that seems easy to solve, just get off the phone?

@copper.hat ikr, kids these days and their phones ..

But fr I don't have access to my laptop right now

@X4J have u looked at the structure of $Aut(S_M)$

"Let K be a nonempty compact set and $x\notin K$. Prove that there are two open sets U and V, which are disjoint $(U\cap V = \varnothing)$, such that $K\subset U$ and $x\in V$." I am having trouble with understanding this problem

Aren't $(-\infty,0)\cap(0,\infty) = \varnothing$ disjoint open sets

Are u in a metric space?

In that case u can find this

oh wait singletons aren't compact sets

@nickbros123 haven't covered metric spaces. We're in $\mathbb{R}$

I was thinking $K = \{0\}$ would be a counterexample but yeah then it's not compact

Actually I don't get why this must be true

Finite sets are compact

@Obliv this is related to the normality of your space. For R just look at the distance between x and K. More over, K is closed and bounded

Yes but why must $K\subset U$?

U need to find such a U

You * need to find such a U

Oh

Open sets aren't conscious

Can't I have $x \in U$

Let $K = [0,1]$ with $x\notin K$. I can define $U = (-\infty,2)$ and $V = (2,\infty)$ why can't I have $x \in U$ or $x \in V$. Why must $x \in V$?

Its just labelling. You need to just find two open disjoint sets, let's call them 😀 and ☹️ such that 😀 contains K and ☹️ contains x

Oh so basically

@Obliv if X is not in K, it's inside an open set. I.e, there is an epsilon ball around x that is fully outside K. Can u go from here?

I define an open cover on $K$ as $U$

U take the epsilon/2 ball around x as V, and u take the complement of the closed epsilon/2 ball around x as U. In R the closure of an open ball is just the closed ball.

yeah or we can define $U = (-\infty,d)\cup(d,\infty)$ where $d = \inf\{|x-y|:y\in K\}$ and $V = (x-d,x+d)$

oh wait that's what you meant

@nickbros123 in what sense

@X4J that was a joke nvm

@nickbros123 i may rant about phones, but i rarely leave mine down...

@psie i added an answer to your polar integration question.

paul vs. tyson?

@nickbros123 regularity to be more precise

right, yes

@Obliv no, you're mixing everything up. You are given a compact set and a singleton and your job is to find two non intersecting open set separating them.

sets*

@psie do you know that $|\nu|(E) = \nu(E\cap P) - \nu(E\cap N)$ where $P$ is positive set and $N$ a negative set?

@copper.hat ah. Copper responded too. Sorry

@nickbros123 for this to hold you need at least a Hausdorff space

@Obliv they are

@SoumikMukherjee Hausdorffness, I suppose

regularity wouldn't imply that because its not immediate that compact sets are closed

this is equivalent to Hausdorffness

@Jakobian right, yes. compact ==> closed is not generally true

my metric space muscle memory

@nickbros123 well it would hold in this case

just maybe not immediately

The way to prove this is to use that $\mathbb{R}$ is Hausdorff

that is, if $x, y\in \mathbb{R}$ are distinct then there exists open disjoint $U, V$ with $x\in U, y\in V$

if $x\in \mathbb{R}$ and $K\subseteq \mathbb{R}$ is compact with $x\notin K$ then for each $y\in K$ you find $U_y, V_y$ as above

then find finite subcover of $\{V_y : y\in K\}$, say $V_1, ..., V_n$ and $U_1, ..., U_n$ are corresponding neighborhoods of $x$

You can then take $V = \bigcup_{k=1}^n V_k$ and $U = \bigcap_{k=1}^n U_k$

this is a very important technique to remember

because a similar proof arises in many contexts

@Jakobian oh right

@copper.hat accepted :)

@Jakobian yeah, in fact $\nu(E) = |\nu|(E\cap P) - |\nu|(E\cap N)$, so if $|\nu(E)|<\infty$, then can I simply conclude that $|\nu|$ is finite too?

@Jakobian oh wait, you're right with your formula.

