groups-rings-fields

[sturdy marsh]

and something funny happens when G = H

and if you tensor over Z

then Z[G] \otimes _Z M with the left G action is isomorphic as G-modules to Z[G] \otimes _Z M with the diagonal G-action

I thought this was pretty weird when a friend pointed it out

but this seems to be an instance of a general fact about hopf algebras

[stark sigil]

This smells analogous to the fact that any G-torsor trivializes itself

[sturdy marsh]

the proof is that you have this shearing isomorphism

which sends g \otimes m to g \otimes g^{-1}m

or something like that

[wraith obsidian]

Is there a way to translate this in a purely module-theoretic context? Like with S subs. R
Or does this phenomenon work only for group modules

[sturdy marsh]

you definitely want something that looks like a group ring

[sturdy marsh]

otherwise this wouldnt make too much sense

[wraith obsidian]

Ah, right, else e.g. doubling on the left vs. Doubling on both sides cannot be the same

[stark sigil]

My B \otimes_A B thing shows up here in disguise!

[wraith obsidian]

When we look at Z acting on things eg

[stark sigil]

$\bZ G\otimes_\bZ M=(\bZ G\otimes_\bZ \bZ G)\otimes_{\bZ G}M$

[cloud walrus]

Icy001

[stark sigil]

I believe $\bZ G\otimes_\bZ \bZ G\cong \prod_G\bZ G$ too (nope this isn't true)

[cloud walrus]

Icy001

[stark sigil]

I don't know where I'm going with this by the way

[sturdy marsh]

lol

but yeah, it has got something to do with the fact that Z[G] is a hopf algebra

with comult g to g\otimes g

and antipode g to g^{-1}

[stark sigil]

Hmm I'd have thought that it being a Hopf algebra gives the reason for why $\Spec \bZ G$ is a group scheme

[cloud walrus]

Icy001

[stark sigil]

I can't immediately see how co-stuff proves the original thing

[sturdy marsh]

the antipode is clearly involved

and maybe that's all you really need

[stark sigil]

The map looks like $g\otimes m\mapsto g\otimes gm$

[cloud walrus]

Icy001

[sturdy marsh]

it's a g^{-1} on the right

[stark sigil]

From the module with only a left G-action to the module with the diagonal G-action

I checked h times that for both sides and it works though

The inverse would have g^-1 maybe

[sturdy marsh]

yeah that works

but you need a g^-1 for it to be an iso

[stark sigil]

and you do have it since ZG is the group algebra for a group G and groups have inverses (!)

[sturdy marsh]

yes

[stark sigil]

Hmmm

What is this geometrically

Pullback definitely somewhere

Well Spec ZG only makes sense when G is an abelian group

that's a slight bummer

I still have a possible interpretation in terms of [*/G]

[latent anvil]

iirc affine group schemes/k are dual to commutative hopf algebras/k, so you're smelling something right

[wraith obsidian]

Hopf algebras are a mystery I'd also like to understand at some point :> glad that it actually seems to be something people talk about

[next obsidian]

What the fuck

[golden pasture]

iirc i saw that before

isnt it kinda

by an appropriate definition

[sturdy marsh]

it pretty much follows from the definition

[oblique leaf]

This might be a stupid question but if n|2m and m<n, why does n have to be even?

[chilly ocean]

so n|2m means kn=2m right? for some k

[oblique leaf]

yeah

[chilly ocean]

then m<n implies 2m<2n

so kn<2n

which implies k=1

so really n=2m

[oblique leaf]

oooh got it

tysm!!

[rustic crown]

(or if you like, if n was odd, then n|2m => n|m)

[cyan marten]

If I have an ideal J, how does J tensor R/J look like?

0?

[latent anvil]

not zero in general

Think about the case of Z

[next obsidian]

It's J/J^2 yeah?

Pretty sure M (x) R/I = M/IM

[latent anvil]

Yeah, that makes sense to me

so almost never zero

I was thinking of how J ≈ Z in Z

So it'll be R/J always

(but this isn't true in general rings)

[next obsidian]

Aluffi Corollary 2.8 in chapter VIII

[latent anvil]

wait what

Isn't this just what you said??

[next obsidian]

The formula I said

[latent anvil]

Oh lol

[next obsidian]

Juggabalooga

[cyan marten]

Yeah, but I didn't know how to "restrict" it

Ohh nevermind

[cloud walrus]

randleS

[chilly ocean]

what does ptr mean?

[restive star]

oops

supposed to be $p^t*r$

[cloud walrus]

randleS

[restive star]

and I believe t is just a positive integer

any ideas?

[chilly ocean]

maybe "pth roots always exist in a field of characteristic p"?

er

no, maybe this is only true for finite fields

[restive star]

Yeah I don't believe that's true generally

I'm very confused about this problem

[cyan marten]

Let F be the field F3(x), and consider the extension K of it by adjoining a square root of x.

Oh, nevermind. I was hoping to construct a clever counterexample

[mild laurel]

hm, not entirely sure if the details of this works but

you have that a satisfies the polynomial f(x) = x^m - a^m, which has coefficients in F

since m = p^t * r, this factors as (x^r - a^r)g(x^p)

So g here is not a separable polynomial, but since the extension is separable, this must mean that the minimal polynomial of a divides (x^r - a^r)

and this should imply that a^r is in F I think

[restive star]

I'm sorry I don't quite get the final step

[mild laurel]

yeah I'm not sure it works

[restive star]

I think it uses some weird property of the characteristic, but even looking through my notes from the class I can't find it

[cyan marten]

I think it does

Formulated slightly differently: If a^r is not an element of F, but a^pr is, then F(a^pr) is obtained by adjoining a pth root, which should not yield a separable extension.

(We can take t = 1 for the moment)

[cyan marten]

Then F(a^r) is...*

[rustic crown]

(can we not say that [K:F] divides deg(f)!)?

[restive star]

hmm

maybe?

Is that necessarily true if it's reducible?

[rustic crown]

okay ig thats wrong then

ig this works

so say a1, a2, ..., ap are the roots of the f in K

and we know K = F(a1, a2, ..., ap)

What can we say about the degree [F(ai):F]?

the idea is [K:F] divides the product.. hence p has to divide one of [F(ai):F]

This means degree of minimal polynomial of ai, is atleast p. but the minimal polynomial has to divide f, which shows f is the minimal polynomial, which in turn means it is irred.

i hope this is right... i need to confirm that [L(alpha):L] divides [F(alpha):F] if L is an extension

[restive star]

hmm from what my prof said I think she expected us to do it roughly the other way

assuming it's reducible, then showing some contradiction comes when considering the fact that each of the components have to divide it

[rustic crown]

okay that works too, and maybe it also avoid the thing i'm not sure of...

