groups-rings-fields

[white grotto]

ah right, its R or C

not just any field

[chilly ocean]

also

"let B be the matrix of I"

did you mean \Phi?

if B is the matrix of \Phi, then it is going to have dimensions n^2 x n^2 (because the dimension of M_n(K) over K is n^2), so how are you multiplying by the n x n matrix M afterwards?

there are just too many details missing

but it's a good first step to look at the matrix of this map

[south patrol]

btw uh so like we have this hom(V,W) and V* otimes W isomorphism thing (for finite dimensional spaces) and the map V* \otimes W -> hom(V,W) is nice but not so much the other round - is there any nice way to show this map is injective without appealing to bases though?

i've heard it should always be injective regardless of dimension

Eh okay I think because elements of V* \otimes W are finite sums of elementary tensors it is fairly straightforward to do by hand (just checked it) so nvm

just was curious if there's a like nonsense way to show this is injective and then we can appeal to the dimensionality to finish it off lol

[chilly ocean]

the nonsense way (i.e. without writing anything down) is to that it's easy to check

which proves it without appealing to bases 🙂

[rotund aurora]

Can you soulve P(X)^2+Q(X)^2=D(X)^2 where X is some set of variables for P, Q, D polynomials in X over Z?

[wooden ember]

There’s a pretty nonsense way to do it via tensor hom if I’m not mistaken. Take Hom(M,Hom(V,W)) = Hom(M(x)V,W) = Hom(M^(+)n, W) = Hom(M,W)^n = Hom(M,W^n) = Hom(M, V(x) W)

Then by Yoneda you get Hom(V,W) = V(x)W

Though I suppose if we’re saying that V(x)W = W^(+)n its even easier to prove cause you get the isomorphism straight away from that lmao

[wooden ember]

Seems like a very strong thing to ask

[viscid pewter]

let R(X) = P(X)^2 + Q(X)^2 - D(X)^2

then just solve R(X) = 0

[rotund aurora]

doesnt Brahmagupta work?

[gritty sparrow]

No he’s been dead for a while

[rotund aurora]

taking c=a and d=b, you have (a^2+b^2)^2=(a^2-b`2)^2+(2ab)^2, substitute a and b for some polynomials

[wooden ember]

I have no clue what brahmagupta is so maybe

[rotund aurora]

https://en.wikipedia.org/wiki/Brahmagupta–Fibonacci_identity

[chilly ocean]

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

[rotund aurora]

its an identity, only assumes associativity and communism I guess

[chilly ocean]

communism

[rotund aurora]

xd idk why I said that

I meant commutativity

[wooden ember]

Always assume communism

[viscid pewter]

ok but then what

how do you solve that

[wooden ember]

It still seems strong to me since you’re taking any set of variables

[rotund aurora]

its solved

[wooden ember]

And any polynomial

[viscid pewter]

what do you mean it's solved

[rotund aurora]

it works for all a and b that commute with each other and satisfy associativity

[viscid pewter]

i mean ok do it right now

[wooden ember]

So what

[viscid pewter]

P(X) = x^2 + 3

Q(X) = 2x

solve it

[rotund aurora]

oh no

[viscid pewter]

oh

D(X) = 1

[rotund aurora]

I meant a solution to the equation x^2+y^2=z^2 where x,y,z are polynomials

not like given p,q find d

[viscid pewter]

i know

[rotund aurora]

I dont understand what you mean then

[viscid pewter]

i gave you P, Q, D, find a solution

you said you solved it?

[wooden ember]

I still don’t see how you can do it cause in the end you still have to solve the polynomials

[viscid pewter]

what x solves it

[wooden ember]

Like

Take Q = D = 0

[rotund aurora]

ah

I dont want to solve it for x

[wooden ember]

And P a complicated polynomial

[viscid pewter]

??

[rotund aurora]

like Im doing transcendental stuff

[wooden ember]

Then you’re basically just asking: solve P(X)

[rotund aurora]

the solutions are the polynomials themselves

[wooden ember]

Which is a non trivial question

[viscid pewter]

what polynomial lmao

[wooden ember]

Ah okay

[viscid pewter]

what do you want to find

[wooden ember]

No I see what he means

[rotund aurora]

polynomials x,y,z such that x^2+y^2=z^2

in integers

[viscid pewter]

ok

[wooden ember]

Find all polynomials x y z such that x^2 + y^2 = z^2

[viscid pewter]

ok fine

[rotund aurora]

I think I found a family of those polynomials already lmao. Also I think the classic solution to Pythagorean equation might actually work

uhmm maybe no

[viscid pewter]

i mean yeah

just take any a, b, c that satisfy pythagoras

[wooden ember]

Yeah but that doesn’t give all such polynomials

[viscid pewter]

then let p = a(x+p1)(x+p2)...

[rotund aurora]

nonconstant polynomials

[viscid pewter]

q = b(x+p1)(x+p2)...

[rotund aurora]

oh well yeah

[viscid pewter]

d = c(x+p1)(x+p2)...

[wooden ember]

Ah but fair point the leading terms have to be a Pythagorean triple

[viscid pewter]

yes

and similarly the constant terms

[wooden ember]

Still are those all the solutions

[rotund aurora]

no

[viscid pewter]

example?

[white grotto]

yeah

its nonsense what i was writing

[rotund aurora]

https://en.wikipedia.org/wiki/Brahmagupta–Fibonacci_identity

[viscid pewter]

give me just an example

[rotund aurora]

substitute c=a, d=b

[viscid pewter]

my brain is small, i don't see how it applies

a concrete example

[rotund aurora]

^

substitute a, b for some polynomials

[viscid pewter]

give me an example

[rotund aurora]

I think it works

??

[viscid pewter]

pick some a and b

do it

[chilly ocean]

i'll give you a hint. write out the basis of \Phi with respect to the standard basis of M_n(K) and apply row/column operations to get a nice looking block matrix whose diagonalizability you can relate to that of A

[rotund aurora]

uhmm... Im not sure what you are trying to prove, like if its wrong just say it

a=x^2+1, b=x^3+1

[viscid pewter]

i'm not trying to prove anything

[wooden ember]

Yeah I’m not sure I see what happens to c then croqueta

[viscid pewter]

i'm just saying, let's see if it works

i don't know either way

[wooden ember]

Like you pick a and b, sure, but how does it give a solution

[viscid pewter]

ok so that's

well then P = a^2 + b^2

Q = a^2 - b^2

or no

D = a^2 + b^2

P = a^2 - b^2

Q = 2ab

[rotund aurora]

cuz thats sum of two squares equaling a square

ok Ill do hand computations

[wooden ember]

Ah i see what you mean now

[viscid pewter]

ok these are too complicated a and b lol, too many terms, can't be bothered

[rotund aurora]

dude

substitute x^2+1=a and x^3+1=b

and show it holds

[viscid pewter]

argh

ok

[rotund aurora]

thats how you would do it, idk why you wanted to get dirty

[wooden ember]

I mean there’s no need to sub here

The identity holds

[rotund aurora]

like that equation is easily seen to hold in any ring where a,b comute

[viscid pewter]

yeah ok

[rotund aurora]

you just want commutativity and associativity between a, b

[viscid pewter]

let me think then

[rotund aurora]

this is like the above

[viscid pewter]

ok

i have something

so this is just this, right

[rotund aurora]

also, I remembered that there was an additional condition on r,s being of different parity when you wanted the solutions to be primitive (in integers), but thats just an identity, so it works for any r,s that are associative and commute

[viscid pewter]

importantly it doesn't generate every triple even in the constant case

just every primitive triple lol

so you can take any solution and multiply everything by a constant

ok so i guess analogously to the integer case you can try to show that if there's a triple that doesn't just fall out of this formula, they share a factor?

