groups-rings-fields

[next obsidian]

tyty <3<3

[snow flint]

why does it read wrong

[rustic crown]

interesting discussion... shouldn't have slept 😶

[celest brook]

matrices die

[untold sapphire]

Matrices are great

[chilly ocean]

matrices are reduntant when you can do everything with tuples

[rustic crown]

i act on matrices... if matrices are gone, then there is no point of my existence 😛

[hidden haven]

You can act on linear transformations

[rustic crown]

[untold sapphire]

I love matrices because you can write them down

I'm a simple man

🤔

[rustic crown]

3d matrices would be so hard to draw

[untold sapphire]

by tuples do you just mean writing down where the basis vectors go?

because that shit is annoying

[rustic crown]

yea just write down all the n x m entries anyway you want

[chilly ocean]

yo

[rustic crown]

yo

[untold sapphire]

seriously I didn't realise how much my lack of drawing ability was going to hold me back in higher level mathematics

I should take a class

[rustic crown]

i was scared when my book started drawing commutative diagrams in 3d

[untold sapphire]

3d commutative diagrams are too much

[kind temple]

holy why

[chilly ocean]

@chilly ocean yo u good?

[rustic crown]

this notation is much better 😛

[untold sapphire]

I prefer stuff like this

so pleasing

[kind temple]

yes. so simple

[chilly ocean]

oh so its like a spiral or sth?

[rustic crown]

yep

[untold sapphire]

it's a staircase diagram

for monomial ideals

❤️

[kind temple]

M.C. Escher has entered the chat

finally

[rustic crown]

looks similar to the young diagrams for partitions

how do you define this?

[untold sapphire]

it's showing you which exponent vectors are in your monomial ideal

[rustic crown]

oh makes sense

[rustic crown]

so how will you write (xy, yz, zx)?

[untold sapphire]

this one is $J= <x^4, y^4, z^4, x^3y^2z, xy^3z^2, x^2yz^3>$ ( I think)

[rustic crown]

or just write the ideal for the above image

[cloud walrus]

bdobba

[rustic crown]

i see

what do we do with the white dots?

[untold sapphire]

ah, this is the cool part - if you connect the white dots and form a graph, you can get a free minimal resolution of the ideal

[rustic crown]

😮

[untold sapphire]

sorry, grey dots even

the white dots are where the edges of the graph must pass through

Combinatorial Commutative Algebra is dope 👍

[rustic crown]

looks pretty fun

(i'll go, i have an exam in 10 minutes)

[untold sapphire]

k later

[unique juniper]

$\mathbb{Q}(\sqrt{a} + \sqrt{b}) = \mathbb{Q}(\sqrt{a},\sqrt{b})$

[cloud walrus]

Yes

[unique juniper]

my struggles are at

showing \sqrt{a},\sqrt{b} is in \mathbb{Q}(\sqrt{a} + \sqrt{b} )

a,b are natural

[chilly ocean]

hm

maybe:

$\bQ(\sqrt{a}+\sqrt{b})$ must be a degree 1, 2, or 4 extension over $\bQ$. you can probably show it is not a degree 1 extension in non-trivial cases, and degree 4 extension would prove that $\bQ(\sqrt{a}+\sqrt{b})=\bQ(\sqrt{a},\sqrt{b})$, so we just have to disprove that degree 2 extension case.

assume that $\bQ(\sqrt{a}+\sqrt{b})$ is degree 2. then $\bQ(\sqrt{a}+\sqrt{b})=\bQ(\sqrt{c})$ for some $c$. then $\sqrt{c} \in \bQ(\sqrt{a}+\sqrt{b})$, and this boils down to the linear (in)dependence of $1, \sqrt{a}, \sqrt{b}, \sqrt{ab}$. so it is a matter of proving that $\sqrt{b}\notin \bQ(\sqrt{a})$, i think

[cloud walrus]

Ultramuck (8da)

[unique juniper]

i see

[rustic crown]

another way

(sqrt(a) + sqrt(b))(sqrt(a) - sqrt(b)) = a-b in Q

sqrt(a) - sqrt(b) in Q(sqrt(a) + sqrt(b))

[final pasture]

what's the book ?

[sinful mirage]

how does the statement about mutliplicities follow from Schur's?

[sour plume]

How is Schur's lemma formulated in your book?

[sinful mirage]

let v_1,rho_1 v_2,rho_2 be irreducible reps. phi \in hom_{g}(V_1,V_2) is either invertible or zero

[sour plume]

I guess it becomes easier once you have the refinement that in the case when $V_1 = V_2$, you can even say that all $\phi \in \text{Hom}_G(V,V)$ are a multiple of the identity; so not just any invertible map, but a very specific one

[cloud walrus]

Lartomato

[sinful mirage]

right

they are a multiple of identity,we proved that

[sour plume]

Once we know this: If, for example, V is irreducible and W is not necessarily, then we need to know one more thing: That we can write W as the direct sum of irreducible representations

[sinful mirage]

right,we also proved that

and also that this decomposition is unique

I don't see though,why W would be $\oplus_{i} V^{n}$

[cloud walrus]

ProphetX

[sinful mirage]

why can't there be more irreps appearing in that

[sour plume]

There can be a bunch of irreps appearing, but they don't contribute because of Schur's Lemma: If $U$ is an irrep which is not isomorphic to the irrep $V$, then $\text{Hom}_G(V,U) = 0$

[cloud walrus]

Lartomato

[sour plume]

Otherwise, if there was some nonzero element in this space, by Schur's lemma it must be invertible, and hence it is a G-equivariant isomorphism of V and U -- but we assumed that they're not isomorphic

[sinful mirage]

from Hom_{g}(V,U) how you conclude that the HOM(V,W)^g is the multiplicity of V in W?

ohhhh danggg

right

[sour plume]

Sounds like you got it! naisu

[sinful mirage]

do you have 2 more mins?

[sour plume]

Ye sure

[sinful mirage]

my prof did it a bit other way and I want to check whether something is a typo or I'm missing something

I agree with all the statements here,but I'm confused why she looks at dim hom_{g}(V_i,W) instead of dim hom_{g}(V,V_i)

since she decomposed W in sum_i V_i

or well sorry,I think I got it

it's the same argument you said

[sour plume]

Yah I think it's the same thing 😄

[sinful mirage]

it's V_i ^{n_I}

but the V_i-s which are not V do not contribute

[sour plume]

Needed a moment to fight through the handwriting

Yeah that's precisely it

[sinful mirage]

so this is exactly mutliplicity

right

thanks

[rustic crown]

Aluffi Algebra Chapter 0

(I like it actually )

[chilly ocean]

i read abit of it and seemed nice

[untold sapphire]

I liked Aluffi

apart from the later parts on homological algebra which are just a mess imo

[rustic crown]

Yea i still have to read from homotopy

[untold sapphire]

I learned a lot from that book though, it's what finally made me 10% understand category theory

[rustic crown]

I really liked the Field Theory in that book (but yea I haven't read that in other books, so shouldn't say this?)

[old lava]

aluffi's order of topics is super cringe

[untold sapphire]

which ones?

[old lava]

all of them?

[untold sapphire]

Yeah I think he does Galois Theory really well

[old lava]

there's no good reason to split group, ring theory and lin alg into non-consecutive chapters

[untold sapphire]

I mean that's literally exactly what Lang does

[old lava]

and I don't like Lang either?

[untold sapphire]

fair enough

what's your favourite?