So say f was reducible...

and again a1,...,ap are roots of f in K

Now look at [F(a1...a_{i-1}, a_i):F(a1,...., a_{i-1})] <= [F(a_i):F] < p

but the product of the things on the left is precisely [K:F]

but none of the factors are divisible by p... and so is the product

(for the second part... could you remind me what did separable 'over F' meant?)

[next obsidian]

I think there’s a more succinct way to do this without roots. Separable means that it has no repeated roots

[restive star]

^

[next obsidian]

Suppose f = gh, with g,h having deg < deg f

So they have degree < p

[restive star]

(to the second part. your proof is better than what I could come up with haha)

[next obsidian]

Let F be the splitting field of f

You can get to F by looking at the splitting field of g over K, then taking the splitting field of h inside that

So you can write [F : K] = [F: L][L: K]

L is the splitting field of g over K

Then both [F:L][L:K] cant contain a factor of p

Because they have to divide deg(h)! and deg(g)!

But both those degrees are < p since we assumed f wasn’t irreducible

[rustic crown]

nice

wait you sure they have to "divide" deg(h)!

[next obsidian]

Yes

Or err

Yes you need to divide that

Because maybe h reduces down a bit inside of L

But either way the degree of the thing you take should be <= deg(h)

And so this is like saying n! < m! If n < m... I think?

Let me think a second haha

Maybe not a strict divide actually

It might split weirdly and you keep just kind of going through one more splitting field in the irreducible components

But at the very least, at each step the poly has degree < p

So a factor of p could never appear

[rustic crown]

yea

[next obsidian]

So I feel like for the other part you want to show inseparable => t = 1 so that the extension is deg p

Since f is irreducible, this says the splitting field is generated by one of the roots

Idk why that would be tho I’m bad at field theory

[rustic crown]

yea that part is easy... because irred, insep => derivative = 0

[next obsidian]

Uhhh

Oh huh

Then how does that say it’s deg p?

Let me use my Chmonkey brain

[restive star]

p is assumed to be prime if that helps

[next obsidian]

We’ve been using that the whole time hahaha

[restive star]

lmao oops haha

[restive star]

I hadn't looked at what you wrote and so I was responding to this. thought you meant one of p's factors

[next obsidian]

No haha

[rustic crown]

so derivative can't be 0 if characteristic wasn't p. And if char is p, then the polynomial is forced to look like x^p - a. now just adding one root is enough to get the splitting field, as if b was the root, then this splits as (x-b)^p

[next obsidian]

Does irreducible plus inseparable imply f’ = 0 as a polynomial?

I thought f and f’ sharing a root was equiv to having a repeated root

Or something like that

[rustic crown]

yea because insep means that f and f' have a non-trivial factor but since f is irred that can't be true

(unless f' = 0)

[next obsidian]

How do they share a non-trivial factor? Is this following from them having the same root?

[rustic crown]

gcd(f, f') would be a polynomial over the same field, and it will have that repeated root

[next obsidian]

Ummm hmmmm

So if we calculate the gcd up in the splitting field I see why it must have that root

But I don’t see why the gcd in the base field has it

This might be me being dumb

Oh wait nvm

[rustic crown]

yea this is a bit awkward, but the way to find gcd doesn't depend on the base field... like a hand wavy argument would be to say doing euclidean algorithm in both fields should give the gcd...

[next obsidian]

Hmm

I was thinking something like

Look at the min polynomial

Actually this suffices

As long as you know the min poly divides any polynomial of which this thing is a root

Since dividing f’ means it has deg < p

Then it also divides f

So as long as f’ isnt 0 then you get some non-constant poly dividing f and it has deg < p

[rustic crown]

yep, this is nice too

[next obsidian]

I see yeah

Now it’s x^p - a

And then it’s generated by like a^1/p*primitive root of unity

[rustic crown]

yea, just one root is enough to do all the work

because char is p

[next obsidian]

Actually yeah what you said about it factoring

[rustic crown]

and we can do weird things like (a-b)^p = a^p-b^p

[next obsidian]

Yeah, that’s preferable

I don’t like “root of unity”

Since I forget if those even exist

Lmao

[rustic crown]

lol

[next obsidian]

I see I see

Field theory rip

[rustic crown]

yea ig in this case the only pth root of unity is 1

[next obsidian]

Right

x^p - 1 = (x - 1)^p

[rustic crown]

btw are those 2 things true? [K:F] dividing deg(f)! or [L(a):L] dividing [F(a):F]?

in the first one K is the splitting field of f in F[x] and in the second one, L is just a field extension of F

these things 'feel' true but i can only easily see the "inequality" instead of "divides"

[next obsidian]

I thought it’s divides

But maybe it’s <=

I was thinking about it and couldn’t convince myself of it either like

I think it could go like...

6,2,2

I guess that still divides 6! haha

[rustic crown]

proof by lack of counterexamples

[next obsidian]

I don’t think it divides

Haha

I think I just made that up LOL

I think it’s true for deg 3

Maybe for deg 4?

By like size concerns

[rustic crown]

its true if the extension is galois

[next obsidian]

Right

Yeah I’m almost certain it’s false

But you’d need to go up to not so small degrees to find one I think

[restive star]

alright thanks so much for the help!

[rustic crown]

[rustic crown]

2nd is definitely false lol... take a = cbrt(2)*w, F = Q, L = Q(cbrt(2))

[old hollow]

What does it mean for something to "act on" something else

[rustic crown]

ig k bar is algebraic closure of k and G_k is Aut(k bar/k)?

[delicate bloom]

yeah, each of the x_i in k are sent somewhere by some isomorphism in G_k is all

[mild laurel]

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

[old hollow]

Oh thank you, that's what I needed

[cloud walrus]

vov&sons

[south temple]

Is this correct reasoning for determining size of orbits?

oh, that should say |Stab(x)| != 49, not 40

[mild laurel]

No, a nontrivial action could act trivially on one of the x's

It just can't act trivially on all of the x's

So it's possible that |Stab(x)| = 49 for some x, just not all of them

[south temple]

hm, but we know that orbits are disjoint sets right

so the other possible sizes for orbits are 1 through 6

but only 1 and 7 divide 49, so this forces the other orbits to also be 1

contradicting that the action of G is nontrivial

[mild laurel]

Yeah, that works

[south temple]

also no orbit has size 49 lol

since there aren't 49 elements in V

so orbits necessarily have size 7

[unique juniper]

Dont exactly understand this

I dont see how both of those cases are the only two possible cases

[carmine fossil]

p either divides index of subgroup or it does not

[wraith obsidian]

I'm curious now. In what context are you reading this? Seems unusual to do this before learning about group theory.

[fading wagon]

a shorter proof is using group actions

[rigid cave]

Hello! I have a very short question. Is every polynomial of the form $x^5 + ax +b$ solvable by radicals? I have tested a few polynomials and all my examples have been not solvable by radicals

[cloud walrus]

おねえちゃん

[rustic crown]

if a = 0 or if b = 0, then it clearly solvable by radicals... there are example such as a = -5, b = -1 where this isn't the case.

