groups-rings-fields

[rustic crown]

but the analogue for injective shouldn't be true

[tough raven]

If this holds, and B is the inverse, then this formula https://en.m.wikipedia.org/wiki/Cauchy–Binet_formula should give us a linear combination of m × m minors is 1

[rustic crown]

ooh right

[lavish spoke]

my algebra is shaky, is this a correct start to the argument that if $A$ is an arbitrary ring, $\mathfrak p$ is a prime ideal, $\mathfrak p_1 \subsetneq \mathfrak p_2$ are prime ideals of $A[T]$ that contract to $\mathfrak p$ in $A$, then $\mathfrak p_1 = \mathfrak p A[T]$? Not sure how to get a contradiction in the latter case

[cloud walrus]

George!

[tough raven]

I probably know no more algebra than you, but this looks correct (assuming A is commutative). I don't think you would get any contradiction though, p1 = pA[T] is a possibility since A[T]/pA[T] = (A/p)[T] is an integral domain and not a field, so pA[T] is prime but not maximal.
(OK, that only shows there has to be p2 bigger than p1 and prime, not necessarily that p2 also contracts to p, but taking s2 to be (X) and working backwards should give you an actual p2 = (p,X) strictly including pA[T].)

[formal ermine]

$\bQ[x]/(x^2+2) \cong \bQ[i\sqrt{2}]$ yes?

[cloud walrus]

yes yes yes no

[tough raven]

[green locust]

Hey everyone, i got a quick question here for exercise III.6.15 in aluffi’s chapter 0. to restate: Let $R$ be a commutative ring. Prove that a commutative $R$-algebra $S$ is finitely generated as an algebra over $R$ if and only if it is finitely generated as a commutative algebra over $R$. I don’t really understand what the difference between being finitely generated as a commutative algebra over $R$ is vs finitely generated as an algebra over $R$. We are assuming $R$ is commutative and $S$ is a commutative $R$-algebra so wouldn’t it automatically be the same thing?

[cloud walrus]

iceberg56

[lavish spoke]

doubtful lol - ah yeah I was hoping for something a bit nicer, so q_1 can be a bit bigger than p A_p[T] but still contract to p A[T]?

that makes sense but idk where to go from s_1 = s_2 (and A is commutative fwiw)

[tough raven]

Oh I think s1 = s2 implies q1 = q2 implies p1 = p2

[lavish spoke]

i was thinking that too but I wasn't sure if it's literally just contracting back

[tough raven]

The first by Lattice Isomorphism Thm and the second by properties of prime ideals of localisations vs original ring

IG p1 = p2 would have to use that both p1, p2 are disjoint from A \ p

[gritty sparrow]

By the way, what exactly are you trying to prove?

[lavish spoke]

uh this is a lemma in proving dim A[T] <= 2 dim A + 1

[gritty sparrow]

I mean, what is this?

[lavish spoke]

what's what lol

the dim is krull dimension

[gritty sparrow]

Oh sorry i didn’t see the message where you explained what you were trying to prove

[lavish spoke]

ah fair

[tough raven]

Niceee

This is very neat

[lavish spoke]

yeah i thought so until i realised my original proof of this lemma didn't work xd

[lavish spoke]

what's the lattice isomorphism theorem btw lol

dont think i've had that term used

[tough raven]

Ideal of R/I <=> ideal of R including I

[lavish spoke]

oh that one

makes sense

well cheers !

[formal ermine]

to find the multiplicative inverse of $x+1$ in $\bQ[x]/(x^2+2)$ I have to find a polynomial $q(x)$ such that $(x+1)q(x) = x^2 + 2 + 1$ because then $(x + 1)q(x) = 1 \pmod{x^2 + 2}$, no?

[cloud walrus]

yes yes yes no

[rustic crown]

not just x^2+2 on the right

you wanna find p and q such that
(x+1) * q + (x^2+2) * p = 1

[formal ermine]

why?

[rustic crown]

the same reason, if you go modulo (x^2+2), the second term dies and you get the inverse x+1

[formal ermine]

are p and q both polynomials

[rustic crown]

yep

[formal ermine]

why though? what's wrong with (x+1)q(x) = k*(x^2 + 2) + 1?

[rustic crown]

there may not be any solutions to that in general

[ruby sundial]

what are some weird sequences that show up in algebra?

sorry im interrupting

[formal ermine]

hmm

[formal ermine]

how would I solve this though

[ruby sundial]

i think you can use euclidean algorithm?

no this is wrong

@formal ermine what i would do is first consider x+1 in Q[x]

then find the inverse

the inverse is some big power series for sure

[lavish spoke]

uh how would you treat A_p[T] as a localisation of A[T]? sorry if that's a stupid question

[ruby sundial]

and then try and compute what it looks like when identifying by x^2+2

[lavish spoke]

oh no I'm being silly nvm

[ruby sundial]

you consider fractions

[formal ermine]

x + 1 has no multiplicative inverse in Q[x]?

[lavish spoke]

it's (A \ p)^(-1) (A[T]) isn't it

[ruby sundial]

it doesnt but it does in power series

[formal ermine]

wut

how does this help

[rustic crown]

wait just to confirm, you don't mean that k is a constant here, right?

[formal ermine]

I did mean that at first, but yeah a polynomial would also vanish :v

[rustic crown]

since you write q(x) and one place and k at the other, i was a bit confused

[ruby sundial]

yes

[lavish spoke]

got it

[ruby sundial]

so the denominators are A\p

and the numerators are elements of A[T]

[lavish spoke]

👍

[rustic crown]

you can use the extended euclidean algorithm

it's just like how you would solve 11x + 13y = 1

[gritty sparrow]

Troll solution:
1/(1+x) = 1-x+x^2-x^3….
= 1-x + x^2(1/(1+x)) but x^2=-2, therefore 1/(1+x) = 1-x -2/(1+x) so 1/(1+x) = (1-x)/3

Now ofc you can verify that this element really does work

[formal ermine]

how is that a troll solution

[gritty sparrow]

Infinite series are not legit in Q[x]/(x^2+2)

[formal ermine]

oh

it seemed so elegant

[ruby sundial]

i forget when you want to use extended or not

[ruby sundial]

it isnt a troll solution

[gritty sparrow]

Well, you can just start the proof by noting that the second equation holds

[ruby sundial]

you can use that and then quotient to get same solution im pretty sure

idk what the quotient map turns into though

[rustic crown]

is there really a way to make sense of that? like actually using infinite series...

since sqrt(-2) has magnitude more than 1, i would have guessed otherwise

[ruby sundial]

its like you have Q[x]->Q[x]\I induces Q[[x]]->idk

[rustic crown]

right, the evaluation maps only makes sense for x = 0?

[ruby sundial]

you need a topology to make sense for other inputs

but yeah

this is in atiyah macodnald i think

i adic topologies

[rustic crown]

you could still do the same thing, but ofc in general this won't work...
(1+x)(1-x) = 1 - x^2 = 3 - (x^2 + 2)
so (1-x)/3 is an inverse

[rustic crown]

that does make sense...

