groups-rings-fields

[molten viper]

I don't see why, really

[south patrol]

Hm

Okay so i should've answered lol

[molten viper]

like, x^2 is non constant rational function

[south patrol]

non-constant polynomials, yes

[molten viper]

so like, basically if f(u) = 0 I can show x is algebraic?

ie, contradiction?

[south patrol]

Yeah

[molten viper]

ok so, f(p/q) = p^n/q^n + ... + a1 u + a0 = 0, in our assumption

how do I find a polynomial for x in all this

I guess like, this is one big giant polynomial g, with g(x) = 0

and that's why if q is a unit we're done

I seeeeeee

so
q^n * f(u) = g(x) = 0

Ok I will accept that that's fine

[south patrol]

no, it's one big rational function

[molten viper]

so, one final question.
K subset F is finite, u in F
[K(u) : K] is odd
prove K(u) = K(u^2)

[molten viper]

now, I guess we have that for this, f(u) = 0, so say deg(u) = n, then we can say u^n = -f'(u) where deg(f') < n

Sort of as a "degree reduction" equation

and I can see why K(u^2) subset K(u)

but the other direction is tough

[next obsidian]

What values can [K(u):K(u^2)] take

[molten viper]

odd value

values*

[next obsidian]

Forget the setup and the rest of the problem for now

There’s a far more elementary answer just based on what’s written

[molten viper]

factors of [K(u):K]

or even just integer values?

[next obsidian]

How do you normally compute the degree of an extension

[molten viper]

the degree of the minimal polynomial of the generating element

[next obsidian]

And what could that be here?

[molten viper]

so it's the degree of u over K(u^2)

1 is certainly possible (and that's what we ant to prove)

It's somewhere between 1 and the degree of u over K

[next obsidian]

Can you write down a polynomial u satisfies

Over K(u^2)

[molten viper]

x^2 - u^2

[next obsidian]

So that’s an upper bound

[molten viper]

so it's 2 or 1

[next obsidian]

Except now

[molten viper]

and because it's odd, it has to be one

[next obsidian]

[molten viper]

I seeeee

Many thanks

[south patrol]

why does this question have K a subset of F

[primal tusk]

when people use the term "classify ____ up to isomorphism" is that a topology thing or an algebra thing

and does that just mean two things are the same other than their symbols representing them?

[vagrant zinc]

Hi guys, I'm back here again

[white oxide]

how exactly did they use the fact that multiplication of cosets by multiplication modulo n is well defined?

is it because they're in the same coset

moreoever how did they simplify the first congruence to the second congruence

number theory bad

moreover how do we know that if a is relatively prime to n then the coset a + nZ of nZ containing a contains an integer b < n and relatively prime to n

[frigid lark]

For this add kn for some integer k such that 0 < a + kn < n, there should always exist such a k as if not, a = n.

[frigid lark]

So if a and b are in the same equivalence class then a = b + kn for some integer k. Then a^phi(n) = (b + kn)^phi(n). Expanding the rhs with the binomial theorem yields, b^phi(n) + ns for some integer s, thus a^phi(n) is equivalent to b^phi(n)

That is another way to think about this

I think what they are saying is that a^phi(n) + nZ = (a + nZ)^phi(n) = (b + nZ)^phi(n) = b^phi(n) + nZ since coset multiplication (and thus exponentiation) is well defined

[spark crow]

I am trying to find an isomorphism that maps $D_4 \to C_2 \wr C_2$. I tried mapping $r \mapsto ((0,1),0)$ and $s \mapsto ((0,0),1)$ but i don't know how to proceed from here. I know that a similar mapping works for the semidirect product $D_4 \cong C_4 \rtimes C_2$ however I don't know how to calculate with the elements of the wreath product. I believe it should work like the semidirect product e.g $((a,b),c)((a',b'),c') = ((a,b)c(a',b')c^{-^},cc')$. I don't see how i can actually calculate $c(a',b')c^{-1}$ even though I know it should result in $(a',b')^{-1}$.
How do I proceede here?

[cloud walrus]

achilles199703

[pastel cliff]

is there an intuitive way to think about the abelianization of a group

[elder wave]

I mean you quotient out the commutator subgroup

So you artificially add the Relation ab=ba

[pastel cliff]

oh that's easy

whoopsy

[elder wave]

Alternatively you may view it as a functor which is what I was talking about yesterday

[south patrol]

Funky

[pastel cliff]

i'll skip the cat theory for now, that was what was confusing me yesterday lol

[elder wave]

:(

[pastel cliff]

was asking bc i have it next to my notes on SVT

but am confus

will ask proper qustions later

[elder wave]

Probably because of pi1 and H1

Wait this isn’t the topology channel

[pastel cliff]

i can move there

[elder wave]

Nono you’re good it was mb

[vagrant zinc]

Guys, how can I get all the subgroups out of Z module 24 quickly?

[lunar path]

can't we take the factors of 24?

[vagrant zinc]

Where can I read about this theorem or definition, I haven't found it in any book, neither as an example xde

[chilly ocean]

Z module 24?

Z/24Z ?

Well its a theorem ig

All subgroups of cyclic group are cyclic and subgroups of Z/nZ correspond to factors of n

[analog zephyr]

Z/24Z is isomorphic to Z_24. Just search for all integer less than 24 such that their gcd are diferent than 1

for example, gcd(6,24)=6 so 6 generates a group or order 4

In Z/24Z we have 6+24Z must generate a subgroup of order 4

[wooden ember]

Aight as usual I’m struggling with some pretty basic stuff but I’m not sure how to tell k[x]_(x) and k[t^2,t^3]_(t^2,t^3) apart

Like they certainly don’t seem like the same rings and I have a natural inclusion from one into the other but I can’t find a distinctive feature that tells them apart

(k is alg closed)

[pastel cliff]

,rotate

[pastel cliff]

what is this proposition saying

and what is alpha

im getting bogged down in symbols

[south patrol]

well the alpha are just coefficients

I guess this is essentially just telling you how to compute "in coordinates"

Like if you fix a basis and just work in terms of that basis, how do you compute φ

[rustic crown]

it saying if w = Av, then phi(w) = det(A) phi(v)

[topaz heart]

what is k[x]_(x)

?

[wooden ember]

Localization at maximal ideal (x)

[topaz heart]

ty

[rustic crown]

one way is to compute the cotangent space

[topaz heart]

i might not understand the definition but do you not have t^3/t^2 in the second

and then sending x to this will give an isomorphism?