[rigid cave]

Okay, but if a or b are not zero and the polynomial is irreducible?

[rustic crown]

I'm not too sure... but i remember this result that if the polynomial has 3 real roots and 2 complex roots, then we would get that the galois group contains a transposition (which is the complex conjugation) and it would also contain an element of order 5. This would be sufficient to conclude that galois group is S5 and that isn't solvable...

(in general degree a prime p, p-2 real roots, and 2 complex roots)

[rigid cave]

Yeah I heard about this as well. But what if x^5 + ax + b has exacly one real root then? Then the rest four complex roots are conjugates, so the Galois group contains two transpositions and an element of order 5. Does this imply that the Galois group is S_5?

Well if one transposition and an element of order 5 is sufficient to show that the Galois group is S_5, then the case where we have two transpositions must also imply this, right? Or am I missing something?

[rustic crown]

no so you don't get 2 transpositions... you get a product of 2 transpositions

if I label the roots 1, 2, 3, 4, 5 then and say 2-3 and 4-5 are the complex conjugate pairs

so conjugating is same as the permutation (1)(23)(45)

you probably just need to check if a 5 cycle and the permutation (23)(45) can generate something of order less than 60

[rigid cave]

But if 2-3 and 4-5 are conjugate pairs, then isn't (23) and (45) permutations in the Galois group?

[rustic crown]

but the conjugation flips both 2-3 and 4-5 simultaneously

[rigid cave]

Why simultaneously? Why can't I just conjugate 4 to get 5 and just stop there and not conjugate 2 to get 3?

[rustic crown]

so lets say F = Q(r1, r2, r3, r4, r5) is the splitting field of f(x)

you're basically saying that you can find an automorphism T that fixes r1, r2, r3, but T(r4) = r5 and T(r5)=r4

and I'm saying, look, we know one automorphism which is complex conjugation which swaps r2-r3 and r4-r5

[rigid cave]

Oh yeah, that's true. So the conjugation is of the form (23)(45)?

[rustic crown]

yep

i'm not an expert at this... so there maybe a lot of people who have something nice to say

[rigid cave]

So the question now becomes, does a permutation of order 5 and a "double transposition" (like (23)(45)) generate S_5?

[rustic crown]

(you can say that permutation of order 5 is a 5-cycle)

[rigid cave]

Oh okay, so something like (12345)?

[rustic crown]

there are a lot of 5-cycles ... so idk what to say

[rigid cave]

Yeah I know, but something of the form (abcde)?

[rustic crown]

yea

like in this the group is A5... so still not solvable

[rigid cave]

So it generates a group of order 60?

[rustic crown]

the specific (12345) does

Maybe we can just run through all 5-cycles and see if we ever get a order smaller than 60

[chilly ocean]

I proved the formula using observation and applying it in general

I didn't use galois theory

I just use the same observation and used inclusion and exclusion

for arbitrary number of sets

[rustic crown]

but this is bound to fail

in the sense that we already know that 5th roots of unity have 4-complex and 1 real root

ah but that is not irred

lesse what happens

[hot lake]

but their galois group doesn't have a 5-cycle

[rustic crown]

okay already failed

[rigid cave]

But by cauchys theorem there is an element of order 5 in the Galois group of x^5 + ax + b

[hot lake]

isn't that the dihedral group of order 10

[rustic crown]

like if all were >=60 we could have surely said that the group is always in solvable

[rigid cave]

Man, that's a shame...

It really felt like we were on to something

[rustic crown]

now that you say... why did this give 60?

[hot lake]

because that was A5

[rustic crown]

its a flip and a rotation

shouldn't that also generate D10?

[hot lake]

no

why do you say you can think of (23)(45) and (12345) as a flip and a rotation

[rustic crown]

imagining a pentagon labeled with 1, 2, 3, 4, 5

oh oops

to make it D10 it should be labeled something like 1, 2, 4, 5, 3

[old hollow]

I’m reading a brief intro to algebraic geometry. The screenshot I sent was something similar online

[rustic crown]

yea but the problem with this is that even if we get small cardinalities, that wouldn't guarantee that the corresponding group is indeed a galois group of something. another problem is that we only know that the galois group contains atleast these many elements

[old hollow]

I haven’t taken a course covering group theory or anything

So I have a lot of holes in my knowledge of this stuff

[rigid cave]

yeah that's true. But maybe there's some sort of restriction somewhere (for example a restriction of the element of order 5) so that the 5 cycle and the double transposition generates a group of order greater or equal to 60

It really feels like x^5+ax+b, with only one real root, should be unsolvable by radicals

[hot lake]

doesn't x^5-1 only have one real root and is solvable by radicals ?

[rigid cave]

but when a and b are not zero

and the polynomial is irreducible

That really feels like some sort of theorem

[rigid cave]

Okay never mind. I just read about quintic polynomials on wikipedia and they present the polynomial x^5+625/4x + 3750 which is solvable with a and b not zero, one real root and is also irreducible...

[kindred mist]

Does anyone happen to know this? https://math.stackexchange.com/q/4063444/677593

I am trying to compute Z(G) where G is the (nonoriented rubiks cube group)

[wraith obsidian]

As the comment says, it's far more efficient to just calculate the effective permutation (i.e. permutation of the stickers) the superflip corresponds to and then check that this permutation commutes with each generator

that can even be done visually, since you don't need to know how the superflip is composed as a term, but what it does all at once

[solar wyvern]

what does it say about the ring R and a proper, nonzero R-submodule I (that is, an ideal) if I²=I?

[chilly ocean]

Hm, that feels almost illegal

Maybe an idea: this sound like it would be true for I a minimal ideal (do these things usually exist? Something something artinian ring?)

[wraith obsidian]

Does this work if the ideal is finitely generated?
PID should be out of the question due to (r)² = (r²) if I'm not insanely stupid

[wraith obsidian]

This does not have to be the case if we look at idempotents btw
Take ℤ³ and the idempotent (1,1,0) for instance, it certainly is not minimal
Of course we're not talking about domains anymore in that case

Also, appearently the following holds: https://math.stackexchange.com/q/255885/91103
(Finitely generated idempotent ideals in a commutative ring are generated by an idempotent)

[next obsidian]

It says R can't be local, for one (I guess this is only in the case I is f.g.). I think you need a pretty messed up ring for this to ever be true, it's certainly rare

[next obsidian]

I did this exercise lol

[solar wyvern]

maybe cursed amendment, but what if I replaced ring with associative unital local algebra

[next obsidian]

isn't that a special case of a ring?

oh

cursed

[solar wyvern]

context: I was wondering when Ω_{B/A} = I/I² is trivial for a map of rings A→B

[cyan marten]

This exercise in Atiyah-MacDonald is probably not very useful, but is interesting: R is absolutely flat (that is, every R-module is flat) if and only if all principal ideals are idempotent.

