groups-rings-fields

[stone fulcrum]

@uneven arch

[sullen flint]

this channel should get a new name

[stone fulcrum]

Not a lot of people know what abstract algebra is, you're probably right.

What's the new name?

[sullen flint]

I don't know lol

groups and rings n shid

[stone fulcrum]

Group/Ring theory? But then nobody will talk about fields

[uneven arch]

Oh thanks Kaynex

Im guessing my question don't belong here

from my understanding this is abstract ?am = n/p?

[sullen flint]

abstract algebra is a field of pure math

that has to do with abstract algebraic structures

and their properties and so on

your question is a practical problem

[uneven arch]

I see

[stone fulcrum]

It's a common mistake. Same with linear algebra, which is actually a study of vector spaces.

Math isn't named well

So, did the hint help? Feel free to ask if you have any other problems with it

[uneven arch]

yeah, but now im trying to fugire out how to convert my sum into a number

I know fractions have a very special process for things like this

[stone fulcrum]

Try cross-multiplying
1/12 + 1/10

= 10/120 + 12/120

= (10 + 12) / 120

= 22/120

= 11/60

Together, they tar 11/60 roofs per hour. So, they will take 60/11 hours to tar a roof.

That's just under 6 hours.

[uneven arch]

im taking note, of your process, I just added them together then tried to round it off, but how did you get = 10/120 + 12/120

Mines (12 + 1) (10+1) = (13 + a = 12) = (13/12 % 2) (12%2=2) (13%2=6.5) 2/6.5

then I converted that to a decimal and rounded the sum

no need to tell me ware I went wrong, my guess I made it to complex

[stone fulcrum]

Note that 1/12 = 10/120
And that 1/10 = 12/120

@uneven arch
If you are allowed to work in decimals, then maybe you can forget fractions.

[uneven arch]

Well thats just how I was tought to do them, i don't get it but thanks

taught

OH i get it

10/12 = 1.2 x100 = 120

life just got a tad bit easier

[sullen flint]

you are doing something wrong

10/12 is definitely not 120

[uneven arch]

i know 10/12 = 1.2

[delicate chasm]

Wrong way up

[uneven arch]

if you do 1.2 x 100

[errant drum]

10/12 is less than 1

[sullen flint]

god hates me

[snow dew]

Maybe they hate each other and they'll take forever to tar the roof

[errant drum]

$\frac{10}{12}$

[sullen flint]

$\frac{1}{12} =\frac{1}{12} \times \frac{10}{10} =\frac{10}{120} $

[cloud walrus]

Cosmicrays:

cardiou:

[errant drum]

Mathbot was better

[delicate chasm]

How? 🤔

[errant drum]

Because you could type =tex and it would just work. Now we got this Texit bullshit.

[delicate chasm]

,$ works

Or you can do ,tex if you set ,tex --alwaysmath

[sullen flint]

,$\frac{1}{12}

[delicate chasm]

Space

[errant drum]

I had an emotional connection to Mathbot okay

[delicate chasm]

Lol

[sullen flint]

wo

[errant drum]

, $ \text{Bring Mathbot back}

[earnest valley]

That's a battle you already lost.

[errant drum]

@earnest valley Can't win with an attitude like that

[chilly ocean]

using $ is more natural

god intended it to be that way

[errant drum]

@chilly ocean But you know who didn't intend it? Mathbot

Checkmate Atheists

[chilly ocean]

the owner and developer of mathbot is not in the server anymore

[delicate chasm]

And that's why mathbot is dead xD

[chilly ocean]

lmao

[uneven arch]

heres another equation same type of problem

working alone it takes dan 10 hours to harvest a filed. julia can harvest the same field in 16 hours. if they work together how long would it take them?

1/10 + 1/16

10%16 = 1.6

1.6 x 100 = 160

26/160 = 16.25

[errant drum]

@chilly ocean One day Mathbot will come back and when he does he'll realise your betrayal

[chilly ocean]

lol I am actually banned from the mathbot server for triggering the owner

I'm too edgy for my own good 😌

[delicate chasm]

LOL

Poor DX

[timber bay]

not sure how to deal with the multivariable division algorithm

i know that x(f2)-y(f1) = x^2 - y^2 but i dont know how to find the positive remainder

oh wait

is it just g(x) = -y(f(1)x+x(f2)?

and +x is your positive remainder?

and -y is like your q?

[thorny slate]

you need a monomial ordering

to carry out the algorithm

any way you do it though, the remainder should be 0

[thorn delta]

I don't know any abstract algebra, but I have a random question. Is there a full proof way to factor polynomials of arbitrary degree known to have rational roots (that doesn't involve guessing)?

[thorny slate]

nope

[thorn delta]

oh okay, thanks lol

[thorny slate]

you just have to guess with the rational root test

[timber bay]

@thorny slate my instructor said the remainder shouldn't be zero

[thorny slate]

maybe I miscalculated then

[timber bay]

the remainder should be -y^2 -y, i have my prof's work (went to office hours) if youd like to see @thorny slate

[thorny slate]

that's fine, I just did it fast

surely I made a mistake

yeah a minus sign

[timber bay]

gotcha

thanks!

[timber bay]

I think I understand the first and third part. still really not sure how to deal with coset addition

by coset, i mean ideal. we are relating cosets to ideals

im still struggling quite a bit with cosets and stuff

anybody know any good resources on that. of course my textbook, but anything thats like exceptionally good>

[earnest valley]

The part i is easy, iii is consequence of I being an ideal.

[thorny slate]

you just need that an ideal is a normal subgroup

[stone fulcrum]

The best way to understand it is to actually use it. Take the cosets formed by the homomorphism Z → mZ. Then the cosets form modular arithmetic.

Take the homomorphism Z → 3Z. The cosets are:
|0| = 0, 3, 6, 9, 12, 15...
|1| = 1, 4, 7, 10, 13, 16...
|2| = 2, 5, 8, 11, 14, 17...

Note that the cosets themselves (which I've labeled |0|, |1|, |2|) form a group of three elements. Some examples:

|1| + |1| = |2|
(any elements from coset 1 added together give an element from coset 2)

|2| + |2| = |1|
(any elements from coset 2 added together give an element from coset 1)

So the cosets themselves form the group Z/3Z.

@timber bay
With that in mind, any questions about how it works?

[timber bay]

Sorry. I'll look in a minute.

Afk for a sec

[stone fulcrum]

So this is enough to make statements like
7 + 11 ≡ 15 (mod 3)

7 from coset 1
11 from coset 2
15 from coset 0

[onyx mirage]

Teach me something

A normal subgroup was a group built out of cycles which could be decomposed into an even amount of subcycles, right

And that had to do with alternating groups?

[timber bay]

i dont think necessarily cycles

a normal subgroup is just a subgroup that is commutative

[onyx mirage]

I might be confused with alternating groups

[timber bay]

maybe a finite group

thanks @stone fulcrum that makes sense. I think I have up to that part understood. its the harder stuff that's throwing me off. stuff for my algebra class

[onyx mirage]

If gH = Hg for all g in G then it's a normal subgroup

[timber bay]

yah

[onyx mirage]

So not only commutative with itself, but also all elements outside the subgroup 🤔

[timber bay]

outside the subgroup

i thik'

[onyx mirage]

Any reason why these boys get a special name

[timber bay]

theyre very useful

[onyx mirage]

outside the subgroup yeah, all elements in the group, hence gH = Hg

[timber bay]

applicable

yeah

[onyx mirage]

I guess because you can faff around with conjugates

[timber bay]

are any <@&286206848099549185> able to help me with part 2 of my problem?

