groups-rings-fields

[south patrol]

The best quotients go both ways

[tender wharf]

that is so good

[spice whale]

i mean it's so you can naturally define a surjection to the quotient

[coral shale]

G/N

[south patrol]

M \ G / N

[open sluice]

setminus gang

[tender wharf]

I was avoiding G\N because that quotients it the other way lol

thats why I used setminus

[coral shale]

just use -

[tender wharf]

weird

[coral shale]

sorry for detracting from the discussion, carry on.

[summer path]

I mean usually it's clear in context

But also with set minus there's often a space between the variables and the setminus

But with quotient there often isn't I think

[spiral scarab]

So for example R/Z would be all a+Z where 0<=a<1?

[tender wharf]

yeah (I'm guessing the z is a typo and you meant a)

[spiral scarab]

Ohhhhhhhh

[tender wharf]

I mean there are more common examples like Z/nZ which is just integers mod n really

[spiral scarab]

It makes sense now

its simple why didnt I understand it

omfg

[tender wharf]

lmao you should've seen the number of times i posted crankery here

(accidentally ofc)

[wraith cargo]

@hidden haven btw very quick here's a challenge for you (I wanna hear your take on this problem lol)
I'm trying to show that Hom_Z(Z[1/p], Q/Z) is isomorphic to the p-adics
Somebody said that it can be done by constructing the isomorphism explicitly but I'm trying to see if there's a cleaner way

[spiral scarab]

so youre the abstract algebra version of the RH crank or the Collatz crank?

jk

[tender wharf]

lol by crankery I meant I posted something incorrect

[summer path]

Derpz (complexvariable)

[hidden haven]

Not sure, seems like constructing an isomorphism would be the cleanest thing possible

Maybe you can show that Hom(-, this) = Hom(-, p-adics)

[south patrol]

by p-adics do you mean Q_p or Z_p lol

[hidden haven]

Both have universal properties with respect to mapping in

[wraith cargo]

I was thinking maybe like
Writing Z[1/p] as lim-> 1/p^n Z then taking out the direct limit

[south patrol]

i mean must be the latter ig

[wraith cargo]

Z_p

[hidden haven]

Might work

[wraith cargo]

Concretely I'm doing this problem from Weibel

[hidden haven]

Try this too maybe

[tender wharf]

I should totally do this

[wraith cargo]

Might work
Will try

[coral shale]

R/Z is also visualized as a circle. (R number line, R/Z 'number circle')

[formal ermine]

yes?

[spiral scarab]

i am bamboozled

is it using euler's identity?

e^ix?

[delicate orchid]

no it's just a circle

lemme open paint one sec

[spiral scarab]

yeah but what is the isomorphism

[delicate orchid]

actually no wait you can see it like this using first iso you're right

[spiral scarab]

from R/Z to presumably S^1?

[delicate orchid]

good idea

the kernel of the map x |-> e^{2pi*i*x} is Z, and so by first iso R/Z \cong S^1

but that's not fantastic for intuition

[coral shale]

i drew it and deleted earlier cus its ugly

note im writing .1, .2, .3, .4, .5 (for 0.1, ...) there

[coral shale]

1=0

[spiral scarab]

oh i understand

[delicate orchid]

literally what we're quotienting by

nerd

[spiral scarab]

but why e^{2piix}

[coral shale]

trivially true

[delicate orchid]

so the kernel is Z and not 2piZ

not like it matters cause they're iso but

trying to keep things intuitive

[coral shale]

there's an i in there.

isnt there

oh \* to escape discord formatting

[empty rose]

$e^{2\pi ix}$

[cloud walrus]

bee [it/its]

[coral shale]

the reason is just because that's how circles are in the complex plane

or more specifically - the unit circle in C

So this particular quotient is a bit like taking the number line and winding it around this circle again and again and again, and gluing points together that coincide. Is my visual for it

[spiral scarab]

Ohhh I see

yeah it makes sense because you wind it again and again for each integer right?

[coral shale]

yeah...

the string is infinite both 'forwards' and 'backwards' directions

[spiral scarab]

Oh I get it

[delicate orchid]

oh my golly gosh it's the wholesome pi_1(S^1) chungus

[coral shale]

shh

[spiral scarab]

now that I did spheric and hyperbolic geometry I finally know what is

[delicate orchid]

the glome

trying to think of the funniest way to describe the 3-sphere now

open ball in the 2-norm on R^4?

SO(4, R)

[teal vessel]

just checking, the groups under which I'm checking subgroup-ness are the sets of numbers from which each restriction is drawn, right? (part a is checking for a subgroup of C, part c is checking for a subgroup of Q, etc)?

[delicate orchid]

yeah

[teal vessel]

aight, (non) challenge accepted.

[delicate orchid]

although I suppose it would only be an issue for showing it's a subset

could just show they're all groups lol

[delicate orchid]

ah no, inherits the operation as well

ignore me

[spiral scarab]

actually

why is it left coset and not right coset

for quotient groups

if the subgroup is normal it doesnt matter sure

but what if it isnt?

[empty rose]

well if the subgroup isn't normal you don't get a quotient group

(you can have the set of cosets but there isn't a nice way to give them a group operation)

[delicate orchid]

the fact that they're the same for a normal subgroup is why it's a group

[coral shale]

what is 'it' 🤔

weird if your defn writes that

but theyre the same, as others have said due to normality

[spiral scarab]

well nA isnt necessarily An if the groups operation is not commutative

[tender wharf]

yeah sure if A ain't normal

if A isn't normal G/A and G\A (right cosets) aren't groups in the canoncial way

[spiral scarab]

oh I see

I see

[tender wharf]

but if it is

calling the factor group

G/A or G\A makes no difference

[tender wharf]

in your case here even if the operation of G is not commutative

if A is normal then they're equal

for example consider D_4 and

<R_90> which is the subgroup of rotations in D_4

[spiral scarab]

Sure

That makes sense

[coral shale]

i usually recommend checking defn of quotient sets before introducing quotient groups btw

idk why its usually not done

as far as ive seen

[autumn kite]

how do they teach the construction of the rationals

[open sluice]

the field of fractions

[south patrol]

of Z

[spiral scarab]

i just realised its the same as quotient spaces in ligear algebra

🤦‍♂️

today is not my day

[open sluice]

that must be why it’s called algebra

[spiral scarab]

[formal ermine]

linear algebra is just commutative algebra with extra steps

[chilly radish]

?

[open sluice]

tbh idk what a quotient space is, I don’t think it was covered in my edition of axler

[barren sierra]

quick group module question

[south patrol]

group module

[barren sierra]

Let G be a group

if M is the direct sum of completely reducible G modules

and H normal in G

then M is the direct sum of completely reducible H modules right?

restricting the G action to H doesn't change reducibility?

