groups-rings-fields

[crystal vale]

How do I prove if F\ subset L \subset E such that they are field extension and L is separable over F and E is separable over L then E is separable over F?

[rustic trench]

For topics that you'd learn in a first introduction to abstract algebra, pertaining to groups, rings, modules, and fields.

what are modules

[rocky cloak]

You can prove that the seperable degree satisfies
[E:F]_s = [E:L]_s * [L:F]_s

[rocky cloak]

Module is an abelian group acted on by a ring.

So essentially a vector space, but where your scalars are taken from a ring

[thorn jay]

vector spaces, except the scalars can be in any ring (so the integers, polynomial rings, matrix rings, etc)

[twilit wraith]

Im surprised ur first course involves them frankly

[rustic trench]

i'm not doing any course

[twilit wraith]

Usually you see those being the hot topic in a second course

Oh I see

[crystal vale]

But the professor has not introduced a separable degree yet

[rocky cloak]

I thought that was the definition you posted earlier

Allrighty then, so splitting field of seperable polynomials, or what definition you got?

[rustic trench]

so like vector(GF(2), [...]) would involve modules?

[rocky cloak]

I mean, a first introduction can consist of two or three courses

[twilit wraith]

Fair enough

I guess thats how mine works now that I think about it

[crystal vale]

E is a separable extension over F is E is alg extension over F and for each x in E its minimal polynomial over F should be separable over F

[rocky cloak]

That sounds like just a vector space, but vector spaces technically are modules

[rustic trench]

maybe i should look at the definition for vector spaces

ohhh

do vector spaces need fields?

and modules don't?

[rocky cloak]

Do you have Galois theory available?

[crystal vale]

No

[rocky cloak]

Can you use that for an irreducible polynomial not to be seperable it must be off the form f(x^p^k) in characteristic p?

[crystal vale]

Yes

[rocky cloak]

Then I would look at the minimal polynomial of x over F and try to factor it over L

[tulip otter]

so there exist polynomials q,r such that f=gq+r with deg r< deg g. We know that r neq 0 since f is irreducible, but how do i show that r=1

[crystal vale]

Suppose K is a field of char p, and K ≠ K^p that means there exists a in K such that x^p-a has no roots in K.

Now how do I show this polynomial is an irreducible polynomial?

[tulip otter]

hmm i thought it should be because otherwise we wouldve found 2 distinct inverses of g in k[[x]] no? the formal power series one and the polynomial in k[x] too. Oh wait, the one i am assuming is in k[x] is actually in k[x]/(f(x)) right? ie is an inverse of g(x)+(f(x)) not of g(x). So there is only one inverse of g in k[[x]] which is the formal power series one and there is no inverse of g in k[x]?

sorry if this is non-sense . Its just that I was confused a bit about this

[rocky cloak]

It's not even in k[[x]], but in something like k[[1 - g(x)]]. I mean just imagine like g=2 or something. Then clearly
1 - 1 + 1 - 1 + ...
is not a finite sum equal to 1/2.

[rocky cloak]

Think about the minimal polynomial of one of its roots

[rustic trench]

vector spaces -> fields
modules -> rings

?

[rocky cloak]

A vector space is exactly a module where the scalars are from a field yes

[rocky cloak]

Further
g = rp + s with deg s < deg r.

Eventually you get something with degree 0

This is Euclids algorithm

[tulip otter]

yes, my problem comes after this. Which is to show that this constant polynomial at the end is actually 1.

[rocky cloak]

It doesn't have to be 1. Any constant polynomial is invertible as long as it's not 0

[tulip otter]

ah right its k[[1-g(x)]] mb

[tulip otter]

ohhh i see. ok i think i understand now. tysm jagr, have a great day/night

[tulip otter]

actually i implicitly used (c)\implies (b) when i was proving (a)\implies (b). So the way you were suggesting was definitely easier

tysm Tiessie, have a great day/night

[crystal vale]

Its roots are only one a^1/p, and it splits as (x - a^1/p)^p

[rocky cloak]

So then how could it factor into irreducible polynomials?

[crystal vale]

I see

[iron arrow]

I can imagine that two finite abelian groups with the same order and with the same number of elements of the same order are always isomorphic due to the classification theorem, but do there exist two non isomorphic non abelian finite groups with such property?

[rocky cloak]

(C3)^3 and the mod 3 Heisenberg group are two examples of order 27 where all nonidentity elements have order 3

[rustic trench]

doesn't "same order" imply same number of elements?

[true bolt]

"same number of elements of the same order" I assume means like "same number of elements of order 5" and so on every n <|G|

[rocky cloak]

That would still imply they have the same number of elements in total, but a little redundancy doesn't hurt

[true bolt]

https://mathoverflow.net/questions/39848/finite-groups-with-elements-of-the-same-order huh interesting

[elfin wraith]

That’s really cool! I guess not unexpected but somewhat surprising

[true bolt]

Groups are evil so you expect it to be false but its cool people have spent some time thinking about this

[elfin wraith]

It does seem like group theory research is just “how broken can we make this random property?”

[true bolt]

As opposed to finite rings which are much easier to work with

common ring theory W

[rocky cloak]

I mean, finite rings can get pretty crazy

[elfin wraith]

I don’t think I actually know anything specifically about finite rings. I think I saw once that they’re all isomorphic to some matrix ring but I’ve never specifically studied them

Other than like, ya know, finite fields

Quotients of Z I guess too

[true bolt]

huh are you thinking of artin-wedderburn.

[elfin wraith]

I presume it’s a consequence of that, I can buy all finite rings are semi simple artinian

But I could also just be making up a result, as I said they’re not something I’ve ever really thought about specifically

[rocky cloak]

Well you shouldn't buy that because it's obviously wrong

They're artinian though

Like Z/4 exists for example

[elfin wraith]

Oh yeah duh lol, that’s like the go to example

I am probably just hallucinating something then

[iron arrow]

Yeah

[iron arrow]

Well yeah that is true xd

[iron arrow]

Hm and what is the property that makes them not isomorphic? And I was right about this property on abelian groups making it so that they're isomorphic right

[rocky cloak]

Well the first is abelian and the second is not for example.

And yes for abelian groups it should be true

[iron arrow]

Oh sorry

I think I said explicitely

That both had to be non abelian

Like I could imagine there existing one abelian and one non abelian

[rocky cloak]

Then just take the product with a non-abelian group I guess

Or there are some examples of order 16

[iron arrow]

Like which?

[rocky cloak]

Loads
https://mathoverflow.net/a/223820/157483

[iron arrow]

Oh that's cool

Thanks

[rocky cloak]

So I guess C4 semidirect C4 (the 4th dicyclic group(?)) and C2xQ8 make the best example for you

[tulip otter]

for part a, view the polynomials $X^4+1$ and $X^6+X^3+1$ as polynomials in $\mathbb{Z}[x]$. If they have any roots in $Frac(\mathbb Z)=\mathbb Q$ then the roots should divide $1$ by the integral root test (since the leading coefficients of each of these polynomials is $1$), ie at least one of $-1$ and $1$ is a root. But plugging in shows that neither of them is a root, hence both of these polynomials are irreducible in $\mathbb Q [x]$.\\Now for part b. I claim that given a field $k$, any reducible polynomial of degree $2n+1$ has a root in $k$ for any $n\in\mathbb{N}$. The proof goes by induction, the case $n=0$ is obvious. Assume that the assertion is true for any natural number $< n$ and let $f(x)\in k[x]$ be a reducible polynomial of degree $2n+3$. Since $f$ is reducible, there exist polynomials $g,h\in k[x]$ such that $f=gh$ with $0<\deg h,\deg g<\deg f$ and $\deg h+\deg g=\deg f$. Now $\deg f$ is odd so it follows from $\deg h+\deg g=\deg f$ that $\deg g$ and $\deg h$ have different parities, ie one is even and the other is odd, say $\deg g$ is odd and then the assertion follows by induction since $\deg g<\deg f$ and so is of the form $2m+1$ with $m<n$.\\The polynomial $X^3-5X^2+1$ is irreducible since otherwise it should have $1,-1$ or both as roots by the integral root test, but neither of them is. Hence it is irreducible by the first part of b.\\I am not sure about part c, so any hints for this part?

[cloud walrus]

ali yassine

[velvet hull]

just because the polynomials have no roots in Q doesn't mean they have to be irreducible

you have only shown that they have no linear factors

[tulip otter]

ah yes i remember seeing this and taking it into consideration but then i forgot about it for some reason

like x^4+4 is not irreducible in Q[x]

[velvet hull]

your claim in part b looks really sus

I don't think it's true

[tulip otter]

but has no roots in Q

[tulip otter]

[tulip otter]

ohhh why

[velvet hull]

you're essentially claiming that F[x] doesnt have an odd degree irreducible polynomial of degree at least 3 and an even degree irreducible polynomial of degree at least 2

because otherwise 2+3 = 5 is odd but has no roots lol

[low cargo]

holy shit, i just watched a video on abstract algebra and it's so beautiful

what are the prerequisites?

