groups-rings-fields

[coral shale]

i still can't see what this looks like

thinking thinking

[delicate orchid]

for free abelian that's true yeah

[pastel cliff]

should i bother with proving that cosets of a subgroup partition the group or just take it as a fact

[coral shale]

prove it

[delicate orchid]

do it do it now

[coral shale]

its not much

[chilly ocean]

ur saying abelian and ab=e?

[coral shale]

yes

no

[chilly ocean]

bruh

[coral shale]

ok i see what you said

[chilly ocean]

what

[coral shale]

anything AND abelian

[pastel cliff]

you were supposed to say

[coral shale]

But I was wondering what about not abelian

[chilly ocean]

any relation and abelian implies there will be finite subgroups

[coral shale]

Can you give me an explicit example

[chilly ocean]

too many coujter examples

yeah

D_infty

[coral shale]

No cus the generator s has order 2

[chilly ocean]

thats my point

[terse crystal]

What is the original question…?

[coral shale]

$$\langle a, b \mid R\rangle$$

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[coral shale]

Look look look.

I want this thing

[chilly ocean]

what lol

[coral shale]

order of a and b are infinite

[chilly ocean]

what do u want

ok

[coral shale]

Question: can there be finite non-trivial subgroups

[terse crystal]

Subgroup of what?

[coral shale]

.

If you kill the order of a or b, yes that makes this trivial, but what if you don't.

[terse crystal]

Of <a,b>?

[delicate orchid]

over the free group its trivial to show but I think they're talking about infinite abelian groups

[coral shale]

im not

R is such that order of a and b are still infinite

[delicate orchid]

good thing I'm not joining in then

[terse crystal]

Subgroups of what…?

[coral shale]

^ this

[terse crystal]

What is R?

[pastel cliff]

is it enough to show it on G \leq G

[coral shale]

Any relations that preserve the infinite order of a and b

[delicate orchid]

literally what are you talking about

are you mental bruv

[coral shale]

For example, $$\langle a, b \mid ab\rangle$$

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[coral shale]

this preserves the infinite order of a and b

are there any finite non-trivial subgroups

[delicate orchid]

"is it enough to show it on G to prove it for G"
durr hurr uehhhh uuehrherrr

[pastel cliff]

i um uhhhi uhhh ummm

[coral shale]

I feel like there can't be in general

[terse crystal]

If Your question is for any R, there exists… then no

Your example is a counterexample right?

[coral shale]

how

[terse crystal]

It’s isomorphic to Z

[barren sierra]

?

how is that isomorphic to Z wut

[coral shale]

a = b^-1

[terse crystal]

<a,b:ab=e> is isomorphic to Z

[barren sierra]

o

[coral shale]

its just <a>

[barren sierra]

it is

[coral shale]

Claim:
Forall R that preserves the infinite order of a and b, <a, b | R> has no non-trivial subgroups

[terse crystal]

Oh that’s false

Let R=abab

[coral shale]

example 🙏

[terse crystal]

Then <ab>/<abab> is finite

Subgroup of <a,b:abab>

[coral shale]

ima draw this, ty

[pastel cliff]

OH WAIt

identity of G must be in a subgroup H

[terse crystal]

Yes

[pastel cliff]

and the set of all cosets is every element of G times H

[terse crystal]

e is contained in <ab>

[pastel cliff]

so their union will have every element of G times 1 so just G

mr wew lads hurr durr me now

https://tenor.com/view/big-brain-markiplier-gif-14835823

me now that ive proven a simple thing

[barren sierra]

Can anyone quickly check a proof?

very short proof

mainly used this fact from my notes which I can take for granted

[pastel cliff]

*disjoint union

[coral shale]

@delicate orchid im sorry to disappoint but for an automorphism f, f(x^G) = x^G does not imply f is an inner automorphism

D:

I asked to be sure

https://math.stackexchange.com/questions/4396020/what-does-the-outer-automorphism-group-act-on/4396053#4396053

For groups where this holds true, it's called the Grossman property

[pastel cliff]

is there anything important about cosets that are equal

we went over two ways to show that that is the case in class so just wondering

[coral shale]

wym

You mean xN = yN

That means x and y are 'equivalent'

[terse crystal]

Two cosets are equal iff they have non-empty intersection

[pastel cliff]

yeah im asking about the usefulness of this

[coral shale]

you want to show the cosets partition G

[terse crystal]

So it defines a equivalence relation on G

Since it’s a partition

[coral shale]

if the cosets did not partition, we would have problems

[delicate orchid]

I'm going to poopy my pants... how is this so?

[coral shale]

The given example in there is good

(just note alpha and beta are switched)

Let $G = \tn{Sym }\bZ$ (symmetric group of $\bZ$).\
Let $H = \tn{Sym}_f\bZ$ be the permutations with finite support (permute a finite number of elements).\

Let $\beta\in G$, with $\beta(n) = n + 1$ for $n\in\bZ$.\
Let $\sigma_\beta\in\tn{Aut }G$, with $\sigma_\beta(\alpha) = \beta\alpha\beta^{-1}$ for $\alpha\in G$\

Claim:

$$(\forall\alpha\in H)(\beta\alpha\beta^{-1}\in H)$$
but $\sigma_\beta\not\in\tn{Inn } H$.

written badly but hope its ok

[delicate orchid]

infinite groups
I literally stopped caring

just like that

[coral shale]

Apparently its not even true for finite groups tho

(grossman property)

[delicate orchid]

gross, man

n -> n+1 is an automorphism of (Z, +)?

hold up

[coral shale]

wrote that wrong

G just has to be a permutation group

Symmetric group of Z

Let $G = \tn{Sym }\bZ$ (symmetric group of $\bZ$).\
Let $H = \tn{Sym}_f\bZ \trianglelefteq G$ be the permutations with finite support (permute a finite number of elements).\

Let $\beta\in G$, with $\beta(n) = n + 1$ for $n\in\bZ$.\
Let $\sigma_\beta\in\tn{Aut }G$, with $\sigma_\beta(\alpha) = \beta\alpha\beta^{-1}$ for $\alpha\in G$\

Claim:

$$(\forall\alpha\in H)(\beta\alpha\beta^{-1}\in H)$$
and for each $\alpha$, the permutations $\alpha, \beta\alpha\beta^{-1}$ are conjugate in $H$,\
but $\sigma_\beta\not\in\tn{Inn } H$.

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[maiden heath]

Does anyone have any advice about finding subrepresentations by hand (not using character theory), is there any way to approach this, or do I just make guesses of what I think could potentially be one?

[delicate orchid]

first thing I try and do is look for subspaces that are invariant under the representation

that implies the representation is a direct sum of some representation over that space and a kernel space by Maschke's (unless you're working over some wacky field)

[coral shale]

https://i.imgur.com/1szJgTi.png

2.5.10, Algebraic NT notes

'Moreover, if f1, f2, ..., fn are the n distinct K-emeddings of K(a), then ...'

The necessary and sufficient conditions for ^ to happen are...

K(a) splits the minimal polynomial of a
or
K(a) : K is normal

And a condition of separability?

Right? (Haven't got to this bit in Galois but want to know 🙏)

[next obsidian]

I think it’s that it’s separable???

[coral shale]

Oh, only separable?