[old lava]

d&f is decent, but has some glaring flaws

mainly that it's exposition is terribly boring for most, and some of the sections are kinda not that useful

[untold sapphire]

yeah it's not bad, the main problem is that if you want to take algebra further I don't think (it's been a while) it uses any category theory

[old lava]

it does use a bunch of category theory

especially in the modules section (specifically when dealing with the tensor stuff, and projective/injective/flat module stuff)

and a bit in the last part of groups

[untold sapphire]

It has been a while since I've looked at it

[old lava]

it definitely doesn't try to frame things in a category theoretic perspective

if it can be done easily without it (for basic algebra)

[untold sapphire]

yeah I know just saying that Aluffi is better preparation for what comes after in Algebra since it tries to shoehorn category theory into everything

but it depends on what you want I guess

some of the stuff in Aluffi does seem kind of forced

[old lava]

ya, d&f takes the opposite approach I suppose

where it avoids category theory unless it's important to the understanding of the topic

like when discussing universal property stuff, I suppose

[chilly ocean]

Guys, i have a question

[rustic crown]

[chilly ocean]

Toeplitz matrixs

Teacher told us we need fourier coeficients of a function

https://gyazo.com/a30bfdc59879fec6e04279ceb29e324b

and this is how the matrix is built

https://gyazo.com/6e80594639ec9b25d0a685d11c4a2432

But now, the problem is the other way

having the Toplitz matrix, how can i guess the symbol?

[old lava]

what do you mean "guess the symbol"

[chilly ocean]

mmm, isnt the symbol the function that generates the matrix?

[old lava]

I'm not sure, I haven't heard of that notation

but the original function can just be reconstructed from he coefficients c_i

with the fourier inverse formula

[chilly ocean]

sure

how?

[old lava]

(also I think this might not be the channel for this)

[chilly ocean]

this is to build the matrix, but once i have the matrix, how can i get the function?

[chilly ocean]

which channel should we move to?

[old lava]

uh you don't even need the matrix, I don't know what that's for

you just do the infinite sum

that's about it

[chilly ocean]

from -inf to inf

this, right?

[rustic crown]

t in T?

not e^(i n theta) ?

[chilly ocean]

t is e^i*theta

[old lava]

ya

[rustic crown]

oh okie

[old lava]

I'm not sure what t \in T means

it should just be e^{i theta}

[chilly ocean]

yeah, but teacher writes e^it as t

[old lava]

also for next time, this seems like a calculus or analysis question

[chilly ocean]

okey

[old lava]

so go to a channel for that probably

[chilly ocean]

do i need to copy everything again? uwu

#advanced-analysis ?

[carmine fossil]

You probably need to use the fact that int_{0}^{1}2πt dt=0

[chilly ocean]

T is sometimes used to mean unit circle

should i stay here or move to analysis?

[shut halo]

Hey I am trying to do this question about the lie algebra of the real symplectic group

a,b,c,d are nxn matrices. But I'm a bit clueless about how to compute the dimension and show that it is equal to 2n^2 + n

[hot lake]

did you do (ii) ?

[chilly ocean]

any hints how can I find the smallest normal field extension L /Q such that Q(i + 2^(1/3)) is in L?

The degree of that element is 6 and it would be pretty hard to find all the roots of the minimal polynomial, so surely there's a better approach?

I was trying to do something of the sort saying that feild is Q(i, 2^(1/3))

but idk

[rustic crown]

if you already showed that degree is 6, then this means that Q(i + cbrt(2)) = Q(i, cbrt(2))
since L/Q is supposed to be normal, x^3-2 and x^2+1 irreducible polynomials over Q definitely have to split over L
so what if we look at the splitting field of (x^3-2)(x^2+1)?

[chilly ocean]

I was thinking about it, but I was not sure why would that work, since that polynomial is deg 5, not 6

so I thought maybe I would miss something, but idk

[rustic crown]

how does the degree matter?

the splitting field of a degree n polynomial can be a large as a degree n! extension

[chilly ocean]

ye true, hmm I dont know why Im confused a bit right now

[rustic crown]

The splitting field will be Q(cbrt(2), i, omega) = Q(cbrt(2), i, sqrt(3))

[chilly ocean]

okay yeah I see the =

[rustic crown]

Q(cbrt(2), omega) is the splitting field of x^3-2 right

[chilly ocean]

yes

I guess Im not sure why we need to split (x^3-2)(x^2+1)

[rustic crown]

yep, so adding these 3 elements will add all the roots of (x^3-2)(x^2+1) and the theorem that splitting fields are normal guarantees that this will be a normal extension

L will split (x^3-2) AND (x^2+1)

which is same as saying it splits the product

[chilly ocean]

Oh okay yeah, since minimal polynomial for like uhh cbrt 2 overQ(i) is the same as just over Q

[rustic crown]

i don't see how this helps in the proof, except maybe showing that Q(i+cbrt(2)) = Q(i, cbrt(2))

[chilly ocean]

I kinda did show it

I guess Im not sure why L doesnt have to split a bigger polynomial

Ill think about it tho

[rustic crown]

I'm just saying if (x^3-2) and (x^2+1) split over L, then so does (x^3-2)(x^2+1). Conversely, if you look at the splitting field of the product, it will give you a normal extension where both (x^3-2) and (x^2+1) split.

And we need these x^3-2 and x^2+1 to split because they are the minimal polynomial of cbrt(2) and i, and since they have a root in Q(cbrt(2), i) they must split completely in any normal extension that contains Q(cbrt(2), i)

[chilly ocean]

oh okay I see

thanks det

[rustic crown]

If your doubt is why i'm not looking at the splitting field of the minimal polynomial of cbrt(2) + i over Q, then the reason is, its more complicated to look at, and finding its splitting field will just be harder.

in the end you'll still end up with Q(cbrt(2), sqrt(3), i)

[chilly ocean]

https://tenor.com/view/kon-fun-things-are-fun-anime-cute-yui-gif-16787421

fun

things

[rustic crown]

are

[chilly ocean]

ruined chain

[junior moth]

.

[chilly ocean]

delete msgs

det you have the power

[rustic crown]

I don't think I do

[chilly ocean]

Ah, I was wondering what is the smart way to do it 🙂

[shut halo]

Yes

[final pasture]

https://tenor.com/view/anchorman-the-legend-of-ron-burgundy-comedy-vince-vaughn-i-hate-you-gif-3520833

[unique juniper]

https://tenor.com/view/yui-hirasawa-hirasawa-yui-yui-hirasawa-kon-gif-6026019

[hot lake]

then the dimension should be easy

it's the dimension of the space of 2n * 2n matrices with the constraints that you found

so b and c have to be symmetric, and d is determined by a

[shut halo]

Yeah I think I got it. 4n^2 - n(n-1) -n^2

[hot lake]

the dimension of the cartan subalgebra isn't as obvious i guess

[shut halo]

Yeah I thought it would be the set of diagonal matrices with those entries

[hot lake]

yeah it's the subspace of matrices that are diagonal

[shut halo]

@hot lake so that would be n right because we can say b = c = 0.