Sure, mine isn't wrong too, but I think yours is better in this case.

So if $\nu(E)$ is finite for any $E$, then in particular for $\nu(E\cap P)$ and $\nu(E\cap N)$. Thus $|\nu|(E)$ is finite.

@copper.hat paul, but only for even money 🥊🥊

And win by KO only, otherwise it's a push :-)

Hi

In this question math.stackexchange.com/questions/52509/…

The $\left\langle d\pi(V), d\pi(W) \right\rangle_{p}$ in the definition is just $\left\langle p'(0), q'(0) \right\rangle_{p}\,$, right? Or am I crazy?

Using the projection $\pi$ just make it confusing imo

Not to bore everyone with set theory axioms again but I was reviewing this definition on wiki and they say if '=' is not included in your first order logic, you can use that following formula for $x=y$. I thought we don't write sets being elements of sets in set theory? Shouldn't it read $\forall z(z\in x\iff z\in y) \land \forall w(x\subseteq w \iff y\subseteq w)$?

> I thought we don't write sets being elements of sets in set theory?

@Obliv what do you mean by that

can you explain what you mean by the sentence jakobian quoted

(people do that all the time, both inside formal set theory and outside of it)

@Jakobian I recall being scolded for writing a set $A \in B$ for example

were you doing that in a situation where A wasn't in B? :)

@Obliv in ZFC, we can write $A\in B$, it makes sense to do so, always

its not always true, but it always makes sense to ask if $A\in B$ is true

I don't remember what the convo was about exactly but something to do with axioms of ZFC and Jakobian&Xander helped me understand some things

what you might actually be recalling is interchanging between $x\in y$ and $x\subseteq y$

Oh yeah that might have been it

and how those are not the same symbols

Can we write $x\in y$ is the same as $x\subset y$ then? Like proper subset

no

So then what is the distinction between subsets & being an element of?

$\{a, b, c\}$ is a subset of $\{a, b, c, d\}$, $a, b, c, d$ are elements of $\{a, b, c, d\}$

oh right so $\{a,b,c\}\notin \{a,b,c,d\}$ but $\{a,b,c\}\subset\{a,b,c,d\}$?

@Obliv I didn't say that

$\{a, b, c\}$ might still be an element of $\{a, b, c, d\}$ if its equal to one of the $a, b, c, d$

well, that wouldn't happen in ZFC for a reason unrelated to this, but yeah

(removed)

@Jakobian Specifically, $d$. By the foundation axiom it cannot be equal to $a$, $b$, or $c$

not sure what (removed) was, but ZFC imposes that certain things have to hold

maybe choose simpler examples rather than arbitrary sets. {1,2,3} is not an element of {1,2,3,4} but is a subset of {1,2,3,4}, in any sane encoding of things like 1, 2, 3, and 4 as sets

I think I get it

depending on how much axioms of set theory we might want to impose we might even have that $\{a, b, c\} = a$

in ZFC this wouldn't happen, but it might happen in a weaker axiom system

i guess you could probably do an encoding in which 4 is {1,2,3}

Numbers are the most primitive objects that contain nothing else in them so I should have used that in the example because letters can represent sets

I guess

never mind any of what i said, the point is the difference between elementhood and subsethood is fairly concrete and intuitive

Or in which $3$ is $\{0, 1, 2\}$, like how you define von Neumann natural numbers

@Obliv oh boy...

EVERYTHING is a set in ZFC

numbers included

Numbers have values.

obliv if you take set theory as it is usually done in ZFC, then sets are the only things that can be elements of sets

there's no ontological difference between set-hood and element-hood

"1" is just an abbreviation for some set

a set is simply whatever you consider in ZFC

so a set doesn't have a definition

i feel like this too is somewhat at odds with elementhood and subsethood being pretty intuitively distinguishable concepts

but in ZFC we define numbers, and we do so using sets, because that's the only thing we have

"1" is "one"

I don't understand how axiom of regularity/foundation applies to singletons?