[rotund aurora]

So I think the questions I was asking before about purely/totally transcendental are nonsense. Because Q<Q(t,sqrt(t^2+1)) is totally transcendental, but not purely transcendental (OK then I think this is just FALSE lol), and if you send t to p(x)/q(x) where p(x)^2+q(x)^2=d(x)^2 for some polynomial, then thats an embedding into Q(x), and so the image will be purely transcendental... as an extension of Q

wait... idk

[wooden ember]

Aight ima just ask this again but this time i can be a bit more precise. I don’t get why we can say wlog rho is unitary for the proof of plancherel. I now am aware that all irreducible reps are equivalent to unitary reps, but I don’t see how that allows us to just choose a different inner product to take the adjoint over without changing the trace (cf my question in #linear-algebra )

[formal ermine]

how do I calculate the minimal polynomial for an element in a field

[coral shale]

calculute? doubt

find, maybe.

its not like theres a one way for all method

wait

[formal ermine]

my homework wants me to "Calculate the minimal polynomial for ... over Q"

[coral shale]

minimal polynomial not within the field i hope

Like for an extension makes sense

otherwise its just x-a for a

F extends K, find minimal polynomial in K[x] for a in F

[formal ermine]

yes

[coral shale]

.

[formal ermine]

hmm

so do I just try out random shit till I figure it out?

[coral shale]

well my point also is

we cannot help if u dont give an example

[coral shale]

but basically, yes.

[formal ermine]

$\sqrt[3]{3} - \sqrt[-3]{3}$ in $\bR/\bQ$

[cloud walrus]

an illuwuminator3

[coral shale]

yh ok, so your first guess hopefully is that its a cubic

also the hell is the -3rd root

[formal ermine]

1/thirdroot

[coral shale]

bruh

anyways, first step is to find a polynomial its a root of

then factor it.

[formal ermine]

I was thinking it's a higher degree than cubic

[coral shale]

it might be

[formal ermine]

because we have 3^2/3

3^4/3

3^10/3

etc

[coral shale]

well consider the corresponding quadratic case

so i dont think so.

indeed, rationalize the denominator.

[formal ermine]

I'll try this later or tomorrow

but thanks for the tips so far

[coral shale]

try a quadratic example and see if u can do something similar

for a quadratic ull likely try a conjugate-like idea idk

[formal ermine]

let $\tau : \bK \to \bM$ and $\sigma : \bL \to \bM$. what does the notation $\sigma_{\mid \bK} = \tau$ mean?

[cloud walrus]

an illuwuminator3

[coral shale]

domain restriction

sigma does what tau does on K

[formal ermine]

and undefined on the rest?

[coral shale]

no

[coral shale]

.

[formal ermine]

ok

[coral shale]

sigma | K creates a new function from sigma with domain K

that does what sigma did on K

[formal ermine]

yeh

"For every field homomorphism $\tau : \bK \to \bM$ there exists a field homomorphism $\sigma : \bL \to \bM$ with $\sigma_{\mid \bK} = \tau$. The mapping $\tau$ can be continued as follows:"

[cloud walrus]

an illuwuminator3

[formal ermine]

I don't understand the last sentence and the diagram

$\bL/\bK$ is an algebraic extension and $\bM$ is algebraically closed

[cloud walrus]

an illuwuminator3

[coral shale]

So domain of tau

is smaller

you can think of sigma as an extension of tau

(thats the formal term anyways)

sigma extends tau

thats what they mean by continue

In particular, sigma is a homomorphism that extends homomorphism tau

Its not obvious this is possible to do

[formal ermine]

ah ok

so it's just like the homomorphism theorem for groups (or whatever it's called in english)

[coral shale]

no

[formal ermine]

so for every homomorphism from K -> M and if we have a homomorphism from K -> L then there exists a unique homomorphism from L -> M

[coral shale]

iso is a different statement

[formal ermine]

not talking about the iso one

[coral shale]

phi is a 'bigger' map

Like the quotient is a 'smaller' domain

so i wouldnt say its like it

[chilly radish]

It's generally not unique

Counting the number of these is important in galois theory

[coral shale]

oh yh that too

[formal ermine]

then why the dashed line

[chilly radish]

And algebraic number theory

Among others

[coral shale]

just implies existence of at least one

[chilly radish]

Dashed line means that there exiets

Usually you see \exists ! If it's unique

[formal ermine]

ah ok

[coral shale]

this is stronger - the iso theorem says its unique
although they didnt distinguish it in diagram

[formal ermine]

yh

oke thanks

what does the notation $m_a(x)_\tau$ mean

[cloud walrus]

an illuwuminator3

[formal ermine]

$m_a(x)$ is the minimal polynomial of $a \in \bL$ over $\bK$

[cloud walrus]

an illuwuminator3

[coral shale]

unsure

[formal ermine]

oh nvm

they explain it in the next sentence

[coral shale]

doesnt your course define notation before its used

bad not to

[formal ermine]

it usually doesn't

[coral shale]

thats terrible

[formal ermine]

yeah...

[coral shale]

what is it

[formal ermine]

tau applied to every coefficient

[coral shale]

t(m_a) makes sense to me

i think

more sense*

so (t(m_a)) (x)

ig more brackety

[formal ermine]

oh wow

I wouldn't have thought that every algebraic closure is isomorphic lol

[coral shale]

its because they have to be algebraic extensions

hmm

like u start adding stuff to try get a closure

and no matter how u do it it ends up being the same

[formal ermine]

yeah

[coral shale]

all the missing roots of polynomials

so i dont think its that surprising

[formal ermine]

I mean

if you phrase it that way

it isn't lol

[chilly radish]

It would be VERY bad if there were non isomorphic algebraic closures

I guess there can be, in the absence of choice

[coral shale]

what really?

[chilly radish]

Choice is used to prove both general existence and uniqueness

Surely, if you don't assume it (Technically you ol ly need boolean prime ideal.thm.but still) you can construct a model where smth like a finite field has several closures

Or maybe smth.more.complex is needed

No.ideahow one would.sbkw.this, tho

[coral shale]

choice doesnt come up in cat theory does it

[chilly radish]

Depends

[next obsidian]

Totally does

[coral shale]

time to quit math

[next obsidian]

Showing an equivalence of categories is equivalently a fully faithful and essentially surjective functor

[south patrol]

a fully faithful functor which is essentially surjective forms

DAMMIT

[next obsidian]

In-fact cat theory may end up requiring even stronger versions of choice, it may well use global choice

[next obsidian]

Chmowned

[south patrol]

:(

[chilly ocean]

CHMMMOWNED!!!

