groups-rings-fields

[sonic current]

@chilly ocean 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.

[worthy kindle]

Definition seems not to be unanimous

[chilly ocean]

is the question missing a symbol

[sonic current]

(yes, it's A except 0 -- for some reason discord ommits the \ )

[cerulean rune]

the condition "aA = A for all a \in A \ {0}" doesn't force A to be commutative (GL(2, Z) satisfies this property for instance), right? [edit just use A = quaternions]

oh iam dumb

ignore me

[worthy kindle]

Double antislash to make the antislash render

[sonic current]

thanks, @worthy kindle

i've thought about aA = A = bA and then try something from there

does anyone have any idea?

[chilly ocean]

I guess you just have to define some ring A to be an integral domain

you are given some headstart by them using A \ {0}

[sonic current]

(thanks @worthy kindle for all the help!)

[worthy kindle]

You're welcome

[sonic current]

well, i'm not sure where to follow from here

I guess I'll ask the teacher when I'm in class with him; his exercises can be very obnoxious sometimes

[chilly ocean]

you just need a nonzero commutative ring where product of two elements is nonzero

basically A \ {0}

[worthy kindle]

Well, I read some stuff, and it seems like, when it comes to ring stuff, no one has got the same definitions

[cerulean rune]

sorry -- regardless of the definitions, the condition in question -- namely, "aA = A for all a" -- doesn't force A to be commutative, in general. the quaternions are a counterexample, unless i am still being crazy.

[worthy kindle]

a somewhat usual consensus seems to be "you only study these things in the case of commutative rings" so I suppose the ring in the exercise was already commutative or something like that

But yea, I would also tend to say that integrity is independent of commutativity

[thorny slate]

@gentle pendant is my proof correct?

I'm having second thoughts

like

the diagram I drew

where I got [E(r):E] | deg(f)

I don't trust that

you can have deg(f) in the [E:K] part

like

yeah

i dont see why not

just because adding a root extends degree by k in some field doesn't mean it extends degree by a divisor of k in a bigger field

adding the roots of some polynomial with galois group Sn or something just add degrees n, n-1, n-2...

depending on the order

[gentle pendant]

well that was the part I was unconvinced by, but you sounded confident enough of it that I assumed it was a property of lifts I had forgotten

[thorny slate]

yeah I think I was just tricking myself

at least it doesn't follow directly

[gentle pendant]

what conclusions can you draw about the degree of the extension of such a lift? I will think about this in a bit but need a coffee first

[thorny slate]

no clue

in principle almost nothing

[gentle pendant]

yeah from your symmetric example hey

[thorny slate]

yeah

[gentle pendant]

That's a shame!

[thorny slate]

:(

yeah gg

gonna try to fix it for a bit but

I glossed over the most important aspect of the proof anyway

so that's bad

that's the essence of the problem

[gentle pendant]

yeah absolutely

the substitute for primality of deg(f), so to speak.

[thorny slate]

yeah

[whole basalt]

I read over the discussion about that problem

Not something I would've found on my own, but definitely interesting

And gives me at least some intuition I can use!

For sanity check purposes

[sick acorn]

does anyone have any suggestions/tips for this?

i already showed that H has a unique 3-sylow subgroup K of order 3

and that K is a normal subgroup of H

So K is a normal subgroup of H, and H is a subgroup of G

I need to show that K is a normal subgroup of G

(or that K is the unique 3-Sylow subgroup of G)

[covert vector]

hint: $K \subseteq H \implies gKg^{-1} \subseteq gHg^{-1} = H$

[cloud walrus]

woog:

[sick acorn]

ahhh thanks! ugh, i forgot that H was a normal subgroup of G

i'll ponder on that

[covert vector]

ok!

[sick acorn]

would one route be to show gKg^-1 is a subgroup of H? We know gKg^-1 has order #(K) = 3, so it would be another subgroup of order 3 in H, i.e., it would equal K because K is the unique subgroup of order 3 in H

i... think i was able to show gKg^-1 is a subgroup of H. And it has order 3, so gKg^-1 is a 3-Sylow subgroup in H. But K is the unique (3-Sylow) subgroup of order 3 in H. So gKg^-1 = K. So K is a normal subgroup of G.

thanks!!! @covert vector (hopefully the solution's right haha)

[covert vector]

👍

[low inlet]

how do i do this one?

[lone hedge]

write the quadratic form out as a matrix and diagonalise it

maybe not what you've learned

[low inlet]

if it works it works my prof doesnt care about the method as long as i get the correct answer

[lone hedge]

I got 6x^2 + y^2 = 1

so it's an ellipse

[raw moth]

(sqrt(5)x+2sqrt(5)/5y)^2 + 8/5 y^2 = 1 is a rearrangement of the original one, I guess this or something similar is what was intended

although maybe you need to write it as a sum of orthogonal squares

which is possibly equivalent to the diagonalisation method?

I'm not sure

[lone hedge]

Here's what I did

Bit more working than necessary

I guess if you multiply PP^-1

You can get the angle

Although I'm not sure

[raw moth]

PP^-1 is the identity surely?

[lone hedge]

oh yeah I'm an idiot lol

[vale folio]

I think the columns of P/the eigenvectors would be the new principal axes so their rotational displacement from the standard axes is the angle?

[lone hedge]

oh yeah

makes sense

[vale folio]

You can represent any vector by its coordinate in the eigenvector basis, the eigenvectors aren't rotated in the quadratic form, so the rotation of an arbitrary vector is purely a result of the rotations of the eigenvectors from the x and y axes, I think?

[lone hedge]

so it's pi/8

I reckon

[raw moth]

looks like pi/6 to me

[lone hedge]

it's y = (1/2)x

probably should've added that

[raw moth]

and arctan(1/2) = pi/6

[lone hedge]

fair enough lol

[raw moth]

or not

lmao I did a dumb

[vale folio]

yeah it's neither one of those

[raw moth]

it's whatever arctan(1/2) actually is

[lone hedge]

not a multiple of pi

confirmed

[vale folio]

it's (arctan(1/2)/pi)*pi

[lone hedge]

lol

anyways glad my algebra is coming in handy

[raw moth]

my guess is a(kx-y)^2 + b(x+ky)^2 = 1 and then equating coefficients for a,b,k for a simple (in terms of what is used rather than actually doing it :^)) method

ak^2+b = 5

a+bk^2 = 2

k(a+b) = 4

[sonic current]

hello! would you say that for a ring S, S^2 is contained in the ring S?

i now this is a rather dumb question, but please bare with me

[delicate chasm]

$S^2 \ceq {a \cdot b \st a,b \in S}$?