[ruby sundial]

i havent ever worked out any examples before

but i hope to when i get older

[rustic crown]

wait so to confirm, we will make the ideal (x^2+2) smol

so in a sense it would be power series in (x^2 + 2)?

and not x

like Q[[x]] is the completion of Q[x] at the ideal (x)

[ruby sundial]

i dont know exactly but you define a metric for the topology by making numbers closer to x^2+2 very small

yes

this is true

[rustic crown]

but then i feel we can't use the identity for Q[[x]] as Q[x]_{(x^2+2) hat} looks pretty different

[formal ermine]

what is Q[[x]]

[ruby sundial]

ring of power series with coefficients in Q and indeterminate x

[gritty sparrow]

Hmm this might be a simple way to do it in Q[[x]]:
Let u = 1/(1+x) = 1-x+x^2(1/(1+x)) as before, so u = 1-x-2u+ (x^2+2)u. Therefore u = (1-x)/3 + (x^2+2)u/3. Now multiplying by 1+x on both sides, 1= (1+x)(1-x)/3 + (x^2+2)/3. But this is now the equation we need in Q[x]

[rustic crown]

nice

[formal ermine]

what's the easiest way of showing that $\bZ_3[x]/(x^2 + x + 1)$ isn't a field? do I just show that the ideal isn't maximal by giving an "even bigger" ideal?

[cloud walrus]

yes yes yes no

[rustic crown]

show that x^2+x+1 isn't irreducible

by factorizing it

(or by simply showing that it will have a root by plugging x = 1)

[formal ermine]

rightttttt

thanks

sounds like a dumb question but in $\bZ_2[x]/(x^2 + x + 1)$ the inverse of $x$ is $x + 1$?

[cloud walrus]

yes yes yes no

[rustic crown]

yep

[coral shale]

-x-1 ?

[rustic crown]

1 = -1
:3

[coral shale]

o

[next obsidian]

Slurp more like not char 2-pilled

[rustic crown]

wait slurp = shuri?

[coral shale]

its slurps fault

[next obsidian]

Oh

Yeah

That’s what I meant

Yup

Hi Cat Bread

[rustic crown]

hewwo Cat Bread

[wet zodiac]

hi

[formal ermine]

hi cat bread

[wet zodiac]

just reading artin rn

[next obsidian]

Ooo

What are u reading about

[wet zodiac]

groups

[next obsidian]

Nice

Name me a group

[wet zodiac]

(R,+)

[next obsidian]

That’s a good group

[rustic crown]

what's a bad group?

[next obsidian]

Kinda schnasty tho

[formal ermine]

{e}, obviously

[rustic crown]

nu

[wet zodiac]

was about to say R+ but forgot that can be read as positive real numbers

[next obsidian]

(R,+) is kinda a bad group it’s like not got enough structure

IMO

[rustic crown]

{e} very good group

[next obsidian]

What’s your favorite group?

What’s my favorite group…

I have pretty favorite rings

But groups hmmmm

[formal ermine]

my favorite group is end(G)

[next obsidian]

That’s not a group

You need to use Aut

[wet zodiac]

the identity element kinda confuses me
if im in a group like (Z,+) how does the identity element work

[formal ermine]

the endomorphisms form a ring

[rustic crown]

end in a groupoid :3

[next obsidian]

I mean okay but like

Then you’re just like talking about adding functions lol

[next obsidian]

So like it would be an element e so that

x + e = x = e + x

For any x in Z

What element does that?

[wet zodiac]

oh so it doesnt have to be 1

[next obsidian]

Yah so

[wet zodiac]

damn you artin

[next obsidian]

For groups

Well here’s the thing cat bread!

1 isn’t “the number 1”

[formal ermine]

1 is the identity in (Q \ { 0 }, 1)
0 is the identity in (Z, +)

[next obsidian]

It’s a symbol in this case

[wet zodiac]

ah

[next obsidian]

Like you can make a group out of automorphisms

[wet zodiac]

so an identity can be whatever

[next obsidian]

So like take the set of bijections of like {1,2} to itself

This is a group but no element like is “the number 1”

[wet zodiac]

as long as 1a=a=a1 holds

[next obsidian]

But usually groups are written multiplicatively

[rustic crown]

me wanna cat pill cat bread

[next obsidian]

Which is why ppl use 1 cuz like our intuition says 1a = a

That makes sense

But when groups are abelian (if you know what that means) often times people write it additicely

So we would say a + b not ab

And in that case ppl usually denote the identity as 0

Cuz we want 0 + a = a

[rustic crown]

(Q*, +)

[next obsidian]

Technically we could use the symbol 1 still, but then we’d be saying 1 + a = a

And that looks weird

[rustic crown]

2 + 3 = 6

🙈

[next obsidian]

Does that make sense tho cat bread?

[wet zodiac]

yes

[next obsidian]

:D

So I guess the last thing is

Because of this

Sometimes people use e for the identity

ea = a = ae

[lunar ledge]

edentity

[next obsidian]

Like when we write 1 or 0 sometimes we think we are dealing with the actual number

But that’s not the case

[next obsidian]

Silence, panda

[lunar ledge]

:'(

Ive seen 1_F, 1_G, 1_R, 1_V to denote units

[next obsidian]

ChmonkaS

[rustic crown]

what's 1_V

(rest i'm guessing are field, group, ring)

[next obsidian]

Vegeta

[rustic crown]

oooh

[elder wave]

vector space

[rustic crown]

1_V = 0 >.<

[elder wave]

No.

[next obsidian]

No.

[lunar ledge]

1_V = 0_W0

[rustic crown]

owo

[chilly ocean]

I get what you mean

vector spaces have no units so...

maybe it's an algebra

[restive birch]

https://cdn.discordapp.com/attachments/496784958430380033/1045026234733707364/image.png how would you do this problem? is the kernel of the action just {1}?

ig what i dont understand is how the bijection that g represents relates to the binary operation of G

so i dont know how id show that the kernel is closed under the group operation of G

[chilly ocean]

No

Action is a homomorphism from G to Sym(A)

And it's a subgroup for standard reasons

[restive birch]

what standard reasons

[chilly ocean]

Kernel of homomorphism is a subgroup

In fact a normal subgroup

[main needle]

How to find the generators of the normalizer of the subgroup <(12), (34), (5,6), (7,8)> in s8 ?

[tough raven]

Yes, (S^{-1}A)[T] = S^{-1} (A[T]) in general for S a subset of A and viewed as a subset of A[T] on the RHS.

[tough raven]

That's actually not really troll, you could phrase it as “if r is an inverse, then r = 1 - X + X^2r” which is true since (1-X^2)r = 1-X.

ofc you got that from the power series, but it's not wrong to use that to guess what to prove

[ionic spade]

I'm reading the proof about how a submodule of a cyclic R-module is also cyclic (when R is a PID) and I want to make sure I understand a specific step.

[cloud walrus]

Volkenborn

[ionic spade]

When they write S=Iv, are they saying that since R is an R-module over itself, every ideal I of R is a submodule, so S=Iv since you are just restricting R to I?

[deft gale]

S=Iv comes from definition of I

S=Iv because S is defined to be cyclic group generated by v so its elements are v^i. I is defined to be the ideal quotient of S. and by its definition multiplying I by an element in S gives you S.

Im guessing you meant to ask Iv=Rsv step?

[gritty sparrow]

Yeah, I know that it is valid to use these sorts of things to guess equations, i meant that the manipulations themselves are not valid. Maybe in the (x) adic completion of Q[x]/(x^2+2) they are legit?

[tough raven]

X is a unit in Q[X]/(X^2 + 2) though…

[gritty sparrow]

Oh right lol

[past path]

Does this seem like an appropriate way to prove the statement?

[rustic crown]

Yee looks great

[past path]

Thanks man!

That stumped me for a long time and your idea really helped me.

I wanted to turn it into my default language to make sure I knew how to do it properly on my own.