How does $\{a\}$ contain an element disjoint from itself

okay let's stick to natural numbers

well, the only one that it can is $a$

So that means $a$ and $\{a\}$ are disjoint, or that $a\notin a$

Okay what about $a$

what does it contain?

doesn't matter

its an arbitrary set

The axiom applies to all nonempty sets, so is $a$ empty or nonempty?

consider $\{1\}$ as a concrete example then

$\{a\}$ is non-empty

$1\in \{1\}$ with $1\cap \{1\}=\varnothing$

but now is the underlying 'set' $1$ empty?

it'll be a waste of my time if you are refusing to listen

I'm listening

Carefully.

@think_meaning_buildß You are free to join the discussion. I'm not sure why you think I'm not listening carefully

the axiom of regularity is a pretty weird axiom to me. when i first learned about it, i found it pretty weird that it isn't a consequence of some of the other axioms

The axiom of regularity says that a non-empty set $b$ contains a set $a$ disjoint from $b$

@Obliv doesn't matter

In here the non-empty set $b$ is $\{a\}$

i don't find it weird as a statement of intuition about sets, just odd that it wasn't somehow covered by other such things

@Jakobian I know, and $a\cap \{a\} = \varnothing$ I'm not disputing anything

I'm just wondering what happens when we now consider the element inside of $a$, (Not $\{a\}$)

$a\cap \{a\} = \emptyset$ is equivalent to $a\notin a$

So $a$ only contains $\varnothing$. I guess that makes sense.

So the axiom of foundation applied to the singleton $\{a\}$ says that $a\notin a$

that's all it says

it doesn't matter what $a$ is in this setting

I don't think that's what it says. Doesn't it say there exists an element of the set $\{a\}$, call it $y$, such that $y\in \{a\}$ and $\nexists z(z\in y\land z \in \{a\})$

...

@Obliv provide me evidence and I might respond

I don't see how $a\notin a$ follows from the formulation of the axiom on the wiki page. It says for any nonempty set $x$, $\exists y \in x$ s.t. $\nexists z(z\in y \land z \in x)$

I was just applying it directly with $x$ being $\{a\}$

well guess what, I explained that

:(

@Jakobian The axiom of foundation applied to the singleton $\{a\}$ says that $\nexists z(z\in a\land z\in \{a\})$

OHh

I see, since $a \in \{a\}$, it must be true that $a\notin a$

Sorry, I didn't see that.

-1

Ty peanut gallery

For your very helpful comments

👂👂

@think_meaning_buildß But he didn't

He said $a\cap \{a\} = \emptyset$ is equivalent to $a\notin a$

Ask him politely to do it again.

I said that axiom of foundation applied to $\{a\}$ implies that $a\cap \{a\} = \emptyset$ which is equivalent to $a\notin a$

No, I understand it now.

Sure, I didn't explain it in detail, but I did explain it

Sure, and you are correct but I didn't see it immediately until I applied it directly myself.

Thank you

Sorry for acting like a peanut gallery.

can't get too flustered with someone who literally chose "Obliv" as their chat name :)

LoL

I do genuinely put an effort to learning btw, even if it might not seem like it over chat. I hope no one ever thinks that I'm purposefully being obtuse

i prefer to be acute

being obtuse is my job

the world needs more left angles to counteract the current shift

Just slow down a bit @Obliv

Im freaking scared of Dummit and Foote. Anyways, the rapture, i.e, algebra exam, in 8 hours 💀💀💀

I know what $\operatorname{ob} \mathrm{tuse}$ is, but what are the morphisms

@think_meaning_buildß I'll try :P

You're losing accuracy by going so fast.

@nickbros123 good luck. What chapters/sections is it on

First 6 chapters, roughly. And some parts of rings

First 6 chapters are mostly group theory stuff?

