groups-rings-fields

[thorn jay]

Yeah I saw somewhere

I love math

[shy fossil]

Here's a snapshot of the last part of the last section of the report I presented in my logic class. Sometimes i just think what is not possible with category theory (I can see a joke up for grabs lol)

Eff is the effective topos btw, sth sth kleene and computability

that haven't should be a have, that's a bad typo right there

[thorn jay]

I wish I could understand this :0 I still have to learn topos theory sometime

Seems fitting tbh

[shy fossil]

Basically I had a categorythat i built with intuitions of turing machines, then upgraded it to a topos by bilding a subobject classifier, which turned out to be just the effective topos! Then I used Lurie's ideas to make an infinit-topos version of it.

[thorn jay]

I agree!

What's an effective topos? Lol

[shy fossil]

https://ncatlab.org/nlab/show/effective+topos

have fun

[thorn jay]

I so will

[inner owl]

I want to find out what is this quotient ring Z[x]/(2_x^2 - 4 , 4x-5) I did some calculations and concluded that x-10 and 7 are elements of this ideal and I know that the ideal generated by x-10 and 7 is maximal therefore the ideal generated by the original two elements is either the same maximal ideal or the whole ring can someone give me the idea how to show that the original two elements are contained in the ideal generated by the newer elements or if not then how to show that the original 2 elements generate the whole ring?

[tribal moss]

Was the underscore in Z[x]/(2_x^2 - 4 , 4x-5) meant to be there? If so, what does it mean?

[inner owl]

Nah sorry it's a typo

@tribal moss

[tribal moss]

By subtracting appropriate multiples of x-10 and then appropriate multiples of 7, every element of Z[x] can become one of {0,1,2,3,4,5,6}.

Also, if you map 2x²-4 and 4x-5 into Z/7Z by setting x=10, then you'll find both of the generator polynomials map to 0, so this map factors through a (surjective) map from your quotient to Z/7Z.

Since the ring has at most those 7 elements, and also at least 7 elements, it must be isomorphic to Z/7Z.

[inner owl]

Okay so modding successively first by (x-10) and then by (7) we would get the field Z_7 and also the original polynomials get sent to 0 so those two are the elements of the ideal generated by x-10 and 7 and therefore the original ideal is the same as (x-10 , 7) therefore the original quotient ring is isomorphic to Z_7.
Thanks @tribal moss

[twilit wraith]

understanding it

[white oxide]

Any hints for (i) $\implies$ (ii)? So far, I have the following: Suppose that $\text{Spec } A = U_1 \cup U_2$ where $U_i \neq \varnothing$ is open in $X$ and $U_1 \cap U_2 = \varnothing$. If $U_i = X \setminus V(E_i)$, then $V(E_1 \cup E_2) = \varnothing$ and $V(E_1 \cap E_2) = X$. Define $\varphi: A \to A/\langle E_1 \rangle \times A/\langle E_2 \rangle$ by $a \mapsto (a + \langle E_1 \rangle, a + \langle E_2 \rangle)$, however I'm not sure this is right (I used the fact that every prime ideal contains $E_1 \cap E_2$, in particular $E_1 \cap E_2 = {0}$, so if we can define a ring homomorphism with kernel $E_1 \cap E_2$ and show its surjective then we're done

[cloud walrus]

okeyokay

[white oxide]

Here <E_1> and <E_2> are the ideals generated by E_1 and E_2

I know that the map is a bijection if and only if <E_1> and <E_2> are coprime and their intersection is trivial, but I don't know if this is the case

What does that mean again

I was able to prove that the map is surjective

Since they're coprime

I want to now show that they have trivial intersection

Shouldn't be too hard?

[sonic coral]

what book is this from?

[white oxide]

Atiyah-Macdonald

[velvet hull]

Same thing different terminology, I got confused

But yeah you’re basically there

[white oxide]

I see, so E_1 n E_2 = (0) implies that if a is in <E_1> + <E_2>, then a is a finite sum of elements in E_1 and E_2. But that implies that a is 0

[velvet hull]

Hmmmm, I’m not sure that is sound

[white oxide]

Porque

[velvet hull]

If every prime ideal contains E1 n E2, what does that tell you it’s in?

[white oxide]

What is it in this case

[velvet hull]

What’s the intersection of all prime ideals

[white oxide]

Well it's in the nilradical

[velvet hull]

So what is <E1> n <E2> contained in

[white oxide]

Uhh gimme a sec lol

Uhh the nilradical as well?

[velvet hull]

But then what does that mean about V(<E1>) and V(<E2>)

[white oxide]

Well I mean if <E1> n <E2> is contained in the nilradical which is contained in (0) then their intersection is (0) right?

Because (0) is a prime ideal

[velvet hull]

The nilradical is not contained in 0 lol

Nothing else is contained in the 0 ideal

[white oxide]

Wait am I trippin

A n B n C \subseteq A no?

Wait I feel like I'm going insane

Because clearly the nilradical is not always 0 because why would we define it that way

So why is the nilradical not contained in (0) if (0) is prime

Or do we not consider (0) prime in any context

Oh I guess we need the ring to be a domain lol

[velvet hull]

Indeed if there are indempotents then 0 is not prime

[white oxide]

Okay so we know that every prime ideal which contains E_1 n E_2 contains <E_1> n <E_2>

[velvet hull]

I think I’m proving i) implies iii), not i) to ii) lol

Hold on

[south patrol]

There is a trick that gets you around the nilradical being nonzero tho

Lemme see what you have done so far lol

[white oxide]

(i) $\implies$ (ii). Suppose that $\text{Spec } A = U_1 \cup U_2$ where $U_i \neq \varnothing$ is open in $X$ and $U_1 \cap U_2 = \varnothing$. If $U_i = X \setminus V(E_i)$, then $V(E_1 \cup E_2) = \varnothing$ and $V(E_1 \cap E_2) = X$. Define $\varphi: A \to A/\langle E_1 \rangle \times A/\langle E_2 \rangle$ by $a \mapsto (a + \langle E_1 \rangle, a + \langle E_2 \rangle)$. If $\langle E_1 \rangle + \langle E_2\rangle \neq (1)$, then $E_1 \cup E_2$ is contained in some maximal ideal, hence prime, contradicting $V(E_1 \cup E_2) = \varnothing$. Thus, $\varphi$ is surjective by Proposition 1.10ii. On the other hand, suppose that $a \in \langle E_1 \rangle \cap \langle E_2 \rangle$.

[cloud walrus]

okeyokay

[white oxide]

I hope surjectivity is okay lol

Working on injectivity

[south patrol]

Sure looks good

[twilit wraith]

not to interrupt, but i need some clarification on this problem

[south patrol]

Also my hint is uh like this map cannot always be an iso

To see this, note that the ideal generated by E1 is not determined by V(E1)

[white oxide]

Wait so this map is not even correct then right

[south patrol]

Not necessarily

But if you change E1 to smth else ur good

[velvet hull]

Hint - Z(G) is always normal

[white oxide]

Uhhh what if I took the intersection of all prime ideals containing E_i

Right

Okay I'll try that thanks

[twilit wraith]

well yes which means G/Z(G) makes sense

i shouldve said what ive already thought about

of course we have to have that |Z(G)| cant be 1 or p^3

as its not trivial and G is not abelian

if Z(G) were order p^2, then G/Z(G) \cong Z/pZ as Z/pZ is the only group with order p up to isomorphism

but im not sure G/Z(G) \cong Z/pZ necessarily exists

[south patrol]

Actually my hint was also a bit bad lol

[velvet hull]

It does by assumption

[twilit wraith]

or rather the surjective homomorphism from G to Z/pZ

[velvet hull]

Now try and play around with cosets in G and see what happens to elements

[south patrol]

Iirc when I did this I showed that i) => iii)

You can even reduce to the case where the nilradical is zero if u want

[twilit wraith]

well im pretty sure it just boils down to proving that if G/Z(G) is cyclic then G is abelian

[south patrol]

Yes

[twilit wraith]

not sure how to do that quite yet

i think its just looking at the cosets

[velvet hull]

Every element in G has the form x • z

[delicate orchid]

I'm gonna be sick

[twilit wraith]

i see

[delicate orchid]

this is true of any group. What you mean to say is that g = x^kz for some integer k and a fixed x

[velvet hull]

Sure, they can work out the details lol

[white oxide]

Is => true? I saw somewhere online that it's not true

And I don't want to try to prove it if it's not true lol

[velvet hull]

There are a few ways to understand A-algebras; one of them being that B is an A-algebra if there exists an embedding of rings of A into B

Then you can verify that multiplication by A (when viewed as a set of scalars) is compatible with the internal addition of B (viewed as a set of module vectors); and B also has internal multiplication that is compatible with scalar multiplication (and hence has an algebra structure)

[twilit wraith]

i think i see it

[velvet hull]

So an A-algebra homomorphism can equivalently be thought of as a ring homomorphism that respects the embedding of A, ie preserves the algebra structure.
But then this is exactly the claim

[twilit wraith]

it was less bad than i thought

just sps G/Z(G) = <aZ(G)> for some a \in G. Then let x,y \in G be arbitrary and suppose x = a^k z and y = a^l z' for some k,l \in Z and z,z' \in Z(G). Then xy = (a^k z)(a^l z') and all these elements commute with each other, so we can rearrage to get yx

[velvet hull]

Oh, I think I see the post you’re talking about, I think you necessarily need to assume h is a ring homomorphism to begin with, and the fact that it commutes with the embedding of A enables it to be a module homomorphism as well, which is not something we initially assume

[twilit wraith]

im still wondering though why it doesnt matter if there is no surjective homomorphism from G to Z/pZ in the first place

would that just mean that |Z(G)| automatically couldnt be p^2

[velvet hull]

Well, there could still be one, the kernel just cannot be the center

[twilit wraith]

i see

so it works to suppose a homomorphism did exist such that Z(G) is the center

and by contradiction show that the homomorphism doesnt exist because Z(G) cant be p^2

so in a sense we are getting two pieces of info out of the contradiction

i guess i should know thats possible from my studies in philosophy

[velvet hull]

I would say that they’re two interpretations of the same piece of knowledge that there is a contradiction

But sure

[twilit wraith]

fair enough

man it feels like im actually getting better at algebra

i dont feel completely stuck on problems anymore

[delicate orchid]

bonus follow up problem, show that for G a p-group, there is always such a homomorphism

[velvet hull]

I was thinking about that, but are we sure that the derived series always terminates at Zp?