[sonic current]

is that so?

maybe

[delicate chasm]

Is that the definition?

[sonic current]

i've got S*S

i don't know -- its a matrix whose elements are rings 😐

so I suppose they are outright multiplicating

if your definition is right, may I conclude S^2 = S?

[delicate chasm]

Yeah, essentially by closure under ring multiplication

And existent of multiplicative identity

It might not be right though

It could be the direct product of rings

[lone hedge]

if it's a matrix then it isn't

[sonic current]

i'll give you the full exercise

[delicate chasm]

Sure

[sonic current]

hmm

let me grab a pic

[delicate chasm]

👍🏽

The above is more usually a product of ideals of rings

When we multiply rings we usually use the direct product, which is essentially the cartesian product with component wise operations

[sonic current]

Number 2

[delicate chasm]

b, you mean?

[sonic current]

No, it's really exercise 2

So what I did is multiply R with a

To prove Ra is contained in A

[delicate chasm]

Yep

I guess you'll want to show that A is an abelian subgroup under addition as well

Although that's easy

[sonic current]

I assumed it is

[delicate chasm]

Yeah

[sonic current]

Didn't want to waste time there lol

[delicate chasm]

Aye, it's pointless symbol pushing lol

[sonic current]

So what about the S squared?

Am I safe to assume it's equal to S?

[delicate chasm]

We don't really get S^2 here right?

[sonic current]

Oh damn.its 2S

[delicate chasm]

Well

[sonic current]

Wait it's not

I'm dumb

Well it's S x S

[delicate chasm]

If we are careful, we multiply things using arbitry elements

S x S is different from S^2 :P

So you are looking for the cosets?

[sonic current]

Of course, yeah

No, I must prove first Ra = A

Which implies proving that SxS is contained in S at least

[delicate chasm]

Well Ra is a coset

[sonic current]

Yes, but isn't it necessary to show A is an ideal first?

[delicate chasm]

Careful with terminology.. $$S \times S \ceq { (a,b) \st a,b \in S$$ as a set

[delicate chasm]

In this case

Take some $a\in A$

[sonic current]

Oh I see

[delicate chasm]

So that's the same as saying take some $$\bmqty{0 & s\ 0 & 0} \in \bmqty{0 & S\ 0 & 0}$$

[sonic current]

Then the element a will have an element of S

[delicate chasm]

Are these right or left sided ideals?

[sonic current]

Ra

So left sided?

[delicate chasm]

Sure

[sonic current]

And now is sS contained in S? I'm so lost in this

[delicate chasm]

Well, is it? given that $s\in S$

[sonic current]

I'd suppose so, but I should probably refer to coset theory no?

[delicate chasm]

And noting that $Ss \ceq { s' \cdot s \st s' \in S}$

[delicate chasm]

Nah, you can do it from first principles

[sonic current]

Yes!

[delicate chasm]

Yeah

[sonic current]

Theorem 1 lol

[delicate chasm]

True

[sonic current]

Alright

Let me finish my notes

[delicate chasm]

👍🏽

[sonic current]

I must now do the right ideal

I might need some assistance in the cosets, I'm not sure

I wish there was a magical book to make me understand all of this

[delicate chasm]

There are a few other resources on the topic

[sonic current]

I was pretty comfortable with group theory but then rings come along

[delicate chasm]

Damn

[sonic current]

OK, I'm through with proving it is an ideal

[delicate chasm]

Good!

If you want another resource btw

I favour http://library.adamwalsh.name/$/28kic

Topics in algebra

I have to sleep now unfortunately :/

[sonic current]

alright

i'll try to advance anyway

[delicate chasm]

There should be plenty of other people who can help with this stuff though! Good luck!

[sonic current]

thanks for the link! i'll check it out

Can anyone give me some help on this exercise?

I must describe the cosets of R/A

[whole basalt]

So

Thank y'all

Being able to talk things out with people in here really helped me as I was getting ready for this final

[halcyon turret]

:0

[upper dome]

I have to show Dn is solvable for n>=3. Do ya'll think induction would be a good path? been grindin on induction for a bit and can't get it, so thinkin maybe this isnt the way?

[upper dome]

different question: if a group G has order 2n, and a subgroup H has order n, is it sound to state that the quotient group G/H is isomorphic to Z/2Z ?

<@&286206848099549185>

[earnest valley]

Is H normal ? Is n special ?

[upper dome]

yeah its normal, i said since H has half the elements of G that means the index of H in G is 2, and thus normal.

and n >=3

[earnest valley]

Is G abelian ?

[upper dome]

no, the G in question is Dn and thats not abelian

[lyric falcon]

you don't need it to be abelian

G/H contains 2 elements

[upper dome]

right! so it is isomorphic to Z_2

[lyric falcon]

yup

[upper dome]

ahh awesome i think my proof is complete then. thank you both 😃

[earnest valley]

You are welcome.

[sonic current]

hello people!

is anyone around to help me? still having trouble with the earlier exercise

https://cdn.discordapp.com/attachments/496784958430380033/521755197781966858/JPEG_20181210_182716.jpg I suppose it is somewhat abstract

it's number two. I must describe the cosets, now

i've made little progress, mostly from trying to adapt an earlier example from the book

so I've tried to establish the form of a given element x in R/A

[solar wyvern]

@sonic current do you understand the cosets of a group

[sonic current]

yes, I suppose so

well, there's one idea thats bugging me

I've read in one or two sources that ideals should be compared to normal subgroups, not cosets

i feel like that is an obstacle

@solar wyvern

[solar wyvern]

you wouldn't quotient by a arbitrary coset

would you

[sonic current]

ok

i have a question

well I have many

[solar wyvern]

I mean you still have things like a+I where I is an ideal.

and a an element

[sonic current]

hmm.

yeah

[solar wyvern]

lemme migrate to computer

[sonic current]

i've actually made some progress in this

on my own but i'm not sure I went anywhere meaningful or if it's correct even

because I was trying to adapt some other example on the book without quite understand the process

i'm sorry, I should learn latex asap...

[solar wyvern]

I think I used the same book as u

(regrettably)

[sonic current]

one example i'm trying to follow is the gaussian integers

lol haha

it's not that I dislike it, I think it's thorough and very well indexed

i was a big fan of a. gallian before, still am

but I've been using this one more

[solar wyvern]

what is this one?