[elder wave]

[chilly ocean]

i m glad chmonkey is back to the original pfp, the other one i didn' t like it much

[next obsidian]

What

That was a long time ago

Are you not relatively new to the server?

[chilly ocean]

i am Carla_'s alt

[next obsidian]

Oh

Why aren’t you on Carla_?

[chilly ocean]

banned

[elder wave]

[next obsidian]

Bruh

How do u even get banned on discord

Except for being 12

[chilly ocean]

banned on the server not on discord

[next obsidian]

Oh

[chilly ocean]

because a moderator has a bad day

[next obsidian]

Very brave to just openly say u got banned and are now on an alt

[elder wave]

Indeed

[chilly ocean]

it doesn't matter

[next obsidian]

I mean if a mod read it they might ban you again for ban evasion or something idk

Just say you’re reformed

“Reformed”

Idk what you got banned for originally so

[chilly ocean]

Anyway, i wonder if the function sending a ring to its lattice of ideals is (contravariantly) functorial

[next obsidian]

Yes

It’s just pullback yeah?

[chilly ocean]

my only doubt is whether it preserves joins

meets are easy to show cuz they commute with inverse images

[next obsidian]

Yeah it does because of how preimages work

They commute with union and intersection

This is better than image

[chilly ocean]

joins of ideals aren't just union tho

[hollow mica]

@tribal moss right here

[next obsidian]

Okay true

Hmmm

I think it’s true

Tho

Like the join of ideals is +

f^-1(I + J) contains f^-1(I) and f^-1(J) by inspection

Wait no

No, no, it doesn’t

Consider Z[x] and the ideals (x) and (x + 1)

Their join is all of Z[x]

But they both pullback to 0

[chilly ocean]

pullback along which map?

[next obsidian]

Sorry to Z

So just intersect with Z

[chilly ocean]

i see

[chilly ocean]

so it's not functorial if you view it as a lattice

it is as a meet-semilattice instead

[next obsidian]

Yeah, but it is for surjections

Which is like the content of the 4th iso or lattice iso or correspondence theorem or some other thing ppl call it which probably exists

[chilly ocean]

yeah ik

was just wondering about in more general what happens

[next obsidian]

Yeah it don’t work good

☠️

[chilly ocean]

i guess you could say that this situation is not ideal

[next obsidian]

I’m on board with this guy now

[tribal moss]

What seems to be the problem?

[next obsidian]

Nothing

No problem

[tribal moss]

Ah, I see.

[coral shale]

just some people aping

[solar glacier]

silly question but if an ideal I is equal so say

$(3,2+i\sqrt{5})$

[cloud walrus]

MyMathYourMath

[solar glacier]

is that all multpiples of 3 and of 2+i\sqrt{5}

and of the two of them multiplied together

[next obsidian]

(a1,…,an) = {Sum b_i a_i}

Where b_i are any elements of the ring

[solar glacier]

ah

[simple mulch]

Need a hint on if $\langle a \rangle \langle b \rangle < G$ then $\langle a \rangle \langle b \rangle = \langle a,b \rangle$

[cloud walrus]

mns

[chilly ocean]

< means that it's a subgroup right?

[simple mulch]

Yeah

[chilly ocean]

yes

a^0 = e

[simple mulch]

I don't see the relation with being a subgroup

[coral spindle]

Observe both a and b are in G. Now consider that G is closed under the operation.

[simple mulch]

It might be because we have no information about the orders

[chilly ocean]

<a, b> contains a and b, and is a group

hence contains <a><b>

since it's the smallest such group, there is equality

[coral spindle]

Haha true, but I suspect the definition given is different

[simple mulch]

this is the generated group by ..

[coral shale]

show containment both ways is probably the way

[coral spindle]

Of course proving that <X> is the smallest subgroup containing X is easy

[simple mulch]

Yeah I thought about that

[knotty frigate]

btw whats the difference between the minimal subgroup with some property and the smallest subgroup with that property?

[coral spindle]

minimal implies not necessarily unique

[coral shale]

'a minimal'

[coral spindle]

smallest implies unique

[knotty frigate]

alright

thx

[simple mulch]

how so?

[coral spindle]

e.g. there are many minimal nonempty sets

there is no smallest set though

[chilly ocean]

minimal means there is no smaller. Smallest means every other is bigger than it

[chilly ocean]

it contains <a>, contains <b>, so does their product

[knotty frigate]

bc <a,b> has all elements of the form a^n*b^m. so it contains(<a> is a subset of <a,b>) <a> (the set which has all elements are of the form a^n)

[simple mulch]

Why am I thinking about <a><b> = {(a^n,b^m)}?

[chilly ocean]

you don't need to think about description of group generated by an element at all

[simple mulch]

but instead it should be <a><b> = {a^nb^m} ?

[chilly ocean]

everything here follows directly from <a, b> being a group, and from definition of <a>, <b> as smallest groups containing a, b respectively

[knotty frigate]

bc notation is confusing

like / being used for set difference, division and quotient groups 🙃

[chilly ocean]

usually it's \ that's used for set difference and / for quotients

[knotty frigate]

i really cant pay attention to details that small. ive tried just doesnt work

[simple mulch]

Where can I read something about this?

[knotty frigate]

about what?

[coral shale]

GH = {gh}

[knotty frigate]

cursed

[coral shale]

what.

[knotty frigate]

but true

[simple mulch]

indeed

xd

about this 😐

[coral shale]

This is the only defn that can make sense, given we want to GH to also be a group

[simple mulch]

Since <a,b> is the smallest group containing a and b, then <a,b> subset <a><b>

right?

[chilly ocean]

yes, assuming <a><b> is a subgroup, because a and b are contained in <a><b>

[simple mulch]

got it

thanks

Final proof: Notice that $\langle a,b\rangle$ is the group generated by $a$, $b$ and products of $ab$. Hence, $\langle a \rangle \langle b\rangle \subset \langle a, b \rangle$. For the reverse inclusion, notice that $\langle a,b\rangle$ is the smallest subgroup containing $a$ and $b$. Since $\langle a \rangle \langle b \rangle$ is a group and it contains both $a$ and $b$, then $\langle a,b\rangle \subset \langle a \rangle \langle b \rangle$.

[cloud walrus]

mns

[chilly ocean]

the first phrase seems a bit weirdly phrased to me

[simple mulch]

I meant powers of ab instead of products

products is a bit redundant in this situation lol

[chilly ocean]

<a,b> is generated by a,b saying anything more is already redundant

[simple mulch]

fair enough

[chilly ocean]

the reason <a><b> is contained in <a,b> is because an element of the former is of the form a^n b^m, which is clearly in <a,b>

[chilly ocean]

I didn't want to bother, but yeah

[simple mulch]

Hum ok, I will change it then

thank you again!