Let me go through my own notes again

Oh wait if it doesn’t terminate at Zp

Then we’re done as well lol because its always an abelian p group anyways

[delicate orchid]

ok that's not how I was going to do it but
A) every p-group is solvable (nilpotent, even)
B) all of the composition factors have to be C_p for obvious reasons

the proof I had in mind is much more elementary

but you're correct in that this method works

[glad osprey]

Is your confusion that you know Z/pZ is the only group of order p up to isomorphism, but you're not sure whether that means there exists an isomorphism from G/Z(G) to Z/pZ?

[twilit wraith]

it was

but i think ive resolved why it works

its mostly just that saying that |Z(G)| = p^2 is equivalent to saying that G/Z(G) \cong Z/pZ is possible

[glad osprey]

Hmm, so if |Z(G)| = p^2 then |G/Z(G)| = p, right? But Z/pZ is the only group of order p up to isomorphism, so G/Z(G) is isomorphic to Z/pZ, which is the same as saying there exists an isomorphism from G/Z(G) to Z/pZ. This is not just a possibility, we know it exists

[twilit wraith]

oh of course

its just that G/Z(G) always exists

so necessarily if |Z(G)| = p^2 then we have that G/Z(G) \cong Z/pZ

[glad osprey]

And if we were trying to rule out |Z(G)| = p^2, knowing that such an isomorphism "may" exist is not enough

[delicate orchid]

well no it always exists

[south patrol]

I'm confused lol

[delicate orchid]

it's a group of order p. There is one of them. They are all isomorphic to one another

ts shit is skeletal

and brother? I got a bone to pick with u

[glad osprey]

I know, I'm trying to point out the fallacy of arguing that an iso "may" exist

[delicate orchid]

true, it is june after all

[twilit wraith]

i was thinking about this so incorrectly that it confused everyone

i was just not recognizing a corollary of a fact i already knew

[south patrol]

Real

[glad osprey]

Yep, I think you got it now tho, good job hiido

[south patrol]

I do wonder if there is any other funny way to do this probpem

Oh yeah lol, the way I knew to do it is different but essentially equivalent

If the centre has size p^2 then consider the centraliser C of any non-central element x. Then C contains x and the centre, hence is the whole group, so x is in the centre, contradiction

Lol

[spice wagon]

the letter p make me think about p-Sylow

worst thing in group theory 🥀🥀

[tardy hedge]

Test

Oh ya. Math Tag.

[small tendon]

What's wrong with p-Sylow?

[inner owl]

So I had to prove or disprove that the Q rings Z[x]/(x^2 + 7) and Z[x]/(2x^2 + 7) are isomorphic to do so I assumed isomorphism that implies there is a bijective homomorphism between the two rings then there would be a unique pre image of the coset for x in the 2nd ring would be a coset of some linear polynomial ax + b in the first ring now 2(ax+b)^2 + 7 = 2a^2x^2 + 4abx + 2b^2 + 7 therefore x^2+7 doesn't divide this => that a non 0 member of the first ring is mapped to 0 therefore contradicting injectivity of the map and therefore these rings are not isomorphic.

Now this is my work regarding this problem . I would like to know 2 things :-

1. Is my proof correct?
2. If (1) then is there any easier way to do the same without using the module structure of the first ring.

[velvet hull]

Your proof looks fine, I’m trying to look for a simpler proof but I haven’t found anything that is massively more so yet

[velvet hull]

In Z[x]/<2x^2+7>, notice that 2(x^2+4) = 1 so 2 is a unit.
But then if f is an isomorphism, this means that 2 must be a unit in Z/<x^2+7> as well (note that f(2) = f(1+1) = f(1)+f(1) = 2).
But then 2(ax+b) = 1 has no solutions for integers a,b, which is a contradiction

This proof isn’t that much faster but it’s the best I can come up with

(Bonus question: why does this reasoning fail when I reverse the two rings?)

[chilly ocean]

what do u mean by reversing

[velvet hull]

Why can’t I say 2(ax+b) = 1 has no solutions in the other ring so 2 can’t be a unit there either

[chilly ocean]

a non unit may go to unit

[velvet hull]

lol, I was thinking of the issue that ||elements in the other ring is not always of the form ax + b anymore, as thats an important subtlety||

But they can figure that out in their own

[chilly ocean]

Doesnt this make Z/n just Z since you only have that equality in Z/0? Or should I interpret = as an element of?

[spice wagon]

p-Sylow theorems are hard to understand

personally

[azure iron]

that just means "we are using the symbol 0 to denote this"

[chilly ocean]

ohh so he just means := ?

[azure iron]

sure

[chilly ocean]

thanks

[candid patrol]

The proof ?

[keen badge]

How can I prove "by definition" that sqrt(2) in Q(sqrt(2)+sqrt(3))?

More specifically, can someone help me find f(x) in Q[x] such that f(sqrt(2)+sqrt(3))=sqrt(2)?

[keen badge]

I found an f(x) in Q(x)(instead of Q[x]) such that f(sqrt(2)+sqrt(3))=sqrt(2).
Is this enough, becuase Q(sqrt(2)+sqrt(3))=Q[sqrt(2)+sqrt(3)]?

[cobalt heath]

Yes, as long as you can see why that equality holds

[chilly ocean]

Q(sqrt(2)+sqrt(3))=Q(sqrt(2),sqrt(3)), you can show this by considering degree argument

[keen badge]

I knoy but im asking a different qhestion

[chilly ocean]

use conjugates

[glad osprey]

I don't think Q(sqrt(2) + sqrt(3)) is equal to Q[sqrt(2) + sqrt(3)]. The former is a field, the latter is just a ring

[swift tundra]

This is true when the adjoined elements are algebraic over the base field.

[glad osprey]

Oh, you're right 👍

Either way, the equality is not important in this case, since you're working with Q(sqrt(2) + sqrt(3)) anyways

[swift tundra]

Ye

[harsh citrus]

Urgent Galois Theory book recommendation needed that does it like Milne's notes but in detail

[chilly ocean]

patrick's book is nice with problems

[harsh citrus]

Patrick morandi ? Does it do it as much as Milne?

Or should I do Lorenz

[chilly ocean]

patrick book has amazing excersises

[chilly ocean]

i dont know about this one

[harsh citrus]

OK bro

[tribal moss]

You can do that (as the subsequent discussion details), but for more "by definition" you can also let $\xi=\sqrt2+\sqrt3$, and then $$\frac{\xi^2-5}{2} = \sqrt6$$ so $$\xi\frac{\xi^2-5}{2} = 2\sqrt3 + 3\sqrt2$$ and $$\xi\frac{\xi^2-5}{2}-2\xi = \sqrt2$$

[cloud walrus]

Troposphere

[white oxide]

Can somebody explain to me how this map D \otimes D \to D was obtained? I don't think 2.14 says much about it (in particular, B \otimes C \otimes B \otimes C \to D is not bilinear, so we can't factor through a map (B \otimes C) \otimes (B \otimes C))

[knotty badger]

Wait I’m pretty sure the tensor product is associative

(Up to coherent isomorphism at least)

[twilit wraith]

How many elements of order 4 does S8 have?

i personally got 744 but im skeptical

since i suck at counting

oh i think i just interrupted sorry

[white oxide]

Yeah I think it is, I'm kind of new to it so I'm just trying to see how the map arises via the universal property

[white oxide]

Ur all good

[knotty badger]

Ah ok sure sure

I think the map is bilinear

At least at the level of tensor product

[white oxide]

Well this is what I had in mind:

[
\begin{tikzcd}
{(B \otimes C) \times (B \otimes C)} \
\
{(B \otimes C) \otimes (B \otimes C)} && {B \otimes C}
\arrow["g", from=1-1, to=3-1]
\arrow["f"', from=1-1, to=3-3]
\arrow["h", dashed, from=3-1, to=3-3]
\end{tikzcd}
]

[cloud walrus]

okeyokay

[knotty badger]

Yeah this seems good

[white oxide]

yeah but how does the map $B \otimes C \otimes B \otimes C \to B \otimes C$ induce $h$ is my question

[cloud walrus]

okeyokay

[white oxide]

Because here we have f inducing h but they don't have the same domain

[knotty badger]

Well you’ve got the bilinear map f right?

