groups-rings-fields

[robust pollen]

Do you know about Hopf algebras? Because, here is a theorem. Let $H$ be a Hopf algebra, over some field $k$ (would probably work just as well over commutative rings). Let $M$ be a free $H$-module. Then $M \otimes_k M$ with the diagonal action is free.

[cloud walrus]

expectTheUnexpected

[sour plume]

oh my that's perfect

[robust pollen]

Do you want the proof? I'm typing itright now 😄

[sour plume]

Yes, I shoulda known that Hopf algebras are the setting for this! Go ahead, I'd love the proof

If it's too complicated I'm also happy with a reference but if you're super excited about typing it out I won

't stop you

[simple valley]

I'm looking at this problem regarding bimodules and morita equivalences and I can't help but think that a bicategory is the natural setting for this

[cloud walrus]

expectTheUnexpected

[robust pollen]

The first sentence in the proof is part of the so-called fundamental theorem of Hopf modules, but I will prove it to you anyway

[cloud walrus]

expectTheUnexpected

[robust pollen]

That's true. 0-cells are rings, 1-cells are bimodules, and 2-cells are given by tensoring bimodules

qiaochu has a post on this: https://qchu.wordpress.com/2012/02/16/morita-equivalence-and-the-bicategory-of-bimodules/

[lethal dune]

[robust pollen]

@sour plume do you agree with the proof?

[sour plume]

Probably! It's blowing my mind a little bit since the Fundamental Theorem you mentioned sounds insane, but I guess Hopf algebras do be like that

So this just uses that my module M is a module w.r.t. the algebra structure, and not the coalgebra structure?

Or is that somehow automatic or so

[robust pollen]

Yeah, just a module

[sour plume]

Cooool

[robust pollen]

So, the fundamental theorem has the following nice formulation:
Let $H\text{-mod}$ be the category of left $H$-modules. Inside this, $H$ is a coalgebra object. Thus, we can consider for example right comodules over $H$ inside of $H\text{-mod}$, denote this category by $(H\text{-mod})^H$. Then the fundamental theorem says that there is a monoidal equivalence
[ (H\text{-mod})^H \cong Vect ]

[cloud walrus]

expectTheUnexpected

[simple valley]

oop that's spoilers for the next hw problem

[sour plume]

Very cool! Thank you so much, I will digest this a little more! This has rekindled my love for Hopf algebras

[robust pollen]

no worries 🙂 I'm working with all sorts of Hopfish structures, I just love them ❤️

[white nymph]

you know when people say the phrase "same difference"... like if they accidentally said nicki minaj and it was really cardi b? is this a reference to nicki and cardi being equivalent/congruent modulo something if it is a "same difference" situation?

[weak oriole]

congruent modulo public's perception of them

[scarlet estuary]

i think youre reading too much into this.

i dont think "same diff" is an equivalence relation, for one

[white nymph]

perhaps people think they're equivalent modulo the genre of music called "stripper rap". this was a suggestion of my friend who did a phd in automorphic forms, shrugs

[scarlet estuary]

surely "stripper rap" would be the equivalence class, and "genre" the modulo

if we're saying theyre equivalent because theyre both stripper rappers

[wise igloo]

stripper rap...

[white nymph]

maybe "same difference" means that cardi and nicki have the same difference when you subtract them with something in the thing you're modding out by... i.e. that theyre in the same class

[gleaming bronze]

how can i represent the cyclic group of Z with addition

with the cyclic group notation

[white nymph]

you mean $(\mathbb{Z},+) = \langle1\rangle$ ?

[cloud walrus]

EricW

[gleaming bronze]

yeah,i think i am not getting what is a generator

[robust pollen]

1

[gleaming bronze]

because i don't know how you get negative numbers with a generator

[robust pollen]

If it's a group, then it must contain the inverse of the generator

[scarlet estuary]

generators include inverses

$\langle 1 \rangle$ means all elements of the form $1^k$ for $k$ an integer

[cloud walrus]

Namington

[scarlet estuary]

k can be 0 (so you get the identity 0) or negative (so you get inverses)

[gleaming bronze]

yeah but because g is a group ,it has he inverse and indentity ?

[scarlet estuary]

alternatively, just apply 1 enough

in ℤ/5ℤ, for example, 1 + 1 + 1 + 1 = -1

since 1 + 4 = 5 = 0

[coarse storm]

Yes. A group has an identity and the inverse of every element (by definition).

[gleaming bronze]

i am also trying to get this definition

[scarlet estuary]

how many times do you need to apply g to itself to get the identity?

it will be a finite amount in any finite group

(why?)

so for example, in the cyclic group of order 4

1 + 1 + 1 + 1 = 4 = 0, so 1 has order 4

2 + 2 = 4 = 0, so 2 has order 2

3 + 3 + 3 + 3 = 12 = 0, so 3 has order 4

and 0 of course has order 1

[gleaming bronze]

okay 0 is the smallest natural such that g^m = e

[scarlet estuary]

theyre taking natural to be nonnegative

so m=0 isnt allowed

[gleaming bronze]

now i think i understand it ,michael penn video doest explain so well this concept ,and the book i am using don't cover it.

[white nymph]

sounds like a rather unusual textbook if they dont talk about orders of elements. perhaps they use the notation o(g) ?

[gleaming bronze]

i am using this one http://abstract.ups.edu/aata/section-cyclic-subgroups.html

[white nymph]

they talk about it in section 4.1 just after remark 4.4

[gleaming bronze]

yeah ,i havent noticed

it

i thought the chapter was only about cyclic subgroups

[white nymph]

the "order" of an element a, is the "order" of the cyclic group generated by a = <a>

i.e the number of elements in <a> is the same as the "order" of a

so they have something in common - hence being in the same section

[lilac trench]

Im stuck- let $A$ and $B$ be subgroups of $G$ and let $A\le N_G(B)$ show that $AB$ is precisely the join of $A$ and $B$ under the subgroup lattice of $G$

[cloud walrus]

please request a new nickname

[lilac trench]

And its suggested I use this theorem:
$$ A,B\le G\text{ and }A\le N_G(B)\Rightarrow AB\le G, B\trianglelefteq AB, A\cap B\trianglelefteq A\text{ and }\frac{A}{A\cap B}\equiv \frac{AB}{B} $$

[cloud walrus]

please request a new nickname

[lilac trench]

the first statement after the arrow is $AB\le G$ not $AB\le B$ by the way

[cloud walrus]

please request a new nickname

[gleaming bronze]

can someone proof why <1>^(order of cyclic group) = e

[lilac trench]

what group is this under?

should I assume that 1 is the identity

[gleaming bronze]

Z with plus operator

<1> is the generator

[lilac trench]

ohh ok

now do you mean that 1 has the order of the cyclic group or the generator of 1 has the order of the cyclic group

for instance, Im not sure what is meant by
$\langle 1\rangle^n$

[cloud walrus]

please request a new nickname

[gleaming bronze]

can you give me the proof for that?

[gleaming bronze]

i will write it in latex

[scarlet estuary]

Let $G$ be a group and $g \in G$. Suppose $g^n \neq e$ for all $n$; then $g^n$ is a distinct element of $G$ for each $n$, since if $g^n = g^{n+m}$, that means $e = g^0 = g^{m}$ (right-multiply by $g^{-n}$). This means $G$ must have at least as many elements as there are natural numbers, i.e. $G$ is not a finite group.

[cloud walrus]

Namington

[scarlet estuary]

so in a finite group, every element g has a natural number n such that g^n = e.

this same proof also justifies that the order of an element is ≤ the order (size) of the group

so you wont have a group with 10 elements but one element has order 13

thats not possible

in fact, with a bit more work, one can prove each element has order 1, 2, 5, or 10

https://en.wikipedia.org/wiki/Lagrange's_theorem_(group_theory)#Applications

[south storm]

Since A is in the normalizer of B you have that AB is a subgroup, trivially it contains A and B. What‘s left to show is that any other subgroup C that contains A and B contains AB, that however is clear as C has to be closed under your group law.

[simple mulch]

any recommendation on a book about group theory?

[south storm]

What‘s your (abstract-) algebra background?