[tired horizon]

Does anyone know how to get the cyclotomic classes ?

Like in here

How did we get those C's

[vagrant zinc]

name of book?

[solar glacier]

can i verify a proof that in a comm ring with 1 R is a field iff the only ideals are (0) and R itself

going forward suppose (0) \neq I \subset R is an ideal not all of R let a \in I {0}

then a \in R

as R is a field then there exists an a^-1

and by property of ideal a^-1a=1 \in I thus I contains a unit and is all of R

going backwards

assume (0) \neq (a)

as (a) is not zero then it must be all of R

so there exists an r \in R such that ra=1

thus r = a^-1 and R is a field.

[south patrol]

yup that is great

i think this technically misses the edge case of the trivial ring but that aside sure

:)

[solar glacier]

thx 🙂

[south patrol]

okay now show that any ring homomorphism between fields is injective

xd

[elder wave]

[solar glacier]

do you do so by showing the kernal is trivial

[elder wave]

perhaps

why is the kernel trivial?

[solar glacier]

idk i have to think about this

[elder wave]

there is a reason why potato gave this exercise after the one you just solved

[solar glacier]

so theyre fields

so the only ideals are the whole thing and the trivial ideal

deos it use first iso theorem

[elder wave]

yes

now use that fact

[solar glacier]

so the only posiblilities for the kernel are trivial or whole ring

if its whole ring its the zero map

[elder wave]

a homomorphism of rings with unit sends unit to unit

[solar glacier]

so the kernel has to be trivial

[coral shale]

the zero map cannot be a unital ring homomorphism unless 0 = 1

[solar glacier]

oh thats right

so then the kernel is forced to be the trivial ideal

[elder wave]

indeed

[solar glacier]

nice argument!

[elder wave]

right?

[solar glacier]

indeed

hoping this question shows up on my algebra final lol

[south patrol]

noice

yeah this is cool cause it shows like

maps between fields are just inclusions basically which is helpful to know when studying them lol

[elder wave]

☝️

[south patrol]

but from the point of view of ring theory it makes it seem a lil boring perhaps

[tired horizon]

Nvm a pdf my teacher gave us

[solar glacier]

if a gaussian integer has prime norm, is it prime

[chilly ocean]

Is the norm multiplicative for any UFD?

[fading holly]

what does the M22(Z2) mean in this?

[chilly ocean]

22x22 matrix with coefficients in Z/2

[solar glacier]

yes

[chilly ocean]

probably 2 x 2 matrices with entries in Z_2

[fading holly]

oh ok

[chilly ocean]

or 22

lol

it could be either

the fact that you're being asked to do it for such an absurdly large number means it probably doesn't depend on the number...

[fading holly]

i looked up the M22 group and got this and was so confused

[chilly ocean]

it is not that

[fading holly]

ok ok

thanks

[chilly ocean]

i'm gonna go with 22

goofy question

[fading holly]

wait so its 22x22 matrices?

[chilly ocean]

So the standard proofs that Z[-sqrt(5)] is not UFD uses this fact, and that if the norm is 1 then the element is irreducible?

[solar glacier]

if the norm is 1 the element is a unit, no?

[chilly ocean]

Oh yea, my mistake

[solar glacier]

arent the units in Z_12 those coprime to 12?

[chilly ocean]

spoilers!

[tribal moss]

22×22 would make sense as a matter of notation, but 2×2 would fit better with the kind of exercise it seems to be.

[barren sierra]

I mean if it's 22 x 22

they may not expect anything more than "matricies with determinant 1"

I can't think of any other characterization tbh

tho one could argue that's not a super useful answer

[zenith seal]

Does anyone know how to solve this question? E/F is a finite Galois extension , Gal(E/F) is a cyclic group and its order is 100. So what is the number of K(a field which is satisfied F \subset K \subset E)(K is allowed to be E or F).

[next obsidian]

the fundamental theorem of Galois theory reduces this to a purely group-theoretic question, how many subgroups does the cyclic group of order 100 have

[zenith seal]

Thank you! I think I know how to figure out this!

[wooden ember]

I know that an integral extension stays integral under localization but does this go the other way?

Ie if T^-1R c T^-1S is integral then R c S is integral?

Im not sure how to say that a Monica polynomial in T^-R admitting an element of S as solution induces a monic polynomial in R with such a solution

[chilly ocean]

Is there a group of order 54 with no subgroup of order 9?

[white grotto]

how to show that the minimal polynomial and characteristic polynomial are equal iif its a cyclic endomorphism

[slender hamlet]

That's not true

For example, take an integral domain R which is not a field, K be it's field of fractions

Take T=R-{0}

T^-1R=T^-1K=K so T^-1R->T^-1K is integral

But R->K isn't

[white grotto]

how can a diagonalizable matrix be cyclic, any examples?

[rotund aurora]

Can you remind me what cyclic meant?

I think I have never heard this terminology for matrices, maybe you mean that theres a basis of the form (v,Av,A^2v,A^3v,...) for the space?

where A is your matrix

[south patrol]

I've not heard of that term cyclic either

Do you mean one with A^n = I for some n perhaps?

Idk

[lethal dune]

actually no

[wooden ember]

I see yeah I expected that much

[lethal dune]

cyclic means it has a basis in terms of v, Av, A²v,... which is not always diagonalizable

[south patrol]

They said any examples

[lethal dune]

a matrix is cyclic iff minimal poly=char poly

[white grotto]

i got this

[rotund aurora]

send $e_i$ to $e_{i+1}$, and $e_n$ to $e_1$

[white grotto]

it turned out if suffices to take the vector that is the sum of all eigenvectors

[lethal dune]

just take charpoly for example (x-2)²(x-1) and min poly (x-2)²(x-1) and that's will give you an example

[white grotto]

now i am struggling to understand why if the characteristic polynomial equals the minimal polynomial then f is cyclic

[lethal dune]

don't stress over it if you don't understand

[white grotto]

cyclic meand there exists x0 so that x0,f(x0),f^2(x0.... is a basis

[white grotto]

i mean i want the answer

[lethal dune]

A quick proof comes from rational canonical form

[white grotto]

i havent learned this yet so i prefer without

[lethal dune]

otherwise cumbersome, look at Hoffman Kunge

[rotund aurora]

You should try to work it out by yourself first...

I think

[white grotto]

if P0f=Xf

then P0 has degree n

[lethal dune]

as a hint, if min ≠ char then v, Av, .. A(ⁿ-1)v can never be LI

[white grotto]

then necessarily f is cyclic cause f,f^2,f^3.... are necessarily indepenedent

can i do this?

[lethal dune]

this is more of LA than AA so you probably should move there

[white grotto]

but is it done?

[lethal dune]

basically that's how to see it

[white grotto]

so this is it ?

[lethal dune]

don't have the time to walk you through it

[celest cairn]

What is $\varprojlim Gal(L/K)$?

[cloud walrus]

Sapphire Gaming

[shell agate]

absolute Galois group of K. The inverse limit runs over all finite extensions L of K.