[sonic current]

i read a brilliant quote today

let me see if I can find it

i think it's in gallian's

nevermind - it's somewhere

this one is nicholson

Abstract Algebra

[solar wyvern]

yep, that's it

[sonic current]

it's big and unconfortable

[solar wyvern]

and unreadable

[sonic current]

but on the other hand feels very serious

haha

[solar wyvern]

it sorta isn't tho

[sonic current]

i wish algebra could be read like a novel lol

it's something really specific to convey I suppose

today i've read about someone's book whose book started on rings

and only then groups

that's interesting

[solar wyvern]

kaplansky has a good ring theory book

ground up

[sonic current]

anyway - do you happen to have any idea that I could think of for the exercise?

let me see

[solar wyvern]

2 right? do you know how to do first part?

that should be straightforward

[sonic current]

well more or less

had a bit of trouble

because I didn't take an element s from S in the matrix; I went for a matrix with S as elements

[solar wyvern]

just multiply on right and left

[sonic current]

yeah, i understand it now, but it was a stupid slump

[solar wyvern]

make sure it's invariant under the action by left/right mult

[sonic current]

hmm

so it's an additive group?

i'm sorry -- keep going

[hushed pivot]

Can anyone explain me why the 6th statement is true?

[solar wyvern]

no u got it, just gotta check it's closed under multiplication by ring elements

[sonic current]

@hushed pivot do you know what linearly independent is?

can you relate that concept to a base of a space?

[hushed pivot]

Yes, but when x=1

Sin X is equal to 1

[sonic current]

x can generate the whole R

[hushed pivot]

I think

[sonic current]

sin and cos can't

I think that's it, I'm not sure?

[hushed pivot]

Ok, thanks.

[sonic current]

hey, i'm not sure! let someone chime in i'm not very good at this

[hushed pivot]

Lol, ok

[sonic current]

@solar wyvern what's the name of the author again?

[stone fulcrum]

@hushed pivot
There is no choice a, b, c such that
asin(x) + bcos(x) + 1c = 0
For all x

[sonic current]

I'm actually curious about this, why are they linearly independent really?

okay

that was very clean

[hushed pivot]

@stone fulcrum okay, so an individual example cannot proof a set is linearly independent, am I correct?

[stone fulcrum]

There's ways, lol. A set like this requires some calculus though

[hushed pivot]

Linearly dependent, I mean.

Ok. Thanks

[sonic current]

linearly dependant I suppose so?

[stone fulcrum]

You could easily prove any set is dependent by finding a, b, c.

[sonic current]

it's a counter example to linear independency

[hushed pivot]

It is clear now, thank you guys so much.

[stone fulcrum]

For example
ax + b(x + 1) + c = 0

Are x, x + 1, and 1 dependent?

[hushed pivot]

Emmm, maybe not

[sonic current]

x + 1 can be dismissed?

[stone fulcrum]

No, since a = 1, b = -1, c = 1 is a solution

[sonic current]

never thought of it that way

[hushed pivot]

[stone fulcrum]

Note that a = 0, b = 0, c = 0 is always a solution. We're ignoring that one though

The zero solution doesn't count. We call it "trivial" and tell it to leave

[chilly ocean]

EXCYOOSE ME WHAT IS THIS TOMFOOLERY

[hushed pivot]

Ok. So if I want to show a set is linearly dependent, I have to proof that there is an non-Zeros solution for all circumstances. Correct?

[stone fulcrum]

Yus.

[hushed pivot]

Ok, non trivial solutions

Thank you~

[stone fulcrum]

Vectors u, v, w are independent if there is no nontrivial a, b, c such that

au + bv + cw = 0

[hushed pivot]

Ok. I see

Ok

I think I am good now 😄

Thanks!

[stone fulcrum]

Or, to use technical terms

A set of vectors are independent if they span zero exactly once

To use less techy terms

A set of vectors is DEpendent if you can create any one of them with a linear combination of the others.

[hushed pivot]

Cool method!

[sonic current]

I have a question

[stone fulcrum]

You can easily create x + 1 using 1 and x. Therefore, 1, x, x + 1 are dependent

[sonic current]

is the information about spanning zero exactly once useful for anything?

to determine solutions?

[stone fulcrum]

It turns out that if a set spans zero exactly once, then it spans every vector in its spanning-set exactly once

[hushed pivot]

I think so, but sometimes you don’t need even one zero, to judge a set is dependent

[stone fulcrum]

There's only one way to represent [5, 7] using a linear combination of [1, 1] and [0, 1].

I know this because [1, 1] and [0, 1] spans zero exactly once

[sonic current]

that's cool

[stone fulcrum]

The matrix that takes these vectors to some vector in the spanning set is invertible.

[sonic current]

i usually just think about reducing the set to a minimum by removing dependents, so I never had to think about the uniqueness of zero in a set

in my head it didn't make sense how cos(x) and (sin) could be on the same set and still be independent

so it was a nice approach

[stone fulcrum]

Note that sin²(x), cos²(x), 1 are not independent

So sin, cos get interesting

[sonic current]

hmm

[stone fulcrum]

Can you prove that one? ;)

[sonic current]

I was thinking about that lol

[solar wyvern]

s^2+c^2=1

that's a linear combination so they're not LI

[sonic current]

"We see therefore that ideal prime factors reveal the essence of complex numbers, make them transparent, as it were, and disclose their inner crystalline structure"

@solar wyvern that's in nicholson

[solar wyvern]

they could have explained something properly in that space

(maybe not)

[sonic current]

haha

[solar wyvern]

do you know what a module is?

[sonic current]

i don't know, I quite like it

[solar wyvern]

different strokes i guess

[sonic current]

hmm - the mathematical operation?

[solar wyvern]

sorta like a vector space, but a little more general

[sonic current]

nope

[solar wyvern]

it's an additive abelian group, but instead of having scalar multiples in a field, they're in a (usually commutative) ring

you know how there are vector spaces over R and C right?

[sonic current]

yes

C is something I'm not as comfortable with

[solar wyvern]

no need to be, just need to know coefficients from there and a vector space is an abelian group with certain stuff

the way I like to look at ideals of a ring A is as their A-submodules

you always have to have 0, additivity between elements in your ideal, and closure under scalar multiplication, so this exactly characterizes an ideal.

incidentally, assume your ring has a multiplicative identity 1_A and consider which A-submodules (i.e., ideals of A) contain 1_A

[sonic current]

hmm

[solar wyvern]

easier task is which contain 0_A

are you comfortable with group theory/group actions

[sonic current]

this is suddenly very much

the ones who are cyclic?

