groups-rings-fields

[oblique river]

the first certainly implies the latter, but the latter is a weaker condition

[woven obsidian]

Yeah sorry I was perhaps a bit unclear, I want to show the first one

I forgot to mention that I also know that none of them has a fixed prime divisor

[oblique river]

my suggestion is to think of gcd's and bezout's theorem

what is gcd(f,g) in Z[x]? (or maybe -- what is it in Q[x] and then clear denominators)

can you do anything with that identity?

[woven obsidian]

Hmm ok so in Q it is 1

[oblique river]

so what does bezout's identity say?

[woven obsidian]

So we can write hf+lg=1 for some pols h,l in Q[x].

[oblique river]

mhm

[woven obsidian]

It would be nice if the denominators were constants

[oblique river]

what do you mean?

what do elements of Q[x] look like?

[woven obsidian]

Ah I'm stupid

Thought of Q(X) for some reason

Ok so then we have hf+lg=C for some new h,l in Z[x]. So if one of them is 0 I guess the other one can't be

Except at prime divisors of C

[oblique river]

neither h nor l will be zero in Z[x]

[woven obsidian]

Yeah because the pols have deg>0

[oblique river]

so what can you use this for

[woven obsidian]

Well if p | LHS it must divide RHS too

So only works for finitely many p

[oblique river]

yeah

[woven obsidian]

Neat

I guess the only problem is the slight nonexplicitness of the C

[oblique river]

what do you mean?

the problem doesn't ask us to determine the size of the set {p | f and g have a common root mod p}

so we dont need to know what C is

[woven obsidian]

The book I'm reading gives an example later and looks at the polynomials (in n) n^2+1 and n^2+3. Then says they cannot have a common root mod p for p > 2

[oblique river]

yes

[woven obsidian]

Btw this problem isn't a homework problem, I was reading a book and got stuck on that detail

[oblique river]

oh

in that case i dont feel bad saying more

you don't need to know anything about C other than that it's a nonzero integer

and by reducing kf + lg = C mod p you conclude that when p doesn't divide C, f and g can't have a common root mod p

[woven obsidian]

Yeah I understand that

[oblique river]

for a fixed f and g, you can find C

just do the euclidean algorithm in Q[x]

[woven obsidian]

So is Euclid's algorithm the way to go?

[oblique river]

that is a quick way to find the gcd of two polynomials, yes

[woven obsidian]

Ah ok, I thought there was another way, since they mentioned it that fast and only in passing

[oblique river]

but that shouldn't really matter here anyway

you know that f and g have gcd 1

so there must be some k and l with kf + lg = 1 in Q[x]

it doesn't really matter how you set out to find those

[woven obsidian]

yeah true, this is trivial to find

So the 2 comes from 3-1=2

[oblique river]

yes, in that particular example, you can calculate the gcd by observation

(1/2)(x^2 + 3) - (1/2)(x^2 + 1) = 1

so when you clear denominators you get (x^2 + 3) - (x^2 + 1) = 2 in Z[x]

so C = 2

[woven obsidian]

Thanks for the help, I would never have thought to use Bezout's identity

[oblique river]

I think the way to get to bezout's identity is just to notice that "f and g have a common root" just means their gcd is nontrivial

so you have the tension between "gcd is nontrivial mod p" but "gcd is trivial in Q"

[woven obsidian]

Yeah I would have perhaps tried it if this was my intro algebra course

[oblique river]

and it's just a matter of formalizing that -- once you've rephrased the problem in that language, bezout is how you actually get something concrete out of a gcd statement

[woven obsidian]

It's a very useful identity

[oblique river]

yee

ok gotta step away for a bit

gl with your reading

[woven obsidian]

Thanks!

[unique juniper]

Prove that in the quotient group G/N, $(gN)^{\alpha} = g^{\alpha}N$ for all$\alpha \in \mathbb{Z}$

[cloud walrus]

Yes

[unique juniper]

could someone help me out on this pls?

idk how to show (gN)^{a+1} = g^a+1N

[next obsidian]

just go by definition homie

[unique juniper]

N isnt normal tho

[next obsidian]

It has to be

G/N isn't a group otherwise

[unique juniper]

Well it doesnt say

oh

[next obsidian]

To form G/N as a group you need N normal

[unique juniper]

oh ok, if N is normal then i got it

[south temple]

If I have a group of order 60 that is not abelian, then is it true that there are 60 inner automorphisms? Because there are 30 pairs of elements of the form $(g, g^{-1})$, and we can switch them around to $(g^{-1}, g)$ to get 30 more?

[cloud walrus]

vov&sons

[carmine fossil]

Consider D_60. I don't think that works for D_60

[south temple]

doesn't D_60 have 120 elements?

[rustic crown]

Some people use the notation D_2n while others like D_n

[south temple]

oh ok

[south temple]

actually it is indeed 60 because $A_5$ is simple, so $|Z(A_5)| = 1$. combined with the fact that for a group $G$, it holds $G/Z(G) \cong Inn(G)$, we can conclude that $|Inn(G)| = 60$.

[cloud walrus]

vov&sons

[carmine fossil]

Consider A_4 x Z_5

center is non trivial

So |Inn(G)|<60

[south temple]

oh my bad! I didn't specify that I was interested in A_5

[tight otter]

I'm trying to find the minimal polynomial of $\sqrt[3]{2}+\sqrt[3]{4}$ over $\mathbb{Q}$ and I'm a bit stuck on showing that the polynomial $p(x)=x^3-6x-6$ is minimal. From this it's clear that $\textrm{deg}(m(x))\leq 3$ but I'm struggling to understand field extension multiplicity. I know that $[\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{4}):\mathbb{Q}]=[\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{4}):\mathbb{Q}(\sqrt[3]{2})]\cdot[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]$ but I'm not sure where to move on from there and show that the lower bound is 3. My instincts are telling me that I went wrong somewhere because that last equation seems like it evaluates to at least 4

[cloud walrus]

panoramatopia

[tight otter]

is it not true that $\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{4})=\mathbb{Q}(\sqrt[3]{2})$? I think I'm overthinking this

[cloud walrus]

panoramatopia

[vestal snow]

Have you heard of Eisenstein's criteria?

[tight otter]

oh my good lord i didnt even think of applying that here

we discussed it in class ~two weeks back but not in relevance to finding a minimal polynomial

[vestal snow]

Yeah, it's a very useful tool for stuff like this

[tight otter]

eistensteins with p=3 would show that p(x) here is irreducible meaning that it's the minimal polynomial

[vestal snow]

Yup

p=2 also works

[tight otter]

damn that's so much more useful than what my professor worked through

thank you banana

[vestal snow]

Np

Can someone help me out with a computation?

Let $k$ be algebraically closed. Let $F'/k(x)$ be an Artin-Schreier extension fedined by $y^p-y=f(x)$ where $f(x)$ is a polynomial. Calculate the degree of the differential $x^iy^jdx$ ($i,j\geq0$) at the place ramified over infinity.

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

In particular, when is this differential holomorphic at the point ramified over infinity?

[vestal snow]

If anyone has any ideas, @ me

[carmine fossil]

The set of commutators is generally not a subgroup

[vestal snow]

What have you tried so far?