[formal ermine]

the splitting field of $f(X) = (X^2 + 1)(X^2 + 4) \in \bQ[X]$ is $\bC$ because $f(i) = 0$?

[cloud walrus]

an illuwuminator3

[formal ermine]

o wait

Q(i) isn't C lol

[shell agate]

indeed

The splitting field is the smallest field containing all roots of f.

[formal ermine]

yeah

so it's just $\bQ(i)$?

[cloud walrus]

an illuwuminator3

[shell agate]

Yes

[formal ermine]

because the roots are $\pm i$ and $\pm 2i$

[shell agate]

exact

[cloud walrus]

an illuwuminator3

[formal ermine]

oke thanks!

[shell agate]

So just adding i is enough

But for say f(x)=(x²+1)(x³+1), you'd need i and a cube root of 1

[formal ermine]

why is $\bQ(\sqrt[3]{2}, \sqrt[3]{2}e^{\frac{2\pi i}{3}}, \sqrt[3]{2}e^{\frac{4\pi i}{3}})$ the splitting field of $x^3 - 1 \in \bQ[x]$? where does the $\sqrt[3]{2}$ come from?

[cloud walrus]

an illuwuminator3

[shell agate]

This is the splitting field of x³-2

[formal ermine]

oh then my lecture script contains a typo

ty

what is a $\bK$-linear field homomorphism?

[cloud walrus]

an illuwuminator3

[shell agate]

any field extension of K can be seen as a vector space over K, so K-linearity is just usual linearity of vector spaces, f(ax)=af(x) for any a in K.

[formal ermine]

ah okay

[formal ermine]

$i$ in $\bC/\bR$ is not separable right?

[cloud walrus]

an illuwuminator3

[formal ermine]

because $x^2 - 1 \in \bR[x]$ isn't

[cloud walrus]

an illuwuminator3

[rustic crown]

it is separable

also you mean x^2+1

[formal ermine]

but x^2 - 1 = (x - 1)(x + 1)

oh

jesus christ

I thought $i^2 - 1 = 0$

lmaooooooooo

[cloud walrus]

an illuwuminator3

[rustic crown]

even then, separability only means no repeated roots

x^2-1 doesn't have repeated roots

[formal ermine]

ok so for x^2 + 1

[rustic crown]

every irreducible polynomial over char 0 fields is automatically separable :p

[formal ermine]

it's irreducible in R[x] and d/dx (x^2 + 1) \neq 0

[rustic crown]

yee

[formal ermine]

so it's separable

right

ok thanks

had a brain fart right there

[formal ermine]

loooool this is the literal next thing I was about to read

[pliant forge]

$M = X \oplus Y = Z \oplus W$, submodules of $M$, where $X = Z$. How do you show that $Y \cong W$?

[cloud walrus]

Eso

[rustic crown]

do you mean X = Z or only isomorphic?

[pliant forge]

im not sure. cuz in the book its an equal sign

while for Y and W its iso

[rustic crown]

if they're equal then use that Y is iso to M/X

[pliant forge]

yeah i was about to do that

[rustic crown]

but if they're only iso, then it's false

[pliant forge]

thx. seems like im on right path

[formal ermine]

is $\bQ(i, \sqrt{3})$ the splitting field of $(x^2 + 1)(x^2 + 3) \in \bQ[x]$ and $[\bQ(i, \sqrt{3}) : \bQ] = 4$?
the roots of the polynomial are $\pm i$ and $\pm i \sqrt{3}$ so we need to add $\sqrt{3}$ and $i$ to the field. for the basis I thought of $\Set{1, i, \sqrt{3}, i\sqrt{3}}$, does that work?

[cloud walrus]

an illuwuminator3

[rustic crown]

simplest example i can think of is Z-modules where M = countable direct sum of Z
then M iso to M ⊕ Z iso M ⊕ 0

[spice whale]

not sure where to start here

[compact needle]

This is correct, though strictly speaking you have not given an argument about why you have a basis (i.e., why the dimension is properly 4 and not something smaller)

[formal ermine]

yuh

oke thanks

[spice whale]

like I'm not sure whether I'm supposed to be looking at global structure of the group or just those two characters

[hot lake]

just those two characters

[spice whale]

hm

interesting

[spice whale]

I'm still not really sure where to start

[hot lake]

a character of dimension 1 is a group morphism into C*

would you know a way to make more of them if you are given one (a nontrivial one)

[spice whale]

oh you complex conjugate

[hot lake]

well it could be real and then that wouldn't be telling you much

[spice whale]

compose with any automorphism

i suppose

[hot lake]

you don't really know of any

[spice whale]

hm

[hot lake]

are you given that G is finite

because I guess complex conjugation could work there

[spice whale]

infinite groups haven't been covered

[hot lake]

so let's assume G is finite

what kind of values can the character take

[spice whale]

we don't need that assumption actually

[hot lake]

we don't

[spice whale]

literally just conj compose chi is another 1d char

[hot lake]

what if chi(g) = 2 for some g

if chi has some nonreal value then yes complex conjugation gives you a third character and that's a contradiction

but what if chi is real

[spice whale]

ok let's assume all the elements have finite order

[hot lake]

alright

[spice whale]

oh he doesn't cover anything non-complex

[hot lake]

uuuuh

you're doing representation theory over the real numbers ???

[spice whale]

no

what?

[hot lake]

ah uh

what do you mean by non complex

[spice whale]

by "only one nontriv 1d char" he means "only one nontriv 1d char over the complex numbers"

[hot lake]

yes

but what I'm saying is

what if chi only took real values

[spice whale]

it has to

that's what we've proven

[hot lake]

oh yeah

but then how do you explain why it can't happen that chi(g) = 2 or some other random real number for any g

[spice whale]

if g has order n then chi(g) is an nth root of unity

[hot lake]

alright

so that shows |chi(g)| = 1

[spice whale]

yes

[hot lake]

so that should be it

[spice whale]

i mean if g has infinite order it gets weird but you can probably say sth like there must be more than one irred 1d char

not sure tho

[hot lake]

the book probably is about finite groups throughout

[spice whale]

this is ch10 of artin

[hot lake]

another proof that would work even for infinite groups

is that given chi : G -> C*

[spice whale]

yeah

[hot lake]

chi² is also a character

[spice whale]

oh

hm

[hot lake]

and if you only have two you get either chi(g)² = 1 or chi(g)² = chi(g) and you get what you want

[spice whale]

true

[hot lake]

you're taking the tensor product of the nontrivial representation with itself

to get a new representation of dimension 1

[spice whale]

i see

[coral spindle]

ALLY'S DOING REP THEORY

[spice whale]

yes

for 1 chapter

[coral spindle]

Hey that's good enough

[spice whale]

it's very fun

[coral spindle]

Lmao nerd

jk I love it too

[rustic crown]

me wanna learn rep theory too >.<

[solar glacier]

is 4 reducible in Z[i\sqrt{7}]

[rustic crown]

2 * 2

[solar glacier]

yeah thats what i was thinking

so it is then

cause taking norms

we can have that 16 = 4 times 4 which is valid in this ring

[rustic crown]

that's not the reason exactly

you need to show that 2 is indeed not a unit

[solar glacier]

yes that i have shown

since 16 = 2 times 8?

an element a is a unit iff N(a)=1

[rustic crown]

or -1

(i.e. a unit in Z)

[chilly ocean]

Can 2 ever be a unit in some Z[a] where a is the root of some monic polynomial in Z?