[white oxide]

Actually I'm not even sure what f is lol

Or how it was induced by B \otimes C \otimes B \otimes C --> D

wait

if $h$ is the map $B \otimes C \otimes B \otimes C \to D$, then $h$ induces a bilinear map $f: (B \otimes C) \times (B \otimes C) \to D$ given by $f\bigl(b \otimes c, b' \otimes c') = h(b \otimes c \otimes b' \otimes c')$ right

[cloud walrus]

okeyokay

[knotty badger]

yep!

you get it by composition :3

[tardy hedge]

Who up ringing they fields

[white oxide]

okay great thanks

I still haven't convinced myself entirely but I'll give it more thought

[knotty badger]

I am very tired so I’m not as sharp as I usually am~

apologies I can’t be more helpful

[white oxide]

nw you're all good

[twilit wraith]

Alright im coming back to ask for help on this

i need to count how many elements in S8 have order 4

ive sorted it down to four possible cycle expressions of the elements

(....)(....)
(....)(..)(..)
(....)(..)
(....)

im just so terrible at counting that idk what to do here

[velvet hull]

I mean it looks good

You just have to count the cycle types and you’re golden

[twilit wraith]

yeah thats the part i have no clue how to do

i dont really understand how counting these would be any different

its just like 8 choose 4 over 4

or smth like that

[velvet hull]

Yeah that’s it

There’s not much to it

[twilit wraith]

fair enough

i mean is it really tho

wouldnt the first one be like

8! divided by 4^2

then the second one is 8! divided by 4 * 2 * 2

so the same thing

[velvet hull]

It’s 8 choose 4 and then 4 choose 4, divide by 2

[twilit wraith]

why divide by 2

[velvet hull]

The two 4 cycles are the same

[twilit wraith]

what?

oh i see

and then i also have to divide by 4

because smth like (1374) has four equivalent forms

[velvet hull]

Wait no you’re right

You also have to multiply by 4 twice then

[twilit wraith]

i mean the way that i thought about it was

i just have an arrangement of all 8 things

but i have to divide by 4 twice

for the first case

to consider the four equivalent cycles

but i think i also have to divide by 2

bc if i have like (1374)(5268)

then i can just flip them

the other ones i dont have to divide by 2 bc im already considering the commutativity of disjoint cycles by only considering one arrangement

[candid patrol]

lcm(x1, x2, …, xk) = 4 iff there exists i € {1, …, k} with xi = 4 and for j != i, xj = 2^a with a € {0,1,2}

Do you have to count :
(. . . .)
(. . . .)(. .)
(. . . .)(. .)(. .)
(. . . .)(. . . .)

Oh, you already wrote it mb

[cobalt heath]

So, DFT corresponds to CRT over C. Meanwhile, DFT also comes from Pontryagin duality as well.
How are the relations of the two?
Is this just coincidence in specific case, or is there more general phenomenon going on?

[long obsidian]

Does anybody know a way to look at a single variable polynomial and say whether it will have more than a single root based off the coefficients?

[fading acorn]

ah i misunderstood the question ignore it

[coral spindle]

Yes. You can calculate common factors of f(x) and f'(x). If f(x) has a root of degree k+1 then f'(x) will have the same root of degree k.

Therefore you can calculate how many repeated roots there are via this method

This works over the algebraic closure of any field

[long obsidian]

So you’re suggesting that I take a known root, then see the number of times I can take the derivative while having this a root, and if that number less than deg(f) then there exists a different root if I assume the coefficients come from an algebraically closed field?

[coral spindle]

You don't even need to take the derivative more than once!

Suppose in the algebraic closure that f(x) = (x-a)^n -- let's just assume monicity for now

Then f'(x) is some multiple of (x-a)^(n-1)

So gcd(f'(x), f(x)) is (up to some coefficient) going to be f'(x)

gcd(f'(x), f(x)) = (up to coeff) f'(x) if and only if we're in this situation, in fact

So all you need to do is calculate the GCD, which is easily done with the Euclidean algorithm

[barren sierra]

I hate group theory man

So I need r: B -> A s.t. r o phi = id_A

I know that without the assumption on the image of phi landing in Z(B) that this sequence splits, so there exists b in B such that <bA> is isomorphic to Z

psi(b) = 1

but then how does this further assumption give me the retraction r?

I know how to do this with the splitting lemma (you can show A, B are abelian) but I want to do this without the splitting lemma

[coral spindle]

If f(x) has a repeated root, then f'(x) will have a root of the same kind

Therefore gcd(f(x), f'(x)) will be nonzero.

Savvy?

[long obsidian]

Hold up lol

[coral spindle]

You can probably find this in any intro abstract algebra book

Heck, it's even on the wikipedia page for the formal derivative!

See: https://en.wikipedia.org/wiki/Formal_derivative#Application_to_finding_repeated_factors

[next obsidian]

This isn’t necessary and sufficient though right?

If you had alpha a root occurring two times and beta a different root, you’d fail this test to be separable but you’d still have more than one root

[coral spindle]

Why not?

But that's not the question at hand

wait my bad I misunderstood what you're saying

[next obsidian]

Didn’t they say “tell if it had more than one root”

This isn’t the same as “is separable”

[coral spindle]

I'm not quite seeing your point

We can definitely detect whether or not there's a unique root in the same way

[next obsidian]

Let’s say you had a cubic which had alpha as a root twice and beta as a root one time, this has more than one root

[coral spindle]

it's just when gcd(f'(x), f(x)) is f'(x) up to scalar multiples

yeah yeah I get that

[next obsidian]

What’s your algorithm going to tell us?

I might not understand what you’re suggesting

[coral spindle]

gcd(f'(x), f(x)) is nonzero when the polynomial is inseperable, this is true

but equally gcd(f'(x), f(x)) = f'(x) up to scalar multiples when there is a unique root

[coral spindle]

wait, the question I was answering here was deleted!

:(

I think I can see how this got lost in the mix

[next obsidian]

Okay but I mean can you tell me how your algorithm tells me that f has more than a single root

[next obsidian]

I thought you were saying this algorithm will answer this question

[coral spindle]

Well let's say there is more than one! That f(x) = (x-a)(x-b)g(x)

[next obsidian]

Sure

[coral spindle]

how do I want to set this up actually

Maybe I will actually say

$f(x) = \prod_{\alpha in A} (x-\alpha)^{n_\alpha}$ let's say instead

[cloud walrus]

Duckt(ji=-k)e

[next obsidian]

Okay

[coral spindle]

Then the gcd of f(x) and f'(x) will be $\prod_{\alpha \in A} n_\alpha(x-\alpha)^{n_\alpha-1}$, up to some scalar multiple I don't care about

Fuck

[next obsidian]

Sure

[coral spindle]

I keep making typos!!!

[cloud walrus]

Duckt(ji=-k)e

[coral spindle]

So in fact if there is a single root -- in which case n_alpha = deg f for some alpha -- we can just read this off from the degree

[next obsidian]

You have issues in char p no?

[coral spindle]

Oh shite good point

[next obsidian]

In char 0 I agree

But it’s still hard to compute this gcd I think

In terms of some sort of algorithm in terms of the coefficients

[coral spindle]

I guess this works out uh if p does not divide the degree of f

[next obsidian]

Nah

If you take like

[coral spindle]

Why not?

[next obsidian]

(x-a)^p•(x-b)

You spit out 0

[coral spindle]

Well that's still fine, we can still read that off from the degree

[next obsidian]

The GCD spits out 0 though?

I mean maybe if you think harder you can still distinguish

[coral spindle]

There is indeed more than one root and the gcd should be deg(gcd of the two) = deg f - 1

So in that case it's fine

[next obsidian]

But the gcd is 0

Isn’t it just p(x-a)^p-1

[coral spindle]

Yeah...? the degree of the gcd is then not p

lol

[next obsidian]

Uhh

Okay

Sure

[coral spindle]

Am I barking up the wrong tree

[next obsidian]

No I see

But what if you were

(x-a)^p•(x-b)(x-c)^p-1

Wait

Nvm

Yeah yeah I agreege

[coral spindle]

one sec someone is pinging mods

Ok nvm people are dealing with it

OK so in general my idea sucks

[tidal schooner]

(this is basically the splitting lemma but I think it's the most natural way to prove this) You've shown (using the fact that B is an extension of Z) that this sequence splits. Equivalently, B is a semidirect product of Z acting on A. So (using the fact that A lies in Z(B)), what's the action of Z on A?

[winged void]

Is there someone who does mathematical logic

[twilit wraith]

#foundations

[valid fox]

If I have an element $\omega \in \mathbb{Q}(\omega)$ for some algebraic extension $\mathbb{Q}(\omega)$ of $\mathbb{Q}$, and I find a polynomial $f \in \mathbb{Q}[X]$ such that $f(\omega) = 0$ and $f$ is irreducible, then is $f$ the min poly of $\omega$ (over $\mathbb{Q}$)?

[cloud walrus]

Tiessie

[shy fossil]

Upto being monic. If f kills \omega, then f factors as a product of the minimal polynomial and something. Since f is irreducible, that something must be a constant. So f is the minimal polynomial if you normalize the leading term.

[valid fox]

yeah that does make sense

thankssss

[shy fossil]

np

[valid fox]

yeah so the minimal polynomial divides any polynomial which evaluates to zero at w

[shy fossil]

Yes. It follows from Q[X] being an euclidean domain

huh... a euclidean domain or an euclidean domain?

[valid fox]

a sounds better

[shy fossil]

I think so too, but rn can't remember why. It's just a gut feeling

[valid fox]

Let $g$ be such that $g(\omega) = 0$. Divide by $f$ the minimal polynomial. Then $g(X) = q(X)f(X) + r(X)$ with $deg(r) < deg(f)$. Filling in $\omega$ gives $r(\omega) = 0$. Since $f$ is minimal we must have $r= 0$. So $g = qf$.

[shy fossil]

okay, because it starts with /ju:'... which is a y-glide

[cloud walrus]

Tiessie

[valid fox]

Cool!

[limber crystal]

guys had my exam and i just realised i got the isomorphism type for the galois group to be klein 4 instead of C4 because im silly is it over

[next obsidian]

Nah

[chilly ocean]

share que

[static mauve]

If we have 2 sylow 2-subgroups, H and K (where each element of H and K has an order of 2^n for various values of n), does HK only consist of elements of order 2^n (for various n)? Was trying to prove this, and not quite getting there. It SEEMS like my book is considering this is true, but Im not convinced. I suppose it possible my book is doing something subtle and Im misunderstanding it.