[next obsidian]

Or maybe normal as well…

I know having n-distinct embedding into k-bar is equivalent to separable

[coral shale]

Ok, so K-embeddings in these notes

mean L -> C

So I think just separable

But if it was L -> L instead

[next obsidian]

Idk, you might want normal too

[coral shale]

then I think we need normal

[next obsidian]

I just am thinking that like

a can only go to one of its conjugates

And that determines the entire automorpjism

So you’re bounded by the number of conjugates

Also these are number fields so separable is automatic lmao

It must be normal then

[coral shale]

yah its ANT so it gives the specific defn

but no - I am after the general defn

[next obsidian]

Or maybe…

Because it’s separable you always have n?

[coral shale]

I think you always have n in the closure

[next obsidian]

So you said embeddings into C?

[coral shale]

so if you consider L -> C

There will be n

[next obsidian]

That’s true cuz of separable extension thingy

Gimme a moment

[coral shale]

But if you instead consider L -> L is also what I'm interested in ig

[next obsidian]

https://stacks.math.columbia.edu/tag/09GZ

If you glance through thi

Lemma 9.2.10 and 9.2.11 is what you want

I think

[coral shale]

Ah thanks !

[next obsidian]

I am bad at field theory

[coral shale]

And amaze book it looks like 👀

Can you define $Mor_F(K, \overline F)$ for me

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[coral shale]

I'm guessing K-hom, F -> Fbar

[next obsidian]

It’s the maps K -> F-bar fixing F

[coral shale]

oh

thanks

[next obsidian]

[coral shale]

Can we say how many maps we lose from inseparability?

I feel like it's precisely the number of repeats in the minimal polynomial

I'm not sure if something like this makes a difference\$(x-\alpha)^2(x-\beta)^5$ vs $(x-\alpha)^3(x-\beta)^4$

🤔

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[chilly ocean]

yeah

oh i remember

seperable extensions are those which elements are representable as linear combinations

[cloud walrus]

Great Icosiheptapeton

[trim grove]

form elementary divisors how can i show two given finite abelian groups are isomorphic?

[lethal dune]

@trim grove having same elementary divisor is not enough

you'll need more info

[trim grove]

can you please tell me what are other conditions we need to show they are isomorphic?

[lethal dune]

u can show they have same invariant factors and free rank, but if you only know the elementary divisors, I don't think there's much you can do

[trim grove]

so one thing is clear that if they don't have same E.D then they never going to be iso?

[trim grove]

^

[lethal dune]

yes

[spice whale]

why is Lagrange's theorem useful
when you could just use the counting formula

[next obsidian]

[maiden heath]

How do you find subrepresentations without using character theory?

[iron vessel]

Hey guys, for this proof my argument was by contradiction. Initially I supposed that there was such an alpha, which in turn meant that there was a polynomial f in K_0 such that alpha was a root of it, and f is also monic. This meant that since alpha is not in K_0, then we must have that f is irreducible of degree at least 2 in K_0, which is a contradiction as K_0 is and algebraic closure of K and thus algebraically closed; meaning that all polynomials must split entirely into linear factors. Is my argument faulty?

[median valve]

Can someone help me understand slices? I am trying to understand these class notes -

@chilly ocean

[iron vessel]

Anyone available for just a quick feedback?

[terse crystal]

I think it’s correct

[iron vessel]

Ok perfect, thanks

[toxic zephyr]

just to clarify, a semigroup is a set with one binary operation which is associative. but there is no guarantee of an identity or inverse element?

and so to prove that a set with that operation is a semigroup i just have to show that the operation is associative?

[coral shale]

yes and perhaps show that function is indeed a binary operation

[toxic scaffold]

it must be closed under the operation too

[toxic zephyr]

cool thanks

[pastel cliff]

two weeks away in syllabus

but i can smell it from here

[delicate orchid]

I'm on the verge of crying in the club once more chat

I hate finite fields so much it's unreal

[pastel cliff]

L

[delicate orchid]

I might have to admit defeat on my funny representations of C_p over F_p excursion

like does anyone have any idea how diagonalizability works for matricies over finite fields?

cause every matrix I try and work with seems to be diagonalizable

[tribal moss]

$\begin{pmatrix}1&1\0&1\end{pmatrix}$?

[cloud walrus]

Troposphere

[delicate orchid]

am I diagonalizing things wrong?

[tribal moss]

P^{-1}AP should be diagonal for some invertible matrix P.

[delicate orchid]

the characteristic polynomial (1-x)^2-1 = x^2+2x definitely has non-repeated roots (except for char 2) so you should be able to diagonalize it?

[pastel cliff]

why phi bar

sorry for interruption, just curious bout deciphering notation here

[tribal moss]

No, the characteristic polynomial is (x-1)^2 + 0·1 =(x-1)^2.

[delicate orchid]

oh yeah lol

god my brain is so frazzled I've been at this for 4 hours

can I just ask a series of stupidly easy and probably obvious yes or no questions

[tribal moss]

Sure, though I have to go in about 5-10 minutes.

[delicate orchid]

firstly, are elementary row/column operations expressible through conjugation by another matrix

like adding one row to another

[tribal moss]

No, row operations is multiplying an elementary matrix from the right left only.

[delicate orchid]

that definitely matches with what I found

and over char 0 it's obvious because the character of the representation isn't preserved so the matricies cannot be conjugates of each other

[tribal moss]

(In principle diagonalization can be described as doing some row operations, and then also doing the same operations as column operations -- but I don't know a way to make that picture actually helpful, intuitively)

[delicate orchid]

oh yeah that's what I've been trying to do

and what I did, before I realised that I can't do it via conjugation

which broke everything

oh I know why I thought that it's because over C you can do a change of basis to diagonalize it

[tribal moss]

Well, P^{-1}AP is a conjugation all right. And since P is invertible, it can be factored into elementary matrices. I just don't know a way to figure out those elementary operations one by one with diagonalization as the goal.

Change of basis to diagonalize still works over other fields.

[delicate orchid]

OH

so I just can just do it directly?

[tribal moss]

If (and only if) you can find a basis consisting entirely of eigenvectors, that will diagonalize the matrix.

[delicate orchid]

yeah, I'm assuming that you can find that

well kind of

[tribal moss]

What you can't do over arbitrary fields is hope to get an orthogonal basis change, because F^n in positive characteristic doesn't have a well-behaved inner product to normalize the eigenvectors by.

[delicate orchid]

yeah the lack of an inner product is... annoying

anyway thanks for your help tropo

[tribal moss]

yw

[sharp turret]

Can someone verify my work for this homework problem? Seems more simplistic than my other problems but worth a lot of credit so just wanted to make sure I wasn't missing something obvious, thanks

[delicate orchid]

looks good to me

[sharp turret]

god bless 🙏🙏🙏

[next obsidian]

One thing to note here is that

If you just adjust this proof accordingly, you can assume that b^n = 0

For any n

This is a useful thing to know sometimes

[delicate orchid]

yeah this is a special case of a more general theorem

[lethal cipher]

Okay, so let's say I have two extension fields: $\Q(\sqrt{2},\sqrt{3})$ and $\Q(\sqrt{2}+\sqrt{3})$, how would I show that these two are the same?

[cloud walrus]

dackid

[lethal cipher]

I do know that the first extension field is the smallest subfield containing those roots, so we have one inclusion, but what about the other?

There is a hint about looking at the inverse of rt(2)+rt(3), which is in the first field, but I am not entirely too sure how that helps.