[simple mulch]

uh not much

[lilac trench]

what do you mean by "C has to be closed"

[south storm]

We required that C is a subgroup of the group one‘s dealing with

Hence it needs to be closed under the group law, meaning that for any elements x and y of C their product under the group law is also in C.

[south storm]

What about your general maths background and book preferences? Are you feeling well reading dense textbooks or do you prefer rather easy to read maths books?

[lilac trench]

ahh, I thought you had derived something I had not seen in the theorem

[south storm]

Does the explanation make sense to you though?

[lilac trench]

so basically you took the definition of the join of A and B, namely the intersection of subgroups containing their union, and said that AB must be within this subgroup since all a\in A and all b\in B must be in these subgroups and hence, their product must be as well, that is, AB

[south storm]

Yes

And also that AB itself is contained in the intersection

[lilac trench]

yes

[south storm]

There‘s some details I didn’t write out explicitly though, you‘d need to check that AB indeed is a subgroup

[lilac trench]

Yeah, it seems that the theorem provided allows me to just say it though

[south storm]

But since A is contained in the normalizer of B you have that AB=BA and then it follows

[lilac trench]

but how does AB=BA imply that it is a subgroup

[south storm]

You need to check that it is closed under the group law and that inverses are contained, if x and y are in AB, so there exist a,a‘ in A and b,b‘ in B such that x=ab and y=a‘b‘ then xy= aba‘b‘, because AB=BA you have that ba‘ is in AB and hence there are a‘‘ in A and b‘‘ in B sich that ba‘=a‘‘b‘‘ and hence xy=aa‘‘b‘‘b‘ which is in AB. Hence xy is in AB whenever x and y are in AB and so AB is closed under the group law.

If x is in AB then there are a in A and b in B such that x=ab, then x^{-1}=b^{-1}a^{-1} which is in BA, since BA=AB we have that x^{-1} also is in AB.

[lilac trench]

oh ok so you can reconstruct the product so that it becomes an element of AB

[south storm]

Yes

[lilac trench]

I see

[simple mulch]

Maybe something in between? I think I could handle most assuming it doesn’t have that many pre requisites

[south storm]

I'd perhaps recommend "Algebra chapter 0" by Paolo Aluffi, not exclusively on group theory but that's the first thing it covers and if you aren't familiar with any much algebra you might be interested in continuing reading it

[robust pollen]

Dummit and Foote is also very good imo

[simple mulch]

interesting, I foudn the idea of the book interesting

[south storm]

I'd agree but I somewhat dislike the style personally

[south storm]

Which book do you mean?

[simple mulch]

the one from aluffi

chapter 0

[south storm]

What idea of it are you referencing? You mean a general purpose basic algebra textbook?

[simple mulch]

Algebra: Chapter 0 is a self-contained introduction to the main topics of algebra, suitable for a first sequence on the subject at the beginning graduate or upper undergraduate level.

[south storm]

Personally wouldn't call this idea original but I suppose why not

[south storm]

There's a quite some books like that

Lang is a famous other

[south storm]

Though I don't recommend it if you have no background in algebra nor advanced maths

[robust pollen]

Try Bourbaki Algebra, it's da 💣

(a joke btw)

[simple mulch]

lol

[next obsidian]

I’m on Bourbaki gang

(When you have a specific result you want a proof for and know it’s in the book, or have a feeling it will be)

The proofs are very clear albeit you have to flip pages a lot, but you won’t be sitting there wondering how they justified a step

[south storm]

Never read Bourbaki algebra but from what Bourbaki I skimmed through it was pretty good

[next obsidian]

It’s not good to learn from

But it’s an excellent reference

[simple mulch]

the kernel of a vector space is ${v : T(v) = 0}$

[cloud walrus]

mns

[simple mulch]

where $T : V \rightarrow W$

[cloud walrus]

mns

[simple mulch]

I am seeing an exercise where the author straight away assume $T(0) = 0$

[cloud walrus]

mns

[simple mulch]

Why can't we have a linear map where for the zero vector in V we have a non zero vector in W ?

[chilly ocean]

T(0) = T(0 + 0) = T(0) + T(0). subtract T(0) from both sides

[south storm]

Given a linear map $T:V\to W$ between two vector spaces over the same field $K$, one has that $T(0)=T(0\cdot 0)=0\cdot T(0)=0$

[cloud walrus]

27182818284tropy

[south storm]

That 0T(0)=0 follows from that 0T(0)=(0-0)T(0)=0T(0)-0T(0)

[south storm]

Perhaps it‘d be better to put this like $$T(0_V)=T(0_K\cdot 0_V)=0_K\cdot T(0_V)=0_W$$

[cloud walrus]

27182818284tropy

[simple mulch]

I see

interesting

thanks

[south storm]

Further one can prove that the kernel of a linear transformation is a subspace of its domain

[simple mulch]

Exactly

I am seeing it

[tame grove]

homomorphisms can be from sets of different cardinality correct? for example Z_10 to Z_25, has homomorphisms right?

[maiden ocean]

Yes

[tame grove]

aight. thats what I thought but then I second guessed myself since the cayley tables are not trivial

[south storm]

Usually the term homomorphism is used in the context of algebraic structures, not simply sets necessarily, but the underlying sets may have different cardinalities yes

[simple valley]

cayley tables are about multiplication within a single group

[maiden ocean]

For an incredibly trivial example take G = Z and H = 1 and let f: Z -> 1 be the map sending everything to 1

or 0 depending on preferred notation

[tame grove]

oh yea, I guess that is true

[tame grove]

Yea, I was trying to visualize everything though. I find that helps a lot

low key, I really wish I could find books that had pictures of everything for high level maths. It just gives more context imo

[south storm]

The way to visualize homomorphisms f:A->B between magmas(or groups if you prefer) by Cayley tables, I’d say is to imagine collapsing some rows and some columns of the Cayley table of A and inserting it into the Cayley table of B.

Personally I prefer Cayley graphs rather than Cayley tables for this kind of visualizations

[tame grove]

fair. I just 2nd guessed myself since I wasnt sure if it was proper to call something a homomorphism when it doesnt map to every element, but I guess it wouldnt really matter since not all sets are well ordered

[south storm]

What do you mean?

[simple valley]

what does well ordering have to do with any of this

[south storm]

A homomorphism in this context is a map that preserves the operations of your structure

[tame grove]

I mean operations on sets can skip values. it wouldnt matter for homomorphisms though since if you can include that skip within your function you would still have a bijection between corresponding elements.

[simple valley]

what bijection

[tame grove]

between the sets

[simple valley]

there's no bijection

[south storm]

A homomorphism needn‘t be a bijection

[tame grove]

oh, am I thinking of isomorphism?

[south storm]

It seems so

Are you talking in the context of groups?

[simple valley]

isomorphism is a homomorphism with a homomorphism inverse

[tame grove]

damn. my prof covered them in the same day so now they get tangled in my head

[simple valley]

isomorphisms preserve cardinality yes

[tame grove]

Yes, that makes sense for what I was trying to say, thanks.

[south storm]

For algebraic structures bijectivity suffices

In what context are you talking about homomorphisms Caesium?

[tame grove]

group homomorphisms

where some operation can map elements in a group to always return the same value in another group

[south storm]

A homomorphism from a group (G,•) to a group (G‘,•‘) is a function f:G->G‘ such that for any a,b in G you have that f(a•b)=f(a)•‘f(b)

An isomorphism from a group (G,•) to a group (G’,•‘) is a homomorphism f:G->G‘ such that there exists a homomorphism g:G‘->G such that g(f(x))=x and f(g(y))=y for any x in G and y in G‘.

[tame grove]

thanks, that makes sense. I just got the definitions messed up in my head a bit

[south storm]

One can prove that a homomorphism is an isomorphism if and only if it is bijective

[limpid edge]

why is x^n + x + 1 only reducible (over Q) for n = 5, 8…

x^2 + x + 1 is a factor for those n

[chilly ocean]

uh

well ig there is way to show it

(x^2 + x + 1)(x^n-2+a1*x^n-1+…+an)=x^n+x+1

and then solve relationship between coefficients

[delicate bloom]

if you look at it $x^{2+3k}+x+1 \mod x^3-1$ then $x^3=1 \mod x^3-1$ so it simplifies to $x^2+x+1 \mod x^3-1$. Since $x^2+x+1$ divides $x^3-1$ we can simply reduce further to see that it's really $0 \mod x^2+x+1$.