[velvet hull]

I can just spoil the answer in this case, the reason why for degree 3 it is enough to check linear factors is exactly because 3 = 2+1 is the only way of splitting up 3 into the sum of 2 or more positive integers

[tulip otter]

yea i got that, but then i tried to generalize

[velvet hull]

so any factorization of a degree polynomial is exactly a degree 2 * a degree 1

the generalization does not work

Imma read through it to see where it goes wrong

[tulip otter]

yea i see your point from here

[rocky cloak]

You're using the induction hypothesis to say that g must have a root because the degree is smaller than f, but your hypothesis is only that reducible polynomials have roots

[tulip otter]

ohhh right

[rocky cloak]

There aren't really strict prerequisites, but most people learn linear algebra first

[velvet hull]

no strict prerequsites, but a knowledge of basic set theory and proof methods is usually expected. as jagr said linear algebra is also extremely nice to have but not as crucial

[rocky cloak]

I guess high school algebra might be a prereq

Well not a very strict one, but meh

[tulip otter]

but is there some sort of a generalization to other degrees?

[low cargo]

i've already studied differential calculus, i've started LA this week and i know set theory and proofs (will brush up the coming weeks)

[rocky cloak]

I guess the most useful generation is that if a degree n polynomial is reducible it has a factor of degree <= n/2

[rocky cloak]

Sounds like you could jump into it

[tulip otter]

I mean it does hold for any odd degree in R for example, I suppose that there must be a similar result for other fields which have similar properties of R that allow for such a thing to happen?

[velvet hull]

over Q it is already kind of a mess, because the polynomials x^n - 2 already guarantees that there is at least one irreducible polynomial of degree n (or literally anything you can make up using eisenstein)

[tulip otter]

true

[velvet hull]

like the R thing works because of the analytic properties of R (it is a complete metric space)

[tulip otter]

hmmm i see

[rocky cloak]

I'm not sure there really are other fields with similar properties to R

It is the unique complete ordered field after all

[velvet hull]

so I'm guessing if you want something similar to work over another field at some point you need to invoke a non-algebraic property of that field

which is just not common for fields that are not Q, R or C

[tulip otter]

i see

[rocky cloak]

Over finite fields I guess you can compute gcd with x^p^n - x. I'm guessing that's not very efficient for large n though, given how big p^n gets

[tulip otter]

yea i see

[velvet hull]

it's the frobenius polynomial

[tulip otter]

because any element of the finite field is a root of this polynomial

[velvet hull]

really useful for working with finite fields mainly due to 2 facts I can think of right now

if |F|=p^n, then |F^x| is cyclic of order (p^n)-1
also Gal(F / Fp) is a cyclic group of order n, generated by the frobenius map x -> x^p

[low cargo]

I'm not fully acquainted with proof, relations and functions yet

[rocky cloak]

Well, it helps detect irreducible polynomial of degree dividing n, but you might need to be a little clever in the relationship between n and the degree of the polynomial. So I guess not a perfect method anyway

[rocky cloak]

Only one way to get acquainted, might as well be through absalg

[tulip otter]

ohhh ok

[tulip otter]

hmm so what do i do for part a, i can cheat by finding the complex roots and showing that x^4+1 cant be factored in Q[x] for example, but there should be another way

[velvet hull]

yeah no thats a valid proof idea

[tulip otter]

ah so there doesnt seem to be a better way 🥀

[velvet hull]

oh no, there are other ways, but yours is perfectly fine

[tulip otter]

eisenstein might work if i apply some kind of transform to the polynomials but imagine expanding deg 4 or deg 6 polynomial using binomial just for something that might work

[tulip otter]

hmm if you have some way in your mind rn can you tell me about it or give me some hint towards it

[velvet hull]

shift + eisenstein is a good trick and it is not hard to do here using the binomial theorem

[tulip otter]

I saw what you deleted

[velvet hull]

the exact statement for non prime n is annoying to figure out but there are divisibility tricks relating the prime factors of n and the prime factors of (n k) (for 0 < k < n) that you can use to avoid expanding out the polynomial if you dont wanna do it

[rocky cloak]

Another approach can be to use Gauss lemma. So if it's reducible it must factor as
(x^2 + ax + b)(x^2 + cx + d) for integers a,b,c,d

Not so many options for what those integers could be

[tulip otter]

so b=d=\pm 1, a=-c and ac+b+d=0 so that a^2=2b which means that b=d=1 and a=-c=\pm\sqrt{2} but \pm\sqrt{2} is not in Q. Hence x^4+1 is irreducible in Q

[tulip otter]

yea replacing x by x+1 should do the job for x^4+1. Because then all the coefficients of (x+1)^4+1 are 0 mod 2 except for the leading coefficient, and the constant coeff is 2 which is not 0 mod 4.

[cursive spindle]

Usually shifting by x+1 works for these problems

[tulip otter]

the other one can be done by finding the complex root ig, its not hard to find the roots after setting y=x^3 which transforms it into a quadratic polynomial

tho i dont like this way but well it is what it is

[cursive spindle]

Many such cases. Soon enough you will be able to tell which are irreducible if you stare at LMFDB enough

[tulip otter]

I see, yea lang also showed that the polynomial x^{p-1}+x^{p-2}+...+1 is irreducible over Q for prime p by doing this same shift

hmmm it remains to deal with part c

just went from k[x] to k[x,y] and its suddenly much more difficult/subtle to deal with things

at least it appears to be like this to me rn

[tulip otter]

btw its rare to see you here

[elfin wraith]

Theres a book (I think just called "Abstract Algebra") by Artin which develops linear and abstract algebra together, that may be of interest. Its a bit slow, but it is a classic

[cursive spindle]

I'm always here I just don't talk much

[tulip otter]

i see

[thorn jay]

always watching

[cursive spindle]

https://tenor.com/view/doakes-im-watching-you-gif-16773151899168519026

[rocky cloak]

You have the same irreducible in k[x, y] iff irreducible in k(x)[y] as the relationship with Z[x] and Q[x]

[tulip otter]

hmmm so for example i can use eisenstein criterion by saying that the coefficient of y^2 is 1 which is not 0 mod x+1 and x^2-1\equiv 0 mod x+1 but not equiv 0 mod (x+1)^2?

because x+1 is a prime of k[x]

[rocky cloak]

Sure, though I guess it would just be easier to use that a degree 2 polynomial is irreducible iff it doesn't have a root

[tulip otter]

ohh i see right

so thats one place where k[x,y]=k[x][y] becomes useful

tysm jagr and HChan, have a great day/night

[boreal flume]

HKR does abstract algebra! (DnF)

[low cargo]

This looks good but with books like these, I fear that I will know less than doing 2 books. Is it reasonable to worry about this?

[cerulean token]

could use eisenstein ? i dont think it's gonna work and this method is better , but good to have that in mind

[tulip otter]

yea i did that too, eisenstein works for x^4+1 by shifting and considering (x+1)^4+1 instead of x^4+1, then the new polynomial satisfies eisenstein's criterion

it also works for the other polynomial if you do the same shift, ie if you look at the polynomial (x+1)^6+(x+1)^3+1 instead of the original one

as HChan noted before, you dont really have to explicitly expand using the binomial theorem, the only thing that matters is to check that the coefficients satsify the desired conditions, you can check that since you know that the coefficients of (x+a)^n are of the form \binom{n,k}

[cerulean token]

Nice ,what’s the argument on the expansion?

You just rewrite as (y)^4 +1 with y =x+1

I guess it’s because it’s bijective with f R to R with x going to x+1

[tulip otter]

yea, it works because this translation is bijective

[tulip otter]

the automorphism is is given by f(x)\mapsto f(ax+b) right because it is an automorphism that extends x\mapsto ax+b and it induces the identity on A. I forgot to check uniqueness tho.The inverse automorphism is given by f(x)\mapsto f(a^{-1}(x-b)).

[crystal vale]

i have to show that for any prime p and for any non zero a in F_p, x^p - x + a is irreducible and separable over F_p. So for separable we can see if y is a root then y+1 is also a root, it implies {y+j | j = 0,1,...,p-1} are roots of that equation and they are p roots and polynomial of deg n over field can have at most n roots hence only these are roots and they are all distinct, therefore it is separable.

i don't know how do i show irreducibility

[velvet hull]

Ah, I recognise this question

It is a very neat trick

[crystal vale]

any hint?

[velvet hull]

It is a rather unconventional method of proving irreducibility

So the polynomial is separable, so let’s look at a splitting field of it over Fp

Try and see if there is anything interesting you can say about the roots

[velvet hull]

Oh wait you’ve shown that already lol

Okay, so what does this say about the splitting field?

[crystal vale]

I think it looks like F_p(y), where y is a root of x^p -x + a

[velvet hull]

Think about y a little more

[rocky cloak]

An automorphism always permutes the roots of the minimal polynomial of something

[crystal vale]

Sorry i don't get it what do you mean?

[rocky cloak]

If f is an automorphism, then y and f(y) share the same minimal polynomial

[crystal vale]

Yes it needs to be an identity on F_p, right?

[rocky cloak]

Sure

As all automorphisms are

[crystal vale]

So you are saying if f : F_p(y) -> F_p(y) and if f is identity on F_p then y maps to one of the roots of its minimal polynomial

[rocky cloak]

Yes, and what is a famous automorphism of finite fields

[crystal vale]

x -> x^p

[rocky cloak]

And what is y^p ?

[crystal vale]

y-a

[rocky cloak]

So how many roots will the minimal polynomial have at least?

[crystal vale]

2

ah, i don't see

[rocky cloak]

So it will have y and y-a, any more roots you get from this argument?