I'm trying to think of the ones that have 0_A

[solar wyvern]

modules required to have 0's (or, equivalently, ideals )

[sonic current]

hmm okay

where do we go from here

[solar wyvern]

hmm

was going to point out a notational thing about cyclic modules not quite being cyclic groups

like

say you have V=R^3 as a vector space

then if you take one element v and all its scalar multiples $\alpha v, \alpha \in \mathbf R$ that'd be a cyclic submodule generated by v
sometimes written $\langle v \rangle$ or $(v)$

[cloud walrus]

flimflam:

[sonic current]

ok, i'm not quite familiar with the definition of submodule

[solar wyvern]

if modules seem weird might be more straightforward to not think about them 😄

[sonic current]

or module by any matter

not that they seem weird, it feels like familiar notation and operation for something I haven't got a nmae

but is it algebra?

[solar wyvern]

yes

might be better to get a bit more grounded with rings for the moment

since you usually use those for the definition of modules lol

[sonic current]

okay

what was that book you talked about?

[solar wyvern]

i looked at it and it wasn't as good as I remembered

lemme see if I can find another one

http://www.math.uconn.edu/~kconrad/blurbs/

everything on conrad's site is great lol

digestible too

I'd check out http://www.math.uconn.edu/~kconrad/blurbs/ringtheory/ringdefs.pdf

yeah, that's a good place to start.

[sonic current]

thanks! i'll give this a look

[solar wyvern]

good luck!

[chilly ocean]

So I've been working with quaternions and I keep hearing the term "normed division algebra" and I don't know what it means. I know it includes real numbers, complex numbers, quaternions, and octonions, but what exactly is a normed division algebra?

[thorny slate]

a division algebra with a norm

[chilly ocean]

I'm 16 and I don't know what that means though.

[thorny slate]

an algebra is a vector space which also has a product

which is bilinear

in this case over the real numbers

[chilly ocean]

Ok.

[thorny slate]

a division algebra has multiplicative inverses for any element

[chilly ocean]

Ok, I see what you're saying.

[lone shard]

a norm is a generalization of the idea of length, and a division algebra is a generalization of the way that you can add, subtract, multiply and divide real numbers

to put it in v handwavy terms

[thorny slate]

we say the algebra is normed if the vector space norm is compatible with the product

that means ||ab|| and ||a|| ||b|| are related

[chilly ocean]

Oh!

Thank you so much. I've been trying to find it on the internet and couldn't seem to find anything that actually made sense. Thanks!

[thorny slate]

a famous result says that the only finite dimensional real division algebras have dimensions 1, 2, 4, 8

[lone shard]

oh is that the frobenius theorem

i was just looking at that like an hour ago lol

[thorny slate]

yeah

[chilly ocean]

Yeah, that's what I saw. I'll have to take a look at that theorem.

[thorny slate]

kinda

frobenius asks for associative as well

which excludes dim 8

why are you looking into this stuff @chilly ocean ?

[chilly ocean]

Mostly for fun. I just really enjoy it.

[thorny slate]

have you learned much higher math?

or not yet

[chilly ocean]

I also am working on a math API, also for fun.

I don't really think I can say, because I don't know what I don't know, but I'm trying to learn some linear algebra. Just not calculus, it's not my favorite thing.

[thorny slate]

yeah you should learn some real analysis and linear algebra

analysis is basically calculus with theory

as opposed to just calculus which is mainly computations

[chilly ocean]

Oh neat.

[thorny slate]

which linear algebra book are you using?

most linear algebra books tend to be quite dry and focused on repetitive application of some computations

[chilly ocean]

I'm not using a book. Call me an e-child but I use Kahn academy.

Lol

[thorny slate]

if you're interested in a book, hoffman&kunze is great

[chilly ocean]

I'll take a look at it then.

[thorny slate]

linear algebra is a rich subject and a nice way to start to get into math

it kinda loses its charm when you reduce it to just memorizing tricks

[chilly ocean]

Yeah, that's what burnt me out on calculus.

[thorny slate]

usually it's recommended to learn calculus anyway because you'll probably need it for late highschool / early undergrad

but if you learn enough analysis then you can do the calculus stuff

while also knowing how it works

[chilly ocean]

Maybe I'll come around to it then.

Ok, I found a preview.

Thanks for the recommendation!

[lone shard]

btw you can find pdfs of most textbooks on libgen

itll save you a lot of money if you plan to go to university lol

[chilly ocean]

Yeah, I actually heard about libgen for the first time earlier today

[thorny slate]

yeah just use libgen

if you feel bad about it when you're in uni you'll have library access

or you can buy good condition books for cheap in alibris

digital distribution for personal, academic use is justified imo

[chilly ocean]

Yeah, I don't think I'd feel bad about it. I've just never done it before.

[thorny slate]

it has every book basically

except really advanced niche stuff

which you can find somewhere else

[chilly ocean]

Oh sweet.

[lone shard]

b-ok is also pretty good

http://b-ok.org/

[chilly ocean]

Oh nice

Alright, I got the book

[karmic moat]

in this, how do i describe translations in a parabola
horizontal shift, vertical shift, vertical stretch, horizontal stretch, reflection on y/x axis, etc
i could google it, but what i find online contradicts what my teachers taught me

$$f(x) = af(bx+c) + d$$

[cloud walrus]

analyst:

[chilly ocean]

That looks like it's more for trignometric functions, but I suppose it could work for a parabola too

[karmic moat]

I'm just going off what's on our finals review sheet /s

[errant drum]

I've been told that in Q[X] you can have irreducible polynomials of any degree is there any way to test if a polynomial is irreducible over Q?

[chilly ocean]

Some random words: If it has irrational roots?

[thorny slate]

you can just take x^n - 2

if you can prove that the nth roots of 2 are irrational

for testing irreducible in general you can use eisenstein

[errant drum]

As in given any polynomial is it possible to test whether it's irreducible over Q

[chilly ocean]

You can use the rational roots theorem <- more random words

[errant drum]

@thorny slate So do all polynomials satisfy the Eisenstein test?

[chilly ocean]

Random words: consider differentiating the polynomial and see what the terms look like (ie, power rule)

[thorny slate]

what

it's straightforward

[errant drum]

Irreducible polynomials

But I mean is it possible to have a polynomial that passes this test but is still irreducible?

[thorny slate]

yeah jupliter maybe clear out denominators and see what you get

it looks like the product rule for derivatives

@errant drum yes

[sterile pecan]

Struck here

[thorny slate]

calculate what f'(x) is

using product rule

and try to relate the stuff

i mean you kinda did

but don't write it like that

clear out denominators

multiply by all the (ai - aj)

yeah but that's wrong

[sterile pecan]

Which part is wrong?