[rustic crown]

wait, i'm dum

you had sqrt(-7)

[left estuary]

Can’t find any good practice for rings

[formal ermine]

I can send you my old homework if you want lol

[woven obsidian]

If 2 was a unit in $Z[i\sqrt{7}]$, then it would also be a unit in $\mathbb{C}$

[left estuary]

Sure with work shown if possible lol

[cloud walrus]

AoiKunie

[chilly ocean]

2 is an unit in C

[woven obsidian]

Yeah woops

lmao

Still, it's inverse in $\mathbb{C}$ would be the same as its inverse in $Z[i\sqrt{7}]$

[cloud walrus]

AoiKunie

[solar glacier]

can you send to me plz i have a final tomorrow lol

[formal ermine]

I'll just send them here

I'm still ingame give me a sec

[solar glacier]

ok

@ me when you do

[chilly ocean]

it would, but you still need to show whether 1/2 is in Z[i\sqrt{7}]

[woven obsidian]

1/2 is not integral

[formal ermine]

@solar glacier @left estuary (all rings are commutative) \

1. Let $R$ be a ring, $N$ its nilpotent elements, $r \in R^\times$ (unit) and $x \in N$. Show that $r + x$ is a unit. \
2. Find all nilpotent elements in $\bZ_{20}$. \
3. Find all maximal and prime ideals in $\bR$, $\bC[X]$, and $\bZ_n$. You may use the fact that $\bC$ is algebraically closed. \
4. Show that $\operatorname{gcd}(3, (1 + i\sqrt{5})) = 1$ in $\bZ[i\sqrt{5}]$. \
5. Let $R$ be a pid and $\Set{0} \neq I \subseteq R$ a prime ideal. Show that $I$ is a maximal ideal. \
6. Show that $I = (X^2 - 2)$ is a maximal ideal in $\bQ[X]$. \
7. Let $R$ be an integral domain with $\operatorname{char}(R) = p > 0$. Show that the frobenius endomorphism defined by $x \mapsto x^p$ is a ring homomorphism. \
8. Let $R$ be a field. Show that the only ideals of $R$ are $\Set{0}$ and $R$. \
9. Check which polynomials of degree $2$ are irreducible in $\bZ_2[X]$. \
10. Give an example of a polynomial that is irreducible in $\bQ[X]$ but reducible in $\bZ[X]$. \
11. Check whether the following polynomials are irreducible in $\bQ[X]$: $3X^7 + 5$, $X^4 + 4X^3 + 2X^2 - 6$, $X^4 - X^3 + 9X^2 - 6X - 3$, $X^3 - 3X - 1$. \
12. Show that $X^{2^n} + 1 \in \bZ[X]$ is irreducible for all $n \in \bZ_{\geq 0}$. \
13. Let $R$ be a ring. Show that $R$ is a field iff every finitely generated $R$-module is free.

[cloud walrus]

an illuwuminator3

[formal ermine]

lmk if you wanna see the solution to any of those

[left estuary]

Alr

[woven obsidian]

Slight typo in question 3, C[X] is not algebraically closed

[formal ermine]

oops

meant C

fixed

[solar glacier]

for number 5 so we let P be prime ideal in PID then P=(p) then assume there is some ideal strictly between P and R

then this ideal is generated by a single element say I = (m) then (p) \subset (m)

thus there is an r \in R such that p = rm \in (p)

which is prime so either r or m in (p)

if m \jn P then I = P

if r \in (p) then r = sp for some s \in R then r=srm so m is a unit thus I = R

[spice whale]

how would you use a character table to find normal subgroups?

[woven obsidian]

There is some general theorem about this

Do you know what the kernel of a character is?

[spice whale]

x such that chi(x) = 0

[woven obsidian]

It's the x that makes chi(x)=1

[spice whale]

ah

wait yeah ofc

[woven obsidian]

So to every irreducible character you can find it's kernel by looking at the character table

It turns out, the set of normal subgroups of G is the set of all possible intersections of these kernels

(the empty intersection gives all of G)

[kindred mist]

Supposedly this v has rank one, I can’t really figure out why, any ideas?

It’s either rank one or it’s value group is (Z^2)_lex

FYI k-rank(R):= trdeg_{k} (R/M)

And I think he meant to say k-rank(R) = 0

(You can probably ignore the stuff about quadratic transforms, this is really just a valuation theory question)

[spice whale]

wait the identity isn't necessarily in the kernel?

[woven obsidian]

Oh sorry I messed up the definition of the kernel

$Ker(\chi) = {g\in G: \chi(g) = \chi(1)}$

[cloud walrus]

AoiKunie

[spice whale]

ah makes more sense

[spice whale]

so only irreducibles?

[woven obsidian]

yeah

[formal ermine]

let $K$ be a field with $\operatorname{char}(K) \neq 2$ and $a,b \in K$. Let $L$ be the splitting field of $X^4 - (a + b)X^2 + ab$. Show that $[L : K] = 4$ iff $a, b, a \cdot b$ aren't squares in $K$

[cloud walrus]

an illuwuminator3

[formal ermine]

it's like pretty obvious to me but I don't really know how to prove it formally

cuz like our polynomial has the roots \pm sqrt(a) and \pm sqrt(b)

without loss of generality if a would be a square in K then sqrt(a) would be in K

and then L : K wouldn't be 4 anymore

[rustic crown]

right

so first it's clear that the splitting field is K(sqrt(a), sqrt(b))

you wanna show this has degree 4.

first since a and b are not squares, the degree is at least 2

so you wanna show that sqrt(b) doesn't lie in K(sqrt(a))

[formal ermine]

because ab isn't square b \neq a so sqrt(b) \neq sqrt(a)

hmmm but what if b = k^2a

[rustic crown]

show that if r * sqrt(a) + s * sqrt(b) = t, then squaring gives you that sqrt(ab) lies in K

[formal ermine]

what about this tho

[shell agate]

then ab=ak²a=a²k² is a square

[formal ermine]

ah right

[proud bear]

i think it's just a typo. like they mean f^*(V(b))=V(b^c)

[celest cairn]

I was trying to find the minimal polynomial of $\mathbb{Q}(\sqrt[6]{2}) / \mathbb{Q}$ using Galois Theory. For example, we can compute the minimal polynomial of $\mathbb{Q}(\sqrt{2}) / \mathbb{Q}$. The minimal polynomial of $\mathbb{Q}(\sqrt{2}) / \mathbb{Q}$ can be computed using the formula $\prod_{\sigma \in Gal(L/K)}(x-\sigma(\alpha)) = (x-\sqrt{2})(x+\sqrt{2}) = x^{2}-2.$ How would I do this with $\mathbb{Q}(\sqrt[6]{2}) / \mathbb{Q}$?
I tried computing this but kept getting different roots and wasn’t getting anywhere.