[cloud walrus]

det

[rustic crown]

the second ring is k[x,y]/(y^2-x^3) and at the point (0,0) the cotangent space has dimension 2

this computation becomes really easy if you know about Kähler differentials

[rustic crown]

because when k is algebraically closed and the ring A is a finite-type k-algebra, the residue field A/m would also be k, and because of this the composite k --> A --> A/m is the identity is etale and so the conormal sequence gives you an isomorphism m/m^2 = Omega^1_{A/k}⊗_A A/m
and computing kahler differentials is really easy. Say A = k[x, y]/(y^2-x^3) then the module Omega^1_{A/k} is (A dx ⊕ A dy)/(2y dy - 3x^2 dx) and for the maximal ideal m corresponding to the point (0, 0) when you base change, the denominator dies completely, so you get k{dx, dy} which is 2 dimensional. but the cotangent space for k[x]_(x) would be one dimensional

oh what have you thought so far?

okie, so one way to define the splitting field is the smallest field where your given polynomial splits completely into linear factors

[rustic crown]

this is a tiny bit weird definition because it assumes that there already exists a bigger field where your polynomial splits (like in our case, we know C is a thing)

but constructing bigger fields is easy, so it's not a big problem

yee so since you already know that C is a thing, consider L = Q(all the roots of x^3-1)

yee, let's call them 1, w, w^2 :p

[wooden ember]

No this only works when you localise at points other than (0,0)

[wooden ember]

I’ll read your message but unfortunately it’s not a tool that we have

[rustic crown]

there are loads of field extensions of Q where the polynomial x^3-1 splits

the first condition in your definition says the polynomial needs to split

and the second says it's the smallest

so like in C every polynomial splits, but that's not true for Q(w)

you're looking for a smallest field where it would split

(and there is a theorem that any two splitting fields are isomorphic :3)

oh so it depends what we mean by smallest right

there isn't any way to compare to arbitrary fields

but when you talk about subfields there is an obvious notion of inclusion

yeeap

okie so say we wanna find a subfield of C, call it L which is a splitting field of x^3-1 over Q.

so the first condition says that x^3-1 needs to split in L

so 1, w, and w^2 must be elements of L

yee, once you know a big field like C, constructing splitting fields is really easy :3

yep

and what does the second condition say?

yee from this do you see why L = Q(w) = Q(1, w, w^2)?

okie so x^3-1 does split in Q(1, w, w^2) right

it's the smallest field containing Q, 1, w, w^2

that's not the definition, but here the elements of Q(w) would turn out to be a+bw

because w satisfies a quadratic

yep

because Q(w) would include w^2 as well

as it's a field

only reason they pluralized it would be if they didn't prove that any two are iso, or if they wanted to you figure out that there is a unique one

what's the degree :3

[solar glacier]

question

let $p(x)=2x^3+7x^2-\frac{1}{2} \in \Bbb{Q}$ show $p(\sqrt{5} \in \Bbb{Q}(\sqrt{5})$

[cloud walrus]

MyMathYourMath

[solar glacier]

is the trick to rewrite $p(x)=x^2(2x+7)-\frac{1}{2}$

[cloud walrus]

MyMathYourMath

[solar glacier]

Then we get

[formal ermine]

what are you supposed to show

[cloud walrus]

MyMathYourMath

[formal ermine]

do you mean p(sqrt5)

[cloud walrus]

MyMathYourMath

[formal ermine]

ah

that's trivially true

if sqrt 5 is in K then sqrt5^n is in K lol

[solar glacier]

just checking i did it correctly

[cloud walrus]

MyMathYourMath

[rustic crown]

oh ooops i got busy, sowwy >.<

[solar glacier]

well theres some work to be shown, no ? like rearranging p to be of that form

[rustic crown]

yep, because w satisfies w^2+w+1=0

[formal ermine]

why doe

every element of the form asqrt5 + bsqrt5^2 + ... is in your field

[strong valve]

@rustic crown

[rustic crown]

pwease no tag me, you need to know the definition of ideal to talk to me

[strong valve]

Can you prove that P is a pid

[rustic crown]

.<

[solar glacier]

which P

[desert hawk]

could i ask a coding theory question here,, the cs server seems dead ah

[formal ermine]

you are setting the requirements too low

[rustic crown]

this was the context

[strong valve]

For any P

[solar glacier]

idk abuot others but i know ZERO coding lol

[formal ermine]

lmaoo

[desert hawk]

its not "coding"

[solar glacier]

whats P though any conditions on P

[formal ermine]

I wanna make a dn joke

[summer path]

i mean the closest thing here is like #computing-software but idk what your question is so

[desert hawk]

we are learning about fields in this course

[strong valve]

P is a thing which we have to show is a pid

[rustic crown]

okie so in general, if you have simple extension k(a) of k, then the degree of the extension is the degree of the minimal polynomial of a over k

[solar glacier]

it usually says let P be .... prove P is a PID

[agile burrow]

I'm PID

[rustic crown]

hewwo walter

[solar glacier]

all your ideals are principal?

[agile burrow]

hi det

[summer path]

ok, just ask your question and we'll see where it actually belongs from there

[agile burrow]

walter is reviewing algebraic number theory while procrastinating on writing his alg top paper

[desert hawk]

okay

[solar glacier]

im studying for field theory procrastinating my alg top hw lol

[strong valve]

You can say het to save time

HET

[desert hawk]

its kind of fill in the blanks, i wished to know what syndromes to select from this table to decode (a) r1 = (01011 00000 00010) to (01011 00000 00001) and (b) r2 = (10000 10110 10111) to (10001 00110 10111)

using error trapping?

[rustic crown]

but yea, you're right... since 1 and w generate it as a Q vector space so the dimension is at most 2. but it can't be 1 since Q(w) is strictly larger than Q

yee

[summer path]

ok maybe ask in #computing-software, but otherwise idk

[desert hawk]

😦

[solar glacier]

anyone have an exercise list for field extension problems like finding degrees of splitting fields and finding min polynomials and inverses in field extensions

[tribal moss]

It looks vaguely like it might be coding theory, but you'll probably need to cough up a lot more context no matter where you post that, 101.