@Obliv It's just group theory yea

A** whopping amounts of it

Honestly if life gave me a time slowing machine of sorts I'd just solve Dummit and Foote for 2 years

6 chapters of group theory 💀

6 chapters of $1=g\cdot g^{-1}$ 💀

pretty sure you could prove Feit-Thompson in 6 chapters 💀

@BenSteffan Dummit and Foote outlines a similar result at the end of 5th 💀

"""outlines"""

Dummit and Foote are grade A savages. I didn't know into what machine I was putting my head into

@BenSteffan if you allow each chapter to have the length of the stacks project

sounds reasonable

Some of the stuffs fine, but things started getting fishy in and around sylow, then progressively became satanic as it went closer to nilpotent and solvable

those are the good stuff!

the student became nilpotent and the exercises unsolvable

Anyone handy with relative de Rham cohomology?

i wear a hat so i look idempotent

@Thorgott I was not ready for it 💀

@nickbros123 Dummit and Foote is not so bad. It is actually a fairly easy text (in my opinion) compared to some of the alternatives, e.g. Hunderford's undergraduate algebra text.

Some of those proof techniques man, I cried I think when I saw you have to decompose G into direct product of its sylow subgroups in order to prove some nilpotent result (I guess it was that if H is proper in G, H is proper in N_G(H), or something like that)

@XanderHenderson it takes a lot of time for me.

the Hungerford Games text is the worst

@nickbros123 Any math text should. I typically allocate about an hour per page when I am trying to learn something new.

Sometimes more.

wow, that's fast

@copper.hat That's the first read. :/

i spend most of my time trying to figure out what the point is. once i get that, is usually fairly straightforward. that's why working with experts makes life easier

Also the proof that the sylow 11 group of order 231 is normal, and the centre of G contains this 11 group, these constructions are straight from hell. I could never come up with these if you have me a millennia

one or two sentences can save a week of head scratching

Once group theorists realized that such long arguments could work, a series of papers that were several hundred pages long started to appear. Some of these dwarfed even the Feit–Thompson paper; the paper by Michael Aschbacher and Stephen D. Smith on quasithin groups was 1,221 pages long

Why do u want to classify groups? Just let them be man

as a mathematician, classifying things is in my nature

surely finding a shorter proof is a goal too?

@copper.hat LONGER proofs!

one can always remove intuition, like Rudin

@copper.hat That's how he can write such short proofs.

there should be two versions of papers, the simplex, and, for real mathematicians, the cryptic version.

@leslietownes I find the axiom of set induction more intuitive than the axiom of regularity. It is a straightforward analogue of the axiom of induction for natural numbers, expressing a notion of "completeness" or "closure". It also has the advantage of working well in intuitionistic settings.

These classification crusades are just abuse of orbit-stabiliser theorem and counting conjugacy classes. Why why why should I look at the group acting on P(G) by conjugating

they probably have some intuition, but do not relay to the reader

@user76284 interesting, thanks

Although I am a fan of orbit stabiliser theorem, tbh

@copper.hat the intuition I have for group theory are just a bunch of techniques I have adopted so that when I look at a new problem I can reapply: i guess I should look at the quotient and inverse project. Not big on intuition

it is like conjugate gradients, as an algorithm it is opaque, but when viewed in terms of Krylov subspaces suddenly makes sense. unfortunately it took me years before i saw the different perspective.

well, it think intuition is an inevitable victim of generalisation. that is why working from specific cases is a necessity for me

then again, sometimes generalisation gets rid of the cruft. who knows...

key say as someone said

To be clear though I really like group theory. Trying to go through it in less than 4 months is, though, asking for a nice beating

my first encounter was high school in the 70's. jst did not get the point then.

I'd say till sylow things were fine, probably the course should've ended there with us spending more time on problems and revision

@nickbros123 a lot of books like dummit and foote are very much written assuming that they will be the basis of a year length course, or at least that an instructor will pick and choose judiciously from the contents. what you are describing feels like a lot of group theory for an intro course.

Yeah, I think so too.