Then a completely determines d

Since a is an nxn diagonal matrix, we can find a maximum of n of those

However, I don't get the next part of the question which says the following:

I found the new constraints to be $a=-a^\dagger;d=-d^\dagger;\ c^\dagger = -b$

[cloud walrus]

snypehype

[shut halo]

I know that dim(Sp(2N,C)) = 8n^2 but I'm a bit unsure on how many constraints there are

[next obsidian]

I'd like to know how to prove (I sure hope it's true since it's a hint in Harthsorne) the following result:
Let A < B be integral domains, such that the transcendence degree of Frac(B) over Frac(A) is finite, equal to n.
Then there are elements t_1,...,t_n in B such that their image in Frac(B) is a transcendence basis over Frac(A)

So, clearly we get a transcendence basis of the form {t_1/s_1,...,t_n/s_n}, so ideally I'd like to clear denominators or something? If I let f = the product of all s_i, then I can get this by just localizing at f, but then I have to look at A < B_f, and I'm not sure this is good enough for my purposes?

I think maybe {ft_1/s_1,..., ft_n/s_n} might still form a transcendence basis?? But honestly I really don't know, I am kind of doubtful about it

Oh, I think it might suffice to do it one variable at a time.

Actually, this is easy. You have B = A[x_1,...,x_n] by finite generatoin, then Frac(B) = Frac(A)(x_1,...,x_n), and it's just field theory you can get a transcendence basis from a subset of the generators

[compact needle]

I don't buy that argument. Finiteness of the transcendence degree of the fraction fields does not imply finite generation of B over A. Imagine A is Z and B is the ring of all algebraic integers adjoin some variables.

[next obsidian]

Oops, I forgot to say that B was fg over A lol

Really both were f.g. over a field k to begin with, but that implies B is f.g. over A 😳

[final pasture]

[next obsidian]

Sometimes I write down what I think is the general statement cuz I don't want to write down the entire situation, then forget a hypothesis because I either think it's superfluous or just forget. Then I realize how to solve it in my own situation

[compact needle]

That's understandable. You just have to watch out for people that look into statements that seem too good to be true 😆

[next obsidian]

Yeah... xD

[compact needle]

For what it's worth, having thought about it more, in total generality, we know that {f,ft_1/s_1,..., ft_n/s_n} is a transcendence set (in the obvious sense) with n+1 elements in B, so some subset of n elements is a transcendence basis.

[charred pewter]

How can I show a group (like the complete graph K_3) acts freely?

[next obsidian]

acts freely on what?

[charred pewter]

Well I'm trying to use Sabidussi's theorem to determine whether or not K_3 is a Cayley graph. The theorem says that Aut(K_3) has to contain a subgroup that acts simply transitive on K_3.

[next obsidian]

O_O

well... uh... idk in that case haha. I assume the action here is just the obvious one where like, given alpha in Aut(K_3) it acts on K_3 by just applying alpha

But to find some subgroup which acts simply transitive is kinda yikes I guess :/, I don't have any suggestions cuz this is pretty far outta the realm of stuff I know much about

[untold sapphire]

does anyone know about Hopf Algebras?

Stuck on something kind of elementary, trying to prove that this is a coalgebra:

[cloud walrus]

bdobba

bdobba

[untold sapphire]

ah wait I think I've got it

[cloud walrus]

bdobba

bdobba

[untold sapphire]

this stuff makes me feel really dumb

[unique juniper]

"Every irreducible polynomial over a characteristic 0 field is seperable"

im struggling to imagine why its not true if it wasnt 0

because the derivate could be 0 and that would mean the gcd doesnt exist?

[wind steeple]

a polynomial P is separable iif gcd(P,P') = 1

[unique juniper]

yeah

[wind steeple]

yeah if P' = 0 the gcd is 0

[unique juniper]

oh

[wind steeple]

oh wait

the gcd is P I mean

[unique juniper]

P ?

[wind steeple]

P divides P and P divides 0

then the gcd is P

gcd(P,0) = P

[unique juniper]

thanks though

[weary terrace]

Hi, I'm currently reading this article which refers to the term "k-eigenvector", what does it mean?
(if needed, I'll add more context)

[unique juniper]

They used the splitting field to show this, But the next theorem is that any irreducible polynomails over a field of char 0 is seperable

i dont get this, if a polynomial could be written like this, then it cant be irreducible?

[carmine fossil]

I think it means (T-cI)^kv=0 but (T-cI)^{k-1}v!=0

Ok,Ignore that

It's probably not that

[weary terrace]

Google says nothing :\

[carmine fossil]

https://stats.stackexchange.com/a/157088

Context would be useful

[compact needle]

In principle, it is possible for the minimal polynomial to have repeated roots. In characteristic zero, this does not actually happen, but in characteristic p it does. In particular, the field $F_p(x)$ does not have a p-th root of x, so the polynomial $y^p-x$ has no solutions. But in characteristic p, something has at most one p-th root, so $y^p-x = (y-x^{\frac{1}{p}})^p$ over the splitting field

[cloud walrus]

Turgul

[unique juniper]

OH

thanks

i see

[weary terrace]

A(Q) is the adjacency matrix of Q. At first I thought A(Q) has k eigenvalues, but I'm not sure it's the case.
The paper is discussing McKay quivers - graphs in which each vertex represents an irreducible representation of a group G (representation theory).
Hope it helps..

The link u shared implies that a k-eigenvector is the eigenvector related to the k-th largest/smallest eigenvalue, did I get it right?

[carmine fossil]

Yes

[weary terrace]

thx 🙂

[weary terrace]

A prof. in the department thinks it's just an eigenvector with k as its eigenvalue

[carmine fossil]

That could work too

[compact needle]

I don't know if I will be able to shed clarity, but knowing what paper you're trying to read might be helpful. I would be somewhat surprised by the k-th largest weight interpretation in this context

[weary terrace]

That's the paper

The full part I'm talking about is this

[weary terrace]

page 3

[compact needle]

I concur with the professor you talked to. It's the eigenvalue

I get that by reading the proof of Corollary 2.4, where they say that the column of the character table given by the identity of G is (dim-\rho) weighted, where in Proposition 2.3, they say that the eigenvalue for a column of the character table is an eigenvector with eigenvalue "the value of \chi_\rho in that column" which I take to mean the value of \chi_rho on that conjugacy class. Going back to the proof of 2.4, a character evaluated at the identity is the trace of the identity matrix of size dimension of the representation, i.e. \chi_\rho(e)=dimension of \rho

So at least when they prove the k-weighted-ness of something, they prove it's the eigenvalue of an eigenvector with all positive entries

[weary terrace]

yeah ,exactly where I'm at.. It just hit me 🙂
Thanks for the patience!

[compact needle]

No worries. It definitely wouldn't have killed them to be a little more explicit in their definition. But when I get confused about how someone defines something, I try to see how they use it to see if that sheds some light. Good general paper reading strategy 🙂

[weary terrace]

Agreed, it's just a bit frustrating for me to keep on reading while something is not clear to me

[compact needle]

For sure

Good luck making some more progress!

[weary terrace]

Thanks again ❤️

[thorn delta]

I asked this question a while back, but didn't get any responses. Still curious about it and haven't figured it out. Let $L,M$ be intermediate fields of a field extension $F/K$ such that $L/K$ is finite Galois. It is possible to show that $LM/M$ is finite Galois as well. I want to show that $Gal(LM/M) \simeq Gal(L/L\cap M)$. There is a homomorphism $$r : Gal(LM/M) \to Gal(L/L\cap M)$$ given by restricting automorphisms in $Gal(LM/M)$ to $L$. This map is injective, but I have been struggling to show that it is surjective, i.e. automorphisms in $Gal(L/L\cap M)$ can be extended to automorphisms in $Gal(LM/M)$. I emailed my professor about this, and he said this: \ \
"Let $G$ be the image of this map in $Gal(L/L\cap M)$ and consider an element x
of its fixed field. This lies in the fixed field of $Gal(LM/M),$ and is
thus in $L\cap M$."