(If this holds, then every f. g. ideal is principal).

[chilly ocean]

In invariant theory, say we have a group action $G$ on a vector space $V$ with dimension $n$), we can identify each $x_i$ in $F[x_1,\dots,x_n]$ with $(0,\dots,1,\dots,0)$ in $V$ (I think choice of basis doesn’t matter), and this extends to a unique mapping from $F[x_1,\dots,x_n]$ to itself, which corresponds to $(gf)(x)=f(g(x))$. This to me feels like the more natural definition than the usual $(gf)(x)=f(g^{-1}x)$ for $g\in G$ and a polynomial $f$. My question is, is the reason why we define it to be $g^{-1}$ because this will result in a group action and the theory stays the same (in the sense that same group action will result in the same invariant using either definition) ?

[cloud walrus]

tina

[oblique river]

What you described is not a group action unless G is abelian

At least, the formula (gf)(x) = f(g(x))

[chilly ocean]

https://youtu.be/RYPt7kGdo7s i’m watching this video btw

[oblique river]

Try to look at how a product gh acts on some polynomial f

[chilly ocean]

Yeah that’s why I said mapping instead of action

my question is is it because reason we define it to be g inverse

[oblique river]

Oh, sorry — in that case I think that’s your answer, yeah

We want it to be a group action because then we can do things with the group

[chilly ocean]

Yeah makes sense

plus this results in the same theory right?

[oblique river]

Otherwise this function doesnt respect the group operation and so it might as well not even be a group

I dont know all the details but yes it should, in general these kinds of things are more just for “bookkeeping”

[chilly ocean]

oh wait not necessarily

but thank you!

hang on

so let $(f\times_1g)(x)=f(gx)$ and $(f\times_2g)(x)=f(g^{-1}x)$, then we have $f\times_1g=f\times_2g^{-1}$, which means that if a polynomial is fixed under $\times_1$ by all $g$ then it will be fixed under $\times_2$ by all $g$, and vice versa. This is what I meant by it will result in the same theory

[cloud walrus]

tina

[chilly ocean]

@oblique river

[oblique river]

Yeah that seems reasonable to me

Also im going to bed now — sorry for the abrupt ending!

[chilly ocean]

it's fine

all my questions are answered now

thank you!

[oblique river]

Np!

[unique juniper]

is there any general way of finding sylow subgroups of any group?

[cyan marten]

I am not sure what do you mean exactly by "finding," but factorization is a special case of this problem (take Z/nZ).

[unique juniper]

ok

Let $G$ be a group of order 294. Show that G has exactly one Sylow 7-subgroup.

[cloud walrus]

Yes

[unique juniper]

So 294 = 2 3 7^2

and then

49 divides 294

i dont know how to show that the number of left cosets the normalizer of P makes with G is 1 tho?

so $n_7 = [G : N_G(P)] = 1 $ mod 7

why does it need to divide 6 ?

[carmine fossil]

That's part of Sylow theorems

[unique juniper]

OH

ok

sorry

missed that last part at the bottom

[shut halo]

Does anyone have any idea why the statement in the exercise here is true?

[chilly ocean]

wdym why it's true

[shut halo]

I mean why is $\mathbb{A}^1(\mathbb{C})$ is irreducible.

[cloud walrus]

snypehype

[chilly ocean]

well, suppose it were reducible, then 𝔸¹_C = V₁ ∪ V₂ for algebraic sets V₁, V₂. then, one of the V must be infinite

now what do you know about the number of roots of a polynomial in C[x]

[shut halo]

They are finite. Oh I see!

[chilly ocean]

yeah

[shut halo]

This wouldn't work in $C[x,y]$ right?

[cloud walrus]

snypehype

[chilly ocean]

yeah it wouldn't

[rustic crown]

(you can prove that an algebraic set is irreducible if and only if its ideal is prime)

[shut halo]

I see, I think at some point my book will introduce this so for now I want to stick with more elementary methods

[plucky flicker]

Hi. Can someone give me an example of a ring R and a subset X, such that the ideal generated by X is not the same as the subring generated by X

[rustic crown]

R = Z, X = {2}

(i hope that subring generated by X, means that smallest subring containing X, also that your rings contain 1)

[plucky flicker]

yes its the smallest subring containing X

[rustic crown]

(so the argument is subrings must contain 1... but then it would be all of Z)

[oblique river]

I think an easier example that doesn't rely on "do you contain 1 or not" is R = Q, X = {1}

the ideal (1) is all of Q

but 1 generates the subring Z

[rustic crown]

ooh nice

[old hollow]

That guy is good

[plucky flicker]

i guess it's okay, cuz there is no such a condition that 1 shouldn't be contained

thanks a lot guys 🙂

[unique juniper]

If $a^n = 0$, $aaaaa\cdots aaa = 0 \Rightarrow (aaaaaa\cdots aaa)a = 0$

[cloud walrus]

Yes

[unique juniper]

so we do the same thing to the inner part of the brackets

until it itself is 0

and then call it b

then b^2 = 0

like that?

idk

seems weird

[chilly ocean]

hmm?

[unique juniper]

uh ok

so a is not 0

so Xa = 0 implies that X is 0

[small bison]

only true if you have no zero divisors

[unique juniper]

oh

[chilly ocean]

isnt X that element?

[unique juniper]

what element

[chilly ocean]

b

[unique juniper]

i thought so?

is it?

idk

[chilly ocean]

oh ok ye what young smasher said tho

[unique juniper]

well

[chilly ocean]

Sup

[unique juniper]

we dunno about zero divisors

[small bison]

do you know what a domain is?

[unique juniper]

what kind of domain

[small bison]

integrla domain

[unique juniper]

yes

[chilly ocean]

Someone know about Octornion number?

[small bison]

yeah so you can only make the conclusion that Xa = 0 \implies X = 0 in an integral domain

for example you can find nonzero matrices which multiply to 0

[unique juniper]

i see

[small bison]

but you're on the right track for the problem

i'm not sure if the problem is stated correctly but i think it means that you should assume that a^k \neq 0 for all 1 \leq k < n

[chilly ocean]

[unique juniper]

i see

[cloud walrus]

Bany, um futuro phideas

[chilly ocean]

Here

[unique juniper]

thanks lol

but then

just let b = a^n ???

[chilly ocean]

a^n-1 I guess

[unique juniper]

a^n-1 isnt 0?