[thorn jay]

if HK consists of elements of order 2^n then it must be contained in some sylow 2-group (as by Cauchy it would be a 2-group), which by size considerations (in finite groups) is impossible

[static mauve]

Well HK isnt even necessarily a group tho, right? I thought one of these groups would have to be normal for that to be the case.

[thorn jay]

ah, yeah that is true

[tidal schooner]

No. Let $G= S_3, H = \langle (12)\rangle, K = \langle (13)\rangle$. Then $H$ and $K$ are Sylow 2-subgroups, but $HK$ contains a 3-cycle.

[cloud walrus]

harmacist

[tidal schooner]

But as enpeace music said, it's definitely true if HK is a subgroup

[static mauve]

Yeah, this seems fair.

Well maybe the detail that Im not understanding can be cleared up specifically:

I have underlined what I currently dont believe based on what we are saying here

TLDR: Trying to show that a Group of order 160 is never simple.

Lemma 37.8 is |HK| = |H||K|/ |HnK|

[tidal schooner]

I thought about it for a while — and ppl can feel free to chime in if I'm wrong here — but I think the step in this proof isn't valid (none of the other assumptions here would make this true about HK).

FWIW, here's the way I'd personally prove this: If G is simple, then n2 = 5. Let G act by conjugation on the set of its Sylow 2-subgroups. This induces a homomorphism G -> S5. It's nontrivial because of Sylow II, and it's not injective because then G would embed into S5, but 160 doesn't divide 5!. So, the kernel is a nontrivial normal subgroup

[static mauve]

Yeah, I was beginning to believe the same thing! It wouldnt be the first time Ive found a mistake in a book. Thanks for your thoughts and you own path forward in this problem. I appreciate your time 😄

[tidal schooner]

Np! Happy studying :)

[inner owl]

So the question is I have to classify the rings of order 10. (I am assuming rings with unity and are commutative since they are the only kind of rings I am dealing with in my coursework)
My work :-

1. The ring can't be a domain since then it would be a field but a finite field has a prime power as it's cardinality.
2. Since a ring also forms a group over addition therefore by lagranges theorem the subgroup generated by unity divides 10 => the ring can have characteristic either 2 5 or 10 (it can't be 1 since then it would be the 0 ring) now if it is 10 then this ring is isomorphic to Z_10 which is isomorphic to Z_5 × Z_2.
   Now if We assume that the characteristic is 2 or 5 then and consider the ring of order 10 as an extension of these fields now the minimal polynomial of the added element can't be monic and linear cuz then we are not really adding any new element to the ring so if it monic and non linear of then it would be a field extension and the cardinality of the bigger field would be either a power of 2 or 5 in the case of 5 the bigger field would have cardinality at least 25 so the sub ring (of order 10 ring) can't have cardinality 5 (or characteristic 5) and also any extension of F2 would have cardinality a power of 2 so again it can't be a subring.
   Now if suppose that the minimal polynomial is not monic then the ring extension would have infinite order therefore the only possible ring is Z_10 or Z_5×Z_2
   Can someone please check my work...if it is correct can you give me more rings of order 10 if we remove the condition for it being commutative and unital also can you comment on my way of analyzing this like if there could have been a better way to do it.
   Thank You.

[kind temple]

Z_5 x Z_10 has order 50

do you mean Z_2 x Z_5?

[inner owl]

Ah yes sorry typo

[kind temple]

i havent checked your work, but there is an easier way
if 1 + 1 = 0, examine what happens to r + r for arbitrary r in R

same for r + r + r + r + r if 1 + 1 + 1 + 1 + 1 = 0

the condition on it being commutative doesn't matter here since the addition is always commutative and multiplication by 0 on the left or right is always 0

[inner owl]

Then that would mean each element has order 2 or 5 depending upon the characteristic of the ring and that would violate the cauchy theorem?

Since if characteristic is 2 then the ring of order 10 won't have an element with order 5

And similarly for characteristic 5

[tidal schooner]

btw, you can show that all rings of order 10 (or more generally a squarefree order), unital or not, are commutative, either directly or by applying Sun's theorem/CRT to the underlying additive group of the ring
also, there do exist nonunital rings of order 10 -- try finding some!

[kind temple]

yea, this reasoning sounds fine. you could also argue that the additive group would have to be Z/5Z or Z/2Z if the characteristic of the ring is 5 or 2, resp

[tribal moss]

It looks right, if perhaps more involved than it needs to be. Do you have the structure theorem for f.g. abelian groups?

At least, instead of speaking of minimal polynomials you could jump directly to saying, if the ring contsins F2 or F5, then it would be a vector space, and it has the wrong size for that (as you already end up arguing in the F2 case anyway).

[kind temple]

how do you start classifying the non-unital ones?

the simplest one i can think of is ||just to make all products 0||

btw, the general fact that troposphere and i are referring to is the fact that if every non-identity element of an abelian group G has order p for p prime, then G is a vector space over F_p

[inner owl]

Thanks for the input will definitely try

[inner owl]

Ah , yes thanks

[inner owl]

Sadly no 😔 this my first course in abstract algebra

[inner owl]

Ah yes , thank you

[tidal schooner]

another one is 2Z/20Z

[inner owl]

Oo nice , will definitely work this out

[kind temple]

is nZ / 10nZ a non-unital ring in general?

[empty kernel]

is it generally true that if phi is a group homomorphism from G to K
and if N is a subgroup of K then phi^(-1)(N) is a subgroup of G?

[tribal moss]

If n is coprime to 10 you just get Z/10Z.

[kind temple]

okay, true

[tribal moss]

(With a nonconventional unit)

[empty kernel]

i think i proved it but since we didnt see it as a theorem i feel like i might have made a faulty proof

[tribal moss]

Yes.

[empty kernel]

thank you

[grizzled crane]

Can you comment on this?

How may I interpret the binomial coefficent as a field element?

It seems like nCk should be a natural number that invokes repeated multiplication of a field element

so only the multiplication operation is involved in the summand in the right?

not the notation k*x = x + x + …+x (k times)

[grizzled crane]

right, I can just express any natural number k as group element (1 + .. + 1 (k times))

It got bit tricky because,

3 = 1 + 1 + 1

but if K = Z_2 , 3 is the group element 1

similar to this

Didn’t click me before

I’ll share hold on

I don’t have the english but it says the same thing

If n is a natural number, then

below is a an element in the field

when K = R, this element is just the natural number, otherwise it changes

[drifting mason]

Not sure if I've come to the right channel. Anyway, I know that if $F[x]$ is a polynomial ring over a field $x$, then each polynomial determines a polynomial function. The map that sends an element in $F[x]$ to the polynomial function is one-to-one if $F$ is infinite. Is it also onto? Why? \

Now I'm studying the Zariski topology very briefly and without going too much into the details. The author lets $k(X_1,\ldots,X_n)$ be the ring of polynomials in $n$ variables over the field $k$. Then also says that there is a one-to-one correspondence between the polynomials $P$ and polynomial functions $p$ if $k$ is infinite. Is the proof of this result basically the same as for $F[x]$?

[cloud walrus]

psie

[next obsidian]

Isn’t it by definition onto lol?

What is your definition of a polynomial function?

I think if you just examine that definition you see it’s onto

[drifting mason]

I don't mean for the polynomial function to be onto (perhaps it is anyway...). I mean the function that takes a polynomial and maps it to the polynomial function. The definition is kind of loose; a polynomial function is simply the function that we obtain by plugging in elements of the field into the variables of the polynomial.

[next obsidian]

That’s what I was talking about

[drifting mason]

ok 👍

[tribal moss]

So it is onto the set of all polynomial functions (because that set is by definition the image of your map) but definitely not onto the set of all functions.

[vivid kestrel]

we proved that for an artinian ring A with jacobson radical J(A), A/J(A) is simple, since, because A is artinian, we can write J(A) as a finite intersection of maximal submodules X_i. The prove was essentially just that the above map is an injective module homomorphism into a semisimple module, so A/J(A) is semisimple as a submodule of a semisimple module. Now my question is, to establish that the above map is an injective module homomorphism, how does the direct sum being finite actually matter

[next obsidian]

It’s not really an exact answer, but the key point here is that J(A) is a finite intersection

This lets you do the CRT to get that A/J is isomorphic to the product of the things which interact to form J (and actually it’s the product of the ideals)

To illustrate why this finiteness matters, in a polynomial ring k[x] with k algebraically closed the Jacobson radical is 0, but k[x]/m = k for any maximal ideal m

[next obsidian]

So if you could still do this you would get that
k[x] = k[x]/J ≈ Prod_a in k k

[vivid kestrel]

sure

[next obsidian]

Where the product has a single entry of k for every element in k

[vivid kestrel]

I don't see what breaks with an infinite intersection, we could still write down the same map and it would be a well defined module homomorphism right

[next obsidian]

Sure

But there’s a couple issues

1: There’s no guarantee the image is almost all 0 (difference between direct sum and direct product)

2: Infinite direct sums of rings are not rings because they lack a 1

[vivid kestrel]

Ah I see so it wouldn't actually be a well defined map into the direct sum

[next obsidian]

Right

[vivid kestrel]

thanks!!

[next obsidian]

Is any element of A in all but finitely many maximal ideals?

The answer is usually no (or at the very least it’s always no for 1)

[spice wagon]

$\Lambda$ solvable iff its derivative sequence is stationary at ${ e}$. does this statement have a name ?