[cloud walrus]

kxrider

[thorn delta]

anyone have an idea of how what my professor said helps?

[next obsidian]

So I think the idea is that you want to show anything in the fixed field of G is in L\cap M. This tells you that fixed field of G is contained in the fixed field of Gal(L/L\cap M), and the other is automatic so that the fixed fields are the same. Then G = Gal(L/L\cap M). So if you take x in the fixed field of G, it is an element of L which is fixed by the restriction of every alpha in Gal(LM/M) (because that's what it means to be in the fixed field of G, G is just the restriction of every alpha in Gal(LM/M)). This implies x is in the fixed field of Gal(LM/M), thus in M. So x is in L\cap M

@thorn delta

does that make sense?

[thorn delta]

i see, so G and Gal(L/L \cap M) having the same fixed field is enough to show that G = Gal(L/L \cap M) by the Galois correspondence?

[next obsidian]

Yup

this is the point of Galois theory so to speak haha

also to tell intermediary fields are the same by looking at the automorphisms fixing them :3

[thorn delta]

ah ye true. Maybe more precisely its the point of finite Galois extensions (this way the correspondence is bijective)

[next obsidian]

yeah

[quaint tree]

Is Stewart a good book to learn Galois theory from, if all I know of algebra so far is from Pinter (i.e. I haven't read Dummit and Foote yet)?

[median tree]

How is that equality derived?

[delicate bloom]

take f(x) and multiply it by x and you get nearly all the same terms

[old lava]

do the multiplication

[delicate bloom]

xf(x) = x^p + f(x) -1

[old lava]

you can also just directly do (x-1)f(x)

and see that all but 2 terms cancel out

[obsidian sleet]

finite jeometric series

[median tree]

Oh yeah, of course, that too

[old lava]

that too, that's how I'd think of it

[median tree]

I worked out the other way, too

[old lava]

but also (x^p - 1)/(x-1) = x^{p-1} + .... + x + 1 is basically just a fact in my brain

at this point

I've encountered it a ton of times

[median tree]

noted! Thanks for the clarification

[delicate bloom]

you can take it a step further to get another handy relation, let x=y/z and you can work out what y^p-z^p = (y-z)(y^{p-1} + y^{p-2}z + ... + z^{p-1})

[cinder bone]

what's an example of a field that the set of all integers generalizes*

the rationals?

[steady axle]

Wdym

[cinder bone]

so like a ring generalizes* fields right

and the set of all integers is a ring

sorry not generate

generalize lol

[delicate bloom]

not exactly, but there's a way you can make that happen sort of like you're thinking maybe

if you have an integral domain, which is a kind of ring with extra stuff, then you can form its field of fractions

the field of fractions of the integers is the rationals

[cinder bone]

ah ok

[steady axle]

If u want a ring that contains field then look at k[x] for any field k

[cinder bone]

so what would be a field $\mathbb{Z}$ alone generalizes

[cloud walrus]

therealjoshua

[steady axle]

Idk what exactly r u asking

[delicate bloom]

what you're asking for is vague, what I said is probably closest to what you're probably thinking of in your head of what you want

every integral domain has a unique field of fractions you can make from it

[cinder bone]

ok, yeah I think that's probably what I'm thinking

[steady axle]

I have a linear alg qn. Consider $n+1$ linear functionals $f_0,f_1,\cdots fn$ on an arbitrary vector space V. I want to prove that if $\cap_{i=1}^{n} ker(f_i) \subset ker(f_0)$ , $f_0$ can be written as linear combination of rest of the linear functionals.

[cloud walrus]

bert

[final pasture]

is this true ? 🤔

take f0 = Id_R² and f1(0,1) = (1,0), f(1,0) = (0,1), Ker(f0) = Ker(f1), but how do you write f0 as a linear combination of f1 ? @steady axle 🤔

[steady axle]

Functionals r linear maps from V to F

[final pasture]

is F the field on which V is defined ?

[steady axle]

yes 🙂

[gentle pendant]

I am sure there are slicker ways, but one way is to define $U=\textrm{span}(f_j)$, $W=\textrm{span}(f_0)$, which are subspaces of $V^* $. Then from the canonical isomorphism $V\cong V^{**}$, we have $$U^\perp=\bigcap \ker(f_j)\subseteq \ker(f) \subseteq W^\perp$$

Then as taking perps reverses inclusion, and again identifying spaces with their double dual, we have $U=U^{\perp\perp}\supseteq W^{\perp\perp}=W$ as desired.

(In the above, for $A\subseteq V$, we define $A^\perp:={\phi\in V^*: \phi|_A=0}$.)

[cloud walrus]

gomez

[chilly ocean]

what if V is infinite-dimensional

[gentle pendant]

https://tenor.com/view/lazy-cat-stairs-gif-13378074

[compact needle]

As far as I can tell, Pinter's book already has some Galois theory. Stewart's book looks like it was intended for someone with less background than all of Pinter, so you should be just fine.

[gentle pendant]

yeah stewart is quite gentle

[quaint tree]

Pinter has Galois theory in like the last 3 chapters; given how big a name Galois is, I very much expect there to be more Galois theory than just the minimum required to prove that there's no general equation for a quintic.

[compact needle]

I'm sure there is more in Stewart than Pinter. I'm just saying if you've read what's in Pinter, you should be more than ready for what is in Stewart

[quaint tree]

Oh, thanks, I misunderstood what you had typed slightly.

[gentle pendant]

@steady axle A second way that does not rely on finite dimensionality:

Assume wlog that ${f_1,\ldots,f_m}$ are linearly independent. Any functional that vanishes on $W=\bigcap \ker(f_j)$ factors through the natural projection $V\rightarrow V/W$. But since $W$ is the intersection of $m$ codimension $1$ subspaces, we have that $\dim((V/W)^*)=\dim(V/W)\leq m$, and so the subspace of $V^ *$ consisting of possible $f_0$ is at most $m$-dimensional. As each $f_j$ is in this subspace, we get that this subspace is spanned by the $f_j$.

[cloud walrus]

gomez

[steady axle]

I like this better

not just bcos it works for infinte case but bcos its easier imo

[gentle pendant]

yeah, same

[celest brook]

[chilly ocean]

[rustic crown]

[chilly ocean]

I enjoy this picture.

[cloud walrus]

mirza

[rustic crown]

Aww this is so sweet

[cloud walrus]

mirza

[rustic crown]

yep you're right

so the thing you noticed is that 2 and 3 commute in Z, but 2 and 3 could be sent to some elements such that the images don't commute...

so the generators shouldn't have any relations among them, and in particular not commutativity

yep

if you generating set has size atleast 2, then the free free group on S is not abelian

i think you're right

the center has to be trivial if |S|>1

i think if you pick something in the center which starts with a generator g in the reduced form then this won't commute with any other generator than g

yea that works

have you seen a construction of free group?

yep

sth?

something?

yea you define the reduction of a word, only way to reduce it is if it contains the subword aa' or a'a

it requires some work, but you can show that if you reduce a word completely then the final reduction is unique

this is because if i an pick a word in the zenter, which is not identity and reduce it, and the conjugate it by a letter which is not the starting letter then you get something different

it was to make this a bit more precise.