[small bison]

a^n doesn't work cause a^n = 0

[unique juniper]

oh yes

ok

ok

$a^n a^n = a^{2n} = (a^{2n-2})(a^2) ~If ~we ~let ~b = a^{n-1}, b^2 = (a^{2n-2}) = (a^n)(a^{n-2}) = 0a^{n-2} = 0$

[chilly ocean]

in integral domain I guess, yes

[cloud walrus]

Yes

[unique juniper]

isnt it done generally

without being in integral domain

because

[chilly ocean]

ok actually its fine if you added that for all k <n a^k isnt zero

[unique juniper]

yes i added the restriction

[chilly ocean]

@unique juniper how you put a white background

[unique juniper]

,ti

[cloud walrus]

The current time for Yes is 10:03 PM (GMT) on Tue, 16/03/2021.

[unique juniper]

hm

,help

[cloud walrus]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[unique juniper]

i forgot

,ti set black

[cloud walrus]

Couldn't find a member matching set black!

Mar 16, 2021, 10:06 PM

[unique juniper]

,black

[proud bear]

,texconfig colour white

[cloud walrus]

You have switched to the white colourscheme.

[proud bear]

$x^2=1$

[unique juniper]

hm yes

@chilly ocean

,texconfig colour grey

[cloud walrus]

You have switched to the grey colourscheme.

[unique juniper]

$x$

[cloud walrus]

Yes

[unique juniper]

,texconfig colour light

[cloud walrus]

You have switched to the light colourscheme.

[unique juniper]

$x=1$

[cloud walrus]

Yes

[unique juniper]

,texconfig colour grey

[cloud walrus]

You have switched to the grey colourscheme.

[chilly ocean]

Hmm

Thx

[chilly ocean]

hi guys, i posted my question on linear algebra but maybe that channel isnt for that since noone replied. i will ask here

working on the vector space of polynomials, i am given p(t) = 1+t

the dot product is defined as <p, q> = SUM(ak * bk, k=0..max(n,m))

ak are the coeficients of p and bk from q

i am asked to find the orthogonal subspace generated by p(t)

This is simply all polynomials whose constant coefficient and linear coefficient are negatives of each other

well

i got that

are all plynomials that the sum of their coefficients is 0

because what i have is

https://gyazo.com/62dd8cf2e94624b8843ad45d59ccc772

@chilly ocean is this correct?

I'm not sure what you are doing, it looks weird

mmm

Inner product of p(t) and 1+t should just be a0 + a1

first line is the definition of my dot product

2 polynomios are orthogonal if dot product is 0, right?

so i need <p(t), 1+t> = 0

Yes

cuz 1+t is the one i am given

and p(t) can be any

so a0 + a1 * t + a2 * t^2 + ... and so on

the definition of my dot product is the sum of the product of the coefficients

so the coefficients of 1+t is b0 = 1 and b1=1

and the coefficients of p(t) are a0, a1, a2, etc

so basically after the dot product i have twice each coefficient (?)

and since it has to be 0, i can divide by 2

o.O

How do you get twice each coefficient? This makes little sense to me

a0b0 + a1b1 should be a0+a1, right?

And the rest of a2b2+ ... Are 0

[chilly ocean]

cuz i have ak * bk (ak is a0, a1, a2) and (bk is 1 and 1 (the 1 from 1 and the 1 from the t)

so i have [a0, a1, a2, a3] * [1, 1] = a0 + a0 + a1 + a1 + a2 + a2

[chilly ocean]

why?

[chilly ocean]

why?

[chilly ocean]

Because b0=b1=1

[chilly ocean]

Because b2=b3=...=0

[chilly ocean]

alright xd sure

[chilly ocean]

oooooooh true true

[chilly ocean]

I'm not sure what is happening here

[chilly ocean]

mmmm i have 2 polynomios: p(t) = a0 + a1 * t + a2 * t^2 + ... + an * t^n and q(t)=1+t

and i have to do the dot product, which is SUM(ak * bk)

But how do you get a0 + a0 + a1 + a1 + ... ? I thought you might be confusing it with polynomial multiplication, but that is not right either

because my ak's are [a0, a1, a2, a3, ...]

so far so good?

No, I am not following what you are saying

and my bk's are [1, 1]

[chilly ocean]

does this make sense to u?

Sure

okey

and my bk's are [1, 1]

Why not [1,1,0,0,0,...]?

okey, yeah

1,1,0,0,0,0..... until n (the grade on my p(t))

right?

Yeah

ah true, i was doing every ak with 1, and then the same again

i wasnt doing k with k

i was doing like a double loop

aaaah my bad

so i have a0 + a1

a0 + a1 = 0

is this my orthogonal vector subspace?

Yeah

So "constant coefficient and linear coefficne tshould be negatives of each other"

okey

and the second part is

prove this inf-dimensional

just because ak belongs to R?

I guess ak belonging to R can be thought to be a part of it, but I wouldn't say this is really the main point

then which one is?

Do you recall a condition for a vector space to be infinite dimensional?

not right now, let me search

okey i have that

for any n € N, we can find any vector linearly independant (sorry if terminology sucks) of dimension n

Yeah

so, how does this apply here?

Can you find n linearly independent vectors (polynomials)?

1/2 - 1/2 * t, 1/3 - 1/3 * t

and so on

this?

But these are linearly dependent, the second one is 2/3 the first one

no no no

yeah

ehm

can i?

i dont think so

since a0 = -a1

Er, you're looking at polynomials of any degree, not just linear, right?

yeah

So as long as a0=-a1, it can be any polynomial

ah,yeah, but thats because a0 € R

no?

no

i mean, i know why but idk how to say

I guess I'm not sure what you mean by that, I agree that a0 can /must be in R, but this works in any field

is because for any n i will get equation = 0 with n terms

like, imagine if my p(t) was 1 - t + 1/6 t^2 - pi * t^3

i will get a0 - a1 + 1/6 * a2 - pi * a3 = 0

and i can make this happen

[chilly ocean]

What does this equation mean? (a0, a1,... Are supposed to be coefficients? But then why not eg a2=1/6,etc?)

what?

sorry i didnt understand that

I don't know how you got that equation

before we did a0 + a1 = 0 cuz my p(t) was 1 + t

Overall you are saying some weird things that I don't understand

what if my p(t) = 1 - t + 1/6 t^2?

Oh, as opposed to 1+t?

yee

no

HAHAHA

But why are we talking about changing 1+t?

finding the orthogonal subspace to 1 - t + 1/6 t^2

instead of the orthogonal to 1+t

This should be infinite dimensional too, if thats what you are saying

yeah

i know

so the same way with 1+t we get a0 = - a1

for this new case coefficients remain like a0 - a1 + 1/6 * a2 = 0

no?

Sure

okey so the reason is this

for any n-degree polynomio

i can do this

but idk how to write it properly as a prove

I'm still not sure what you are trying to say, I don't see what this says about the 1+t case

it has nothing to be with 1+t

it was a new example

why for 1+t, the orthogonal vector subspace is inf-dimensional?