[unique juniper]

do i have the right idea here? the elements of N are (x^-1 y^-1 xy)^n . im trying to show g(x^-1y-1xy)g-1 is inside N

[vestal snow]

I think it means the subgroup generated by the set of commutators

Almost

they don't have to be of the form (x^-1 y^-1 xy)^n

Try and think of what they are

Like how do you get a subgroup generated by a set?

Yeah but try and be a bit more explicit

[unique juniper]

Ok

The closure of <x^-1y-1xy>

[vestal snow]

Hmm

Alright lets do an example

What is the subgroup generated by <a,b>

[unique juniper]

like this

[vestal snow]

What do the elements look like

Yep

That's it

[unique juniper]

they look like 1 a b ab

+-1

[vestal snow]

Now apply that to the problem

[unique juniper]

ok ty

[tight otter]

Let $K$ be a field and $\alpha$ an algebraic element over $K$. Assume that $[K(\alpha):K]$ is odd. Show that $K(\alpha)=K(\alpha^2)$.

[cloud walrus]

panoramatopia

[tight otter]

Really not sure how to go around this--I have that $[K(\alpha^2):K]=[K(\alpha^2):K(\alpha)][K(\alpha):K]$ through field multiplicativity and that $\alpha^2\in K(\alpha)$ but not really sure what to do next

[cloud walrus]

panoramatopia

[mild laurel]

This is not quite the right equation

Which is the larger (or potentailly equal) field? K(\alpha) or K(\alpha^2)?

[tight otter]

Ah it would be K(a) wouldn't it

So the intermediate group would be K(a^2)

[mild laurel]

Right, a^2 is an element of K(a) so K(a^2) is a subfield of K(a)

Yeah

[tight otter]

Ah so instead it would be $[K(\alpha):K]=[K(\alpha):K(\alpha^2)][K(\alpha^2):K]$, which means $[K(\alpha):K(\alpha^2)]$ divides $[K(\alpha):K]$

[cloud walrus]

panoramatopia

[mild laurel]

Right

[tight otter]

i might be making an incorrect assumption here but would it then be true that $[K(\alpha):K]=[K(\alpha^2):K]$, thus meaning that $[K(\alpha):K(\alpha^2)]=1$?

[cloud walrus]

panoramatopia

[mild laurel]

Not necessarily?

Well, it's true, but I don't see how you can immediately conclude that

You haven't used the fact that [K(a): K] is odd at all, and the theorem is false without this assumption so

[tight otter]

hmmm true

I guess that $[K(\alpha):K(\alpha^2)]$ would have to be odd then as well

[cloud walrus]

panoramatopia

[mild laurel]

hint: ||What are the possible values of [K(a) : K(a^2)]?||

Yes that's true

[tight otter]

wouldn't it be 1 and 2? for the possible values?

and if it must be odd then it must be 1

[mild laurel]

And why is that?

[tight otter]

Because the base field contains all multiples of $\alpha$ rather than all multiples of $\alpha^2$ so either $\alpha=\alpha^2$ and the degree is 1 or the degree is 2, because $K(\alpha)$ contains twice as many coefficients for the polynomial as $K(\alpha^2)$

[cloud walrus]

panoramatopia

[mild laurel]

why does K(a) contain twice as many coefficients as K(a^2)?

[tight otter]

Because the coefficients of $K(\alpha)$ contain all the powers of $\alpha$ whereas the coefficients of $K(\alpha^2)$ contain just the even powers of $\alpha$

[cloud walrus]

panoramatopia

[mild laurel]

Okay I'm not really sure what you mean by coefficients here

[tight otter]

hm yeah sorry that logic doesn't work

I was misremembering polynomial field extensions

The way I understand it is that a field $E=F(d)$ is the field of polynomials of $d$ with coefficients in $F$, with $[E:F]=n$ where $n-1$ is the maximum degree of a polynomial in $E$

[cloud walrus]

panoramatopia

[tight otter]

the thing that I'm struggling to understand is what determines that maximum degree

ohhhhh

$n$ is the degree of the minimal polynomial

[cloud walrus]

panoramatopia

[tight otter]

my prof's notes conveniently skipped over that

[mild laurel]

Yeah

So how does that apply to the problem

[tight otter]

Well, if $\alpha\neq\alpha^2$ then the degree of the minimal polynomial of $\alpha^2$ would be at least twice that of $\alpha$, right?

[cloud walrus]

panoramatopia

[tight otter]

ack sorry no that isn't right

[mild laurel]

Maybe just think about K(a) over K(a^2)

instead of thinking about K(a) over K and K(a^2) over K separately

[tight otter]

K(a^2) is just K(a) with every instance of a^n squared

[mild laurel]

Let me put it this way, is there a polynomial that a satisfies, with coefficients in K(a^2)?

[tight otter]

It's a^2 isn't it lol

[mild laurel]

Uh, not quite

[tight otter]

My professor made note of that in the problem statement but I forgot about that

ah

hm

[mild laurel]

So you're looking for a polynomial f in K(a^2)[x] such that f(a) = 0

[tight otter]

$K(\alpha^2)[x]$ being the polynomial ring with coefficients in $K(\alpha^2)$, right?

[cloud walrus]

panoramatopia

[mild laurel]

Right

[tight otter]

$x^2-\alpha^2$ comes to mind

[cloud walrus]

panoramatopia

[mild laurel]

yeah

So this isn't necessarily the minimal polynomial

but now you know that the minimal polynomial must be degree 2 or less right

cause it must have degree smaller or equal to this

[tight otter]

Yeah

Ahhhhhhh

I see

that's the minimal polynomial of $\alpha$ under $K(\alpha^2)$

[cloud walrus]

panoramatopia

[mild laurel]

Well, like I said, it might not be minimal

[tight otter]

so then it would follow that $[K(\alpha):K(\alpha^2)]\leq2$

[cloud walrus]

panoramatopia

[tight otter]

yeah sorry oops

But we know that $[K(\alpha):K(\alpha^2)]$ is odd, thus, it must be equal to 1, which means that $K(\alpha)=K(\alpha^2)$

[mild laurel]

but that does follow yeah

[cloud walrus]

panoramatopia

[mild laurel]

Exactly

[tight otter]

That's a neat solution

thank you Zoph!

[mild laurel]

no problem

[frank fiber]

what is an element of order n in $S_n$?

[cloud walrus]

Go

[mild laurel]

What have you thought about?

[frank fiber]

in $S_1$ the element of order $n$ is $e$, and in $S_3$ the element of order 3 is defined by $x(1)=2$ , $x(2)=1$, $x(3)=3$

[mild laurel]

Uh, is that an element of S_2?

[frank fiber]

sorry, its S_3

in $S_2$ the element of order 2 is defined by $x(1)=2$ and $x(2)=1$

[cloud walrus]

Go

Go

[mild laurel]

Does that element of S_3 have order 3?

[frank fiber]

in $S_3$ the element of order 3 is defined by $x(1)=3$ , $x(2)=1$, $x(3)=3$

[mild laurel]

Uh, is that an element of S_3?