[rustic crown]

this is because if no reductions happen after conjugating, then two reduced words of different length can't be same. more than one reduction can't happen, by the same reason. And one can't happen because they would differ at the starting letter.

its kinda boring when you need to examine everything rigorously 😶

yep

yea

you can just say any relation between the generators can be represented by a reduced word equaling identity

if you have a non-trivial relation, then the group isn't be freely generated by that set

i'm saying say two generators x and y commute, then this can be represented by the word xyx'y', this is a reduced word cause it doesn't contain any subword of the form zz' or z'z

the intuition with "basis" is pretty nice

in linear algebra you don't want any non-trivial linear combination to be 0

here we don't want any non-trivial reduced word to be identity

yep, both are instances of this "free" construction

we'll you need a notion of free and forgetful ness, then you can define it

but yea, the universal property can be used, but you still need to give a construction by hand. universal property doesn't guarantee the existence of a free object

but it guarantees the uniqueness (by a unique morphism) if they exist

yep

and you can also look at free R-modules

they look very similar to free abelian groups as Ab is just Z-Mod

I had my Galois Theory final today, and someone here asked me a very similar question
without it i wouldn't have been able to finish it in time

it was: if [F bar : F] is finite and char F = 0 then show the degree is either 1 or 2

[paper flint]

HSCT intensifies

[rustic crown]

Yep

[urban acorn]

I'm wondering if there's an embedding of the free group with countably many generators in F2

my bet is there is

I heard that there's an embedding of F3 into F2, and this gives you an embedding Fn -> F2 in a pretty straightforward way by induction

the intuition really is that you should be able to find "sufficiently independent" words

which is easy to formalize, of course

[golden pasture]

smt like x^nyx^-n should work

if i didnt messss up

[urban acorn]

hmm

so, imagine the countable free group as freely generated by N, then any set map N -> F2 gives us a homomorphism Finf -> F2

like, for example, n -> x^nyx^-n corresponding to your suggestion

so imagine we have a nontrivial word going to 1

can we somehow simplify the form that word is gonna be in?

[urban acorn]

okay i'm pretty convinced that this works now

given some some word in N, write down the word in x,y this creates, then look at the first coefficient of x, that tells you the first entry in your original word, look at the second coefficient and add the first entry, that tells you the second entry in your original word, etc

[delicate hawk]

ik how to prove that R is an integral domain

but how to prove its a PID not sure

[eager yew]

Hello

Is a set of complex 2x2 matrices of the form [] a vector space

Any help

[hidden haven]

Try proving the contrapositive

[hidden haven]

Check the vector space axioms one by one

[eager yew]

I tried them but nothing appears to disprove... probably I've missed something

[hidden haven]

If all of them are satisfied then it is a vector space

[eager yew]

I'm kind of doubting that it is

[hidden haven]

Ok, which field are you working on

well answer depends on the field

So maybe problem intended R vector space

[eager yew]

I'll go over once more

[old lava]

your brain

wait sorry, that doesn't work in your case

[hidden haven]

[old lava]

no

people can use whatever

I'll just mock them for choosing the inferior option

[nova plank]

finally i can respect mirza

For not liking DF

[chilly ocean]

d&f is p good tho

[old lava]

they're haters who've dug themselves too deep in the d&f hatred hole

and can't dig themselves out

sadge

[nova plank]

It's so bad though

DF is literally short for dumbfuck

[viscid pewter]

dragon-fox, durian flesh, distilling furniture

[chilly ocean]

dblocked f*ck you!

[untold sapphire]

Lol

[untold sapphire]

If in doubt just remember that everything is a vector space

If you squint hard enough

[final pasture]

lol

[modest wedge]

Maybe this is a dumb question, but is there a set of properties shared by every $2^n$-dimensional Cayley-Dickson algebra ?

[cloud walrus]

anøk

[eager yew]

Yh, thanks for checking 👍

[modest wedge]

I mean, are there any binary operations that are preserved by each and every cayley dickson algebra ?

[old lava]

addition

I suppose

power associativity perhaps

apparently each of them satisfies the flexible identity

seems about it

power associativity and that are about the few things preserved in every cayley dickson algebra

[modest wedge]

Oh ok, I expressed myself poorly again 😅 , I was thinking about a property of a binary operation, not of a binary operation in itself. Thank you very much 😄

And power associativity seems to be what I was looking for

[jagged dune]

How do I show that an operation holds when I'm showing that this is a ring homomorphism?

C -> M2(R)

a+bi -> [a b ]
[-b a]

Yes I get that

in particular, how do I show f(x+y) = f(x) + f(y), and f(xy) = f(x)*f(y)

Yes, but for some reason the matrix is throwing me off

so I'm not sure how to write it out

We can just do addition and I can do multiplication on my own

[next obsidian]

Mirza is expensive

😳

[cloud walrus]

expensive group

expensive group

expensive group

[jagged dune]

uh huh

[delicate bloom]

think of it this way, you can add the complex numbers first then turn it into a matrix, or you can change them into a matrix and then add them, then you check that the answers are the same

[chilly ocean]

u good?

[cloud walrus]

expensive group

[jagged dune]

That last picture put it all together for me

I think

ok I actually get it now lol

thank you!

tbh I think it was the factoring in the f() that threw me off

once I got what you meant it was a lot easier

so thank you, again

so for multiplication

am I just FOILing both functions?

gotcha

[jagged dune]

How do I find all possible ring homomorphisms from Z10 to Z12?

[chilly ocean]

I guess look what can 1 map to?

[jagged dune]

?

[chilly ocean]

?

[jagged dune]

May you elaborate?

[chilly ocean]

no

[jagged dune]

Alright.

[chilly ocean]

ok I may

just see what happens when f(1) = x for x in Z12

[jagged dune]

1

[chilly ocean]

ye for example f(1)=1, check if it's a homomorphism.

If it is, you can determine image of any element of Z10

[jagged dune]

Well 1-10 is gonna map directly to 1-10

From Z10 to Z12

[chilly ocean]

yeah but 5+5 = f(5) + f(5) = f(10) = 0 right

or am I talking nonsense? idk

[jagged dune]

oh you're correct about that

mod 10

[chilly ocean]

I guess you need f(1) =x where x is such that 10x = 0 (mod 12)

[chilly ocean]

f(1)= 0?

If you map f(1)=6 then it's also homo I think.

[untold sapphire]

No?

[chilly ocean]

Why?

[delicate bloom]

f(x)=4x a ring homomorphism from Z/6Z to itself, since 4^2=4 mod 6

[delicate bloom]

evaluate f(1)=f(1)*f(1)

[chilly ocean]

Oh yeah lol

[untold sapphire]

If you allow f(1) not to be 1 then weird things happen

https://math.stackexchange.com/questions/270883/in-a-ring-homomorphism-we-always-have-f1-1

[viscid pewter]

um

[delicate bloom]

nope

[viscid pewter]

4*1 == 4 !== 1?

[chilly ocean]

So the only ring homo is the zero one I think, right?

[viscid pewter]

i feel like you should be able to map to Z2

in Z12

[delicate bloom]

it is because f(xy)=f(x)f(y) and f(x+y)=f(x)+f(y)

[viscid pewter]

am i a fool

i literally only know group theory lmao

[chilly ocean]

I mean there aren't many group homomorphisms, so maybe just look at those.

a ring homo requires f(1)=1 but a rng homo doesn't

[nova plank]

If you are unbased and do not require rings to have unity, then there is no condition f(1) = 1 and even if you have unital rings, you might have a ring homomorphism that doesn't satisfy f(1) = 1. If you are based and require rings to have unity, usually you add a third condition to the definition of a ring homomorphism which is f(1) = 1. So stop fighting. Mero is just unbased.