[crystal vale]

I don't know how I get more from this argument?

[rocky cloak]

Well y-a is also an element of the field

[crystal vale]

Yes

Oh I see

[rocky cloak]

So what are some roots of its minimal polynomial?

[crystal vale]

so now y-a maps to y-2a

[crystal vale]

I got it, so minimal polynomial has atleast p roots so it will be exactly x^p - x + a, because that polynomial has degree n and minimal polynomial divide x^p - x + a and monic, so it will be equal to that

[crystal vale]

so in Lang, separable degree of E over F is number of extension of embedding F to C to E to C, where C is alg closure of F, right?

[tardy hedge]

I had that problem on homework last year too

[hidden kite]

Is it worth learning SageMath in tandem with learning intro algebra? Of course, I don't expect there to be necessarily a universal answer.

[elfin wraith]

Sage is cool, and very easy to pick up assuming you know python (which is also easy to pick up). I don’t think it’s necessary at all but knowing some CAS can be useful for computations you can’t/don’t want to do.

My UG algebra course taught us it and we had homework’s where we basically wrote code to get an idea of how something behaves, made a conjecture based on that and tried to prove it (and these were generally really hard things to notice and also prove…) which was nice. A lot of people actually do research like that, I know Eisenbud often just computes a bunch of free resolutions, looks at it for a while and works something out from that

Also I guess depending on the flavour of algebra you like there could be better options, GAP if you like groups, Macaulay2 if you like alggeo, Magma if you’re 100 years old etc

That being said I reckon sage is the best general purpose option if you’re just getting into it, plus it’s reasonably well documented and easy to learn

TL;DR you don’t have to but it can be a nice thing to know

[hidden kite]

I see

[rotund aurora]

magma has built in functions that sage doesn't have

[hidden kite]

This week I tried using Julia's Symbolic.jl to implement a little script to help me prove that a specific bijection I constructed is an isomorphism between groups. But turns out, its more lackluster than I thought: it can't even correctly check the equivalence of something simple.

I thought fuck it let's reimplement it in Sage and spent the whole of yesterday trying to install Sage, which I eventually managed.

Programming has just been a pain in the ass whenever I tried it sadge... well sometimes it can be quite satisfying, but eh. The previous time I got hooked on a programming project, which was also my first time, I spent 6 months writing a LaTeX template... its still not done. Other times I had to fumble my way to success, like writing some Julia scripts; one was a sanity check. Well, kinda a skill issue ig lol

[hidden kite]

Would it be possible for you to share those homeworks?

[elfin wraith]

I can’t unfortunately, they were Jupyter notebooks hosted on the unis server and I’m not there anymore and didn’t download them :/

[hidden kite]

Ah rip, that's unfortunate.

[finite turtle]

can anyone help with b

showing that root -n is not prime

tired now will check solution tommorrow ty

[tardy hedge]

any formula for (I+J : K)?

I see one for (I : J+K)

colon ideals

[thorn jay]

is this like x where xK in I + J?

[tardy hedge]

Yeah

[thorn jay]

probably not

joins dont tend to play well when considering elements smaller than it

[tough raven]

In general (comm) ring, I think ideal arithmetic is so finicky that (heuristically) nothing is true unless it has to be true by "pure ste/order theory". So apart from (I : K) + (J : K) \subseteq (I + J : K), nothing can be said.

[thorn jay]

well ideal lattice is modular and if you look hard enough, distributive behavior can be found

[tough raven]

In a Dedekind domain, you can calculate + as gcd and : as "numerator of / in lowest terms" using prime factorisations: for any prime p, v_p(I + J) = min(vp(I), vp(J)) and vp(I : J) = max(vp(I) - vp(J), 0). So you get something (unilluminating, at least to me).

[thorn jay]

and I * J is given by the UA-theoretic commutator so there is something :3

[tough raven]

Sure, but : isn't meet; it's (relative) complementation.

[thorn jay]

yeah true true

[crystal vale]

Which problem?

[tardy hedge]

x^p-x-p or somethint

[crystal vale]

I see

[tardy hedge]

My galois theory class was badly organized. I did learn some field theory but the actual galois theory was rushed and we didnt even cover it mostly

[crystal vale]

Ah

[tardy hedge]

😢

[crystal vale]

I am interested in Galois theory i want to see the connection between the fundamental theorem of Galois theory and the Covering space

Galois theory seems interesting, rn I am trying to learn functional analysis and Galois theory

This semester I have also commutative algebra in my algebra course

[tardy hedge]

Nice

I was liking it until symmetric polynomials or whatever

That was pretty confusing to me

[crystal vale]

Me too

[tardy hedge]

Yeah im not sure what about it was confusing

Its been a while

[crystal vale]

Lang is a good book for Galois theory i think

[tardy hedge]

Cool, i was using dummit and foote

[crystal vale]

I am using both

First I used Patrick Morandi

Then I switched to Lang

I liked the construction of alg closure in Lang, it gives the set theoretical construction not identification of field

[thorn jay]

lwk arithmetic topology

[crystal vale]

What is that?

[fading acorn]

tfw etale fundamental groups

[thorn jay]

im definitely not anywhere near studying that stuff but i believe its the study of connections between number theory and topology (in particular manifolds? iirc)

like there are analogies between prime numbers and knots (legendre symbols and linking numbers for example)

[thorn jay]

yes that too

[thorn jay]

it's unique

the monoid of isotopy classes of knots is a unique factorisation monoid with iirc trivial unit group

most algebraic knot invariants i can imagine play well with connected sums but @tidal schooner probably knows a little more than i do

maybe (k)not

and in general its nice to be able to reduce a problem

I love connections between geometry and algebra so I'm especially fond of like quandles

braid groups seem cool too but they're a huge topic so I'd have to set time away for it

[tidal schooner]

Yeah, for certain algebraic invariants (e.g., Vassiliev invariants, Heegaard Floer homology, fundamental quandles), there are explicit formulas for connected sums of knots, though the proofs of some of these are quite nontrivial

This is not true for non-algebraic invariants; recently it was proven that the unknotting number isn't additive wrt connected sums

[thorn jay]

this is so peak

[cursive spindle]

Vassiliev invariants mentioned

[copper kestrel]

why do some definitions of an ideal have closure under subtraction, some having closure under addition, and others using cosets? i know that theyre all the same but like is there any benefit to which one you use?

[thorn jay]

they all kinda give different perspective in a sense but like honestly it doesn't matter at all

I've never heard cosets before though

[copper kestrel]

its very interesting

[thorn jay]

that's just the usual definition lol

[copper kestrel]

o

[thorn jay]

only slightly obfuscated language

[copper kestrel]

ah

[tough raven]

Honestly IDK. I don't feel like the subtraction one is ever really useful, you can almost always verify addition, negation and zero separately.

[glass sinew]

closure under subtraction js seems like an attempt to reuse the one-step subgroup test for conciseness

like in an additive group yeah a subgroup is closed under subtraction but who says that

[tough raven]

Yes, that's the test whose usefulness I'm questioning.

[copper kestrel]

ahhh

[kind kindle]

why does this group have a huge permutation and linear degree(over C) for it's size?

https://www.lmfdb.org/Groups/Abstract/30720.c

[marble hinge]

sphynx studies Abstract Algebra

[twilit wraith]

ok im going crazy

i have to show that if |A| = |B| with A,B sets then F(A) is isomorphic to F(B) as R-modules

im trying to do it categorically using the universal property

but the equalities of the compositions of functions im getting dont work out the way im wanting

f is a bijection between A and B

but i just keep getting that $i_A \circ f^{-1} \circ f = \Psi \circ \Phi \circ i_A$

[cloud walrus]

hiidostuff

[twilit wraith]

but since i_A isnt always surjective i cant show that the composition of the induced morphisms is the identity map

wait

i could just create a natural surjection s_A from F(A) onto A

then $s_A \circ \Psi \circ \Phi = \text{id}_A \circ s_A$

[cloud walrus]

hiidostuff

[twilit wraith]

but ofc that map on the right is just sA circ idA

[south patrol]

As soon as you know F is a functor you are done

[twilit wraith]

nah but thats still not what im going for

what

F just means free doesnt it

oh i shouldve clarified

im looking at the free modules on A and B

[south patrol]

I know and I am saying this construction gives a functor

Just like cause you said "categorically"

[twilit wraith]

fair enough

i mean im just using the bare minimum category theory that my prof has discussed in class

he didnt mention functors but they seem simple enough

[south patrol]

Ye sure

[south patrol]

So this is great and now you should show the two maps F(A) -> F(B) and F(B) -> F(A) compose to identity

Hint: use the universal property again to prove the composition is the identity

[twilit wraith]

well i know that theres an identity isomorphism from F(A) to F(A)

[south patrol]

Yes

[twilit wraith]

ohhh duh

wow

[south patrol]

This is the unique map induced by the identity map from A -> A

Yeah exactly

It's fun

[twilit wraith]

but by the universal property of F(A) that map from F(A) to F(A) has to be the unique morphism making the diagram commute

[south patrol]

Yee

[twilit wraith]

but we know that Phi circ Psi or whatever makes the diagram commute

so theyre equal

then other way around

[south patrol]

This is how uniqueness via universal properties generally goes which is fun

[twilit wraith]

i see

[kind temple]

there is a "one-line" proof if you can use the Yoneda lemma

the proof i have in mind makes use of the adjoint relationship:
||hom(FA,M) = hom(A,UM) = hom(B,UM) = hom(FB,M) for all modules M, so by the Yoneda lemma, FA and FB are isomorphic||

but uh

i guess this is complicating things. functors preserve isos lmao

[twilit wraith]

@south patrol wait how is it exactly we know that the composition of Psi and Phi still make the diagram commute

[kind temple]

the subdiagrams commute

[twilit wraith]

Yeah I guess id just have to prove that makes the whole thing commute

We've done essentially no category theory

Just basic diagrams to show universal properties

Maybe theres a simpler way to do this

[kind temple]

yea. you can reduce this to two sets of equations:

• \phi_A \circ i_A = i_B \circ f
• \phi_B \circ i_B = i_A \circ f^{-1}

[twilit wraith]

oh i see

[kind temple]

they correspond to the top and bottom squares

so now take path along the upper right side of the big rectangle

\phi_B \circ \phi_A \circ i_A

and apply the first equation

then the second

[twilit wraith]

Alright

Ugh maybe I shouldnt have done this categorically

[kind temple]

what are you unsatisfied with?