[desert hawk]

okay the context is this Let C be a binary (15,10)-cyclic code with canonical generator g(x) = 1 + x2 + x4 + x5. It is known that C is a 2-cyclic burst error correcting code. Decode the following received vectors using error trapping:

why did i take this course

[rustic crown]

Sapphire Gaming might know :p

[summer path]

im procrastinating catching up on a alg top lecture, while trying to decide when to eat lunch

[solar glacier]

same i need to catch up to lecture we already started first homology group am im still stumped on first homotopy group lol

[solar glacier]

thx!

[rustic crown]

(that was a joke >.<)

was that a series question then gomen >.<

[desert hawk]

why is coding theory seemingly obscure i assumed it would be more popular

[tribal moss]

It's not generally considered to be a part of abstract algebra (though much of it applies abstract algebra).

[desert hawk]

i see

[solar glacier]

how many elements are in Z_3(r) if r is a root of x^3-x+1 which is in Z_3[x]

is it 8?

[rustic crown]

not really

[solar glacier]

for irreducible do i use the fact that degree 3 or less polynomials are irreducible iff they have no root from the field

[tribal moss]

(The most fitting channel for coding theory is probably #combinatorial-structures. You could get away with asking in #discrete-math too, but I wouldn't be particularly optimistic about finding someone who can answer there).

[cloud walrus]

MyMathYourMath

[topaz heart]

ω in Q(ω) but not in R so can't be in Q maybe

[cloud walrus]

MyMathYourMath

[rustic crown]

i mean you're just showing that w doesn't lie in Q... i didn't think that was what your prof wanted to say

[topaz heart]

that's normally the reason i see

i thought they meant that you can't just state Q(ω) =/= Q without anything else

[desert hawk]

okay thanks Troposphere

[solar glacier]

is this true

[rustic crown]

yea that is true. it's not as easy to say why cbrt(2) can't be written in terms of 1 and sqrt(5)

one way is to use that degree of the extension Q(cbrt2, sqrt5) woudl be both divislbe by 2 and 3, and since they're coprime the degree is 6

the coprimality is the important tool here

yee

w is a generator

yep

but you can write Q(w) = Q(w^2) = Q(w, w^2)

since 1+w+w^2=0

right, when talking about degree you only look at the Q-vector space structure

but when talking about generating as a field, you're allowed to multiply as well

yep

say z = exp(ipi/3)

oh not quite

the extension this time contains sqrt2 * z^n for n = 0, ..., 5

wait i didn't catch that

oh wait nvm, you're right :p

right

so your field needs to contain these

in particular since it contains sqrt2 and sqrt2 * z

it also contains the quotient z

(w is usually reserved for exp(2ipi/3) but sure :p)

yep

this is also equal to Q(sqrt2, w)

nah, it's 6

[rustic crown]

oh oops i did a dum dum

it's 4 mb mb

cause as a vector space you need 1, sqrt(2), w, and sqrt2 * w

nope

write sqrt2 * w like that

Q(sqrt2, zeta_6) = Q(sqrt2, zeta_3) = Q(sqrt2, sqrt-3)

as a field it is generated by those two elements

can you can multiply

but as a vector space, you wanna write each element as a sum of things so you'll need 4 generators

namely 1, sqrt2, sqrt-3, sqrt-6

the last generator is redundant

you can get it by multiplying the first two

but you can't get it from adding the first 2 with Q-coefficients

you'll have to be comfortable with the definitions for that >.<

when you write k(a, b, c ..) it's the smallest field containing k and the elements a, b, c etc

if a field contains a and b then it also contains ab

and 1/a if a is non-zero etc

because degree is a different question

if L/k is a field extension

then L is automatically a k-vector space

so you can use tools from linear algebra to understand field extensions

one such in the dimension of a vector space

here the question is, "can you find a list of elements such that k-linear combination can be used to write every element of L uniquely?"

so you're only allowed to multiply with stuff from k (scalar multiplication) and add stuff together

b is a scalar from Q

this is very subtle if you're not comfie with definitions

okie lets to easy examples

Q(sqrt2) is generated by one elements namely sqrt2

but as a Q vector space a basis is {1, sqrt2}

similarly

Q(cbrt2) is generated by one element

but as a Q-vector space a basis is {1, cbrt2, cbrt2^2}

didn't get you

w^2 = -1 -w

so you can write it as a Q-linear combination of 1 and w

so degree is 2

but you can't write cbrt4 as a Q-linear combination of 1 and cbrt2

as a Q-vs you need 3 generators

as a field you can just do with 1

you'll soon prove that any finite degree extension of Q can be generated by a single element as a field

so w = sqrt2 and z = exp(ipi/3)

right?

yep that's the splitting field

and Q(w, wz) = Q(w, z)

they're the same thing, so no need to say mb :p

yeep

and i'm saying a Q-basis is {1, w, z, wz}

oh oops :p

yep

choose the words carefully >.<

you have a polynomial in K[x] and over K

else it's confusing >.<

are you saying if L is a splitting field of f over k, and E an an intermediate field, then L is also a splitting field of f over E?

yee, that's true

wait those are the exact same propositions

ah no, that maybe different

for example C is the splitting field of x^2+1 over R

but the splitting field of x^2+1 over Q is Q(i) not C

you're looking at subfields of C containing Q where the polynomial splits

and then you pick the smallest one among those

since you know there is a bigger field where any polynomial splits, you can just lok at the subfields to find the splitting field

but for example, consider the field C(t) of rational functions in t. what is the splitting field of x^2-t?

x^2-t is the polynomial in C(t)[x]

and you want it to split in some extension of C(t) say L

see so you don't already know a bigger field where that polynomial factors

which makes it a little bit different

here the splitting field would be C(sqrt(t)) whatever that means :p

what are your thoughts :p

yep so in that theorem can you see how to go from <= to |

sure will

det tired >.<

illu can you take over. me tired >.<

L = K[t]/(p(t)) is the standard construction

and this has degree n

as a basis is 1, t, t^2, ..., t^n-1

[formal ermine]

yee

oke so the proof for leq is just: induction & tower law

the proof for divides is exactly the same except that you change one or two things

[rustic crown]

https://tenor.com/view/blazedpx-pokemon-eevee-gif-23902862

[formal ermine]

good night det

[rustic crown]

(oh i still need to eat dinner >.<)

[formal ermine]

like you look at [K(a1, .., an) : K] = [K(a1, ..., an) : K(a1)] * [K(a1) : K]

by induction the first thing is leq (n - 1)!

can you see why the second thing is leq n?

ai are the roots of f

yes

you need to change and argue for one tiny thing to make it work for divides

ye

no need to tag me

our first induction step would be to only look at a degree 1 polynomial

then we look at a degree n polynomial and assume n - 1 works

yes

you're thinking too complicated

I mean

can our polynomial have a root that is not in K

yee because if it did then it would no longer be a polynomial over K

oke good

now the general case

hint: you have to look at [K(a1, .., an) : K] = [K(a1, ..., an) : K(a1)] * [K(a1) : K]

now apply our induction hypothesis somewhere...

no, we did n = 1

our polynomial has no root ouside of K thus the splitting field has degree 1 over K

huh?

what would our splitting field be?