[south patrol]

You can use minimality in the other direction I believe, too

[coral shale]

Theres 2 ways I've heard of

[south patrol]

In fact I feel you've proven the harder inclusion lol

[coral shale]

The naive way is to create a system of equations

[lethal cipher]

That first inclusion luckily follows from a lemma we have done

[coral shale]

wait what

You already know this?
Q(rt2, rt3) sub Q(rt2 + rt3)

[south patrol]

Basically i just mean both are extensions containing both square roots and their sum, conclude by mininality

[lethal cipher]

Yes

[coral shale]

Q(rt2 + rt3) sub Q(rt2, rt3)

This is meant to be the trivial inclusion ^

huh

[lethal cipher]

Okay, so what about the other direction makes it trivial?

[coral shale]

So let me get this straight

You have shown

Q(rt2 + rt3) subset of Q(rt2, rt3)

Or the other one

[lethal cipher]

Other way

[coral shale]

wut

how does that lemma show it

I don't understand

[lethal cipher]

Q(rt2,rt3) is a subset of Q(rt2+rt3)

[coral shale]

How does that lemma show it tho

[south patrol]

The latter contains rt 2 and rt 3 with a tiny bit of work

[lethal cipher]

actually yeah, good point. Gotta put a little extra bit of work to show Q(rt2+rt3) actually contains rt 2 and rt3

So not quite there yet

[coral shale]

right but the other way round

is surely trivial

[lethal cipher]

Anyways, what makes the other direction trivial?

[coral shale]

you can show rt2+rt3 is in Q(rt2, rt3)?

[south patrol]

If a field contains sqrt(2) and sqrt(3) then it contains their sum

[coral shale]

closure under addition

[lethal cipher]

And their products....

Yea that direction is easy

[coral shale]

right, so the challenge is the other direction

So the first general method of showing this (if you are told it is true):
$$\sqrt[n]a, \sqrt[m]b \in K(a + b)$$
if $n, m$ aren't too big is to consider powers of $(a+b)$. This generates a system of linear equations with variables you need to eliminate

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[coral shale]

If you call x = a^{1/n}, y = a^{1/m}

Then the a+b is a 'constant'

x, y are unknowns you need to find
the powers of x and y are unknowns you need to eliminate

[lethal cipher]

Isn't this trivial since 2 and 3 are already in Q?

We already know the field contains Q

[south patrol]

Sorry, sqrt(2) and sqrt(3), typo

[lethal cipher]

So you end up with 5+2rt(6)

[coral shale]

oops not quite linear (but same idea)

5 + 2xy

[lethal cipher]

So where does the multiplicative inverse of rt(2)+rt(3) come in?

[coral shale]

in this method it doesn't.

Let k = a+b
Let x = rt2
Let y = rt3

k = x + y
k^2 = 5 + 2xy

[south patrol]

Oh ignore me

[coral shale]

I feel like im missing something here

🤔

[coral shale]

no nvm im not. 2 equations 2 unknowns, proceed.

[lethal cipher]

Oh, we can show rt(2) and rt(3) are in this field. Since we are in a field, the element rt(2)+rt(3) has an inverse: Namely rt(3)-rt(2). So subtracting both of these and dividing by 1/2 gives rt(2), and adding both of them together and dividing by 1/2 gives rt(3)

[coral shale]

whoa whoa

[sturdy marsh]

Z is a PID so Noetherian

[coral shale]

Oh rt(2)+rt(3) and rt(3)-rt(2) are multiplicative inverses

yes that is another way

[lethal cipher]

So since we have shown the field contains both elements, we can apply the lemma

[coral shale]

not sure it works out so nicely in other examples 🤔

[lethal cipher]

There is one small kink, say you had $\Q(\sqrt{a}+\sqrt{b})$, then [ \frac{1}{2} ((\sqrt{a}+\sqrt{b})+(a^2-b^2)(\sqrt{a}+\sqrt{b})^{-1})] will give one of the roots

[cloud walrus]

dackid

[lethal cipher]

And then subtracting instead of adding will give the other one

[coral shale]

looks ok for sqrts I guess 👌

[lethal cipher]

And that's all I need for the moment

[sturdy marsh]

no

it's true

[coral shale]

you probably meant a - b

[sturdy marsh]

but not entirely obvious

[lethal cipher]

Ye I did. My bad

[coral shale]

If you wanna look ahead the 'Primitive Element Theorem' generalises this idea

$$K(a + cb) = K(a, b)$$

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[coral shale]

It tells you precisely when this occurs for algebraic a, b

[lethal cipher]

Out of curiosity, what would be a situation where this fails? Apart from when c is equal to 0

[south patrol]

Another way ig is that also Q(sqrt(2) + sqrt(3)) is a deg 4 Q- extension by considering its min poly

[lethal cipher]

Oh actually I didn't. I just wrote (rt(a)+rt(b))^-1 instead of the rationalized form

[lethal cipher]

The min poly is going to come from the next part, so I am trying to avoid that (since it wants us to use info from this and that [Q(rt(2),rt(3)]=4

[south patrol]

Sure, just considering the inverse is faster too

[lethal cipher]

Yea, this inverse trick got me there fast. In the end, I just needed to show that the lemma holds in both directions, and that I didn't quite understand right off the bat.

[sturdy marsh]

Oh rip I thought they were talking about abelian groups

the problem is about abelian groups

[coral shale]

I believe its this statement but im not 100%

Let $L:K$ be an extension. Let $a, b\in L$ algebraic and separable.
$$(\forall a', b')\quad c\neq-\frac{a - a'}{b - b'}\implies K(a + cb) = K(a, b)$$

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[coral shale]

where a', b' are conjugates of a, b (not equal to them), respectively

conjugate meaning root conjugate

The roots of an irreducible polynomial are considered to be 'conjugates' of each other

You probably know what algebraic means. For now, you can consider stuff in Q, R, C separable 🤔

[sturdy marsh]

tG isnt even a group in general if G isnt abelian

Z is noetherian, submodules of finitely generated modules over a Noetherian ring are finitely generated

[delicate orchid]

That an incredible proof lol

[next obsidian]

You sully that proof

But this is immediately what I thought of

[lethal cipher]

Thanks for the help ryc and shuri. Much appreciated

[tribal moss]

Yes.

[next obsidian]

Does it??

[upper pivot]

this is just <ab,b>

[next obsidian]

I’m pretty sure that’s the entire thing

[upper pivot]

yeah it is oops

[next obsidian]

You need to do something more clever

[upper pivot]

the correct way to reason about this is by covering spaces, find a countable cover of the rose on 2 elements

hatcher has good pics

[tribal moss]

<ab,b> is too large -- it contains b, which the subgroup defined by Leaf doesn't.

[upper pivot]

(ab)^{-1} (abb)

[tribal moss]

Oh, bah, you're right.

[upper pivot]

here are some subgroups of F2

(10) and (11) in particular are infinite generator free groups

[delicate orchid]

love these diagrams

[next obsidian]

Okay yeah b^nab^-n is the one I was thinking of

[uncut girder]

Very pretty

[lethal cipher]

Okay, I need help. I somehow need to show that $\sqrt{3-\sqrt{-2}}\in \Q(\sqrt{3+\sqrt{-2}})$.

[cloud walrus]

dackid

[lethal cipher]

Well, I need to show that field is a splitting field, but this condition needs to happen in order for that to work

[coral shale]

I think try figuring out the minimal polynomial

essentially 'guess' a quartic

That would be my appreciation ach at least

[lethal cipher]

Yea, I know the min poly

[coral shale]

sure then can you show the other root is in it?

factor theorem - you can just plug.

[lethal cipher]

Nevermind I got it!

The issue is not that it is the minimal poly, the issue is making sure that element is in the field.