[twilit wraith]

My having no idea whats going on

Also I think constructing an explicit map and showing its a module isomorphism could potentially be simpler

Okay i mean I guess I do know whats going on

[kind temple]

i guess, how have you defined the universal property of the free module generated by a set

this doesn't even need to be categorical

maybe even think of it as a characterising property of the free module instead

[twilit wraith]

for any map from A to an R-module M, there exists a unique morphism from F(A) to M making the expected diagram commute

which, letting g be the map from A to M and f be the induced morphism from F(A) to M, would mean that f \circ i_A = g

[kind temple]

yea, exactly

you can think the homomorphism FA -> M as the extension of g along i_A

so my question is, given an element of FA, how do you compute f at that point?

[twilit wraith]

i could send \sum r_i a_i to \sum r_i g(a_i)

i mean i did know thats exactly what the morphism is

[kind temple]

right, so you extend g by linearity

okay, now, apply this in the case that M = FB

given that you have an isomorphism A -> B

[twilit wraith]

ok now i have to rename my functions since i accidentally reused variables lol

uhh with f the bijection, iA and iB the inclusions, phi the map from FA to FB

[kind temple]

cool

[twilit wraith]

i mean the explicit map would just send sum ri ai to sum ri f(ai)

[kind temple]

yea. and you can repeat this process with f^-1 to get a morphism \psi : FB -> FA

what does that look like

[twilit wraith]

oh i see

just like the same thing

but with extending f^-1 by linearity

[kind temple]

yea

so what happens when you compose them?

say, \psi \circ \phi : FA -> FB -> FA

[twilit wraith]

i mean ofc the identity

[kind temple]

yup!

[twilit wraith]

so i guess there we go

[kind temple]

sometimes its good to be more concrete and get your hands dirty to see how everything is working

[twilit wraith]

yeah

[supple oracle]

oh shit i forgot how long ago that convo was

[hidden kite]

That's hilarious

[vapid vale]

if a book is specialized on this topic its probably better to read it after an introductory chapter from a first course in algebra anyways

it depends on your interests. idk the actual contents of this book but looking at the table of contents, you could go into a commutative algebra textbook immediately after chapter 2

but i'd say 1-4 are more or less standard fare for a first course in algebra

are you self-studying?

[hidden kite]

My interests are not too specific now: analysis/functional analysis/differential geometry/etc; comm alg/hom alg/alg geo/etc; mathematical logic/set theory are all things I would like to explore in the future

[hidden kite]

Yes

[vapid vale]

yes

[hidden kite]

As for background, I have read an Axiomatic Set Theory text at the ug level, almost finished FIS up to the chapter of inner products, and have completed intro analysis (except sequences of functions which I still need to get back to).

[vapid vale]

although you shouldn't skip the galois theory bit

[hidden kite]

I see

[vapid vale]

maybe you don't need to be super thorough with it

but chapters 1-3 can get you into a commutative algebra book like atiyah-macdonald and/or eisenbud

[hidden kite]

I always have an strong stubborness to prove every theorem myself and do all of the interesting exercises

I have been doing this for a few years. Progress definitely isn't very fast; but not proving theorems myself or not attempting the interesting exercises just makes everything feel kinda boring, and I kinda feel that I haven't really taken enough ownership of learning

I have tried stopping trying to prove all theorems myself, but everytime I find myself going back.

I see

[vapid vale]

i don't think proving every theorem in the book is totally productive, algebra is about learning about certain structures and the techniques that come with them, the text can be helpful with setting your mind straight with the correct way to think about a theorem

its certainly a good practice, but textbooks often have some pedagogical reason for the theorems being proven in the text so there is some merit to learning from them (obviously this varies theorem to theorem)

[hidden kite]

Right

[vapid vale]

enjoy the math

[dire wren]

I was trying to develop an intuition for groups, so i was considering associative magmas, and was thinking if there was a name for the equivalence classes we get when we consider two objects (a and b (in the magma)) the same, iff ai = bi for all i in the magma

[dire wren]

i can tell wehere this came from, i was consider members of the magma as functions on the magma, and associativty turned into composition of two elements a and b, is in a sense the same as the equivalence class of ab (under the equivalence relation i gave)

[thorn jay]

this has to do with smt called greens relations

[spare cove]

I saw this one the other day xD

[supple ice]

<@&268886789983436800>

[nova holly]

it's called R-classes in green relations

[delicate orchid]

BLEEEEEHHH! I like orbits under left translation

[tough raven]

Just say right cosets like a normal person

[delicate orchid]

Tbf what I said was a misnomer anyway it’s more akin to taking fibres under the obvious magma morphism M -> End(M)

[thorn jay]

Let the n-hypercube be the set {0, 1}^n. Is there a neat characterisation of bijections f : {0, 1}^n -> {0, 1}^n which are the reflection symmetries of the hypercube?

the ones parallel to a face are neatly given by f_k : (x1, ..., xn) |-> (x1, ..., 1-xk, ..., xn)

[tough raven]

Define reflection symmetry?

[thorn jay]

if you realize the n-cube in R^n with the centre at the origin, then a reflection symmetry is a choice of hyperplane such that reflecting along that hyperplane doesn't change the resulting cube

you can also think in terms of the automorphism group of the hypercube graph

but I don't know how you'd characterize a reflection symmetry there

[tough raven]

Ah, so specifically any reflecton in a hyperplane.

[thorn jay]

are there others?

[tough raven]

x |--> -x is sometimes called reflection in a point in 3 (odd?] dimensions

Also I thought you might mean group generated by reflections

[thorn jay]

oh yeah I remember reading abt that in some alg top book lol

or at least that this map has nice properties for a sphere in odd dimensions

[tough raven]

I think this is just reflection normal to the vectors \pm e_i, \pm e_i \pm e_j

(ie the reflections of the Weyl group of type B_n or C_n)

What about (1, 1, 1) in 3 dimensions?

[thorn jay]

in 3 dimensions it's

[tough raven]

Nice

Matches my first guess then

[thorn jay]

yee

3 + (3 * (3 + 1)) / 2 = 3 + 6 = 9

wait no

ugh

combinatoricsss

oh how I despise you

[tough raven]

The number is n + nC2 = n(n+1)/2

so 6

[thorn jay]

oh so it is that

[tough raven]

Yes

[thorn jay]

in 2d though, we do have that there is a reflection wrt (1, 1)

and (1, -1)

[tough raven]

2(2+1)/2 = 3

Although I think I see 4

[tough raven]

Sorry, n + 2nC2 = n^2

:facepalm:

[thorn jay]

yes

[thorn jay]

oh that makes more sense

lol

[tough raven]

Consider reflection normal to n, ie v |--> v - 2(v.n)/(n.n) n. If this preserves the cube it preserves the integer lattice so applying to v = std basis, 2n_i n_j/(n.n) are integers so when n is normalised to length sqrt(2), n_i n_j are integers; but then sum n_i^2 is a sum of nonneg integers to 2, so either n is \pm sqrt(2) e_i or \pm e_i \pm e_j.

Interesting that all reflections preserving the lattice preserve the cube.

[thorn jay]

I suppose because lengths are preserved and origin is fixed

[tough raven]

But the cube is not all lattice points with length sqrt(n)

[thorn jay]

i see

well lucky us

[tough raven]

(2, 0, 0, 0)

[tough raven]

lol

[thorn jay]

I'll have to formally go through this tmr when I'm a little less sleep deprived

thank you very much though, this is going to be a great help

[tough raven]

If it helps tomorrow, the idea is "preserving integer coordinates implies by bashing that you (almost) have integer coordinates with a fixed normalisation => finite list of possibilities".

[thorn jay]

what genre of math is this

it feels like a mix of combinatorics group theory and euclidian geometry

[tough raven]

Depends on where it goes from here I think

I'm guessing it could extend to some combinatorics, some kind of lattice theory, root system theory at least

[thorn jay]

Im messing around with higher dimensional generalizations of equivalence relations (actually, congruences on algebras) called Malcev-complexes. These are sets of n-cubes (maps Γ : {0, 1}^n → A) closed under higher dimensional generalisations of the congruence relation axioms. The sets of these Malcev complexes for some algebra A form a simplicial set and so one can take the homology of this.

since everything is based on hypercubes i thought it would be interesting to see if i could poke it a bit and see if some nice group representations on the homology groups fall out

[tough raven]

You're replacing an (internal) equivalence relation with an (internal) cube complex kind of thing?