[cloud walrus]

tm

[next obsidian]

A definition?

[spice wagon]

solvable groups aren’t defined by derivative sequence in my book so 🤷‍♂️

[next obsidian]

They’re equivalent definitions

[spice wagon]

okay

[next obsidian]

Usually people say “X is Y if it satisfies one of these equivalent definitions”

And the proofs that they are equivalent aren’t given names

[neat beacon]

how did they directly conclude that these are primitive elements of F_16?

[topaz solar]

Well, conjugates sounds like the image under an automorphism

So proving that any one of them is primitive would be nice

[neat beacon]

yeah but they've introduced that interpretation after this example

[topaz solar]

Curious order but nothing saying you have to only interpret examples in light of what comes before

Not too hard to show that map is nice algebraically

So all that would be left is showing that they’re primitive

So why is alpha, a root of that minimal polynomial, primitive

[neat beacon]

yeah, also the fact that the degree is 4

and f is over F_2, with splitting field F_(2^4)

[topaz solar]

Well you can make it out of that one f so I’d sure think at least one root is principal, and they’re all conjugate

(Which isn’t too hard to see is an automorphism, I’m pretty sure)

[neat beacon]

hmm.. i dont understand what you mean, Im sorry 😭

the conjugates of $\alpha \in \mathbb{F}_{q^m}$ with respect to $\mathbb{F}_q$ are the the images of $\alpha$ under the automorphisms $$\sigma_i (x) = x^{q^i}, i\in {0, 1, \dots, m-1}$$

[cloud walrus]

omeganebula

[neat beacon]

how does the fact that they are automorphisms imply they are primitive?

[vast stump]

if alpha wasn’t primitive, it would have order 3 or 5, which is not the case

[neat beacon]

ah... im dumb

thanks!

also can we generalize this? the root of any irreducible polynomial in $\mathbb{F}_q [x]$ is a primitive in the splitting field of the polynomial?

[cloud walrus]

omeganebula

[vast stump]

answer is no, you can find a small counterexample over F_2 (for instance)

in particular, ||irreducible x^4 + x^3 + x^2 + x + 1 = (x^5 - 1)/(x - 1), roots all have order 5||

[spice wagon]

let $\bigoplus G_i ={ (g_i) \in \prod_{i \in I} G_i,}$ where $(g_i)$ has a finite support, this thing is the external direct sum or the internal one ?

[cloud walrus]

tm

[spice wagon]

im so lost with all these direct sum

[thorn jay]

External, as you're constructing it directly

internal direct sum is more of a property of a group, rather than a construction

As you can always build an external direct sum from groups, but not every group is an internal direct sum

[spice wagon]

so H and K are in direct sum (internal) if $H \cap K={ e }$ ?

[cloud walrus]

tm

[thorn jay]

No

[spice wagon]

0 mb

[thorn jay]

Are we talking about abelian or nonabelian groups here

[spice wagon]

H and K subgroups of G abelian

[thorn jay]

Yeah ok

[spice wagon]

the subgroup generated by $H \cup K$ is $G=H+K$ and is said to be the direct sum of the groups H and K when $H \cap K={0}$

[cloud walrus]

tm

[spice wagon]

that’s what it says in my book

[thorn jay]

G is the internal direct sum of K and H, if K and H are subgroups of G, and K + H = G and K ∩ H = 0

[spice wagon]

yeah ok so this is the internal direct sum

and what if G is non abelian?

[thorn jay]

Then you get the direct product

[spice wagon]

how ? the direct product is a set of tuples

[thorn jay]

You have an internal direct product

[spice wagon]

which does not correspond to the cartesian product then

[thorn jay]

It's naturally isomorphic

Same way that an internal direct sum is naturally isomorphic to the external direct sum

[spice wagon]

yeah but it’s not longer like H \times K, it’s HK

[thorn jay]

It's naturally isomorphic

Formally, the internal direct sum is also not equal to the external direct sum

[spice wagon]

vro i don’t ask if they are isomorphic or not 😭😭😭😭

just wanted to know if the elements of the internal direct product are tuples or not

like H \times K

[thorn jay]

They aren't

[velvet hull]

they're isomorphic to tuples

[spice wagon]

okay thx that’s what i needed

[velvet hull]

but since we're working within a group they're necessarily just elements of that group

[thorn jay]

Naturally isomorphic means that they are almost literally the same object, and is even stronger than just isomorphic

[knotty badger]

Usually natural isomorphism means the isomorphism used is “polymorphic”

In the computer science sense

It can be constructed in a uniform way, often as a result of being purely syntactic

[velvet hull]

hmmmm, mind explaining a little more? I dont know any cs

im interested

[knotty badger]

So for example, there’s a “natural” way to turn elements of a type T into a list of type T

You send x to [x]

You could also do x to [x, x, x]

Or even just x to []

For specific types T, you can do fancier things

E.g. send a natural number to its list of prime factors

[velvet hull]

hmm, so natruality in the comp sci sense literally boils down to the function accepting an entire type as input instead of specific values of the type

[knotty badger]

But the constructions I described above are “polymorphic”

[knotty badger]

It’s being able to define a function that works on arbitrary types

We’re only using the syntax inherent to lists here, nothing else

And so these functions make sense for any type T

[velvet hull]

hmmmm

so what's a syntactic function

mind giving an example

[knotty badger]

Or for example, you can define a function from List[T] to List[T] that reverses lists

This is also “polymorphic” in the sense I described above

[knotty badger]

All the ones I gave above, except for the “send natural to list of prime factors”

[thorn jay]

Actual natural transformation in the category of types

[knotty badger]

Also yes that

Though polymorphism is technically stronger than naturality

[thorn jay]

Call the type theorist

I can imagine

[jade mason]

Are there any naturally appearing artificial natural isomorphisms

[thorn jay]

What do you mean by naturally appearing artificial lol

[jade mason]

Like natural isomorphisms which do make choices

[knotty badger]

Maybe those you use in proving full, faithful and essentially surjective => equivalence?

[thorn jay]

Ah yeah, of course

Extending the classical AG duality from coordinate rings to finitely generated reduced k-algebras, for example

[knotty badger]

I see

[thorn jay]

(coordinate rings are quotients of polynomial algebras)

[jade mason]

I'll get back to you on that someday lol

[thorn jay]

Basically you have to choose a generating set for each algebra

But it works out

[hot goblet]

how do I know which field extensions are Galois? I keep getting them backward in my head

[next obsidian]

Do you want a definition or a heuristic?

[spice wagon]

so ur saying $H \oplus K \cong H \cdot K$ ?

[cloud walrus]

tm

[hot goblet]

both

[spice wagon]

the internal direct sum is isomorphic to the internal direct product

[next obsidian]

Definition: Normal and separable
Heuristic: Idk

A splitting field is going to be Galois in char 0

[hot goblet]

definition is just |Gal| = [E:F]

[next obsidian]

That’s probably the most common situation you will find

[hot goblet]

these are not the definitions or words that were in my course lol

[next obsidian]

Okay, but that’s not very verifiable

[hot goblet]

although i think we only did galois theory for polynomials over Q

[next obsidian]

Separable means the min poly of any element in the extension has no repeated roots

This is automatic in char 0

[hot goblet]

hence why i find it frustrating

ah right yes makes sense

[next obsidian]

Normal means that the field extensions contains every root of every minimal polynomial

[thorn jay]

That's not what I was saying

[next obsidian]

So if alpha is in E and beta is another root of m_alpha, then beta is in E

This is true for a splitting field

[hot goblet]

so for the last one

[thorn jay]

I don't think with abelian groups I've ever heard of internal direct product being used

[hot goblet]

theres an extension thats not Galois

[hot goblet]

mm makes snes

[spice wagon]

so what ur saying about the isomorphism

[hot goblet]

sense

[next obsidian]

Sure

I don’t remember off of the top of my head to be honest

But I think yes

[hot goblet]

but im not really sure where or why

[next obsidian]

But the Galois correspondence tells you that you can do this group theoretically

[thorn jay]

I'm saying that the internal and external versions are isomorphic

[next obsidian]

Once you compute the Galois group you find a non-normal subgroup

[hot goblet]

so the roots are all the 4th roots of 2

[thorn jay]

Of both direct sum and direct product

[spice wagon]

in the case non abelian?

[next obsidian]

And this corresponds to a subfield which is not Galois over the base

[hot goblet]

and the splitting field size is 8 since 4*2

[thorn jay]

Why are we switching to nonabelian groups all of a sudden

[spice wagon]

because i was talking about non abelian groups

[hot goblet]

I know that the galois group is dihedral from looking at the answers but again not really sure why

[next obsidian]

You need to figure that out

Write down automorphisms

They are determined by where the generators = roots of the polynomial get sent

And then you find some relations

[hot goblet]

hmmm interesting

[next obsidian]

The dihedral group is generated by something of order 4 and order 2

Which have some relation like

rfrf = e

[thorn jay]

Well, yes

[next obsidian]

If you can verify those sorts of relations you can either come up with an explicit isomorphism

[thorn jay]

I guess?

[hot goblet]

yeah but how do we know that the Gal is nonabelian?

[thorn jay]

Its easy to verify

[next obsidian]

You just verify it lol

[hot goblet]

like, why isnt it Z/4ZxZ/2Z

[hot goblet]

ig this why

[next obsidian]

the elements are automorphisms which you can explicitly write down

[thorn jay]

But anyways, internal direct sum and direct product are the same

[next obsidian]

So you just find two of them that don’t commute

They’re functions you can just compose and plug stuff in

[hot goblet]

ok well ig the instructions are clear mess around with them and see what happens

thanks

[thorn jay]

It's just that for abelian groups you'd tend towards saying direct sum while with groups you'd tend towards saying direct product

[next obsidian]

You gotta just start writing stuff down and play with it

[hot goblet]

yeah ill give that a go

thank you

[next obsidian]

I’m not the best at Galois theory so I can’t give any heuristic better than that

Yeah np

[hot goblet]

no worries still helpful

[hot goblet]

actually wait sorry

this is the step i was confused on

oh wait

you can use that crappy definition

[hot goblet]

we already computed size is 8 and we know this is degree 4 so it cant be galois

nevermind problem solved thanks anyway

[spice wagon]

it coincides if we take the addition for the direct product law right ?