[onyx trout]

1st isomorphism theorem if you need to cite it specifically

[rotund aurora]

By decomposition you get that M=Z/(q_1)+...+Z/(q_n) where q_1,...,q_n are prime powers that divide the order of M, say p^k-1. But you still don't know if the q_i are coprime, so you cannot conclude that this is cyclic straight away. It seems to me there's no point in looking at the structure theorems, because you have to do some extra work, which is like proving it without the structure theorems in the first place

Am I missing something?

[rustic crown]

say n = least common multiple of all the q_i

then n kills every element of M

[rotund aurora]

sure

[rustic crown]

for M = F*, this means x^n - 1 has |F*| roots

[rotund aurora]

yes

[rustic crown]

so n >= |F*|

:3

[rotund aurora]

but you do not need the structure theorems for this argument

that's what I was saying

Let n be max(n | n is the order of some element of G), then x^n=1 for every element in G, but G is a field, so x^n-1=0 has at most n solutions

[rustic crown]

yea, structure theorem is a general result about modules over a pid, so ofc if you're looking at any finite subgroup of F*, then you definitely have to use something about the field

the first part isn't obvious

[rotund aurora]

but its from group theory

no?

[rustic crown]

yep

[rotund aurora]

like you don't need to talk about modules

you just use the euclidean algorithm or something I think

[rustic crown]

nah, it's a litlte bit more complicated than that

[rotund aurora]

I must have encountered this at some point

let me think about it

[rustic crown]

you first have to prove that for if there is an element of order n and and element of order m, then there exists an element of order lcm(n,m)

and the standard proof of this first shows this when gcd(n,m) = 1

and then given general n, m uses the prime factorization to find divisors n' and m' of n and m such that gcd(n', m') = 1 and lcm(n, m) = n' * m'

[rotund aurora]

oh I see, so the structure theorem immediately gives you that

[rustic crown]

yep

[rotund aurora]

I was just assuming that and saw no point

[rustic crown]

and even more directly if you use the invariant factor decomposition

M = Z/d1 + ... Z/dr where d1 | d2 | ... dr

the element of maximal order is (0, 0, ..., 1)

which is dr

[rotund aurora]

then M is just Z/dr

no?

[rustic crown]

yep

because by the field theory stuff you will have dr >= |M| = d1*...dr

which is only possible when M = Z/dr

[rotund aurora]

thanks

was there an expression to say that a^n=1 in the case where n need not the order of a? Like saying it with words

something like a belongs to the exponent n or something, but idk if that meant that the order of a was n

[rustic crown]

n annihilates a

[rotund aurora]

ah k

[rustic crown]

or just n kills a

[rotund aurora]

yeah I guess that's cool

[rustic crown]

very violent terms >.<

[rotund aurora]

that's standard terminology of a module

and in my context, im working with Z-modules, though defining (n,a)-> a^n

one of my professors uses the word "survives" when talking about stuff that doesn't get annihilated xD

[rustic crown]

[white grotto]

topology/algebra question

B' is the closed ball in C, center ak,k and radius the sum on the right

why are eigenvalues of a matrix A contained in Dk

I have no clue, this seems out of the blue

[chilly ocean]

Killing forms

[chilly ocean]

what's a proper value of a matrix?

[white grotto]

eigenvalue

sry🤦‍♂️ i used the french name

[chilly ocean]

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

[white grotto]

niiice

thank you

[chilly ocean]

np

[white grotto]

any clue why it makes sense

nvm i think the wiki article explains it

[white grotto]

another question... if i show you this equality between endomorphisms, is there some property you can deduce quickly

with alpha a number

[chilly ocean]

that if I take G(x) = x o g - g o x then G^n(f) = alpha^n f ?

[white grotto]

i did not get the discussion part

[chilly ocean]

left side is [f, g], right side is alpha f and G(x) = [x, g]

[white grotto]

why do they have small norms... we have no condition on the norms

[chilly ocean]

they don't in general

but if they do, then the Gershgorin circle theorem tells you that the eigenvalues are close to the diagonals

[white grotto]

Ah they said if ok

what axes?

[white grotto]

how did you deduce this?

[white grotto]

and wdym by [f,g]

[chilly ocean]

by induction

[chilly ocean]

commutator

so basically, this G is a linear transformation in the vector space of endomorphisms

and your equality tells us that f is an eigenvector of G with eigenvalue alpha

[iron vessel]

Does anyone have an idea for the first part? I was thinking about inducing a representation from the subgroup H = <r> to Dn. Although I'm a little confused about how I should think about this basis.

[white grotto]

[fg] is gfg^-1f^-1

no?

[chilly ocean]

wel... not really

you're just saying that because you are thinking about groups

and commutator is a more general concept

[white grotto]

whats commutator for a function

[white grotto]

Ive never heard of it i just wikied it rn

[chilly ocean]

it's more a commutator for a ring, you know

the ring of endomorphisms of a vector space

or more precisely an algebra

[white grotto]

Ah i see

[chilly ocean]

anyway, if you think about this in terms of matrices A, B

[white grotto]

[a,b]= ab-ba

[chilly ocean]

then commutator [A, B] = AB - BA

[white grotto]

and whats it significance

[chilly ocean]

and it measures the commutativity of A and B

as the name suggests

[white grotto]

how :p ... ok if its 0 then theyre commutative... otherwise do they indicate something?

[chilly ocean]

space of matrices with [A, B] forms a Lie algebra so that might be somehow significant

[chilly ocean]

not really, but think as if they do

[rustic crown]

I would suggest to do it directly. just show that the images of r and s in GL(V) satisfy the usual relations, R^n = S^2 = (RS)^2 = id

[chilly ocean]

that's what I do

[white grotto]

how did you see this quickly?

or did you see it before

[chilly ocean]

I didn't see it before, I was improvising

because I didn't know what to tell you

[white grotto]

lool well hoowww

youre a genius

[iron vessel]

hmm ok ill give it a try.

[white grotto]

ok so what do they mean by axes?

[chilly ocean]

I don't like when people call me a genius after I haven't done literally anything

but out of politeness.. thanks I guess

[chilly ocean]

It's kind of vague and I wouldn't sink into what they're saying.
They're just stating a point that one disk can have multiple eigenvalues

[white grotto]

fine... its coming more in context tho not for this question only

[iron vessel]

How do I even find the images of r and s?

[rustic crown]

they're given to you

i just denoted it by R as pi_e(r) was too much to type :p

that's what generators and relations are good for. you can very easily construct maps out of groups defined via them

[iron vessel]

yeah I'm not sure I understand too well how I would obtain the image from R v{m }= v{m+1}

and the other equation given

[rustic crown]

so you need to check that R^n is the identity map on V

this can be done by checking that on basis elements

where you could use the definition of R on v_{k}

does this make sense?

[iron vessel]

Yeah it does

[rustic crown]

R is the image of r under the group homomorphism

recall that constructing a group homomorphism <X|R> --> G only requires you to find a set function X --> G such that the relations are satisfied

[iron vessel]

this is what I had forgotten

[white grotto]

how does the second row cover both a and b?

[white grotto]

.

if a and b are eigen values

this is the example in the wiki page

[iron vessel]

I managed to show what you said, thanks alot!

[rustic crown]

[iron vessel]

How does uniqueness follow tho?

[rustic crown]

it's because D_n is generated by r and s

if you tell where r and s go, image of everything is determined!

[iron vessel]

ahhh and thus any other such ones will give the same homom

understood. I need to freshen up my group theory

[sweet echo]

Am I being stupid, how is V isomorphic to $V \oplus V$? They have different dimensions no?

[cloud walrus]

ΣAC

[rustic crown]

2 * infinity = infinity

🙈

[sweet echo]

are they taking countable to mean not finite?