Sylow is a misspelling of slow

it was pretty common for profs of the beginning one semester algebra class at my undergrad to get close to a point where one could introduce the sylow theorems, and then cover them hastily or just not cover them, depending on how much time they had left. it was generally understood that the course needed to cover very basic ring theory and even sometimes fields and that did not leave time for "interesting" group theory

then the game you could play as the teacher of the second semester undergrad algebra course was "as you all know from the previous class, ..." and assume whatever you wanted of group theory :D

Lol

@leslietownes That was my experience with undergraduate algebra, as well.

I didn't actually see the Sylow theorems until my phd program (my masters algebra focused more on fields and algebras, with an eye toward linear algebra and programming).

@XanderHenderson this number varies wildly for me but on avg somewhere around that

Most of the time goes into me trying to prove the theorem myself. That's the source of the beating- once you see the proof it's either soul crushing cuz I missed something trivial or its soul crushing cuz I would never come up with it in a million years

xander: i don't know if i was in a course that covered them. the thing above of "as you all know from the previous class" certainly happened in my second undergrad semester where it went from maybe on the syllabus to part of a background we were assuming (and did not ultimately use). we may have covered them in grad school, maybe??? too long ago.

@leslietownes Yeah, I've never taken a course which covered the Sylow theorems (I avoided algebra as a phd student, and didn't take those classes). So I have only heard of those theorems from others in my cohort who were being introduced to them in a phd program.

@Ben I don't know anything about the operad stuff, but for some reason I felt compelled to comment and now I'm out of my depth again... math.stackexchange.com/questions/4998643/…

oh, fun story

so I know the person that asked that question, they're in the same seminar as me

they asked me yesterday and I didn't know the answer, so they brought it here

so we've come full circle I think

oh lol

but I don't understand your last comment

why would we be free to change $\mathcal{C}$ up to equivalence

?

you should disregard what Qi said

In Lurie's setting everything is defined strictly at this point

cause Def. 4.2.1.12 only requires an equivalence $\mathcal{C}^{\otimes}\simeq\mathcal{O}_{\mathfrak{a}}^{\otimes}$, no?

the word Lurie uses is isomorphism

I believe this means isomorphism of simplicial sets

not just equivalence

in any case the relevant use case should be Example 4.2.1.16, and there you have an isomorphism, not just an equivalence

I suppose it has to mean isomorphism of simplicial sets (even though I find this weird, but it's not inconsistent with how he uses the word in HTT on second thought)

very little in the $\infty$-operad section is given up to homotopy

somehow everything needs to be strictly defined to make things work

yeah, I'll go with this

second question: is the Alg_{/LM}(O) thing closed under equivalence as a subcategory of the functor category?

I doubt so

hmm, is the map between Algs a pullback of the map between functor categories?

otherwise I don't see how to conclude the map on Algs is a categorical fibration from the map on functor categories being a categorical fibration

(the latter being the claim from my initial comment that I can actually demonstrate)

elements of $\mathrm{Alg}_{/\mathrm{LM}}(\mathcal{O})$ are sections of the fibration $\mathcal{O}^\otimes \to \mathrm{LM}^\otimes$ that are also $\infty$-operad maps, but being an $\infty$-operad map is not homotopy invariant a priori

yeah, that seems fair

so perhaps it is a pullback? I suppose that comes down to asking whether the upper horizontal map in such a pullback diagram of fibrations of $\infty$-operads reflects inert edges?

hm

which pullback would this be?

the one involving O over LM and C over Assoc

but this seems dubious too, being a map of $\infty$-operads should not be determined by a restriction...

Hey guys

hi :)

hey

gathering of men of culture

perhaps it's worth noting that the functor $\mathrm{LM}^\otimes \to \mathrm{Assoc}^\otimes$ has a section

@Frusciante Hay is for horses.

@BenSteffan LIES!

NONSENSE!

I've just wanted to chime in. I've added a small comment to my question, but I'll repeat it here: Lurie actually proves that the map $\fgt: Alg_{LM}(C) \to Alg_{Assoc}(C)$ has all Cartesian lifts, so it is also a Cartesian fibration (since it's an inner fibration, as we have discussed), so I guess this gives us it is an isofibration as well. But it is not super satisfactory I guess

FALSEHOODS!