[cloud walrus]

Sapphire Gaming

[delicate bloom]

I would just use x^6-2 and eisenstein's criteria.

[celest cairn]

Well yea ik how to do that, but is it possible to use that formula?

[delicate bloom]

not entirely sure what you're looking for, but if w is a 6th (not necessarily primitive) root of unity, then for f(x)=x^6-2 you have that if f(r)=0 so is f(wr)=0. so your roots are all r*w^k, which are all the conjugates of r.

[celest cairn]

Ah interesting, ok. Thanks

[ruby sundial]

yea

but the contraction of q is in A

yes

this is the goal

so we want to show f*(V(b))=V(X) in A for some set X

so from here do you know how to expand out the definition

|| for any q in V(b) then q contains b and q^c contains b^c so f*V(b) contains b^c. Correct? what happens after this? ||

[hollow mica]

Any complex vector space equipped with a non-degenerate, skew-symmetric bilinear form must be of even dimension.

Let the vector space be V and its dimension n, and denote evaluation of the bilinear form by (x, y)

if n = 1 then (x, x) = 0 -> (v, x) = 0 for all v, contradicting the fact that the bilinear form is non-degenerate

but I don't know how to do this for n = 3, 5, 7, ...

non-degenerate means that (·, x) \neq 0 for any x \in V

and skew-symmetric means (x, y) = -(y, x) for all x, y \in V

[chilly ocean]

if you have an invertible and skew-symmetric matrix B, what happens when you take the determinant of both sides of B^T = -B?

[hollow mica]

you get det(B) * det(B) = -1

[chilly ocean]

do you?

[hollow mica]

det(B^T) = 1/det(B) and det(-B) = -det(B) right

[chilly ocean]

i don't think so

no

neither are true

[hollow mica]

💀

oh the transpose is the same determinant

[chilly ocean]

det(B^T) = det(B) and det(-B) = (-1)^n det(B) where n is the dimension

[hollow mica]

I see

so n must be even

[chilly ocean]

why

[hollow mica]

for the equality to be true

det(B) = (-1)^n det(B)

and det(B) \neq 0 by assumption

[chilly ocean]

yes

a more interesting way to prove this is to use a "skew-symmetric gram-schmidt" to get a basis e_1, ..., e_n, f_1, ..., f_n in which the form looks like the standard symplectic form

this necessitates even dimension

you can prove it by inducting on the dimension iirc

[hollow mica]

gram-schmidt is a way to generate an orthonormal basis given an arbitrary one

wdym by skew-symmetric gram-schmidt

[chilly ocean]

i think you should write out what "e_1, ..., e_n, f_1, ..., f_n in which the form looks like the standard symplectic form" means

i put "skew-symmetric gram-schmidt" in quotations because it's not exactly gram-schmidt

but the aim is similar

[raw totem]

how is (1/n) Z usually interpreted? In the context of a discretization of the d-dimensional torus, I have $[(1/n) Z]^d$ as its discretization.

[chilly ocean]

actually, more generally, if B is any skew-symmetric bilinear form (not necessarily non-degenerate) then you can find a basis e_1, ..., e_n, f_1, ..., f_n, g_1, ..., g_r in which B is a block diagonal matrix whose top 2n x 2n block is the standard symplectic form wrt e, f's and the rest are zero

non-degeneracy is the condition that r = 0

[hollow mica]

this is what you're saying right

[chilly ocean]

yes

[cloud walrus]

AdrianV

[formal ermine]

let $f \in \bK[x]$ be an irreducible polynomial. how does $\operatorname{char}(\bK) = 0$ imply that $\operatorname{deg}(f') = \operatorname{deg}(f) - 1$? and how do we thus conclude that $f' \neq 0$?

[cloud walrus]

an illuwuminator3

[chilly ocean]

what is the leading coefficient of f'

in terms of that of f

[formal ermine]

deg(f) * leading coefficient of f

[chilly ocean]

yes

so what is the degree of f'?

[formal ermine]

deg(f) - 1

[chilly ocean]

there you go

[formal ermine]

but where are we using the fact that it's char 0

[chilly ocean]

hmm

i retract "there you go"

i think if i say any more i'm gonna spoil the entire thing

so i'll let you work on it some more

[formal ermine]

can't seem to figure it out

[chilly ocean]

you said that the leading coefficient of f' is deg(f)(leading coefficient of f)

is this zero?

non-zero?

it's the coefficient of x^{deg(f) - 1}, so, because you want to prove that deg(f') = deg(f) - 1, you should prove that this is non-zero

you are in a field my man

[formal ermine]

oh-

hmm

[chilly ocean]

the leading coefficient of f is non-zero

so actually, maybe it's more accurate to say "show that the leading coefficient of f' is deg(f)(leading coefficient of f)" instead

but this just ends up with you showing it's nonzero

so do that

the leading coefficient of f is non-zero, so is deg(f) times that non-zero?

really, deg(f) here means 1 + ... + 1 (deg(f) times)

and char(K) = 0, so...

[formal ermine]

but if the coefficient is zero then K wouldn't be a field

because ab would be 0 for nonzero a,b

[chilly ocean]

do you have non-zero a, b?

[formal ermine]

the leading coefficient is nonzero by definition and deg(f) is nonzero because f is irreducible

[chilly ocean]

...

even writing deg(f) as if it were an element of K is a bit of an abuse of notation

it really means what i wrote there

is that non-zero? why?

[formal ermine]

but constant polynomials aren't irreducible so deg(f) can't be 0?

[chilly ocean]

i don't think my point is getting across

the highest degree coefficient of f' that could possibly be non-zero that of degree deg(f) - 1, and the coefficient here is (1 + ... + 1)(leading coefficient of f), where 1 + ... + 1 means you take 1 in the field and add it to itself deg(f) many times

is this element of the field non-zero?

[formal ermine]

yes, because char is zero, I get what you mean

but what is wrong with my reasoning

[chilly ocean]

good

nothing is wrong with your reasoning, but i needed to make sure you understood what was actually going on

as in what "deg(f)" actually meant as an element of K

[formal ermine]

where are we using char 0 here then

[chilly ocean]

nothing is wrong with your reasoning up to proving that you're not adding 1 to itself zero times

i.e. getting zero in the field

you still have to go from "the degree of f is non-zero" to "deg(f)(leading coefficient of f) is non-zero in K (where, to drive the point home, deg(f) here means 1 added to itself deg(f) many times)"

your reasoning is not wrong but it is also not a complete proof of the statement

[formal ermine]

if deg(f) is nonzero then deg(f)(leading coefficient of f) is nonzero because K is a field

what's wrong here

[chilly ocean]

we are going in circles

i'll say it one last time

go from "char K = 0" to "deg(f) is non-zero in K"

deg(f) in K being 1 added to itself deg(f)-many times.

prove that $$\underbrace{(1 + \cdots + 1)}_{\text{$\deg(f)$ many times}}(\text{leading coefficient of $f$}) \neq 0$$