I actually need to show $\sqrt{2-\sqrt{-5}}$ is in $\Q(\sqrt{2+\sqrt{5}})$ and I can do this by showing the inverse of $\sqrt{2-\sqrt{-5}}$ is in the field.

[cloud walrus]

dackid

[lethal cipher]

And I was having trouble seeing how to do it, but I just got to be a bit smarter with how I simplify things since $(\sqrt{2-\sqrt{-5}})(\sqrt{2+\sqrt{-5}})=3$.

[cloud walrus]

dackid

[plush wasp]

Quick question, why is Z/nZ not a subring of Z for n ≥ 2?

Ig it has sth to do with the definition or I'm just tripping

is it because, say for Z/3Z, 2 x 2 = 4 is not in Z/3Z?

[paper flint]

It would be, unless you require the rings to have unity.

Wait hold on

Z/nZ

They don't even have the same operations

You're adding mod n in Z/nZ

It's not the operation inherited from Z

[plush wasp]

right, so 2 + 2 = 4 is also not in Z/3Z

[paper flint]

Correct

[plush wasp]

wait 2 + 2 = 1 is perfectly in Z/3Z so it is closed under addition

or is it not ._.

[paper flint]

True, but you need to change the operation for that

At that point it seizes to be a "substructure" of Z in a sense

[plush wasp]

makes sense

the elements should inherit the operations as they would in R

[paper flint]

Yes

Otherwise you could take any subset and force a ring structure on it

[plush wasp]

Got you ✓ 2 + 2 = 4 and not 1 🙅

[paper flint]

Probably slightly more pedantic but Z/nZ is also not a subset of Z, strictly speaking. Z/nZ consists of equivalence classes of integers, not integers themselves.

[subtle ivy]

um, also more obviously, subrings are additive subgroups, so... it may not even be embedded in Z. lol.

[plush wasp]

why not

{0, 1, 2} is not a subset?

[paper flint]

0 is representative of the equivalence class of integers that are congruent to 0 mod 3

[plush wasp]

okayyyyy- so 0 is not really an integer per se

[paper flint]

In this context, yes. To not cause confusion some textbooks write equivalence classes with an overline

Like a bar above 0 to denote the equivalence class of integers that are congruent to 0 mod n, or so on.

[plush wasp]

Daaaamn, that really changes my outlook on Z/nZ by a hell lot. Ty for warning me before hand

[wooden ember]

You can also consider Z/nZ as a structure that lives apart from the integer (so that you can call its elements things that please you better instead of equivalence class) that inherits a “less detailed” form of its operations

Though I guess this is better for groups than rings

[plush wasp]

Also, can you hint me to proving:

A finite division ring is necessarily a field.

[wooden ember]

Since cyclic groups are sometimes better to think about without reference to Z

[wooden ember]

I think this is a rather hard problem if I’m not mistaken

The easier one is showing that a finite commutative integral domain is a field

Lemme check

[plush wasp]

Yeah I just showed that any finite integral domain is a field

[wooden ember]

Yeah this is wedderburn’s theorem

It’s non trivial I’m pretty sure

[plush wasp]

commutative wasn't necessary, finite was sufficient

[wooden ember]

Commutative is necessary

Otherwise you just show it’s a division ring

And wedderburn finishes it off

But wedderburn is not trivial to show

[paper flint]

Hmm, apparently this is Wedderburn's little theorem: the main theorem proves the result for finite domains, the little theorem proves it for division rings.

[wooden ember]

Yeah the wiki proof seems complicated

[wooden ember]

Huh okay

[paper flint]

Or maybe I got confused, nevermind

[wooden ember]

Yeah no it’s the other way around

Finite domains is easy

[paper flint]

Right

[plush wasp]

Hmm. That thing's exactly where I need to be (ヘ･_･)ヘ┳━┳
Unfortunately, I need to go through the whole thing to reach there smh

[wooden ember]

Why do you need to be able to prove wedderburn?

[plush wasp]

our batch's already done with Field Theory and they're half way into the Galois Theory

It's not like I need to be able to prove it, but at the bare minimum.. I should be able to understand it..

Anyways, I'm starting with like Ring Theory ... hopefully can cover some stuff before I'm totally overwhelmed

[wooden ember]

Oh I mean I’m sure you should be able to understand it yeah

It’s seems like the background material isn’t too intense

It’s just a bit long

[plush wasp]

tysm

[lethal dune]

the feeling of being left out

[plush wasp]

the feeling of having midsem up your ass

[lethal dune]

end sem in weeks,

[plush wasp]

in 3 days-

[lethal dune]

my modsem just ended

[plush wasp]

Lmao gl

[lethal dune]

stat and combinatorics

[plush wasp]

Umm,

If R is a subring of a field F that contains the identity of F then R is an integral domain.

say 'a' is a zero divisior, that is, there exists b such that ab = 0

1(ab) = 0 = (ab)1

1 is a zero divisor in R, which which is :pain:

hence, R must be an integral domain?

[wooden ember]

If R has zero divisors so does F

And zero divisors aren’t invertible

[wooden ember]

So yeah this works

Well no

[wooden ember]

This doesn’t work

This doesn’t show that 1 is a zero divisor

Since ab=0

[plush wasp]

bruh, ryt

[wooden ember]

All you have to do is suppose there is a b such that ab=0, then a^-1 * ab = a^-1 * 0 => b=0

So a is not a zero divisor when invertible

[plush wasp]

umm, fields are integral domains right?

[wooden ember]

Yeah

Integral domain is an inherited property

[wooden ember]

And you show this by what I said earlier

[plush wasp]

I'm very concerned atm

What is a field norm?

I'm following Dummit n Foote, it keeps giving me facts first and proofs later on smh

[wooden ember]

A field norm is an application N: F -> R such that N(ab) = N(a)N(b) and N(0)=0 iirc lemme check my definitions again I always mix it up with vector space norms

[plush wasp]

is reading on Vector spaces compulsory to follow up with field theory?

I've came across the term bases several times, so I kinda feel like if I'm gonna read on splitting fields n extensions, vector spaces would be the prerequisite

[wooden ember]

Yeah no this is so wrong lol

An integral domain norm N takes R -> \bN with N(0) = 0

[wooden ember]

You definitely ought to be familiar with vector spaces yes

I mean

Surely you’ve had a linear algebra course before field theory 🤔

If you don’t know what a basis is that’s a problem I’d say yeah

[plush wasp]

I mean, yes I do understand what basis is, on a very beginner level but understanding something like this.

[wooden ember]

What is it you don’t understand

[woven flume]

reposting from getting-help due to no response. how do I find the generator(s) of GF(2^m)? for example, let's take GF(2^3) with irreducible polynomial 1011. I know that it its order is 8. I know that I can generate all the elements by exponentiating the generator 0 to 7 times, e.g. f^0 = 1st element ... f^7 = last element. however, what is the general approach of determining the generator to use? also, does it make sense to talk about order of each element (polynomial) in GF(2^m)?

[hidden haven]

Is GF(2^m) the Galois group of the field with 2^m elements over 𝔽_2?

[woven flume]

yes

[hidden haven]

Then it is cyclic of order m, generated by the frobenius automorphism x ↦ x²

[woven flume]

I do not even understand what you said

the second part

[hidden haven]

x ↦ x² is an element of the Galois group

[woven flume]

so, how do I find the generator(s) of GF(2^m)?

[hidden haven]

That generates the whole group

[hidden haven]

You can prove that this automorphism has order m

And since the Galois group has order m as well, this must generate the entire group

[woven flume]

so is it x or x^2?