[cloud walrus]

Merosity

[delicate bloom]

although for the 3k and 1+3k case it remains to show that those really are irreducible, maybe try translating to get some sort of eisenstein criteria to show up

[limpid edge]

yea i understand why n=3k + 2 is divisible by x^2 + x + 1 lol

[delicate bloom]

what have you tried so far

[limpid edge]

showing they’re the only reducible ones js the hard part

[delicate bloom]

what ways do you know of to show irreducibility

[limpid edge]

how tf will eisenstein work

[delicate bloom]

translating a polynomial doesn't change whether it's irreducible or not

[limpid edge]

uhh

[delicate bloom]

so by looking at f(x+n) it's possible to make it satisfy the eisenstein criteria

[limpid edge]

x^2 + 2x + 1 is reducible

[delicate bloom]

this is a standard trick, you can use it to prove cyclotomic polynomials are irreducible

[limpid edge]

x^2 + 2x + 2 is not

maybe i’m misinterpreting

oh

i thought tou meant shift

[delicate bloom]

you're not plugging in correctly

yeah

another trick that's easy to try is map f(x) to x^n*f(1/x)

really just change the order of the coefficients from left to right to right to left, might help you achieve making eisenstein criteria

[limpid edge]

yeah i saw this too https://math.stackexchange.com/a/800835

[delicate bloom]

where's your problem come from, I haven't started looking at MSE cause I figure it's just a textbook problem

you're being kind of evasive, you haven't answered my questions so you're forcing me to pull teeth here

what have you tried? what methods do you know of to show irreducibility?

[dull root]

So I have the following problem. Let a_0,a_1,a_2 be algebraic integers. Then if I have some number b\in C with the property that b^2 a_2 + ba_1 + a_0 = 0, then b must also necessarily be an algebraic integer. That is, there should exist a monic polynomial with integer coeffficients which have b as a root. My current attempt was to esentially set up a system of 4 equations all = 0 since we also know a_0,a_1,a_2 are all algebraic integers, so together with b^2 a_2 + ba_1 + a_0 = 0, we get 4 equations which = 0. However, I am not seeing where to proceed from here.

Another idea I tried was showing that if b^2 a_2 + ba_1 + a_0 = 0, then a_0,a_1,a_2 must be rational. Then if they are rational and algebraic int, they must necessarily be integers and we would be done. This one seemed promising, but I quickly realized it was not true at all. In fact, we can even look at a simpler case, where a_0,a_1 are algebra int and b has property b = ba_1 + a_0. We set a_0 = a_1 = \sqrt 2

Does anyone think this is a good idea or have a hint for other approaches?

[hidden haven]

The discriminant of the quadratic polynomial you have written is a square root of algebraic integers, so is an algebraic integer

[dull root]

So, I see any root of algebraic int is necessarily algebraic int, but why is looking at the discriment enough? Because algebraic int's are closed under additon and multiplication?

[hidden haven]

Yeah, that was what I was thinking but I just realised that you need to divide by 2a_2 to get b

So I don't think this works

[dull root]

yea, its not closed under division

I think a approach should work in a more general setting, where we can have as many a_0,a_1,....a_n-1 all algebraic int and b is a root of a monic polynomial with coefficients a-s.

[dull root]

A third idea might be that I would want to show Z[b] is finitely generated, which implies b is an algebraic integer. So I know a_0, a_1, a_2, are algebraic integers thus Z[a_i] is finitely generated. If b is a root of a monic polynomial with coefficeints a_i -s, why does that imply Z[b], all polynomials with integer coeffieicnts taken at x = b, is f.g....

[next obsidian]

Ugh…

Man this stuff really sucks

If you were looking at the notion of integrality which is basically algebraic plus satisfies a monic polynomial (okay the latter condition implies algebraic so not particularly useful lmao) I would refer you to a commutative algebra textbook

[next obsidian]

OH I CAN SOLVE THIS FOR YOU

b satisfying a monic polynomial says for some n you get a relation like b^n = a1b^n-1 + … + a_{n-1}b + an

So now take a polynomial in b of the form
cmb^m + … + c0, we want to show you can write this as Z-linear combinations of 1,…,b^n-1

If m < n, you’re done

If not, do induction on m, so all you have to do is write b^m = b{n-m}b^n

[next obsidian]

Then use the relation here to turn b^n into a poly of degree n-1 in b, then you’ve turned
cmb^m + … + c0 into
d_{m-1}b^m-1 + … + d0

All you did was turn b^m into some polynomial in b of degree m-1

By turn some b^n factor into a poly of degree n-1

So once you just collect all the coefficients you get a poly in degree m-1

[dull root]

Hmm, I have seen this trick before, but I dont see where the fact that the a_i -s are algebraic integers were used anywhere

[next obsidian]

Well… it doesn’t

[dull root]

Doesn't what you have above require b to be an algebraic integer?

[next obsidian]

Wait

What does algebraic mean to you

Satisfying a monic polynomial?

[dull root]

algebraic int means that b is the root of a monic polynomial with interger coeffiecients

[next obsidian]

Oh!

Okay swag!

So I proved algebraic => Z[b] is finitely generated

[dull root]

That is what I want to show, and I know that is equivalent to Z[b] being finitely generated

[next obsidian]

The other direction is easier

If Z[b] is finitely generated then you can’t have {1,b,…,b^n} linesrlynindependent forever

So eventually you can write b^n as a Z-linear combination of 1,…,b^n-1

So it’s algebraic

Now we can finish

So this time we assume b satisfies a monic poly b^n + … + a1b + a0 = 0

Where all the ai are algebraic

So all you need to do is see that Z[a0,…,an] is finitely generated

Cuz all the ai are algebraic

Then Z[a0,…,an,b] is fg over Z[a0,…,an]

[dull root]

Yes, that is clear since Z[a_i] is f,g

[dull root]

wait, why is this?

[next obsidian]

Cuz

The former is

[dull root]

oh yea

nvm

[next obsidian]

Yeh

So joe use transitivity of finite generation

Z[a0,…,an,b] fg over Z

So you can again look at 1,b,…,b^n

Eventually these are linearly dependent

So b^n = a0b + … + an-1b^n-1

Where all the ai are in Z

[dull root]

What I understand Z[b] to mean is that Z[b] is the group (under addition) of all polynomials over Z taken at x = b. Is Z[a,b] meaning all polynomials over Z taken at x = a OR x = b?

[next obsidian]

No

It’s like

Take multivariate polynomials

And plug in the vector (a,b)

So like a^2 + 3b^2a + 17b^9

Exists in there

[dull root]

So a polynomial like f(x,y) = x^2 + 2y + 0?

[next obsidian]

Sure

Then you plug in (a,b)

[dull root]

Oh I see.

[next obsidian]

And if you want to adjoin more numbers you just increase the number of variables

So you probably want to switch to x1,…,xn to scale up

[dull root]

Then Z[a0,a2,a3,...a_n-1 b] is f.g over Z, and this must imply Z[b] is f.g too

[next obsidian]

Yeah

This is a little subtle

One way to get it is to use Noetherianness but that’s kinda overkill

[next obsidian]

I’d use this to just see that b is integral directly

The reason it’s subtle is thinking along the lines of “subobject of finitely generated thing is finitely generated” isn’t always true

[dull root]

Doe it also work to notice Z[a_1,a_2,...a_n,b] is a f.g abeliean group, with Z[b] a subgroup, hence Z[b] is f.g

[next obsidian]

Yes that works

If you just happen to know that :)

[dull root]

I think there is a simple reason for this, but I can't see: Let cos x be an real algebraic number of degree at most 2, so its minimal polynomial has degree at most 2, then I want to show
cos x + isin x has degree at most 16 as an algebraic number

[rustic crown]

[Q(cos x) : Q] <= 2,
[Q(sin x) : Q] <= 4 as Q(sin x) = Q(sqrt(1 - cos^2 x))
[Q(i) : Q] = 2
combining everything gives that [Q(cos x + i sin x) : Q] <= 16

[dull root]

So if cosx has degree at most 2, this implies sin x has degree at most 4, and noting i has degree 2(the polynomial x^2 - 1), we have the desired result

[rustic crown]

x^2+1, yep

we're using that degree of composite is at most the product of degrees

[dull root]

Hm, I have not seen that before, is it a trivial exercise, or is it involved?

