groups-rings-fields

[delicate orchid]

Ok whatever

[crystal vale]

I want to show that if R has no non-trivial nilpotent element and (0) is prime ideal in R, then R has no zero divisors.

I showed it but use prime ideal definition that aRb subset of I then a in I or b in I.

[delicate orchid]

Ok I give up trying to prove that these two are equivalent for non-commutative rings because I don’t believe the result anymore

I’m thinking about the ideal (x1, x2, …) in the free algebra R<x1,…> and I just can’t see it working

[delicate orchid]

This seems weird. The existence of no zero divisors means you can’t have nilpotent elements so that’s not a necessary condition, and if you’re allowed to use the “a in I, b in I” definition or prime ideals, then (0) being prime gives you ab = 0 <=> a = 0 or b = 0

[crystal vale]

But this definition I can use when R is commutative but here R is commutative not given

Nilpotent condition necessary for non commutative ring

[delicate orchid]

I’ve googled it. It works for non-commutative rings you just replace everything with their principle ideals

[delicate orchid]

And I don’t buy it. If R has a nilpotent a^n = 0 then a^(n-1) if a zero divisor, regardless of commutativity

A two sided zero divisor no less

[crystal vale]

But if I define prime ideal definition in non commutative ring as aRb subset of I implies a in I or b in I then I need nilpotent condition to show that if ab=0 then aRb =0 which implies that aRb be a subset of (0) then a = 0 or b = 0

[delicate orchid]

Show me why?

[crystal vale]

Because if you take any br_1a ≠0 then (br_1a)^2 gives you 0 which contradicts that it has no non-trivial nilpotent element

Then bRa be a subset of (0) and then b in (0) or a in (0)

[delicate orchid]

Ok I see now

God i hate rings

Wait so what’s the actual problem? It sounds like you’ve completely solved everything

[crystal vale]

But I used the definition of a prime ideal in a non-commutative ring(without unity) which is not consistent with the commutative ring

[boreal inlet]

Rings without unity is already a bad news..

jokes aside

[delicate orchid]

Ok I’m just going to say it. What?

Have you just never divulged to me the definition you’re using for non-commutative rings? The commutative one is STRONGER than the one for non-commutative

So if you’ve shown an ideal is prime using the commutative one, you’re done

(I think those that satisfy the commutative one are called totally prime ideals or something like that)

[crystal vale]

But R is not a commutative ring so which definition should I use here ?

[boreal inlet]

Yeah like for every ab in P => a in P or b in P that's similar to like saying (a)(b) \subset P implies (a) \subset P or (b) \subset P

[delicate orchid]

Honestly I’m done thinking about this.

You’ve proven a stronger result than necessary, so you’re done

[coral spindle]

Ur a king wew

[crystal vale]

Not (ab) here I need aRb be subset of I

[boreal inlet]

First one is for commutative rings, and if that holds, the non-commutative one also holds

[delicate orchid]

Yeah, that’s the definition for non-commutative rings I know

You multiply the principle ideals together rather than the elements

[boreal inlet]

Sorry, (ab) was a typo, I meant to say (a)(b)

[delicate orchid]

Actually, no not just principle. It’s like

[boreal inlet]

For non commutative any subideal which satisfy these, works

[delicate orchid]

I J P ideals of R, if IJ subset P implies I or J subset P then P is prime

[boreal inlet]

Yeah so it doesn't need to be principal or whatever

[delicate orchid]

I tried like for 10 minutes to prove this is equivalent to the two sided nonsense but I couldn’t do it

It’s almost certainly very obvious

[boreal inlet]

I mean probably there's a counter- oh

[crystal vale]

I proved this for the unity ring, hopefully it will be hold without unity

[delicate orchid]

If you don’t have unity I am indeed just going to delete my discord account

[boreal inlet]

I don't think unity would matter for atleast this case

[delicate orchid]

How did you prove it with unity?

[crystal vale]

First I proved a->b->c

[delicate orchid]

I see, thanks

[crystal vale]

Is it correct now?

[delicate orchid]

Sure!

[crystal vale]

Without unity I can not use Ra notation for an ideal, so since for any r_1 br_1a = 0 so left ideals generated by a and b , so (b)(a) = {0} so it is subset of (0) hence (a) subset of (0) or (b) subset of (0).