Like this a posteriori proves that the map is a categorical fibration, but Lurie claims it a priori

sup Xander

$\mathrm{Assoc}^\otimes$ is actually a full subcategory (sub-$\infty$-operad?) of $\mathrm{LM}^\otimes$

@SineoftheTime this gathering is like a joke set-up: "How many mathematicians does it take to understand a remark by Jacob Lurie?"

@Frusciante wait where does he prove this??

@Frusciante sorry if I caused confusion, but to be clear, the induced map on the functor categories is also a categorical fibration a priori

@Ben 4.2.3.2

the only thing that isn't clear to me is why its restriction to the Alg-categories is still a categorical fibration

(a priori, that is)

@Thorgott $i$?

@Thorgott "Bread but in French" - Ben

@Thorgott ah okay, I didn't understood this point

@SineoftheTime $\limsup$, please.

@Frusciante fantastic

@XanderHenderson noted :|

boo, chat aligns to the left

@Thorgott but isn't this due to the following: the Alg categories are full subcategories of Fun_{LM}(LM,O). We have argued that the map induced on Fun_... are categorical fibrations (isofibrations + inner fibrations). All the solution to the lifting problems have to stay inside Alg_... since well, they are full subcategories.

Dang...

@Thorgott I'm trying to fix it...

Does this make sense or am I just saying silly things?

this is true for lifting against inner horns, but not so obvious for lifting equivalences, you'd have to know Alg is closed under equivalences for that (and Ben suggested that it isn't)

if it is it isn't clear to me, and I don't recall Lurie mentioning it

@Thorgott right...

@XanderHenderson effort appreciated

@XanderHenderson multiline messages kill formatting

Okay, multiple lines breaks things.

?!

magic

is it clear what the inert edges of LM are?

@Thorgott I think so

inert maps in N(Fin_\ast) are just maps of pointed sets

Maps in LM are similarly defined but with an ordering on every fiber

But the fiber of inert maps are just singleton, so an ordering is no data at all

So you basically can always lift inert edges in N(Fin_\ast) to edges in LM and these should be the inert lifts

hmm, the question is basically whether a section LM -> O is a map of operads if its pullback Assoc -> C is a map of operads, but it's not quite clear to me if that's true

as there are inert edges of LM not coming from Assoc

(unless I'm misunderstanding)

Yes there should be inert edges of LM not coming from Assoc; this is my understanding too

(just to make clear, with "not coming from Assoc" I mean that the inclusion Assoc \to LM doesn't hit all the inerts)

Btw I think I'll go to sleep now. But I really appreciate your help @Thorgott, I'm quite satisfied by also understanding why LMod_A(\Ccat) is an infinity category, leaving alone the whole isofibration part

I'll think a little bit more in the next few days and if some idea comes to my mind I'll ping you here I guess

sounds good, good night

(there have to be more inerts in $\mathrm{LM}^\otimes$: it has more objects than $\mathrm{Assoc}^\otimes$)

@copper.hat T — 2 hours :-)

@BenSteffan I gotta ask about an idea of symmetry as in Galois theory but I prolly will not be formally accurate

ok

🕒

If we'd countably order in a multiplication table all the polynomials in Q[X] vertically and all the elements in some simple extension of Q horizonally, then for an automorphism f we'd expect that the entry corresponding to the i'th element in the extension, Ai, and the j'th polynomial, Pj, it satisfies Pj(Ai) = Pj(f(Ai))?

I'm trying to motivate myself for the idea of symmetry

I don't see how labeling everything matters to this question

maybe it's not

I'm just trying to understand

well, I'm not sure what exactly you're getting at, but the property that holds is that f(P(x))=P(f(x)) for any polynomial P and element x in the extension

I'm not sure why you need this to motivate symmetry

because of the invariant of the solutions

I may be inaccurate

in any case there's your statement

what I just said implies that f permutes the roots of P

why

if P(x)=0, then P(f(x))=...

field morphisms are injective