[rustic crown]

yep

ah so do you define countable as finite or countably infinite?

[sweet echo]

right that clears things up, i was like uh 1+1 is not 1

yeah my courses have usually said finite sets are countable

[rustic crown]

yea, the book should have just wrote countably infinite

[sweet echo]

whats the obvious map here? v goes to (v,v)? but this isnt surjective surely

[rustic crown]

there is no "obvious" map

you have to choose basis and then define the map

[sweet echo]

ah so can i send a basis of V "diagonally" to V+V?

[rustic crown]

pick a basis (v_0, v_1, ...) and define the map
v_n --> (v_{n/2}, 0) when n even
v_n --> (0, v_{n-1/2}) when n odd

[rustic crown]

diagonal won't be enough, as that map isn't surjective

[rustic crown]

here by "obvious" i meant a natural map

[sweet echo]

i meant as in like cantor diagonal argument

[rustic crown]

you would do that for to get a bjection between N^2 and N

and not N and {0, 1} x N

but yea, you could do that :p

[sweet echo]

oh yeah lol you wouldnt have a grid just two perpendicular lines

of basis elements

[sweet echo]

and so it would just be this

got it thank you

[maiden heath]

I have two versions of the density theorem here. My question is why does the first one only state surjective, and the second one has isomorphism. I'm not sure what extra condition makes the second one an isomorphism

[rustic crown]

first one is for more general algebras

second is only for the group algebra C[G]

[maiden heath]

Ohhh, ok thanks. Is it easy to see why the additional condition for group algebra C[G] implies isomorphism?

[iron vessel]

Any chance you have a hint I could use for part (b) of the same question?

My idea was to show the same relations hold for the induced representation. However I haven't found a way to.

[lavish spoke]

another q - I want to calculate the height of (T, 2) in Z_(2)[T], I guess you just need to argue that every prime ideal strictly contained in (T, 2) is principal (then principal ideals are of height 1 since we're in noetherian ring that's an integral domain) but it's not coming to me why that's true

we obviously don't have anything like bezout which was my first thought

[rustic crown]

sorry i was in a lecture

Let G = D_n and H = <s>

so you have the one dimensional representation. Then this induces an [G:H] dimensional representation

[cloud walrus]

det

[rustic crown]

so you already know that this is a representation of G, no need to check that the relations are satisfied

to show that this is isomorphic to the old representation, find a careful basis of this such that the action of G matches exactly like above!

[vocal patrol]

can someone help me with the injection proof?

I dont quite understand what they want with injection from S -> S

Ive only had like injection from a set on the natural numbers before for example

[chilly ocean]

If f(z_1) = f(z_2), then z_1 = z_2.

Show this for all z_1 and z_2 in S.

That's what it means to be an injection.

[vocal patrol]

ahh okay

[formal ermine]

@vocal patrol and not expressed in notation: a function is injective when every output only ever is mapped to by at most 1 input. i.e. two inputs can't map to the same output

[formal ermine]

any hint on how I would do this? \
\
Let $R$ be euclidean and $x, y \in R \setminus \Set{0}$. Let $z_0 = x, z_1 = y$ and $$z_{j + 1} = \begin{cases}z_{j - 1} \text{ mod } z_j, &\text{if } z_j \neq 0 \ 0, &\text{if } z_j = 0\end{cases}$$. Show that an $n \in \bN_0$ exists such that $z_{n + 1} = 0$. Show that $z_n = \operatorname{gcd}(x, y)$ if $n$ is minimal.

[cloud walrus]

yes yes yes no

[rustic crown]

first, for general euclidean domain the notation a mod b makes no sense

for a notation to make sense, it should represent something and only one thing

here a mod b represents a whole class of elements

but i'm sure you want a particular element which is small with respect to the euclidean valuation

but the problem is, in general this is not unique!

[formal ermine]

it's on my homework sheet

[rustic crown]

even for Z it is not unique, but we usually fix that be demanding it to be non-negative

[rustic crown]

okie, then how did they define it?

[formal ermine]

they didn't lol

[rustic crown]

.<

[formal ermine]

but I suppose it's modulo as in programming

fwiw

[rustic crown]

yea i understand what you're trying to say

but i feel one should be careful about what they write

like when we do integers we can say 100 mod 11 = 1 or we can say it is -10

no one stops us

but anyway, that's just a rant :p

the euclidean algorithm doesn't really depend on the choice

and it will give you the gcd (upto a unit) whatever the choices you make

[formal ermine]

oh is that a different version of the euclidean algorithm

[rustic crown]

for Z, we usually insist the remainders to be non-negative

but other than that, it's cool

for example, if you don't want that assumption, then you can often finish the algo faster

98 = 9 * 11 + (-1)
11 = (-11) * (-1) + 0

so gcd(98, 11) = -1

as compared to
98 = 8 * 11 + 10
11 = 1 * 10 + 1
10 = 10 * 1 + 0

in either case we get that gcd up to units is 1

[formal ermine]

okay so for the first thing we needa show can I just show that $z_{j + 1} < z_{j}$?

[cloud walrus]

yes yes yes no

[formal ermine]

(unless z_j = 0 of course)

[rustic crown]

< doesn't make sense in general euclidean domain

[formal ermine]

oh

[rustic crown]

you need to convert to N where < does make senes

for Z we use the function |.|

[formal ermine]

ah

degree function

euclidean function

[rustic crown]

and polynomials k[x] we use degree

[formal ermine]

whatever it's called

[rustic crown]

ig its called euclidean valuation

[formal ermine]

d(z_{j +1}) < d(z_j) ?

[rustic crown]

yep!

more precisely z_{j+1} = 0 or that

some people define d(0) = -infinity to avoid saying this

so anyway, after each step the size of the remainder reduces

and since size is a natural number it can't reduce forever

this would show that the procedure ends

now you need to show that it produces the gcd at the end

give that a thought, else i'll spoil you :3

[formal ermine]

only thing I could think of that it's literally the euclidean algorithm which we've shown to give out the gcd

like we have $x_{n - 2} = q_{n - 1}x_{n - 1} + x_n$ for the euclidean algorithm

[cloud walrus]

yes yes yes no

[formal ermine]

if we take that mod $x_{n-1}$ we get

[cloud walrus]

yes yes yes no

[formal ermine]

$x_{n-2} \text{ mod } x_{n - 1} = x_n$

[cloud walrus]

yes yes yes no

[formal ermine]

we don't needa mod the right side because d(x_n) < d(x_{n - 1})

[rustic crown]

pwease no use the notation "a mod b = c" unless you define it >.<

[formal ermine]

$x_{n-2} \equiv x_n \pmod{x_{n - 1}}$

[cloud walrus]

yes yes yes no

[formal ermine]

yes

[rustic crown]

better :3

[formal ermine]

and that's the recursive function w/e we're dealing with

[rustic crown]

we just showed that this algorithm will end

but we will have to show it's gives out the gcd of the original two things

[formal ermine]

I mean like

we've shown that it's "equivalent" to the euclidean algorithm (at least the one we introduced in the lecture)

we also showed that it terminates in the lecture (gotta read up on the proof again later)

so can I just use those two facts?

[rustic crown]

i didn't get you exactly

are you saying you've already proven this?

[formal ermine]

yeah we've already proven that the euclidean algorithm terminates a couple of weeks ago

I mean

[rustic crown]

but did you show that the last non-zero remainder is the gcd

[formal ermine]

no not terminates

I mean

goes to the gcd

yes

[rustic crown]

ah, then nice

[formal ermine]

epic

thanks for the help!