[onyx mirage]

and "normalize" o: them

[timber bay]

did i use that wrong?

yeah, and some prime stuff

[onyx mirage]

Like gHg^-1

[timber bay]

yeah

[onyx mirage]

Prime groups?

[timber bay]

i think if a normal sub is finite then it is of prime order or cardinality or something

[onyx mirage]

That's fucking sick yo

Prime order?

[timber bay]

smallest n such that g^n = e

e is identity

[onyx mirage]

That's definition of order of an element yeah

[timber bay]

n is prime

[onyx mirage]

That's so weird

[timber bay]

im not sure if any of this is correct

[onyx mirage]

You sure though?

[timber bay]

im doing it based on memory

so i may be confusing stuff

i think z/pz is normal

not sure if its iff

[onyx mirage]

Might just be that you're thinking of lagrange's theorem

And the fact that subgroups divide the order of the group

[timber bay]

i dont think so

idk

[onyx mirage]

Do we call abelian groups normal too?

Well, makes sense, because gG = Gg

And G is a subgroup of G

[timber bay]

shit

hm

normal only applies to subgroups

i think

yeah

[onyx mirage]

Proper ones

[timber bay]

left cosets = right cosets => normal

[onyx mirage]

Or all

[timber bay]

probably all

[onyx mirage]

Can't really find anything about the order thing you said

But what you said can't be true then

Because suppose we have a group of order 4, that's abelian

Then trivially it's a subgroup of itself

Suppose it's generated by 1 element, like r

Then we'd have {I r, r², r³} which trivially are all commutative

Hence H is a normal subgroup

But |r|=/=p

@timber bay can I look at your problem?

[timber bay]

its in the caht

my screenshots

okay maybe it is proper then

idk

i could look at it later, but right now im pretty busy with this hw

we should talk later about it though

have you not had an algebra class?

[timber bay]

nobody able to help with my problem?

[thorny slate]

what porblem

[stone fulcrum]

@timber bay
This comes from Fraleigh's abstract algebra

[timber bay]

Yes @stone fulcrum i dont know how to deal with addition

Should it just be the same

[stone fulcrum]

@timber bay
Sorry, what I posted was for a group being quotiented by a normal subgroup

[timber bay]

I'm afraid I don't understand. Isn't this just like part 3 of my question? @stone fulcrum

[stone fulcrum]

Closer to 2, since addition forms a group

Basically, you know that the addition of cosets of a normal subgroup is well defined.

Think of an ideal as an additive normal subgroup (because addition is commutative in rings). Then the above is the exact proof

[timber bay]

Oh.

Hm.

Okay

[stoic citrus]

x + 1 = x + 2

[onyx mirage]

When do we call things an additive group or a multiplicative group if their binary operations aren't addition or multiplication?

[earnest valley]

Additive because it reminds us of commutativity and multiplicative if it isn't commutative.

[onyx mirage]

strange dichotomy 🤔

It holds for matrices, and I guess similar structures, so for those it "makes sense"

[earnest valley]

Yes.

[toxic plinth]

@placid egret DON'T LOOK SOLUTION SPOILER

https://media.discordapp.net/attachments/360581765661982720/482052171765121034/Screenshot_from_2018-08-23_01-03-02.png

i put it in this channel cuz this proof works for any ring

whoops i forgot to add the binomial coefficient in the last sum

[stone fulcrum]

What's up? Have an example?

[thorny slate]

chances are it's not

[sick acorn]

Anyone have any suggestions? I'm supposed to show:
Let G be a group such that G != {1}. Show that G is simple if and only if, for every group H and homomorphism f : G -> H, either f is trivial (i.e., Im f = {1} ) or f is injective (i.e., Ker f = {1} ).

I already showed that simple => trivial or injective

having a harder time showing the converse

so far, i have:
(<=) Assume either Ker f = {1} or Im f = {1}.
Im f = {1} <=> f(G) = 1 <=> Ker f = G.
So either Ker f = {1} or Ker f = G. Know Ker f <| G.

[thorny slate]

Hint: Assume G is not simple, so that it has a normal subgroup H. Can you construct a map from G to something with kernel H?

[sick acorn]

ok, thanks! i'll try to think about and do tah

so a proof by cont

H is a new subgroup in G that’s unrelated to the H in f:G->H, right?

I’m also not really sure what Kernel H means

[thorny slate]

yeah it's a new subgroup

it means a map f : G -> K with Ker(f) = H

think about normal subgroups and what you can do with them

[sick acorn]

oh okay got it

[sick acorn]

for uh

i'm checking to see if i'm thinking of the problem correctly. the goal is to show

(x^n)h1 * (x^m)h2 = (x^m)h2 * (x^n)h1, where h1 and h2 in H?

[north tide]

Yes

[errant drum]

@sick acorn I mean you're almost there

[sick acorn]

oops got it

Thanks!!!!

By closure, x^n, x^m in G.
Know H <= Z(G).
So we have h1h2 = h2h1,
h(x^n) = (x^n)h, and
h(x^m) = (x^m)h.

Then (x^n)h1 * (x^m)h2
= (x^n)(h1 * x^m)h2
= (x^n)(x^m * h1)h2
= (x^n * x^m)*h1h2
= (x^(n+m)) * h1h2
= (x^(m+n)) * h1h2
= (x^m * x^n)*h1h2
= (x^m * x^n)*h2h1
= (x^m)(x^n * h2)h1
= (x^m)(h2 * x^n)h1
= (x^m)h2 * (x^n)h1.

[open wraith]

I'm trying to prove a result in Galois theory:
Let f(x) be a polynomial in F[x], where F is a field. Let E be the splitting field of F. Any automorphism in Aut(E/F) is defined by the way it permutes the roots of f(x), thus inducing a group embedding Aut(E/F) -> S_k. PROOF: Since E = F(a_1, a_2, ... , a_L) = F(a_1, a_2, ... a_(L - 1))(a_L), it may be shown by induction that E is the set of multivariable polynomials in a_1, a_2, ... a_L over F, and therefore any automorphism of Aut(E/F) is determined by where it maps those roots, and since it also permutes those roots, we have that it is uniquely determined by the way it permutes the roots.

I don't feel like this is a legitimate proof, can anyone check ?

[thorny slate]

dunno looks fine

[dusky oxide]

Hi, I'm having a little bit of trouble with inner and outer semidirect products
Based on the wikipedia article it appears that they are different things, but I'm not entirely sure. Is the inner semidirect product a way to write G as the semidirect product of N and H when you already have G where as the outer semidirect product is a way to define G as the semidirect product of N and H when you already have N and H, and (you choose which one you want to be normal in G)?

[thorny slate]

yes

[cerulean rune]

the short answer to that question is "yes" -- some remarks:

• just a reminder that you need an action of H on N by automorphisms (or the other way around, but by convention N is the thing we call the normal subgroup), which typically ("in the wild") distinguishes between H and N. Like, you don't normally get to "choose" which subgroup is normal, because typically you'll only have an action of one on the other

• as you suspect the difference between "inner" and "outer" is mostly "pedantic" here. like, the claims "G is the (inner) semidirect product of H and N" sort of implies that H and N come to you as subgroups of a group G that you already have, whereas "G is the (outer) semidirect product of H and N" suggests that G has been constructed from groups H and N and an action of H by automorphisms on N.

[thorny slate]

the definition for inner is easier because the action of H on N is just conjugation

that's the only difference

[cerulean rune]

(i worry that is misleading to call that a "difference". the action of H on N becomes conjugation inside the (outer) semidirect product. one might just say that the outer semidirect product is an inner semidirect product of the (obvious inclusions of) H and N)

[thorny slate]

yeah it's not a real difference

[dusky oxide]

@covert seal @thorny slate Ah, thanks guys!