Is it correct?

[south patrol]

I don't really understand what you are saying

[crystal vale]

I want to show that if R has no non-trivial nilpotent element and (0) is prime ideal in R, then R has no zero divisors.

Now I use the prime ideal definition that if IJ is a subset of P then I is subset of P or J is a subset of P.

I and J are both left ideals or both right ideals

[dull ginkgo]

RAAAAAA

Hard problem

[dull ginkgo]

(0) is a prime ideal is equivalent to it being a domain

A is prime is equivalent to ab in A <=> a in A or b in A. So if ab = 0 <=> a = 0 or b = 0

Wait noncommutstive

[rocky cloak]

(0) is prime iff R is a prime ring

[rustic crown]

is prime ring the non-commutative analogue of integral domain?

[south patrol]

Sad

[rocky cloak]

Maybe, though Id argue domain is the noncommutative analog of integral domain

[dull ginkgo]

TLDR if aRb = {0} then a = 0 or b = 0

[rustic crown]

(a)(b) ⊆ 0 then (a) ⊆ 0 or (b) ⊆ 0 is easier to parse

[rocky cloak]

Anyway, thinking of this problem, I would guess a good approach was to assume you have a prime ring with zero-divisors, then try to construct an idempotent

[dull ginkgo]

Idempotent or a nilpotent?

[rocky cloak]

Ah, I misread the problem

Then it might be easier to solve

Then it's really easy actually

[dull ginkgo]

Noncommutative

[crystal vale]

This works, right?

[dull ginkgo]

(bRa)^2 = bRabRa

If ab = 0 then it’s nilpotent and thus 0

so b = 0 or a = 0 by ring primality

Might try to prove that if R is a ring (not necessarily with unity) such that each element x has an o(x) > 1 such that x^o(x) = x that R is commutative

[dire siren]

that's Jacobson's theorem

[dull ginkgo]

I don’t care who’s it is

I must try

[dire siren]

I just mentioned it as a random fact

however, I don't know if it has an easy proof

even the particular case when o(x) is constant is not easy

[dull ginkgo]

I think I can extrapolate an additive order

x^N = x then x^(n(N - 1) + 1) = x

x^o(x) = x
(nx)^o(nx) = n^o(nx) x^o(nx) = nx

I want to chose a k = a(o(nx) - 1) = b(o(x) - 1)

k = (o(x) - 1)(o(nx) - 1) + 1 kinda just does the trick lmao

[dire siren]

I'd recommend giving a try to the particular problem "x^3=x for all x => commutativity"

it's tricky enough and might give some insight

[dull ginkgo]

My immediate inclination is to dice it up into a direct sum of subrings

[delicate orchid]

That one’s a classic but I have no idea how it goes

[dull ginkgo]

But I want to do it more algebraicly

[delicate orchid]

How is that not algebraic

If u were taking K(-) of the mf then yeah maybe

[dull ginkgo]

Assume x^3 = x for all x
(nx)^3 = n^3 x^3 = n^3 x = nx
thus (n^3 - n)x = 0, and x has an additive order

In particular, n(n - 1)(n + 1) must be divisible by this order for each n

Now n and (n +1) (or (n-1)) are coprime, so that order must divide n or (n+1) for each n

[delicate orchid]

Feel that? That’s true

Are we showing it’s char 2 btw is that where this is going

[dull ginkgo]

Bro I was just fucking around trying to extrapolate properties about this ring I wasn’t expecting that

[delicate orchid]

Yeah and all char 2 rings are commutative because I said so ok? Cry about it

[dire siren]

what about F_3?

[dull ginkgo]

Oh shit yeah

[rustic crown]

for n = 2, this only gives you 6x = 0

[delicate orchid]

It’s char 2 or 3 then? You kind of neglected n-1

[dull ginkgo]

True

[delicate orchid]

It has to divide one of n, n-1, n+1

[rustic crown]

why we assuming char = prime?

[dull ginkgo]

n i think is coprime to n^2 - 1?