[rotund aurora]

Over a field F of char p, if the polynomial P(x)=x^p-a has no roots, for some a in F, then it is irreducible right? This polynomial is purely inseparable, so it splits as (x-z)...(x-z) where z is the pth root of a. Say P(x)=f(x)g(x) with deg f=n and deg g=m, f,g polynomials over F. Then z^n and z^m are both in F, with n+m=p and so gcd(n,m)=1, so by Bezout, we can write xn+ym=1, from where it follows that z is in F

sorry this was very sketchy and even reused variable names lmao

[rustic crown]

wait why gcd(n,m)=1 and why does that imply z is in F?

also i don't immediately see why z^n and z^m are in F

[rotund aurora]

n+m=p, so gcd(n,m)=1 or p. If its equal to p then n=0 and m=p, so the factorization is trivial

[rotund aurora]

Vieta

like (x-z)...(x-z) n times is in F[x]. Multiply everything out

[rustic crown]

ah okie

[rotund aurora]

if (n,m)=1 then there exist integers x and y such that xn+ym=1, and since z^n and z^m are in F, z^{xn}z^{ym}=z^{xn+ym}=z is in F too

[rustic crown]

hmm makes sense

[rotund aurora]

[rustic crown]

but ig you can also give a slightly shorter proof by looking at the coefficient of x^{n-1} instead of the constant term

[rotund aurora]

its easier if you focus on the constant term

[rustic crown]

say P(x) = f(x) * g(x) where f and g are monic

then f(x) = (x-z)^n

and this should be in F[x]

[rotund aurora]

i think its the sum of the roots

no?

[rustic crown]

this means the coefficient of x^{n-1} which is nz is in F

and 0 < n <p so n is invertible in F

which means z is in F

[rustic crown]

ig oen benefit is that i don't have to worry about reducing the power back to 1

but i have to clear out the coefficient instead

[rotund aurora]

yeah yours is easier

idk I just thought about the constant term

its always fun to apply Bezout tho

[rustic crown]

yea it was a nice argument

[rotund aurora]

[formal ermine]

is $31 - 2i \text{ mod } 6 + 8i = 1 + 8i$? (det pls don't yell at me for using that weird mod notation )

[cloud walrus]

yes yes yes no

[rotund aurora]

lool emotes in latex go crazy

tbh idk

just do euclidean algorithm

[formal ermine]

that's what I'm trying to do

euclidean algorithm in the gaussian integers

[rotund aurora]

I think you can check this stuff in wolframalpha or something similar tho

[formal ermine]

couldn't find it anywhere

[rotund aurora]

write a program

if you can

[formal ermine]

I found the formula $a + bi \text{ mod } c + di = a + bi + (c + di)\ceil{-\frac{a + bi}{c + di}}$

[cloud walrus]

yes yes yes no

[formal ermine]

but apparently it doesn't work 100% of the time or something?

[rotund aurora]

then you will also learn programming meanwhile and you will make sure you know every logical step

[formal ermine]

I already know programming

[rotund aurora]

so write it out, it doesn't take that much

it should be the lowest integer

wait idk what you just wrote

[rustic crown]

just do the same thing like you would do for integers

to find a mod b

[formal ermine]

for integers I would just repeatedly subtract b from a until I can't no more

[rustic crown]

you will compute a/b and then approximate it with an integer, say q and then compute a - bq = r

here you have two gaussian integers, which i'll call alpha and beta

so you wanna compute alpha/beta

this is same as (alpha * beta conjugate)/|beta|^2

now approximate this to the closest gaussian integer q, and compute alpha - beta * q

try computing (31-2i)/(6+8i) for like in your example

[wooden ember]

just to make sure im not making an oopsy (my proof seems legit but im not trusting myself right now), in a UFD a prime ideal is of height 1 iff it is principal right?

[rustic crown]

hewwo @wooden ember

[wooden ember]

hellu

[restive birch]

thought i already did this, but i cant remember how

[agile burrow]

You need the ideal to be nonzero, but yes

[wooden ember]

right ofc

k thanks @agile burrow

[rustic crown]

is this true in general btw?

[wooden ember]

naw

[rustic crown]

that a non-zero prime ideal contains a non-zero prime element

[wooden ember]

you can have principal ideals that arent of height one

lemme think of an example if im not being stupid

[rustic crown]

ig i assume that factorizations exists, then every non-zero prime ideal will contain irreducibles

but they may not be prime

[wooden ember]

wait im being dumb

isnt there krull's ideal theorem or whatever

[rustic crown]

me no know commie algebruh >.<

[wooden ember]

which in particular says a principal ideal must be of height 1 in a domain

i dont know the theorem either but ive heard it pop up a few times

[restive birch]

just to be clear, phi is a homomorphism here

[rustic crown]

[rustic crown]

yee this is true

you need to prove that phi(g)^-1 = phi(g^-1)

which is same as showing phi(g) * phi(g^-1) = 1

[restive birch]

ahhh thats what i was missing

thanks

[rotund aurora]

Does the group $\textup{Aut}(\overline\bQ/\bQ)$ have a name?

[cloud walrus]

Croqueta

[next obsidian]

Lmfao

That’s the absolute Galois group of Q

It’s like, the central object of study in NT

[rotund aurora]

yeah ik

[coral spindle]

And we know very little about it

[next obsidian]

Maybe you

Not me tho

[coral spindle]

🧠

I mean generally ofc

[rotund aurora]

but like i just didnt know the name lol and tried searching for a wikipedia entry and didnt succeed

[rustic crown]

hewwo

[rotund aurora]

Let 5^{1/7}=x and 7^{1/5}=y. To show Q(x,y)=Q(x+y) isnt the following argument sufficient? Consider the tower Q<Q(x)<Q(x+y)<Q(x,y). The full extension has degree 35, [Q(x):Q]=7, and since Q(x)!= Q(x+y), if Q(x+y)!= Q(x,y), Q(x+y) will lie strictly in between Q(x) and Q(x,y), but [Q(x,y): Q(x)]=5 is prime, so that's impossible (by multipilcativity of the degree again).

is this argument fine?

[south patrol]

Maybe I'm missing something, but isn't showing that e.g. Q(x,y)/Q is of degree 35 and that Q(x+y) isn't Q(x,) harder than just explicitly showing that x and y are in Q(x+y)?

[rotund aurora]

but if I assume that the argument is correct, right?

I dont know how you would explicitly show that anyway, if you mean by that an expression involving x+y that will equal x or y

[south patrol]

Yes, sure, it seems fine otherwise

owo

[rotund aurora]

uhm maybe its not that bad actually

[rustic crown]

uwu

[south patrol]

UwU

[rustic crown]

[coral shale]

0ᆺ0

[solar inlet]

OwO

[rustic crown]

^.^

[coral shale]

^oᆺo^

[rustic crown]

^w^

[south patrol]

OwO

[rustic crown]

.<

[rotund aurora]

where can I find lang algebra pdf with bookmarks?

[chilly ocean]

You could bookmark it yourself. It doesn't take that long.

[rotund aurora]

I did bookmark the first part like some time ago when I was bored

but it does take time lmao

or i don't know how to do it then

i cant believe theres not a circulating pdf of lang with bookmarks tbh

[chilly ocean]

Is the definition of the free group functor such that F(id_X):=id_FX or can this be shown?

[rustic crown]

you need to prove that

to be a functor you need send objects to objects and morphisms to morhpism such that composition and identity are preserved

so if you're claiming that taking free groups forms a functor, you have to show all that

[chilly ocean]

So to show that is suffices to show that it makes this diagram commute.