[hidden haven]

It is the squaring automorphism

That takes any element to it's square

[woven flume]

as I said, I do not speak mathemateese

[hidden haven]

oof

The automorphism is

f(x) = x²

For all x in the field

[woven flume]

alright. thanks

[terse crystal]

Oh a generator of Gal(F_2^n/F_2)… I thought you meant generator of (F_2^n)*😂

Yeah Gal(F_p^n/F_p)=<φ:x—>x^p> and Any m|n we have a galois correspondence Inv(φ^m)=F_p^m

[trim grove]

Can someone please give me a simple definition of Derived series of a group, i am confused in definition available on Google. Thankyou

[delicate orchid]

you just keep applying the commuter to the group to get new terms in the series

[wooden ember]

The dérivée series is just an iteration of taking commutator subgroups

Derived*

So the factors of the series gives you a measure of how your group fails to be abelian

[delicate orchid]

like, say $G = S_4$
then $G_1 = [S_4, S_4] = A_4$, $G_2 = [A_4, A_4] = C_2 \times C_2$, $G_3 = [C_2 \times C_2, C_2 \times C_2] = {e}$

[cloud walrus]

Wew Lads Tbh (200 🍇) ✓

[delicate orchid]

so the derived series is $S_4, A_4, C_2 \times C_2, {e}$

[cloud walrus]

Wew Lads Tbh (200 🍇) ✓

[delicate orchid]

this is a good way to think about why we'd care about the derived series, for example the derived series of all abelian groups immediately terminate at the identity group after 1 step (although in general the series does not have to terminate at the identity group)

[trim grove]

Thankyou ,now this is very much clear to me by this example.

[trim grove]

So basically this is a tool to verify the given group is abelian ( can we say this)?

[delicate orchid]

a group is abelian if and only if it's commuter is trivial yeah

[wooden ember]

To be absolutely precise a derived series tells us whether a group is solvable or not

Ie it terminates iff the group is solvable

The case where it terminates after one step tells us the group is abelian

[coral shale]

I thought terminates at {e} iff solvable?

[delicate orchid]

S_5's derived series terminates at A_5 an S_5 is not solvable, so yeah

[coral shale]

oh so quintics do have a formula ig

[delicate orchid]

there's a specific way you can go from a derived series to the nice funny field extension pretty easily but I cannot remember it

[trim grove]

Looks like something related to Galois tower

[coral shale]

which quintic has corresponding galois A5

[delicate orchid]

reverse galois problem

[coral shale]

is it not easy to write down 👀

[delicate orchid]

I can barely compute galois groups from polynomials let alone the other way around

[distant rain]

I have likely a silly claim with a likely silly proof for the claim. Suppose $R$ is a non-unital ring of real-valued functions (the details are probably not so important, but correct me if I am wrong). Suppose I have a generic element $\sum_{i=1}^nf_i\otimes g_i$ in the tensor product $R\otimes_RR$ such that $\sum_{i=1}^nf_ig_i$ is a zero function, namely $\sum_{i=1}^nf_i(x)g_i(x)$ is identically zero on the domain $D$.

Is it correct if I claim the tensor $\sum_{i=1}^nf_i\otimes g_i$ is
then zero in the tensor product?

My (silly) reasoning goes as follows, since $\mathbb R\otimes_\mathbb R\mathbb R=\mathbb R$, then $\sum_{i=1}^nf_i(x)\otimes_\mathbb R g_i(x)$ is identically zero on $D$. We can actually view (this is where my doubt lies in) $\sum_{i=1}^nf_i\otimes g_i$ as a function on $D$ with target $\mathbb R\otimes_\mathbb R\mathbb R$: $x\mapsto \sum_{i=1}^nf_i(x)\otimes_\mathbb R g_i(x)$ hence the claim follows. Is there anything wrong with my reasoning here?

[cloud walrus]

τφκ

[wooden ember]

That’s what I meant yeah

I don’t consider it terminates if it gets into a loop

But fair your terminology makes more sense

[delicate orchid]

I mean, you can keep computing more terms of the series even if it reaches {e}, it just loops back to {e}

Same goes for every perfect group

[wooden ember]

Yeah that’s fair

[median pawn]

I'm proving (1 implies 2)

and I need help understanding the meaning of (2)

What is a maximal element under inclusion?

[chilly radish]

Isn't that just zorn?

[lavish nexus]

what is $Cl^0(V,Q)$? Is it the subset of clifford algebra on which the grading automorphism evaluates to 0?

[median pawn]

The proof uses AOC

[chilly radish]

It's just a maximal element when looking at the partial ordering of inclusion. That is, it's a submodule in the set such that no other submodule in that set contains it

Yea I just mean 1 implies 2 is a direct application of zorn

[cloud walrus]

Iteribus

[median pawn]

I see

So if some submodule is not a maximal element

then there necessarily exists a larger submodule containing it?

[chilly radish]

Yes

[median pawn]

Great, I can see a proof by AOC now

[chilly radish]

Otherwise it would be maximal

Great!

[lavish nexus]

wait you can't evaluate to 0
evaluates to 1?

[median pawn]

Thanks!

[chilly radish]

Np

[median pawn]

I don't understand the part in yellow, because the definition of a Noetherian ring says that it is not possible to have an infinite ascending chain of ideals

[next obsidian]

No

Well… yes

But these are separate chains

For each n, there exists a chain of length > n

[median pawn]

Oh okay wait, so "arbitrarily long" means that we can find a chain C_n of length >n for every n

Gotcha, my bad. Thanks!

[next obsidian]

Right

People didn’t realize this was a thing for a while I think

And Nagata constructed an example of a Noetherian ring with infinite Krull dimension

[tribal moss]

Even in Z we have (p^n) subsetneq (p^{n-1}) subsetneq (p^{n-2}) subsetneq ... of arbitrary length.

[wooden ember]

That’s weird wording, you’d expect “arbitrary” to include any ordinal

[maiden heath]

are all non trivial irreducible subrepresentations isomorphic to the trivial, standard or sign representations? I can't think of a easy counter example

[delicate orchid]

What is the standard representation

Also of which group lol

Cause I can give you a degree 2 irreducible of D_3 acting on a triangle and a degree 3 irreducible of A_4 acting on a tetrahedron

[delicate orchid]

ah ok

[maiden heath]

ahh cheers

[delicate orchid]

in that case definitely not, the number of conjugacy classes in S_n is the number of paritions of n as Todd said and they're in a 1 to 1 correspondence with the number of irreducibles - so for n > 3 there are more than 3 different irreducible representations

[lapis gyro]

Hello. I am not sure if this is the right channel to ask but... how can a quintic equation have only 3 solutions? why aren't there 5 solutions?

[chilly ocean]

A quintic polynomial has exactly 5 roots, but some of them might not be real roots

also some roots can have multiplicity > 1

[lapis gyro]

I see. The reason I am asking is because I'm working with one equation like this that I'm solving numerically in Python via newton and it only gives me back 3 roots

[rapid bramble]

are the other 2 roots complex?

[lapis gyro]

They probably are tbh but my code doesn't give me 5 roots, only 3, so I probably have to tweak it until something happens and I get all 5.

[coral shale]

this isnt what #groups-rings-fields is about btw

[desert marsh]

If two of the remaining roots are complex, then you won't get them. Similarly, if the multiplicity of a root is 3, or two roots have multiplicity 2 each and the rest 1, then also you won't get more roots. Consider expressions like (x-1)³(x-2)(x-3) or (x-1)²(x-2)²(x-3).