But 1+t case is the one we are interested in, right?

yes we are, but i wanted to generalize it

[chilly ocean]

this ^

[chilly ocean]

@chilly ocean u here? uwu

I am here, but I should technically not be on discord, I am at work

ups sprry

if someone else can tell me i will appreciate ^^

[chilly ocean]

this one

ye

[cloud walrus]

slimvesus

[chilly ocean]

oh

is that called direct sum or something?

so k[t] is inf-dimensional

and 1+t is 1

so inf = 1 + x

so x inf too?

this works here? nice nice, ty 😄

yeah, the dimension of the orthogonal

okey okey

thanks ^^

no no

i think we saw this but i forgot

we are working

on a prehilbert space

so yeah, it is

cuz pre hilbert are closed, right?

oh u edited ur msg

this one

yeah, here the norm is induced by inner product

[cloud walrus]

slimvesus

[chilly ocean]

https://gyazo.com/24e040b44007834deb7b17ba66ca6be5

dot product and norm

[latent anvil]

sorry is the claim that the orthogonal complement of the span of 1+t is not finite dimensional?

if so, you can see this more directly. Let B = {1-t, t^2, t^3, ...}. By direct computation, each element of B is orthogonal to 1+t. Also clearly B is linearly independent. I claim B spans (and so is a basis for) the orthogonal complement of span(1+t). Say f(t) is orthogonal to 1+t. Write f(t) = a + bt + g(t), where g(t) has no degree 0 or 1 terms. Then a+b = <1+t, f(t)> = 0, so a+bt = a + (-a)t = a(1 - t). Clearly g(t) is in the span of {t^2,t^3,...}, so f(t) = a(1-t) + g(t) is in the span of B

I like this a little better because you get an actual basis for the orthogonal complement

I should note that by basis I mean Hamel basis, not Schauder basis

[small bison]

not finite = infinite

[chilly ocean]

i guess this is harder for me than the direct sum

[chilly ocean]

Are U and V subgroups of G and u is a homomorphism from G to some other space? Overall I'm not familiar with this notation

And HK={hk: h in H and k in K}?

Hopefully I am not doing anything wrong, but I think it's something like: that subgroup of U/u is the set of cosets au (a in U) such that there exists x in au with x in U cap V. Then this condition boils down to something like a in (U cap V)u, which is should be(?) equivalent to u(U cap V), I think

What is even V/v cap U/u? This is like a compilation error, makes no sense

Also (U cap V)/u is also a compilation error, unless we can show U cap V contains u

What do you mean by "homomorphic to some subgroup of G"? Homomorphic=isomorphic?

Er, sorry, sec

I guess you meant homomorphic as in, is a homomorpic image of a subgroup of G

But yeah, surely I think it is saying that U/u is the homomorpic image of U, under the canonical map

And "subgroup of U/u consisting of blah blah blah" is cosets au such that au cap U cap V is nonempty

They mean something like "the subgroup of U/u consisting of images of elements x such that x in U cap V" right?

Or I guess maybe "consisting of cosets au such that au=bu for some b in V"

But so b should be in U, so this should be cosets bu with b in U cap V, right?

So this is something like (U cap V)u/u, and probably something like (U cap V)u=u(U cap V)

Or maybe there was a commutativity error somewhere

I think the "right" explanation is via some isomorphism theorem, relating subgroups of U/u to groups A with u < A < U, I'm not sure I can recall the details off the top of my head.

Something like that, (U cap V)/u is illegal in the first place, the top group doesn't need to contain u

Yeah, I guess that's true

I think there are just some annoying details to work through in terms of what precisely is meant by these notations, and it will make sense

[unique juniper]

if R doesnt have identity, then the subring cant?

so we assume R has identity?

??

[chilly ocean]

yes, we dont talk about rings without identity

Imagine assuming rings do not need to have identity

[unique juniper]

idk i feel like it was emphasised that we should assume that rings dont have identity and if they did, it would be explicitly said

[chilly ocean]

Does it say at the top of the exercises list or the beginning of the chapter? I recall d and f usually put it in those places

[shut halo]

Hey, I have to show that $wx^3 + zy^3 = 0 \in \mathbb{C} \mathbb{P}^3$ is irreducible but not exactly sure on how to proceed. Does anyone have any idea?

[unique juniper]

ah yes it does

i didnt see lol

[cloud walrus]

snypehype

[delicate bloom]

like, show it has solutions?

[cyan marten]

I think that's what is meant. Alternatively, we can interpret it as "contains the identity if it exists."
Let E be the set of all identity elements of R. What is said here is thag Z(R) contains E. Sometimes E is empty, sometimes it's not.

[shut halo]

@delicate bloom oh sorry I forgot to fully write the question

The actual question is:
Hey, I have to show that $wx^3 + zy^3 = 0 \in \mathbb{C} \mathbb{P}^3$ is an irreducible surface but not exactly sure on how to proceed. Does anyone have any idea? i.e. the polynomial is irreducible

[cloud walrus]

snypehype

[rich ravine]

Use Gauss's lemma and the Eisenstein criterion or reduction mod a certain polynomial, probably

[chilly ocean]

yo quick question cuz Im blind: how do they expand this f_p(x+1)

[viscid pewter]

looks like binomial theorem

[chilly ocean]

oh ok just binomial right

ye

[rich ravine]

As a polynomial in x, it is primitive, and then it is an Eisenstein polynomial at the prime z

[whole basalt]

Would Axler be better to ask about in here than in #linear-algebra since it's not a "row reduce your way to the promised land" book? 😛

[chilly ocean]

is that what you think #linear-algebra is?

[whole basalt]

I know when it was shoved into Calc II at Georgia Tech aaaages ago when I took it, that's what it turned into, lol.

[magic owl]

(to phrase tterra's response in a different way, arguably Axler is more "linear algebra" than matrix operations are)

[whole basalt]

And when I used to tutor students in it before they finally separated the classes.

[magic owl]

either way it would be fine to talk about axler in #linear-algebra

[oblique river]

@magic owl is your name an lgbtq joke?

[magic owl]

not u too

thats a joke right

[oblique river]

yes I was there

when it happened

the first itme

[magic owl]

okay okay

[whole basalt]

Ah. I just didn't wanna clutter the room for people who need more computational help.

[oblique river]

sorry ashura I'm just meming

[magic owl]

i was about to have an aneurysm if not

[whole basalt]

LOL it's fine 😛

[oblique river]

I agree that you could talk about it in #linear-algebra

[whole basalt]

Cool cool.