[delicate bloom]

that's not necessarily in the definition

[viscid pewter]

oh well that's just brilliant

[untold sapphire]

Yeah but who likes rings without identity

[viscid pewter]

feruchemists

[untold sapphire]

I've never seen them come up

[viscid pewter]

wait for era 3

[delicate bloom]

how about f(x)=0 do you consider this a ring homomorphism

[chilly ocean]

yes

[viscid pewter]

rings don't exist

[chilly ocean]

depends on the codomain

[untold sapphire]

this is why we use modules because rings are horrible

Not a good category in general

[delicate bloom]

ok I think I see how we're looking at it, my example from earlier can be fine if I call f(x)=4x as a ring homomorphism from Z/6Z to Z/3Z

I'm being careless with what the codomain is

[thorn delta]

why does the product of those ideals have to meet S as well?

[oblique river]

S is multiplicatively closed

So pick elements from P + xA and P + yA which are in S, then their product is as well

[thorn delta]

ah yea, duh. thanks

[cloud walrus]

expressive group

[final pasture]

just assume it commutes

Not it definitely doesn't

[cloud walrus]

expressive group

expressive group

[final pasture]

I think the whole computation is wrong

But I may be mistaken

But I may be misunderstanding something

wait

but that would give the wrong result

I'm confused

wait

does $\sigma\tau$ means $\sigma(\tau(x))$ or does it mean $\tau(\sigma(x))$ ?

asking just in case the book follows weird conventions

[cloud walrus]

Shika-Blyat

[final pasture]

just read rotman

Oh

Wait

pi(sigma)pi(tau)f(x_1, .., x_n) means
pi(sigma) ( pi(tau)f(x_1, .., x_n) )
So you get pi(sigma) ( f(x_tau(1), .., x_tau(n) )
And then you apply pi(sigma), so you get f(x_sigma(tau(1)), .., x_sigma(tau(n)), which is is equal to pi(sigma tau)f(x_1, .., x_n) right ?

I guess that's what the book meant ?

Ok

But I'll 'cause I don't have nothing else to do

that's coherent with what I wrote isn't it ?

let me latex it

just to be sure

'cause latexing something means it's right

$\pi(\sigma) \pi(\tau) f(x_1, \cdots, x_n)$ means $\pi(\sigma) (\pi(\tau)f(x_1, \cdots, x_n))$\
By applying $\tau$, you get $\pi(\sigma) (f(x_{\tau(1)}, \cdots, x_{\tau(n)}) = \pi(\sigma)g$.\
Now we get, by applying $\sigma$, $(f(x_{\sigma(\tau(1))}, \cdots, x_{\sigma(\tau(n))}) = \pi(\sigma\tau)f(x_1, \cdots, x_n)$, and we're done, right ?

[cloud walrus]

Shika-Blyat

[final pasture]

yeah, I think so

But idk, its kinda confusing, we may be missing something 🤔

I mean

if it is a typo

there's only one typo

like just one expression with 2 typos

[cloud walrus]

expressive group

[final pasture]

here, if you replace pi(tau) with pi(sigma) and replace the x_sigma(i) with x_tau(i), the computation becomes the same as what we did, right ?

lol

I mean the other lines don't change

so that wouldn't be such a big typo

may be in some errata

do you have the last version ?

Also, read Rotman

[junior moth]

just swap sigma & tau in rhs

[final pasture]

that's exactly what we were saying

[junior moth]

typo

so it's wtvr

[junior moth]

oh, ya this

[final pasture]

basically Lang's not clear, but if we're being nice he's not completely wrong, and it also says that you should read rotman

[past temple]

let's say that i have a group of order 8

and in that group i exhibit elements s,r that satisfy s^2 = 1, r^4 = 1, rs = sr^-1

is that sufficient proof that my group is isomorphic to D8?

[carmine fossil]

If r,s satisfy no other relations yes

Alternatively,if there are 8 elements in <r,s>,That works too

For example if we take the usual D_8 and denote a=r^2,then
a^4=1,s^2=1 and as=sa^-1 is true

But a^2=1 is also true

[past temple]

hm

well in the case of my problem

i know that |s| = 2, |r| = 4

s,r denoting the elements i exhibited in my group

not the usual s,r in D8

[carmine fossil]

Yea,in that case a group G=<r,s|s^2=1,r^4=1,rs=sr^-1> would be D_8

[past temple]

well i havent shown that G is generated by r,s

i just know that the relations are satisfied and the order is 8

[carmine fossil]

If H is a subgroup of G having same size as G,then H=G

[past temple]

right

[carmine fossil]

So,Since order is 8 you are done

[past temple]

do we know that my group G is a subgroup of D8?

how do we know that

[carmine fossil]

We know <r,s> is a subgroup of G

And |G|=8 and |<r,s>|=8

[past temple]

ahhh okay

[carmine fossil]

So G=<r,s>

[past temple]

thank you thank you

oh fuck

okay basically what i've shown so far is that

the galois group of the splitting field of x^4 - 7 over Q is isomorphic to D8

now i have to use the fundamental theorem of galois theory to find every intermediate subfield by corresponding them to subgroups of D8

this is actual death

is there a shortcut to doing this??

[carmine fossil]

https://math.stackexchange.com/a/1748931

[past temple]

that's just a forum post descrbing how to show that the galois group is isomorphic to D8

my goal rn is to use that fact to provide every intermediate subfield

[carmine fossil]

mb

There are exactly 6 subgroups of D_8

That might be helpful

[past temple]

D8 only has 6 subgroups?

[carmine fossil]

mb,no

9 subgroups

[past temple]

rip

[chilly ocean]

i see 10 in that image

[old lava]

This is algebra channel not arithmetic channel

[chilly ocean]

arithmetic is just ring theory over Z

[rigid cave]

This might be an odd question, but do any of you know any exercises that wants you to show that for example -(-x) = x? I am pretty certain that some person posted a exercise consisting of "subexercises" that wants you to prove "basic" things (including that -(-x) = x), but I can't find it now.

[smoky verge]

tao's analysis I?

anything that builds up the integers from peano axioms

[rigid cave]

well I actually don't have that book

[old lava]

or you can just use good old ring theory axioms

assuming x comes from a ring

[rigid cave]

Yeah that is what I am searching for, but I can't find any exercises

[smoky verge]

it's freely available online (legally)

[old lava]

I mean you can just write down the things you take for granted

[chilly ocean]

just about any book about an algebraic structure should have it either as an exercise or theorem?

[scarlet estuary]

are you looking for exercises that ask you to prove this for rings

or for naturals

[old lava]

and try to prove them using the ring theory axioms

[scarlet estuary]

or for groups

or what

[rigid cave]

for rings

[old lava]

I mean I assume it's from a ring

considering the channel

[scarlet estuary]

okay yeah, any intro algebra text should have this then

unless its omitted for being "obvious"

[old lava]

I think they're kinda just treated as obvious facts

but you can just think of the obvious facts you usually take for granted

and prove them yourself?

idk

[rigid cave]

Yeah that's true

[scarlet estuary]

-1 * x = -x is an important one

[old lava]

(-1)^2 = 1 is a good one too I think

[scarlet estuary]

and requires more than 2 seconds of thought

[old lava]

0 is unique?

as well maybe

[chilly ocean]

how do you prove an axiom

[old lava]

they're not axioms?

sully?