Also, as others have mentioned, Abstract Algebra is different. You can head over to #prealg-and-algebra, #precalculus or #computing-software depending on what you intend to ask.

[pastel cliff]

not exactly alg but i got nowhere to ask and im doing alg hw so - what does $\mathbb{R}^\times$ usually mean

[cloud walrus]

μ₂

[chilly ocean]

the invertible elements of R ?

what's the context exactly

[chilly ocean]

I don't exactly know why the codomain would be R star, but that usually denotes the set of invertible elements in a ring

[pastel cliff]

yeah i dunno ¯_(ツ)_/¯

[south patrol]

The multiplicative group R^x I assume here but then there's a typo lol

[pastel cliff]

maybe author meant $x$...?

[delicate orchid]

no that should be the set of invertible real numbers

$\langle \bR, + \rangle \mapsto \langle \bR^{\times}, + \rangle$

[cloud walrus]

Wew Lads Tbh (200 🍇) ✓

[delicate orchid]

dunno how this is a group homomorphism though

[pastel cliff]

classmate thinks it's ordered pairs of reals

but then if it was it'd just be \R \times \R and not this buffoonery

[devout crow]

yeah (R^x, +) isn't even a group, the most likely thing is they meant to write f(x)f(y)

[chilly radish]

You should.probably just ask your prof

[delicate orchid]

yeah I even initally wrote \times as the group operator out of habit

[chilly radish]

Additive functions f(x+y)=f(x)+f(y) from R to R^x can't even exist

Since we must have f(0)=0

[pastel cliff]

not a group with this definition right?

[delicate orchid]

not under + no

[pastel cliff]

[delicate orchid]

it might be the old "hahaha lets use + to denote any abelian group operation" in which case I'm jumping off of a bridge

[pastel cliff]

this isnt even my prof's fault for once, it's textbook lol

[spice whale]

how do you even do a proof for this lol

it's literally the definition

[pastel cliff]

i had to prove this for my last homework

[lavish nexus]

well it's closed under multiplication

closed under inverses

that's it

[spice whale]

and contains id

[pastel cliff]

yeah i just mean it's the most obvious thing, ive been getting weird hw's

[lavish nexus]

with the two conditions it automatically contains id

[sharp sonnet]

its the kernel of det, which is a homomorphism

[pastel cliff]

loch is shiny and blue now

[pastel cliff]

the operation of two groups doesnt matter in an isomorphism right

like one group can have addition and the other can have fucking concatenation or some shit idk

what i mean is there can be an iso between those right?

[sharp sonnet]

there can be, yes

[pastel cliff]

[sharp sonnet]

the free group is usually constructed via concatenation

but on e.g. 1 generator its just Z

[lavish nexus]

what are the weight spaces here

I'm totally clueless

[mighty spade]

Unless I’m being stupid, shouldn’t it be all the products of the eigenspaces generated by the diagonal entries of the matrices? That is, the weight space of the subgroup of G generated by one element is just the eigenspace of that element, and the group itself is an infinite product of such subgroups. You can then say that the corresponding sub-representations are all products of eigenspaces.

[lavish nexus]

I thought a weight space is some eigenspace of the corresponding Lie algebra?

[mighty spade]

An eigenspace of a Lie algebra is a weight space, but you can have weight spaces that aren’t obviously eigenspaces of Lie algebras. A weight space of an algebra is just the space generated by the one dimensional representations of the algebra over its corresponding field.

That doesn’t require your algebra to be Lie, although it may be true that you can create a Lie algebra with an identical weight space given a weight space. I’m not sure.

How does your book define it?

[lavish nexus]

I'm using whatever Brocker tom Dieck is using as it goes along and it hasn't been defined in the general case yet
my prof never defined it rigorously

[hybrid island]

can someone look over one of my proofs?

[rapid bramble]

Better to just post it and ask, rather than asking to ask

[latent anvil]

looks good!

[hybrid island]

really?

i felt super unconfident while writing it ngl

[delicate orchid]

topology in abstract-alg

no worries boss lol, you might be able to get a quicker answer in #point-set-topology

[chilly ocean]

guys

what is a group presentation?

im kinda confused on how to approach this problem

[coral shale]

,, \langle a \mid a^n\rangle

[coral shale]

Have you seen this before?

[chilly ocean]

im getting help on #help-4

[chilly ocean]

isnt that the def of cyclic group

?

[coral shale]

Yes, but this is known as a 'group presentation'

[chilly ocean]

i put C_3 = <a,a^3 =1>

[coral shale]

Alright.

Similarly all groups can be presented in such a way.

You list the generators

Then you list the relations they follow.

[chilly ocean]

would this be a correct

[coral shale]

(look up group presentation for a better explanation than mine - im just trying to give an idea)

[chilly ocean]

this is what i did

[coral shale]

reading

$$C_3\times C_3 = {(a, b):a,b\in C_3}$$

[coral shale]

with the usual operation

now.........

I'm just thinking to myself here

if it is right

[chilly ocean]

take your time bro

[coral shale]

My approach would be to identify generators of this

(a, e), (e, b)

As you have implicitly done sounds like a good approach

Their order is indeed 3

and ab = ba is necessary

So the question is if you have left out anything else (I feel you haven't for now)

What would be convincing is if you construct an isomorphism between the two

To show a morphism in general, if you decide where the generators map, it will fix everything else

So I would go about showing an isomorphism between that group you have written down and $C_3\times C_3$

[coral shale]

Decide where the generators map to, then show the homomorphism property holds.

[chilly ocean]

hmm

so all i have left to do is prove injective in this case?

[coral shale]

You haven't named S

so I'm not sure

Hmmmm what you've written doesn't sound complete to me

[chilly ocean]

okay what parts of my attempt is good then

[white jackal]

how do you prove relations

[coral shale]

Let $G = \langle a, b \mid a^3, b^3, aba^{-1}b^{-1}\rangle$\
Let $H = C_3\times C_3 = {(c, d):c,d\in C_3}$\

Let $f:G\to H$ with $f(a) = (a, e)$, $f(b) = (e, b)$

This would be my start

[coral shale]

Then I go further and say

Let $f(a^nb^m) = (a^n,b^m)$

[coral shale]

And you show this is a well defined homomorphism on all members of G

and show bijective

(I feel like I could have written this explanation better but )

[coral shale]

(this is for any n, m, and you note the abelian relation in G makes this well defined)

[chilly ocean]

so H would be my group representation right

[coral shale]

This is the group presentation

https://i.imgur.com/gPvv7Zh.png

Equivalent to what you wrote

Either is fine.

[chilly ocean]

do i just do this by showing bijective?

one more question, when it says find a group presentation what exactly am i supposed to do? I kinda just guessed to do mine

like how do i know if its a valid group presentation

[coral shale]

show homomorphism

show bijection

[coral shale]

By constructing an isomorphism explicitly

If you want to prove it beyond a doubt

Perhaps if you look up presentations of common groups, though

you will get some intuition for it

Your guess wasn't just any old guess

You listed the order relations. And the commutativity relation --- both are necessary in that direct product

The question then is a matter of convincing yourself you haven't left anything out

[chilly ocean]

which is why we create a homomorphism

then make sure that is isomorphic?

and how do i know which homomorphism to make?