K(a)

but what is a?

yes, it's in K

so K(a) = K

[formal ermine]

nu

it depends on the way you prove that one lol

things you need to answer to prove this:

• why does [K(a1, ..., an) : K(a1)] divide (n - 1)! ?
• why does [K(a1) : K] divide n?

you need to stop citing proofs and definitions over and over again

you've posted it once already e

[formal ermine]

why is it equal to 1

this a1 is not the a1 from before

lol?

n = 1

[rustic crown]

you could wlog assume the poly is irred >.<

idk how far you into the proof

me just had some food

okie so say you have the poly f

if f = gh for smaller degree stuff

actually let me give them names

say splitting field of f is L/k

oh i'm assuming the induction hypothesis for all degrees smaller than deg f

that's sometimes called strong induction

oh okie

so say L:K is the splitting field of f, E:K is the splitting field of g

so you can check that L:E is the splitting field of h

so you agree that L:E is the splitting field of f = g*h right

and g is already split over E

L is the splitting field of the polynomial h in E[x] over the field E

yep

did you say this proposition a while ago

if L is a splitting field of f over K, then for any intermediate field E, L is also the splitting field of f over E

yea i'm using that

so L is the splitting field of f = g*h over E

but since g already splits over E, it won't contribute anything towards the splitting field

so L is the splitting field of h over E

yep

this is going to be simple definition chase, that's why i said "you can check :p"

[rustic crown]

yee so by induction hypothesis you nkow [L:E] divides (deg h)! and [E:K] divides (deg g)!

so [L:E] divides the product

which further divides (deg f)!

because (deg f)!/(deg g)! * (deg h)! is an integer

it's just (deg f) choose (deg g)

because combinatorics >.<

n choose r has that formula right

.<

yea and that product divides deg f!

also you mean [L:K]

[rustic crown]

because this

this expression is counting number of some stuff

so has to be an integer

have you not seen binomial theorem >.<

because it's a pretty common thing that people learn in high school >.<

the number of ways to choose r objects out of n objects is n!/r! * (n-r)!

because that coefficient is an integer

[rustic crown]

i was claiming this, and that's one reason for this

yee

maybe i read this wrong >.<

"no of course" vs "no, of course"

if you expand (1+x)^n the coefficients of x^r is on one hand an integer

and on the other is n!/r!(n-r)!

so this other guy is also an integer

in particular r! * (n-r)! divides n!

[white oxide]

Why does the subgroup <(2, 3)> not have order 4 in Z4 x Z4? Since the order of 2 in Z4 is 2 and the order of 3 in Z4 is 4, isn't the order of (2, 3) equal to the lcm of the orders, i.e. 4?

I've tried computing <(2, 3)>

And I got {(0, 0), (2, 3), (0, 2), (2, 1)}

in Z4

[rustic crown]

it's pretty standard to say that (a+b)!/a!b! is an integer

nobody will be mad at you for just stating that without proof

.<

[rustic crown]

so that has size 4 which you want

[white oxide]

oh

ignore that

LMFAO

that's embarassing

[rustic crown]

hehe :p

[white oxide]

lmao i got to the fourth element and didn't even think of adding it again LOL

oops

[rustic crown]

anyway, so [L:K] divides (deg g)! * (deg h)! which further divides (deg f)!
so nwo you only need to show thsi when f is irreducible, in this case look at E=K(a) as the intermediate field where you attach one root of f which i called a. Say f = (x-a)*g, then L is the splitting field of g over E whibh by induction hypothesis has degree dividing deg g!

that is (deg f - 1)!

and [E:K] = deg f

so here as well you get [L:K] divides deg f!

.<

[white oxide]

Also, why exactly is <(3, 1)> = <(1, 3)> in Z4 x Z4?

is that just a coincidence

[rustic crown]

because in general <g> = <g^-1>?

[white oxide]

certainly not

ohhh

that makes sense

thx!

[rustic crown]

[white oxide]

hm i'm curious as to why i listed 6 subgroups of order 4 in Z4 x Z4 by giving a generator

but the problem says there's one group of order 4 that is iso to the Klein V group

and the Klein V group isn't cyclic

[empty rose]

well apparently the subgroup of order 4 of Z4 x Z4 that's isomorphic to the klein v group isn't any of the 6 subgroups that you've found

[white oxide]

I found (1, 0), (0, 1), (1, 1), (2, 1), (1, 2), (3, 1)

too lazy to put in the <>

but those are all distinct i believe

[rustic crown]

i wouldn't have put the (,)

[white oxide]

haha fair enough

[rustic crown]

yee so you find 6 cyclic subgroups of order 4

now you need to search for non-cyclic subgruops

[white oxide]

ye but it says that there are only 6 subgroups of order 4

or maybe i'm interpreting what they said wrong

[chilly radish]

Can you post a sc

[rustic crown]

it does say "cyclic"

[white oxide]

oops

bruh how would i even go about finding this shit i'm too lazy for trial and error

[rustic crown]

how many elements of order 2 are there in klein-4?

what the elements of order 2 in Z4xZ4

[white oxide]

nvm

i got it

(0, 0) (3, 1) (1, 3) (2, 2)?

oops

that's cyclic

[white oxide]

oh right

nvm i got it

00 02 20 22

[rustic crown]

you're doing a strong induction here, since you don't know if deg g or deg h is exactly n-1

also at the end n! not just n.

a linear poly is always irred

hm?

if k is a field then ax+b is always irred in k[x] as long as a is non-zero

because if you try to factor it, one of the factors would have degree 0

[delicate orchid]

elements of this ring are not polynomials although they smell like them

[rustic crown]

also C[x]/(x^2+1) isn't a field

it's iso to C x C

[delicate orchid]

brutal

[rustic crown]

lol that's the least interesting irreducible polynomial :p

yea

a in L

f in L[x]

x-a is already split over E >.<

because E is isomorphic to K[x]/(f)

and the right has K-basis given by {1, x, ..., x^deg f -1}

nah it doesn't

this is not a statement about splittign fields

[white oxide]

do multiplicative cancellation laws hold in a field?