[delicate orchid]

yeah I think so - irreducible H modules will be "smaller" than irreducible G modules

so it should all still work

[coral spindle]

I’m not sure. An irr G-mod may split up when restricted to H

I’m not sure if one of those H-mod constituents may be nonsimple

[barren sierra]

Well completely reducible

Not irreducible

[coral spindle]

Unless you’re talking only about the case when Mashke holds

[coral spindle]

Doesn’t matter, still happens

If any of the simple constituents split up into something indecomposable then we have issues

[barren sierra]

Hm

[coral spindle]

I can’t think of an example, but I know zilch about modular reps

I have a vague idea but I’ll have to wait until I’m in front of a keyboard

[delicate orchid]

ah yeah, good point about modular reps
the only irreducible representation of a finite p-group over a field of char p is the trivial

so if H is a normal p-group (like Q_8 in SL_2(3) or something) then it could break

[rocky cloak]

So at least if H is not normal, then (over a field of characteristic 2) the standard representation of S3 is irreducible, but applying restriction of scalar to C2 < S3 does not give a completely reducible representation

[glossy crag]

If A is a simple central artinian algebra, is there an elementary way of seeing M_n(A) is artinian (e.g. without appeal to Hopkins-Levitzky or Morita equivalence or whatever else)?

[glossy crag]

Actually it's enough to show M_n(A) has a minimal left ideal (since M_n(A) is simple)

So i need to come up with one...

I was thinking matrices with 0 columns except the first, with entries in a minimal left ideal of A

or the same with rows

[wooden ember]

Maybe I’m being dumb but can’t you use that A^n is artinian and multiply a descending chain of ideals on the left by orthogonal idempotents of this to get stabilizing chains in A^n?

And then combine those to show your original thing stabilizes?

Probably doesn’t work

[glossy crag]

Lol it's actually even easier than I thought

A simple artinian => A\cong M_k(D) with D a division algebra => M_n(A)\cong M_nk(D), a simple artinian ring

[wooden ember]

Noice

[rocky cloak]

You don't really need all that fancy machinery of simple central algebras. M_n(A) is a finitely generated A module, so if A is artinian so is M_n(A). Same for noetherian.

[glossy crag]

True dat, didn't cross my mind.

What is the most accepted definition of a subfield of a K-algebra?

Subalgebra that is a field or subring that is a field?

i.e. should it contain K or not

[static yew]

1. I wish discord would fix keyboard support on mobile

2. I'm up to polynomials over a field in my book. It is strewn with
   "exercises" for which no answers are provided

One of these is "show that F[x] is not a field" (for some unspecified field F)

Is that because, without a polynomial for modular reduction, many polynomials have no multiplicative inverse? (And if you don't allow negative exponents, which this author does not seem to do so, then only constant polynomials would have inverses)

[glossy crag]

Precisely so, no polynomial of positive degree in F[x] has an inverse (because multiplying adds degrees).

[static yew]

Is something going to get to why negative exponents aren't valid?

Would I need an F[x,x'] where xx' = 1 by definition?

Hopefully my syntax makes sense

[glossy crag]

Because polynomials are literally defined not to have negative exponents.

You would need to take F[x][x^-1], i.e. adjoin an inverse of x, then every polynomial has an inverse

[static yew]

Is there a term for an expression where you can have negative exponents? It seems odd to think that nobody has ever done anything useful with one of those

[glossy crag]

Yes, there is, these are called formal Laurent series (or sometimes formal Laurent series with finite principal part) and they are important, just not at the beginner level.

[static yew]

Huh. Thanks for the info

Now I got one last edge case to check 🙂

[glossy crag]

Sorry, I misspoke a little actually

I forgot you were talking about the polynomial ring, not formal power series

For the polynomial ring you have the field of rational functions F(x)

Literally defined as sets of fractions f/g

There every polynomial is invertible

[static yew]

Right but this is a field not a field of fractions

[glossy crag]

If you were to talk about formal power series, i.e. expressions of the form \sum_n a_nx^n (with infinitely many a_n non-zero), THEN the place where everything gets an inverse is the field of Laurent series that I talked about

In particular, it contains F[x] and F(x)

[glossy crag]

A field of fractions is an example of a field

And this IS a field of fractions, namely of F[x]

[static yew]

Ahh yeah I'm not interested in a polynomial with infinite terms

[glossy crag]

Field of fractions just means the smallest field containing your integral domain of choice. In the case of F[x], the FOF is F(x)

[static yew]

Yeah but even if you have a field of fractions FF
The ring created by adjoining x FF[x] would itself not be a field

[glossy crag]

Ofc

E.g. F(x)[y]

This is not a field

[static yew]

Not ready for that stuff yet

[glossy crag]

But it's FOF is F(x,y), i.e. the FOF of F[x,y]

[static yew]

I'm thinking about F[x] where defining x as a root of unity, e.g. xx=x^2=1. Is that even a valid concept?

[static yew]

Wait before you used F[x][y] for F plus two variables, is this F[x,y] different?

[glossy crag]

It is but not entirely the way you imagine it.

[glossy crag]

No, they're isomorphic

You can either construct F[x,y], i.e. formal expressions in two variables with coefficients in F

[static yew]

Mmmmmh this seems like an interesting avenue of exploration

[glossy crag]

or you can construct F[x][y], i.e. formal expressions in varibale y with coefficients in F[x]

it's the same thing

[glossy crag]

If you pick a root of unity z in C, then Q[z] is what you're looking for. Note Q[z] is a subring of C, not some new concept entirely.

Sorry, i gotta get back to studying

[static yew]

So if it's a square root I get polynomials with coefficients in Q with degree at most 1? Cuz x^2=1

[glossy crag]

Yes, Q[-1] is Q, but just to clarify, you don't get "literal" formal polynomials.

[static yew]

Although I might need 4
I^4=1

[glossy crag]

Q[...] means smallest subring of C containing Q and ...

If you take something like a 3rd root of unity w^3=1, then Q[w] will be of the form a+bw

[static yew]

Since my focus is on finite stuff anyway I'll deal with that later

Gl studying

[glossy crag]

It's isomorphic to a quotient of Q[x] by the ideal (x^2+x+1)

[south patrol]

TIL every extension of Q is contained in C

[static yew]

Ahh i see
Next section says that the analog for absolute value is the degree

Absolute value of x^(-32) might make some heads pop

[delicate orchid]

my favourite subring of C is Q[S_3]

[glossy crag]

He's reading Stillwell's Elements of Algebra, it just deals with finite extensions of Q.

I don't think it's necessary to confuse him with more complicated definitions, no?

But feel free to rectify my explanation, if you wish

[static yew]

I wonder if this is how someone with access to the multiverse would feel

Polynomial rings F[x] which have coefficients in F, when F is GF(p^k) k>1, where the traditional representation is a polynomial, also usually in x

Keeping which field I'm in is not easy

Gotta head back to work

[sly rain]

When does one think of the elements as a quotient of Q[x]/(x^2+x+1) instead of Q[w]? Q[w] seems nicer

[delicate orchid]

Stevie-O is constructing these rings via polynomial quotients iirc?

might be wrong

[sly rain]

Its a general question, I didnt really follow the conversation, but im learning this stuff now too

[delicate orchid]

ah I see, well it's easier to implement in software via a polynomial quotient

also it's directly showing the use of the universal property of polynomial rings

[glossy crag]

When one doesn't know Q[w], I'd say. Same with finite fields and stuff like Z_2[x]/(x^2+x+1)

[delicate orchid]

but once you have the isomorphism I'd just switch to Q[w]

[glossy crag]

You first learn about constructions "ex nihilo", i.e. out of nothing

E.g. you take a field K, an irreducible polynomial f and see that K[x]/(f) is a field

[formal ermine]

huh

[glossy crag]

Then some time later, once you know that such fields exist, you simply view the latter as a field L containing K such that L=K(a) for a root a of f

It's just a matter of juggling viewpoints

[delicate orchid]

read

[glossy crag]

https://tenor.com/view/boondocks-read-rap-fans-exorcism-uncle-ruckus-gif-23038600

[formal ermine]

oh kek

[lethal dune]

[alpine island]

this construction is honestly one of the coolest things in algebra for me so far. It's a simple concept and yet it leads to almost any field you can think of

(almost)

[delicate orchid]

it's because polynomial rings are free objects

that's the sauce of the swag

[lethal dune]

freedom is an illustion

[glossy crag]

If D is a division K-algebra, do these definitions agree?