Oh, no, a simplicial set.

[thorn jay]

uhhh maybe?

id have to check

basically you have transitivity and symmetry in all k directions

rather than just "linear" symmetry and transitivity

this kinda idea

[tough raven]

I cannot handle the notation

but I get the idea

This doesn't "look" simplicial, though, it looks cubical.

I will read a bit about Malcev complexes.

[thorn jay]

yes but the sets of such malcev complexes form a simplicial set

its a bit weird and the homology acts as if 1-dimensional complexes (so congruences) act as points lol

[tough raven]

Nvm I cannot find a definition of this

[tough raven]

This isn't saying not to try, but vibewise I'm sus. The n-simplex has an action of S_n but arbitrary homologies made from simplicial complexes/sets do not have actions of any symmetric group.

(In fact, one usually has to order all the simplices and break the symmetry to define boundary maps. The same probably happens in the definition of homology of s "cubical complex".)

[thorn jay]

yeah thats the issue i ran into when i was trying to do it naively by permuting the inputs of the squares

however! i have found there is at least a C2 action

idk, im gonna try

[thorn jay]

there's like three papers or smt about them

[sullen salmon]

Algebra is cool.

kool

[elfin wraith]

Speaking of, what kinda maths do you do lol? You seem to know a lot about a lot of things

[ivory anvil]

Is it impossible for the additive identity in a ring to have a multiplicative inverse?

[plush elbow]

Yes, by definition of inverse

In fact, if you allow an inverse to be zero, the only ring that satisfies this is the zero ring (0 = 1)

[elfin wraith]

First prove that 0*a = 0 for all a in R (this is striaght forward and a good exercise to get comfortable with the axioms). Then suppose that there was some element 0^{-1} so that 0*0^{-1} = 1. Then by the proof before,, you must have that 1 = 0 so this only holds in the trivial ring

[heady pasture]

Okay this last step is driving me mildly insane.

In this context, acting means acting via conjugation,
acting trivially on a group thus translates to centralizing it,
x^g is conjugation: x^g = gxg^-1 and Q^# is Q without the identity.

The reasons left out for the first few conclusions I can fill in:

• For any g in Q^# it follows that [g,A] is a subgroup of Q as Q is normal.
• g is not in [g,A] as [g,A] is a subgroup of <A,A^g> and we assume g is not in <A,A^g>.
• Thus [g,A] is a proper subgroup of Q and A must centralize it or it would be a smaller counterexample.
• To say that A centralizes [g,A] is equivalent to saying 1 = [[g,A],A].
• Since this works for any g, 1 = [[Q,A],A] =: [Q,A,A] as [[Q,A],A] is simply generated of all the [[g,A],A].

How do you get that [Q,A,A] = [Q,A]? Google states that this has something to do with coprime action and uses other lemmas that I also don't know much about to justify it. I can't really find anything useful about those either. Is there some kind of simple divisibility trick I am not seeing to justify this?

[heady pasture]

Groupprops seems to have the statement but with no proof or references 😭

https://groupprops.subwiki.org/wiki/Commutator_of_finite_group_with_coprime_automorphism_group_equals_second_commutator

[tough raven]

I'm learning geometric representation theory these days

[tough raven]

I remember something similar in a theorem proof on GroupProps or ProofWiki or something similar, where they characterised the possible orders of an automorphism of a p'-group of order coprime to p'. I can't recall more but maybe it will be helpful at some point.

[obtuse reef]

Oh sounds interesting

[elfin wraith]

That would explain the seemingly wide interests!

[heady pasture]

Okay after searching through several layers of references, I found a proof of this in I. M. Isaacs' Finite Group Theory. The author notes that the proof ultimately relies on another lemma by Glauberman whose proof fills multiple pages Did not expect it to be this bad...

[tough raven]

Hey I can have wide interests of my own accord ;p

But yes lately many of my questions have come from there in some way

[slim kayak]

Have you heard of the springer correspondence?

[thorn jay]

well itll all come together in the end in some way

lately ive taken an interest in algebraic number theory (blame Deltoid for that) and im sure there is some way to connect it to universal algebra
i just gotta find it

[tough raven]

That's the topic of my oral presentation so I'd hope so

[slim kayak]

Oh cool! I considered doing a seminar talk on it next semester, but I'd have to get into the necessary rep theory. Can you recommend something to read into for someone with more of an AG background?

[tough raven]

I'm not sure, but I will think about it/ask around. I learnt it largely from Chriss--Ginzburg, which is approximately the opposite.

[glossy crag]

Say I've proved the symmetric polynomial theorem for Z (every symmetric polynomial uniquely a polynomial in the elementary ones), how does this then generalise to arbitrary commutative rings (w/o using the general lexicographic proof)? The first part is easy, just write f as sum over permutation orbits and apply the case of Z, but how would I prove uniqueness?

[tough raven]

OK I just found a wacky (but probably most abstract and general) solution partial solution that I'm pretty sure can be completed.

So you have an algebra map Z[y1, ..., yn] -> Z[x1, ..., xn] mapping yi to ei(x1, ..., xn). You proved that this map is injective with image the Sn-invariant subring of Z[x1, ..., xn] and want the same to hold after extension of scalars to R.
But this is equivalent to its being the equaliser of two maps from Z^Sn (x) Z[x1, ..., xn] = Z[x1, ..., xn]^Sn (this is a pro&uct of Sn copies of Z[x1, ..., xn]), the first maps p(x) to 1 (x) p(x) = (p(x), ..., p(x)) and the second maps p(x) to the tuple (p(sigma(x)) : sigma in Sn).
Or equivalently, the kernel of their difference.
(And the analogous statements over any R are also equivalent.)

So the question is reduced to: how do we show that some exact sequence 0 -> A -> B -> C remains exact when we change scalars? I think this follows from B, C being free Z-modules, so that the image in C is free so the ses 0 -> A -> B -> im -> 0 splits, but I got stuck after this.

[rocky cloak]

I guess you would need
im -> C to also be split to really seal the deal

[tough raven]

Yeah that's why I got stuck. In fact I guess for the base change to work im -> C has to remain injective.

[rocky cloak]

Which should be equivalent to showing that C/im is flat. Maybe that's an approach...

Like say you have an element
(p1, p2, ..., pn) such that there is a non-zero integer z with
z(p1, ..., pn) = (q - q(sigma) : sigma)

Then you want to be able to choose q divisible by z... Doesn't seem totally obvious

[brittle wharf]

im about to start reading artin's abstract algebra book. the college ill be attending this fall has some higher-level algebra courses that come after a standard algebra sequence. after taking two courses of abstract algebra, the following opens up (some undergrad courses, some grad that id be able to take after the necessary prereqs):

commutative algebra -> algebraic geometry 1/2, algebraic groups
noncommutative algebra
intro to representation theory

there's also some courses on lie groups and lie algebras, then a course on methods of representation theory..?

my question is that when i finish reading artin, would i be able to know which courses would appeal to me? the descriptions for those courses fly way above my head and i don't really know what their contents are

[long obsidian]

Ya probably

[copper kestrel]

"Suppose R is a ring such that x^3 = x for all x \in R. Prove that 6x = 0 for all x \in R."

i'm really not sure how one would go from x^3 = x to somehow get 6x = 0? my best guess is to just. multiply x^3 = x by 6, but that doesnt help with anything at all

bc we're only given its a ring and x^3 = x, so then i guess we can try to calculate 6x, which is 6x = 6x^3, but then wow what a shock i cant do anything again

[quiet pelican]

Calculate (x+2)^3 and (x+1)^3

[copper kestrel]

x^3 + 6x^2 + 12x + 8 and x^3 + 3x^2 + 3x + 1

[quiet pelican]

Now calculate them “another way” (using the equality you’ve got)

And note that ||8 = 2^3 = 2||

[copper kestrel]

wdym calculate them another way? do you mean like just simplify with the property?

[quiet pelican]

Yes

[copper kestrel]

i got 6x^2 + 13x + 2 and 3x^2 + 4x + 1

[quiet pelican]

I disagree

Wait no I don't

But you haven't done what I expected

Simplify (x+1)^3 = x+1

[copper kestrel]

oh

idk why i didn't recognize that

i still dont see how that helps us finding 6x = 0

[quiet pelican]

Ok so you get like
6x^2 + 12x = 0
3x^2 + 3x = 0
And you want something that looks line nx = 0
How do you get that?

[copper kestrel]

solve the system of equations to cancel 6x^2

[quiet pelican]

Yes

[copper kestrel]

right that makes sense, but does this just give us x + 2 = 0 and x + 1 = 0?

or something?

i'm trying to get to the point where you got 6^2 + 12x = 0

and 3x^2 + 3x = 0

[quiet pelican]

(x + 2)^3 = x+2
x^3 + 6x^2 + 12x + 8 = x + 2
x^3 = x, 8 = 2^3 = 2
So 6x^2 + 12x + x + 2 = x + 2
And cancel

Then do the same with x+1

[copper kestrel]

oh

that makes sense

idk how you thought of that

[tulip otter]

is this actually meant to be τ(f(α))?

it doesnt make much sense to me

i think it should be f^σ(τα) or something like that

my reasoning is that the coefficients being a_k^σ means that σ is applied to f here, so we are talking about f^σ and not just f. Then we are evaluating at τα

[rocky cloak]

tau(f(x)) = f^sigma(tau(x))

[tulip otter]

ohhh is that by definition or something like that?