[rustic crown]

it's very simple. say E/Q and F/Q are finite extensions, define EF as the smallest extension containing both E and F.
(try to see if you can show EF = {e_1f_1 + ... +e_rf_r | e_i in E, f_i in F})
pick spanning sets for E/Q and F/Q, find a spanning set of EF/Q.

oh but you don't need to use this btw, since you're only working with simple extensions we can make remove this

[cloud walrus]

det

det

[dull root]

I am trying to work out the fact that the minimal polynomial for sin has degree at most 4. The use of x was bad... so instead use g. Say cos g has minimal polynomial ax^2 + bx + c. Then I know sin g = \sqrt(1 - cos^2g), but I am not seeing a clear way to come up with the minimal polynomial. Namely, the issue is that we have some cos^4 g's that pop up that I don't know how to get rid of

[rustic crown]

don't work with the actual minimal polynomials, use degrees, they make life easy

can you see why the second extension has degree at most 2?

[dull root]

We took cosg such that it has degree at most 2

[cloud walrus]

det

[dull root]

So the 1st extension is clear, but I cant see the 2nd one immediately

[cloud walrus]

det

[dull root]

Oh, I see now.

[cloud walrus]

Gewisser Fler

Gewisser Fler

[chilly ocean]

But what about the second equality?

[upper pivot]

I'd guess we should do something like, X= X'+iY' and Y= X'-iY' but that requires your field K to have a square root of -1

[cloud walrus]

Gewisser Fler

[chilly ocean]

Ok lets just assume $K = \mathbb{C} $

[chilly ocean]

Yes that works, thanks!

[upper pivot]

right if K is alg closed then there is a square root of -1

so it works

[dull root]

So I am trying to show the following: If a is an algebraic integer with minimal monic polynomial f \in Q[x]. Then necessarily, f \in Z[x]. Since a is algebric int, I know there exists a monic g(x) \in Z[x] such that a is a root of g. Moreover, since f is irreducible, we have g(x) = f(x) * q(x) for some q \in Q[x]. However, I am not sure how to proceed. The minamality of f is used, but I don't see how to use f is also monic. All that is telling me is that q must also be monic, right?

[winter thorn]

strict subgroups of rationals under addition are either trivial or integer multiples of a particular rational number, right?

[hot lake]

no

[hot lake]

shouldn't it be g(x) = f(x) * q(x) ?

[dull root]

Yes, mb I switched them accidently

But it should still be true that q is also monic right, else g could not be monic

[hot lake]

yes

[winter thorn]

if you are replying to me, can you at least correct me?

[dull root]

Then, I think we can turn them into polynomials with integer coefficents by taking the lcm

[hot lake]

I'm not sure how that helps showing that f has integer coefficients

[rustic crown]

you can either use gauss lemma to do it or use that other roots of f are also going to be alg int

[hot lake]

yeah I have a feeling gauss lemma has to be involved

[rustic crown]

If b is another root of f, then notice that g(b) = 0, this means that b is an algebraic integer

this will avoid gauss lemma and would give you a simpler proof

hint ||coefficients of f are algebraic integers and are rational||

[dull root]

This gives that all roots of f are algebraic integers

[rustic crown]

yep!

[hot lake]

oh yeah that works

[dull root]

Then f can be factored as a product of (x - roots) all of which are algebric integers, and alg int are closed under multiplication, so all coeffiicents are alg int and rational, thus integers

[rustic crown]

[dull root]

This is more of a definition thing. But if f(x) is an irreducible monic polynomial with a root at x=a, is that the same as saying that saying f(x) is the minimal polynomial of a.

[rustic crown]

not quite, but you can prove it in your case. in general minimal polynomials are not irreducible

[dull root]

Is there a easy example to see the difference?

[rustic crown]

yep... it can happen if the element doesn't lie in some integral domain

[cloud walrus]

det

[rustic crown]

we physically adjoined alpha = coset of t that satisfies x^2 = 0

in particular it's minimal polynomial is x^2 and is not irreducible

in our case, if you had a non-trivial factorization of the minimal polynomial f(x) = g(x)*h(x) then we can plug in the x = a and get that either g(a) = 0 or h(a) = 0 from the integral domain-ness. This will give a smaller degree polynomial contradicting minimality of f.

same thing doesn't work there because Q[t]/(t^2) isn't an integral domain

[lethal dune]

Is there a way to show that the quadratic integer ring $\mcal{O}(\sqrt{D})$ is E.D. where $D=-2, -3, -7, -11$ without bruteforce calculation of $r,s$ ?

[cloud walrus]

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

[rustic crown]

what's r, s?

[lethal dune]

remainder and quotient

[rustic crown]

you need to find the circum radius of the lattice triangle and show it's <1

[lethal dune]

what is it? can you link any resources?

[rustic crown]

this is directly showing that the usual norm gives a euclidean valuation

like consider alpha and a non-zero beta. look at alpha/beta. need to find a close enough lattice point.

in case of Z[i], you can choose the lattice point to be at most 1/sqrt(2) away

this will give you that N(alpha/beta - gamma) <= 1/2

the same thing, but as -3, -7, -11 are 1 mod 4, those rectangles become triangles

[lethal dune]

I understood the proof for Z[i] but what is the 'rectangles become triangles'?

[dull root]

So if I changed the condition of f from instead of being the minimal polynomial of a to just some irred monic polynomial with a root of a. Should I be showing the reason f is minimal is because otherwise let q be the minimal poly of a. Then f = q*r for some other polynomial r, contradicting the irred of f?

[rustic crown]

so if you did the same thing for Z[sqrt(-2)] we'll have to find the longest distinct from a point inside the rectangle with sides 1 and sqrt(2) from the 4 vertices. this is half of the diagonal = sqrt(3)/2 < 1

for -3, -7, -11, we need to add that middle point, so now the question is to find the largest distance from a point inside the triangle to it's 3 vertices

we can cover the plane with these triangles and it's upside down image

[lethal dune]

oh nice, what about D=-19, -43, -67 tho? the proof in the book is given in terms of Universal side divisors which I didn't quite understood.

[cloud walrus]

det

[rustic crown]

for D = -3, -7, -11

so if you look at our calculation for D = -19, we get that maximum distance is 5/sqrt(19) which isn't < 1, so our proof fails. but the thing is they are not even euclidean domains. we need to show that no euclidean valuation will work

[rustic crown]

yep! this is true

[rustic crown]

and to do this, we need to get some criterion which doesn't depend on the euclidean valuation as it's very hard to work with all the possible valuatoins

[dull root]

I think this argument works: Let f be a irred monic poly with root at a and g be a minimal poly of a. I claim f = gr. If not, then f = gk + r by divison, and r has a root at a with degree less than g, which contradicts the minimality of g. Since f = g * r, we can let h be a monic poly over Z[x] with a as a root. Then I want to say h = f * k where k \in Q[x], but thats not necessarily true. I only know h = g * k' for some k' in Q[x]

wait, nvm f is irred implies f = g

[rustic crown]

and one criterion is this universal side divisor. if it were a euclidean domain, then look at all non-zero non-unit elements and pick the one with minimal valuation. call it u.
now if you divide anything by u, the remainder will have valuation smaller than that of u, so will be 0 or a unit.

if we show that for D = -19 and others that these have no such u, then it will show that there isn't any possible euclidean valuation on it.

only units in those rings are +-1 so it's not a hard thing to check

[lethal dune]

Let me read about the universal side divisors a bit more

[rustic crown]

okie

[weak oriole]

I find universal side divisors p weird

in the sense that Idt I've actively thought about those properties in familiar rings like Z, Z[i] etc

[rustic crown]

yea, only time i ever used it is showing the quadratic ring of D = -19 isn't ED xD

[weak oriole]

lmao

[lethal dune]

feels like the ESD thing came out of nowhere

[rustic crown]

yea it felt pretty weird at start... but there was one similar thing a prof said in one of our lectures

there is a correspondence between valuations and like some partition of subsets idr

[lethal dune]

[chilly ocean]

det I have a question

do u know about graded rings and hilbert/poincare series?