The former are subfields that contain K, so my question is if every subfield necessarily contains K.

[lethal dune]

what's the question?

[glossy crag]

I was wondering about the definition of a subfield of an algebra, whether it's better to have it mean "subalgebra that is a field" or "subring that is a field"

And i was wondering if there are any situations when the definitions agree

e.g. in a division ring finite-dimensional over its centre

or when talking about maximal subfields (in the former or latter sense)

[white oxide]

this might be a little silly, but i remember briefly reading something about galois groups being closely tied with the symmetric group on n letters. so $|\mathrm{Gal}(\mathbb{Q}(\sqrt[3]{2}, i\sqrt{3})/\mathbb{Q})| = 6$, which directly corresponds with the permutation of 3 roots, the order of $S_3$. is this what they were talking about, or something of those lines?

[cloud walrus]

okeyokay (analysis is MID)

[glossy crag]

Yes, that's exactly what it is. If f is a separable polynomial, then the Galois group of its splitting field acts as permuttaions of its roots (not necessarily transitively)

[lethal dune]

without any context if I have to guess, I would guess the same

if your algebra is a field extension, it could mean the prime subfield

[glossy crag]

What about situations when the definitions agree?

[lethal dune]

I think it needs to be clarified what the author means in what context

[glossy crag]

It's not one author, i've seen the definition used by different authors to mean different things and i was wondering when they might agree

[coral spindle]

Does anyone know of a reference for the (possibly false!) statement that chi(1) divides |G| for characters over non-algebraically-closed fields of characteristic zero? I think it might be in Curtis–Reiner's tomes but christ I cannot find it.

N.b. I mean only irreducible characters, ofc I realise this shan't hold for reducible ones.

[rocky cloak]

Well, they should agree iff the field is prime. If the field isn't prime then its prime field would distinguish the two definitions

[coral spindle]

Very nice counterexample

As expected a semidirect product works

I had something like this in mind. Glad I saw your answer and saved myself the effort lmao

[rocky cloak]

I haven't been able to come up with a normal counterexample though. Probably mostly because I don't want to think about groups of order more than 6

[formal ermine]

when is a field prime?

[coral spindle]

Yeah I think such an example might force one to look outside of Z/p

[white oxide]

Why is it that $$\mathrm{Gal}(\mathbb{Q}(\sqrt[3]{2}, i\sqrt{3})/\mathbb{Q}(i\sqrt{3})) \simeq \langle \mathbb{Z}_3, + \rangle$$? The two complex roots are not in $\mathbb{Q}(i\sqrt{3})$, so they don't have to be fixed; therefore shouldn't it be of order 6 and hence can't be isomorphic to $\mathbb{Z}_3$?

[cloud walrus]

okeyokay (analysis is MID)

[rocky cloak]

When it has no subfields, i.e. when it is equal to Q or Z/p

[formal ermine]

Ah

[rocky cloak]

An automorphism is determined by where it maps cbroot(2)

There are 3 choices for where that could be

The thing to observe is that Q(cbroot(2), isqrt(3)) is the splitting field of x^3 - 2.

Or at least that it contains the roots

[white oxide]

oh i thought it was the number of permutations of all of the roots of x^3 - 2

[rocky cloak]

Not every permutation gives you an automorphism necessarily. Here the extension is generated by cbroot(2), so an automorphism is determined by where it is mapped

It's enough to define a map on generators

[white oxide]

got it that makes sense

thanks

[white oxide]

sorry to post this example again, but i'm not quite following how alpha_i being a zero of q(x) of multiplicity 1 implies that F(alpha_1, ... alpha_i) is separable over F(alpha_1, ..., alpha_i - 1). i know that it's separable over the latter if and only if q(x) has all zeros of multiplicity 1, but we only showed that alpha_i is of multiplicity one, and i'm struggling to see which part implies that the rest of the zeros of q(x) are of multiplicity 1

i know it's there somewhere but my poor slow brain

also how is this expression true? for any f(x), g(x) with coefficients from a field F, it is not true in general that (f(x)g(x))^n = f(x)^n times g(x)^n

[south patrol]

Yes it is

This is just how multiplication works

In any commutative ring (ab)^n = a^n b^n

[white oxide]

yeah nvm im so stupid i tried to produce a counterexample and i did the computation wrong

[white oxide]

oh yea right oops

🤦‍♂️

[formal ermine]

real

[rocky cloak]

For an irreducible polynomial all roots have the same multiplicity. You can see this because you have an automorphism mapping and root to any other.

[glossy crag]

Frobenius theorem (division R-algebras) via Skolem-Noether and maximal subfields is neat.

That is all, carry on.

[coral spindle]

I noticed that while I was flipping through Curtis–Reiner! I might give it a read.

[white oxide]

ah ok got it thanks

yo i'm not going insane here right it should be $E = F(\alpha)$

[cloud walrus]

okeyokay (analysis is MID)

[white oxide]

or am i trippin

[south patrol]

Yes

E = F(alpha)

[formal ermine]

hey guys uwu

this is a \circ right??

$\chi \circ \lambda$

[formal ermine]

lambda is C -> T and chi is T -> C so it would only make sense

[untold turret]

can any isomorphism of two sets be extended to a module isomorphism on the free modules generated from these sets?

[formal ermine]

what is an isomorphism of sets lol

[untold turret]

set-isomorphism, a function bijection

[coral spindle]

Yes, dare I say obviously

Any map induces a homomorphism which restricts to that map, and if we have a pair of bijections that are mutually inverse their induced homomorphisms must also be mutually inverse.

You don't even need the universal property for this – it follows from functoriality.

[untold turret]

noted, thanks a lot!

[lime badge]

I just wanted to check something that seemed deceptively simple (I'm sure I'm wrong). This problem I'm doing in Aluffi (problem 5.23 in chapter 7) is as follows: Suppose there are no irreducible polynomials of degree n in k[x], where n > 0 and k is a field. Prove there are no separable extensions of k of degree n.

If there was such an extension, it would be a finite separable extension and therefore simple. Wouldn't this immediately give a contradiction or am I missing something?

[south patrol]

That's the solution I'd give too

[lime badge]

Ok, thanks!

[white oxide]

how does one get the abstract algebra role

[ashen heron]

#get-advanced-access

[formal ermine]

@white oxide based name

[ashen heron]

(Analysis is HIGH btw)

[white oxide]

thx ur the goat

[white oxide]

indeed

[glossy crag]

What are some non-commutative infinite-dimensional division algebras over R and C?

[formal ermine]

what is a flat extension of rings? one being flat as a module over the other?

[rustic crown]

yee an injection R --> S such that S is a flat R-mod

[formal ermine]

yeah ok thx

another question

since you're here now det

what does it mean for a polynomial in C[x, y] to be y-generic?