[vestal needle]

can someone help explain to me part c

[stone fulcrum]

(1) is the identity, the permutation where nothing changes.
(23) is the "switch 2 and 3" element.

These make a subgroup because (23)² = (1)

You can find the left cosets by taking elements in S3 and multiplying them on the left of {(1), (23)}

This example is a bit more interesting, since S3 is non-abelian

[vestal needle]

yeah i think i did that part already

hold on i can show you my work

[stone fulcrum]

The number of left cosets you may find is the index. This number should divide 6

[vestal needle]

So are they all not left cosets

[stone fulcrum]

You can't reduce a subgroup of two elements into one element

[vestal needle]

i thought left coset just meant you take an element in the group G and left multiply it by the elements in the subgroup H and that is your left coset

[stone fulcrum]

Yes

[vestal needle]

so if there are 6 elements in G that must mean there are 6 left cosets since each element creates its own left coset

[stone fulcrum]

For example
(123)H = {(123)(1), (123)(23)}
= {(123), (12)}

[vestal needle]

oh but wait

yeah cause two of them can be equal to each other

so is it only the unique left cosets

[stone fulcrum]

So {(123), (12)} is a left coset of {(1), (23)}

[vestal needle]

oh

[stone fulcrum]

Left cosets should partition the group. So you will have 3 left cosets of 2 elements each

To cover all 6 elements of S3

[vestal needle]

ok i think im starting to get it

[stone fulcrum]

Cool cool, feel free to ask if you have any more questions!

[vestal needle]

thanks

[civic linden]

how do i work out the elements in H?

[thorny slate]

compute the composites of that cycle

to understand how they work

[civic linden]

i dont know what to composite what with :/

[stone fulcrum]

<(12345)> means the group generated with powers of (12345).

Keep multiplying by (12345) until you get the identity

[civic linden]

oh right ok

[pallid ginkgo]

you should also be able to spot a "pattern"

[civic linden]

is there a quick way of doing this, or do i just keep plugging the vals

oh right ok

[pallid ginkgo]

as in, knowing how many times you have to multiply is something you can figure out easily

but is the question really asking for all the elements?

[thorny slate]

just keep plugging

[civic linden]

so (1,2,3,4,5)^2=(1,3,4,5,2)?

or have i done this wrong

[stone fulcrum]

(12345)² = (13524)

[civic linden]

can explain how u got the ans pls?

ive forgot permutations havent done them in so long ;x

[stone fulcrum]

(12345)(12345)

We read the right one first. In that, 1 → 2. Then, we read the left. 2 → 3. In total, 1 → 3

[civic linden]

oh right

[stone fulcrum]

Then look at 3. It is sent to 4, then is sent to 5

Then 5 is sent to 1, which is sent to 2

[civic linden]

ah okay, thanks, i think i remember it now , great

[dusky oxide]

Hey, I just got back from an algebra exam and I'm feeling a little bit unsure of myself

I was wondering if my proof is correct for this problem

Let G be a group with normal subgroups H, K such that H intersect K = 1. Then show that hk = kh for all h in H, k in K

I said that due to H, K normal and H intersect K = 1, G ~= HK ~= H x K

And since in H x K, elements (h,1) and (1,k) commute, the corresponding elements h, k in G must commute

[thorny slate]

that's true

but it's kinda circular

when you prove that due to H, K normal and H intersect K = 1, G ~= HK ~= H x K

one of the steps is actually proving that hk = kh

maybe look up the proof of this fact

[dusky oxide]

ah shit

I was also wondering another thing

if you are trying to find the number of groups of order 5 x 7 x 11 up to isomorphism

can you deduce that 7 is normal, find the semidirect product of 7 and any sylow 5-subgroup, deduce that the only result is Z_35, then since 11 is normal, find the semidirect product of Z_35 with Z_11 and use the result from that as your answer

I think its standard to do Z5 semidirect product with Z77

[thorny slate]

you're definitely supposed to use the sylow theorems like that

let's see

the 7 groups divide 55 and are 1 mod 7

so yeah that's 1

so 7 is normal

and the 11 groups divide 35 and are 1 mod 11 so that 1 too and 11 is normal as well

but Z5 and Z7 don't have semidirect products

Aut(Z7) = Z6 doesn't act on Z5 nontrivially

so yeah the only result is Z35 there

[dusky oxide]

Yeah thats what i did haha

[thorny slate]

and Aut(Z11) = Z10 can act on Z5

[dusky oxide]

Then you find homomorphisms from Z35 to Z10?

[thorny slate]

so you have what, Z7 x Z5 lx Z11

[dusky oxide]

Yeah

[thorny slate]

and that should be it

yeah you have to classify actions of Z10 on Z5

which is just actions of Z5 on itself

[dusky oxide]

Its only trivial and then x -> a^2, a^4, a^6, a^8 (all of which correspond to same group up to isomorphism) right

[thorny slate]

yes

[dusky oxide]

So 1 abelian and 1 non abelian group

ok

[thorny slate]

not sure if all of them are the same group

2 and 8 probably are the same, same for 4 and 6

nor sure if 2 and 4 are the same

you'd have to compute it

[dusky oxide]

Yeah I put into wolfram it said all them are the same

[thorny slate]

yeah I think so

[dusky oxide]

I just wasnt sure about my approach

[vestal needle]

guys how do i do this

[thorny slate]

you need a matrix which is its own conjugate transpose and inverse

can you think of one?

[vestal needle]

huh we havent done matrices in this class that im in

[thorny slate]

what

what's U(n)

I thought it mean the unitary group

[vestal needle]

all the numbers that are relatively prime to n and smaller

[thorny slate]

but it probably means (Zn)x?

[vestal needle]

smalle than n

[thorny slate]

lmao ok

[vestal needle]

i found a source online but i only understood the explanation up to a point

https://math.stackexchange.com/questions/877524/prove-that-the-order-of-un-is-even-when-n2

what do they mean when they say phi(2) = 1

[thorny slate]

they're using the multiplicativity of the euler totient

phi(2) = 1 clearly right?

in general phi(p) = p-1 for primes

[vestal needle]

oh ok

[thorny slate]

and you can show that phi(p^k) = (p-1) p^{k-1}

but you don't need to do anything so complicated

the question just asks you to find an element of order 2

the second answer in that link has a solution in that way

[vestal needle]

oh i see it

On the other hand, for n>2, the order of n−1 in U(n) is 2.

wow i didnt know that cool

[thorny slate]

just compute it

[vestal needle]

ok thanks i think i got it

[thorny slate]

If you notice that's just -1

And clearly -1^2 = 1

N=2 fails cuz 1 = -1

[inner acorn]

Quick question, looking at the definition of an algebra (vector space A over a field F with a bilinear map m), and noticed:

1. The definition here also requires a unit 1 such that m(1,a) = m(a,1) = a http://math.ucr.edu/home/baez/octonions/node2.html

2. Wiki however omits this unit
   https://en.wikipedia.org/wiki/Algebra_over_a_field#Definition

Is this just an author preference thing like with ring theory (e.g. subrings may or may not require unity?)

[thorny slate]

you call it unital if there's a unit

a lot of people assume their algebras to be unital associative

[inner acorn]

ah, thankyou 😃

Ah, I see wiki says this lower down!