Yeah, n is coprime to n + 1 and n - 1

[delicate orchid]

I’m cold and scared in this dark frightening world

[rustic crown]

[delicate orchid]

Also I wasn’t

[dull ginkgo]

So the additive order of element x must divide n XOR n^2 - 1 by coprimality

[delicate orchid]

Which is where we get F_3

[dull ginkgo]

Elements might have differing additive orders

[delicate orchid]

Might yeah, but we’ve accounted for the char 3 exception to our char 2 hypothesis is what I meant

[dull ginkgo]

I proved this. Hm

[delicate orchid]

I hope to the Heavenly Father I don’t have to think about this

[dull ginkgo]

Of note is that if ord(x) = n, then n[x,y] = 0

[delicate orchid]

So we just need n = 1 le troll face

[dull ginkgo]

if ord(x) = n, and ord(y) = m, then we know n[x,y] = 0 and m[x,y] = 0. Therefore if we can show n is coprime to m, then n and m both are divisible by ord([x,y]), which then therefore must be 1, thus x and y commute

So x and y commute if ord(x) and ord(y) are coprime

[dire siren]

I don't think it's iff

[dull ginkgo]

Good point

[delicate orchid]

The unit in question:

1

[dull ginkgo]

1 is coprime to every n because the only element dividing both 1 and n is 1

[delicate orchid]

1 also commutes with everything chucklebucks

[dull ginkgo]

Rings not necessarily with unity

[delicate orchid]

I leave now

[dire siren]

luckily the solution doesn't use the unity anyway

[dull ginkgo]

Might fuck around with x and y with disjoint principal ideals

[dire siren]

try to think multiplicatively rather than additively

[dull ginkgo]

I.e there does not exist an n and m such that nx^m = y or ny^m = x

[dire siren]

@dull ginkgo what are the nilpotents in the ring?

[dull ginkgo]

Oh shit

Let me think

[dull ginkgo]

I think I also can use that coprimality argument on this maybe

[dire siren]

I don't think coprimality is of any use

because of the Boole rings (in which the characteristic is 2)

[dull ginkgo]

are they not commutative always?

[dire siren]

they are

but I mean you can't use coprimality for them

[dull ginkgo]

It’s not a two way

[dull ginkgo]

I am trying to use it for this case possibly by teasing out some relationship between o(x) and ord(x)

By assuming k divides ord(x) and ord(y)

Of which both divide n^3 - n for each n

I wonder if I can manipulate that to show x = my^n or vice versa or something

[delicate orchid]

Find the ideypotents and fartin wedderburn it

[dull ginkgo]

this is for AW

Well an alternate proof

[delicate orchid]

Cancel the x so you get x^2 = 1. Simple as that

[dull ginkgo]

x and y commute if nx^u= my^v

I think

Actually

Just nx = my^k or vice versa

[dire siren]

@dull ginkgo if an element satisfies a^7=0, can you say something about it?

[dull ginkgo]

a^3 = a
0 = a^7 = a^3 a^3 a = a^3 = a => a = 0

[dire siren]

good

can you generalize?

[dull ginkgo]

Is there a name for the nilpotent “order”

[dire siren]

idk, I just call it nilpotence index

probably nilpotency index from what I could find

[dull ginkgo]

Assume nil(x) is the minimum n such that x^n = 0. If nil(x) = 3n + 1, then 0 = x^(3n+1) = (x^3)^n x = x^n x = x^(n + 1). n + 1 < 3n + 1 contradicts the minimality

[dire siren]

yes

[dull ginkgo]

Wait

If nil(x) > o(x)

then nil(x) = qo(x) + n

which creates the contradiction unless o(x) = 1, which is impossible unless x = 0 by definition

if nil(x) <= o(x)

Then o(x) = qnil(x) + n

which implies x = 0

as x^o(x) = (x^nil(x))^q x^n = 0 = x

So the only nilpotent element is 0

[dire siren]

yes, 0 is the only nilpotent

[dull ginkgo]

Please let me try from here :)

Or spoiler please

[dire siren]

just an alternative proof for "the only nilpotent is 0"

that can also be observed like this: if x^n=0, then x^(2n+1)=0, but x=x^3=x^5=...=x^(2n+1)

[dull ginkgo]

x^((o(x) - 1)n + 1) = x
set n = nil(x)

Immediate

Damn

I have an idea

I need to get some stuff done first but I’ll be back soon-ish. Please spoiler any further hints :)

[dire siren]

||find a property of x^2||

||then calculate (xe-exe)^2 for x arbitrary and|| ||e idempotent||

[delicate orchid]

If x is a Nilpotent then we have to have x^2 = 0 cause odd powers of x are equal, and then we have that x is 0 because x^3 = x = 0x = 0

Just throwing my nilpotent hat into the Jacobson ring

[south patrol]

Real

What is the problem

[south patrol]

Ah this

[dull ginkgo]

Very difficult

[delicate orchid]

Trivial jacobian so it’s semisimple

[long obsidian]

Do the p adics numbers have an order on them?