[rustic crown]

you define the map K[x] --> E extending the inclusion K --> E and sending x --> a this is surjective and the kernel contains the polynomial f(x) and hence the ideal (f) which is maximal as f is irred, so kernel must be exactly this

which gives you the iso K[x]/(f) = E

[rustic crown]

yea, as long as thing you're cancelling is non-zero

[white oxide]

oh ok cool thx!

[rustic crown]

show explicitly that {1, a, a^2 ..., a^n-1} is a basis

its not, its teh same proof lol

it's linearly independent because if any combination was 0 then you get a smaller degree polynomial which has a as a root

and that's not possible because f was irred

and if h is a poly not divisible by f, then can write pf + qh = 1

this shows that h(a) has the inverse q(a)

so E is generated by 1, a, a^2, ....

but you don't need stuff from a^n onwards because f(a) = 0

yea

[white oxide]

This question only applies to rings with no divisors of 0, correct? Because in the ring Z12, 4 and 8 are inverses to 3, since (3)(4) is congruent to 0 mod 12 and (3)(8) is congruent to 0 mod 12

[rustic crown]

try on your own a little >.<

[white oxide]

oh wait oops

unity in that case would be 1

was thinking of additive inverse

nvm

[white oxide]

how does this cor follow? is it because a^(p-1) = apa^(-1) so you can multiply both sides by a, or how do congruences even work lol

like how specifically did they get to the corollary

I don't know the rules of modular arithmetic, but I'm thinking since $a^{p - 1} = a^pa^{-1}$, $a^pa^{-1} \equiv 1 \text{ mod } p$

[cloud walrus]

okeyokay

[white oxide]

And so you can multiply both sides by $a$ to get $a^p \equiv a \text{ mod } p$

[cloud walrus]

okeyokay

[white oxide]

<@&286206848099549185>

[next obsidian]

You just multiply the equation a^p-1 = 1 mod p by a

[white oxide]

oh so you can multiply both sides

okay thanks!

[rustic crown]

pwove it!!

[white oxide]

oh god number theory gives mea headache

i wonder if i should just skip this section

but if i must i must

[white oxide]

oh equivalent modulo n right

and thank you!!!!

[frigid lark]

Yeah

[white oxide]

that makes complete sense tysm

[next obsidian]

This is the point of the operation being well-defined

Regarding your question from a few days ago

[white oxide]

ohhh ok

[frigid lark]

Beyond the snake, butterfly and horseshoe, are there any more animal themed lemmas?

[next obsidian]

There’s a salamander lemma

Which is this super-powered homological lemma which implies a lot of the other ones

[rustic crown]

so what we're proving is this, say you have an algebraic simple extension E/K with E = K(a) then [E:K] = degree of minimal polynomial of a

say f is the minimal polynomial of a

and say it has degree n

we want to show E has {1, a, ..., a^n-1} as a K-basis

since f is the minimal poly you know that these cant be linearly dependent

so only thing you need to show is that they generate all of E

but elements of E can be arbitrary quoteitns p(a)/q(a)

where q(a) is non-zero

but the fact is every element of E is actually just a polynomial in a

and to see that you need to show that given any such q(a) such that q(a) is non-zero the inverse of this is also just a polynomial in a

for that look at the polynomials q(x) and f(x), their gcd divides f(x) so is either 1 or f, but f doesn't divide q as q(a) is non-zero... so gcd must be 1 which means you can write a linear combi q(x)Q(x) + f(x)F(x) = 1, plugging in a, you get Q(a) is an inverse of q(a)

so every element of E is a polynomial in a

but you don't need to use all the powers because f(a) = 0 means that a^n can be expressed as a linear combination of smaller powers

and thus any elements can be written as a linear combi of {1, a, ..., a^n-1}

oh hi zan

[tribal furnace]

hi det

[rustic crown]

anyway, if you develop the theory nicely, this proof is much easier to keep in head

it is

why do you keep quoting theorems instead of actually understanding the statement and thinking about it on your own >.<

if it's included in your notes, you need to understand it well

you can't just skip/blackbox it

yep cause it is root of some poly

that's how we defined a

nooo >.<

over K

not over K(a)

it's a root of f in K[x]

use irreducibility

nah, doesn't matter over fields because non-zero scalars are units anyway

[prime sundial]

if i have a field of characteristic p \ne 0, then i can do (a+b)^p = a^p + b^p
i believe this also holds for the polynomial ring F[x] and arbitrary (ax^n + bx^m)^p, but does this extend to arbitrary polynomials as well?
does (p(x) + q(x))^p = p(x)^p + q(x)^p in F[x]?

[rustic crown]

the first equation holds in any commutative ring of char p

even more specifically any ring of char p where a and b commute

[prime sundial]

ok thank you

this is in relation to finding the equivalent coset of a polynomial x^large power, e.g. x^2020 in E[x], E = F[x]/(x^3 + x^2 + 2x + 1), F = Z_3

i’m honestly a little stumped, don’t know what i’m supposed to be going for

[rustic crown]

i think you mean F[x] and not E[x]

and the idea is to do division with remainders with x^2020 and x^3+x^2+2x+1

[prime sundial]

i want to find x^2020 in E[x] represented by something of the form ax^3 + bx^2 + cx + d

[rustic crown]

say
x^2020 = (x^3+x^2+2x+1) * q(x) + r(x)

then in E = F[x]/that thing

[prime sundial]

oh

sorry yes

i meant E

[rustic crown]

you get x^2020 is congruent to r(x)

[prime sundial]

right i understand this

but going to 2020 seems impractical

what’s the trick i’m missing

[rustic crown]

okie maybe we can exploit that E is a field

here i didn't use that at all

[prime sundial]

i didn’t notice it was

[rustic crown]

yea that cubic poly is irred since it has no roots

[prime sundial]

ok i see

right

[rustic crown]

so E is the finite field of size 3^3 = 27

[faint fractal]

is there a good way to see that two dihedral elements are the same?