[inner acorn]

Hmm.. I feel like I've gone down a rabbit hole, but trying to show for an algebra A over a field F:
if ab = 0 implies a = 0 or b = 0 for all a,b in A...
THEN
A is a division algebra (i.e. all a,b in A, b non zero, there exists a unique x,y in A such that a = bx AND a = yb)

I've shown uniqueness:

Suppose a = b(x1) & a = b(x2) with x1 =/= x2
Then x1 - x2 =/= 0
Hence b(x1) - b(x2) = b(x1 - x2) = 0 due to our multiplication being bilinear
But then b =/= 0, so our assumption fails

[cerulean rune]

is that even true? isn't k[x] a simple counterexample?

[inner acorn]

Hmmm, maybe the statement requires the algebra A to have finite dimension as a vector space?

if that's so I think I know the proof

I think you're right https://proofwiki.org/wiki/Division_Algebra_has_No_Zero_Divisors
This threw me for a loop n.n;;

[thorny slate]

yikes

[inner acorn]

Okay, looking around the internet for definitions concerning Algebras is starting to get annoying - anyone know a source/textbook which has most the definitions I'm looking for?
-algebra
-unitary/associative/commutative algebra
-division algebra
-normed division algebra
-*-algebra/involutive algebras

[thorny slate]

what for

[inner acorn]

Trying to gather enough of a background to play with the Cayley-Dickson construction

but I'm constantly fighting against an author's preference to use/omit units xD

and the last thing I want are definitions were.. don't actually work :c

[thorny slate]

https://math.stackexchange.com/questions/1213599/algebra-book-that-covers-algebras

[inner acorn]

Hmmm, the first book they listed mainly goes over intro group/ring/field theory without much focus on algebras by themselves. And the second book seems to be locked behind a paywall T~T

but thanks for looking

actually, looks like my university library might have that one :DD

[snow dew]

What results are there in algebras that are more than results on their vect space/group/ring?

[errant drum]

How can I show that even though 1 is a gcd of 3 and 2+√(-5). There are no a,b in Z[√-5].

Where 3(a) + b(2+√(-5)) = 1

I think I got it

[thorny slate]

essentially b has to be in Z times (2 - sqrt(-5))

and then you check what happens

[rocky dagger]

Hi, I'm hoping someone can help me with some of my group theory problems?

"Prove that the action of the symmetric group S_n on the set of its transpositions is primitive when n != 4"

[thorny slate]

check that the stabilizer is a maximal subgroup

for any transposition

[covert vector]

@marsh rose what's up?

[marsh rose]

@covert vector I'll send you a picture of the proof in a sec

But basically it's a proof that a kxk matrix is invertible if rank = k

But when I look at it I just get so lost

It's probably not that hard it's just I've been slamming my head against it for too long

Now it won't send the picture

Ignore everything below example

There we go

[covert vector]

wait

what does AA^t have to do with anything

usually the definition of rank is the number of linearly independent columns in the matrix = dimension of column space

and invertible means there exists an inverse matrix B such that AB = BA = I

[thorny slate]

yeah it's very weird proof

looks bogus

[covert vector]

indeed

[marsh rose]

I get why it's the case too

Like I understand how rank and invertibility works

But this proof

It makes no sense

And what the hell does it have to do with QR Factorization

[marsh rose]

@covert vector so I'm not alone

It makes sense why it would be invertible

But this proof makes no sense

[covert vector]

yeah, the "proof" is dumb

[marsh rose]

Cool thanks

We have 1 more class before the final. I'm going to ask him to explain it

[drowsy meteor]

What does it mean for a ring or element in a ring to be unique?

[thorny slate]

each element is unique?

it doesn't mean anything

that there is a unique element satisfying P means that there's only one element satisfying P

[lone shard]

you'll often see this in the context of like "the identity element is unique" meaning that any element satisfying the properties of the identity must be the identity element itself

[drowsy meteor]

I'm looking at this theorem that says
"If a ring has a unity, it is unique. If a ring element has a multiplicative inverse, it is unique"
Is that the same context?

[thorny slate]

yeah

means there's only one unity

and each element has exactly one inverse

[drowsy meteor]

Ah okay, I misread the wording, I thought it was saying that the ring and the element was unique

Thanks so much

[inner acorn]

Is this actually true?
https://math.stackexchange.com/questions/1253237/

Lots of sites I'm reading require the algebra A to be unital, finite dimension, AND associative (see https://ncatlab.org/nlab/show/division+algebra )

[thorny slate]

it's not only true, there's several proofs there

oh

algebra means unital associative

and they say finite dimensional too

[inner acorn]

see now, in their proofs I can't see associativity used o:

[thorny slate]

surely it must be used

implicitly at least

[inner acorn]

that's what I'm struggling to find

I mean, the fact the algebra multiplication is bilinear is used, many times..

maybe I'm interpreting associativity wrongly as bilinearity somewhere :c

For simplicity, I'm trying to find where/if associativity is used in Robert Lewis's response
https://math.stackexchange.com/q/1253297

[thorny slate]

what

it's used a bunch of times

explicitly

in (1), in (7)...

7 might be bilinear

dunno

[inner acorn]

(1) is associativity of the ring/field elements
(7) pretty sure is bilinear + maybe power associativity

[thorny slate]

yeah

I guess it works then

dunno how non associative algebras work

is a^2.a = a.a^2 even?

maybe you don't have that

so a^n doesn't make sense

[inner acorn]

That's power associativity.. which I guess is implied here?

But what worried me is:
https://ncatlab.org/nlab/show/division+algebra
The first bullet point in the definition, contradicts the proofs on math.stackexchange

[thorny slate]

maybe power associativity implies associativity

or maybe it's enough

[inner acorn]

power associativity is a weaker condition o:

[thorny slate]

I wouldn't know

I've only ever used associative algebras

[inner acorn]

the octonions are power associative but not associative

fair enough :<

[thorny slate]

I see

but maybe it's enough for the proof

since it's about polynomials in that particular element

oh you need more

you need like

a(a^2 + a) = a^3 + a^2

oh

this is distributivity

yeah I don't know how algebras work

but you get the point

[inner acorn]

I see that power associativity is required for the proof,
however what concerns me is that the ncatlab site says:
"It's easy to construct nonassociative unital finite-dimension algebras over R, such that A has zero divisors BUT each element a in the Algebra, there exists a^-1"

[thorny slate]

so what

we didn't prove it's iff

just the other direction

[inner acorn]

which contradicts the proofs on the math.stack exchange.. which say
If A is a nonassociative unital finite-dimension algebra
A has no zero divisors iff every a in A has a left and right inverse

it is an iff o:

the otherway is easier to show

[thorny slate]

show the other way then

and check where you use associativity

[inner acorn]

Let A be an unital finite dimensional Algebra

Statement 1:
For all u, v in A, uv = 0 implies u = 0, v = 0

Statement 2:
For all a,b in A, b nonzero, there exists a unique x & y such that
a = bx
a = yb

The proofs on the stack exchange show (existence) Statement 1 => Statement 2

Proof of Statement 2 => Statement 1:
ab = 0
if a = 0, then we're done
if a =/= 0, then there exists a unique b in A such that ab = 0
we also know by the zero vector a0 = 0
so by uniqueness b = 0
Hence ab = 0 => a=0 or b=0
QED

Proof of (uniqueness) Statement 1 => Statement 2
for a, b in A, b nonzero
suppose there exists an x1, x2 in A (with x1 =/= x2) such that
a = b(x1)
a = b(x2)
then x1 - x2 =/= 0 
also 0 = b(x1) - b(x2) = b(x1 - x2)
then b = 0
Hence by contradiction x1 = x2
(same argument for y1 & y2)
QED

<@&286206848099549185>

[thorny slate]

I don't think you're gonna get much

it's unlikely that nlab is wrong

so the proofs use associativity somewhere

I really don't wanna comb through the full proofs line by line

but you probably have to

[inner acorn]

I've tried combing through the proofs line by line though qwq and I honestly don't see where associativity is used...