[rocky cloak]

I mean, not really by definition. It's just that
tau(ab) = tau(a)tau(b)

So for example
tau(ax + 1) = tau(a) tau(x) + 1
etc

[tulip otter]

and tau is sigma when restricted to F

i see

tysm

[limpid hemlock]

hey, this is an exercise of mine. any tips for what characterization of projective modules might be best to use? and would there be any vital corollaries or propositions to use? (dummit and foote, 3rd ed)

[quiet pelican]

It should be easy from any definition

[limpid hemlock]

it isn't easy to me yet 😢

[quiet pelican]

Just unwind all the universal properties

[limpid hemlock]

Ok yeah it becomes immediate using the commutative diagram one, im just slow at setting things up, thanks

[rocky cloak]

I guess from the Hom(P, -) exact or the P(+)Q free definition it's just immediate.

So one of those would be best

[limpid hemlock]

Wow yeah the hom(P, -) one is even quicker

[copper kestrel]

i know im supposed to verify that psi is a homomorphism, but i'm not quite sure how i would apply psi to a polynomial?

i dont believe i'm allowed to do psi(a_0 + a_1x + a_2x^2 + ...) = psi(a_0) + psi(a_1x) + ...

but if i cant do that i'm not sure how i would do it without that

[rocky cloak]

So first you would need to define psi.

You know that a ring homomorphism needs to satisfy
psi(a0 + a1x + ...) = psi(a0) + psi(a1)psi(x) + ...
so maybe just define psi that way

[copper kestrel]

oh i didnt realize i had to define psi

i'll try that, ty!

[rocky cloak]

I mean, it would be hard to ask about properties of something that doesn't exist I guess

[copper kestrel]

yeah, i just never know when i'm supposed to define a map or not

bc i thought the map was psi(a) = phi(a) and psi(x) = x

[slim kayak]

Can't you just argue that since 2^3=2 gives you 6=0, that you have 6x=0. In particular this forces your ring to have characteristic 2, 3 or 6.

[copper kestrel]

i guess that should've made sense to me with "show that there exists a homomorphism"

[copper kestrel]

the 6=0 makes me very hesitant

[slim kayak]

A ring is still a group under addition, so in particular 2^3=8 gives you the equality 8-2=2-2. But I get what you mean, it's awkward to phrase with only the elementary language

[copper kestrel]

like i guess it makes sense

right

[slim kayak]

The other argument still requires you to how to make sense of general integers as elements in your ring (defined by consecutive addition of the 1 element)

[copper kestrel]

these problems are really hard but of course they will be

[copper kestrel]

wdym?

[slim kayak]

like when you evaluated (x+2)^3 = x^3 + 6 x^2 + 12 x + 8 presuppposes that, say, 6 x^2 is even inside your ring, or 8.

The binomial theorem would actually need some more care, since it only holds in commutative rings. Or in particular when (a+b)^n has a and b commuting with each other (as is the case with integers)

[copper kestrel]

ah true!'

[copper kestrel]

that makes sense!

how do you recognize stuff like that when doing a proof?

because i just tend to think i can do something, but then boom i made an assumption i didnt realize was an assumption

[slim kayak]

By banishing all non-commutative rings unless they are sufficiently nice in some other way

[copper kestrel]

LOL

[slim kayak]

Or well, I am not sure if you've even seen the binomial theorem as a result for rings before

[copper kestrel]

i have yeah

[slim kayak]

Well, what does the statement say?

Part of the pedagogy of naming it a theorem is to seperate it into its own little "tool" box. When you pull it out to apply it to something, remember the requirements.

[copper kestrel]

well i havent seen it used a ton, but we used it that binomial is only valid in a commutative ring

[slim kayak]

One way to formalize it a bit more is just to construct a canonical ring homomorphism Z \to R, for R any unital ring. n is then a shorthand for the image of n. If need be, you can call it say phi and then involve it in your notation

phi(8) = phi(2^3) = phi(2)^3 = phi(2) and so phi(8)-phi(2)=phi(6)=0.

[copper kestrel]

what does canonical mean?

[slim kayak]

In this case it means unique

[copper kestrel]

ah

[slim kayak]

Otherwise its kind of a vibe word, use it when there is a natural choice/obvious map.

[copper kestrel]

ah

back to the other problem, i'm allowed to do this right?

like assume the map has those properties?

i would imagine yes but i'm second guessing myself

and i know in a polynomial ring we have component wise addition and multiplication so this problem seems pretty simple

[slim kayak]

what properties?

[copper kestrel]

that psi(a) = phi(a) and psi(x) = x

like i dont think i have to prove that the map has those properties

[slim kayak]

First define a map, see that it is a ring homomorphism and then check if it matches the requirements

[copper kestrel]

wait what

okay i'll do that then

[rocky cloak]

A good strategy to prove that something exists is to define a thing and show that it is the thing you're claiming exists

[marble hinge]

If I started a thread on my studies (of group theory, hopefully followed by other algebraic topics) and posted a question there, should I just wait until someone discovers it there or should I cross-post or link it here?

[copper kestrel]

i dont think im doing this right

bc we dont know if the R's are commutative

so i cant really do anything from here

maybe i defined the map wrong?

[rocky cloak]

I don't see R being commutative having anything to do with anything, but you should probably finish defining psi.

Like
psi(a0 + a1x) = psi(a0) + psi(a1x) doesn't actually tell you anything unless you know what psi(a0) and psi(a1x) is

[copper kestrel]

yeah that makes sense

i thought i finished defining psi

[rocky cloak]

Doesn't a link to the thread get posted here when you create it?

Test thread

[copper kestrel]

i think it works if i put the x's on the outside of psi, so like psi(a_ix^i) = psi(a_i)x^i, which gives us psi(x) = x

and then if i say psi(a_i) = phi(a_i)

[rocky cloak]

Yeah, then you get a complete definition in the end at least

[marble hinge]

Yeah, I created it a week ago or so

It’s supposed to be a long running thread :) But that’s fine, Kerr is already there, trying to help me

[rocky cloak]

I see. I don't know the etiquette I guess

[marble hinge]

It’s here: https://discord.com/channels/268882317391429632/1464743615208227010

[copper kestrel]

shoot i actually dont know if i'm able to do this last step (sorry for constantly talking over yall)

[marble hinge]

I guess I won’t break too many etiquette rules by posting it here

[rocky cloak]

Why not? Or what do you mean?

[copper kestrel]

like i dont know if i'm allowed to do the very last step of gathering up similar terms

doesnt that count as commutativity?

[rocky cloak]

Well addition is commutative

[copper kestrel]

OH WAIT RIGHT

OMG

i definitely know what im doing

bc addition is abelian in every ring

wait no i believe i have to define my map in terms of multiplication too

[slim kayak]

Can I sell u on sums? (\Sigma)

[copper kestrel]

shoot i probably should've done that

i'll change it

i think i need to take a break from this problem and work on another bc im dying bc of this problem

[slim kayak]

Maybe make some bullet points for future you

[copper kestrel]

I DID IT

i will do uniqueness not right now LOL

[next obsidian]

Uniqueness is virtually already done

[copper kestrel]

huh

i know its supposed to go suppose there is a homomorphism phi' s.t. psi'(a)=phi'(a) and psi'(x)=x

[next obsidian]

Wha does a homomorpjism do to a polynomial?

It just gets to be applied to everything individually

[copper kestrel]

right

[next obsidian]

So what happens to ax^n?

Mapped to the same thing

What happens to any sum of them? Mapped to the same thing

GGez

[copper kestrel]

oh right we can say since each psi'(a)=phi(a)= psi(a) and psi'(x)=x=psi(x), each polynomial gets mapped to the same as if it was from psi, making psi'=psi

[next obsidian]

Yeh

You can say something fancy if you want

Maps are determined by what they do to generators, and as a ring R[x] is generated by x and R, and you’ve said the maps agree on these

If that’s meaningful or not is up to debate

[copper kestrel]

very fancy

are we talking about the generator of psi and psi'?

no that doesnt make sense nv

nvm*

[next obsidian]

As a ring

You know what it means for a group to be generated by stuff right?