[rustic crown]

not really sure

[formal ermine]

oof

I can't find anything online either

[rustic crown]

what about something like H[x]?

and that ⊗_R C

[south patrol]

Uhh don't think that is a division algebra

[rustic crown]

oh oops

whut about H(x) then

does this make sense?

what if i just say R(x) ⊗_R H

[south patrol]

H(x) ought to make sense right

[rustic crown]

and C(x) ⊗_R H

are these division algs

yea feels like it

idk how to see it lol

[south patrol]

Yeag I looked it up and they gave same example as you lol

[rustic crown]

i see non-zero elementary tensors are invertible

what about sums

i can think of it as an R(x) algebra with basis {1, i, j, k}

so shouldn't the same f0 + f1i+f2j+f3k behave nicely with f0 - f1i -f2j - f3k

yee this looks legit

[south patrol]

Ye

[rustic crown]

at least it multipies to give me a elementary tensor f⊗1

and this is invertible

okie maybe it wont be as nice over C though

[glossy crag]

Nice

[rustic crown]

because stuff like i⊗1 - 1⊗i

just like how C⊗_R C doesn't stay a field anymore

[glossy crag]

Thanks @rustic crown, good example.

Centraliser theorem via Skolem-Noether is just HNNNNG

SO fucking clean

I feel dirty just copying it to my notes

[glossy crag]

One question though

If K/R is an arbitrary transcendental extension (e.g. R(x) like you said), how do we know a^2+b^2+c^2+d^2\neq0 in K?

I'm not that familiar with the topic, but I believe such fields are called real-closed or something like that (sums of non-zero squares non-zero)

Do we know that every extension of R is real-closed?

[south patrol]

C

[rustic crown]

yea that's exactly the problem here

[glossy crag]

Yeah, so is R(x) real closed?

and which extensions of R are?

[rustic crown]

feels like it lmao

maybe clear denoms and argue a lil

yea i think arguing at the highest degree works

[glossy crag]

Yeah it's fine looks like

[rustic crown]

det no familiar either >.<

[rustic crown]

or maybe just think of it as real valued functions, that gives it too

at each value of x, you must have all are 0s

and polys are determined by their values

(as R is infinite)

[glossy crag]

Looks like this ought to be true in every ordered field, so the question is which extensions of R can be ordered.

[white oxide]

yo theorem 8.6 is just lemma 37.5 extended by induction right

and i guess generalized

[open sluice]

wait what’s H v K?

[white oxide]

the subgroup generated by the union of H and K

[open sluice]

spook

[white oxide]

no way ur in #groups-rings-fields

[open sluice]

this specific notation isn’t covered in my books

[white oxide]

yeah hungy has some weird ass notation

oh

you were talking about fraleigh right

interesting i thought that was the universal notation for the join of two groups

[open sluice]

I don’t think I’ve covered that as a concept at all

but I’m not an algebraist so whatever

[sweet echo]

I looks to be arbitrary so its not necessarily just extending by induction

[open sluice]

what book is this from?

oh hungerford?

[ashen heron]

looks like it

[white oxide]

yup

[open sluice]

wait

oh it’s the big boi hungerford

[white oxide]

ye

it's actually surprisingly very good

lots of examples

sure it's terse but in a good way

not like a rudin type of way

(or so i've heard)

[open sluice]

I’ll do it, um, eventually

I was wondering why you were 37 chapters in

[white oxide]

yee lol fraleigh was pretty good for a first semester algebra class imo

[untold turret]

Let $R$ be an integral domain, let $R^n$ be a free $R$-module with basis ${v_i}i^n$, let $\phi:R^n \to R^n: v_j \mapsto \sum_i^n m{ij}v_i$ be an $R$-module homomorphism, let $F$ be $R$-module isomorphic to $R^n$, and let ${\sum^n_i m_{ij}x_i}j^n$ be a basis for $F$. How would you argue $\phi$ is an isomorphism (in order to establish ${\sum_i^n m{ij}v_i}^n_j$ is a basis)?

[cloud walrus]

nrs

[untold turret]

ah, we need basis under module isomorphism is basis

[white oxide]

what does he mean that a "for the trivial group, a G-invariant subspace is nothing more than a subspace"? aren't all G-invariant subspaces subspaces lol for any group G

[glossy crag]

If A is a finite-dimensional K-algebra, then it has maximal subfields (subalgebras that are fields), since the set of subfields is non-empty (contains K), thus has a maximal element by the noetherian condition, right?

[south patrol]

Yessirree

[frigid lark]

If A is a G module, $N \in \mathbb{Z}^+$ what does $A_N$ mean?

[cloud walrus]

parrottea

[white oxide]

yeah so why did he make that remark

is my question

[frigid lark]

Also what does Z(p^n(p)) mean?

From Lang Page 304-305

[tough raven]

Is the following argument correct?

Let K be a p-adic field (i.e. finite extension of Q_p) with valuation ring A and uniformiser π. Let f in A[X] be monic.

Suppose f is irreducible and the constant coefficient a_0 of f has valuation equal to that of π (i.e., it is divisible by π but not π^2). Then f is an Eisenstein polynomial (i.e., all the intermediate coefficients are also divisible by π).

Proof: let e = deg(f) and α be a root of f in an extension of K, then it is clear from the norm formula for valuation in an extension that v(α) = v(a_0)/e = v(π)/e; but K(α) is of degree e over K. Thus K(α)/K is totally ramified and α is a uniformiser for K(α), so its minimal polynomial f is an Eisenstein polynomial.

It seems too good to be true that just because the constant coefficient is divisible by π exactly once that all the other coefficients are also divisible by π. Is there a more direct argument? It would be nice to see how irreducibility plays a role in it, for example.

[real sparrow]

an Injective homomorphism only maps to a group from one of its subgroups.
a surjective homomorphism only maps from a group to one of its quotients.

A surjective injective homomorphism is an isomorphism.

[lethal dune]

what's the question

[sly crescent]

Hi

[delicate bloom]

irreducibility played a role when you found the valuation based off the degree, if it weren't irreducible, it couldn't have a norm of that form

maybe a different perspective to look at is to look at the newton polygon, although I don't know if that helps other than to give a visual perspective on what it looks like

[tough raven]

Well yes, and now that I think about it, in assuming that the norm is the constant coefficient (upto sign).
But what I meant is that it would be interesting how irreducibility would be used in a “direct” proof.

[delicate bloom]

when you say "direct" can you give a fake example to help me understand what you mean by that

like what would that look like

[tough raven]

I don't know; something that doesn't feel like it used black boxes, I guess.

Say, an argument algebraically manipulating coefficients and evaluating the polynomial at some values, using the norm and the ultrametric inequality, or something concrete like that.

[delicate bloom]

hmm, I think I might be able to come up with a version like that, I'll think about it

also maybe thinking about your assumption earlier: "assuming that the norm is the constant coefficient (upto sign)." might help too, since that sorta contains the black-boxiness I feel

idk if this is more convincing to see, but the fact that in the extension field we have $v(\alpha)=v(\sigma(\alpha))$ for each galois conjugate means it factors as, $$f(x)=\prod_\sigma(x-\sigma(\alpha))$$ which once you expand it you can see all the intermediate coefficients are divisible by $\alpha$, but since they actually lie in $A$, they must be divisible by $\pi$

[cloud walrus]

merosity

[delicate bloom]

the main trick is seeing that if you have v(x)>0 after we expand the linear terms out in the extension field, and if our original irreducible polynomial were in, say Q_p, this means the smallest v(x) can be is 1. So it kind of forces it upwards magically.