[dull ginkgo]

I think the completion has an element such that x^2 = -1

[delicate orchid]

\sqrt(n) is in Z_p iff uhhh the minimal poly in Z splits in Z_p seomthing something Hensel ring something something

[coral spindle]

Only some do. Not for all p.

However, all the p-adic fields contain higher roots of unity anyway

(Just not necessarily a primitive 4th root)

In particular for odd p they contain all the (p-1)st roots of unity

And for p=2 they contain the 4th roots of unity

I think that should be enough to preclude them

[south patrol]

If x^2 = -1 in Z_p then x has this property in Z/p^n

But 4 doesn't divide (p-1)p^(n-1) unless p=2 or p = 1 mod 4

Converse is left to the reader

[delicate orchid]

yeah that's what I said

[south patrol]

Ok

😭

[coral spindle]

UR A NERD WEW

[south patrol]

I think a funny thing you can do is use p-adic analysis

like

okay that trick i know only works for sqrt(n) not roots of unity lol

[delicate orchid]

https://media.discordapp.net/attachments/173087249741643776/1216079515307085854/rapidsave.com_dhfg2d7nbvfb1.gif

[south patrol]

for roots of unity i'd just hensel it up

[delicate orchid]

I understand Brauer lifting way more than Hensel's lemma

[coral spindle]

Brauer? I hardly

[wise helm]

Q1 i) I've proven that the statement is true in the left direction (that if gcd(m,j) = 1, then <j> = Z_m). I'm hitting a brick wall on proving it in the right direction (if j generates Z_m, gcd(m,j) = 1).
I've also attached the Hint question and the proof provided for that, though I haven't been able to make any progress with the hint: I don't see how any property to do with gcd(n, |a|) would be helpful here.

[sly rain]

Consider a j such that <j>=Z/mZ then I [j]_m I = I [1 + ... + 1 ]_m I = I [j*1]_m I = [1^j]_m = m

I think thats how you can go forward with the hint

[wise helm]

I don't know whether I've interpreted your notation correctly or not, but the <1^j> does it I think.

[wise helm]

Thank you!

[winged void]

I have small question regarding the last point

the question is whether the identity for the group is actually contained in the subgroup

what do they exactly

mean

can somoen elaborate on that please

[coral spindle]

The group G has an identity: e.

A subgroup H must also contain e.

That's it.

[winged void]

but it should containt the same identity

right

i assume

but is that not always the case since its a subgroup of some group then the identity must bein the subgroup

[coral spindle]

That is what I said

[winged void]

sure thx alot but is it not always the case that a subgroup contains an identity

so we do not need even to check that

[coral spindle]

No, it is ALWAYS the case.

[winged void]

can you give an example if possible

[coral spindle]

I am so confused by your question

OK, take any group G with identity e

Then {e} is a subgroup of G

There's a subgroup that contains the identity

As, after all, all subgroups must

[winged void]

yes i get that but i mean that all subgroups contains an identity so we do not need to check if the subgroup has an identity element to verify that its a subgroup

that is what i mean

we should only check closure

and inverse

[coral spindle]

No.

Because the empty set is not a subgroup.

If your subgroup is nonempty, then indeed you need only prove closure under the group operation and inverses

But you must have nonemptiness.

[winged void]

oh sure so i should verify that its not empty first

[coral spindle]

Indeed

This is usually most easily shown by showing that the identity is in the subgroup

[winged void]

ah i see well that make sense

thanks

i see also now some pictures in the book

i find it amazing so im going to share it

is H a subgroup

this is my answer

im not really sure if that is sufficient

[coral spindle]

Your proof is incorrect

You incorrectly assume a = 1/b

You cannot do such a thing.

You must prove closure for all a and b, not specific ones of your choosing.

[winged void]

well here a and b are arbitrair

right

[coral spindle]

No they are not

You specifically choose b to be the same value as a.