[rustic crown]

wait our alg geo prof talked about hilbert series on thursday, dunno about poincare series

but don't think i understand the hilbert series well either, too much language

[chilly ocean]

[rustic crown]

wait it was hilbert polynomial actually

are they same?

[chilly ocean]

yes I think actually

moldilocks I have a question

do u know about graded rings and hilbert/poincare series?

[dull root]

Let z be an algebraic integer on the unit circle with degree n (ie its minimal polynomial is degree n), I'm having a hard time understanding what does that have to do with z being either a root of unity or not a root of unity. I don't see the connection between the degree of a minimal polynomial and being a root of unity. So if z is a n-th root of unity, I know the minimal polynomial has degree at most n, since z^n-1 does the job, ie all roots of unity are algebric integers. However, there are more algebraic integers on the unit circle which are NOT roots of unity. We can for example find algebraic integers on the circle with has degree at most 16, so how does knowing the degree of an algebraic integer help us determine if it is a root of unity or not

[final pasture]

What exactly is the statement you are referring to ? @dull root

[hidden haven]

[chilly ocean]

shika I have a question

[rustic crown]

😂

[hidden haven]

oh I didn't see the question

I have seen the definition of a hilbert series but nothing beyond that

[dull root]

I am trying to show that there are infinitely many algebraic integers on the unit circle which are NOT roots of unity. The hint I was given was to think of the degrees of such algebraic integers, ie their minimal polynomial

[chilly ocean]

I just cant find any sources that cover that stuff well imo my course is all around the places

[rustic crown]

do you know that cyclotomic polynomials are irreducible?

[hidden haven]

is it a topological combinatorics course?

[dull root]

Nope, I have not heard of that term

[chilly ocean]

not sure if asking me or masaka but neever heard of that

[hidden haven]

you was asking because that's where I encountered them so could send a resource in that case

[dull root]

So eariler, we were able to find infinintely many points on the unit circle which had degree at most 16 by taking cosf + i sin f. Then I want to show there are a finite amount of roots of unity which have degree at most 16.

[chilly ocean]

this is intro comm alg class first month in .

[hidden haven]

oof

[rustic crown]

ah right

[final pasture]

that basically boils down to prove that if alpha is a primitive n-th root of unity for n > 16 then alpha is an algebraic integer of degree n > 16

Because there is finitely many primitive n-th root of unity for n <= 16, and those clearly have degree atmost n (because X^n - 1)

[rustic crown]

primitive nth root of 1 has degree phi(n), so that can be <= 16 even when n> 16

but idk how to avoid using that and phi(n) >= sqrt(n/2) for (n >= 2)

[dull root]

So the euler toitent function counts the degree of a n-th root of unity right?

[rustic crown]

yep

[dull root]

Yea, so the above claim is not true...

[rustic crown]

that's because the primitive roots have minimal polynomial = cyclotomic polynomial

and proof of its irreducibility requires some work

[rustic crown]

it's true but we need to look at nth roots where n > 512 or something

there might be some elementary way

[dull root]

But maybe I am misunderstanding: the goal was to find infintely many algebraic intgers on the circle which were not roots of unity. We have infinitely many such points which had degree at most 16. We know there are finitely many primitive n-th roots of unity for n up to 16. So the current issue is that some of our points might be primitive roots of unity for something > 16?

[lethal dune]

btw for D=-15, is it E.D. tho? the method we used previously won't work because the circum radius >1

btw is it enough to conclude that it's not ED? ig yes

[rustic crown]

wikipedia says it's not a pid as well

[final pasture]

Oh yeah fuck

I was thinking of prime numbers

but this is still not true

[lethal dune]

lol should have checked, mb

[final pasture]

phi(p) = p-1

Sorry

[lethal dune]

is it yes?

[rustic crown]

don't think so... but there might be some weird theorem saying that some weird class of number rings are Euclidean if and only if usual norm is an euclidean valuation

[final pasture]

But the phi(n) >= sqrt(n/2) bound fix the idea anyway

[lethal dune]

Intuitively saying that if R > 1, then it is possible that we can find one point which is more than 1 dist apart from all the vertices then we cannot find a lattice point.

[rustic crown]

Q(sqrt(D)) is dense in C when D < 0

so should be true

we'll need to pick a point very close to the cirucumcenter

and hope that other triangles don't interfere

[dull root]

Hmm, is there another really tirivial bound other than that sqrt bound? We just need the lower bound to be > 16 after a certain point right?

[final pasture]

yeah

I mean, you can just do it by hand

using the explicit expression of phi

[lethal dune]

yes cool

[final pasture]

phi(p_1 ^e_1.. p_k^e_k = n)

Show that whenever any p_i > 16, phi(n) > 16

and then cases assuming each is p_i <= 16

but this is ugly

not hard at all but ugly

[rustic crown]

yea lol so ugly

i had a problem in one of our endsems to show that if [K:Q] is finite then it contains finitely many roots of unity

i had to prove that bad bound phi(n) >= sqrt(n/2) to make writing a little nicer

what's the simplest bound that you can give without any ugly work?

like a very bad upper bound for number of divisors is 2sqrt(n)

[lethal dune]

@rustic crown you are doing masters rn?

[rustic crown]

nah, 3rd year of ug

[lethal dune]

[rustic crown]

i don't understand the sotrue emote lol

[lethal dune]

we never covered this much in our undergrad

[hidden haven]

it means you are lying

[rustic crown]

no i mean how does and sotrue connect

[lethal dune]

they don't

used them randomly, they don't necessarily have to mean something

[rustic crown]

[hidden haven]

bruh

have you not seen the so true meme

[lethal dune]

we barely touched EDs and PIDs in our undergrads

[rustic crown]

nope, stopped using insta a year ago, that was my only source of meme

[hidden haven]

F

[rustic crown]

damn

[lethal dune]

[rustic crown]

why tho? isn't that like taught in first ring theory course

[lethal dune]

we were taught only the basic rings, hell our professor even skipped the whole part in the class in our masters class and gave a load of assignments

that's how things are done here, lol

[rustic crown]

our ag prof does most things in assignments, and classes are so hard to follow

[lethal dune]

yeah, the whole group theory from the syllow was given as an assignment, so was the field extension part

[rustic crown]

[lethal dune]

I'm just reviewing these topics because I forgot most of the things

need them for entrance

[rustic crown]

i have to review each thing 3-4 times until i can prove them in my sleep

[lethal dune]

yeah, our sem 1 was rushed like hell, all had to be covered in 3 months

[rustic crown]

else i just forget it in a month

how long was your ug course?

[lethal dune]

due to covid

it's supposed to be 5-6months,

[rustic crown]

our sem are usually 4 months

but sem5 they made it 2.5 months

this ag course is too rushed up

[lethal dune]

I forgot most of the things we were taught in group theory, had to so a full run through. It's a pain, ig I need to be more in touch with the subject

[rustic crown]

i have studied so little analysis in the past year that i feel i forgot all ofit

[lethal dune]

yes I mostly do those, good thing I found the server to help with abstract stuffs

[fossil shuttle]

i never studied this either no worries

but as you go deeper into AG you need this stuff more

[tame grove]

Im not sure what I need to show for this.

Write down the formulas for all homomorphisms from $Z_{10}$ into $Z_{25}$.

[cloud walrus]

CaesiumIsFake

[tame grove]

I understand all the ideas independently but I am having trouble putting them into a form where I can easily show a homomorphism

[fossil shuttle]

#groups-rings-fields message

try proving these two lemmas as an exercise and then applying them to solve your problem

[lethal dune]

thank god

[fossil shuttle]

this stuff is pretty accessible imo, if you read Gallian's book on algebra ED's and PID's are both covered thoroughly

[lethal dune]

I read dummit foote

[rustic crown]

I read Aluffi, but our algebra prof did a very good job in the ring theory course

[lethal dune]

wish I could say the same. Also there is another problem I have about group theory. any subgroup of R ( Q) is either cyclic or dense in R. How do I prove it?

wrt •

[plucky flicker]

Hi! Any hint for showing that the dimension of any irreducible representation is a divisor of |G:Z(G)|?