[cloud walrus]

TTerra

[chilly ocean]

since this is the highest possible non-zero coefficient f' could have, this proves that f' has degree equal to deg(f) - 1

no more notation abuse

[chilly ocean]

nothing is wrong as long as you understand the notation abuses you're using, and it's hard for me to tell if you do or not

[formal ermine]

if $\underbrace{1 + \ldots + 1}_{\text{$\deg(f)$ many times}}$ were zero then we'd have a constant polynomial

[cloud walrus]

an illuwuminator3

[formal ermine]

but constant polynomials aren't irreducible

[chilly ocean]

no, it's non-zero because char(K) is non-zero and deg(f) is a positive integer

if i take a field of characteristic two, then x^2 is not a constant polynomial, but 1 + 1 = 0

i'm really, really trying to stress the difference between the degree of the polynomial as an integer, and "1 added to itself degree many times" as an element of the field

[formal ermine]

isn't x^(1 + 1) = x^0 = 1

[chilly ocean]

NO

that's not how polynomials work

[next obsidian]

ChmonkaS

[formal ermine]

ohhhhhh

ok I get it now

thanks

[chilly ocean]

the powers of x in an element of K[x] are nonnegative integers, not elements of the field

when you call an integer an element of a field, you're actually referring to the multiplicative identity of the field added to itself that many times

"char K = 0" means that you never get zero by doing this

(except for adding 1 to itself "zero times" which is by definition just zero)

[formal ermine]

yes

[formal ermine]

I didn't understand this part

but now I do

thank you

[chilly ocean]

i probably should have said it like that a while ago

i had thought you understood this

[rotund aurora]

I think Q(x,y, cbrt(x^3+y^3)) cannot be embedded into a field of the form Q(x,y,z,...). See why?

I think something like Q(x,sqrt(-x^2-1)) already works

[left estuary]

For the first one is the exponent in R^x the same as x element of N?

[formal ermine]

R^x means units of R

so r is a unit in R

[left estuary]

Do you know the symbol for that or is that the symbol?

[formal ermine]

that's the symbol for it

$R^\times$

[cloud walrus]

an illuwuminator3

[left estuary]

Ah ok

So if you treat N as a group then with addition being the operation the solution should be contained in R right?

[formal ermine]

x is not necessarily in R^x

[left estuary]

But I thought n is where it = 0

[formal ermine]

just because x^k = 0 for some k doesn't mean x is a unit

[left estuary]

So R^N may not exist in R?

[formal ermine]

R^N?

[left estuary]

Sorry I get a little confused when you change the variables in an example

[formal ermine]

$\mathcal{N}$ is the set of nilpotent elements of a ring

[cloud walrus]

an illuwuminator3

[formal ermine]

we have an $x \in \mathcal{N}$

[cloud walrus]

an illuwuminator3

[formal ermine]

i.e. there exists a $k$ such that $x^k = 0$

[cloud walrus]

an illuwuminator3

[formal ermine]

we also have a unit $r$ meaning that there exists an $r'$ such that $r' \cdot r = 1$

[cloud walrus]

an illuwuminator3

[formal ermine]

show that $x + r$ is a unit

[cloud walrus]

an illuwuminator3

[left estuary]

Wait

Where did r’*r=1 come from

[formal ermine]

definition of a unit

[left estuary]

Oh shit I mixed up unit and element

So what’s the first step in solving

Cause idk where to go from here

r is a unit of R and x exists in N so then I guess you would need to show that N exists in R?

[formal ermine]

uh if I told you the first step to the solution I have in mind then it'd already spoiler the rest, kind of

[left estuary]

Ok

[left estuary]

Is this correct does N exist in R?

[formal ermine]

what do you mean with "N exists in R"

[left estuary]

The elements of N exist in R

[formal ermine]

by definition

[chilly ocean]

to prove that something is a unit you check it against the definition of a unit

[formal ermine]

,, \mathcal{N} = \Set{x \in R | x^k = 0 ~\text{for some k}}

[cloud walrus]

an illuwuminator3

[left estuary]

What’s the pipe for?

[formal ermine]

why are you doing abstract algebra without basic set theory knowledge

[chilly ocean]

https://en.wikipedia.org/wiki/Set-builder_notation

[left estuary]

Bordem

[formal ermine]

lol I feel ya

I thought I was a bad example, doing aa without la

but also the constant pain and agony without la knowledge

[left estuary]

I mean I’m getting the hang of it

[chilly ocean]

since r is a unit you might as well write r + x = r(1 + xr^{-1}). now you're showing that 1 + y is a unit for y nilpotent

(because xr^{-1} is nilpotent and the product of units is a unit)

to show that it's a unit you should write down an element z of the ring with z(1 + y) = 1

[left estuary]

So this is saying x exists in R where x^k=0 for some k?

[chilly ocean]

N is the set of all x in R such that x^k = 0 for some k

[left estuary]

What’s the difference between saying such that and where

[chilly ocean]

it's more formal, i guess?

[left estuary]

So they mean basically the same thing though?

[chilly ocean]

the problem in your phrasing is not the "where" part but the "x exists in R" part. you're reading the set builder notation wrong

you gotta be fluent in this stuff to do abstract algebra

[left estuary]

But doesn’t it mean is an element of

[chilly ocean]

fine

it's technically fine but no one ever says it like that

[left estuary]

Well that’s how I understand

[chilly ocean]

and the grammar feels off

[left estuary]

When I see is an element of

[chilly ocean]

and here's a (big) hint: ||what's the power series of 1/(1 + y) in terms of y?||

[left estuary]

Isn’t that the derivative of a log?

[left estuary]

Oh wait distribute r into the ()

[chilly ocean]

please don't undo the simplifying step

first prove that 1 + y is a unit for any nilpotent y, and use this to deduce that r + x is a unit for any unit r and nilpotent x

[left estuary]

So to do that I would have to test it as if it were not a unit

[chilly ocean]

.

[left estuary]

y is a unit or set?

[chilly ocean]

first prove that 1 + y is a unit for any nilpotent y

y is a nilpotent element

y is in N

[left estuary]

ah

[chilly ocean]

y is an element of R such that y^k = 0 for some k

etc.

we're doing ring theory and not set theory so you should not think of y as a set

[left estuary]

ah well then I see the problem here

I don’t know what a ring is

I know what a set and a group is

But not a ring

[chilly ocean]

this may not sound good but i don't know if you should be tackling this problem yet

you should be doing much more basic problems involving rings before this one

[left estuary]

I was trying to see if I could figure it out by looking at the problems

Well I have already wasted too much of your time to just not do the problem

[chilly ocean]

you're not wasting my time

[left estuary]

Plus I still think I can do it

[chilly ocean]

i'm just saying you should probably read the definition of a ring and learn a few basic things about them before doing an exercise about rings

[left estuary]

I’ve read the definitions

[chilly ocean]

but you said you don't know what a ring is

[left estuary]

Cause I didn’t understand the definitions

So this is plan b