Krasner's lemma is also a good one that's a bit peculiar like this I feel

idk, ping me if you come up with something better or if you think about this and feel like that helps later would be fun to think about more

[rocky cloak]

It's the "nothing more" part that is important. When G is trivial every subspace is invariant. So the only irreducible representations are those without subspaces.

[fervent gyro]

Do you have book recommendations for the basics of exterior algebra? I need to reference some basic properties to include this proof in my work:

[untold turret]

A is a basis for free module F_1, B is basis for free submodule F_2 of same rank as F_1. How is the determinant of the change of basis A to B matrix related to the quotient module F_1/F_2?

[lethal dune]

what's the relationship of F_1 and F_2? How's the quotient is defined then?

[rocky cloak]

Are you saying you have a homomorphism F_2 -> F_1 and you write it as a matrix with respect to the bases A and B?
Then at least you can say that if the determinant is a unit the homomorphism is an isomorphism, so F_1/F_2 = 0.

[untold turret]

oh! I see! I'm trying to figure out if we can infer something about the size of F_1/F_2 from that determinant and your result makes a lot of sense

you're right, it isn't a change of basis matrix, but a matrix expressed in two bases

[rocky cloak]

For commutative rings you might be able to use the adjugate matrix to say something more. But I'm not sure how much you can say

[wraith cargo]

Appendix in Eisenbud's commutative algebra book

[formal ermine]

this lecture we did lie theory stuff

like the classification of finite dimensional complex irreducible representations of sl and how newton okounkov bodies help you find bases

[hidden haven]

@gritty sparrow look, your favourite topic in the wild

[gritty sparrow]

lol nice

[agile burrow]

That's cool

Rep theory of sl(2, C) is cool

I want to understand geometric interpretation via action of PGL_2(C) on rational normal curve but I do not understand projective geometry

[delicate orchid]

C[x, y]

I do find it funny how you can just like

for sl(n, C) you just consider 2n elements acting on C[x_1, ..., x_n] it's so easyyyy

[elder wave]

What

None of the things you showed me required any of this apart from basic terminology

[white oxide]

ah ok that makes much more sense, thanks!

[summer path]

im getting the feeling more and more that whatever class illumi is taking doesn't actually teach much of anything and just does an overview of the material instead

[formal ermine]

I learnt some new stuff so I didn't see it as basic terminology

[elder wave]

The course using the word functor does not mean cat theory is a prereq

[delicate orchid]

edrmmm mbut how can you ermmm uhhh how cna you eemrmm how can you map between categoryies without huhhh carateogries bieng a hting???

[summer path]

carateogries

[lethal dune]

my algebra 1 was: cat theory, groups and solvability of groups, rings, modules over pid, projective injective modules, quadratic forms, homological algebra and rep theory

[summer path]

in that order

[lethal dune]

yeah

algebra 2 was simpler, galois and comm algebra

[barren sierra]

Anyone know what this fancy y thing is

Don't just stare

[delicate orchid]

never seen that before in my life

[barren sierra]

yea me either which is why I'm asking

[delicate orchid]

nobody has ever seen that before

[barren sierra]

wunderbar

[delicate orchid]

it is an artefact of antediluvian origins long forgotten to modern man

[coral spindle]

You know what, I saw that symbol in the TeX symbol list recently

If you find out what the hell it means let me know

I think it was \ydown or something?

[lethal dune]

I'm willing to call it a typo

[delicate orchid]

maybe we can work it out

[lethal dune]

[delicate orchid]

what's G_1 and what's E

[barren sierra]

E is an extraspecial group

[barren sierra]

this is used many times

[lethal dune]

and they never mentioned what that is?

[barren sierra]

I mean this is for my REU so like there's alot of BS moving around elsewhere

[delicate orchid]

ok just an extra special group right

[lethal dune]

not even on the symbols section,

[barren sierra]

this is purely a notation question in case anyone here knew lol

[delicate orchid]

do you know the order of G_1 so we can at least determine if the base set is E x SL(2, 3) x C_12

[barren sierra]

specifically order 27 exponent 3 if you care

[delicate orchid]

3_{+}^{1+2} my... neutral

no strong feelings on that one

[barren sierra]

I have no strong feelings on most anything stated here

[delicate orchid]

now if it was extraspecial of exponent 7 of order 343 we might be in business

[coral spindle]

Not sure it's a typo since they state it twice

[delicate orchid]

it's obviously not a typo

[barren sierra]

it's def not a typo

it appears multiple times

[lethal dune]

could it be the wreath product?

[barren sierra]

but none of the appearences define it

no that's a different symbol

that they do bother to define

[south patrol]

Wreath

[lethal dune]

there's not many group products known to mankind

[coral spindle]

S'not wreath

[barren sierra]

I've learned of them than I want to

[lethal dune]

idk what wreath product is, just saw on wiki

[barren sierra]

groups suck man gimme back my rings

[delicate orchid]

ok so the extra special group p_{1+2}^+ is the sylow p-subgroup of PSL_3(p) which does not help because it's SL(2, 3) not SL(3, 2) FUCK

what's the action I don't get itttt

[coral spindle]

Perhaps this? https://en.wikipedia.org/wiki/Central_product They mention extraspecial groups on this page

[delicate orchid]

no way it's the central product

[barren sierra]

¯_(ツ)_/¯

it wouldn't surprise me

[delicate orchid]

\ast is standard for central product

when it's clear from context that \ast isn't the free product

[coral spindle]

Well I mean

[delicate orchid]

before you quip

[coral spindle]

quip

damn

[lethal dune]

is there a similar notion of attaching map in group theory?

[barren sierra]

but the only time they state something is the central product here they do so without a symbol

[formal ermine]

do YOU know what a flavorable module is

[wraith cargo]

do they cite a source for the results

[lethal dune]

f: H ( ⊆ G) → N and define G×N/ ~

[delicate orchid]

that's basically the form a central product takes

[lethal dune]

oh lol

[barren sierra]

this is the result lol

OH IT MAY BE CENTRAL PRODUCT?

I think it is

[delicate orchid]

you take a subgroup Z of Z(G) with an isomorphic image in Z(H) then the central product is GxH/{(x_i, x_i'^{-1}) : x_i \in Z} lol

[coral spindle]

😏

[barren sierra]

I hate this

[lethal dune]

spam is getting spam answers rather than a helpful one

[delicate orchid]

so if that symbol is commonly used for an attaching map then it would make sense

[barren sierra]

I appreciate the efforts to cheer me up

[delicate orchid]

proceed by assuming it's the central product

[lethal dune]

what's the map tho

[delicate orchid]

wdym what's the map

[barren sierra]

current location

[lethal dune]

the isomorphism you need for a central product

[barren sierra]

ngl idk what central product really is

[delicate orchid]

there's a copy of Z in Z(G) and one in Z(H)

[lethal dune]

there's no canonical one unless Z is itself the product on the right

[delicate orchid]

you just map them

sure ok it's up to an automorphism but like

[barren sierra]

it kinda keeps popping up and so far I've been able to brush it under the rug

but now I cannot

[lethal dune]

yeah

[hollow mica]

is (2) somehow equivalent to completeness since it characterizes the real numbers?