But it's unclear to me why F(id_X) couldn't map a generator x_i to a different one x_j

[rustic crown]

the commutativity ensures x_k maps to x_k

[chilly ocean]

I am trying to show it commutes

[rustic crown]

okie, so first lemme ask, what does the functor do on morphisms?

say you have a function f : X --> Y

what's Ff : FX --> FY?

[chilly ocean]

I guess you would want to define this to be the map induced by f.

[rustic crown]

and what is the map induced by f?

[chilly ocean]

The map that linearly extends f to words on X

[rustic crown]

"linearly" is a weird word :p

[chilly ocean]

I think so as well, but I see a lot of people use it

[rustic crown]

for group, i would assume something like "homomorphically" extend, but idk

anyway

[chilly ocean]

So I guess it's a consequence of that that the identity maps to the identity

[rustic crown]

the point is, by the universal property of free groups, to define a map from FX --> G, you just need to give a set function X --> G

and like you say, the map FX --> FY is obtained by extending the map X --> Y --> FY

to get the square

[rustic crown]

yee

[rustic crown]

so what i was saying was, that by definition of F on morphism, this diagram is forced to commute

[chilly ocean]

I understand, I was being stupid and not thinking of what the definition is for mapping general functions between sets.

[south patrol]

Well I think one way to do it is to relate the min poly of y over K with the min poly of y over L [possibly after passing to a Galois closure of M]

(at least that's what I did when this was on my homework as well xd)

[chilly ocean]

Is it true that the only representable G-action is the action of G on itself by left multiplication ?

[wind steeple]

you can define the separability degree of an extension L/K [L:K]sep as the number of ways you can extend the inclusion K→Ω (where Ω is an algebraically closed field containing K) to L. You can check that [L:K]sep = [L:K] iif L/K is separable and [L:K]sep is multiplicative

[wind steeple]

by representable do you mean that you are viewing a G-set as a functor 𝓑G→Set ?

you have only one representable functor

[chilly ocean]

If BG is the group as a category, then yes

[wind steeple]

yes

so you have only one representable functor (spanned by the point of 𝓑G)

which is indeed the right G action on G

but you have many representable functors

a representable functor here would be a G-set isomorphic to the right G-action on G

I believe that these G-sets are exactly G-torsor (it means that they are satisfying the property that the action of G is free and transitive)

what

doing what

[south patrol]

this is a common characterisation of separability basically

Yes

[wind steeple]

yes, this is a strategy of proof

your idea using a primitive element might work as well

[south patrol]

I think you don't need primitive element theorem though like

[wind steeple]

(sorry here I meant one functor represented)

[south patrol]

If you can show y is separable over K then you can show any element in M is separable over K

[wind steeple]

yes, the strategy I suggest doesnt use primitive elements

[south patrol]

Indeed

Mine didn't either

[wind steeple]

oh ok

[south patrol]

Yours would be my preferred way too XD

Tbh I am now scared there is a flaw w my argument lol

Tbf I think it is okay

[south patrol]

okay so yes it turns out the question I had required a specific type of field extension lol

well it assumed the smaller extension was galois

not sure why lol

it does :(

i think the key thing is basically given an irreducible polynomial in K[x] then like you can like turn it into an irreducible poly in F[x] by making it stable under the action of Gal(K/F) lol

but obviously if you don't have a galois group that doesn't work

anyway you made me realise an error i made w my hw (basically a step i didn't justify which turned out to be fine but needed justification lol) so i have just corrected my submission XD

[smoky ivy]

Do you want to show it only for finite extensions? It also works if we don't assume so

Do you know anything about the degree of separability?

If not, I would personally use the primitive element theorem

$L = K(\alpha), ; M = L(\beta) = K(\alpha)(\beta)$ for some $\alpha \in L, ; \beta \in M$. We get $\operatorname{irr}(\beta, L)$ divides $\operatorname{irr}(\beta,K)$ and you know that $\operatorname{irr}(\beta, L)$ is separable by assumption. Suppose that $\operatorname{irr}(\beta, K)$ is not separable, then the characteristic of $K$ must be greater than zero since every irreducible polynomial over a field with characteristic 0 is separable. Now being inseparable is equivalent to: $\operatorname{irr}(\beta, K) = f(x^p)$ for some irreducible $f$ and the divisibility from before implies that $\operatorname{irr}(\beta, L) = g(x^p)$ for irreducible $g$ and the characteristic for $L$ is also $>0$, so again $\operatorname{irr}(\beta, L)$ inseparable, which is a contradiction, so $\beta$ is separable over $K$

[cloud walrus]

lewis

[smoky ivy]

oh

that's the minimal polynomial

prob had different notation

If you didn't have the theroem I used yet, it is quite easy to show:
f irreducible over a field K: if char(K) = 0, then f is separable and for char(K) > 0: you have the equivalence that f is not separable iff f' = 0 iff f = g(x^p) for some g

[rotund aurora]

Say K<L=K(a), then [L:K]sep is very explocitly counting the number of distinct roots the minimal polynomial of a has

Embeddings and their extensions are a recurrent theme in gield theory

And the degree of the minimal polynomial of a (over K) is the degree of the extension

[tender bough]

I'm reading this:
https://en.wikipedia.org/wiki/Root_system

doesn't 2 entail 3?

oh I see..

[unique valve]

Hey I’ve been trying to teach myself Galois Theory, does anyone have any advice on how to approach this?

[formal ermine]

how can a polynomial be irreducible in $\bQ[x]$ but reducible in $\bZ[x]$? like, if it's irreducible $\implies$ its roots lie outside of $\bQ$, but since $\bZ \subset \bQ$ they'll also lie outside of $\bZ$, no?

[cloud walrus]

yes yes yes no

[quiet pelican]

It can’t be?

[formal ermine]

my homework wants me to name one lol

[pliant raptor]

Roots lying outside of $\mathbb{Q}$ only means that the polynomials doesn't have linear factors in $\mathbb{Q}$

[cloud walrus]

ImHackingXD

[quiet pelican]

Any factorisation of a polynomial in Z[X] is also a factorisation in Q[X] so reducible in Z[X] => reducible in Q[X]

[pliant raptor]

No

[quiet pelican]

Oh wait

[formal ermine]

yeah and since Z subset Q it doesn't have linear factors in Z either, no?

[quiet pelican]

||2(x - 1)||

[pliant raptor]

Ye

[quiet pelican]

||Factors in Z[X], but 2 is a unit in Q[X]||

[pliant raptor]

I know that $x^5+x^2+1$ is irreducible in $\mathbb{F}_2 [x]$, so $\mathbb{F}_2 [x]/x^5+x^2+1$ is a field. Can it be written as a direct sum of fields?

[cloud walrus]

ImHackingXD

[pliant raptor]

I don't know if the fundamental theorem of finite abelian groups could be applied here

[chilly ocean]

I don't think there is a direct sum of fields...what definition are you using?

[pliant raptor]

I'm not saying there is, I'm asking if there is or if not if there is a reason why it can't be written like that

[pliant raptor]

Definition of which part exactly?

[chilly ocean]

direct sum of fields

[chilly ocean]

A better way to say this is that that the direct sum of fields is not a field.

[pliant raptor]

I guess the usual, the field can be written uniquely as a sum two elements of other fields

[chilly ocean]

If you do the usual Cartesian product with the operation defined pointwise, then it won't be a field since (0,1)(1,0)=(0,0), and thus not even a domain.