[coral shale]

You created your presentation in such a way that the isomorphism you want is obvious

map the obvious generators to each other

(there really isn't much choice of homomorphism that has a hope of working you could try here)

[hidden haven]

You can read this explanation

[chilly ocean]

okay i think i just need to study a lil more

appreciate the help guys

[coral shale]

@hidden haven can I have an opinion 🙏 #advanced-lounge
#advanced-lounge message

just a 'its possible or impossible' would be plenty 😄

[hidden haven]

I didn't quite get it, you are considering doing a project in cat theory and you want to know if that would be possible for you?

[coral shale]

Yes

Like I have no idea how steep the learning curve is

[hidden haven]

Ye it should be good

[hidden haven]

That differs

But I think you are at a point where it is reasonable to learn cat theory

And you should learn cat theory as soon as possible, it is very worth it

[coral shale]

👌 thanks. Will speak with my tutor then

[terse crystal]

Just read those two pictures…

I can send them again if you can’t find that channel…

π here is your f, C_3 I denote it Z/3Z

[coral shale]

bruh

they started group theory relatively recently (last few months?) iirc

[terse crystal]

Subgroups of commutators might be at beginning sections of their text book, since he is reading presentation , he must have learned subgroups of commutators already.

[coral shale]

I highly suspect this lives in #real-complex-analysis

I haven't seen horseshoes in algebra yet

[next obsidian]

This is complex analysis yeah

[chilly ocean]

Hello
I was wondering if a Dihedral Group $D_n$ is actually a union of the rotational symmetry group and the mirror reflection group?

$D_n = R_n$ U $M_n$

[cloud walrus]

Pencil

[delicate orchid]

By mirror reflection group do you mean the group of just the identity and the reflection or the set of all symmetries with a reflection in them?

[chilly ocean]

Identity+reflection

[delicate orchid]

Then no it’s not the union, it’s the semi direct product of the two

[chilly ocean]

Oh

And what about the latter?

[delicate orchid]

It is the union of the latter, they’re the cosets of the group of all rotations

[chilly ocean]

Ohh

Thank you 👍

[delicate orchid]

No worries

[agile stratus]

I learned about the dual numbers...

maybe a month ago

and I find them very cool 🙂

For you all who have never heard of the duals, a dual number z is an ordered pair of real numbers (x, y) where ε² = 0 and ε =/ 0. A dual number is usually denoted in the form a+bε

[cloud walrus]

Wyatt The Baguette

[delicate orchid]

Woah R[x]/x^2

[lapis trail]

So a field has no proper ideals. Can we still mod something out?

[coral shale]

e no 0

[coral shale]

trivial ideal

[lapis trail]

Si

[agile stratus]

a very powerful property is it's property of automatic differientiation

[cloud walrus]

Wyatt The Baguette

[agile stratus]

For example,

f(x)=x^2

(x+e)²

x²+2ex+e²

now, we know that e² = 0

so, x² + 2ex

now, we divide the terms with e by e to get the derivative

2x

so, when we plug a dual number into a real function, we get that function plus it's derivative (times whatever e is multiplied by)

[delicate orchid]

Dividing by a zero divisor 🤨

[spice whale]

yeah I don't think that actually works

1/epsilon doesn't exist

[delicate orchid]

This is definitely true and is cool

But you can’t have a zero divisor be a unit

Pop quiz prove that a zero divisor can’t be a unit

[agile stratus]

I just meant that that the term with e is the derivative when e isn't there

[spice whale]

yeah definitely

[spice whale]

1/epsilon = 1/epsilon + 0/epsilon = 1/epsilon + epsilon
canceling out 1/epsilon we get
epsilon = 0
which is false

[delicate orchid]

1/ for multiplicative inverse

[spice whale]

i shall be better

[delicate orchid]

Also not sure about that proof

[spice whale]

multiplication is distributive

you could put it more explicitly

[delicate orchid]

a^-1 + 0a = a^-1+a?

[spice whale]

no

[delicate orchid]

That’s what you wrote

[spice whale]

• 0a^-1

not +0a

[delicate orchid]

ok but why is aa^-1 = 0

It’s 1

[spice whale]

what

lemme do it in latex

$a^{-1} = 1 a^{-1} = (1+0) a^{-1} = a^{-1} + 0 a^{-1} = a^{-1} + a\$
(cancellation) $\to 0 = a$ which is a contradiction

[cloud walrus]

all functions are alison

[delicate orchid]

I still do not agree with $0a^-1 = a$

[cloud walrus]

Wew Lads Tbh (200 🍇) ✓

[agile stratus]

epic latex fail

[delicate orchid]

lemme do mine

[spice whale]

$aa = 0 \to a = 0a^{-1}$

[cloud walrus]

all functions are alison

[delicate orchid]

Why does a^2 have to be 0

[spice whale]

oh

[delicate orchid]

It could be any element of the ring times a equal to 0

[spice whale]

my head was stuck in dual numbers

i can just slightly modify it then

$a^{-1} = 1 a^{-1} = (1+0) a^{-1} = a^{-1} + 0 a^{-1} = a^{-1} + b$ for $b \neq 0\$
(cancellation) $\to 0 = b$ which is a contradiction

[cloud walrus]

all functions are alison

[delicate orchid]

$ab = 0$ with a and b non zero
Assume there exists a $c$ such that $ac = 1 \Rightarrow ac+ab = 1 \Rightarrow a(c+b) = 1 \Rightarrow c+b = c \Rightarrow b = 0$ contradiction

[cloud walrus]

Wew Lads Tbh (200 🍇) ✓

[spice whale]

i think we said the same thing

in the end

effectively

[delicate orchid]

I have no idea what your b is cancelling with

Oh well

[spice whale]

you additively cancel a^-1

[delicate orchid]

Ah never mind I see it now

[coral shale]

e is a 0 divisor indeed

[delicate orchid]

Had to look all the way from one side of my monitor to the other that’s craszzzyyy

[coral shale]

please dont do hacks

[coral shale]

what hast thou donest here
me no get

f(x+e) = xx + 2ex

i follow to here

what next

[agile stratus]

thats kinda just it

plug in any x

[coral shale]

huh?

[agile stratus]

this is the last step

f(x+e) = x² + 2xe
x² is function, 2x is derivative, so yopu get function and derivative with one calcluation

[coral shale]

define ef'(x) to be

f(x+e) - f(x)

is this what ur saying or did i misread

[agile stratus]

ye

[coral shale]

show e inverse exists and i may be convinced 🤔

https://media.discordapp.net/attachments/496784958430380033/950420600311132241/698212792905367553.png

[delicate orchid]

You don’t need to actually take the inverse

[coral shale]

this definition alone doesnt convince me

[spice whale]

wikipedia beats both of us

[coral shale]

we dont?

[delicate orchid]

I’m going to cry

[agile stratus]

by e inverse you mean 1/e

okay

so

[delicate orchid]

It’s a 2 dim vector space over R, if you’ve got a pure real part plus a pure e part you can be sure they’re not linearly dependent

[delicate orchid]

Allowing you to clearly separate f’(x) from f(x)

[coral shale]

lemme thinking

oh

[agile stratus]

you can't divide by e, but can you divide by a dual number where a =/ 0

[coral shale]

0 + ef'(x) := f(x+e) - f(x)

[delicate orchid]

Another cool thought I’ve had is that you can kind of separate the nth derivatives by taking f(x+e) with e^n = 0

[coral shale]

is it this?