[errant drum]

@sterile pecan Try write out an nth degree polynomial and differentiate it. See what you get

@thorny slate I did some digging and there are a bunch of tests so does that mean there's no one test that can capture all irreducible polynomials?

Because all of them seem to have exceptions

[thorny slate]

what kind of test

like

there's no easy, simple way

[errant drum]

https://math.stackexchange.com/questions/1935/methods-to-see-if-a-polynomial-is-irreducible

Basically there are a bunch of tests to see if a polynomial is irreducible but it seems like there's no one comprehensive test that works on every irreducible polynomials

[raw moth]

yes

there isn't a known comprehensive test

[thorny slate]

yeah that's the correct one

[sterile pecan]

This?

Then how should i proceed🤔

[thorny slate]

just write it

write the expression in terms of the (ai - aj)

everywhere

and collect terms

it has to work

it's just gross

[errant drum]

@raw moth Is there a reason that you can always generate a counter example. Other than you can always generate a counterexample?

[raw moth]

I don't know

but all the known irreducibility tests have counterexamples

[sterile pecan]

@thorny slate Sorry don’t get what you mean. How to write the expression in terms of (ai-aj)

[thorny slate]

no because you have to evlauate at a_i

and also you are using two different indices as i

'they're not the same

[craggy robin]

Jupliter, struggle more by yourself

you have the solution in your hand

remember you are evaluating f' at a_i

[sterile pecan]

But (a_i-a_i) should be zero

[craggy robin]

yes it should

but remember the i in that expression you found to calculate the derivative is not the same as the i of the a you are evaluating the derivative at (one is an index to iterate through the list and the other is the index of a particular value)

[sterile pecan]

I see

If plug in a_i in to f’(x)

[craggy robin]

just use a_k for the point you want to evaluate and then a_j and a_i to iterate through the sum or the product

it'll become quite clear how the expression simplifies if you stare at it for a while

glhf

[thorny slate]

no

this is just what you wrote before

are you doing this on purpose

[sterile pecan]

No

[thorny slate]

you need to just carry out the algebra slowly

[sterile pecan]

@thorny slate is this the right step?...

[thorny slate]

yes

oh that's wrong

the product is just the top part

the thing on the bottom should have exponent 1

not n

[sterile pecan]

I see

[thorny slate]

and, just carry out the computation

you're second and third guessing every step

there's nothing of interest to this computation

you just write the terms in that way, open it up, and check

it's long and annoying

but that's how it is

[sterile pecan]

But if i open up terms,it will eventually become this form

[thorny slate]

yeah

now clear denominators

please stop pinging me and asking for each step

just try it for yourself

it's gonna take you a while

[inner acorn]

Okay, I'm still stuck trying to find a proof of this xD (Edit: when A is finite dimensional)

Def: An algebra A is a division algebra if given a,b in A with ab = 0, then either a = 0 or b = 0. 
Equivalently, A is a division algebra if the operations of left and right multiplication by any nonzero element are invertible.

[thorny slate]

division -> domain is clear

[inner acorn]

I understand the rough reasoning, I just can't formulate/find a proof

[thorny slate]

for the hard part https://math.stackexchange.com/questions/1253237/prove-that-any-non-zero-divisor-of-a-finite-dimensional-algebra-has-an-inverse

you ned finite dimension

[inner acorn]

I think we talked about this the other day xD

These assume unity

[thorny slate]

omg

[inner acorn]

??

[thorny slate]

not again lol

[inner acorn]

haha

[inner acorn]

Yeah here we go:
The second bullet point from here https://ncatlab.org/nlab/show/division+algebra

Says that for A a non-associative unital algebra with finite dimension, then it's possible to find a case (for R) where
A has no zero divisors, but there exists a non-zero element in A that has no inverse (i.e. nonzero x, where xa = ax = 1)

However the mathstack exchange here https://math.stackexchange.com/questions/1253237/prove-that-any-non-zero-divisor-of-a-finite-dimensional-algebra-has-an-inverse

Says that for A a non-associative (although power-associative) unital algebra with finite dimension
Then if A has no divisors implies every nonzero element in A has an inverse
(particularly looking at the proof by Robert Lewis)

I'mma post this is the Advanced Maths too since this is bugging me n.n;; sorry if it's considered spam, but that server needs more traffic anyway xD

[pallid ginkgo]

$ k \geq 2, n \geq 2, P_k = { \sigma^k, \sigma \in S_n } $

[cloud walrus]

notsykhro:

[pallid ginkgo]

any pointers on finding k and n such that P_k generates S_n?

[covert vector]

at the very least, k=m(n!)+1 works

as the order of σ divides order of S_n = n! for all n

then P_k = S_n lol

pretty sure k=3 works too

for all n

every element of S_n can be written as a product of 2-cycles

and every 2-cycle, ^3 is itself

actually for this reason u just need k=2m+1

so every odd natural number works too 🤔

and every even integer won't work, cuz the sign of each element in P_k will be even

so it'll generate at most A_n

nice

[dusky oxide]

Hi, I need a little bit of help with a problem

Let G be a finite group and let N be a normal subgroup of G. Suppose S is a simple subgroup of G. Prove either S is a subset of N or |S| divides |G/N|.

I was thinking I could use the 2nd isomorphism theorem with the intersection of S and N, but I'm getting nowhere :/

I believe that the intersection of S and N is trivial if S is not a subset of N

Ah, then since NS is contained in G, |NS| divides |G|, so |N||S|/|N intersect S| divides G, so |N||S| divides G, hence |S| divides |G|/|N|

is this correct?

[thorny slate]

yes the intersectoin is trivial since otherwise N n S is normal in S

seems good

[dusky oxide]

ok thanks

[dusky oxide]

I have another question

nvm

Im going to think about it some more

[sonic current]

hello! how do I prove Z/5Z is isomorphic to Z5?

I feel i've came across this already somewhere

should I go through the isomorphism theorem?

If G is a cyclic group, then G is isomorphic to Z or to Zn

Z/5Z is cyclic, generated by 5Z + a, a € {0,1,2,3,4}

and so is isomorphic to Z5?

hmm

maybe this is wrong? I can just argue that Z/5Z is a ring, and since its order is prime then it's isomorphic to Z5?

[stone fulcrum]

You could suggest there's only one group of five elements

[sonic current]

yes

[stone fulcrum]

That is, of course, because 5 is prime

[sonic current]

well at this point I want to check the validity of my two arguments

the first is that Z/5Z is cyclic (generated by 5z + a, a between 0 and 4)

and by the theorem of isomorphism, its isomorphic to Z5 because, as you say, there's only one group of five elements

actually the question is 'Is Z/5Z' a integral domain?

and so I'll conclude my argument by stating Z5 is a ring of prime order which makes it a field

which makes it an integral domain

my problem is: is Z5 a ring? feels dirty to use the isomorphism theorem and then translate Z5 from a group to a ring without any justification

[stone fulcrum]

Are we talking about groups or rings?