[thorn jay]

What do you mean direct product law?

I was talking about internal direct product too

[spice wagon]

the law on $H \times K$ is noted multiplicatively in my course but nvm

[cloud walrus]

tm

tm

[thorn jay]

Oh you mean the operation

[spice wagon]

yeah

[thorn jay]

Internal direct product doesn't have a choice of operation

(nor does the external direct product, for that matter)

[thorn jay]

Because the internal direct product, nor the internal direct sum is an object

[spice wagon]

but when you say that the two are the same, you’re talking in terms of isomorphism, not equality right

[thorn jay]

It's a property

[thorn jay]

internal

You're not constructing anything

A group is the internal direct sum of N and M iff it is the internal direct product of N and M

Where N, M < G

[knotty badger]

(are these abelian groups?)

[thorn jay]

Not necessarily, you technically have a notion of direct sum which is just the product but all the elements with finite support

[cobalt heath]

Yeah, "direct sum" is not a thing for usual groups

[spice wagon]

alr i’ll try to prove it

[knotty badger]

afaik this usually refers to the free product

[thorn jay]

I'd use coproduct

For that

[knotty badger]

yes that's the free product in Grp

[cobalt heath]

Ah, do they call that direct sum as well?

I guess it makes sense

[knotty badger]

in general direct sum is coproduct i think

[thorn jay]

Well there's really nothing that you're summing so the terminology doesn't really fit anyways

[knotty badger]

mhm, i don't think it gets used often for groups

[thorn jay]

Nothing really to prove

The definitions are the same

Direct sum is just for abelian groups and direct product for nonabelian groups

[cobalt heath]

Wellllll

[cobalt heath]

Depends on the definition they learned

[spice wagon]

my brain need it

[thorn jay]

For internal direct sum / direct product?

[cobalt heath]

Like, you could certainly prove that a direct product construction is actually product

And direct sum construction is actually coproduct (in abelian groups)

[thorn jay]

Thats not what I was talking about

Augh nevermind this conversation is useless this is why we should teach category theory to preschoolers so they understand the difference between direct sum and direct product

[cobalt heath]

I mean, it's not like everyone is taught the same definition. You do not need full category theory to talk about universal property, either

[thorn jay]

Yeah, that is true

[delicate orchid]

💔 ts is dire. Direct product for both groups and abelian groups is the same it just so happens that in Ab it's a biproduct 📉

[thorn jay]

Shiver me timbers 🥶

[spice wagon]

what bothers me is that one corresponds to a product and the other to a sum

[thorn jay]

Because one is used for abelian groups and the other for nonabelian groups

[spice wagon]

okk

[jade mason]

I need to prove that if F is a left exact functor Mod_A -> Mod_B, then for every f: M->N, F(kerf) is naturally isomorphic to kerFf. Which naturality diagram do I need to check here?

For morphisms of short exact sequences of the form ker f-> M->N?

[south patrol]

This is pretty much the definition of left exact

[jade mason]

Yes the whole problem is kind of weird

[south patrol]

Like if you consider any f then consider 0 -> ker f -> A -f-> B

Well also any short exact sequence is of this form

[jade mason]

I'm just wondering in what sense the word natural is meant

[knotty badger]

categorical i would assume

oh wait that's not helpful lol

[jade mason]

Is ker a functor

[knotty badger]

apologies

not as far as im aware..

(perhaps on the arrow category...?)

[mighty kiln]

A natural iso of functors [[1], ModA] → ModB?

[jade mason]

Maybe I'll just write "there's no choices made so the isomorphism is plenty natural"

[south patrol]

ker f is a limit and you are saying F preserves that limit

[knotty badger]

ah sure

[mighty kiln]

There's still a naturality diagram for "preserves limit"

[south patrol]

Yes

[jade mason]

We've barely defined categories and functors

Normally the lecturer writes naturality claims out explicitly

[mighty kiln]

The functor F maps kerf → M → N to a kernel diagram F(kerf) → F(M) → F(N)
(Note this is stronger than just saying F(kerf) ≈ kerFf)

I think that's all you have to check

[mighty kiln]

By abstract nonsense they will automatically assemble into such a natural isomorphism

[jade mason]

I decided to prove that F kerf satisfies the universal property of the kernel

[tardy hedge]

Are u in a category theory class?

[jade mason]

No, commutative algebra

[tardy hedge]

Oh nice

Whats the syllabus for that class?

[tardy hedge]

Can u send me the problem set or whatever this comes from cuz this is the sort of problem i want to study atm too

I was thinking about exactness of functors and still not exactly sure in what sense they preserve kernels or cokernels or whatever

[jade mason]

A little bit of everything. So far we've done Zariski topology, some homological algebra, localizations, noetherianity/artinianity, Nakayama&PIT, integral extensions and now discrete valuation rings

[tardy hedge]

Wow what a cool class

Thats nice

[jade mason]

here's the whole statement

I don't know why we're suddenly doing this now because we did homological algebra a while ago already

All the other problems in the sheet are about our current topics

[vast verge]

Hi, could you explain why I got the second part wrong?

I'm not very familiar with the elements of D4 so I derived everything off of what I know of square symmetries

I don't think I should be getting that st=ts

[tidal schooner]

The second part's asking for the elements of order 2 in [D_4=\langle \sigma,\tau\mid \sigma^4=\tau^2=1,, \tau\sigma\tau=\sigma^{-1}\rangle.] The identity $\tau\sigma\tau=\sigma^{-1}$ is crucial to working with dihedral groups; try proving it geometrically if you're not already familiar with it. Note that it's equivalent to the identity $\tau\sigma=\sigma^{-1}\tau$. In particular, for $D_4$, we get $\tau\sigma=\sigma^3\tau$. I think that's where the error in your calculation happened (it appears that you wrote $\tau\sigma=\sigma^2\tau$). \

So, here's a worked solution: It turns out that \emph{all} elements of the form $\sigma^k \tau$ (equivalently, $\tau\sigma^{-k}$) are reflections and thus have order 2. (This actually holds for all dihedral groups $D_n$.) Indeed, [(\sigma^k\tau)^2=\sigma^k\tau\sigma^k\tau=\sigma^k\sigma^{-k}\tau\tau=1,] where we've used the identity $\tau\sigma=\sigma^{-1}\tau$ repeatedly. So, in this problem, we should actually have [L={1,\sigma^2,\tau,\sigma\tau,\sigma^2\tau,\sigma^3\tau}.]

[cloud walrus]

harmacist

[vast stump]

correct; it should be sr = r^3s, whereas you wrote sr = r^2s

[vast verge]

Okay thank you

I'll have a look at it another time

[vast verge]

This should be correct now

[tidal schooner]

Looks great! Nice work :)

[vast verge]

Thank you

I got confused when doing the inverse for σ^(-1)

There's also a more general rule for any dihedral group:

For any Dihedral group $D_n$, one has: $\sigma^{-m} \tau = \tau \sigma^m$

[cloud walrus]

Tropical Greens

[vast verge]

I think this is correct

[tidal schooner]

Yeah! That follows from applying $\tau\sigma=\sigma^{-1}\tau$ repeatedly

[cloud walrus]

harmacist

[glad osprey]

Are all the classical Lie groups of the form $Aut(V, \beta)$ where $\beta$ is a bilinear form? ie. automorphisms of a vector space preserving a certain bilinear form? (except the special variants I guess, where we intersect with SL(V))

[cloud walrus]

sheddow

[dull ginkgo]

<@&268886789983436800>

[tidal summit]

Is this correct??

Can anyone suggest how to proof this using definition??

[twilit wraith]

sorry for the terrible crop

but im really terrible at combinatorics so i dont fully how to explain whats going on here

intuitively, im guessing theres initially 8 elements in the centralizer in S4, and then each symmetric group above that introduces (n-4) times more disjoint cycles to consider

but how do i know something like (12)(345) doesnt commute with (12)(34)

[crystal vale]

Do you know orbit stabilizer theorem?

[twilit wraith]

yes

[crystal vale]

So now if you count the how many there are elements in orbit of (12)(34), then you will get your answer, action is conjugation

Do you know what g(12)(34)g^-1 looks like?

[twilit wraith]

not really

i think thats the biggest problem

im struggling to understand conjugation in the symmetric group

[crystal vale]

I think you are following Dummit and foote , so in the Dummit and foote there is a theorem which states that a and b are conjugate to each other iff a and b have same cycle type

[twilit wraith]

no, this is a homework exercise from my prof

though he likely pulled it from D&F

[twilit wraith]

even if they arent disjoint?

so anything of the form (..)(..) is conjugate

[crystal vale]

Yes

[twilit wraith]

i see

[twilit wraith]

so i have to show that the size of the orbit of (12)(34) is n!/8(n-4)!