[cloud walrus]

Go

[mild laurel]

Still not quite

[frank fiber]

$x(3)=1$

[cloud walrus]

Go

[mild laurel]

again, still not quite

the elements of S_n are bijections

You can't send two different elements to the same number

[frank fiber]

i think this is right $x(1)=3$, $x(2)=1$, $x(3)=2$

[cloud walrus]

Go

[mild laurel]

Yes that's right

Now try S_4, or maybe you can start to see what it looks like in general for S_n

[obsidian sleet]

hi i was wondering about the ways i can prove that (1 2 3 4) is not the product of 3-cycles

the way i thought of showing this was that

(1 2 3 4) is an element of S_4

and then i can show that the 5 cycle structures that permutations in S_4 can take on are never gonna be only 3-cycles

which concludes proof

the other way my friend was talking about was to just do contradiction and use the fact that products of 3-cycles are even so that is not compatible with (1 2 3 4) being an odd permutation

both are fine right

oh are ppl still using this channel

so sorry

go ahead

[frank fiber]

in $S_4$ the element of order 4 is $x(1)=4$, $x(2)=1$, $x(3)=2$, and $x(4)=3$

[cloud walrus]

Go

[mild laurel]

Right

[frank fiber]

In S_n I think this is the element of order n

[mild laurel]

Right

[frank fiber]

And I suppose I can prove it with induction

[mild laurel]

uh, I'm not sure induction would help

You can just prove this directly

[frank fiber]

i solved it, thanks

[thorn delta]

Your friend’s is the first way I thought of. I don’t think I understand your solution tho

[obsidian sleet]

so in S_4, the only structures that cycles can take on are as follows

Let (k) be a generic k-cycle. Then the only cycle structures that permutations can take on are as follows:

(4)
(3)(1)
(2)(2)
(2)(1)(1)
(1)(1)(1)(1)

None of these are products of only 3-cycles

therefore (1 2 3 4) cannot also be a product of 3-cycles

is this legit?

[thorn delta]

by cycle structure, you mean representation as a product of disjoint cycles, right?

[obsidian sleet]

yes

[thorn delta]

how does this reasoning rule out that (1 2 3 4) is not a product of not necessarily disjoint cycles?

in fact, (1 2 3 4)'s cycle structure is just itself since its order is a prime power, so if im understanding you correctly, this doesn't say much of anything useful unfortunately

[summer geyser]

What is the second part of this expecting? Only thing I can think of is because we have sqrt(2), but I don't really see this as a massive problem.

("Answer the same question" refers to finding the degree of their splitting fields over Q)

[scarlet estuary]

i'd imagine that's what they're looking for, yeah

adjoining a solution to X^4 - 4 is just adjoining a solution to X^2 - 2

[summer geyser]

Why would this be a problem though?

since it's basically the same thing

[scarlet estuary]

i mean i wouldnt really consider it a problem? its just we prefer more "simplified" notation where possible to keep things consistent

[summer geyser]

Yeah I guess that's fair

[scarlet estuary]

its a weird question though

i think they just wanted you to realize that this is isomorphic to Q(sqrt(±2))

[summer geyser]

Indeed, I'll go with that then only thing that seems reasonable

[scarlet estuary]

and phrased it in a weird way

to force you to recognize that

not sure though

maybe there's something in your text/course notes that talks about the notational conventions? idk

[summer geyser]

Hm I don't think so, but I'll take a look thanks

[scarlet estuary]

i mean maybe theyre looking for like

the fact that you can solve X^4 - 4 by adjoining a solution to X^2 - 2 or X^2 + 2

so its ambiguous

actually yeah

thats probably what they want

come to think of it

@summer geyser

that makes more sense than what i said

i mean arguably Q(sqrt(2)) notation is similarly ambiguous in that we can affix either sqrt(2) or -sqrt(2), but from a ring theoretic perspective it doesnt matter

whereas the distinction between X^2 - 2 and X^2 + 2 certainly matters since it affects its algebraic behaviour

(the same argument applies for X^4 + 4 with a complex factorization)

[summer geyser]

Oh I see, so it's just more confusing to have it as 4^(1/4)?

[scarlet estuary]

well less "more confusing"

let me take back what i said earlier

and ask you this question: how many unique solutions to there are X^4 - 4 = 0?

[summer geyser]

4

[scarlet estuary]

right

now with x^2 - 2 for example, the thing is that these solutions behave algebraically the same

no matter which we affix

if we affix sqrt(2) then we "automatically" get -sqrt(2)

which behaves the exact same

so this multiple-solution thing doesnt cause problems

but when we pick a solution for X^4 - 4

we run into issues, since the solutions behave algebraically different

only two of tehm solve X^2 - 2 = 0

the other two dont

instead they solve X^2 + 2 = 0

so the notation Q(4th root (4)) is ambiguous

are we affixing the solution to X^2 - 2 = 0, or to X^2 + 2 = 0?

which one you pick changes the algebraic behaviours of your ring

[summer geyser]

ahh I see what you mean yes

that makes sense

thanks so much

[scarlet estuary]

i'm pretty sure that's what they're looking for

[summer geyser]

yeah that sounds like it

[scarlet estuary]

basically, whenever we affix a solution to a polynomial with multiple complex (or whatever) solutions

we need to pick one arbitrarily

sometimes this doesnt matter; picking one "automatically" includes the other by closure under addition/multiplication

but in this case, it DOES matter

and the notation doesnt tell us WHICH solution to pick

which is A Problem ™️

[summer geyser]

yeah I see that now

thanks again

[chilly ocean]

If a and b are commuting elements and their product is a unit, must a and b both be units

[next obsidian]

just prove it my man

:)

answer is affirmative tho

[inner acorn]

https://www.youtube.com/watch?v=DJEabykk8j8&ab_channel=Zero-DimensionalSymmetryResearchGroup

Some exciting progress in mathematics on Kaplansky's conjectures

[chilly ocean]

[sturdy marsh]

[chilly ocean]

wow, i am envious of this role

[sturdy marsh]

what is kaplansky's conjecture

[obsidian sleet]

ah okay. So you're saying that my method doesn't cover the cases where i might be able to express (1 2 3 4) as a product of not necessarily disjoint 3-cycles, so it doesn't prove what i want.

[inner acorn]

wait what

okay, that's a role xD

[obsidian sleet]

eevee

[inner acorn]

why lewd tho o.0

[obsidian sleet]

😈 you know

[chilly ocean]

@inner acorn you know what this means, you must now post hentai pics in #chill

[obsidian sleet]

h m m

[inner acorn]

Kaplansky's conjectures basically talk about the group algebra
RG := { finite sums r₁g₁ + r₂g₂ + ... + rₙgₙ : rᵢ in R & gᵢ in G} for R and ring & G a group
and moreover whether it has any zero divisors/idempotents/interesting units

For example you have CZ (you can think of this as the laurent polynomials C[x, x⁻¹]) which is an integral domain
For example you have C(Z/2Z) (you can think of this as C[x]/(x² - 1)) which has zero divisors e.g. (x+1)(x-1) = x² - 1 = 0

Kaplansky basically conjectured for a torsion free group G, and R is the integers or a field, then:
(1) every unit in RG is λg where λa is a unit in R
(2) RG contains no zero divisors
(3) 1 & 0 are the only idempotents

It turns out (1) => (2), and (2) => (3) iirc
I dunno how straight forward (1) => (2) is, but I think (2) => (3) can simply by shown by x² = x implies x(x-1) = 0, and if x is neither 1 or 0 then you have a zero divisor etc...