[chilly ocean]

but if you assume something to be true

its an axiom

[old lava]

[prisma ibex]

[old lava]

that's just not how it works at all kekw

[chilly ocean]

that's not usually an axiom

[old lava]

you can in fact assume things

without them being axioms

I know I just blew your mind

but it's true

[chilly ocean]

oh like

for implications?

[scarlet estuary]

"assume" was used informally here

i have no clue why youre nitpicking the wording so much

it seems fairly clear from context what was meant

"Things you intuitively assume are true and obvious, but havent sat down and checked"

[rigid cave]

-1*x = -x seems soooo easy, but I just can't prove it!!!

[old lava]

the trick with 99% of them

is to use distributivity

somehow

[rigid cave]

Okay okay don't spoil it yet

[scarlet estuary]

distributivity basically always comes up for these little identities since

its the only thing relating addition and multiplication

in a ring

the ONLY thing

without distributivity theyd be totally unrelated operators

but -1 * x = -x is "translating" a multiplication statement (-1 times x) into an addition one (the additive inverse of x)

so it makes sense youd need distributivity for that

(And similar identities)

[rigid cave]

yeah that's a good point

[old lava]

it's actually insane how important some things you take for granted for like the first 12 years of schooling are

like associativity and distributivity

they give so much

learning algebra has really renewed my appreciation for math

ngl

[rigid cave]

Okay so maybe like this: Assume that it is false, then -1*x + x does not equal 0. Multiply both sides by -1 to get x-x does not equal to 0 which is a contradiction?

But then I have used that -1*-1 is 1

[old lava]

can't just multiply both sides by -1

yes, because you haven't proved -1* -1 = 1

which you can do before hand, but just use distributivity

you will HAVE to use it at some point

in the prove of -1 * -1 = 1

[chilly ocean]

and also used that -1*x = -x which is what is trying to prove

[old lava]

use distributivity

use distributivity

use distributivity

[rigid cave]

But why can't I use this?

[old lava]

use distributivity

[old lava]

.

[chilly ocean]

you can multiply both sides by -1. but then you don't know what -1*(-1*x) or -1*x is so its not very useful

[old lava]

yes, that why

use distributivity

[chilly ocean]

and it's easy to construct a ring-like structure that doesn't have distributivity where -1*x is not -x

if you could prove without distributivity, it would be true for those..

[rigid cave]

But isn't 1x the additive inverse of -1x?

It became italic for some reason

[paper flint]

I think you're supposed to prove that in some texts(Friedberg's LA made me prove that for sure)

Or yeah

[rigid cave]

Man, I really feel like a moron

[paper flint]

This could be axiomatic

The axioms say for every x, you can find a y such that x+y=0, then you define this y to be -x

[rigid cave]

But how am I supposed to use distributivity?

[rigid cave]

lmao

[carmine fossil]

Hint: a+(-a)=0 and inverse is unique

[old lava]

there's quite literally only a single thing to distribute

[rigid cave]

okay so -1x +x = 0 so then -1x must -x?

[old lava]

like saying "use distributivity" is basically solving the entire thing

[carmine fossil]

nvm

Hint: 0x=0

[old lava]

you can prove that as well

if you want to

[lavish gale]

onesan likes groups?

[old lava]

other fun facts to prove:
1 is unique
0 is unique
inverses are unique

[rigid cave]

Writing those things down now

[chilly ocean]

you can prove 1 and 0 are unique at same time if you show identity of a semigroup (set with associative binary operation) is unique (if exists)

[rigid cave]

HOW DO I USE DISTRIBUTIVITY????

[rigid cave]

Wait, are you referring to me?

[chilly ocean]

here's an hint: ab = (a+0)b = ab+0b

[old lava]

there's literally only a single thing to distribute

like I cannot help you any more than that, or I'd be giving you the answer

[rigid cave]

Okay so like -1x = (x+0)-1 = -x?

[old lava]

how does that accomplish anything

[carmine fossil]

Ok,one more hint 1+(-1)=0

[rigid cave]

Well that's why I don't know why distrubutivity will help

[old lava]

bruh

I can't

[rigid cave]

okay so 1+(-1) = 0 so then x +(-1x) = 0 so -1x = -x

It this is not it then I am quitting

[chilly ocean]

ax+bx = (a+b)x

also note that x = 1*x

[rigid cave]

But is my solution correct?

[chilly ocean]

how did you conclude that x + (-1x) = 0?

[rigid cave]

because 1+(-1) = 0 so when I multiply with x on both sides I get x+(-1x) = 0

[chilly ocean]

hmm yea if you multiply both sides on the right by x you get x+(-1)*x = 0

[rigid cave]

yeah and -x is the additive inverse and it is unique so it must be the case that -x = -1*x

right?

[chilly ocean]

hmm let me make sure im a bit suspicious

okay yea cuz

1+(-1) = 0 => (1+(-1))x = 0x => x+(-1)x = 0

so you use distributivity in 3rd step

[rigid cave]

yes that is what I was doing

[chilly ocean]

sounds good then i think

[rigid cave]

finally. Took me awhile lmao

[chilly ocean]

you could also just

x+(-1)x = 1x+(-1)x = (1+(-1))x = 0x = 0

[rigid cave]

yeah that is true. My approach is a bit longer I think

[paper flint]

Have you proven -1x=-x though

[rigid cave]

Well I think so?

Since inverses are unique

[paper flint]

Okay I hate this stuff

[rigid cave]

Well to my defence I am still in highschool. This is not a good excuse but...

[chilly ocean]

i wish i learned this during highschool

[paper flint]

Same

[old lava]

do this, but x1000 and spend your life on it

foundations moment?

[rigid cave]

So 0x = (1-1)x = x-x = 0?

using that -1*x = -x

[paper flint]

[lavish gale]

yes @rigid cave

[rigid cave]

Well I think that I like groups... I don't know a lot of group theory tho

[molten silo]

Can some help me out on how to prove part b if statement.

[oblique river]

Hint: N(s + t*sqrt(d)) = (s + t*sqrt(d))(s - t*sqrt(d))

[old lava]

uh

[junior moth]

use part (a), definition of unit, & N(1) = 1

[molten silo]

hmmm

[oblique river]

Oh sorry I was assuming d is negative

[molten silo]

I dont know what you mean 213

[oblique river]

Which is why i didnt write abs value

[chilly ocean]

paul, to begin with, can you prove a is a unit ⇒ N(a) = 1

[oblique river]

In any case, you can use that formlus for N to go the other way, that if N(a) = 1 then a is a unit

[molten silo]

yes i can prove that part

[oblique river]

Which part?

[molten silo]

only if

[oblique river]

For the “if” direction, use the mutiplicavity of N

[oblique river]

What happens if you take the norm of both sides of ab = 1

Yes that was my hint for the only if direction

[molten silo]

N(a)N(b) =1

[oblique river]

Great, now how many positive integers are there which divide into 1?

[molten silo]

i think i proved this part

its the other one

[oblique river]

That’s the if oart

[molten silo]

i need to prove only if then

[oblique river]

“Only if” is N(a) = 1 implies a is a unit

[molten silo]

yes

[oblique river]

Then use my first hint

[molten silo]

this one is hard

right

[oblique river]

But put absolute values around it cuz i forgot those

[molten silo]

thank you brother

[oblique river]

If N(a) = 1 then that product is equal to +/- 1, so can you use that to find an inverse for a?