[crystal vale]

they have to be disjoint

Yes exactly

[twilit wraith]

alright cool

i missed that theorem

i think my prof went over it very quickly

but i was skeptical about if i was allowed to pull it off in my proof

[crystal vale]

So now you have to calculate number of elements of the form (ab)(cd), all are distinct

[twilit wraith]

that should be easy

[crystal vale]

Yes

[twilit wraith]

its n permute 4 divided by (2!)^3

so yeah i see it now

since we select four elements from n elements to permute and then say that swapping certain pairs doesnt change anything and swapping the order of those pairs doesnt change anything

[minor lance]

I imagined treating the properties themselves that algebraic structures have as open spaces and try to put as many algebraic structures as possible as point in the topology

Things like associativity, existence of unit, commutativity becoming an open space.

I dont know if this is the right place to put this idea, but I found it nice.

[thorn jay]

These aren't quite open sets

But this is very close to the idea of a lattice of deductively closed theories over some signature and infinite set of variables

Although this idea makes sure everything is still set based

(putting a closure operator on the collection of "all algebraic structures" is a bit iffy because that's at least going to be a class if not worse)

[minor lance]

I just can feel lattice is super close with this idea

[minor lance]

But still, why?

I mean, properties themselves can be AND'd and OR'd and it will generate another property.

Kinda close to the whole idea of open spaces being closed to finite intersection and union under a topology.

[thorn jay]

Well, depends on whether or not you want infinitary propositions or not

[minor lance]

Mhm nice point

[thorn jay]

And for the OR to work nicely, you have to make sure there are no free variables, else you'll get some weird meddling interference

Finally, due to negation, every set would be clopen anyways

And at that point it's better to just treat it as a lattice

[minor lance]

Thx a lot

[thorn jay]

(though you can still ask questions like "what are the conditions for a class of structures to be axiomisable by so and so type of propositions")

[minor lance]

Actually when I first rly thought of this it was during the convo with ex and it was chu space instead of topological space

But kinda felt bad pushing a slightly niche concept I was obsessed with to mathcord so I js said topology

Anyways im acc drnk rn buhbye

[thorn jay]

Lmao

[thorn jay]

Huh, never heard of them

Lore drop when

[minor lance]

That shit is TOTAL fire bro

[thorn jay]

Reminds me of fuzzy sets

[minor lance]

You drop og requirements of open sets having to closed under finite caps and cups and somehow still can construct a sensible definition of continuous maps? Absolute cinema

[thorn jay]

Cuz take K = [0, 1] lol

[minor lance]

YEAH

[thorn jay]

And then you've got like a "fuzzy state machine"

Ah chu spaces are used in automata

Who could've guessed

Is there any algebraic chupology?

[minor lance]

You can make one

Also Pratt (from the legendary gangster shit, KMP algoeithm) did something called Stone Gamut

Which is category theory based heavily and also the basic of chupology apparently

But my shallow knowledge couldnt get it

It looked like fire though

[thorn jay]

Stone? Same Stone as Stone duality?

[minor lance]

YUp

[thorn jay]

Actual spectrum moment

Boolean algebras are crazy

Like wtf do you mean your only subdirectly irreducible member is the only simple member is the algebra generating the variety as a prevariety??

[minor lance]

Dude,,, igtg see u man

[thorn jay]

Cya!!

I've got work anyways

[chilly ocean]

wtf

[balmy vector]

sorry for the basic algebra question but uhhh

is it true that if H, K are subgroups of G, the index of K in G would be divisible by the index of H intersection K in H?

(that is, is $[H:H\cap K] \vert [G:K]$?)

[coral spindle]

maybe you want \mid there, not \vbar

[cloud walrus]

bracket

[coral spindle]

Let me think if there's a simple proof for this

[balmy vector]

i know that this works if you replace divides with < then that's a well known result

[coral spindle]

My instinct says it's true and it's easy to prove when H is normal but give me a moment just to think

Just need to write this out on my blackboard

[balmy vector]

oh yeah that is true when H is normal

[coral spindle]

Wow I am struggling so much to prove this

OK so quick observation: the only coset in of $G/(H \cap K)$ which is contained in both $H/(H \cap K)$ and $K/(H \cap K)$ is $H\cap K$ itself. To see this, observe that if $h(H \cap K) = k(H \cap K)$ then $k^{-1}h \in H \cap K$ so both $k$ and $h$ are common to both $H$ and $K$.

[cloud walrus]

Duckt(ji=-k)e

[coral spindle]

So I think we can do this

wait

Yeah OK

Ach no this isn't going to work

Let $a_i$ and $b_j$ be coset representatives of $H$ and $K$ in $G$ respectively. Then:
\begin{align*}
G
&= \big(\bigsqcup_{i} a_iH\big) \cap \big(\bigsqcup_{j}b_jK\big) \
&= \bigsqcup_{i, j} a_iH \cap b_jK \
&= \bigsqcup_{\substack{i, j \ b_j^{-1}a_i \in H}} b_j\big(b_j^{-1}a_iH \cap K\big)
\end{align*}

[cloud walrus]

Duckt(ji=-k)e

[coral spindle]

I can add this b_j^-1a_i \in H condition since otherwise the intersection is zero

and indeed this just becomes:

$= \bigsqcup_{\text{ditto}} b_j(H \cap K)$

[cloud walrus]

Duckt(ji=-k)e

[coral spindle]

So we need to think about how many of these cosets there are

Ah, wait. This isn't right.

[coral spindle]

If H is a subgroup of K then my idea up here is wrong

Darn.

I thought I could argue by the presence of the identity but I cannot

[thorn jay]

Take H = { (1), (12) } and K = { (1), (23) } and G = S3

The intersection of H and K is the identity, so we get [H : H ∩ K] = 2/1 = 2 and [G : K] = 6/2 = 3

I think that suffices as a counterexample

[thorn jay]

It is not, I am pretty sure

[coral spindle]

Nice counterexample

Very nice

[thorn jay]

Thank you :3

Subgroups are nice

But not that nice

Because why would they be :P

[coral spindle]

This is slander!

Wait enpeace aren't there even simpler counterexamples

[thorn jay]

S3 is the smallest nonabelian group

[coral spindle]

P sure cyclics work

Oh wait no ofc not

[thorn jay]

As it works for normal subgroups, this is quite literally the smallest example

[coral spindle]

Yeah

[thorn jay]

Lol

[coral spindle]

Nice

[thorn jay]

It fails IMMEDIATELY, math really told us "no lil bro, this is NOT it"

[coral spindle]

And I didn't even have to do any search work

I can just sit back and let the mathcord do its thing

[thorn jay]

Happy to serve

[distant summit]

Is (i) a well posed question?

If O(3) is a subgroup of O(3,1), that means an element of O(3) is in O(3,1)

But an element of O(3) is a 3x3 matrix, while O(3,1) consists of 4x4 matrices

Or are we treating elements of $O(3)$ like block matrices
$$\begin{pmatrix} 1 & \vec{0}^T \ \vec{0} & Q \end{pmatrix}$$

[cloud walrus]

Douglas

[chilly ocean]

whats ur sign convention

[distant summit]

Mostly positive

[chilly ocean]

t axis with -1

[distant summit]

What?

I don't understand what your point is?

[chilly ocean]

use selvester

[distant summit]

What is that

[chilly ocean]

sylvester law of inertia

[distant summit]

What does this have to do with my question?

[distant summit]

I'm asking about the validity of this

[chilly ocean]

okay sorry didnt read that

[knotty badger]

Yes so strictly it’s not a subgroup

However there’s a subgroup of O(3, 1) isomorphic to O(3)

Another way of saying this is that there’s an injective group homomorphism O(3) -> O(3, 1)

[knotty badger]

The image of this is then this subgroup

[thorn jay]

If the injective homomorphism is obvious enough then it might as well be a subgroup

[knotty badger]

Well it’s not a strict subgroup afaik

[thorn jay]

What is a subobject of X but an object in the skeleton of the category of monomorphisms into X

[distant summit]

So what do you actually need to show (ignoring the subtle point about isomorphism)?

Can you assume O(3) is a group in its own right and you only need to show it is contained in O(3,1)?

Or do you need to prove both that O(3) is a group and it is contained in O(3,1)?

[delicate orchid]

pseudo has already said that this question is equivalent to finding an injective group homomorphism from O(3) to O(3,1). I think it's safe to assume at this point that O(3) is a group

[distant summit]

Right so it just comes down to showing that O(3) is a subset of O(3,1)?

[delicate orchid]

well no because as you correctly said, O(3,1) as a set only consist of 4x4 matrices and O(3) consists of 3x3 matrices, so it definitely isn't a subset. This is where I agree that the phrasing of this question (which I assume is coming from a physics based course) is a bit misleading. Just read it as "find a subgroup of O(3,1) that is isomorphic to O(3)". I am just repeating what Psuedo has said atp

[distant summit]

so the block diagonals diag(1,Q) where Q is in O(3)?

[delicate orchid]

sure that would work

[balmy vector]

oh yeah thanks!

[thorn jay]

:3c

[quartz wind]

what can i say about $a - b^{-1}ab$ if $a,b\in\text{Hom}_{\mathsf{Vect}}(V)$? im trying to prove that for $\mathfrak{sl}(V)$, $[V,V]= V$. my first step would be showing that is indeed a vector space, and i have obtained the following
$$
[u_1, v_1] + [u_2,v_2] = [u_1+u_2,v_1+v_2]-([u_1,v_2]-[v_1,u_2])
$$
$$
[u_1,v_2]-[v_1,u_2]= u_2(v_1-u_2^{-1}v_1u_2)+u_1(v_2-u_1^{-1}v_2u_1)
$$
is this a dead end or could i somehow invoke some conjugacy thing?