We know that the dimension divides the order of G

[robust pollen]

I don't think that's true: consider R_>0

[hidden haven]

denseness of a subgroup is equivalent to having elements of arbitrarily small magnitude

[lethal dune]

hm I think it was wrt + then

[hidden haven]

oh ye should be wrt +

[lethal dune]

yeah so like if there are elements with arbitrary small magnitude then we can say it's dense because of the + right?

[hidden haven]

idk what you mean by because of the +

[lethal dune]

like we would be able to translate it to any nbd of x ∈R by adding it sufficient times

[hidden haven]

yes

[lethal dune]

so how do I show that if it is not cyclic then there is an element of arbitrary small magnitude?

[wooden ember]

You suppose that it isn’t cyclic and that you don’t have arbitrarily small magnitudes and derive a contradiction I suppose? Cause you should be able to take two elements such that no Z linear combination gives a divisor of either and I think you should be able to argue that by making coeffs big enough you can get a magnitude small enough?

Just speculating here haven’t tried this

[hidden haven]

show that smallest magnitude positive element generates it

[lethal dune]

I was speculating something similar as well

[hidden haven]

showing contrapositive of what you said

[lethal dune]

seems like I was going the other way around

[wooden ember]

Right kind of like when you show the only subgroups of Z are nZ

[hidden haven]

yes

exact same argument

[chilly ocean]

uh, i dont think thats right? consider 1/x and 1/y where x and y linearly independent over Q

[hidden haven]

I am not sure what that means

subgroup generated by those 2?

[chilly ocean]

yea

[hidden haven]

can you give a concrete example

because I don't see how this is a counterexample

[lethal dune]

i think it'll be dense not cylic

[hidden haven]

ye it will be dense

we need to show that if there is a smallest positive magnitude then group is cyclic

[lethal dune]

honestly it feels so weird that it's true

[hidden haven]

[chilly ocean]

oh it reminds of bezout

[wooden ember]

.

[chilly ocean]

yea it makes sense

[wooden ember]

Yeah so I suppose you can argue similarly for any Euclidean domain no?

[lethal dune]

[hidden haven]

yes whenever you have long division

[wooden ember]

Pretty nifty

[hidden haven]

though here it is a specific kind

because you only want quotients to be integers

since this isn't a ring

[wooden ember]

Yeah fair enough

[lethal dune]

so similar true for C as well?

[wooden ember]

Is there a notion of being a Euclidean domain as a module instead of a ring

[lethal dune]

C is also an ED

[wooden ember]

So that we could say that “R is a Euclidean Z module” or whatever

[hidden haven]

I think narwhal is referring to ideals

[wooden ember]

Like in this case you can get quotients in Z

[hidden haven]

ED implies PID, and the argument is pretty much this

[wooden ember]

But in general if you’re an R module you may or may not be able to get quotients in R if you know what I mean, and get Euclidean division under a certain norm?

[hidden haven]

C has non cyclic non dense subgroups

[lethal dune]

yeah R itself

[wooden ember]

Yeah you can’t perform the type of Euclidean division I’m referring to in C can you?

As a Z module

[hidden haven]

Yeah you can't

extra dimension

[wooden ember]

Even as an R module for that matter

[hidden haven]

you could by Z[i]

[wooden ember]

True

Btw totally unrelated but I’m proud to announce we’ve finally started looking at matrices in my linear algebra class lmao

After 8 weeks

[hidden haven]

😌

[chilly ocean]

"It is my experience that proofs involving matrices can be shortened by 50% if one throws the matrices out."
E. Artin

[wooden ember]

Lmao probably true

[chilly ocean]

"There is hardly any theory which is more elementary [than linear algebra], in spite of the fact that generations of professors and textbook writers have obscured its simplicity by preposterous calculations with matrices."

[wooden ember]

Lol

Honestly matrices is a breather after the dual space

It’s much nicer to look at an array that indice and star hell

[hidden haven]

matrices aren't that bad

[wooden ember]

Yeah

They make your notation more readable artin is kinda based

[chilly ocean]

i'm just semi-memeing

moldilocks is relieved

[hidden haven]

is there a way to understand upper triangular matrix without matrices

like as a linear map

in a nice way

[chilly ocean]

yea

[hidden haven]

how do

[wooden ember]

Not really to be fair it would seem quite unmotivated

[chilly ocean]

it means that the basis vector b_n 's image is a linear combination of b_1, ..., b_n

[hidden haven]

lol

[wooden ember]

“Pick two bases and orders such that phi(e_i) = sum_j=0 ^ i lambda_j f_j” or smth

Bit gross

[hidden haven]

[chilly ocean]

it kinda relates to the definition of linear independence

[hidden haven]

[lethal dune]

matrix good

[chilly ocean]

matrix just means you pick a basis

[hidden haven]

work in category of pairs of vector spaces and bases

[lethal dune]

as moldilocks said, try explaining upper triangular "transformations" without matrices

[hidden haven]

this is equivalent to category of vector spaces

[wooden ember]

You can’t cause there are none

[lethal dune]

[hidden haven]

and matrices now coincide with maps

[wooden ember]

You can be upper triangular in one but not another basis so eh

[hidden haven]

makes the change of bases formula extremely easy to remember

[lethal dune]

I still mess it up tho

[hidden haven]

you can't mess it up if you view maps as maps of pairs

[lethal dune]

tho matrix makes it easier to calculate EVs

[wooden ember]

Gross moldi

Go back to the category theory hole

[hidden haven]

[lethal dune]

say the operator f ↦f+f'

[hidden haven]

This is how I cat pilled first years in my college

[lethal dune]

[hidden haven]

without mentioning the word category

[wooden ember]

Ew

You’re a monster

[lethal dune]

I need to read the category part ig

[wooden ember]

Our Lin alg teacher does that a bit by subtly mentioning cat theory

[hidden haven]

They thanked me and said it was a really good tutorial

[lethal dune]

category word is foreign to me

[chilly ocean]

i think categories are nice it helps keep things organized

[lethal dune]

[chilly ocean]

not a huge fan

you need to change your name then

it's very misleading

[hidden haven]

not all fans are huge carla

[chilly ocean]

are they supercompact fans?

[rustic crown]

cats are nice

[fossil shuttle]

it's always risky to give a negative answer to a question like this but i would expect a property of matrices to have a good interpretation in terms of linear maps iff it's preserved under similarity

[hidden haven]

That is fair

[fossil shuttle]

however perhaps we could answer in terms of a filtration on the space

like

[hidden haven]

Diagonalisability though it also has a basis dependent description but is way simpler than the upper triangular one

[fossil shuttle]

we have subspaces $V_0\subset V_1 \subset V_2\dots V_n$ where $V_i4 is $i$-dimensional

[cloud walrus]

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

[hidden haven]

actually you can remove basis dependency by saying all generalised eigenspaces are 1 dimensional

[fossil shuttle]

and $T$ carries $V_0$ into $V_0$, $V_1$ into $V_1$, and so on

[cloud walrus]

diligentClerk

[fossil shuttle]

it should be the case that $T$ is upper triangular iff such a filtration exists

[cloud walrus]

diligentClerk

[hidden haven]

this is nice yeah

[fossil shuttle]

sorry im just spitballing here

idk if this is actually true i'm just like,

handwaving

[hidden haven]

its like composition series

[fossil shuttle]

ok, sure

[hidden haven]

of k[T] modules

[fossil shuttle]

yeah and the induced map on $V_{n+1}/V_n$ is the $n$th diagonal element

[cloud walrus]

diligentClerk

[hidden haven]

right

[fossil shuttle]

in particular, the determinant is zero iff the induced map on $V_{k+1}/V_k$ is the zero map

[hidden haven]

I am happy with that description

[cloud walrus]

diligentClerk

[fossil shuttle]

for some k

[ionic sparrow]

I'm struggling with a problem, could someone give me a hint?
"If (G:H)=m, the order of every element of G/H is a divisor of m)"
Any help is much appreciated. Thanks

[wooden ember]

The order of G/H is m

Use Lagrange

[ionic sparrow]

Oh, I see. Thanks 😄

[chilly ocean]

Hi, I am working on this problem and having some struggle.

There is a hint my professor gave, and it is that since the product HK is not necessarily a group, we would need to explain carefully why |HK/K| = |HK|/|K|

I do see that $HK/K = {hK(K):h\in H} = {hK:h\in H}$, that is, $|HK/K|$ is the number of distinct left cosets of K

[cloud walrus]

Mega Euler

[chilly ocean]

<@&286206848099549185>

[waxen hedge]

What's question (i) ?