[delicate orchid]

oh oh I remember

[barren sierra]

I don't think you get completeness from either of these

[delicate orchid]

it has to factor through the projection into G x H

what would the diagram be

[coral spindle]

This is model theory more than anything, just so you know. But as it happens, it certainly does not characterise the real numbers! Skolem–Lowenheim tells us otherwise.

[rocky cloak]

The real algebraic numbers should satisfy (2), and they're not complete.

[delicate orchid]

CRINGE!

no wonder no one uses this DOG construction bro

[hollow mica]

oh for some reason I thought if (1) was true for some field F then F must be isomorphic (as a field) to the reals

[coral spindle]

No that's definitely not true.

Elementary equivalence is actually quite a weak condition.

[barren sierra]

it's all over my REU project and rn I'm semi sweeping it under the rug lol

[coral spindle]

As I alluded to, Skolem–Lowenheim tells us that there are infinitely many distinct structures that satisfy this property, at least one for every infinite cardinality.

[delicate orchid]

it's well defined for extra special groups :trollshiro:

[lethal dune]

Wtf is an extra special group

[delicate orchid]

Z(G) = C_p and Inn(G) is

uhh

elementary abelian

yeah that's the one

thinkin bout p^3 🤤

[lethal dune]

Cringe af

[delicate orchid]

actually they're quite cool

[barren sierra]

they are neat

[delicate orchid]

cool fusion systems on them :pack:

[barren sierra]

they're p groups with center = Z_p

[lethal dune]

They are made up terms to create more jobs

[barren sierra]

so true (I just got paid today)

[delicate orchid]

yeah I suppose "p groups with minimal centre and nice inner automorphisms" is quicker

I don't get paid

[summer path]

you won't be homeless today! (i hope)

[lethal dune]

me yesterday for shittalking on the internet

[barren sierra]

I will be eating good today*

*I will not be, I will be eating the free dining hall food

[summer path]

is the dining hall food good there

[formal ermine]

free food????

[open sluice]

free at the price of 80 years of debt

[barren sierra]

fuck no

[summer path]

[barren sierra]

aaaand I'm trying to lose weight and all the food is like

pasta, pizza, fried chicken in some sauce, mashed potatoes, that type of stuff

but it's free

[formal ermine]

you're making me even more hungry now

I'm like constantly hungry

[barren sierra]

no no but it's like

shitty versions of that

[barren sierra]

me too since i'm on a heavy calorie deficit now

[summer path]

you're gonna need to show us

also #1100503863586455632

[formal ermine]

still better than bütterken

[lethal dune]

is that a misspelling of butter chicken

[small bramble]

been reading bout localization for two days and I just realized localizing by a prime means making the ring (R-P)^(-1)R rather than P^(-1)R . It's not like dividing by 0 shouldn't make sense

[lethal dune]

happens

[formal ermine]

it's dull bread with butter

[lethal dune]

shit

[formal ermine]

I would rather have chicken right now

[delicate orchid]

yeah this is why I've seen some authors take the localisation "away from" the prime

which makes much more sense to me

[small bramble]

yeah sounds less bamboozling

[teal vessel]

oh, @delicate orchid you remember when I said my brother calls fibers pullbacks? it's because his textbook never defined fiber and calls them all pullbacks.

[rocky cloak]

I see some authors use "localize away from" when talking about elements and "localize at" when talking about prime ideals.
That way you can be totally confused when they talk about the localization at 2 and the localization away from 2.

Geometrically it makes sense though.

[south patrol]

Lol this just reminds me of when I emailed a professor about some notes cause I confused R_(0) and R_0

[rocky cloak]

Localization at 2, zooms in on 2 and localization away from 2 zooms into the other primes

[south patrol]

Was embarrassing

[glossy crag]

In a product set $\bigcap_{i,j}A_i\times B_j=(\bigcap_iA_i)\times(\bigcap_jB_j)$ for arbitrary index sets, right? There are no infinite nuances I'm missing?

[cloud walrus]

leave_no_norm

[coral spindle]

Seems so.

[glossy crag]

Just checking.

[delicate orchid]

very funny

[delicate orchid]

geometry is non-rigorous cope

[rocky cloak]

Geometry = coming up with weird names for algebra

[alpine island]

Geometry is just naming shapes

but scaled up from pre-k to uni

[delicate orchid]

if it's in a non-hausdorff topology I will not call it a shape

[south patrol]

This works for arbitrary products too

I guess the question now is what the funniest way to prove this is

[delicate orchid]

are A_i B_j just sets

[south patrol]

ye

[delicate orchid]

ok lol "product set" is a weird term

[south patrol]

why

[delicate orchid]

find the indecomposable sets

[south patrol]

the empty set

uhh

Well that'd depend what you mean by indecomposable right lol

[delicate orchid]

what else could I possibly mean you fool

X is indecomposable iff there are no doodads such that Y \coprod Z = X or whatever

[south patrol]

coprod different tho

well usually decomposability is in situations with biproducts right

but ye

[delicate orchid]

yeah normally with biproducts

but if you'd ask me to generalise it I'd choose the coproduct

[south patrol]

me too

[delicate orchid]

wholesome!

[glossy crag]

Is the assumption that R has non-zero maximal left ideals used anywhere here? I don't see it, looks like the proof works just as well without it.

[south patrol]

I agree yeah it seems the proof would go through fine with Q being 0 for everything lol

[formal ermine]

what is the mapping for $V \tensor V^\vee \iso \on{End}{V}$

[formal ermine]

my prof said v tensor phi maps to phi(v)v but that just sounds like nonsense to me

[lethal dune]

(v, f) ↦f(_)v

[delicate orchid]

yeah that's right

[lethal dune]

noo

[formal ermine]

okay this makes mure moch sense lmao

maybe he typo'd on the blackboard

I was so confoosed

[formal ermine]

wait huh?

[delicate orchid]

I assumed phi was like

the iso between V and V^\wedge

[formal ermine]

nono

[delicate orchid]

yeah I got it now

[formal ermine]

phi was the vector in the dual

okay wew

I got a rep theory question

[delicate orchid]

I am not ok

I am about to be trolled on the internet by a child

[formal ermine]

why for lie algebras we consider endomorphisms instead of just automorphisms

[delicate orchid]

lie algebra
don't care

I know next to nothing about lie algebra/lie group reps

[formal ermine]

amazing

me neither

[lethal dune]

because Aut don't form a lie algebra

[formal ermine]

oh because we want a lie algebra morphism

got it

thanks

[delicate orchid]

I know about root systems and how they relate to lie groups and their reps but I'm way more into the associated Weyl group than the actual reps

[formal ermine]

$e_1 \tensor e_3^* = E_{31}$ right????

[formal ermine]

uhhhh in $\bC^3$

[formal ermine]

over C

[delicate orchid]

I got no idea what you're talking about

[formal ermine]

this

[lethal dune]

no

[formal ermine]

wait I mean 0 0 1 ; 0 0 0 ; 0 0 0

I forgor if that was E_31 or E_13 lol

[lethal dune]

try plugging e_3, you're supposed to get e_1

[formal ermine]

I do??

wait

[lethal dune]

trick

[delicate orchid]

troll

[lethal dune]

[formal ermine]

???