Indeed, there isn't much of a connection there

[sonic current]

okay

glad I pointed out my own mistake

my other alternative would be to prove Z/5Z is a ring outright

and hopefully have it with a prime order

[stone fulcrum]

If all you're looking to do is show that Z5 and Z/Z5 are isomorphic, it's actually just enough to note that they both order 5, and that there's only one group of 5 elements

[sonic current]

(i actually must prove Z/Z5 is an integral domain)

[stone fulcrum]

Walk through the axioms of an integral domain. Do you have to prove this from complete scratch?

[sonic current]

not really

but this was my professor's argument and I'm trying to fill in the blanks of his arguing

if ab = 0 then either a = 0 or b = 0

could I describe the operation in Z/5Z and argue that no two cosets could multiply to return 0?

[stone fulcrum]

This is actually only because 5 is prime. In Z/6Z, 2 × 3 = 0

[sonic current]

yes I suppose so

okay so this is my strategy at this point: since the exercise states Z/5Z is a ring (lol, I'm cheap)

i'll describe its elements

5z + 0, 5z + 1, etc.

state that no product will return zero unless one of the elements is zero

thus proving it is an integral domain

(and I can do this because Z/5Z is an ideal!)

[worthy kindle]

was that irony?

[sonic current]

what do you mean?

[worthy kindle]

state that [...]
thus proving [...]

Was that some kind of sarcasm?

[sonic current]

err no? why?

[stone fulcrum]

Yes that works, but that's 25 multiplications

Maybe there's an easier way

[worthy kindle]

I don't think it's okay to "state" that something happens to "prove" it does happen

[sonic current]

well, it's not 25 multiplications, but I see your point that it's not very elegant

yeah, poor choice of wording on my end

i won't state - i'll show

[stone fulcrum]

Oh yeah it's only 16 lol

[sonic current]

if you take out the ones it's only 9!

[stone fulcrum]

9! is huge!

Okay, that's not bad at all then, you could use that on a test

[errant drum]

The symmetric group of 9 things

[sonic current]

yeah

i'm still not satisfied, if someone could help me out with an alternative route that'd be great

my teacher bluntly stated that Z/5Z is isomorphic to Z5

[stone fulcrum]

Yes, a ring isomorphism

[errant drum]

It's via the first isomorphism theorem isn't it?

[sonic current]

but wait: to be honest, I haven't yet got to the ring isomorphism chapter

so maybe that's why I can't figure it out

[errant drum]

You just need the first isomorphism theorem for groups

[sonic current]

@errant drum I've tried that! Z/5Z isomorphic to Z5 because they're both order 5 and there's only one group of order 5

(also could be because Z/5Z is cyclic of order 5)

but then I must argue Z/5Z as a ring is an integral domain

[stone fulcrum]

Note we are talking about two seperate things here that aren't to be confused:

The groups Z5 and Z/5Z have a group isomorphism.

The rings Z5 and Z/5Z have a ring isomorphism.

[errant drum]

^

[sonic current]

okay. but I haven't yet got to the ring isomorphisms

[stone fulcrum]

It's easy to state that there's only one group of 5 elements. It's a bit more involved to say there's only one ring of 5 elements

[sonic current]

what bugs me is that my professor stated it very simply that they were both isomorphic

anyway, I'm somewhat satisfied with this workaround we've figured

[errant drum]

You'd need to show it's well defined but I don't think you need to resort to using the idea it's an integral domain

I suppose you'd say if aK = bK then you'd need to show that f(a) = f(b)

But that's not implicitly assuming it's an ID is it?

[sonic current]

that's already too convoluted for me, I suppose -- i'm still not very comfortable around this

ok here's another one

that I solved in a not very elegant way too

I must prove 12Z is not a maximal ideal in Z

so I know an ideal is maximal in R if R/A is a field; but to be a field, every element must be a unit

errr nevermind. i'm dumb

could I argue a given class is not a unit without exhausting all the 12 multiplications?

I suppose 2 + 12Z won't be a unit but I don't want to do the 12 cases

this is my argument

also, 12Z is contained in 2Z, so it is not maximal?

[thorny slate]

yes, that's a fast argument

if you wanna do it the other way, consider numbers mod 12

there's something bad with 2

which is that 2*6 = 12 = 0

think about why this implies it can't be a field

[sonic current]

because it's not an integral domain?

i'm sorry - this is definitely the last one, as I'm bombarding the channel with very basic questions.

ring ZxZ, ideal ({0},7Z); I must prove the ideal is not prime.

[stone fulcrum]

All you need is one element ab, where neither a or b are in the ideal

[sonic current]

I've figured that since (0,1)*(1,7) belongs to the ideal without any of them being elements of the ideal, so that's done.

my question is: can I go the long way and prove it's no prime because the ring (ZxZ)/({0},7Z) is not an integral domain?

i'd just have to analyse the ring elements and multiply two elements to return the zero of the ring

I guess all I want to do is see if I can accurately describe the elements of (ZxZ)/({0},7Z)

[stone fulcrum]

Any two elements of Z×Z are equivalent if their difference is in {0},7Z

[sonic current]

alright!

see, that was one of the problems I was having

[stone fulcrum]

Or, find a ring homomorphism on Z×Z with a kernel {0},7Z

[sonic current]

how many elements would this ring have?

[stone fulcrum]

I'm confused, is 0 not an element of 7Z?

[sonic current]

anyway - (0,0) + ({0},7Z) is the addictive identity

yes it is

[stone fulcrum]

Oh derp I'm understanding now.

This is isomorphic to Z×7Z

Still infinitely many elements

[sonic current]

okay

so, say

the elements of (ZxZ)/({0}, 7Z)

[stone fulcrum]

So if you can prove this isn't an integral domain, then the other isn't a prime ideal

[sonic current]

are of the form (z1,z2) + ({0},7z)?

that's really my question at this point

I'm worried that i may not be able to generate elements randomly, so to say

in this case, (1,8) and (1,1) would be on the same coset? because (1,8) - (1,1) = (0,7)?

[stone fulcrum]

For example, (12, 15) ≡ (12, 8)

Because their difference, (0, 7) belongs to {0}×7Z

[sonic current]

ok but, tracing back: are all the elements of the form (z1,z2) + ({0},7z)?