[molten silo]

yes

you can

[oblique river]

Yes I can but can you ;)

[molten silo]

not really

In both cases a must be a unit, but with different inverse in each case

[oblique river]

So youve proven it’s a unit then

That’s what you were trying to do

Because it has an inverse

[molten silo]

danker

[past temple]

how do you show that f(x) = x^4 - 7 is irreducible in F5

so far i know that just by explicit calculation

f has no roots in F5

so it has no factors of degree 1 or 3

this leaves factors of degree 2

and i'm not sure where to go from here

(any help woudl be appreciated - ping me)

[nova plank]

@past temple You can try supposing it has a factorization as a product of two monic quadratics, multiply it out, compare coefficients and get a contradiction

[delicate bloom]

I guess you could say for the sake of contradiction, x^4-2=0 and since F_5 contains all the 4th roots of unity you know x=2^{1/4} times those is all the possible roots

but 2 doesn't even have a square root in F_5, so it has no roots

[past temple]

2?

[delicate bloom]

7=2 in F_5

[past temple]

oh right right

okay wait hm

i still dont get it

how does assuming by contradiction that f is reducible lead to f = 0?

for some x

[hidden haven]

Say r is a root of f in some splitting field, then f(x) = (x-r)(x-2r)(x-3r)(x-4r) in that splitting field because kr is also a root of f for k = 1,...,4

[past temple]

right

[hidden haven]

Product of any 2 of these factors is in F_5 only if r² is in F_5

But there's no element in F_5 whose square is 2

[past temple]

ahhhh okay

that makes sense tyty

[chilly ocean]

This "proof" that sqrt(2) is irrational is bs right?:
Every field automorphism of a field extension of Q fixes Q;
Q(sqrt(2)) js a fjeld extension of Q;
f: a+bsqrt(2) -> a-bsqrt(2) is an automorphism of Q(sqrt(2)) that doesnt fix sqrt(2);

to show that f an automorphism you need to show 1 and sqrt(2) lin independent over Q which is the same as sqrt(2) being irrational

[latent ingot]

yea this assumes sqrt(2) is not in Q

[chilly ocean]

What did you try?

[candid pier]

First choice = ** No **

[nova plank]

Why no?

[chilly ocean]

So your first idea was to flip the coin?

[candid pier]

Is this correct or not ?

[nova plank]

Yeah, that looks good.

[cloud walrus]

squirtlespoof

[chilly ocean]

I'm not sure about an inductive argument working but

I think one way you can do it is, so first linear independence is equivalent to no solutions to sqrt(pn)=c for c in the base field Q(p1,...,.. pn-1), then write down a basis for the base field and work with that

Idk, I guess just intuition. sqrt(pn) not being in the previous field doesn't really feel like it is about induction

pn is not very related to p1,..., pn-1. What can you say about it? I guess pn is not divisible by any of those, in particular it is the smallest one not divisible by those. But this still feels kind of a weak relationship, or a relationship that is not too relevant to the problem

[cloud walrus]

squirtlespoof

[chilly ocean]

Hm, that doesn't look right....

What did you get for the basis of Q(stuff only up to n-1)?

Basis should be size 2^n or so

Yeah

Actually hm, maybe this doesn't make the problem easier.... (?) Let me think...

Idk, it is getting too complicated for me to do it without a notebook

But eg n=2 case is pretty simple, p2=(a+b*sqrt(p1))^2 obviously leads to contradiction

Erm, you are not saying that (x+y)^2=x^2+y^2 right?

Yeah

Hopefully n>2 case is similarly easy 🙂

[cloud walrus]

expressive group

[chilly ocean]

Hm, n=3 case is hard

[old lava]

why are they defining shit with exact sequences

lol

also yes

$\iota$ is injective

[cloud walrus]

F[x]-module

[old lava]

so you can identify N as a subgroup of G

in fact that's basically assumed

when you name your function iota

technically it's defining the action of Q on the isomorphic copy of N in G

[rustic portal]

Took the final for my first class in AA yesterday :D

[old lava]

nice

[rustic portal]

Super interesting stuff, considering going deeper into AA2, but we'll see

Currently applied math & cs, so not sure what I'd do with it

[acoustic lodge]

is a polynomial ring of the integers allowed to have $1x^{0}$ in it? this is how my prof defined polynomial ring.

[cloud walrus]

Panda_Bear59

[acoustic lodge]

so just the element 1 is in R[x], meaning we are allowed to have terms without x in it.

nvm i think clearly we are, since $a_0$ can be 1

[cloud walrus]

Panda_Bear59

[rustic portal]

yeah thats fine

$r\in R \implies r\in R[X]$

[cloud walrus]

aditya

[cloud walrus]

squirtlespoof

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

[restive star]

Can someone help me with a problem? So we are given a field K and a subfield F, and have to figure out when the set {a+bx | a,b in F} is a field for a fixed x in K. And prove that it is necessary and sufficient

clearly x^2 has to equal cx for some c in F, but that's not sufficient

[delicate bloom]

hint: if r is a root of f that appears more than once, then it's also a root of f'

yep

not sure what your approach is

but I was assuming you were looking at defining it by looking in the algebraic closure of F_p

[delicate bloom]

not necessarily could be x^2 = cx+d for c,d in F

[restive star]

oh yeah true

But I'm having an issue with the sufficient part. I can't find inverses for even one of the form x^2 = cx

[delicate bloom]

you can take the irreducible poly over F and divide through by x

like in our case it's x^2=cx+d, divide through you get x=c+dx^-1 so you have x^-1 = (x-c)/d

write this as -c/d + 1/d * x to get it of that form if you want

or what are you wanting to invert, just a general element you mean a+bx?

[restive star]

is this sufficient though?

[restive star]

yeah

that's where my trouble is coming.

[delicate bloom]

you should try with x^2=cx+d since you won't succeed with x^2=cx as it's a reducible polynomial

[restive star]

ah ok. I will try

[delicate bloom]

x=0 or x=c there

so it doesn't correspond to a field extension, try to see if you can use what you know about like simpler quadratic extensions you've played with in the past

[restive star]

yeah this part is stumping me

I cant seem to find a general inverse for a+bx

@delicate bloom Do you have any other tips?

[delicate bloom]

sure, what's the inverse of a+bi?

[restive star]

a/(a^2+b^2) - b/(a^2+b^2)i, right?

[delicate bloom]

yeah good

the thing I want you to recognize is you're using the conjugate

what's the relation(s) between the conjugate roots of X^2-cX-d?

I put capital X to distinguish between x the root lol

[restive star]

I don't think I understand the question. I know they will be conjugates of each other, but Im not sure what else

[delicate bloom]

well ok specifically can you represent the conj(x) in the form a+bx?

since then our inverse of a+bx becomes the same trick as computing the inverse with complex numbers, you multiply and divide by the conjugate

[restive star]

ah ok. But is that possible since our x is arbitrary?

[delicate bloom]

it is, since we know there's some irreducible polynomial it's a root of

expand out (X-x)*(X-conj(x)) = X^2-cX-d

[restive star]

Sorry this is where I don't quite get what you're saying

Wouldn't conj(x) literally just be -x?

[delicate bloom]

no

not generally

[delicate bloom]

do this and show me what you get

compare coefficients, you will get two relationships between x and conj(x)

[restive star]

Oh wait. I think I understand

since the roots of that polynomial will be x and conj x

sorry idk why i misunderstood at first. it's been a long day

[delicate bloom]

yeah, really you could write them out as the roots of the quadratic, they differ by the sign on the square root

but there's no reason to try to do that since we have x+conj(x)=c here

so we do indeed have conj(x) = c-x in our field