[delicate orchid]

oh that's what you meant

yeah then you're right

[lethal dune]

vectors as rows

[formal ermine]

ok thank you wew uwu

[lethal dune]

[delicate orchid]

yeah we got rows on this mf

stacks

[lethal dune]

well

okay

[formal ermine]

when we define an algebra what do we mean with "calculation rule"

[delicate orchid]

yeah so e_i \otimes e_j* is the matrix that's e_i evaluated on e_j and 0 elsewhere

[formal ermine]

in my case U(n^+)

it says "U(n^+) is the algebra with generating system blabla and calculation rule XY - YX = [X, Y]"

what do they mean with this

[delicate orchid]

that's a lie bracket

who the fuck calls that a calculation rule on the nine celestial spheres

but then again who calls M_{3\times3}(C) "C^3"

[real sparrow]

there is 1 group of order 15 and 2 groups of order 21

[formal ermine]

yeah but like

why do we care

[delicate orchid]

wdym why do you care

it's not a lie algebra without a lie bracket

[formal ermine]

what is the calculation rule supposed to mean

[real sparrow]

One of those groups of order 21 is non abelian

[delicate orchid]

yeah it's C_7 \ltimes C_3 with the obvious embedding of C_3 into Aut(C_7) \cong C_6

[real sparrow]

The smallest odd order non abelian group.

[formal ermine]

oh lmao

[tribal moss]

It's the (non-commutiative) multiplication operation in the algebra. Calling it "calculation rule" instead of "operation" is very nonstandard.

[real sparrow]

C7 has automotphism group of C6

[formal ermine]

how do you define multiplication in terms of itself?

[delicate orchid]

that's matrix multiplication dude

cmon now

[tribal moss]

I ... don't?

[formal ermine]

oh

ok that makes more sense

[rustic crown]

(thats rtimes )

[delicate orchid]

anticommutivity is BASICALLY commutivity cmon now

[delicate orchid]

my supervisor gets annoyed at me because I have literally never once got it the right way round

[formal ermine]

also calculation rule might be lost in translation

[delicate orchid]

and I will never learn which way round it goes

[real sparrow]

The symbol points in the direction of the automotphism

[delicate orchid]

wow... cool

[rustic crown]

it points in the direction of the subgruop that's going to be normal

[delicate orchid]

"the triangle is the normal subgroup tirangle sotrue"

[rustic crown]

.<

[real sparrow]

Not the vest way to see it because then it just reduces to the same problem but for normal subgroups

Instead I just look at the direction of the homomorphism

[tribal moss]

Wikipedia https://en.wikipedia.org/wiki/Semidirect_product helpfully informs us that

some sources[9] may use this symbol with the opposite meaning.

[real sparrow]

$$\varphi:A\to\text{Aut},B$$
$$A\ltimes_{\varphi} B$$

[cloud walrus]

s5

[delicate orchid]

I don't see it at all I just write \ltimes because that seems to be more correct more often than not

[real sparrow]

Triangle points in the direction of the honomorphism.

[delicate orchid]

it's usually obvious anyway as who split extends anything other than cyclic groups ykwim

[rocky cloak]

The way I always think about it is that the normal group is on the open side. Because for $A\times B$ both sides are open, and both are normal.

[cloud walrus]

jagr2808

[delicate orchid]

hit mfs up with that $\ltimes !!!!!\rtimes$

[cloud walrus]

wew ladz

[real sparrow]

Isn't that just the Zappa–Szép product

[teal vessel]

we love that group theory literally just said "subgroup? we can't use the subset notation, so..... hmmm.....meh, just call it less than"

[coral spindle]

It's good notation imo.

[teal vessel]

it's definitely not bad notation, just funny

[real sparrow]

Or just say there is an injective homomorphism.

[coral spindle]

Sure lad

sure

[delicate orchid]

tbf set theory is the odd one out on this

literally every other poset in existence is \leq \geq but nooo

or the weird \underset{\sim}{<} the analysists use

[coral spindle]

It's good that we have specific \subseteq notation. Not to open a can of worms of course, but distinguishing between a mere subset of a group and a subgroup of a group is helpful

[delicate orchid]

put which category you're talking about as a subscript on the \leq

[teal vessel]

I just like the fact that the following is true

[real sparrow]

why is there no symbol for a surjective homomorphism

[delicate orchid]

\pi lol

[coral spindle]

Yeah there is, it's a double-headed arrow

you see it in diagrams all the time

[teal vessel]

$\forall x,y \in \mathbb{N} x < y \iff x \in y \iff x \subset y$

[coral spindle]

Yup

[cloud walrus]

.goldenphoenix

[teal vessel]

that makes things very nice™️

[coral spindle]

Yup that is indeed why we construct it like that.

[summer path]

is $\subset, \subseteq$ preferred or $\subsetneq, \subset$?

[cloud walrus]

tubularcat

[delicate orchid]

ew

the former

[teal vessel]

I prefer the former pair, but will on occasion use neq if I've fucked up in my notes

[summer path]

(personally i dislike the second one)

[ashen heron]

$\subseteq > \subset$, $\subset < \subsetneq$

[coral spindle]

It's a matter of preference. I use \subseteq and \subsetneq.

[lethal dune]

⊆ ⊊

[cloud walrus]

pikapikapikapikachu

[real sparrow]

$$A\hookrightarrow B$$

[teal vessel]

now for the cursed \eqnsubset

[coral spindle]

Unfortunately people use \subset to mean various things so it's best avoided.

[cloud walrus]

s5

[summer path]

well i went through half of my topology class before i realized my prof was using the second one

[coral spindle]

Also, #groups-rings-fields is getting clogged. Move to #math-discussion pls thanks xox

[summer path]

#1100503863586455632

[teal vessel]

I don't particularly like the element xox, I prefer the element xox^-1

[real sparrow]

What's that

[lethal dune]

[teal vessel]

some arbitrary commutator

[coral spindle]

That's not a commutator

[teal vessel]

(to use the cubing terminology cuz I can't remember the actual word)

[delicate orchid]

speaking of which

[summer path]

you'd need another o\inv on the right

[lethal dune]

o=e

[summer path]

broken glasses

[delicate orchid]

[g, h] = gh(hg)^{-1}
[X,Y] = XY-YX
really... makes u think....

[summer path]

ok maybe we should stop shitposting

[real sparrow]

This is how I like my commutators
$$a^{-1}bab^{-1}$$

[cloud walrus]

s5

[south patrol]

Pain

[lethal dune]

cursed

[coral spindle]

There's a nice problem actually

Where this form is good

The problem is: can you find a way to hang a frame on two pegs such that if either peg is removed, the painting falls?

Big spoiler I guess but the solution is the commutator.

And if you use this form, it ends up being a symmetric solution.

[lethal dune]

yeah

[teal vessel]

conjugation.... dangit. why do all my interests have to use the same words in very different contexts??

[real sparrow]

[a,b] = b^-aba^-

[delicate orchid]

what other use of the word conjugation is there

other than the stupid field automorphism of C

[teal vessel]

linguistics

[delicate orchid]

ah

le verbs

[teal vessel]

a subtype of declension, yeah

[hidden haven]

les* ☝️

[teal vessel]

time to show some strange meta morphism between linguistics and math actually that's too much work

[alpine island]

declension, when applied to verbs, is conjugation

[real sparrow]

verbing

[teal vessel]

I'm so glad my second and third languages have minimal declension

[delicate orchid]

MODS!!! FRENCHIE!!!