[chilly ocean]

Show that the map is well defined

[oak grove]

can i ask about this question i feel like its dumb but i dont understand the notes

How does a polynomial represent a finite field?

[next obsidian]

they kinda give it to you there

a quadratic over F_2 is like

ax^2 + bx + c

and each of a,b,c are either 0 or 1

so it's like making 3 independent binary choices, so we see that the size is 8

So as a set let's just associate these polynomials with (a,b,c)

so when we add two polynomials together it's like doing this

(a,b,c) + (a',b',c') = (a + a', b + b', c + c')

wait omegalol how do you multiply these. You can't do it component-wise

Okay well I'm gonna just reduce them mod x^3 and see what happens

so if we took (ax^2 + bx + c)(a'x^2 + b'x + c') we get
aa'x^4 + (ba' + a'b)x^3 + (ac' + bb' + a'c)x^2 + (bc' + cb')x + cc'

So I'm going to just kill all the terms of degree > 2 so this is actually a quadratic

so we obtained
(a,b,c)(a',b',c') = (ac' + bb' + a'c, bc' + cb', cc')

So we in particular see that (0,0,1) acts as the identity here

Now we want to show that there's inverses for everything

uh, but this doesn't work lol

okay I dunno

This problem sucks

I am sorry jan Niku

[oak grove]

@next obsidian its okay, thank you 🙇‍♂️

i actually vaguely remember now saying she doesnt remember how multiplication works in class but not knowing what she meant

so its vague how exactly these represent the field since its hard to capture what the operations are doing?

she did say something like

[next obsidian]

I mean like there’s a hacky answer where like

[oak grove]

thinking of F_p^n as isomorphic to an additive group of

direct products

[next obsidian]

You grab a random ass bijection to F_8 which exist cuz both are size 8

[oak grove]

that whole thing

that works okay

but explaining multiplication is harder or something

[next obsidian]

Then you can just artificially import the field structure of F_8

[oak grove]

[next obsidian]

Via that bijection

But this is like a shit answer

[oak grove]

thats okay

i can ask her tomorrow

[next obsidian]

Cuz any size 8 set can then be cooked up to become F_8

[oak grove]

im guessing since its the last class before break ill be the only one that shows up

[next obsidian]

Lol

[oak grove]

i appreciate u tho

[next obsidian]

Why do you have a break at end of November

[oak grove]

thanksgiving

[next obsidian]

👑

[oak grove]

its a dumb reason and a dumb time for a break

[next obsidian]

Oh

I still have class Wednesday

We get Thur,Fri off

Lol

[oak grove]

do people even celebrate thanksgiving anymore

[next obsidian]

Yeh

I eat da turkey with my family

[oak grove]

😮

retro

[next obsidian]

Also America has very very few national holidays

Relative to other countries

So don’t try to get rid of Thanksgiving too lol

[oak grove]

replace it with something

like students day

[delicate bloom]

every element of F_8 satisfies T^8-T=0 so you should factor this in F_2 to get irreducible polynomials

that's how you know what to mod out by

[oak grove]

always on the second wednesday in november

[oak grove]

[delicate bloom]

$T^8-T=T(T-1)(T^3+T+1)(T^3+T^2+1)$ so you have two choices there, either of those cubics

[cloud walrus]

Merosity

[oak grove]

why only the cubics? since theyre above the given degree?

[delicate bloom]

$T^3=T+1$ or $T^3=T^2+1$ are how you handle reduction, depending on your choice

[cloud walrus]

Merosity

[delicate bloom]

there's only one finite field of every order, so they're isomorphic

[oak grove]

im not sure i follow

how do you know T^8-T=0 for all the elements in F_8 without checking

well brute force checking

oh

duh

scratch that

[delicate bloom]

ok scratching

[next obsidian]

Idk why i said that

I am not doing that actually

[delicate bloom]

lol

[oak grove]

and

so this reduction

is it just really basic algebra or am i missing something

how can you be sure these describe an element in the field

this is maybe less trivial than the last question

[delicate bloom]

well there's more than one approach to getting at understanding this stuff

based on the question you asked you would have to take a bit more of a diversion to answer this if you don't already know, so I was kind of assuming some of this had been covered previously

at the very least, the fact that everything is being represented by quadratics means you must have some irreducible cubic that you're using

[oak grove]

no the last class was the first time weve seen this kind of thing

weve seen polynomial uhh

rings before i guess

[oak grove]

? how do you mean

the diversion isnt necessary

if its really a big gap

[delicate bloom]

well you're quotienting by the polynomials

effectively x^3+x+1=0 let's say

if it weren't a cubic and it were some quartic let's say, then you'd end up with nonzero cubic terms

contradicting the statement that you are representing things with purely quadratics

sounds like you're being asked a question that's further ahead than your class has come to yet

[oak grove]

quotienting what by the polynomials?

im not sure i understand what it means to quotient something by polynomials

[delicate bloom]

polynomials with F_2 coefficients

that ring

[oak grove]

were quotienting all of those by some cubic?

[delicate bloom]

yup

[next obsidian]

Oh lmfao so this is almost what I did

[oak grove]

i dont really understand what that would even do

[delicate bloom]

by sending x^3+x+1=0 we're saying x^3=x+1 is a way to simplify cubic and higher polynomials down

[next obsidian]

I see, you’re associating F_2[x]/(x^3 + x + 1) with degree 2 polys

[next obsidian]

It’s kind of what I tried to do here

[delicate bloom]

also maybe I'm overassuming, -1=1 here btw

[next obsidian]

Except I was quotientinf by x^3 = 0

Which doesn’t give us F_8

[oak grove]

idk if this is what were supposed to do

[next obsidian]

I think it is, this makes sense

[delicate bloom]

sounds more like they're saying it as a passing remark, not an exercise

[oak grove]

since im not sure what youre talking about and i think im one of the more on-top-of-it people in the class

well we have a homework that relates to this

[next obsidian]

You’re reinterpreting the definition of F_8 as F_2[x]/(x^3 + x + 1)

As the right being degree 2 polynomials with a bit of a fucked up multiplication

I think Mero is doing a better job explaining this anyway

So I will let Mero do their thing

[delicate bloom]

haha no it's fine, learning through teaching is good

redeeming yourself for trying to quotient by x^3 🤣

[next obsidian]

:(

[oak grove]

oh i dont understand either explanation so whatever works

[next obsidian]

So like do you know how to construct F_8?

[oak grove]

idk she said you find the prime factorization of the number in the subscript

[delicate bloom]

sounds like you have some missing material to cover, what book and where are you in the book

[oak grove]

we dont have a book

[next obsidian]

[delicate bloom]

oops

lol

[oak grove]

its just teachers notes

[next obsidian]

Okay well

[delicate bloom]

can you send those or

[oak grove]

[delicate bloom]

it'd help to see what you're expected to make this from

[oak grove]

like heres what im wanting to cite

[delicate bloom]

cause sounds like you're missing a lot of prior stuff that's necessary but maybe you just don't realize how stuff you know applies

[next obsidian]

Did they define F_p^n tho?

Like you kind of have to make it via quotienting F_p[x] by an irreducible poly of degree n

[oak grove]

im not sure how quotienting would work here

[delicate bloom]

they define the multiplicative group here as being Z_{p^n-1} so in our case Z_7

kind of tricky cause you need to then pick some generator polynomial and show that it has order 7 I guess, but even then you'll still need to find a way to quotient

[next obsidian]

I don’t see how you can define F_p^n just via this tho

[delicate bloom]

it's enough to define it up to isomorphism

[next obsidian]

Can you guarantee the addition and multiplication distribute?