That is not arbitrary at all.

[rapid junco]

n and m have to differ

To be totally arbitrary

[winged void]

well i thought that was the subgroup criterion

[coral spindle]

No they don't have to differ, that's not quite right

But one cannot assume they are equal

Since you have to prove it for every such pair

[rapid junco]

Well yeah this is what i mean

[coral spindle]

The criterion is:

For all a, b in H, we have ab^-1 in H.
You have proved:
For all a in H, we have aa^-1 in H.

This is because you assumed (wrongly) that a = b.

[winged void]

well i will try again

i guess i get what you mean

[coral spindle]

Is this unclear?

[winged void]

no its clear for sure but i want to make sure that i understand it. Thats why im going to write the proof again

[coral spindle]

OK

[winged void]

this is what i have

[coral spindle]

This proof is correct.

[winged void]

sure because here a and b are arbitrair

the first proof was a actually also arbitrair but when i choosed b to be a^{-1} it broke the arbitrairity

am i right

[coral spindle]

Yes.

You made an unfounded assumption.

[winged void]

Exactly thx a lot

!

[winged void]

i have a small question regarding isomorphism i know that its a bijectvie homorphism

i was trying to prove the following

i prove that is homorphism which was easy

but not so sure how to prove that its isomorphism

i know we need to check if its bijective

but is it enough to say well the function has an inverse

[coral spindle]

That is equivalent to being bijective, yes.

[winged void]

i do not know why i had the idea that its something else

but that is true

[hidden wind]

hi frens have:

[hidden wind]

[vivid tiger]

what group is this

[delicate orchid]

Looks like U(34)

[south patrol]

34x34 unitary matrices

[delicate orchid]

Or Z/34Z*

Because I KNEW

I KNEW

exactly what was coming

[vivid tiger]

how'd you know

[vivid tiger]

how'd you know

[hidden wind]

yes

[winged void]

i have question regarding this but not so sure if my solution is correct

this was my solution

[celest furnace]

Where did u show that f(a*b)=f(a)*f(b)?

[winged void]

It’s the second picture

The f(a+b) = f(a)+f(b)

But in the first picture is the way how I proved that the operation is well defined correct is a correct way

[cobalt heath]

I think you need to put in more details on kernel and image

[celest furnace]

You never verified that f(a*b)=f(a)*f(b) you just said “ we need to show that…”

[winged void]

Well I verified

That operation is well defined

[winged void]

This picture

[winged void]

Image I did not solve it yes

Kernel

[cobalt heath]

Yeah, like, completely characterize kernel

[winged void]

Sure but does the well defined part

And homorphism mathematically correct

Because I want to make sure that I understand those two parts good

[cobalt heath]

Ye those seem good

[winged void]

Sure thx I will do now the other two parts

[tough hedge]

https://math.stackexchange.com/questions/3185444/number-of-different-quadratic-functions-mod-12

in the reply, can someone explain what is $(\mathbb{Z}{12}) ^ {\mathbb{Z}{12}}$ and where the $a\equiv b \equiv 0 \mod 3$ comes from?

[cloud walrus]

M8 of 48

[chilly ocean]

Z_12^(Z_12) is the set of functions from Z_12 to Z_12. in general, when X and Y are sets, Y^X denotes the set of functions from X to Y

[tough hedge]

ah ok, never seen that notation before

[crystal vale]

In abelian Group G if we have a finite order element a and b , then what can we say about ab 's order , it is finite. But can we say more accurately?

[sonic coral]

the order of the product divides the lcm of the orders

[crystal vale]

Yes

I want to show that if G is an abelian group and H and K are finite cyclic Subgroups of G of order m and n respectively, then it has cyclic subgroup of order of lcm (m,n).

I don't want to use the converse of Lagrange theorem

[tough hedge]

consider the subgroup HK

[crystal vale]

Yes but it has order rs/ | H and K | and then how can I find | H and K |

[tough hedge]

well do you have a guess as to what |H and K| should be?

[crystal vale]

| H and K | divides gcd(r,s)

In abelian group G if a has m order and b has n order then ab has lcm(m,n) order when m and n are relatively prime?

[tough hedge]

well yes, but weren't you trying to show that result

[crystal vale]

In case of relative prime, can I use this one ?