Finally, the talk I posted is by Giles Gardam, who found an interesting unit in F₂G, where G is the Promislow group
This is a group with presentation < a, b | (a²)ᵇ = a⁻², (b²)ᵃ = b⁻² >

So the unit conjecture is false

[latent anvil]

doesn't that show (3) => (2)?

[inner acorn]

oh wait yeah xD

lemme correct

[median tree]

I'm curious why he specifies that the intersection is a normal subgroup of H and not G. Can anyone give me an example of a case where the intersection of a normal subgroup and another subgroup is not a normal subgroup of G? What about the intersection of two normal subgroups?

[sturdy marsh]

Let G be a group with a non-normal subgroup H

then H cap G = H is not a normal subgroup of G

[median tree]

What about the intersection of two normal subgroups?

Seems like it would be normal subgroup of G

Actually, yes, I think I'll prove this

I suppose I don't' have a very good intuition for what it is that would determine if the intersection of a normal and non-normal subgroup would be normal or not

[chilly ocean]

you could take H a subgroup of N that is not normal in G. say, G=S5, N=A5, H=the subgroup generated by (1 2)(3 4)

[brittle hemlock]

anyone know how to do this?

[median tree]

In the proof of the first isomorphism theorem, or whichever isomorphism theorem you want to call it,

The last part where we say it's surjective since hnN = hN

why is this so?

I understand that this is a quotient with respect to N, but is it just that the extra factor of n just gets divided out? It could be n^2, n^3,... and it would amount to the same thing?

[chilly ocean]

are you asking why n can be gobbled up into the big N?

[median tree]

yup!

[chilly ocean]

so it suffices to find a bijection hnN->hN. an element hnm in hnN is obviously in hN, so the identity map works. now, it must be injective because hnm1=hnm2 implies m1=m2 by simple cancelation. and it must be surjective because hn(n^-1 m) maps to hm

[carmine fossil]

Aka nN=N for all n in N

aN=bN iff b^-1 a is in N(nvm this is a consequence of nN being equal to N)

and (1)^-1n=n is in N

[median tree]

Ah, I see. Because n is in N multiplication by n does not change the group, it's the same group.

[carmine fossil]

To show nN=N you need to show $nN \subset N$ and $N \subset nN$

[cloud walrus]

DrunkenDrake

[median tree]

ah, hence 8da's answer

[carmine fossil]

That works too

[carmine fossil]

That part proves only the first inclusion

[median tree]

Okay, so because N is a subgroup, then it's closed under multiplication, so nN \subset N is the first direction

[carmine fossil]

Note 1=nn'(for some n') by that condition then an element of N,p will be nn'p which is in nN

[median tree]

ah, so you're just taking the identity element in N, and multiplying it by some other element p in N, and then factoring out the n to give nN?

[carmine fossil]

Yea

For the general case,i.e. b^-1a is an element of N,you can say a=bn(for some n in N) and hence conclude aN is a subset of bN and write an^-1=b to conclude bN is a subset of aN

Here a=n and b=1

[median tree]

oh, this is just modular arithmetic. I guess the only lingering question in my mind is how do I know that for example n'p = N? I get that 1*p = p and p is arbitrary, but it seems strange to me to be switching between a specific arbitrary n'p and N so freely

[carmine fossil]

n'p is an element of N

[median tree]

yeah, exactly

[carmine fossil]

nn'p is an element of nN

That just showed if p is an element of N it is an element of nN

[median tree]

ohhh, of course, and we've taken an element of

yeah

Okay, it all makes sense! Thanks for the clarifications and the input everyone!

[vestal snow]

Let $P$ be a place of a function field and let $k$ be a unit element of the local ring $O_P$. Let $u$ be a uniformizer at $P$. Can we say anything meaninful about $\frac{dk}{du}$?

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

It would be really nice if the order of it at P would be 0 (or anything fixed really)

But that seems too good to be true

[chilly ocean]

tfw there are not enough people in the server doing algebraic number theory 😞

[vestal snow]

😦

Too many goomers in this server

I just realized I never defined k

Let $k$ be algebraically closed. Let $F'/k(x)$ be an Artin-Schreier extension fedined by $y^p-y=f(x)$ where $f(x)$ is a polynomial. Let $P$ be the ramified place at infinity with uniformizer $u$. Let $k$ be the element such that $\frac{1}{x}=u^pk$. Can we say anything meaninful about $\frac{dk}{du}$?

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

It would be really nice if $ord_P(\frac{dk}{du})$ equals some fixed number.

[cloud walrus]

Have a Banana, Bitch

[unique juniper]

why does $h_1K = h_2K \Leftrightarrow h_2^{-1}h_1 \in K$

[cloud walrus]

Yes

[carmine fossil]

This

[unique juniper]

dont get it

$h_2^{-1}h_1K = K$

[cloud walrus]

Yes

[unique juniper]

how can we be sure that h^1_2h_1 is in K

[carmine fossil]

That's our starting assumption

[unique juniper]

okekkk

I got it

[fringe lagoon]

Is this correct at all? If not can anyone point out what went wrong? Would i be missing key concepts if i ignore the functor way of defining this? Thanks.

[oblique river]

It's just a definition

I'm not sure what you think could be correct or incorrect about it

[fringe lagoon]

"B's vector space structure shall be ignored" is that good enough to capture that in F(R^2) 2(1,1)!=(2,2) for example?

is span{B} really isomorphic to the direct sum of the field or is it actually equal?

[simple valley]

when we talk about the group of symmetries of a shape A \subst R^n we're talking about the stabilizer of A in specifically SO(n) right?

not O(n)

[viscid pewter]

wild guess in a topic idk enough about but i feel like it'd have to be some weird fractal for there to be things in O(n) that worked that weren't in SO(n)?

[simple valley]

O(n) includes mirror symmetries

for a lot of shapes this is an additional factor of two, and I'm not sure if I'm supposed to be including it or not

[viscid pewter]

oh

wait, right, yeah

[simple valley]

looks like it is O(n) in this context at least

[viscid pewter]

yeah i mean dihedral groups are for polygons and they have reflections so i'd have to guess O(n)

i should not be speaking lmao

[golden merlin]

okayyyyyy

can someone please give me the precise definition of Aut(G) the set of all automorphisms of G

in set notation

i need to check something

[carmine fossil]

An automorphism T is a bijection from G to G,such that T(ab)=T(a)T(b) for all a,b in G

[golden merlin]

yeye in set notation though like {x| x blah blah}

[carmine fossil]

how do you write a function in set notation?

[golden merlin]

I was given this

is this correct?

[carmine fossil]

Yes

[golden merlin]

are you sure???

[carmine fossil]

Assuming S_G is the set of bijections

[golden merlin]

S_G is the symmetric group of G

[cursive temple]

ye

[golden merlin]

the group of all set permutations

[carmine fossil]

So,set of bijections

[cursive temple]

the set of all bijections from G to G is the set of all permutations on G

becaue thats all that bijections are

[golden merlin]

yeye

okay

so if that definition is correct

can u prove that

[cursive temple]

prove what

[golden merlin]

im tryna latex it :P

1 sec

[cursive temple]

okok

[golden merlin]

can u prove that $f_1 * f_2 \in Aut(G)$

[cloud walrus]

Prefix

[carmine fossil]

Just apply the defns?