[chilly ocean]

Oh but what you said is different

[pliant raptor]

Ok but, how do I know there are no two specific fields such that, somehow, their direct sum becomes $\mathbb{F}_2 [x]/x^5+x^2+1$?

[pliant raptor]

?

[chilly ocean]

OK never mind, maybe you meant what I described

[chilly ocean]

I'm not sure I understand your question here.

[cloud walrus]

ImHackingXD

[chilly ocean]

If I'm understanding you correctly, what you wrote is a field so it can't be the direct sum of two fields.

[pliant raptor]

Ohhhhh I get it

okok

makes sense

[chilly ocean]

Or even more simply, it's a domain, but the direct sum of two fields is never a domain so that can't be the direct sum of two fields.

[pliant raptor]

I was thinking way too hard because of the specific example I had

Thank you

[mighty spade]

go through chapters 5 and 6 of lang, and do all the problems

berkeley's math 250 has problem sets for you as well, if you want a specific order

that's how i did it, and that implies it's the Best (TM)

[rotund aurora]

What are good texts that focus specifically on multilinear algebra?

[lethal dune]

can I get some help proving this?

or any reference to read it up

[next obsidian]

Based logic

[next obsidian]

What’s L_i?

Is this the left-derived functor?

[lethal dune]

yes

[next obsidian]

I don’t think it’s true?

You need like one of F or U (can’t remember which rn) to like

Send projectives to projectives

Assuming that like you basically see that applying U sends a projective resolution of any object A to a projective resolution of U(A)

Oh nvm

This is fine

So I think this falls out of a Groth Spectral sequence for one

But you can just do it by hand

So fix a projective resolution P^• of an object A

Then the left-hand side is U(H^n(F(P^•)))

But note that an exact functor commutes with homology

So this is the same as H^n(UF(P^•))

[lethal dune]

it does?

[next obsidian]

But this computes L_i(UF)(A)

Yes

Write down a SES

0 -> A -> B -> B/A -> 0

Apply an exact functor F

Then you know that
0 -> FA -> FB -> F(B/A) -> 0 is exact

But this identifies F(B/A) with the cokernel of FA -> FB which is FB/FA

[lethal dune]

but ker(FB->F(B/A)) = F(ker(B->B/A)) due to exactness

ok got it

[marsh basin]

For $G$ a group, $H \subseteq G $ and $ K \triangleleft G$ subgroups, is it an abuse of notation to write $H/K$ for $\phi(H)$ with $\phi : G \to G/K$ the canonical epimorphism?

[cloud walrus]

...........

[marsh basin]

Because something like $1/K$ seems funky

[cloud walrus]

...........

[next obsidian]

This is standard notation

1/K doesn’t make sense here because you want H > K

Oh I see

You want H > K

[marsh basin]

Cool thanks

[formal ermine]

asking this again because I didn't quite understand it last time

how do I do modulo in the gaussian integers?

e.g. $(31 - 2i) \text{ mod } (6 + 8i)$

[cloud walrus]

yes yes yes no

[formal ermine]

so

I have to find the smallest $s \in \bZ[i]$ such that $\frac{(32 - 2i) - s}{6 + 8i}$ is also in $\bZ[i]$?

[cloud walrus]

yes yes yes no

[formal ermine]

if that's the case, then this program should do the trick, no?```py
num = complex(input("number: "))
modulo = complex(input("modulo: "))
upto = 1000

for rp in range(upto):
for ip in range(upto):
next = complex(rp, ip)
divided = ((num - next) * modulo.conjugate()) / (modulo * modulo.conjugate())

    if divided.real.is_integer() and divided.imag.is_integer():
        print(next)
        exit()

nvm it gets stuck in an infite loop between 25i and 1+8i

[lethal dune]

Complex division is available in python you don’t have to do conjunction

[formal ermine]

epic

I'm getting a different result now lol

[lethal dune]

F

[formal ermine]

well

my original one was totally broken

so let me try this one

[deft gale]

division algorithm

is a good coding exercise as always

[formal ermine]

I'd say I'm fairly decent at coding already

NOOOOOOO

IT'S STUCK IN AN INFINITE LOOP AGAIN

[deft gale]

do you know euclidean algorithm?

[formal ermine]

that's what I'm trying to do right now

[lethal dune]

Also

[formal ermine]

to find the gcd of two gaussian integers

[deft gale]

yessir

[formal ermine]

wait

shit

[lethal dune]

/ gives you float value

[formal ermine]

it's 31

nvm

still stuck in an infinite loop nonetheless

[lethal dune]

Why are you only considering positive ones?

One positive one negative could also work

[formal ermine]

yea

-5 + 0j

I'm sorting based on magnitude now

[lethal dune]

Also I’m tired otherwise I would have tried

[formal ermine]

YES

IT WORKS

FINALLY

[formal ermine]

thanks

this was the issue

I've been struggling with this for the past 2 hours

[lethal dune]

Just part of coding

[warm wyvern]

7.7

Is that true if G isn't commutative?

[next obsidian]

Yah

[chilly ocean]

if G is commutative, then this is trivial, no

[wooden ember]

Not much of a question when G is abelian since then all subgroups are normal

[chilly ocean]

any subgroup is normal

[wooden ember]

Sniped

[warm wyvern]

Yeah, i was asking if it was true when G isn't commutative

[wooden ember]

I mean if the question is asking you to prove it

But yeah fair enough

[chilly ocean]

it's true and the proof is pretty simple

I could give you a hint

[warm wyvern]

Gimme a minute

[chilly ocean]

Hint: ||If H is a subgroup generated by x_i, then H is normal iff g^-1 x_i g is in H for each i and g in G||

no need to, it's in spoilers

[next obsidian]

It’s “easily” seen to be a characteristic subgroup

Which means it’s normal

[chilly ocean]

what's a characteristic subgroup again. Invariant under any automorphisms?

for that, what I said still works tbh. Just replace inner automorphism with arbitrary one

[next obsidian]

Its description tells you it’s characteristic

[warm wyvern]

Oh, conjugation shuffles the generators

Or, wait

What's it called to multiply some element with another element on the right and its inverse on the left? lol

[chilly ocean]

conjugation

[warm wyvern]

ok, I'm not bludgeoning any terminology then lol

But my idea is that if g has order n then xgx^-1 also has order n

So xHx^-1 has the same generators as H which means they must be the same

[chilly ocean]

yeah, good direction

[chilly ocean]

well, if you have to, you can just look at what H looks like in terms of its generators

and that conjugating has the property that product of elements conjugated by the same element individually, is a product of those elements conjugated as this one big element

so basically, that it is a homomorphism

that's why Chmonkey mentioned characteristic subgroups, maybe it's a bit clearer to look at this in terms of automorphisms in general

am I confusing you?

[warm wyvern]

One second, I'm just parsing that lel

Oh, so a conjugation is an automorphism

And H is a characteristic subgroup

How do we know it is tho?

[chilly ocean]

Yep. The map which associates to an element its conjugate is called an inner automorphism.

a special kind of automorphism

[chilly ocean]

well, $h\in H$ is of the form $h = x_1^{n_1}\cdot ...\cdot x_k^{n_k}$ where $n_i\in\mathbb{Z}$ and $x_i$ have order $n$

[cloud walrus]

Blitz

Blitz

[warm wyvern]

Oh, yeah

Lol

That's makes sense

Thanks

[wooden ember]

Is there a non commutative artinian ring which is noetherian?

Just had an exercise showing artinian implies noetherian in the commutative setting

[south patrol]

i assume you mean not noetherian?