[tough hedge]

it would just be a direct result of that fact, since lcm(m,n) = mn when m and n are coprime

[crystal vale]

Yes, but we need to show that ab has order lcm(r,s) when r and s are co-prime

If it is not co-prime then the order of ab is not lcm(r,s), right?

[lime junco]

For some ring, R

what does R(u) actually mean

i understand what it means for a field

not sure for rings

[tough hedge]

yes, i suppose i used "direct" a bit loosely but you should still show that both of them are true

[lime junco]

oh sorry yall were in the middle of sth my bad

i typed this before it fully scrolled down

[tough hedge]

as for the rest of the proof, i'll leave this here for you to look at if you get stuck

you have |H and K| divides gcd(r,s).
if you can show ||gcd(r,s) divides |H and K| then you basically finish it off||
since you get |||H and K| = gcd(r,s), and |HK| = rs / gcd(r,s) = lcm(r,s) ||

[crystal vale]

And this case for general, right?

[tough hedge]

it's only true for these conditions

think about under what circumstances HK isn't a subgroup of G for arbitrary G, H, K

[crystal vale]

Not Abelian

[tough hedge]

yep

i think the most general is when they are not normal subgroups

[crystal vale]

I don't get it how can I show gcd(r,s) divides | H and K |

[tough hedge]

hint - use division algorithm and divide |H and K| by gcd(r,s). show that the remainder is 0

@crystal vale

[crystal vale]

If I let c in H and K then I will get an order of c divides the i , where 0<= i < gcd(r,s)

[chilly radish]

Actually the most general condition is this: AB is a group iff AB=BA as sets

This is always true if even one of the subgroups is normal

[tough hedge]

you also get that $e = c ^ i$, then use the fact that the order is the minimum element satisfying c^k = e$ to conclude i must be 0

[cloud walrus]

M8 of 48
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[tough hedge]

is it not the case that AB = BA iff A,B are normal subgroups?

[chilly radish]

It is, as I said, it holds even if one of them is normal

It's just more general

This can happen with non normal subgroups also

[tough hedge]

ah yes yes

[crystal vale]

Order of c is greater than or equal to gcd(r,s) ? |c| divides both r and s then |c| divides gcd(r,s) , but how can I show that the order of c is greater than or equal to gcd(r,s) ?

[winged void]

Good afternoon i have a question regarding this

what do they mean if sigma is a transposition then e(sigma) = -1

i thought that is something with determinant

but thats not true

[rustic crown]

do you know what sign of a permutation means?

[winged void]

not really i thoguht that it has to do with swapping elements

or something

[rustic crown]

yep

it's the parity of the number of swaps required to make the permutation the identity

if you have any transposition, then you just need 1 swap to make it the identity, so it's an odd permutation

you define e(sigma) = 1 if it is even and e(sigma) = -1 if it is odd

[winged void]

but what has that to do with determinant

[rustic crown]

det also has this property that if you swap two rows/columns, then it changes the sign

if A is a matrix and σ is a permutation. Then let A_σ be the matrix whose columns are obtained from A by permuting using σ.

then det(A_σ) = sgn(σ) * det(A)

sgn(σ) = ε(σ) and the two popular notations

[winged void]

oh i see

[rustic crown]

using this observation (and multi-linearity), you could give an explicit formula for the det of a matrix

[winged void]

that is true im going to try to apply it on a concret example

2*2 matrix

[rustic crown]

okie :3

[winged void]

to get more intuition thx

(:

[rustic crown]

c:

[rustic crown]

strictly speaking, this is not a good definition. this is a characteristic property of sign of a permutation. people don't like definition with choices usually, there are many ways to start from a permutation and end at the identity using swaps. one needs to check that any two will give the same parity.