Or a vector space

Etc etc

[copper kestrel]

yeyeye

we havent gone over generators of rings yet

[next obsidian]

You probably won’t

[copper kestrel]

ah

[next obsidian]

But you can just see by observation that it’s true

Kekw

[copper kestrel]

truthnuke

[next obsidian]

https://tenor.com/view/truth-nuke-truth-true-soyjak-wojak-gif-14828627546768133163

[copper kestrel]

I LOVE THIS LMFAOOOO

thank you all for reminding me how to prove things exist

my memory when it pertains to proofs and logic is similar to that of a black hole, whatever goes in is probably not coming back out bc i forgot it already

[tardy hedge]

I love morgan freeman true

[novel star]

what is the intuition behind homotopy operators

like i understand what a homotopy is in the topological sense

but not in a category of chain complexes

[tardy hedge]

Homotopy preserves something of the objects weaker than isomorphism i guess

[novel star]

but where does that f-g=dK+Kd thing come from

[south patrol]

Wdym by homotopy operators

Do you mean homotopies

[novel star]

i suppose i could take it on faith

[novel star]

i think some people call them chain homotopies

[tardy hedge]

@south patrol i forget what they preserve

[south patrol]

One perspective is you take the "interval" J which is the complex Z -> Z in degrees 0 and 1 (with the identity map) and then a homotopy between two maps C-> D should precisely be a map C (x) J -> D which restricts to your two maps on each edge

[elfin wraith]

It’s because not all quasi isomorphisms/weak equivalences are isomorphisms right

[south patrol]

Homotopy types

But also the maps induced on homology

[tardy hedge]

Oh theyre still isomorphic

[south patrol]

Wdym

[tardy hedge]

The homologies of the two complexes?

[south patrol]

Well ig here you are talking about homotopy equivalences rather than homotopies between maps

[karmic moat]

if you can convince yourself that two morphisms between topological spaces being homotopic is nice, then a homotopy operator (i prefer chain homotopy) feels like a (somewhat) natural generalization to chain complexes

[south patrol]

But yes, homotopic maps induce the same thing on homology and hence chain homotopy equivalences induce isomorphisms on homology

[karmic moat]

hatcher has some stuff about this in his algebraic topology book

[south patrol]

Moreover if you take the e.g. simplicial chain complex for an interval with a simple triangulation then you get this situation after applying chains to a simplicial homotopy

[karmic moat]

you can look/work through those examples and you'll see where the f-g = dk+kd thing comes out

i suspect one can also find concrete examples in the context of smooth manifolds

but im not very good at those so dont take my word for it

[south patrol]

Yes

https://en.wikipedia.org/wiki/Cartan_formula?wprov=sfla1

[karmic moat]

W

everything comes back to cartan

[south patrol]

Both of them

Though both are goated

[karmic moat]

all roads lead to the cartan family

[thorn jay]

well homotopy equivalence is still strictly stronger than quasi isomorphism

[south patrol]

H. Cartan did so much goated AT tbh like I was reading the Cartan seminar where he computes the (co)homology of Eilenberg–Maclane spaces

[thorn jay]

but a homotopy is a more general concept (iirc its equivalently a quasi isomorphism of mapping cones??)

[south patrol]

Idk what you mean by more general

[karmic moat]

didn't h cartan like invent projective modules or something

[thorn jay]

a homotopy is a relation between chain maps

[south patrol]

Well okay sure I see this just don't see how it relates to the parenthetical remark lol

[thorn jay]

it doesnt im just tired and my thought stream is all over the place

lol

[south patrol]

It's ok

[karmic moat]

wtf look at h. cartan's students

holy hell dream team

[south patrol]

Yes

French maths is the best

[karmic moat]

yea

what a lineage omg

[south patrol]

Goats

Sharma got done dirty there

[karmic moat]

tbh idk who that is

[karmic moat]

omg imagine being alive in the 50s and 60s doing algebra with all these people

i pity them though. they didnt have tiktok

kind of essential

[elfin wraith]

Yea ofc I was saying I’m I think the concept of chain homotopities is motivated from wanting a notion of weaker equivalence here

But also it’s late and I’m brain rotted at this point I could be waffling

[karmic moat]

Holy shit

That would be awesome

[thorn jay]

weak equivalence between maps yes :3

[karmic moat]

Wtf can i have your autograph please

[tulip otter]

here you go

[karmic moat]

Thank you 🙏🙏🙏

[tough raven]

In my head the advantage of chain homotopies is that they're guaranteed to be preserved by any additive functor, whereas the failure to preserve quasi-isomorphism (or equivelently, the special case of quasi-isomorphism with zero = exactness of a complex) requires the invention of the whole theory of derived functors/categories just to fix.

[thorn jay]

holy shizzle that's why they're called derived categories

[elfin wraith]

Derived categories seem cool, I might end up learning about them as part of my diss but I’m not sure. The ideal of localising a category sounds equally sick and terrifying

[thorn jay]

calculus of fractions and all that

lol

apparently sheafification is given by a localisation too

I saw somewhere vaguely on an nlab page

[crystal vale]

so today we done fundamental theorem of Galois theory

[knotty badger]

ooh what is it?

[slim kayak]

Uh, why?

[rocky cloak]

Named after derived functors

(not joking)

[slim kayak]

Oh, yeah sure

That's where they live as reasonable functors which you can compose.

[crystal vale]

this is from Riehl's category book

[copper kestrel]

ooh orbit category

[tall igloo]

its funny how i took diffgeo last term and my brain already reset "Cartan formula" to mean the steenrod squares thing

cooked

[south patrol]

Good

That's what cartan formula means for me and I'd say homotopy formula otherwise

[elfin wraith]

I had to prove cartans formula in my curves and surfaces exam which was kinda painful

I’m pretty sure I got close but the two sides didn’t quite match up, but it was such a mess it was hard to tell, and I still got full marks lol

I also then got a 0/10 for the next question and simply a comment of “no” ( iirc a special case of the Poincaré lemma)

In retrospect it’s wild that this course, officially anyway, comes before linear algebra

[south patrol]

How can anything come before linear algebra

Actually even Cambridge kinda does this ig

[knotty badger]

Wait it does?

Vectors and matrices is first term of first year

[south patrol]

Well I joke but like vectors and matrices doesn't even have the notion of a vector space right

Or at least barely

[elfin wraith]

Edi does (or did, they’ve fixed this stupid fucking system now) like computational LA sem 1, then the actual LA module is in the second semester of third year

[south patrol]

I guess you can argue otherwise that linear algebra starts in high school given the matrices stuff

[elfin wraith]

Which just means you already know linear algebra by the time you get there, because all your other courses have already taught you it

[south patrol]

I guess I mean "abstract" linear algebra

[elfin wraith]

Thankfully as of this year they’ve scrapped that nonsense and you just do it properly right away

They also did group theory in semester 3 and again with slightly more content in semester 7 for uhh some reason

[rustic trench]

(unpopular opinion, maybe?) matrices cringe

[south patrol]

Idk why that is unpopular

[slim kayak]

Nothing managed to instill such pure disdain of matrices quite like my lie algebras course

[elfin wraith]

More abstract = more betterer

[elfin wraith]

I wasn’t ready for how much grinding through matrix computations I’d be doing this term

[rustic trench]

then again, idk linalg

until now I've just LLL'd my way out of all of it

[slim kayak]

Also had a class about it this term?

[elfin wraith]

Yeah haha, though I haven’t done much for it, the contents pretty chill and I don’t like the prof so I’ve kinda been ignoring it

There’s also no homework which never helps for my motivation

[slim kayak]

There was also no homework with mine, modulo repressing the traumatic memories

I might just have to accept that my blue collar linear algebra computational skills peaked in my first semester and it's never gonna be the same

[elfin wraith]

I’ve long since come to accept my metal maths peaked at like age 16, while still being solid until 18, and since then I’ve just been hopeless

[karmic moat]

i can send you the homework questions i had from last semester if you want

[rustic trench]

imagine using matrices when you could gröbner instead

[elfin wraith]

There are problems, I’m just bad about doing them if they’re not for submission lol

I just don’t make the time

[karmic moat]

ah okay

[elfin wraith]

And tbh I’m not stressed nothing we’ve done so far is particularly taxing

[rustic trench]

idk I just find polys more intuitive than some weird shit I haven't really seen before

[rustic trench]

(ik what matrices are n such these days, but it's still not "intuitive enough" for me to not call it bs)

[thorn jay]

there's a sweet spot

[karmic moat]

if the notion of a vector space is intuitive then i would say matrices are equally as intuitive

you write out a linear transformation between two vector spaces in terms of a matrix and idk that's already pretty awesome

[rustic trench]

i experienced that very differently

until I played with matrices over SR

[karmic moat]

SR?

stanley-reisner?

[rustic trench]

(idk if over is the right word, SR -> symbolic ring, just sagemath abuse)

[karmic moat]

oh idk what symbolic ring is

[rustic trench]

all the non trivial math ik is because I abused sage enough

[rustic trench]

idk if it's an actual math thing

basically just a polynomial ring

[rustic trench]

(that's the best description I can come up with)

https://doc.sagemath.org/html/en/reference/calculus/sage/symbolic/ring.html

[karmic moat]

tbh idk any sage

[rustic trench]

sage ist just glue

its like pari, singular, flint and a shit ton of other libs glued together

[karmic moat]

i only recognize singular from that list

anyway whenever i need to compute something

macaulay2 always calling my name

[rustic trench]

that one is wrapped in sage as well

[thorn jay]

not a standard math term afaik but they probably mean "ring with an adjoined indeterminate (=symbol)" which is indeed the polynomial ring

[karmic moat]

yeah makes sense

[rustic trench]

there's like 4 different implementations anyways

sagemath code is "special"

but it's the best FOSS thing we have

🙂

[elfin wraith]

Sage is nice until I need to read the docs then I want to pull my hair out

Most of the time anyway, sometimes they’re well written. Just not often

[rustic trench]

what docs are you reading?

usually I just end up with ?? it which gives me the source code for the function

[elfin wraith]

I don’t have like an example to hand (I’ve not written any sage in quite a while) but I’ve often found a lot of the documentation to be really opaque

[rustic trench]

I mean, a lot of things aren't really documented past doctests

[copper kestrel]

i know for a polynomial to be a unit it needs to cancel every other polynomial's coefficient except for the a_0 and b_0, and a_0 * b_0 = 1 i think?

but im lost yet again

like i know we want to somehow reason from the product of two polynomials that they cancel to get 1

like we have this thing

ad we want it to equal 1 somehow

[noble nexus]

hint: think about the highest power of x

[copper kestrel]

im guessing the units are only the constant polynomials but i have to justify it 💔

[tough raven]

I think it's a "free" division ring on all variables introduced so far.