[rustic crown]

which means E* is a finite group of size 26

and so x^26 = 1

in E

[prime sundial]

ok that’s a lot of info

thanks

a ton

[rustic crown]

yea but still this is a pain

because it reduces x^2020 to x^18

[prime sundial]

it’s ok, it’s actually quite helpful already

thanks det

[rustic crown]

i mean you're only assuming it's true for n-1

so in the reducible case, you can't use the induction hypothesis for g and h

unless one of them is linear and other is degree n-1

just change the induction hypothesis to "true for all degrees up to n"

n = 1 is still the same

earlier it was a bit wrong since you couldn't use induction hypothesis on the reducible case

now that's fixed

we note that it is the splitting field of E over K

whut?

so x - a by definition has splitting field E
x-a is not even a polynomial in K[x]

it should be "... L is the splitting field of h over E"

why do you have "a" in the reducible case

you had f , it's splittting field over K was L

and f = gh

E was the splitting field of g over K

and you noted L is the splittinge field of h over E

idk where a comes up here

first correct the reducible case

we note that the splitting field of h over K is E.

h over E is L

E was by definition for g over K

So, f = (x-a)*g, L is the splitting field of E by the intermediate splitting field thingy, so x - a by definition has splitting field E, L is the splitting field of g over E,

L is the splitting field of f over E

and since x-a splits already over E, L is the ...

yea that's the def

but by your intermediate splitting field thingy, it's also over E

f over E

or g over E

splitting field of x-a over E is E

yes

because x-a is not in K[x]

a is in E, but not necessarily in K

you need to somehow say that splitting field of g over E is L

you can give any reason you like >.<

no

f is irred in K[x]

(x-a)*g is it's factorization in E[x]

g not necessarily irred

this factorization doesn't exist in K[x]

else f would be reducible

and we wouldn't have to deal with a second case

yea

m (or whatever it's called). Now since L is the splitting field of h over K, we note that the splitting field of h over E is L.

L is the splitting field of f over K

actually change that a little more

Now since L is the splitting field of f over K, we note that the splitting field of h over E is L.

L is the splitting field of f over E, and since g already splits over E, the splitting field of h over E is L

rest is okay modulo some very minor stuff which i will ignore

(or whatever it's called). L is th
stuff like that should be a , and not a .

.<

vides (deg g)! and [L:E} divide
} should be ]

etc

i'll ignore

replace the period "." with a comma ","

you're saying "so by theorem A. Statement B"

.<

that took a while >.<

i can finally sleep

https://tenor.com/view/pokemon-eevee-eevee-sleeping-gif-20027623

[solar glacier]

question about splitting fields

splitting field of $x^4+1$ over $\Bbb{Q}$

[cloud walrus]

MyMathYourMath

[solar glacier]

does it have basis

${e^{2 \pi i \frac{k}{n}}: k = 1,2,...,n}$

[cloud walrus]

MyMathYourMath

[solar glacier]

i know the split field of $x^4-1$ is 2 because the $1,-1 \in \Bbb{Q}$

[cloud walrus]

MyMathYourMath

[next obsidian]

What does it mean for the splitting field to be 2

[south patrol]

2 = {0,1} which is the field on two elements

[solar glacier]

2 basis elements

[solar glacier]

Oh is that how that works

[south patrol]

What lol

[solar glacier]

The order of the extension being n then the extension is iso to Z_n?

But generally Z_n isn’t a field

[agile burrow]

that is in fact not how it works lol

[solar glacier]

Was gonna say…

[agile burrow]

potato was joking

[solar glacier]

Ok 😅

[agile burrow]

presumably you mean to say that the degree of the splitting field is 2

[solar glacier]

Am I correct tho mt@logic for why it’s 2

Yes the degree of splitting field

Cause it factors

And 1,-1 are in Q already

[agile burrow]

Then yeah

It suffices to adjoin a root of x^2 + 1, which yields a degree 2 extension

[solar glacier]

Ok and what about x^4+1

[agile burrow]

You could either do the factoring thing or just see that this is the 8th cyclotomic polynomial if you want to apply more general theory

[solar glacier]

It’s irreducible for certain over Q

How do you get that it’s the 8th cyclotomic poly and what’s a cyclotomic poly again

[agile burrow]

The nth cyclotomic polynomial is the minimum polynomial for a primitive nth root of unity.

[solar glacier]

Why is it 8 and not 4

[agile burrow]

because a 4th root of unity is i, and the minimal polynomial for i is x^2 + 1

[solar glacier]

I’m lost so this one has degree 8 split field. ?

[agile burrow]

Ok maybe I shouldn't have brought up cyclotomic polynomials

[solar glacier]

Or degree 4

I thought I had the. Basis elements correct up top

[solar glacier]

This with n =4

[agile burrow]

They might not all be linearly independent though

And in your case, they aren't linearly independent. e^pi i = -1 and e^2pi i = 1 certainly aren't

If you adjoin some root of x^4 + 1 to Q, can you show that the polynomial then splits over this extension?

So first of all, F_p[x] is a PID. Thus, an irreducible polynomial f of degree n generates a maximal ideal so the quotient F_p[x] / (f) is a field. To see the order of the quotient, note that every element has a unique representation a_0 + a_1x + ... + a_{n-1} x^{n-1} where the a_i are coefficients in F_p. This follows from, say, the division algorithm in F_p[x]. Finally, f has a root in F_p[x] / (f) by considering the coset of x

Oh lol

Yeah, the coset of x serves as a root of f

Essentially by construction

Right, so if we let $\bar{x}$ denote the coset of $x$ in $F_p / (f)$ and we have $f(x) = a_0 + a_1 x + \cdots + a_n x^n$, then in the quotient, we have $f(\bar{x}) = a_0 + a_1 \bar{x} + \cdots + a_n \bar{x^n} = \overline{f(x)} = 0$

[cloud walrus]

walter

[agile burrow]

The second to last equality follows because each of the coefficients just lies in F_p

In words, I guess you could say that f evaluated at the coset of x is equal to the coset of f(x) which is zero in the quotient

I'm not sure off the top of my head, but I'll let you know if I think of something

[white oxide]

what exactly do they mean by "read modulo m1"?

also how exactly do those numbers reduce to s modulo m1? is it because s + dm1 modulo m1 gives s, and not s modulo m1?

also why would they reduce to s modulo m1 in Zm, shouldn't they be reducing to s modulo m

[north sand]

• is not the operation in that group

[atomic mesa]

Is it addition?