[thorny slate]

damn

maybe try to find a counterexample as nlab says

construct one or look for it

and then try your proofs there

[inner acorn]

I wouldn't know where to start looking for one I'm afraid n.n;;

[thorny slate]

like google I mean

but yeah it might not be easy to find

dunno man I feel like your pain

but I don't wanna get into it because it looks very annoying lol

[inner acorn]

T~T

[inner acorn]

omg xD why do so many sources say different things...

https://en.wikipedia.org/wiki/Cayley–Dickson_construction
https://ncatlab.org/nlab/show/Cayley-Dickson+construction
Both have (a,b)(c,d) = (ac - d*b, da + bc*)

http://math.ucr.edu/home/baez/octonions/node5.html
Has (a,b)(c,d) = (ac - db*, a*d + cb)

Hmmm, nvm, turns out there're 8 equivalent definitions for the Cayley Dicksons product o.o
https://www.researchgate.net/publication/290219081_The_Eight_Cayley-Dickson_Doubling_Products

[marsh rose]

I figured out the proof

I'm an idiot

Rank is equal to number of linear independent rows

A^tA is a kxk matrix

So rank = k means k linearly independent columns

So the whole thing is linearly independent

But to prove that we need to show that A and A^tA have the same Nullspace

Since if they do that means they have the same nullity

And if they both have a nullity of 0 then they both have to have a rank of K

[marsh rose]

@covert vector I figured it out

IDK if your interested

Oh shit the username changes on every server lol

[covert vector]

are u a pickle

[marsh rose]

It's Pickle Rick Dabbing

It's a joke on another server

We generalized that all athiests are probably Rick and Morty fans

That's why it was Pickle Rick and my username was God Is Not Real

[plain solstice]

all rick and morty fans are atheists but not all atheists belong in the subset of rick and morty fans 😉

[inner acorn]

Okay umm, Galois professor is mean and wants us to find all the normal subgroups of the dihedral group order 12:
D12 = { (r^n)(a^m) : n=0,1 & m=0,1,2,3,4,5 }

Obviously D12 & {1} are normal
Also every subgroup of index 2 is normal i.e. <a>, <a^2, r>, <a^2, ra>
And I'm pretty sure <a^2> & <a^3> are normal too

Am I missing any? n.n;;

[inner acorn]

<@&286206848099549185>

[inner acorn]

https://in.answers.yahoo.com/question/index?qid=20091014113730AA7KJDt

I think this answered it c:

[sonic current]

Hello guys! I have a problem in a question. May I ask for you help?

Number 9

So far, I've used the Lagrange theorem to establish that the order of a group element must divide its group order; and then (Kg)^n = K(g)^n = K

So Kg has order n.

Is this correct? Then obviously n divides n

But I think it's too easy and I might be falling short somewhere

[lyric falcon]

(Kg)^n=K

but that doesn't imply that the order of Kg is n

it does directly imply that the order of Kg divides n though

[sonic current]

Hmm

[lyric falcon]

in general if $g^k=e$ then $k|\ord g$

[sonic current]

Can the order of Kg be smaller than the order of g?

[cloud walrus]

Gonzo17:

[sonic current]

I thought about that Gonzo!

But other than that, do you agree with the reasoning?

Can I conclude the exercise?

[lyric falcon]

yeah it's good

I'm just trying to think of an example where the order of Kg isn't n

oh

[sonic current]

Yeah, I tried that but didn't go too far in my thinking

[lyric falcon]

you can just have g be any element of K

then the order of Kg is 1

[sonic current]

But the order of g is fixed as n

Wait.

Hmm

I don't see it

My question is: if g has order n, can the order of Kg be less than n?

[lyric falcon]

right, but for example if $2\in\mathbb{Z}_6$ and we take the normal divisor {0, 2, 4}

whereZ6 is {0;1;2;3;4;5} with addition

[cloud walrus]

Gonzo17:

[lyric falcon]

then {0, 2, 4}2={2;4;6}={2;4;0}

so the order of Kg is 1 because Kg is just K

and the order of g is 3

[sonic current]

Oh I see! Thanks

I'll add that to the exercise as a side note

Let me just check if the exercise is nice

I don't have to relate to G/K in any way?

I don't think so. It's good

Thanks for you help @lyric falcon

[lyric falcon]

no problem 😁

[sonic current]

(what do you mean by normal divisor?)

[lyric falcon]

oh sry I meant normal subgroup

[sonic current]

Okay! Thanks

[lyric falcon]

they do mean the same thing

[sonic current]

I havent mad much progress in the ex 10. @lyric falcon

Do you have any tips?

I know there are m distinct cosets of K in G

[lyric falcon]

sry I was afk

[stone fulcrum]

G/K is a group with |G|/|K| = m elements. Each element's order in G/K divides m, so g^m = e in G/K.

That only happens if g^m = K in G

[lyric falcon]

the end of the first line

shouldn't it be gK^m=e in G/K?

[stone fulcrum]

Maybe, that makes sense

[lyric falcon]

after all $g\in G$ and $gK\in G/K$

[cloud walrus]

Gonzo17:

[stone fulcrum]

Can we salvage my line of thought?

[sonic current]

I agree with Gonzo here

[lyric falcon]

since $g_1K=g_2K$ implies $g_2 {g_1}^{-1}\in K$

[cloud walrus]

Gonzo17:

[lyric falcon]

and we have g^mK=(gK)^m=K

[stone fulcrum]

I mean to take the coset that g belongs to, that coset^m = e

[lyric falcon]

^

that seems to work

[stone fulcrum]

I was thinking to put φ(g)^m = e but eeh

Might have been confusing

[sonic current]

Hmm. seems ok

g^mK = K implies g^m belongs to K

Cool! Thanks you both

[uncut girder]

Example of nontrivial field of fractions?

Can you make an integral domain where the multiplication is function composition?

If so what does the field of fractions look like

[covert vector]

if the multiplication is composition, what is the addition? @uncut girder

and what is the additive identity

[uncut girder]

IDK you tell me

For example consider the set of linear functions F(x) = ax +b.
Then I think this is a field.

@covert vector

[covert vector]

but let f(x)=ax+b, g(x)=-ax+b

f+g=2b

constant

but this is not invertible under composition

and is Nonzero

[uncut girder]

Oh right

So is it an integral domain

[covert vector]

take f(x)=x-2b
g(x)=2b for all x

then f(g(x))=0 for all x

f,g nonzero

so not an integral domain either ^^

[uncut girder]

Oh I see

Unlike groups, the nicer rings are more numbery than transformationy

[timber bay]

im trying to find the left cosets of a subgroup of S_4. i don't really know how to find them without going through lots of mindless computations. anybody got any tips?

[thorny slate]

which subgroup

probably gonna be ugly anyway

but you can exploit symmetry to make it faster

[timber bay]

here's my question. i have a and the first part of b

[lyric falcon]

do you understand all the definitions?

@timber bay

[timber bay]

im very rough with cosets, but i understand the definitions

hardly know how to use them

[lyric falcon]

so you were propably told a specific type of groups, which are always normal subgroups (if subgroups)

do you recall anything like that?