[wooden ember]

(Or a weaker question, we can allow the ring to be left artinian but not left noetherian, so say right noetherian)

[wooden ember]

Oh yeah mb

[south patrol]

so apparently a right artinian ring is right noetherian and similarly for left

[wooden ember]

Really huh

I assume the proof is more complicated than in the commutative setting

[coral spindle]

It is fairly tricky, but I forget how it goes. I may come back with a sketch proof in a bit.

[pliant raptor]

I would like to prove that the kernel of the norm map in finite fields $N(a)=\prod_{j=0}^{m-1} \sigma^j(a), \forall a\in \mathbb{F}{q^m}^{\times}$ is of the form $\left{\frac{x}{\sigma(x)}: x \in \mathbb{F}{q^m}^{\times}\right}$. I have already shown that all the elements of this set are contained in the kernel, and that the kernel has $\frac{q^m-1}{q-1}$ elements. How could I show that the set $\left{\frac{x}{\sigma(x)}: x \in \mathbb{F}_{q^m}^{\times}\right}$ also has $\frac{q^m-1}{q-1}$ elements?

[cloud walrus]

ImHackingXD

[coral spindle]

A brief summary of the proof. We have a canonical maximal nilpotent ideal called N(R) of any artinian ring. Then R/N(R) is semisimple and has very nice structure, in particular is noetherian. By considering the layers R/N, N/N^2, etc (there are only finitely many) we can argue that the whole thing is Noetherian.

N(R) is called the nilradical if you want to look it up

[wooden ember]

Yeah I know the nilradical

That doesn’t sound too bad but are you sure this works for the non commutative case?

[coral spindle]

Yes.

[wooden ember]

Alright neato

Thanks

[next obsidian]

This is in-spirit the same as the commutative case

[coral spindle]

You just have to fiddle with the nilradical and how it works

[wooden ember]

Yeah the proof we had to do was prove the nilradical is a nilpotent ideal, say the ring has finitely maximal ideals then use that to show it’s of finite length as a module over itself

Still fiddling around with the last part

[south patrol]

Classic devissage

Jk

[next obsidian]

Ur too young to know this word

[south patrol]

Pretty sure this isn't even a devissage

But smh my prof mentioned devissages last year

[next obsidian]

I mean

How could you ever hear of devissage? In how I know devisssage it’s a technique when you’re proving stuff about coherent sheaves lol

[latent island]

i dont get this, namely the x * y in (xN)(yN) and the line below that
why does that need to happen for the cosets to act like a group
also how is that equivalent to (xN)(yN) = xyN

[unique valve]

You want to make sure that your binary operation in the group is unaffected by cosets

So the cosets xN and yN need to multiply to another coset

Naturally this should contain x and y which are in the two original cosets, which multiply to the coset with xy

[latent island]

sry wdym unaffected by cosets

[unique valve]

If you multiply two elements in coset A and coset B

You want to get something always in the same coset

x is in coset xN and y is in coset yN

xy should be in the product of the cosets

[latent island]

hm

i think i got it now

by relating it to modular arithmetic

thanks

ok next

whats S3

[chilly ocean]

A_n is a (nontrivial) normal subgroup of S_n

[next obsidian]

[south patrol]

idk

[latent island]

whats A_n

[chilly ocean]

the subgroup of even permutations

[wooden ember]

To argue this, I’m looking at a descending product of maximal ideals that eventually reaches 0. To conclude that this is a composition series can I say that a successive quotient is a R/m vector space for some maximal ideal of R (since it’ll annihilate the quotient) and this is artinian and northerian so of finite length

And thus I can extend my previous chain to be a composition series?

It seems correct to me but a hint given on the exercise is to use that the property of being artinian is inherited by sub and quotient modules, which I’m not using here

[next obsidian]

Yes you are

In each subquotient

You need that to be of finite length

Otherwise you’re pasting together things which aren’t finite

So you need to use that an Artinian vector space is finite dimensional

[wooden ember]

Ahhhh

Right my vector space could be infinite dim

Okay thanks

[next obsidian]

Yeah

[patent girder]

When looking at the Schur Orthogonality relations. I am a looking for a few clarifications on whats actually happening.

1. Is basically what happening is (i) all inequivalent irreducible unitary representations such that all their entries are orthogonal and (ii) for a given inequivalent irreducible unitary representation, the only non orthogonal entries are when they the same in which case the inner product is 1/n?

2. Would this imply that all irreducible unitary representations of degree d have an orthonormal set ${ k\phi_{ij} | 1\leq i,j \leq d}$. And it would make sense for k to be the square root of d since then since all equivalence classes can be defined by representatives $\phi^{(1)}, \ldots, \phi^{(n)}$ with orthonormal sets for each phi in L(G) which implies that the sum of dimensions is less than or equal to the order of G due to linear independence in L(G). So there are finitely many equivalence classes?

[cloud walrus]

ohNoiAmHere

[next obsidian]

The Galois conjugates of any x in L over F are a subset of the conjugates of x over K

Like, the roots of the min poly of x over F are a subset of the roots of the min poly of x over K

And by assumption all of the roots of the min poly over K live in L

Just ignore that

If you don’t know the term

[unique valve]

S3 is the group of permutations of 3 things

[next obsidian]

They showed that every root of the minimal polynomial of x over F lives in L

[smoky ivy]

This might be a really random trivial question, but I need this to show something way bigger: \
If we have a ring $R$ and a prime ideal $P \in R[x]$ and intersect $P \cap R$. Is this prime in $R$?

[cloud walrus]

lewis

[next obsidian]

Yes, go prove it, it’s very easy

[smoky ivy]

okay, nvm, i answered my own questioi

oh god...

For some reason, I couldn't prove it, using the definition of being prime

but you can just consider the embedding $R \to R[x]$ and the preimage of prime ideals are prime

[cloud walrus]

lewis

[next obsidian]

Lmfao

[smoky ivy]

i feel so dumb rn lmao

[next obsidian]

You really ought to be able to prove it directly

Like this is how I view it

But surely it isn’t that hard?

Like assume x,y in R such that xy in P\cap R

[smoky ivy]

I'm just super lost right now 😂 For some reason I had problems with showing that it is in R

[next obsidian]

Well then xy is in P itself

So by being prime in R[x] one of x or y in P

[smoky ivy]

Oh yikes...I realize now what I've done wrong

[next obsidian]

Now x and y are also in R, so whichever one is in P is in P\cap R

[smoky ivy]

well yeah...

I've been doing too much commutative algebra that I forgot how to do basic things like these

Thank you haha

[next obsidian]

Hahahaha np

[uncut girder]

Gal group of Splitting field

[hollow parrot]

How does one go about proving that $(\mathbb{Q}(\sqrt{2}))(\sqrt{3}) = \mathbb{Q}(\sqrt{2},\sqrt{3})$ without assuming one knows that $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is of the form ${a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}|a,b,c,d \in \mathbb{Q}}$?

[cloud walrus]

Skid

[next obsidian]

Defintion? Like no matter what your definitions are it should pop out near immediately

[hollow parrot]

We are using that Q(s) is the smallest subfield containing Q and element s which is isomorphic to Q[x]/((p)) where p is an irreducible polynomial with root in Q(s)

[next obsidian]

Subfield of what

[hollow parrot]

Well in this case it is R

[next obsidian]

And then how do you extend that to Q(sqrt(2),sqrt(3))

[hollow parrot]

I don't know

[next obsidian]

Well you have to figure out what the definition of that is that you’re using

[hollow parrot]

Our definition is only for a single element

Not a set of elements

[next obsidian]

Or you won’t be able to make any progress, idk what to say