[stone fulcrum]

What are z1, z2, and z here? I may not understand your notation

[sonic current]

if so, then what you mean is that (12,15) ≡ (12, 8) and so they're contained in (12,8) + ({0},7z) ?

z1 and z2 are two elements of Z; and z is a random element from Z as well

the elements of (ZxZ)/({0}, 7Z)

[stone fulcrum]

Yes, they are part of the same coset

[sonic current]

okay

hmm

let me see if I got the hang of this

well, I can take the elements (0,1) + ({0},7Z) and (1,0) + ({0},7Z)

they'll return (0,0) + ({0},7Z), right?

[stone fulcrum]

You mean if you add them together?

Or multiply them?

[sonic current]

multiply

[stone fulcrum]

Yeah, it would.

[sonic current]

cool

thanks, @stone fulcrum

[stone fulcrum]

Looks like two zero divisors to me

[sonic current]

your help has been invaluable

[stone fulcrum]

Hey, I'm glad! Feel free to ask if you have anything else.

[chilly ocean]

what does the notation B^A mean? where A and B are sets. apparently its the set of all relations between A and B?

[bleak abyss]

Functions A->B

[sick acorn]

it's the set of all functions from A to B

BUT

the cardinality is also similar

[chilly ocean]

ok. its very specific. why is it notated like a power? do you know

[sick acorn]

the # of functions from A to B is also |B|^|A|

[chilly ocean]

dont remove

[sick acorn]

i'm not sure why it's notated like a power, but

it's useful in thinking of it that way because

the # of functions from A to B

is basically

|B|^|A|

[chilly ocean]

right its just a notation quirk. at least how i understand it

[thorny slate]

we use it that way though

R^2 is just the set of functions from 2 = {0,1} to R

aka the pairs (x0, x1)

[chilly ocean]

ah thanks. thats helpful

[dense hull]

Can anyone explain the solution to b) to me?

In particular, why p(x)q(x) = bx ?

[thorny slate]

what

i mean

x times b is bx in R[x]

x is central I think

[covert vector]

Ya usually no one ever does polynomials over noncommutative rings, but when u do, x and other transcendentals/indeterminate will commute with coefficients

at least in my experience that's how it's done

so I guess the idea of the argument is that as polynomials, xb = bx, but when u plug in a, ab ≠ ba

but maybe they should have made it more clear what conventions are used

[dense hull]

Why do I assume they commute in the polynomial?

this sounds like a non-obvious assumption to make

[covert vector]

it is

[dense hull]

😦

[covert vector]

my conclusion is that this question sucks

I guess you could conclude that evaluation is only a homomorphism if evaluating at a central element

which always passes in commutative ring

[thorny slate]

R[x] centralizes x

R{x} is the one that doesnt

it's not a non-obvious assumption to make

and the question has nothing wrong

@dense hull @covert vector

just look at the construction of R[x]

x commutes with b because 1 commutes with b

x b = (0,1,0,...) (b,0,0,...) = (0,1b,0,0,...) = (0,b1,0,0,...) = bx

there's nothing wrong with polynomials over a noncommutative ring

[dense hull]

Thanks for the input

Ill look at it more

[covert vector]

oh ok !

that makes sense, thank you jaco

[dense hull]

Ok, I think I understand now

So (p(x)q(x)(c) = p(c)q(c) iff c is in the center of R?

[thorny slate]

yeah

[dense hull]

It makes sense looking at multiplication how R[x] is constructed

[sonic current]

hello guys. can anyone help me understand the need of a norm function in the study of irreversible and prime elements?

[errant drum]

Norms allow you to go from the ring you're working in to the integers. This makes it easy to work with its. For example if c is irreducible. N(c) is prime. It's a lot easier to go outside of the ring and calculate the norm to prove its prime rather than working inside your ring

[sonic current]

but where is this proven?

my algebra texts are proving irreducibility and so on with the norm function

without ever giving much into the explanation of it

anyway: just so I can understand this a little better

say you have two elements x and y of Z[√d], such that x = a + b√d and y = c + e√d

how do you multiply them? is it full distributive?

I want to prove N(xy) = N(x)N(y)

[errant drum]

Just calculate the norms manually

Remember when you take norms you can work in the integers

For calculating xy. Z√d is a integral domain

[sonic current]

yes, but

[errant drum]

Also all rings are distributive

[sonic current]

in N(xy)

i must multiply x and y, x = a + b√d and y = c + e√d

[errant drum]

Yes

I don't see where you're having trouble

[sonic current]

well, i'm not sure if it's full distribution

because if it is so, then I'm having trouble with the proof

because ac + ae√d + cb√d + bed

[errant drum]

You know that it's a ring which gives you distributivity

[sonic current]

yes

but when I apply N to ac + ae√d + cb√d + bed it gets very messy

way more messy than the simple N(x)N(y)

that's why I wonder if its component-wise multiplication

[stone fulcrum]

What's N here?

[errant drum]

Try doing N(x)N(y) and using the difference of squatlres to factorise it in to N(xy)

[sonic current]

N(a+b√d) = |a^2 -db^2|

so the N(x)N(y) case is very straightforward with very few calculations

[errant drum]

@stone fulcrum Norm

[stone fulcrum]

Yus but what is the norm? Does |a² - db²| work here?

[errant drum]

Yeah

Trying to show N(xy) = N(X)N(y)

I'm pretty sure you just multiply it out

[sonic current]

let me recheck my calculations

yeah, I don't see where I went wrong

N(x)N(y) = (ac)^2 - d(be)^2

I stopped N(xy) at (ac^2) + 2abcde + (bed)^2 - d(e^2(a^2+2ab+b^2))

also having a bit of trouble in the x is a unit iff N(x) = 1

[stone fulcrum]

It's messy, but done

[sonic current]

let me take a look

what a stupid slump to be into

[stone fulcrum]

I like to do it as LS = RS as that reduces the work a little bit I feel

[sonic current]

still having trouble with it

i'm getting all worked up because of the 2ab factor of the (a+b)^2

it should be obvious why it disappears

[stone fulcrum]

Left side, bottom line?

[sonic current]

ok I see a problem

[stone fulcrum]

Both terms produce a 2abcd term that cancels out

[sonic current]

nevermind

i'm dumb

I used a+b√d and c + e√d

...and my c and e are very similar...