[twilit wraith]

when are two polynomials equal

[copper kestrel]

i did it i think

[twilit wraith]

nice

[copper kestrel]

yayyyyyy

[copper kestrel]

first abstract algebra 2 midterm is tmr :') terrified

hoping to get a C or above, which will be an improvement from last semesters fiasco of exams and finals

[tardy hedge]

gl

[jaunty magnet]

good luck!!!

[thorn jay]

you can do this!

[copper kestrel]

ty yall! i'm taking the exam in a few hrs

[maiden crater]

I'm sorry what is this

[rapid cave]

this is written more complicated then it is

[maiden crater]

what's teh actual theorm 😭

I'm unsure of how to even read this

[knotty badger]

presumably $o(a)$ is the order of $a$?

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[rapid cave]

yeah

[knotty badger]

for this it may be helpful to use the universal property of o(a)

$o(g) | n \iff g^n = e$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[maiden crater]

😭

I don't even know how to read this

like when is | saying a divides b

[knotty badger]

$\frac{\text{lcm}(\text{ord}(a), \text{ord}(b))}{\text{gcd}(\text{ord}(a), \text{ord}(b)) } | \text{ord}(ab) | \text{lcm}(\text{ord}(a), \text{ord}(b))$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[maiden crater]

okay lemme rephrase

what does a|b|c mean

[knotty badger]

ah so

[tulip otter]

a divides b and b divides c

[knotty badger]

$a | b | c$ just means $a | b \text{ and } b | c$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[maiden crater]

AH

🤦

[knotty badger]

you also might want to prove the following result

$\text{ord}(a^k) = \frac{\text{ord}(a) }{\text{gcd}(\text{ord}(a), k)}$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[maiden crater]

I have

[knotty badger]

ok then this should help

[maiden crater]

yea, I'm done :)

[knotty badger]

btw how did you prove 1

[knotty badger]

hm hang on

i can understand the first implication

how do you get the second

oh i think i see

it's cause n / gcd(n, k) and k/gcd(n, k) are coprime right

that's very nice

[maiden crater]

Yea

[knotty badger]

my way was more long-winded

[maiden crater]

This is how the book proved it

I. Had a longer idea in mind too

[knotty badger]

so, hm

$\text{ord}(a^k) | m \iff (a^k)^m = e \iff a^{km} = e \iff n | km$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

so you need to show that $n | km \iff \frac{n}{\text{gcd}(n, k) } | m$, and then you can apply baby yoneda

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

and i guess that's a general number theory fact

that's quite interesting actually

so you can define a virtual object under $\mathbb{Z}$ by $(n, k) \prec m \iff n | km$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

and that's represented by $\frac{n}{\text{gcd}(n, k)}$

fascinating

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

i wonder if a similar thing happens in reverse

[mystic ether]

would you say (x^{n})^m = x^{nm} is the universal property of exponentiation perchanse?

[knotty badger]

Yes this is essentially how exponential objects are defined

$\text{Hom}(A, B^C) \cong \text{Hom}(A \times C, B)$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

I mean this isn’t something i came up with myself

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

[mystic ether]

no I know lol

it's just funny to say it for numbers

or multiplication in a group

i understand the perspective though

[copper kestrel]

me when the exam is 50 minutes long, we started early, and the first person handed in their exam a minute after our sub said the exam was over

welp

[low mural]

how do i go about finding the order of $G$? i think i have $<a>$ and $<b>$ found out but i'm not sure what to do with the fact that $(ab)^2=e$? i distributed the exponent and got $(ab)^2=a^2b^2=a^2e=a^2$

[cloud walrus]

Ushy✨

[quiet pelican]

You can’t necessarily distribute the exponent

Because the group isn’t necessarily abelian

[low mural]

ah ok

[rocky cloak]

I guess the most straight forward thing is just to start listening elements, noting which are equal and counting how many you get

Another approach if you've seen some examples of groups could be to think of an example that has an element of order 2 and 3 and see if this could be it

[rapid cave]

@rocky cloak by any chance do you know of witt vectors over a ring? and any sort of motivation for them?

[rocky cloak]

I think maybe @south patrol does

[formal laurel]

The main idea is that the witt vectors gives a equivalence of categories between rings R which are p-adically complete, p-torsion free rings on which R/(p) is perfect, and perfect rings in positive characteristics

The construction can be given in an explicit manner or you can use deformation theorem which in this case is controlled by the cotangent complex.

In the case of the finite fields, for instance, this just unramified extensions of the p-adic numbers. For general extensions this ring becomes sligthly more complicated

[rapid cave]

p-adic complete ring?

like complete in the topology of (p) when you embed Z into R?

also how hard is it to prove this equivalence?

[formal laurel]

If you have a ring R and an ideal I, you can give R the I-adic topology.

[rapid cave]

right

[formal laurel]

Not that hard

[rapid cave]

what is a perfect ring?

[formal laurel]

There's a good introduction in Serre's book "local fields" and there's a really good article by Kedlaya called New methods in phi, gamma modules

[low mural]

i ended up getting G having an order of 6

[formal laurel]

Is a ring on which frobennius is surjective

[rapid cave]

sounds correct.

[rapid cave]

hmmmm where does this come from?

[low mural]

let's gooo

[formal laurel]

https://link.springer.com/article/10.1186/s40687-015-0031-z

[rocky cloak]

In fact G is isomorphic to S3

[rapid cave]

The construction can be given in an explicit manner
this is what I am working through

[formal laurel]

I mean, this is just the usual terminology for perfect fields of positive characteristic

[rapid cave]

perfect isn't that every extension is seperable?

[formal laurel]

On which any finite extension is separable

Yep

[rapid cave]

how is that the same?

[formal laurel]

So, what does this imply for a field of positive char?

[rocky cloak]

And I guess completely unrelated to perfect ring in the "every module has a projective cover" sense (?)

[formal laurel]

Ah yep, that's completely unrelated

[rocky cloak]

I guess that's a nice field theory exercise

[formal laurel]

Remember that separability in the positive char context means that you don't have irreducible polynomials like x^p-u

[rapid cave]

ok.and then we extend this definition to rings?

[formal laurel]

But yeah, I'll leave it as a homework

[formal laurel]

ye

yes

You can defined it in non-perfect rings but then you won't get an equivalencw

[rapid cave]

if frobenius is surjective then these polynomials have a root

[formal laurel]

And if you define them in zero char you just get a product of the ring you are studying

[rocky cloak]

Hmm, I wonder if there is some condition such that R is perfect if every such-and-such extension is a seperable extension of rings

[formal laurel]

mmm

[rocky cloak]

Maybe integral and reduced would work...

Probably too much to hope for

[rapid cave]

I definetly should read this book sometime

[formal laurel]

I guess that a bijective frobennius already implies reduced

[rocky cloak]

I meant for the extension to be reduced

[formal laurel]

But I think there's some relationship between frobennius and the smothness of some ring, so that you can study singularities by using the frobennius map.

This is at the heart of Kunz theorem. But in general like perfect positive char are, in general, very non-finite so this is not useful definition, I guess

(by non-finite I mean non-noetherian)

[rapid cave]

can you elaborate on this?

[formal laurel]

What part?

[rapid cave]

this sorry

the example

[formal laurel]

Well, an unramified extensions of Q_p is just a finite extension generated enterely by a finite extension of the residue field. So basically given a finite field extension F_q/F_p you have an extension which we can call Q_q/Q_p

I know you have study some algebraic geometry so remember that for a number field and a prime you have three cases, you can have ramification, a residue field extension or you can have that the prime splits

This case is identical but you can't have splitting primes since Qp is henselian and therefore if a polynomial reduces in the residue field it already reduces in Q_p

So, you basically can have only two case, one which ramifies and one which doesn't. Using the fact that basically every unramified extension have to be the splitting field of x^p^n-x then you can show that the ramified part and the unramified part are linealy disjoint

[rapid cave]

wdym by unramified and ramified parts?

[formal laurel]

In one sentence you have that each finite extension of F_p correspond uniquely to one unramified extension of Q_p

[formal laurel]

For a finite extension L/Q_p you have ef=[L:Q_p] with f being the degree of the finite field extension and e the power for which m^e=p with m being the maximal ideal of the ring of intergers of L.

So, you have an intermediary field, let's say K such that L/K is totally ramified, K/Q_p is unramified and therefore you have [K:Q_p]=f. And K is the maximal unramified extension of L

[rapid cave]

ah ok

how I am imagine linearly disjoint extensions confused me

[formal laurel]

Wait, I don't think that is precise.

mmm, that wasn't the right word.