[timber bay]

what, the orbit?

or isnt stab(x) always normal

no, im thinking of the center

[lyric falcon]

it's cyclic groups

[timber bay]

oh okay

[lyric falcon]

oh wait

I think I might've fucked up

yeah @timber bay so sry forget what I said

[timber bay]

lol

okay

[stone fulcrum]

There's only 6 such left cosets. May be a little ugly, but you can find them I'm sure

[timber bay]

yeah i dont know how to begin this

would it be fair to say that the integers over addition is a finitely generated infinite group?

you can keep adding 1 to get all integers but its infinite

[chilly ocean]

Infinite group that is finitely generated, yes

[timber bay]

sick

[stone fulcrum]

Note that the identity is (always) in your normal subgroup, so the left coset will include the element that you multiply with.

Don't use (12)(34), (13)(24), (14)(23) or else you'll get that normal subgroup back. When you find a new coset, all of the elements in that coset are then "used".

You only need to perform 6 multiplications to find these 6 cosets.

[timber bay]

yeah i just dont know how to do it unless i try all of the other 18 elements of sv

s4

[stone fulcrum]

No. Try an element that is not in the subgroup. You'll find a new coset. Every element in this new coset is "used". Don't multiply with these.

You only need to do 4 multiplications, actually.

You already have one coset
4 multiplications to find 4 more
Whatever is left is also a coset.

[timber bay]

oh okay

so ill choose one element outside H, then multiply all elements of H by that element, and that will give me another coset, and rinse and repeat

[stone fulcrum]

Yus yus. But don't use an element that you already know the coset of, or else you'll just get that coset back

[timber bay]

cool thanks! that's a good start

[timber bay]

i remember doing the group operation tables before with these permutation groups. do you know what it means by asking me to write the multiplication table for the quotient group? like what would I have on the y column and the x row? @stone fulcrum

[stone fulcrum]

You can multiply any two cosets together. Take any element from coset a, any element from coset b, multiply to get an element from a new coset, which you can call ab

Note that coset multiplication is only well defined if N is a normal subgroup

The six cosets will form a group which you can call S4/H

[timber bay]

oh okay so these will be on my x and y columns/rows

got it thanks

[stone fulcrum]

Note the lovely fact that there's only two groups of 6 elements, Z6 and S3. So S4/H is isomorphic to one of those.

[timber bay]

is Z6 the same as Z/6Z?

[chilly ocean]

whats good book on abstract algebra

[cerulean rune]

i don't have a wide breath of experience, but: as an undergrad i used Gallian, which is fine but certainly missing a lot of important stuff that you would want to know, like, coming out of first-year grad school (he barely touches on modules and has not much on galois theory); as a grad student we used Dummit & Foote which was fine and has later served me as a fine reference

i have the vague impression that Dummit&Foote is the "standard", for whatever it's worth. both gallian and d&f have a lot of exercises so that's nice

[timber bay]

is there a way to find fixed points and stabilizers of elements in S4 without going through all the multiplication?

doing conjugation

[stone fulcrum]

I'm used to Fraleigh's abstract algebra, it was pretty good to me.

@timber bay
Yes those are the same thing

[timber bay]

what are the same thing?

oh, the fixed points and stabilizers of conjugation are the same thing, arent they

hmm

[stone fulcrum]

No no, Z6 = Z/Z6

[timber bay]

is the only way to find them just to try them?

ohh

[stone fulcrum]

I don't know off hand what stabilizers and or fix points are

[timber bay]

oh okay

[chilly ocean]

@cerulean rune but im undergrad

[solar wyvern]

@chilly ocean hodgy?

[hot wyvern]

,$$\bmqty{1 & 1& 1 \ 0 & 1 & 4 \ 0 & 1 & 1 & a+1\0&0&0}$$

can anyone fix this

[whole basalt]

So, question about field extensions. Suppose I have a polynomial of degree 5 that is irreducible over Q. Now suppose that in some field extension of Q, it partially splits onto a quadratic and a cubic, but that's it. What if anything would that tell me about the degree of the field extension in which this occurs?

(Assuming this could happen at all)

I'm studying for my final Monday, and trying to fill in some remaining holes on my understanding.

[thorny slate]

wow I have no idea

@bleak abyss @gentle pendant @delicate chasm ?

[whole basalt]

Oh hey @thorny slate

This is an effort to try to generalize that question I asked about a few days ago.

[delicate chasm]

Interesting

[whole basalt]

Get an idea how partial splitting works in general, if that's even a thing.

If it helps, the original problem that motivated this was to factor x^16 - x over F_4 and F_8. When I did it over F_2, there were some irreducible quartics, which over F_4 all split into quadratics. When I did it over F_8, none of those quartics split.

[thorny slate]

But this is different

Cuz the degree 5 thing splits

But into no linear factor

[bleak abyss]

So if you take that field extension F where you split into a quadratic and cubic, and then look at the splitting field E, I think E should be the composite of the splitting fields of the quadratic and cubic over F?

[gentle pendant]

Well, you guys have probably already thought of this, but you should get an elementary bound for the degree of the partial extension F by observing that to split the polynomial requires a further extension of degree at most 3!*2!=12.

maybe that doesn't work out so well idk

[bleak abyss]

If that's the case, then E over F should have degree divisible by 6 and at most 12, and E has degree at most 120 over Q. So this gives that F has degree at most 20 over Q

Partial snipe lmao

[gentle pendant]

nah I was writing out more details but then deleted when you started typing along similar lines lol

[bleak abyss]

There's probably something more you can say if you note that picking up the two roots of the quadratic in the decomposition you get in F is gonna be an extension of degree at most 20 which strictly contains F. So actually that suggests F has degree at most 10?

[whole basalt]

Okay so remind me... what do you mean by composite

That was VERY briefly mentioned in my class

But not in any detail

[bleak abyss]

So basically, let's say you have two subfields of a given field, their composite is the intersection of all fields containing both

[chilly ocean]

Wouldn't that be just the intersection of the two subfields themselves?

[gentle pendant]

No

The intersection of all fields containing both in particular will contain both of the subfields.

The intersection of the two subfields themselves only has this property if the subfields coincide.

(So its the smallest common extension he is talking about, whereas you are talking about the biggest common subfield.)

[solar wyvern]

[thorny slate]

so any ideas on the polynomial which splits with no roots added?

really nags my noggin

[chilly ocean]

Gomez, you're right. Because the fields themselves don't contain each other.

[gentle pendant]

What exactly are you looking for an example of @thorny slate ?

[thorny slate]

irreducible polynomial in field K which in an extension E|K factors into things but has no linear factor

[gentle pendant]

x^4-2 in Q

adjoin sqrt(2)

and you can factorise into the product of two quadratics

(difference of two squares)

but you still don't get any of the roots

[thorny slate]

oh sorry I want it to factor into degrees p+q coprime

[gentle pendant]

ah

[whole basalt]

Well I didn't realize I'd asked such a bizarre question. XD

[bleak abyss]

@thorny slate so, let's say you take an irreducible polynomial of degree 5, adding a root requires a degree 5 extension of Q, so if you were to have a degree 6 extension of Q inside the splitting field I feel like the polynomial will have to partially split, but it can't have a root

[thorny slate]

is there such a thing?

I would think the degree 6 extension only becomes available after you did the degree 5 one

oh

that can't be, can it

[bleak abyss]

I mean, if you have a polynomial whose Galois group is S_5

[thorny slate]

yeah I know what you mean

let's try one of those

[bleak abyss]

Then you'd be looking for an index 6 subgroup I think

[thorny slate]

so

quintics with exactly 3 real roots

x^5 - 2x + 1

that's one

what do

wtf this isn't one

idiot blogpost

[bleak abyss]

Lol 1 is a root

[thorny slate]

yeah

I grabbed it out of a blogpost

ugh

2x^5 - 5x^4 + 5

ok

so we try to get degree 3 and 2 factors somehow after adding roots

then add only roots of those to Q

and that should do what we want?

well no

well yes?

cuz they are prime

and coprime to 5

can you build the extensions of this guy

:))))

[bleak abyss]

It'd be tricky to pull this off explicitly, this guy can't be solved in radicals

[thorny slate]

@spark plank

yeah but you can just add a special root

and express things in terms of that

simpleart knows how to do that

[bleak abyss]

But its complex roots come in conjugate pairs. Let's say the complex roots are $\alpha$ and $\overline{\alpha}$. I think $\mathbb{Q}(\alpha + \overline{\alpha})$ would do it

[cloud walrus]

Daminark:

[spark plank]

Solving quintics?