[warm shoal]

Sorry dumb question - answered

[delicate orchid]

actually deleting ur posts.......

[warm shoal]

my question was embarrassing

[delicate orchid]

there are no embarrassing questions

[rustic crown]

wew what's in your pfp

[warm shoal]

In 2, can we infer that ord(G) = m?

[delicate orchid]

pic of me

[delicate orchid]

which part of 2 are you using to infer that

I'm leaning towards no though

[warm shoal]

im trying to do the first part

not sure how to begin

[coral shale]

Clearly not.

"Let m be a fixed integer"

[delicate orchid]

right in that case definitely not, take {0, 2} as a subgroup of Z_4, every element squared is in that subgroup but |Z_4| = 4

[coral shale]

If we let it be something other than the order of G...

(and you're given G and H apriori)

[warm shoal]

i see

right right

so H normal in G

and so

(Hx)^m = Hx^m = H

😄 ?

[delicate orchid]

that's convinced me

[warm shoal]

?

[warm shoal]

Can I straight up infer that |Hx| / |m| ?

[warm shoal]

Could someone help me w the converse?

[warm shoal]

If every element of G/H has a square root, and every element of H has a square root, then every element of G has a square root.

Because every element of G/H has a square root, Hx = Hy^2 for all x, y in G

and then Hx = Hy^2 implies that xy^-1 in H

Though now I am stuck - any insight would be super helpful ❤️

[south patrol]

This is not true

[warm shoal]

darn okay

how would u approach this problem then ?

[south patrol]

Okay there are probably a couple of ways you can go about it

But basically you should take some g in G and yeah consider a square root of Hg which i think is what you were doing

Just you got the quantification a bit off and stuff

They say for some

You say for all

[warm shoal]

ah dangit

well that wont work then

[south patrol]

Yeah so like try this

Essentially try to unfold definitions

[weak flower]

Is there an example of a bilateral ideal $I$ of a ring (a.k.a. unital associative algebra) $A$, which is nil, meaning that every element of the ideal is nilpotent, but for which the ideal $M_n(I)$ in some matrix algebra $M_n(A)$ is not nil?

[cloud walrus]

koaaa

[weak flower]

So the motivation is the lifting of projective modules modulo a nilpotent (or complete) ideal instead of nil. These two properties are easily seen to be inherited by the matrix algebra/ideal, and one has the existence of lifting by Newton iteration.

Now for the lifting of idempotents to exist, a nil ideal (or complete) is sufficient. However I cannot determine how nil ideals behave under the construction of matrix algebrae.

Now this is not a problem if $A$ is commutative, since there the coefficients of a matrix generate a nilpotent ideal, so you do have nilpotence of the matrix.

[cloud walrus]

koaaa

[long nebula]

should the | be bigger here? what do y'all think

[lethal dune]

Wedderburn's little theorem?

It's fine, assuming of course m_i < n

[long nebula]

Yeah I was just wondering about the typesetting though

I think it looks kinda weird lol

[solar glacier]

this might be a REALLY stupid question

but as charZ_p is p for all primes p

isnt charZ_n just n for any n >=2

then is char(Z_m x Z_n) just lcm of m,n

[formal ermine]

yes

[solar glacier]

yes lol typset \bigg first

ok just makin sure

[formal ermine]

you don't even need a restriction on n 😉

[solar glacier]

question

is $\Bbb{Q}(\sqrt[n]{2}) \subset \Bbb{Q}(\sqrt[m]{2})$ iff $n \mid m$

[cloud walrus]

MyMathYourMath

[formal ermine]

does your \subset mean proper subset?

[solar glacier]

unsure honestlyl

[south patrol]

it can't otherwise take m=n

[solar glacier]

i wanna say so

well if n <m

and n |m

is it proper in that sense?

is n strictly less than m an it divides it

[formal ermine]

lgtm

[solar glacier]

i was assuming tbh that n<m

whats lgtm lol

[formal ermine]

programmer slang for looks good to me

[solar glacier]

ahh

[celest furnace]

jesus slang

[solar glacier]

lol

so it is true then

sweet

thanks

given n < m

[formal ermine]

or just do \subseteq lol

for n = m

[solar glacier]

true

i was just gonna say that lol

[long nebula]

hmm if i do that, I have to change the \mid to | but then the spacing isn't right

[formal ermine]

isn't there

,, \divides

[cloud walrus]

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

[formal ermine]

oh

I have a custom macro for it then

in overleaf

lemme check

nvm lol I also just use \mid

[rotund aurora]

Let R<S be an extension of rings, I an ideal of R and s in S. If I·s^k subset I for all k>=0, when can we conclude that s is integral over R?

so ofc s is in the field of fractions of R

when R is noetherian that's clear I think, but idk for other cases? Like that condition seems super strong lol

that's maybe trivial but Im a little confused nevertheless

[lethal dune]

oh that's what you were commenting on

$\Phi(q) \Big| \frac{q^n-1}{q^{m_i}-1}$

[lethal dune]

can't you just say s•I ⊆ RI so by nakayama there is monic f(x) ∈ R[x] with f(s)=0?

[chilly ocean]

help

just help

ill take my leave after somebody tells me wtf this is

also no this isnt 9 year old work i just suck at maths

[formal ermine]

this is still the wrong channel

[coral shale]

Remove the Advanced and Pending G+ roles from yourself - you clearly aren't studying math at undergrad (uni) level

[flat tree]

Please read #❓how-to-get-help

[chilly ocean]

dude

im not here to stay

im here to ask for help

[south patrol]

Having the advanced role doesn't require being an ug tbf lol

[chilly ocean]

yeah uk education system is ass

and ai cant translate answers from images yet

[lethal dune]

try $\frac 1 2 \frak{I} \left( \int _{ \partial T} \overline{z}, \dd z \right)$

[long nebula]

see I just think the spacing is off if you don't use \mid

[\Phi_n(q) \mid \frac{q^n-1}{q^{m_i}-1}]

[cloud walrus]

tietzERIC extension

[long nebula]

yeah using | instead makes the spacing smaller

eh I'll live with it

[coral shale]

i think its generally recommended not to | for anything

[long nebula]

oh I usually use | for absolute value

[coral shale]

try latexhelp though, theyll know.