[delicate orchid]

YurrrrrR

[spice whale]

and you can take arbitrary derivatives in a formal power series ring

[coral shale]

formal derivative lets go

[delicate orchid]

True but I don’t CARE

I just do not CARE

[spice whale]

wew you have hurt me

[delicate orchid]

I mean sure you can take those derivatives... but you can’t do it with the f(x+e) swaggg

[spice whale]

f(x + e + e² + e³ + e⁴ + ...)

wait

that doesn't work does it

[delicate orchid]

Yeah it’s not a clean separation anymore

[spice whale]

it's just f(x + e)

[delicate orchid]

Instead of R[X]/(X^n) we probably will have to take R[X_1, ... X_n]/(X_1^2, X_2^3 ..., X_n^(n+1))

[agile stratus]

e^a when a > 1 is 0

[delicate orchid]

Which is... painful

Although that looks pretty Chinese remainder theorem-y

[agile stratus]

I am tryna think of some cool properties

[spice whale]

for $R[[X]], f(a + X) = \sum_{n\geq 0} X^n D^n (f(a))$

[cloud walrus]

Wyatt The Baguette

all functions are alison

[spice whale]

ye this one is cool

[agile stratus]

e^(a+b(varepsilon)) = e^a + e^a * b(varepsilon)
which means that e^(a+(varepsilon)) = e^a + e^a * varepsilon, which defends the fact that d/dx e^x = e^x

[spice whale]

yep

[agile stratus]

there is also a matrix representation

[cloud walrus]

Wyatt The Baguette

Wyatt The Baguette

[spice whale]

$(a + b\eps)^n = a^n + a^{n-1} b \eps$

[cloud walrus]

all functions are alison

[agile stratus]

ye

I am thinking of writing a paper on this 🙂

[coral shale]

n

[agile stratus]

does not equal 0

for one

[spice whale]

so
$\sqrt{\eps} = (0 + \eps)^{\frac{1}{2}} = 0^{\frac{1}{2}} + 0^{-\frac{1}{2}} \eps$ is undefined

[cloud walrus]

all functions are alison

[spice whale]

uh

hm

[coral shale]

u missed the binom

[spice whale]

yeah sorry

$(a + b\eps)^n = a^n + na^{n-1} b \eps$

[cloud walrus]

all functions are alison

[agile stratus]

this doesn't apply to epsilon by itself

[spice whale]

yeah

[agile stratus]

e^n = e or 0 (for positive integers)

[spice whale]

(sqrt eps)² = eps

neither eps nor 0 satisfies this

[agile stratus]

*vareps

[spice whale]

\eps is what i use for varepsilon

[agile stratus]

when what do you use for epsilon

[spice whale]

I don't

no

it's not

[delicate orchid]

Yeah this is what I was getting at

[spice whale]

$f(x+\eps) = \sum_{n \geq 0}^k \eps^n D^n f(x)$

[cloud walrus]

all functions are alison

[delicate orchid]

I’m not sure this is true though

No it is

Sorry

[spice whale]

yeah

[delicate orchid]

I was thinking about not being able to wretch them out via linear independence

[spice whale]

yeah so

if you let k limit to infty

[delicate orchid]

Although maybe you can... lemme think....

[spice whale]

you get a formal power series ring

[trim grove]

What are maximal ideals in C[x,y] , any hint?

[waxen hedge]

It's related to the Nullstellensatz

Maximal ideals of C[x,y] are the ideals (x-a, y-b) for complex numbers a and b

[trim grove]

This will also help ,but if you can guide me how to reach to this conclusion?

[waxen hedge]

To be honest I don't remember the proof

[sharp sonnet]

you do not need the nullstellensatz, you can use the fact that C[x, y] = C[x][y] and C[x] is a PID

then classify all maximal ideals in R[y] for a PID R

you use something like contraction of maximal ideal is maximal

and gauss lemma

(you maybe saw this done for Z[x] once, the proof should be similar)

(nullstellensatz works too though)

[trim grove]

I have come accross this word alot of times , so it will be better to check this out

[trim grove]

Thankyou

[sharp sonnet]

the nullstellensatz is a lot more powerful

[delicate orchid]

weak nullstellensatz my beloved

[sharp sonnet]

it classifies all maximal ideals in C[x_1, ..., x_n]

(or any algebraically closed field)

[trim grove]

Yes somewhere it is mentioned that it is like a bridge between algebra and algebric geometey ( dont know i am right)

[sharp sonnet]

yes

[trim grove]

I can't even pronounce it correctly

[sharp sonnet]

it is german

[delicate orchid]

it's german if that helps

[sharp sonnet]

"theorem of the roots"

[delicate orchid]

I thought it's "zero place theorem/statement"

[trim grove]

Looks like related to zeros of multivariale functions 😅

[delicate orchid]

I trust your translation more than mine though, loch

[delicate orchid]

welcome to alg geo

[sharp sonnet]

well, nullstelle is zero/root (of a polynomial)

literally "place of zero"

you can probably make some connection to places (also called stellen in german) if you want

[delicate orchid]

yeah I split it up all the way into "zero place statement"

my german is... bad

[sharp sonnet]

mine is pretty ok, dont worry about it

[coral shale]

Is there some quick intuition for why Char 0/Char 0 extension => separable

or wishful thinking 👀

[delicate orchid]

something something every irreducible over char 0 is separable something something so every field extension of char 0 is seperable

[terse crystal]

Because f’ doesn’t equal 0

f irreducible, (f,f’) divides f, (f,f’) has to be 1, f and f’ are coprime, so f and f’ doesn’t have common root in the splitting field of f, so f is separable

(nx^(n-1) doesn’t equal 0 since characteristics is 0)

[chilly ocean]

What's the analogue of semidirect products for rings? Like what can you do, if anything, to present, say, Z[x]/(p(x)g(x)) nicely where those two polynomials are not coprime and chinese remainder theorem doesn't hold?

Is that the best you can do? Just leave it like that?

[lavish nexus]

In any ring is it true that a non unit must have a prime factor

[latent anvil]

no, take Z/6Z and the element 3. The only ways to write 3 as a product (up to ordering) are 3 = 3*1, 3 = 3*3, and 3 = 3 * 5. But 3 is a zero divisor and 1,5 are units, so none of them are primes

In general the idea of "prime factors" doesn't work very well outside of PIDs. Sometimes factoring into irreducible elements makes sense, but there's a lot of rings where factoring just isn't sensible

In the ring Q[t^2, t^3] the elements t^2 and t^3 are both irreducible, which is like a generalization of prime elements, but t^6 has two irreducible factorizations, as t^2 * t^2 * t^2 or t^3 * t^3

[lavish nexus]

I'm thinking about what these P_\alpha are

and I feel like they're principal ideals generated by a prime element

[latent anvil]

Not in general, no

[lavish nexus]

or I can just claim any x in UP_alpha is in a prime ideal

[latent anvil]

So if S = R \ (union of P_α), what can you say about the relationship between S and a single P_β?

[lavish nexus]

nvm there's a proposition saying xR can be embedded in a prime ideal P such that S\hat P = empty

ty

[latent anvil]

yep!

[coral shale]

+1 curious

[terse crystal]

Cross product of a group and a ring sounds like an analogue

But I never really used it, only saw it sometime when reading about second homological groups or something

[coral shale]

Thanks, that was enlightening

So I'm thinking about the other case --- if char = p

Then I reckon f is inseparable only when (f, f') = f. This happens only if f' = 0

And f has to be of the form

$$\sum^n_{k = 0}a_ix^{kp}$$

[coral shale]

(^oops, a_k)
If this happens, then f' = 0

But in general... is it easy to tell if this polynomial is inseparable? 🤔

I can think of examples where this happens, but not if it must happen or anything