[cloud walrus]

esca

[quartz wind]

of course everything here has trace 0, if thats necessary

also char F = 0 i forgot to mention sorry

[quartz wind]

i have to use the trace 0 fact, otherwise its not true

[tidal schooner]

I'm not sure what the notation in Sentence 2 means---I'll assume that you're trying to show that $\mathfrak{sl}(V)$ is closed under the Lie bracket, and that you've defined $\mathfrak{sl}(V)$ to be the set of traceless linear operators $f:V\to V$. You've written the Lie bracket as $[f,g]=f(g-f^{-1}gf)$. Note that this exact expression doesn't necessarily exist because $f$ might not be invertible. However, what we \emph{can} do is write [ [f,g]=fg-gf,] the standard way of writing a commutator. Now, showing that $\mathfrak{sl}(V)$ comes down to showing that $tr(fg)=tr(gf)$. And the fact that $\mathfrak{sl}(V)$ is a vector space follows from the linearity of trace

[cloud walrus]

harmacist

[tidal schooner]

@quartz wind

[quartz wind]

im trying to show that the bracket is surjective, i.e. ${xy - yx: x,y\in V}=V$. but yeah thats a good point that the matrices arent necessarily invertible, i forgot this is the algebra and not the group

[cloud walrus]

esca

[quartz wind]

thanks

basically the closed under linearity problem reduces to "given $v,u,w,x$, when can we say there exist $s,t$ such that $u_iv_j+w_ix_j=s_it_j$?"

this is via the structure constant formulation of the commutator

[cloud walrus]

esca

[inner owl]

Suppose I have a ring homomorphism from C[x,y] to C[t] which is identity over C and maps x to x(t) and y to y(t) and it is given that not both x(t) and y(t) are constant I have to show that the kernel of this map is a principal ideal.
My work :
Suppose that f(x(t) , y(t)) = 0 then since not both the polynomials in t are constant therefore f is 0 at infinitely many points in C^2 and f is not irreducible (in the source ring) then it factors uniquely into irreducibles(upto associates) in the the ring C[x,t] since it is a UFD... now if under the map this goes to 0 then one of the irreducibles goes to 0 since the target ring is also a UFD now I claim that this irreducible polynomial generates the kernel of the map. For this I will use the result that if f and g are elements of C[x,t] then the set of common zeroes of these two polynomial in C^2 are finite unless they have a proper factor in C[x,y].
Now if any other polynomial is 0 under this map and it is not a multiple of our candidate polynomial then it would violate the above mentioned result therefore the kernel is generated by our candidate polynomial.
Can someone please check my work?

[tidal schooner]

Gotcha. If you can use Shoda's theorem, then that leads to a short proof. Otherwise, I'd use the fact that $\mathfrak{sl}n(V)$ has the following basis: Let $E{i,j}$ be the matrix with 1 in the $(i,j)$ entry and 0 everywhere else, and define [H_i:= E_{i,i}-E_{i+1,i+1}.] Then a basis for $\mathfrak{sl}(V)$ is
[{H_i\mid 1\leq i\leq n-1} \cup{E_{i,j}\mid i\neq j}.]

[cloud walrus]

harmacist

[quartz wind]

thanks, ill try that

[inner owl]

.

[next obsidian]

It sounds good, the zero set any f mapped to zero vanishes on is at least as large as {x(t),y(t)} is what you’re using right?

I like the proof

[long obsidian]

Say R =k[x,y,z] and I and J are ideals of R. Say g1,…,gs is such that the congruence classes in R/J span the vector space R/J. Now if the classes of g1,…,gs inside R/I span R/I does this imply J contains I? What if gi are instead an ideal basis?

[harsh citrus]

Can someone give example of a finite field extension that is not normal and not seperable

[rustic crown]

you could compose examples to get an example where neither works. if E / F / K is a tower with E / F is not normal and F / K is not separable then E / K is neither.

since we want separability problems, we need to work in positive char. taking F = F_5(t) and K = F_5(t^5) would be nice. now we need to have a non-normal extension E / F, and we can add a root of the irreducible polynomial X^3 - t. and any other root would be cbrt(1) away, which doesn't lie in F_5.

so F_5(s) / F_5(s^3) / F_5(s^15) works.

[harsh citrus]

@rustic crown can you help me with another problem bro

Its galois theory too

[rustic crown]

me can try

[harsh citrus]

@rustic crown problem no e

Uses normal basis theorem supposedly

[rustic crown]

hmm

maybe you just use linear indep of chars?

[harsh citrus]

@rustic crown how
I am blank in this one

[rustic crown]

hmm maybe this works

so assume φ is the 2-frobenius, and we wanna know if there is an 'a' such that a, φ(a), φ^2(a), ... are a basis

this suggests to look at F_64 as a F_2[x] module

with x acting by φ

if there was such an element a then we get the map F_2[x] --> F_64 sending 1 to a.

since φ^6 = 1, this tells us that x^6-1 is in the kernel and by hypothesis this is surjective.

so we get the map F_2[x]/(x^6-1) -->> F_64.
which must then be an module-iso

maybe this doesn't say much

me brain eepy eepy

[harsh citrus]

Same

[harsh citrus]

@rustic crown OK
Tell me number of elements of Gal(F8 / F2)

[cobalt heath]

Ah, F_2[x]-module

[ashen hill]

any books for ring theory i can find online?

[fading acorn]

I only a know commutative algebra one(which learns about commutative rings)

but general ring theory youre finding abt? I dont know

[tardy hedge]

dude lol just search up any textbook name and then "pdf" and then youll find it

if youre new to ring theory dummit and foote is good

chapters 7-9

[chilly ocean]

i just dropped

[ashen hill]

ive had a breif intro, i read algebra by artin but i wanna get deeper into the subject

[tardy hedge]

commutative ring theory by matsumura is good

intro to commutative algebra by atiyahmacdonald is good but i prefer matsumura a lot more

[velvet hull]

if you choose to approach commutative algebra I'll give you the warning that it is going to be very dry

[chilly ocean]

starting with ringd and modules

[velvet hull]

the subject is very heavily motivatived by algebraic geometry, and unless you have an understanding of the big theorems that you're going to prove it's going to be very aimless

[chilly ocean]

yeah you are right , no motivation.

[candid patrol]

Hello ! Given a semidirect product $N \rtimes_{\phi} H$, I would like to understand how to determine $\phi$ knowing one of the possible sections of the corresponding short exact sequence, and why $\phi$ does not depend on the choice of the section.

[cloud walrus]

UGOBEL

[inner owl]

So I had this question that suppose that U and V are algebraic varieties for distinct set of polynomials in the complex polynomial ring in n variables and also that U u V = C^n now I have to comment about the algebraic meaning of this fact.
My approach suppose that f_1 f_2 ...f_n are the polynomials whose variety is U and g_1 g_2 ...g_m are the polynomials whose variety is V then f_ig_j are the polynomials whose variety is U u V now I claim that all the these "product" polynomials are 0 polynomials since if these polynomials are 0 for all pts in C^n then it would imply that they are elements of the kernel of each and every evaluation map from the polynomial ring to C this would imply that each and every maximal ideal of the polynomial ring contain these polynomials=> their intersection contains these polynomials=> these polynomials are contained in the jacobson radical of this polynomial ring => that 1 - xy is a unit in the polynomial ring for x belonging to J(R) and y in R but this not possible unless x is just 0.
So the ideal they generate is the 0 ideal.
Now my question is this the algebraic meaning of the condition in the problem.
If not then what is?
If no then what else is?

[inner owl]

Yes

[velvet hull]

the implicit claim you've made is that the ideal corresponding to UuV is the product of the original ideals

which does not sound right

otherwise your reasoning looks fine

[inner owl]

I think I had proven this previously suppose that U is the variety for the polynomials (f1 , f2 ...fn) and V is the variety for (g1 ,g2 ,...gm) then the variety for {f_i g_k} should be U u V since the variety for this set of polynomials contains U u V and suppose that it contains some x belonging to C^n and is not in U or V then this would imply there is a f_i such that it's root is not x and some g_j such that x is not it's root then f_ig_j doesn't have x as it's root therefore it cannot be in U u V.

[velvet hull]

I mean, you can just take two parallel lines in C^2, say <x-z1> and <x-z2>. then you're claiming that their product is 0?

[velvet hull]

you've only proved that any element in C[x1, ... , xn] - I(UuV) is not inside I(U)*I(V),
i.e. I(U)*I(V) is inside I(UuV). I don't see how you've shown equality

[inner owl]

For anyf_i g_k since U is a subset of the set of roots of f_i and V is the subset of the set of roots of g_k U u V is the subset of the set of roots of f_i g_k.

[velvet hull]

again it looks like you've only shown inclusion, you've repeated the same argument to tell me that I(U)*I(V) is inside I(UuV)

how are you sure that there isn't some weird polynomial that cannot be written as the sum of f_ig_j's, that also happens to be 0 everywhere on UuV?

[inner owl]

Let me clarify my thought process here cuz to me everything looks fine if I not misunderstanding something fundamental

1. If U is an algebraic variety then there is a finite set of polynomials whose set of common roots in C^n is U call it (f1,f2...fn) similarly for V to be a variety I have set of polynomials (g1,g2..gn) (Now I don't know if these sets are unique or not)
2. To show that UuV is a variety I have to show that there is a finite set of polynomials whose set of common roots is U u V and for that I constructed the set {f_i g_k} I proved that the set of common roots of this set of polynomials is U u V.
   I think I get what you are saying now that there could be more polynomials other than these so that U u V is the set of their common roots but this fact had no use for me in proving that UuV is an algebraic variety since the "product" polynomial set does the job

[velvet hull]

you have shown that it is a variety, that's correct, but we do talk about "the" ideal that corresponds to a particular variety

because if I is an ideal that vanishes on U, then any ideal inside I also vanishes on U

so we usually talk about the biggest ideal

and that's what I've been talking about

let me go over your proof and look at it again

and see if it does the job