[magic owl]

honestly is it worth making separate channels

[oblique river]

someone the other day came in and asked max if his name was supposed to mean that there are only 2 genders

because 2 is the only even prime

and "2 only" means "there are only 2 genders"

[chilly ocean]

[oblique river]

yeah lmao

[magic owl]

that is a stretch that can't even be detected by homotopy

[oblique river]

hahaha

yeah that was really off the wall

they said some other really questionable things

[chilly ocean]

huh moh tuh pee
or
hoe moe toe pee

[oblique river]

homo-top-yuh

jk dont say homo-top-yuh

[magic owl]

we've ruined this channel

in like

2 minutes

[whole basalt]

(Alternatively if anyone wants to tackle it in here feel free since it's been like half an hour 😛 And it apparently it's much harder because it's not true over all fields?)

I do know why F₂ is a counterexample to it always being true ... Because if you consider F₂², then the union of the spaces generated by <1,0>, <1,1>, and <0,1> gives you the entire space of a whopping four points.

[small bison]

what is "it"

[latent anvil]

homo top YUH

[latent anvil]

Sorry, what's the original problem?

[oblique river]

#linear-algebra message

@latent anvil

[latent anvil]

Ahh okay

[cyan marten]

If a group is a union of three subgroups, then each of them is of index 2. But that's probably even harder to prove than this problem.

[latent anvil]

@oblique river I'm thinking about geometry

And how like

This feels like a statement you could overcomplicate with AG

It's lines in space!

[oblique river]

oh boy yeah hahaha

[latent anvil]

Actually I feel like

It's a dimension thing

At its heart

I tried to use inclusion exclusion

(this is only in the finite dimensional case ofc)

[oblique river]

yeah

and what is inclusion-exclusion but combinatorics

and what is combinatorics but AG over F_1

[latent anvil]

So like, $\dim(U+V+W) = \dim(U) + \dim(V) + \dim(W) - \dim(U \cap V) - \dim(V\cap W) - \dim(W \cap U) + \dim(U \cap V \cap W)$

Oh no

I've tricked myself

Into doing combo

[oblique river]

hahaha

[cloud walrus]

s_n -> Shamrock weakly

[latent anvil]

The formula I wrote above?

Hmm, you might be right. I don't see how it can be true over a finite field...

I thought it was true for some reason

So we have an exact sequence

No, any field

[oblique river]

actually yeah

its false

3l ines in R^2

[latent anvil]

Hmm

[oblique river]

y = 0, x = 0, y = x

[latent anvil]

Yeah

So like

You have a complex

0 <- V+W+U <- V (+) W (+) U <- K <- 0

But I guess K is complicated

[oblique river]

yeahhh

[latent anvil]

shakes first sygzygys

Wait hmm

So if v+w+u = 0

Then v+w is in V+W cap U

Right I see

Intersection doesn't distribute over sums

https://mathoverflow.net/a/26735/136523

Oh no, it's cech stuff

So I'm going to bullshit

And say

It works in general position, somehow

And that this condition is equivalent to solidangles' thing failing

Now I'm curious lol

[magic owl]

does dimension games make sense though

whatever proof you use can't work over F_2

[latent anvil]

Yeah ik, I thought it might be some weird formula that fails if stuff is a power of two or something

[magic owl]

ah ic

[latent anvil]

Also yeah I'm assuming we're in finite dimensions

I'm just looking for like

A geometric proof

It feels like a statement about dimensions to me

[oblique river]

yeah that would be satisfying

[magic owl]

okay

bold conjecture

that i have 0 basis for

[latent anvil]

👀

[magic owl]

it is not true for all fields of characteristic 2 but there are infinitely many fields of characteristic 2 where you can do it

[latent anvil]

Hmm

[magic owl]

why do i think this?

pure vibes

[latent anvil]

Lol

So the complex in that post is exact iff U cap (V+W) = (U cap V) + (U cap W)

I have decided this is not a productive use of my time the night before my analysis final

But that I should think about it Later

[magic owl]

lmao i have a final tomorrow too

ping me if u think of something

[latent anvil]

Will do

[magic owl]

i have decided this problem is Out of My Wheelhouse

[chilly ocean]

,w define wheelhouse

[ripe basalt]

can i have some assistance on this 1

H is finite-dim hilbert space

[ripe basalt]

it seems pretty straightforward, the LHS is the set of vectors which are mutually orthogonal to all P_i, since $$(\sum P_i) x = \sum P_i x$$

[cloud walrus]

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

[ripe basalt]

yeah nvm i guess i get it

[unique juniper]

I dont know how to do this one

Can someone help pls

[golden pasture]

sooo

hows induction

[unique juniper]

lol

i think my struggle is

at

concluding a_ng(x) = 0

but ig that gp = 0 shows a_ng = 0

[chilly ocean]

I wouldn't say a_ng=0 is about gp=0, that sounds like a weird thing to say

[ivory cosmos]

they don't ask you to show that a_n g(x) = 0

they ask you to show that a_n g(x) p(x) = 0

oh wait nvmd

[unique juniper]

but

i mean

$(a_nx^n +\cdots)(b_mx^m + b_{m-1}x^{m-1} + \cdots ) = \cdots + (a_nb_mx^m + \cdots a_nb_0) = \cdots + a_ng(x) = 0$

[cloud walrus]

Yes

[ivory cosmos]

is this about why a_n g(x) = 0?

[unique juniper]

yes

[carmine fossil]

Did you get a_n b_m=0?

[unique juniper]

atleast why i think it is?

[unique juniper]

yes

[ivory cosmos]

so do you get why b_m a_n = 0?

[carmine fossil]

What's the degree of a_n p(x)

[ivory cosmos]

• a_n g(x)

[carmine fossil]

<m

But m is the nonzero polynomial such that if a polynomial satisfies that condition, degree of that polynomial >= deg(m)

See minimal degree

[unique juniper]

wait

a_n b_m = 0 because all the coefficients must be zero right?

[carmine fossil]

Yes

[unique juniper]

ok

and the degree of a_ng(x) is atleast of degree m-1 or less right?

[carmine fossil]

*atmost

[unique juniper]

yes

sorry

[ivory cosmos]

since multiplying by a_n "knocks out" the top coefficient

[unique juniper]

ye

ok

so im at the part of concluding why a_ng(x) = 0

[carmine fossil]

See the minimal degree part

[unique juniper]

i see

so that means

theres no other nonzero polynomial that has degree less than g, that it multiplied by p, equals 0

[carmine fossil]

*nonzero

[unique juniper]

yes

[carmine fossil]

So that implies a_n g(x)=0

[unique juniper]

yeah

i can do the rest

the last part implies that b_mp(x) = 0 because its commutative right

[carmine fossil]

Yes

[unique juniper]

cool, thank you

[soft skiff]

Guys does there exists a commutative simple ring which is not a field?

[delicate bloom]

quaternions

shit not commutative 😢

[cursive temple]

wikipedia seems to think that

In particular, a commutative ring is a simple ring if and only if it is a field.