[thorny slate]

yeah can you solve 2x^5 - 5x^4 + 5?

it has 3 real roots

2 complex

[spark plank]

Let u = 1/x

You easily get Bring Jerrard form

[thorny slate]

@bleak abyss if the product is in Q yeah

it probably isnt

lst me check

[spark plank]

Well busy rn ttyl8rs

[thorny slate]

:(

yeah I think their product isnt in Q

so what, add their product too?

Q(a + a*, aa*)

I wanna say that works

one of them should be degree 5 and the other degree 2

no

probably not

I mean this is definitely gonna split it into a degree 2 and something else

but I suspect it's a degree 2, a degree 2 and a degree 1

https://www.wolframalpha.com/input/?i=roots+of+2x^5+-+5x^4+%2B+5

wanna check?

:^)

[bleak abyss]

Lol I've been toying around to see if there's an easier way to do it but I can't think of any nice explicit example

Anyway it turns out some people I know have acquired Smash Ultimate so I'm prob gonna head over and play some to procrastinate even more on grad apps

But at least this gives us an inexplicit way to pull this off, in general if an irreducible quintic has a Galois group that has a subgroup of index 6, then the fixed field of that subgroup is gonna field extension of Q of degree 6. If I'm right that the polynomial can't remain irreducible, then it'll factor there and not have root. If I'm wrong then 😭

Lol @whole basalt your question led to a fun talk for sure 😛

[whole basalt]

Yeah apparently 😛

It's been a fun read though!

[thorny slate]

someone compute Q(a + a*, aa*) for 2x^5 - 5x^4 + 5 where a, a* are the complex roots

it might be an example

[spark plank]

[thorny slate]

do it

like

factor the polynomial

into the degree 2 factor made of the complex roots

and the rest

[bleak abyss]

Yeah it's work let SA do it

[upper dome]

Why are pi and sqrt(pi) not independent transcendental numbers over Q?

[thorny slate]

pretty sure they are Q-linearly independent

but they're algebraically dependent

cuz x^2 - pi = 0 is satisfied by sqrt(pi)

so that sqrt(pi) is algebraic over Q(pi)

[chilly ocean]

Because Q(sqrt(pi)) contains pi by multiplicitive closure?

[thorny slate]

I mean sure

certainly pi isn't trascendental over Q(sqrt(pi)) because it's contained in there

[upper dome]

oh totally!

thanks 😃

[upper dome]

the zeros of a cubic polynomial over a field F of characteristic 0 can always be obtained by means of a finite sequence of operations of addition, subtraction, multiplication, division, and taking square roots starting with elements in F

True or false. i know its false because my professor told me, but I just cant think of a counter example. I was damn sure it was true until he corrected me

[thorny slate]

taking square roots means adding a root of an equation x^2 - a = 0

which means an extension of degree 2

adding a root of an irreducible cubic is an extension of degree 3

do you see the problem?

[upper dome]

maybe. are you implying that if that question said "taking cube roots" instead then it would be true?

with T/F questions that are false, i always feel like theres a little tweak that could be made that would force it to be true, so im tryin to figure that out too

[thorny slate]

yes

I am

well not really

only irreducible cubics

and uhh

not even then

you might still need square root too

the degree of the splitting field of a cubic can be 1, 2, 3, 6

depending on which it is you need square root, cube root or both

[upper dome]

so yea need square

thanks jacobian, that really helped

[thorny slate]

np

[upper dome]

i don't understand immediately why every subgroup of a solvable group is solvable? My professor mentioned it kinda like an aside, and looking it up i'm seeing all these intersection proofs, which i dont quite understand. can anyone shed some light on this?

i want to just think that if G is solvable, then there is a series of normal subgroups which has abelian quotient groups(that are adjacent)... and so a subgroup H would just land somewhere in that series, and thus all subgroups below it would still be normal and all that

[stone fulcrum]

It's a consequence of the second homomorphism theorem. Maybe a proof in a book

[upper dome]

okay i'm staring at that theorem in my book right now, and i like how it has all the pieces like a subgroup H and a normal subgroup N of G.... but hell, I'm just not seeing how to apply that

[thorny slate]

take a composition series for G

intersect everything with H

this is a composition series for H

basically if N is normal in G then N n H is normal in H

so you can go down with the same sequence of normal subgroups

[upper dome]

damn that is slick

okay, thank you @stone fulcrum and @thorny slate

[stone fulcrum]

Jacobian did the heavy lifting, I just barely remembered

[thorny slate]

smallest possible case for the incomplete factoring problem: deg(f) = 8, factors into degrees 3 and 5

[Frac(f):K] divides 3! 5! = 5 3^2 2^4, where K is the field with the splitting 3/5, 8! = 7 5 3^2 2^7, and we would have for a root r of f, [Frac(f):K(r)] divides 5 3^2 2
we had 3 and 5, and they must split further, otherwise we can keep adding roots of f until they do, here we can't add any more roots
we also have that 3 5 = 15 divides [Frac(f):K] since we have polys of degrees 3 and 5, therefore these degree 3 and 5 polynomials can't split, contradiction

wanna say it's impossible in general, let me generalize this proof

oh

adding a root of f here doesn't necessarily give an extension of degree deg(f)

just dividing deg(f)

but that's ok we can keep adding roots nvm

Statement: let f be an irreducible polynomial in K[x] such that in some field extension E|K f splits as f =gh. Show that deg(g), deg(h) cannot be both coprime to deg(f).

Observe that deg(g) deg(h)| [Frac(f):E] | deg(g)! deg(h)!
Let r be a root of f and consider E(r), which is an extension of degree dividing deg(f) as no roots of f are in E, that is, [E(r) : E] | deg(f). The factorization of f in E(r) includes a linear factor, so that we have added a root of g or of h, giving [E(r) : E] | deg(g) deg(h), a contradiction

@gentle pendant

In fact, if either deg(g) or deg(h) is coprime to deg(f), add a root of that one, so that [E(r):E] | deg(g), say, a contradiction

[gentle pendant]

nice 😃

[thorny slate]

therefore both deg(g) and deg(h) have common factors with f

so I suspect that the only way of it happening is your example

which means something like

hm

@whole basalt

some generalizations of your question btw

[whole basalt]

Oh awesome

I'll read when I get up

Like I said I didn't expect it to be such a vibrant discussion XD

[thorny slate]

the general statement is the one up there

Statement: let f be an irreducible polynomial in K[x] such that in some field extension E|K f splits as f =gh. Neither of deg(g), deg(h) can be coprime to deg(f).

[gentle pendant]

Wait a second, throughout this it appears you are using:

If f is irreducible over F and has a root in E then [E:F]|deg(p). The order should be the opposite no? deg(p)|[E:F].

@thorny slate

[thorny slate]

I'm using the fact that adding a root of a degree k polynomial is an extension of degree k

[gentle pendant]

why is that true?