[golden merlin]

by showing that $f_1(f_2(g_1*g_2)) = f_1(f_2(g_1)) * f_1(f_2(g_2))$ where $g_1,g_2 \in G$ and $f_1, f_2 \in S_G$

[cloud walrus]

Prefix

[golden merlin]

can u prove tht

[carmine fossil]

$f_1(f_2(g_1g_2))=f_1( f_2(g_1) f_2(g_2))=f_1(f_2(g_1)) (f_1(f_2(g_2))$

[cloud walrus]

DrunkenDrake

[golden merlin]

tthis part is right ^^^

but how is this part right

[carmine fossil]

It's the same thing

[golden merlin]

no because f_2 is not an element of G

it is an element of S_G

and by definiton you cant apply the homomorphism to it

[carmine fossil]

f_2(g_1) is an element of G

[golden merlin]

really how

how

i dont understand tht part

[carmine fossil]

See defn of S_G

f_2 is in S_G

Which means f_2(a) is in G for all a in G

[golden merlin]

why does this mean that f_2(a) is in G

[carmine fossil]

What do you get when you apply a set permutation to an element in the set?

[golden merlin]

another set permutation

[carmine fossil]

Some element in the set

[golden merlin]

some element of the set that underlies G

hmmmmmm

so if the group we're talking about is C_3, then the symetric group we're talking about is S_3,right?

they are both groups on the same set {1,2,3}

but S_3 is bigger it is of order six

[viscid pewter]

no, an element of S_3 looks like (123)

right

?

[scarlet estuary]

that's one way to write them

[golden merlin]

yea that is an element of S_3 the GROUP, but not the set

[scarlet estuary]

despite what the notation of S_3 may have you believe, it is not a group on the set {1, 2, 3}

[viscid pewter]

they're not groups on the same set

the group is the set and the operation bundled together

[scarlet estuary]

it is the group whose elements are permutations of {1, 2, 3}

i.e. bijective functions on {1, 2, 3}

[golden merlin]

huuh since when all my life has been a lie

[viscid pewter]

(123) is an element of the set and so an element of the group

[scarlet estuary]

there are 3! = 6 such bijections

[viscid pewter]

S_3 isn't a group on {1, 2, 3}, it's a group on the set of ways that the set {1, 2, 3} can be mapped to itself

+- some rigor i think

[golden merlin]

i thought S_3 was a triangle /\

and triangle has 3 sides

[viscid pewter]

no

[golden merlin]

so i thought that meant it was a set of 3

no?

[scarlet estuary]

S_3 is isomorphic to the symmetry group of a triangle

a triangle has 6 symmetries

3 rotations, and 3 reflected rotations

[viscid pewter]

so C_3 would be just the rotations

there's not too much intuition here, it's all jumbled together a bit

[golden merlin]

:P

[viscid pewter]

ok...?

[scarlet estuary]

these are the elements of $D_3$ which is isomorphic to $S_3$

[cloud walrus]

Namington

[scarlet estuary]

(some text write D_6 instead of D_3 for this group, whatever)

more accurately put, they are the image of a triangle under the symmetries (elements) of D_3

[golden merlin]

so lets get this straight

is C_3 a subgroup of S_3

[rustic crown]

yep!

[scarlet estuary]

yes.

take any element of order 3

the group generated by that element is C_3

[golden merlin]

if G is C_3, then S_G = S_3 ??

[scarlet estuary]

and S_3 has an element of order 3: in particular, the element corresponding to a single rotation

[unique berry]

What's the difference between C_3 and S_3?

[scarlet estuary]

right, the symmetry group on 3 elements is S_3.

[scarlet estuary]

the former is the cyclic group of order 3, the latter the group of permutations of {1, 2, 3}

hence the former has order 3, the latter order 3! = 6

and the former is abelian whereas the latter is not

[golden merlin]

bro

one day ima have to sit down with a giant caylee table

and go through all of group theory intuitively

[unique berry]

Oh ok so S_3 is specifically symmetry whereas C_3 is just the permutation of 3 things

[viscid pewter]

it doesn't work, i've tried

[scarlet estuary]

uh

no

you misunderstand me

[viscid pewter]

you can't get any intuition out of it

[unique berry]

Oop

[scarlet estuary]

S_3 represents permutations, C_3 does not

"the former" means the first element in a list

[unique berry]

Ooh yeah that's what I thiugh

[viscid pewter]

you can watch all the symbols shuffle but the cayley table doesn't actually help

[scarlet estuary]

"the latter" means the second

[unique berry]

Thought

[scarlet estuary]

C_3 is just the set {0, 1, 2} under addition modulo 3

i fyou prefer to think of it like that

[golden merlin]

how so? if i spend all day staring at a caylee table for a big group then im bound to get some intuitions

[viscid pewter]

lol

[scarlet estuary]

S_3 doesnt have such a nice description

if we view S_3 as acting on the set {0, 1, 2}

then its elements are these:

[carmine fossil]

Classify groups of order 101

[unique berry]

Oooh ok

[carmine fossil]

[golden merlin]

okay

is 101 small?

:3

[viscid pewter]

you don't want to work with cayley tables

[scarlet estuary]

1 ↦ 1, 2 ↦ 2, 3 ↦ 3
1 ↦ 1, 2 ↦ 3, 3 ↦ 2
1 ↦ 2, 2 ↦ 1, 3 ↦ 3
1 ↦ 3, 2 ↦ 2, 3 ↦ 1
1 ↦ 2, 2 ↦ 3, 3 ↦ 1
1 ↦ 3, 2 ↦ 1, 3 ↦ 2

oops i wrote 1 2 3

instead of 0 1 2

[viscid pewter]

to really understand a group you want to use like generators and relations

[scarlet estuary]

but you can replace as appropriate

the point is that those are the (six!) permutations on a set of cardinality 3

[viscid pewter]

so like various elements that create the whole group, and the rules about it

[scarlet estuary]

hence the elements of S_3.

[viscid pewter]

imo

[carmine fossil]

Try classifying groups of order 2^282,589,933 − 1

[golden merlin]

wdym by generators and relations

[viscid pewter]

generators are the elements where if you multiply them by themselves and each other, the whole group comes out

[scarlet estuary]

and their inverses.

[viscid pewter]

yes, ok

[golden merlin]

and wat is a relation then

[viscid pewter]

so like any rotation of a pentagon apart from doing nothing will give you all the rotations if you do it enough

a relation is a rule that a generator element or elements have to follow

uhhh what's a good one

[golden merlin]

and wat does that look like

[viscid pewter]

ok so for the pentagon

let r be any rotation that isn't doing nothing

well basically r^5 = e

if you rotate 72 degrees, or 144 degrees, or 216 degrees, or 288 degrees, any of those 5 times you'll get back to the start

so you can generate C_5, the group of rotations of a pentagon, by saying <r | r^5 = e>

[golden merlin]

hmmm interesting

[viscid pewter]

massively faster than staring at tables

[scarlet estuary]

similarly, S_3 has the following presentation:

$S_{3} = \langle a, b, \colon a^3 = 1, b^2 = 1, aba^{-2}b = 1\rangle$

[cloud walrus]

Namington

[golden merlin]

if i spend all day looking at generators for say A_5, then will i get deep intuition for how groups work?