[soft skiff]

suppose the ring doesn't have Identity

[cyan marten]

The zero ring has an identity, is commutative and simple, and is not a field

[bleak crystal]

can you divide (x^2 +2x +3) by (6x+4) modulo 8

(I'm in the middle of the euclidean algorithm and this is an intermediary step)

if not, how would i find the gcd?

[cyan marten]

Call them f, g

In Z, we have f = qg + r. We can then reduce this equation to get the result.

[bleak crystal]

i don't think I'm picking up what you're putting down

would you be a little more specific

[cyan marten]

I might be wrong..

Let me think about it

[bleak crystal]

6x isn't ever gonna go into x^2 (IN MOD 8)

so what do i do?

6x=1mod8 has no solution

[cyan marten]

Oh..

And I forgot that Z[x] isn't a Euclidean domain

[bleak crystal]

also true

[cyan marten]

We want to find q, r so that f = qg + r, such that r has degree at most 1?

[bleak crystal]

let me show you my work so far

ignore the top part

the other division

gcd[(1,4,1,6,3,4,3),(1,2,3)]

so i got to what i asked, (1,2,3)÷(6,4)

now I'm stuck, since 6x=1mod8 has no solution, I'm not sure how to find the gcd

(this isn't for hw or a test or anything, just test prep with random polynomials and modulo)

[mild laurel]

I don't think you really need to divide here actually, you can just go through all the possibilities for the gcd

[bleak crystal]

oh?

[mild laurel]

the gcd must divide 6x + 4 so it must be degree 1 or less

[bleak crystal]

that's true

1 and 2 divide it,

[mild laurel]

you can just list out the polynomials that divide 6x + 4 and check which ones divide the other polynomial

[bleak crystal]

so what, they're relatively prime?

[mild laurel]

yeah

[bleak crystal]

so gcd=1

as a polynomial

now comes a bigger problem

how do i make them a linear combination?

[mild laurel]

You can't always do this in Z[x]?

For example, if you take the polynomials 2, x, their gcd is 1, but there's no such linear combination that makes them 1

[bleak crystal]

shit, you're right. is that because ZmodX is not relatively prime with the leading coefficient of the given polynomial? or is it because ZmodX is not prime?

[mild laurel]

what?

[shut halo]

Can someone help me interpret this question?

What is F here? I'm struggling to understand why the domain of phi* is the function field K_v.

[green scroll]

what book is that from

[mild laurel]

F is just an element of the function field? I'm not really sure what you're asking

[mint seal]

it seems phi* is a pullback from functions on the variety to functions on the ground field

[shut halo]

Algebraic Geometry: A problem solving approach

[shut halo]

Thanks, that's what I thought. So an element of a function field acts on a point on the variety?

In other words, by $F \circ \phi$ we simply means the polynomial in the function field evaluated at a point (a,b) on the variety?

[cloud walrus]

snypehype

[shut halo]

@chilly ocean unfortunately the author doesn't define what k(t) is, but he consistently uses the letter k[x1,...,xn] to denote a ring of polynomials with coefficients in field k

[mild laurel]

k(t) is just the function field on k, or the set of rational polynomials in one variable

So I'm not sure what your interpretation means

I see this as saying that an element of the function field of V can be viewed as an element of the function field of k = A^1

[shut halo]

@mild laurel what I mean was that $\phi(t) \in V$ so $F(\phi(t))$ is a function with domain in V?

[cloud walrus]

snypehype

[mild laurel]

I mean, F is a function with domain in V yes

but F(phi) is a function with domain in k = A^1

[shut halo]

Ok I see now. So essentially phi* allows to "call" functions in K_v via the function phi right?

[mild laurel]

I'm not sure what you mean by call

It allows you to transfer functions on K_v to functions on A^1

[blissful ice]

has anyone used zariski's commutative algebra? The list of topics seems intriguing and a preliminary examination of a pdf seems extremely promising too, but I'm afraid if the treatment is too old. Is that the case?

[next obsidian]

... Zariski has a commutative algebra text? O_O

[blissful ice]

sure thing, it just got a dover reprint so it's cheap!

2 volumes, by zariski and samuel

[next obsidian]

Ohhhh

Okay I’ve heard of Zariski-Samuel

[blissful ice]

haha

[next obsidian]

I think this is one of the classical classical texts

[blissful ice]

yeah that that's exactly what scares me: is it too classical, to the degree of ignoring much of the modern development

[next obsidian]

Glancing at the first volume

It covers a lot of like... basic abstract algebra it seems

[blissful ice]

yeah first chapter does have. I mean eventually it gets to the more advanced levels of CA too, specially with 2nd volume

[next obsidian]

Looking at the first volume, and its table of contents

It seems to not have any focus on modules

Which is like... a pretty big shift in how commutative algebra went IIRC

It used to be more focused on ideal theory, and the rings itself

[blissful ice]

isn't chp3 like all about modules

well and ideals

[next obsidian]

Then it shifted to being more about modules

Idk, I don’t see a mention of say associated otimes

Primes*

Which I think was like a more modern replacement for primary decomposition and primary ideals

[next obsidian]

Hmm okay

The other main thing I got looking at both volumes

And again, maybe this is hidden away in a chapter not with that title

It seems to not make use of a lot of homological machinery

I don’t see them mention depth, Cohen-Macaualyness, etc

Maybe this is in the local algebra part of Volume II

But this could be because of the age, homological algebra developed a lot in like...

The 50s,60s,70s

But this is moreso like a “second course in commutative algebra” material anyway

[blissful ice]

hmmm, possible
I know I've seen depth for certain. But idk if this book includes the homological treatment

[next obsidian]

It could work?

It might very well have everything

But at least compared to the more modern books I’ve seen the modern tools don’t appear to be stressed, in that modern books will be like

Uhh... they’ll like have chapter titles indicate

“Now we do Cohen-Macaulay stuff”

Or something like that, so maybe the material is there, but it certainly seems focused on different aspects than modern treatments

I think a large part of Volume II is about polynomial and power series rings which is, don’t get me wrong, a huge part of commutative algebra and algebraic geometry

But my limited understanding of the subject’s development is that these subjects were more at the center of it in the past then it is now

That being said, I doubt the book would be a waste of time

[blissful ice]

I don't see "exact sequence" in the index mentioned
I do see cohen-macaulay ring though, also cohomological dimension

[next obsidian]

Hmmm, interesting

My not very informed opinion is maybe it’s a little dated, but not enough that it’s irrelevant

[blissful ice]

okay okay, cool cool

[next obsidian]

I think if you like the way the books written more than other contemporary sources

And this makes it easier to work through it, I’d say go for it

[blissful ice]

yeah from what I read I like it significantly more than eisenbud, for example. And more than macdonald, although that one wasn't that bad at all

[next obsidian]

Ah

I will always shill for Matsumura

The new one, commutative ring theory