MODDSSS

[real sparrow]

Le frenche

[hidden haven]

La* ☝️

[coral spindle]

Move to #serious-discussion for discussion thanks.

[real sparrow]

Phroge

[alpine island]

words are elements of the free monoid over the set of phonemes

[real sparrow]

🐸

[teal vessel]

blursed. never speak to me again

just kidding that's really funny

now I need to choose: continue working through my abstract algebra textbook, or finish doing my slide rule competition practice problems

[delicate orchid]

your.... what...

[teal vessel]

abstalg it is

[coral spindle]

That's nice, but can you keep discussion in #discussion or #serious-discussion , please.

[formal ermine]

[lethal dune]

Ok here's a serious question

[delicate orchid]

serious face

[formal ermine]

uhhhhhhh trivial

[delicate orchid]

oh it's like Tensor-Hom adjunction but backwards

nifty

I believe it at least

[formal ermine]

so trivial that I will in fact not mention the solution

[hidden haven]

Not sure why R is required to be a PID

[lethal dune]

tor hom? tensor hom

[delicate orchid]

that's what I mean yeah

soz

[lethal dune]

becaise NO SPECTRAL SEQUENCES

[rustic crown]

it true for higher tor/ext?

[hidden haven]

[hidden haven]

Thimkin

[lethal dune]

it was given in our basic homological algebra class

[hidden haven]

I assume first you can check that
Hom(Tor(A, B), C) = Hom(A, Ext(B, C))

[lethal dune]

tor and ext as well?

[rustic crown]

should you just blindly pick free resolution of B of length 1 or something and play

[lethal dune]

wait

[hidden haven]

No

[delicate orchid]

simply apply Hom to a resolution... the rest will follow... you must trust...

[hidden haven]

Not Tor and Ext wait

I mean the total right and left derived functors of Hom and Tensor

Idk how to denote that

[lethal dune]

Ext(Tor, ).. whatevery

[hidden haven]

Hom(LTensor(A, B), C) = Hom(A, RHom(B, C))

This is true

[delicate orchid]

I buy it

[lethal dune]

LTensor?

[delicate orchid]

^

[hidden haven]

Like you tensor a projective resolution of A with B

Or vice versa

The entire chain complex

[gritty sparrow]

I hardly know her

[hidden haven]

And RHom is

[lethal dune]

oh so you mean the tensor in Derived cat?

[hidden haven]

Yes

Then you need to pass to homology

[lethal dune]

you know Reisz, that's enough

[lethal dune]

that complicated solution for a basic HA question

[hidden haven]

You can put R outside both the Homs?

And then taking first homology should do it assuming R is a PID

[delicate orchid]

surely SURELY

wait one mo

[hidden haven]

oh was it supposed to be basic

[delicate orchid]

no nvm

[hidden haven]

Take resolutions of all things I am sure it works

[lethal dune]

so adjunction preserved in derived cat?

wait that doesn't sound right

[hidden haven]

Yeah LTensor is left adjoint to RHom

[lethal dune]

I mean all adjoint pair, but I need to specify category carefully

[hidden haven]

Not all adjunctions are preserved when you derive them

They need to be Quillen adjunctions

[lethal dune]

of course

[hidden haven]

Tensor-Hom is a Quillen adjunction

[lethal dune]

if anyone wants to torture themselves

[hidden haven]

Epic

[lethal dune]

ahem basic homolgoical algebra ahem

[hidden haven]

[delicate orchid]

I got like no intuition for anything in that image except for the direct sum

[lethal dune]

I still don't know enough HA to read the entire thing

my eyes just fly right off

[hidden haven]

If you don't use derived categories then I am guessing taking projective and injective resolutions of all modules involved should allow you to just work through it

[hidden haven]

looney tunes

By any chance

Is HA short for higher algebra

[formal ermine]

high school algebra

[hidden haven]

[lethal dune]

Is this true? Ext(A \otimes B, C) = Ext(A, Hom(B, C))?

I suppose not

anything to do with Quillin adj

[hidden haven]

Should be

[lethal dune]

Interesting

[hidden haven]

Ye because that is true at the Hom level

[lethal dune]

so somehow all derived functors also preseve the adjunction

[hidden haven]

So it is true if you take RHom on both sides

And then it is true if you take first homology on both sides

[lethal dune]

cool cool let me read up on Quillin adj

[hidden haven]

Quillen adjunctions are homotopy theory terminology btw idk if there is simpler terminology for derived cats

Like derived cats are older theory

I guess it is just called a derived adjunction

For Quillen adjunction you would need to look up model cats

[lethal dune]

I've been wanting to read those for a long time

just unmotivating definitons set me off

[hidden haven]

Look at Quillen's original manuscript or my thesis

[south patrol]

So cool

Why is homotopy theory so long to learn

[lethal dune]

send your thesis

[hidden haven]

[south patrol]

I was told Quillen's stuff wasn't too well-written lol

or is it fine

[hidden haven]

Divided opinions

[south patrol]

Fair

[hidden haven]

The standard axioms are slightly modified from what he took as his axioms

[south patrol]

Every opinion has a factorisation into a weak fibratory opinion and a

[hidden haven]

I think the standard one now is Hovey's

Except maybe sometimes functoriality of the factorizations is not assumed

[lethal dune]

cool notation for lift

[hidden haven]

[alpine island]

damn I opened this chat and thought heavy from tf2 had made up some axioms

[hidden haven]

Actually this is kinda silly not sure if it works

At least not immediately

Because I want to take right derived functors of both sides

Not just of the outermost Homs on both sides

@lethal dune I might be messing up here be careful

Mathing while gaming

[lethal dune]

take ^op on one category? I kinda had a rough solution of the first one using Grothendieck SS

[hidden haven]

How will ops help?

[lethal dune]

idk I forgot

[hidden haven]

lol

[lethal dune]

it was long ago and I'm too lazy now to retrace the steps

[hidden haven]

[rocky cloak]

If A=B=Z/2 and C=Z, you get Z/2 on LHS and 0 on RHS

[lethal dune]

I see

@hidden haven

[hidden haven]

Ye makes sense

My argument had that flaw with right deriving only parts of both sides

[rocky cloak]

There should be some universal coefficient theorem stuff relating them though probably

[lethal dune]

wym by weak equivalence of functors here? what's the model structure on Fun(Top, Top)?

do you mean each \eta(x) : Gamma(x) \to 1_Top(x) is a weak equivalence?

that would make sense

[rocky cloak]

That would be my guess

[hidden haven]

Yes

[formal ermine]

whom are you talking to

[lethal dune]

the author

[hidden haven]

Natural somethingmorphism means a natural transformation that is level wise somethingmorphism

In general

[south patrol]

Tbh kinda amazing you can turn it into a functor lol

[lethal dune]

everything is

[hidden haven]

Small object argument op

[south patrol]

I guess this probably requires like a proper class of chocies right

[south patrol]

Yeah sure

[delicate orchid]

I presumed a natural swagmorphism was a natural transformation such that each of the maps were swagmorphisms

[summer path]

It's swagmorphism time!

[hidden haven]

Lmao I just realised

You're very quick

[south patrol]

Lol nice

[void cosmos]

whats Ext

[formal ermine]

homology of something something

[formal ermine]

@void cosmos this is from my ha notes

[void cosmos]

oh