[thorny slate]

by definition kinda

K[x]/p(x) has dimension deg(p)

[gentle pendant]

sure degree k yes

sorry, you edited

[thorny slate]

oh

my bad

the reason we don't have equality and only [E(r) : E] | deg(f) is because the equality is achieved in [K(r) : K] = deg(f)

instead we recover only divisibility from:

however the second one [E(r) : E] = deg(g) is equality

[gentle pendant]

ah okay, I see

[thorny slate]

so we actually have deg(g) | deg(f)

and deg(h) | deg(f)

[gentle pendant]

nice, that makes sense

[thorny slate]

cool

[gentle pendant]

👍

[thorny slate]

so we have like, deg(g) and deg(h) divide deg(f) and sum up to deg(f) so the only case is (2,2)?

that's pretty intense

oh it can be (3,3)

or (n,n) for deg(f) = 2n

but they have to be the same I guess

is this general? like a bunch of numbers add up to n and divide n

what do they satisfy

no 1 allowed

I guess not much you can have like 2, 2, 4

but do they have like a comon factor

I wanna say the come from a cheap x -> x^k

like in your example

[sonic current]

hello! I need a bit of guidance in a question.

Show that, if A is a non-empty ring such that aA=A, for every a in A{0}, then A is an integrity domain.

so far, I've concluded that the multiplicative identity belongs in A, since aA=a implies a1=a; and also that there is only one element such that a0=0, and so A is a domain with identity. For it to be integral domain I must prove A is commutative

that's where my trouble is: how can I prove commutativity? I've thought about having aA=bA, but I don't know how much further I can go with that

[worthy kindle]

I don't really get what's your plan

to prove integrity, you usually start with two elements a and b of the ring such that ab=0 and show that one of them is 0

[sonic current]

hmm

well, if there is only one element such that the product can be 0, then it is necessarily a domain, right?

my thought process is a bit convoluted, i'll admit to that

[worthy kindle]

I don't even get what you mean by "there's only 1 elem such that the prod can be 0"

[sonic current]

since we have aA=A, my point is that there is only possible element such that a*x=0, and it is x = 0; or else, the two sets wouldn't be equal

let me see if I can rephrase that

if there was any other element x such that a*x = 0, other than x = 0, then the two sets wouldn't be equal

I'm having some trouble explaining what I mean, but say A has 5 elements, if there was any other element such that a*x = 0, then aA would have 5 - 1 elements

because there would be two different elements such that a * x = 0. one would be zero, because it necessarily exists, and the other would be some other zero divisor

[worthy kindle]

Who said A was finite? let alone countable?

[sonic current]

hmm

okay

well, there goes my thinking

[worthy kindle]

The exercise is simpler than what you think

[sonic current]

I see what you mean though

thanks for pointing that out

can you point me in some other direction?

[worthy kindle]

start with a and b such that ab = 0
you want to show that at least one of them is 0

if b is 0, well no problem
if b isn't 0...

show that a is 0 :p

[sonic current]

Show that, if A is a non-empty ring such that aA=A, for every a in A \ {0}, then A is an integrity domain.

a will never be 0; lets see what I can work out

[worthy kindle]

mhmm should've used other letters

Start with x and y in A such that xy=0
Show that one of them is 0

if y is 0, no problem
if y isn't 0, show that x is 0

[sonic current]

and where in your process does the aA=A hypothesis come in?

[worthy kindle]

Well in the case y≠0, you have yA=A

And there's a very particular element of A that should be of some use

[sonic current]

which I suppose would be the multiplicative identity element

[worthy kindle]

Yes

[sonic current]

and where is your x in all this?

i'm sorry, I'm a bit confused

i need only one element y in A \ {0} to prove there is the multiplicative identity

[worthy kindle]

1 is an element of A and A=yA, so there exists some element z of A\{0} such that yz=1

Do you see where it can be useful?

[sonic current]

ermm

let me think

i'm completely lost here and I'm afraid i'm wasting your time. let me think for a bit

[worthy kindle]

No problem x')

[sonic current]

I suppose your last point is only possible after concluding the 1 element is in A, right?

[worthy kindle]

the 1 element is obviously in A, otherwise A wouldn't be a ring

[sonic current]

damn. i'm so dumb

I thought the multiplicative identity was not necessary. allright

i mean, I thought it didn't come along in the ring definition.

so we can jump straight into

1 is an element of A and A=yA, so there exists some element z of A{0} such that yz=1

[worthy kindle]

Yep

[sonic current]

z is the inverse of y in this case?

[worthy kindle]

Not really, it's just an element that happens to have the property yz=1

There aren't always inverses in a ring

[sonic current]

ok so let me see if I follow

we take the property A = yA, for some y in A \ {0}; and in particular, there is z in A such that

1 = y*z

[worthy kindle]

Yes

[sonic current]

okay

[worthy kindle]

Oh well, more like because y≠0, yA=A

[sonic current]

yes

and what can I work from there?

[worthy kindle]

You have xy=0 and yz=1, there's something you may want to do to the first equality

[sonic current]

x must necessarily be 0?

[worthy kindle]

That's what you want to prove

[sonic current]

hmm.

[worthy kindle]

Multiply by z.

on the right

[sonic current]

xyz = 0*z <=> x = 0?

[worthy kindle]

xyz = x(yz) = x1 = x
xyz = (xy)z = 0z = 0
x = 0

[sonic current]

hmm, allright

so x is 0

[worthy kindle]

Yes

[sonic current]

i know I couldn't go with my first process (since A is not necessarily finite nor countable)

but it seemed way more straightforward

I'm not sure what was the point here...

so we take two elements x and y in A; we must prove x*y=0 iff one of them is 0.
additionally, from the hypothesis yA = A for some y ≠0, we take yz = 1.
now we have xy=0 and yz = 1

xyz = x(yz) = x1 = x
xyz = (xy)z = 0z = 0

then x = 0

so we proved that for any x and y in A, the product will be zero only if one of the elements is 0

well I can see that. but is it not necessary to prove it is commutative as well?

[worthy kindle]

Something's missing

You forgot the y=0 case, without which the proof has a hole

the whole yA=A thing only takes place in the y≠0 case

[sonic current]

well, but y will never be 0 right?

because we only consider y in A \ {0}

[worthy kindle]

Ugh

, tex property given by the exercise \
$\forall a\in A\setminus{0},\ aA=A$

[cloud walrus]

Tuong:

[sonic current]

(what a nice bot)

[worthy kindle]

, tex what you want to show\
$xy=0\iff x=0$ or $y=0$

[cloud walrus]

Tuong:

[worthy kindle]

well actually only $xy=0\Longrightarrow x=0$ or $y=0$

[cloud walrus]

Tuong:

[worthy kindle]

'cause the other implication is quite clear

[sonic current]

yes

[worthy kindle]

so you study different cases that are possible, namely,
y = 0 and y ≠ 0
at least one of these cases happens

In the case y = 0, we're all good because we wanted to have x=0 or y=0

In the case y ≠ 0, you can use the property given by the exercise, i.e. yA=A, etc.

and in this y≠0 case, you find x=0, which is good because we wanted x=0 or y=0

hence the implication

[sonic current]

alright

I kind of see what you mean by covering all possible cases

[chilly ocean]

brute force is what it's called

[sonic current]

what about the commutativity?

haha

[worthy kindle]

It doesn't give commutativity :p

[sonic current]

but isn't it necessary for having an integral domain?

[worthy kindle]

nope

You really only need
xy=0 => x=0 or y=0

[sonic current]

hmm

my book even goes as far to say xy=0 => x = 0 or y =0 guarantees domain only

[worthy kindle]

O wait

[chilly ocean]

well it's under multiplication so it's implied commutative

what is the question

[sonic current]

are you sure? I suppose I must have ab = ba somewhere