[viscid pewter]

oh, a presentation is just like the generators and the relations together

[golden merlin]

im looking for alternative ways for learning :P

[viscid pewter]

the whole of that angled bracket

<r | r^5 = e> is a presentation

you'll get deeper intuition for A_5

if you use the relations to play around with the generators, hopefully

[scarlet estuary]

im not sure thats the best approach.

i think the best way to learn how groups work is to do exercises

and focus less on specific examples

[golden merlin]

hm probably right

[scarlet estuary]

and more on groups as a whole and how they behave

idk how much i think about, say, A_56

[viscid pewter]

practically anything is better than staring at a table really

[golden merlin]

:P

[rustic crown]

could I ask something 👉 👈 (i didn't wanna stop this discussion)

I needed some help in showing the reverse direction of first part

say f = minimal polynomial of alpha over k

Then we have the exact sequence
0 --> k[x] --> k[x] --> F --> 0
where the map k[x] --> k[x] is multiplication by f

since these are vector spaces over k, tensoring by F is exact, and this gives the sequence

0 --> F[x] --> F[x] --> F tensor F --> 0

hence F tensor F = F[x]/(f)

now say f = product of f_i^a_i where f_i are irreducibles in F

then we can use CRT to break up F[x]/(f) and if f was separable then each a_i = 1 and hence each F[x]/(f_i) is a domain, proving that the product is reduced

how do you go backwards?

if F tensor F is reduced, then each a_i = 1 else we'll have a non-trivial nilpotent element in F[x]/(f_i^a_i) namely f_i and this will give a non-trivial nilpotent in F tensor F

is there a nice way to see why each f_i is separable?

one thing i can see is that if there was one inseparable f_i then all f_i would be inseparable.

[rustic crown]

<@&286206848099549185>

[rustic crown]

Have to sleep now... 😦

[chilly ocean]

maybe sleeping will help you figure it out

[rustic crown]

[sturdy marsh]

the product of the f_i is the minimal for alpha

and alpha is separable

[rustic crown]

no we're doing the other way... given F tensor F is reduced then alpha is separable

[sturdy marsh]

oh lol sorry

I hope the brainfarts stop this week

[rustic crown]

lol issokie

[sturdy marsh]

if some f_i was insep

then it has a repeated root \beta in an algebraic closure

let g be the minimal for beta over F

[rustic crown]

then g is some f_i

[sturdy marsh]

g is f_i ye

[rustic crown]

oh i see

Say E = F(beta)

then E = F[x]/(g) and now break this with CRT

[sturdy marsh]

g is irred over F

[rustic crown]

oopsie

[sturdy marsh]

I wasnt doing anything I was just trying to restate the problem

[rustic crown]

what if i keep going

now look at E tensor E = E[x]/(g)

meh... 2am brain can't think

[sturdy marsh]

this shouldn't be hard, let me try again from the start

assume alpha isnt separable

[rustic crown]

okie

[sturdy marsh]

then f has repeated roots in some algebraic closure

[rustic crown]

yep

[sturdy marsh]

does alpha need to be a repeated root

[rustic crown]

doesn't seem so... apriori

i feel like we need to do some stuff with f(x) = g(x^p) where p is the char of k

but i can't really say f = (something)^p

frobenius need not be isomorphism on k

[sturdy marsh]

ye alpha needs to be a multiple root

[rustic crown]

too sleepy to see why 😦

Anyway arigato! I'll try again tomorrow...

[sturdy marsh]

use the derivative condition

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

[rustic crown]

oh yes

f' = 0

alpha is repeat

[sturdy marsh]

yup

so F[x]/f is not reduced

[rustic crown]

what is the nilpotent you constructed?

x-alpha?

[sturdy marsh]

f factors as g \times (x-alpha)^k

for some k

[rustic crown]

ah makes sense

[sturdy marsh]

k > 1

[rustic crown]

nice

[sturdy marsh]

so (x-alpha)g is nilpotent

[rustic crown]

oh uwu

[sturdy marsh]

the k^th power dies

[rustic crown]

Arigatou!

[simple valley]

is there a name for $\bigcap_{g \in G} gHg^{-1}$ where $H$ is a subgroup of $G$

[cloud walrus]

mniip

[chilly ocean]

conjugate closure of H?

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

[simple valley]

I don't think so

[chilly ocean]

did i forgot how quantifiers work again

[rustic crown]

its the kernel of the action of G on G/H via left multiplication...

[simple valley]

exactly

[rustic crown]

maybe just call it that 😛

[simple valley]

but what is it

my hw question asks what is the kernle of the action of G on G/H by left multiplication

[rustic crown]

ig the expression with the big intersection is enough...

[simple valley]

[rustic crown]

ig not satisfactory

[gritty adder]

wait how is the kernel nontrivial

doesn't every element of G not in H act nontrivially?

[simple valley]

yes

some elements of H will act nontrivially too

[gritty adder]

unless i'm thinking of the wrong group action

[uncut girder]

i think its the largest normal subgroup contained inH@simple valley

[simple valley]

$l : G \to S(G/H)$, $l(g)(g'H) = gg'H$

[cloud walrus]

mniip

[uncut girder]

so its the normalizer maybe?

[simple valley]

it's not the normalizer

I thought so too at first

but the normalizer need not be normal

@gritty adder for H normal, the kernel will be H. For other H the kernel will be smaller

[gritty adder]

I see, I was thinking of the case where H was normal

[next obsidian]

Yea, this is called the core of H

I believe

The “normal” core to be specific

It’s the largest subgroup of G which is normal in H

[simple valley]

well shit

how do I translate this into russian now...

we have the same word for kernel and core

[next obsidian]

https://en.m.wikipedia.org/wiki/Core_(group_theory)

Maybe there’s a Russian version of the page

[simple valley]

there is not

[next obsidian]

Rip nope

[simple valley]

that's the first place I checked

hmm looks like the same word is used

[sturdy marsh]

rip

[warped bane]

$$ Z/abZ ; and ; Z/aZ * Z/bZ$$

[cloud walrus]

Shura

[warped bane]

can anyone tell me why those two groups are isomorphic

was reading a textbook and they just go with "since 'the two groups shown above' are isomorphic then phi(ab) = phi (a)* phi(b)

[gritty adder]

This isn't true unless a and b are relatively prime

[warped bane]

such that phi(n) = card(integers less n and prime with n)

[gritty adder]

if they are, this is the chinese remainder theorem

[warped bane]

@gritty adder yea they are

I'm worried about the first problem

[gritty adder]

you can see it by finding x and y such that ax + by = 1

[warped bane]

why are they isomorphic?

[gritty adder]

there's a natural map Z -> Z/a x Z/b whose kernel is the subgroup generated by ab. so that gives us a map Z/(ab) -> Z/a x Z/ b. so it remains to show that the map is surjective

To prove surjectivity you use ax + by = 1

[warped bane]

i see

[latent anvil]

Just a warning, * means something different than × in group theory

[chrome hinge]

Yup, the notation above is usually used for the free product of groups

[shrewd geyser]

What does R^n mean in the context of R treated as an R module?