[stone fulcrum]

No, lol. It's fair you're trying to understand, which I'm glad you're trying to do

[sonic current]

yeah, I have the second test coming up in january

i'm pretty sure I'll pass the subject since I had a decent grade in the first one

but since a colleague of mine is paying me to tutor him privately (lol) i want to be sure I know the way around stuff

plus algebra is nice when you understand it

[stone fulcrum]

Yeah, this is an awesome field

[sonic current]

but I was more familiar with the first part, groups and all

factorization in ID is new to me

ok, I'm good

but what about x is a unit iff N(x) = 1?

I started by assuming x is a unit

what's should it imply in this case? I know there is x^-1 such that xx^-1 = 1

N(xx^-1) = N(x)N(x^-1) = 1 but I don't think this is very useful

[errant drum]

It is because you know that the only factors are integers

What's the only way you can have the product being 1?

[stone fulcrum]

Note you did just prove something interesting:
N(x¯¹) = N(x)¯¹

[sonic current]

yes! i was wondering if I could have that

ok let me see

[stone fulcrum]

That's true in general
N(xⁿ) = N(x)ⁿ

I think anyway

Homomorphisms and stuff

[sonic current]

why should it be?

i am given nothing on N

[errant drum]

Try proving it by induction

When you take the norm of a+b√d on Z√d remember that you're doing a^2 -db^2

a is an integer

b is an integer

d is an integer

So what can you conclude about a^2 - db^2

[sonic current]

it's an integer as well? lol

I'm currently at N(xx^-1) = N(x)N(x^-1) = 1, so their product is 1

[errant drum]

Right so for any x=a+b√d

[sonic current]

if only I could have N(x^-1) = N(x)^-1

[errant drum]

N(X) is an integer

[sonic current]

because if I don't have that, then N(x) could be 1/7 and N(x^-1) = 7

[errant drum]

So N(x)N(x^-1)

[sonic current]

nevermind!

1/7 is not an integer

[errant drum]

Remember your norms only spit out integer values

[sonic current]

yes, i'm sorry - lol

[errant drum]

Dw it happens to everyone

[stone fulcrum]

Oh, we are only interested in integers! Then this becomes clear

[sonic current]

nice

[stone fulcrum]

Smart

[sonic current]

now I gotta prove the other way around

N(x) = 1 proves x is a unit

let me see.

i think this one is easier

I must use the norm definition for something

maybe if I can prove x multiplies by another element to produce 1, then it will be a unit I suppose

[errant drum]

Proving the norm function is multiplicative will be useful here

Which you have done

[stone fulcrum]

You do have that N(x¯¹) = N(x)¯¹
Try contradiction. What happens if x is a unit, and N(x) ≠ 1?

[sonic current]

where does that come from @stone fulcrum? the inverse

[errant drum]

Oh that's a nice way

[stone fulcrum]

You proved it earlier with
N(x)N(x¯¹) = 1

[sonic current]

oh yes

let me try contradiction

I love the idea of a contradiction but never am able to carry it out

[errant drum]

Usually it's useful when you have very little information to work with

[sonic current]

oh wait: i must now suppose N(x) = 1 to prove x is a unit

[errant drum]

Yeah

[sonic current]

is it still okay to use the contradiction? you were doing it the other way around, Kaynex

[stone fulcrum]

Any way you can prove it is fine.

Are you not trying to prove that:
x is a unit → N(x) = 1

?

[errant drum]

No

[sonic current]

no

[errant drum]

It's the other way

[sonic current]

we've done that

[stone fulcrum]

Ah

[errant drum]

Hint: N(x)*1 = N(1)

N(1)=1

Also use that norm of a unit is 1

[sonic current]

i cannot use the norm of a unit = 1

i mean, I could, but I don't think I should

[errant drum]

You proved it didn't you

[sonic current]

...yes but

😁

i wouldn't have it available if I didn't prove it first

i'm quite lost

is contradiction that straightforward?

[errant drum]

I have a way to prove using that norm of a unit is 1

[stone fulcrum]

I said that thinking we were trying to prove the other way

[errant drum]

I'll show you that way and then I'll think about another way

Is that okay?

[stone fulcrum]

I'm not sure how to go about
N(x) = 1 → x is a unit

[errant drum]

N(x) * 1 = N(1)

Let u be a unit

Then N(x) * N(u) = 1

N(xu) = N(1)

Does that work?

[stone fulcrum]

N(x) * 1 = 1?

[errant drum]

We're trying to show that if N(X) = 1 X is a unit.
N(X) = 1

So N(X) = N(1).

Then N(X) * 1 = N(1)

Let U be a unit then N(U) = 1

So N(X)N(U) = 1

N(XU) = N(X)N(U) = N(1)

So XU = 1

[sonic current]

can't we conclude anything with |a^2-db^2| = 1 ?

[errant drum]

Hence X is a unit

[stone fulcrum]

Oh, duh, N(x) = 1 is the premise. Brainfart.

[sonic current]

yes

[stone fulcrum]

You've shown that N(XU) = 1

This doesn't mean that XU is unity. Presuming it to be a unit is what we're trying to prove

[errant drum]

N(X)N(U) = 1 = N(1)

[sonic current]

but aren't you supposing N(x) = N(y) => x = y? @errant drum

[errant drum]

I'm using the fact it's multiplicative in reverse

N(X)N(U) = N(1)

[sonic current]

i was trying to find another element of y Z[√d] such that x*y = 1 always

but didn't go too far with it

[errant drum]

If d is negative it's pretty easy to show

[sonic current]

x is of the form x = a + b√d

by the hypothesis, we have |a^2-db^2] = 1

isn't this enough to go somewhere?

[stone fulcrum]

Actually, it might be worth checking out which elements of a + b√d are units

I'm thinking we need a property of units for this

A unit × a non-unit = a non-unit

[sonic current]

i just love when book authors leave proofs to the readers

[stone fulcrum]

Since any non-unit is contained in an ideal

[sonic current]

well

i'll try to move forward a bit

I smell trouble in this section for me

well it helps to thing of all the N(x) as integers

[errant drum]

I didn't work this out but I looked in my lecture notes. It says N(x) = x x * where x * is the conjugate of x.

So N(x) = 1 => x x* = 1

So x divides 1 @sonic current

[sonic current]

is that still regarding the proof?

[errant drum]

Yeah

[sonic current]

hmm

i never heard of the conjugate in this context

[errant drum]

I said conjugate but it just means if x = a + b√d then it's conjugate is a - b√d

[sonic current]

i'm going to take a break from this

jesus

[whole basalt]

Anyone have any idea if the split complex numbers form a local ring?

As in, Q [x] / (x^2-1)

[thorny slate]

what

do you mean x^2 + 1

[whole basalt]

No that would be the usual complex numbers.