[delicate bloom]

because it turns out there's only one field of every power of a prime order up to isomorphism

yeah like you can write up a table and just guess stuff until you get something consistent

[next obsidian]

For some specific p and n sure

More so I just think this isn’t the definition they’re given

[delicate bloom]

yeah which is maybe what they intend for this question

[next obsidian]

I feel like this is moreso a result

And they defined F_p^n a different way

[delicate bloom]

yeah true, they're calling this a theorem

[next obsidian]

I think the easiest way to do this

Is to assume that finite fields are unique (this is kinda what this theorem says)

Then just note that F_2[x]/(x^3 + x + 1) is a field because the polynomial we quotiented by is irreducible

So it’s a maximal ideal

Then we can take F_8 to be this, then you can just do the magic of replacing x^3 with x + 1 enough until you’ve written every element as a quadratic

But overall this seems really… wonky. I don’t like writing F_8 unless you already know finite fields are unique and to do that you usually just show they’re all isomorphic to some quotient of F_p[x] lol

[oak grove]

im very confused 😄

[delicate bloom]

weird

start reading from day 34

your teacher walks through that example in detail

[oak grove]

day 34 is tomorrow

oh

[delicate bloom]

I realize that

[oak grove]

that was nice of her

[delicate bloom]

I think your teacher is expecting you to time travel to tomorrow to do the exercise today

[oak grove]

can i ask why

why the definition should be reasonable

[delicate bloom]

I think you're probably supposed to get stuck so that when you see the stuff tomorrow you'll be ready

[oak grove]

i think shes leaning into vector intuition that most of the class has but i do not maybe

[delicate bloom]

sure, which definition

[oak grove]

like why the integers

just one to one correspondence?

[delicate bloom]

what do you mean integers

[oak grove]

the product of the integers mod p

[delicate bloom]

it's kind of hard to answer "why" in this context, it's just how it is

if I say "write down a cayley table for a field with finitely many elements" you'd be forced to do this

[oak grove]

[delicate bloom]

primes are not divisible by numbers less than them, so when you mod them out you're not getting any zero divisors

[oak grove]

okay

that gets you the isomorphism

build it into a box rather than a line

chunk out rows or columns

[delicate bloom]

I guess one common misconception is people think a field with p^n elements is just mod p^n

but that can't be true for n>1 since p^n=0 but individually p is not 0, so it breaks field axioms

that sort of answers why you might take this polynomial approach

[oak grove]

broken by this?

[delicate bloom]

yeah, I guess you could see it that way

I was thinking specifically it has no inverse

$p^n=0$ so if $p^{-1}$ existed you'd have $p^{n-1}=0$ when you multiply both sides by it, but that's a contradiction

[cloud walrus]

Merosity

[delicate bloom]

whatever works to show it, I don't think there's one unique or best way to show it breaks lol

[oak grove]

so it has to be mod some prime

[delicate bloom]

I wouldn't draw that conclusion directly from that

[oak grove]

well at least that it cant be mod the full p^n

[delicate bloom]

but at least once you know F_p are fields you can at least think other fields will be extensions of this and so F_p[a] will have to be what it looks like, so if you look at it as a polynomials in a of degree n, then there are p^n possible elements to make

[oak grove]

but in the problem given we arent extending anything right

unless we are?

we shouldnt be

the definition is saying like

we can take a finite field with 8 elements, and then use some kind of mapping that reduces Z2 + Z2 + Z2 from 24 to 8

im sorry i know im slow i appreciate your attempt lol

[delicate bloom]

yeah we're extending F_2 to get F_8

we're adjoining a root of x^3+x+1 to get the splitting field

[oak grove]

which the problem is saying

think of a polynomial a+bx+cx^2 with a or b or c either 0 1 2

and the problem is we have too many of these so we assume we can find some factor to mod these out by to get 8

right?

[delicate bloom]

yeah, like your teacher is kinda skimming the details on this to assume you could answer this with what you know so far

[oak grove]

bold of her to assume

[delicate bloom]

I mean we can prove x^3+x+1 is irreducible if you want to right now

[oak grove]

i know it is 😄

RZT

[delicate bloom]

how do you know? lol

[oak grove]

oh wait

i guess i only know its not rational then

i dont know

i mean im not sure what significance irreducibility would even be here

i get the quotienting because thats gonna make the orders match

[next obsidian]

Quotienting a poly ring over a field by an irreducible polynomial gets you a field

[oak grove]

4 days ago

im just gonna ask the teacher tomorrow

i appreciate yall

[delicate bloom]

sounds like your teacher is planning on discussing it tomorrow anyways lol

I think you've done enough to prepare for what you need, sleep well @oak grove

[chilly ocean]

an algebraic structure is basically an n-tuple right?

an n-tuple that has a non-empty set and a bunch of operations defined on that set that obey a bunch of axioms?

[lethal dune]

[scarlet estuary]

not "basically", literally

this usually isnt the most useful way to think about them

but its the formal way

and (ℤ, +) or whatever is a very common way to specify them in cases where the set alone isnt enough

though caveat: some algebraic structures come with info about multiple sets

[chilly ocean]

(Z, +) would be a abelian group?

[scarlet estuary]

like vector spaces have a set of vectors and a field of scalars

and that field itself carries info about a set

[chilly ocean]

ahh right

[scarlet estuary]

add a norm and you also need info about ℝ, if you didnt have it already

[chilly ocean]

so how would a vector space be represented using an n-tuple

[scarlet estuary]

yes, the abelian group of integers under standard addition

[chilly ocean]

(F, +, *)? for any field F?

[scarlet estuary]

something like $(V, (F, +_F, \cdot_F), +V, \cdot{V})$

[cloud walrus]

Namington

[scarlet estuary]

$+_F$ and $\cdot_F$ are your field operations, $+_V$ is vector-vector addition, $\cdot_V$ is scalar-vector products

[chilly ocean]

oh so the subscripts indicate that that's scalar addition and vector addition

[cloud walrus]

Namington

[scarlet estuary]

though perhaps this notation is a bit overengineered

you could have one $\cdot$ operator that accepts scalar-scalar and scalar-vector products

[cloud walrus]

Namington

[scarlet estuary]

is domain and codomain would look weird though

[chilly ocean]

right

[scarlet estuary]

$(F \times V) \cup (F \times F) \to V \cup F$

[cloud walrus]

Namington

[scarlet estuary]

¯_(ツ)_/¯

[chilly ocean]

lol it does look weird but makes sense

[scarlet estuary]

this is just a theoretical concern in any case

youll rarely see vector spaces defined like this

[chilly ocean]

even in pure maths?

[scarlet estuary]

its typically not useful to refer to everything by its hyperformal set theoretic definition

we care much more about how it behaves than how its constructed

[chilly ocean]

normally they just use F^n right?

for a n dimensional vector space over the field F

[scarlet estuary]

F^n is the most common type of vector space but certainly not the only ones you see

[chilly ocean]

okay thanks for all that info

[chilly ocean]

if you defined addition and mutiplication like that would the algebraic structure be (V, F, *, +)?

[scarlet estuary]

your field still has an additive structure and its not really compatible with the vector additive structure whatsoever

so its very weird to bundle them into 1 operation

like you COULD

but its strange

[chilly ocean]

oh lol alright

[scarlet estuary]

bundling the different multiplications together makes sense as (ab)v = a(bv)

for scalars a, b and vector v

and similarly a(uv) = (au)v

[chilly ocean]

so how would a vector space be defined in abstract algebra?

[scarlet estuary]

but we dont really have (a+b) + u = a + (b + u) because... you cant add scalars to vectors lmao

[chilly ocean]

in linear algebra most people use F^n

[chilly ocean]

geometric algebra go brrrr

[scarlet estuary]

the standard way, a vector space is composed of two sets, a set of vectors and a set of scalars, satisfying 10ish axioms

my LA course always worked with V over an arbitrary F

where possible ofc

[chilly ocean]

given finite field F

and p in F^n

make a polynomial f in F[x1,…,xn] with f(p)=1 and f(q)=0 for q in F^n\p

[hidden haven]

x-a is zero at a and non zero everywhere else for any a in F

Try multiplying such factors

Oh you might need slightly more complicated factors because this makes things zero on hyperplanes parallel to the axes

You want to make all hyperplanes not through p 0, and there are finitely many such

[lethal dune]

can't it be done by doing something like (x^|F|^n-x)/(x-p)?

[hidden haven]

in one variable?

oh you are viewing F^n as a finite field itself

I think you would have to prove that x^n in this field structure may be expressed as a polynomial in the original field structure

which seems kinda legit

[plucky flicker]

Hi! Can someone help me with this problem?

[robust pollen]

Actually, what field are you working over?

[plucky flicker]

C