[uncut girder]

R is a ring?

[shrewd geyser]

yeah a ring

[next obsidian]

Its just n-tuples

and the action of R is given by multiplying all indices by the thing

so like

a(a_1,...,a_n) = (aa_1,...,aa_n)

[shrewd geyser]

ah ok

[rustic crown]

How exactly am I supposed to go about this? Only way I know how to "compute" Tor1 is by finding a free-resolution. But if the ring is arbitrary then its hard to approach it...

if R was a pid then it would be easy as I = (a)
and we have an easy free resolution 0 --> R --> R --> R/I --> 0 where the middle map is multiplication by a

[carmine fossil]

Here's a better way to phrase that:
Let S_(Omega) be finite then |S_(Omega)|<|S_(n)| for some natural number n

But,S_(n) is a proper subset of S_(Omega) implying S_(n)<S_(Omega)

[sturdy marsh]

there's a pretty nice injective set map \bb{N} \rightarrow S(\Omega)

[carmine fossil]

Ok,I don't think that's very clear

Take Omega and consider the elements of S_(Omega) that fix everything other than the first n elements

This will be iso to S_(n)

So |S_(n)|<=|S_(Omega)|

[ivory cosmos]

(1) (12) (123) (1234) (12345)...

[cloud walrus]

mirzathecutiepie

[ivory cosmos]

You can enumerate a Countably infinite number of elements

So it's at least that big

[sturdy marsh]

n ---> (1n)

[cloud walrus]

mirzathecutiepie

[carmine fossil]

You can have |A|=|B| even if A is a proper subset of B in case of infinite sets

Yea that works,but then you have to show a finite set is never isomorphic to an infinite set which may be a bit of extra work

nvm,it is

"By computation"

(12)(34)=(34)(12) let's say you pick 3
(12) is an identity wrt 3(since it fixes 3) so for a(12),(12)a 3 ends up in the same position a(3)

Apply that logic for every other element

A product of 2 permutations $p_1$ and $p_2$ is basically a permutation p such that $p(a)= p_1(p_2(a))$ for all a in domain

[cloud walrus]

DrunkenDrake

[carmine fossil]

product of 2 cycles is the cycle corresponding to the product of the corresponding permutations

Think of cycles and permutations as matrices and linear operators

[cloud walrus]

mirzathecutiepie

mirzathecutiepie

[lime skiff]

we consider A/I = Z_2[x]/(x^5 + x^2 + 1) and I look at the class [f_1] where f_1 = x^4 + x^2 + 1

I have to determine whether f_1 is invertible or a zerodivisor

if it's invertible, I have to give the inverse

I suppose I need to use the Euclidean algorithm but I admit I have a hard time making sense of the computations because it's a finite field

could someone help me out?

[crisp badge]

Note that you can do the euclidean algorithm bit-by-bit, by matching the highest power of x

[lime skiff]

What do you mean?

[crisp badge]

Well to start off with, subtract x*f_1 from x^5 +x^2 + 1

[lime skiff]

yeah I get:

f = x*f_1 + (x^3+x+1)

then f_1 = x(x^3+x+1) + (x+1)

(x^3+x+1) = x^2(x+1) + (x^2+x+1)

but now I get:

(x+1) = ...?

[crisp badge]

The problem here is that you haven't done that stage of the euclidean algorithm correctly

[lime skiff]

ah

[crisp badge]

Because x^2 + x + 1 has higher degree than x+1

[lime skiff]

ahhhh

[crisp badge]

And here's where you can do it bit-by-bit

You then divide x^2+x+1 by x+1 (with remainder), and add the quotient you got there to x^2 (the quotient you already have)

[lime skiff]

wait so what step is wrong?

[crisp badge]

this one

[lime skiff]

the (x^3+x+1) one?

ok

so here what do I do exactly

I'm not sure I understood what you meant

[crisp badge]

x^2 + x + 1 = (x+1)^2 + x

[lime skiff]

oh I can do that?

huh

but it's x^3 not x^2 right

so (x+1)^3?

[crisp badge]

To get that x^3 + x + 1 = x^2(x+1) + (x+1)^2 + x

[lime skiff]

ah, my bad

[crisp badge]

To get x^3+x+1 = (x^2 + x + 1)*(x+1) + x

[lime skiff]

okay so far so good

and now I have to do (x^2+x+1)(x+1) = h * x for some h

is that right?

[crisp badge]

No

[lime skiff]

huh

[crisp badge]

You have a quotient of x^2 + x + 1

It's also clear that x won't divide the LHS as it has non-0 constant term

[lime skiff]

in the sense that x^3 + 1 = (x^2+x+1)(x+1)?

I've never used the euclidean algorithm for polynomials so I have a bit of a hard time figuring out what I need to do

[crisp badge]

Wait, why are you doing that? Just continue with Euclid's algorithm for 1 more step

[lime skiff]

yeah but isn't that what I wrote above?

and now I have to do (x^2+x+1)(x+1) = h * x for some h

you said no?

[crisp badge]

x+1 = q*x + r

[lime skiff]

ah woops

OH

[crisp badge]

With deg(r) < deg(x)

[lime skiff]

(x^2 + x + 1) is the "h" here

so I get x+1 = q*x + r

okay

so q = 1 and r = 1?

[crisp badge]

Yes

[lime skiff]

okay

now I see

[crisp badge]

So now you can answer the question of "is it invertible?"

[lime skiff]

how do I do that? backtrack my progress?

[crisp badge]

Yes

Well, the "1" you got is telling you that the polynomials are coprime

[lime skiff]

which means it's invertible right

otherwise it would be a zerodivisor

[crisp badge]

Yes

[lime skiff]

since it would divide zero

well, [0]

[crisp badge]

Ehh, more it's invertible because you got a 1, so it's not a zero divisor

[lime skiff]

yeah sorry

[crisp badge]

You can now backtrack to find the inverse, as you would in normal euclid's algorithm

[lime skiff]

err I admit I don't remember how to do this

is it by the factoring we did in the steps?

[lime skiff]

okay I tried substitution

but I end up with f_1(2+x) + f + (x^3+x)

so I guess I messed up somehow

[unique juniper]

anyone read aluffi algebra chapter 0?

[sturdy marsh]

some of it, yes

[unique juniper]

any good?

[sturdy marsh]

yes

[ivory cosmos]

Very good

[next obsidian]

It is good

but the naysayers will tell you it's bad

do not listen

become based

[mild laurel]

hello i am naysayer

[quaint ivy]

how interesting are the exercises compared to, say, D&F?

[scarlet estuary]

bad

aluffi's exercises are piss easy compared to most other algebra texts

if youre gonna read aluffi, supplement its exercises with another source

[sturdy marsh]

D&F + Aluffi is a great combo

[ivory cosmos]

interested in some critiques, since I like the book

[scarlet estuary]

bad exercises, really fluffy text

[ivory cosmos]

what don't you like about the exercises?