\abs{ . }

[south patrol]

Lol

What is mi

[lethal dune]

∈ ℕ

[long nebula]

hm maybe I'll start doing \DeclarePairedDelimiter{\abs}{\lvert}{\rvert}

anyways ty

[south patrol]

Oh lolc

I didn't realise domain was even a term lol

I assumed it was just an abbreviation for integral domain lol

[lethal dune]

it isn't?

[south patrol]

Hm idk wikipedia has smth different

[coral shale]

[south patrol]

I say this because i looked up wedderburn lil theorem and they stated it differently to how I was used

[lethal dune]

non commutative is it?

[south patrol]

To me it is like every finite division alg is a field

But they state it as finite domain is a field

Which to me seemed trivial lol

But yeah

[lethal dune]

yeah non comm bs

hint: all α → α ^p → α^p^2 are the roots of f, then coeff of x = ∑ a_ia_j (i ≠ j)

[lethal dune]

yeah, that's one way to see that

I'm assuming you are familiar with Galois theory

what's the coefficient of x in (x-a)(x-b)(x-c)?

if alpha is a root than is any σ (alpha)

what are the automorphisms of a finite field F_p³?

over F_p

ok for any a ∈ F_p, what's a^p?

[next obsidian]

WOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO

[lethal dune]

ok now f is a polynomial in F_p

with root alpha

now f(α)^p=0 = f(α^p)

verify this remember it's in char p so (a+b)^p=a^p+b^p

ye

trying writing f(α) explicitly

[lethal dune]

use this next

[lethal dune]

.

already given all the details, just read through it again

bye GN

[zealous forum]

Reading a proof that Z/(m) \otimes Z/(n) = Z/(gcd(m,n)) and I can’t figure out how they came up with the inverse map here?

[molten viper]

Quick notation question, what does the m here refer to?

It's a maximal ideal yeah?

[barren sierra]

yea

the unique maximal ideal

[molten viper]

gotcha

That maximal ideal being unique basically being the definition of Artinian

[barren sierra]

Isn't that from the definition of local?

[molten viper]

I guess that's kinda what I'm trying to figure out

I'm a little bit out of my depth

[barren sierra]

Artinian means that R satisfies the descending chain condition

So every descending chain of ideals eventually stabilizes

[molten viper]

Avery chain of ideals has a "least" element?

[barren sierra]

Yea basically

so if we have $I_1 \supseteq I_2 \supseteq I_3 \supseteq \cdots$ a chain of ideals in $R$, there is some $n$ such that $I_n = I_{n + 1} = \cdots$

[cloud walrus]

Spamakin🎷

[molten viper]

Ok yea that's about what I thought

[barren sierra]

But that has nothing to do with locality

You should find a local ring that isn't Artinian and an Artinian ring that isn't local (shouldn't be too hard)

[barren sierra]

At least that's my understanding lol

[molten viper]

It seems like a really strong condition when you have a ring which is Artinian and local

Probably why this author starts with that

Yeah wish me luck teaching myself a lot of very complex stuff over the next year

[barren sierra]

I feel that

I'm self studying some hellish algebraic stuff rn

[molten viper]

Woot woot

[delicate orchid]

algebra hell gang

[molten viper]

The ultimate goal here is producing a result related to totally reflexive modules 😅

A result regarding a tiny special case of a special case, but related nonetheless

[white oxide]

sigh this requires the binomial theorem right

[void cosmos]

yes

why sigh

[white oxide]

because number theory bad

idc if it's not number theory

feels like number theory

number theory bad

[void cosmos]

its just a formula

[molten viper]

It's not really number theory

[void cosmos]

u can come up with it by your self

if you try it

and u can then prove it by induction

[white oxide]

ok this isn't so bad

[void cosmos]

yea with all respect

[white oxide]

but still number theory bad

[void cosmos]

this is just you being a pussy

[white oxide]

nah that's true

i'm too lazy

[rotund aurora]

[void cosmos]

well if ur too lazy about something and ur mind is telling you to procrastinate about something

then ig its a good indicator that that's something u should do

lmfao

[rotund aurora]

bruuhhh literally too lazy to procrastinate

[summer path]

or take a nap and come back to work on it after

c:

[white oxide]

wait isn't this only true in the case that a = 0 or b = 0

since all terms in the binomial expansion include a or b

like for instance

say b^5 = 0

but in the binomial expansion of (a + b)^5, the term 5a^4b is nonzero if a^5 = 0

since being nilpotent is the smallest positive integer that yields the identity

[chilly ocean]

yes.

[white oxide]

good man for that you know that right

[void cosmos]

being nilpotent says that there exists an integer such that a^n=0

[chilly ocean]

if a^n = 0 and b^m = 0 compute (a+b)^n+m using binomial expansion

[void cosmos]

^ yea

[chilly ocean]

i didnt mean it offensively, just that your idea works.. as you asked

[white oxide]

right but what if that integer is not 1, 2, 3, 4, 5

[white oxide]

no i didn't mean it offensively eitehr

like i genuinely call people good men as compliments

my fault

[chilly ocean]

ah my bad

[white oxide]

say in the binomial expansion (a + b)^5, if a^6 = 0 and b^6 = 0

[white oxide]

all g

[white oxide]

or does it have to be the same integer n

[void cosmos]

yes it wont work

but try (a+b)^12

thats good however

[white oxide]

ohh right there exists some integer

my fault

[void cosmos]

getting ur hands dirty with problems u do not know how to do is perfec

it makes the choice of n+m less adhoc

assuming it was in the first palce

[white oxide]

moamen is a good man despite him calling me a pussy

love moamen

[void cosmos]

😦

[white oxide]

why sad face

[south patrol]

is R commutative

lol

[chilly ocean]

alternatively, if R is comm ring then the set of nilpotent elements is an ideal so in particular it is closed under addition

[south patrol]

how do you prove it is an ideal though

[chilly ocean]

zorns lemma

😉

[white oxide]

no clue what an ideal is

[south patrol]

pretty sure that proof requires it to be an ideal right?

[white oxide]

nor zorns lemma

[south patrol]

I was gonna suggest the same thing though aha

But yeah I feel like you appeal to the nilradical being an ideal to prove that

Oh maybe you don't

Anyway, the optimisation isn't n+m right lol

it's n+m-1 i guess

[chilly ocean]

i thought first that it was n+m+1

but then n+m works

[white oxide]

the solution has been spoiled for me

im ruined