[scarlet estuary]

as i said, super easy relative to most other texts

its comparable to something like Pinter except Pinter bestows on more immediate intuition, whereas aluffi's fluff is mostly spent on categorical considerations that realistically dont matter until a student's first alg top course

and category theory is easy enough to pick up when a student actually needs it

[mild laurel]

I don't see the value at all of introducing the category stuff honestly. It was my book to algebra and all the category stuff felt super unmotivated and dry to me

[ivory cosmos]

I'm in the odd position of knowing more category theory than algebra

[scarlet estuary]

you mean youre CS pilled

[ivory cosmos]

unfortunately, yes

so idk, the categorical presentations feel like shortcuts to me

"Oh polynomials are initial objects in R-Alg, got it"

"Oh, free groups are right adjoints, got it"

the exercises being too easy is interesting though

you recommended Dummit and Foote?

[scarlet estuary]

d&f has a ton of exercises so if thats what youre after, it works as a supplement yes

but honestly any common algebra book should have exercises worth doing

for what its worth i dont like, hate aluffi

i think its approach makes sense in a vacuum

i just havent seen it actually work out that well

[ivory cosmos]

yeah, I mean, the only thing that personally worries me is the idea that its exercises are too easy

[scarlet estuary]

i mean meh

i think i mightve exaggerated a bit too much

[ivory cosmos]

I can see why the categorical approach might not vibe so well, but that's not the case for me

[scarlet estuary]

in my explanation

the exercises arent like

trivial

theyre fine

[sturdy marsh]

spawn point for ugcts

[scarlet estuary]

there's just a bit of a dearth

of actually difficult ones

they all sit around the "moderate" difficulty level

[ivory cosmos]

that is kind of true

[scarlet estuary]

if that makes sense

[ivory cosmos]

yeah I get you

[scarlet estuary]

so if you idk

[ivory cosmos]

they're never all that difficult

[scarlet estuary]

grab another text and just do like

the hardest 2-3 exercises of each chapter

should be okay

so yeah im not trying to say aluffi is like EASY

[ivory cosmos]

my usual approach is that I try proving statements in the text before reading them

[scarlet estuary]

it just spends a lot of time building up really sophisticated machinery and then not actually making great use of it

[ivory cosmos]

so that gets me bashing my head against the wall

[scarlet estuary]

oh if youre doing that, thats good

[ivory cosmos]

which is a necessary component of the learning experience

ime

[ivory cosmos]

how much is that really the case though?

Like, you only get to functors in chapter 8

[scarlet estuary]

well thats kind of my point

[ivory cosmos]

you do have a lot of category theory in the chapter on groups

[scarlet estuary]

it spends a lot of time setting up the bedrock for full-on category theory

but doesnt actually DO full-on category theory

for quite some time

[ivory cosmos]

like, showing that products and coproducts are the same for abelian groups

yeah I can sort of see that

but there are gems littered throughout

like defining the characteristic through Z being initial

I liked that

or that Z module structures are unique

because Z is initial

It could have gone further though

like you can define R-Mod as a functor category

with rings seen as single object abelian categories

[scarlet estuary]

yeah i mean my point is that aluffi's category theory is mostly spent just to define things nicely

rather than to let you do really powerful thigns with it

since the exercises never really require super novel insights

[ivory cosmos]

does CT ever really let you prove powerful things?

[scarlet estuary]

honestly, in introductory abstract algebra? not really

[ivory cosmos]

I feel like the total amount of leg work is constant

[scarlet estuary]

but a ton of arguments in alg top are category theoretic

[ivory cosmos]

CT just lets you spread it around

[scarlet estuary]

thats kinda why aluffi's approach is... weird

[ivory cosmos]

where the hard work is diluted to proving simple facts about sophisticated objects

[scarlet estuary]

in the sense that like

okay

it's certainly more approachable if youre already familiar with category theory

dont get me wrong there

but if not, it feels like theres a lot of emphasis spent on making the DEFINITIONS themselves as nice and clean as possible, rather than what you actually do with them

some people might describe this as "elegant" or "the most natural perspective" or whatever

[ivory cosmos]

yeah I can sort of see that

[scarlet estuary]

but personally i see it as "overengineered" and "focusing on definitions rather than mathematics"

ymmv though

[ivory cosmos]

but that kind of dichotomy is present throughout math though right?

frogs versus whatever

like, is math about problems, or is math about nice objects?

[scarlet estuary]

well obviously mathematics is about reasoning from definitions

its just a matter of where you place the focus

and i think aluffi's focus is a bit too... "far back"

at least for my preference

i dont think its a horrible text or anything though

[ivory cosmos]

Well yeah, I mean more so like, should the development of mathematics serve to provide tools to solve hard problems, or to develop nice theories

[scarlet estuary]

like im being really negative about it

in the way i talk

but dont think im saying like

[ivory cosmos]

the CT approach is generally more so in the "nice theories" realm

[scarlet estuary]

if you read aluffi youll never undersatnd algebra

or whatever

im just pointing out my personal criticisms

it's not as bad as my constant negativity may make it sound

[next obsidian]

one criticism I see

is ppl saying he builds up a lot of category theory for no reason

Like... he doesn't really build it up that much until the last two chapters

he introduces words

but it isn't that heavy, you won't become UGCT from anything in all but the last 2 chapters

[scarlet estuary]

chmonkey thats what i was saying earlier

[next obsidian]

Oh

[scarlet estuary]

he gives you the foundations for category theory but never actually does much with 'em

[next obsidian]

Tbh I read the start

then went through

you see tho like

I don't see an issue with that

Like... I found it easiest to have that mindset

it made jumping into hartshorne easy

It's not like I had to rewrite the foundations I had built all my thinking on

¯_(ツ)_/¯

Also, tbh I haven't donea lot of exercises from places other than the last two chapters

but I felt pretty happy with those, it introduced a lot of stuff I felt was necessary. It wasn't exhaustive, but it's pretty solid

the only thing it doens't approach is like Commutative Algebra or rep theory, but I think the former is better learned from a CA book

and the latter... well I suck at it so idk

But idk, maybe intro algebra isn't supposed to exist as a funnel into AG

[ivory cosmos]

well, isn't "Algebraic X" usually pretty categorical, at least nowadays?

[next obsidian]

yeah

I mean the two main examples are geometry and topology

for which the answer is yes

[ivory cosmos]

fwiw, the "Group Theory" course at my school uses Category Theory heavily as well, as do many other undergrad math courses

but that's because of the preferences of the teachers 😛

[next obsidian]

My class definitely did not

[latent anvil]

shoutout to Julia being like "he hasn't introduced functors yet??"

[next obsidian]

yeah

[ivory cosmos]

lol, the EPFL group theory course defines group actions as Functors

based

[next obsidian]

what is that??

EPFL

[ivory cosmos]

https://www.epfl.ch/en/

[latent anvil]

Oh I think that's one of the universities I was looking into studying abroad at

Pre covid

[next obsidian]

Switzerland

[latent anvil]

Yeah it is

[next obsidian]

huh

[ivory cosmos]

One of Switzerland's two polytechnic schools

[latent anvil]

I visited the campus last Christmas

[ivory cosmos]

wow they define group actions in the general context of a category

not just Set

based

And they define products on G-objects through the existence of a forgetful functor, and the fact that right adjoints preserve limits

which uuuh

that's like, way more categorical than aluffi ever gets