[winged void]

i see

i tried it now on a matrix

this is what i got

[rustic crown]

one way to give a choice-independent definition is by making all choices simultaneously. so for a permutation σ, look at the set of pairs Inv(σ) = {(i, j) : i < j but σ(i) > σ(j)}

[winged void]

well actually i always get the same result back so det (A) = det(A_{$\sigma$})

[rustic crown]

and define sgn σ := (-1)^|Inv(σ)|

[cloud walrus]

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

[winged void]

ah i seee

and if you look at my example when calculating the determinant after swaping two coloumn i get the same determinant

what is the intuition behind that result

[rustic crown]

there is a tiny calculation error

[winged void]

why does it work all the time

[winged void]

oeps where

[rustic crown]

det(A) = 3*4-2*2 = 12 - 4 = 8

det(A_σ) = 2*2 - 3*4 = -8

[winged void]

well but we need to multiply it as well with the sign which is -1

right

then we get 8 as well

[rustic crown]

yep, so sgn(σ) * det(A_σ) = det(A)

[winged void]

true but what is the intuition behind it

why does it work all the time

like the guy who found this property he must have seen something (intuition behind it )

[rustic crown]

to give a satisfactory answer, i would need you to know the definitions of det and sgn

these are both characteristic properties of det and sgn, so some people literally use these to define them.

so in some sense, that statement is true "almost by definition"

[rustic crown]

intuition is just this: swapping two elements in a permutation should change its sign. similarly swapping two columns of a matrix should change the sign of a det.

so if you carry this out slowly, instead of getting A_σ directly. do it one swap at a time

with each swap you change both the sign of σ and the sign of det

eventually the σ becomes the identity and you've changed the sign of det by sgn(σ)

[winged void]

ah i see

[dull ginkgo]

Split the NxN matrix into N^3 determinantlets through multilinearity

[rustic crown]

you mean N^N

[dull ginkgo]

Oh yeah lmao

Even worse

[winged void]

there is a question in the book that says prove that $S_3 = {(1),(1 2), (2 3), (1 3), (1 2 3), (1 3 2)}$ what does that even suppose to mean

[cloud walrus]

Mootje

[dull ginkgo]

S_3’s underlying set is that

[winged void]

i did not get what you mean

[dull ginkgo]

[winged void]

S_3 is just all possible permutation

[dull ginkgo]

Y e s

[winged void]

with three elements

1 2 3

so how am i suppose to prove something that is so defined

[dull ginkgo]

So you have to show a bijection from Sym({1,2,3}) to that set

[winged void]

sure

[dull ginkgo]

So do that

Honestly it’s probably easier to do the reverse

e.g. map that set to S_3

[winged void]

what do you mean with reverse

[dull ginkgo]

Show a bijection from that set to S_3

[winged void]

that takes a lot of time cannot i just compute all the elements of S_3 and then finish

because does not even take that long

and its easier than showing a bijection

because bijection means i need to show surjuctivity and injectivity

but not so sure if thats even possible

i find it a stupid quesiton

[dull ginkgo]

Do you have to do it

[winged void]

not really

so thats why i skipped the question

[last spoke]

it's "defined" as the set of permutations of three elements but it takes further computation to show that that's all of them

actually you don't really need to compute all the elements of s_3, you can use a counting argument to show that that's all of them

[winged void]

sure thats true

i guess im going to do that

Let $d, n \in \mathbb{Z}_{>0}$ with $d | n$. \textbf{(a)} Prove that there exists a group homomorphism $f: \mathbb{Z}/n\mathbb{Z} \rightarrow \mathbb{Z}/d\mathbb{Z}$ such that $f(a \mod n) = (a \mod d)$ for all $a \in \mathbb{Z}$. Is $f$ surjective?

this is what i did but i could not reach a conclusion

to see if its well defined

[cloud walrus]

Mootje

[winged void]

however i could not reach a conclusion

[lament bough]

Z -> Z/dZ is an epimorphism for which nZ is in the kernel, therefore it induces an epimorphism Z/nZ -> Z/dZ

[winged void]

Well I’m trying to prove that it’s well defined

Where did I make a mistake

[sly rain]

I think youre done after:
assume a=b mod n hence n divides a-b so d divides a-b so a=b mod d.

baire cows approach gives you an epimorphism by universal property , so you dont even have to check its well defined by hand.

[winged void]

how did you reach the conclusion that since n divide a-b then d divides a-b

and we did not yet get epimorphism

and i want to do it the way that i was doing it here

and know if my approach is correc

or should i change something

[sly rain]

Because we know d divides n, maybe try to write it out to convince yourself.

Yep, you still have to show that its surjective If you want to do it by hand.

I cant properly read what you wrote, but you want to Show that If a=b mod n then f(a)=f(b) mod d

So you start by picking a and b in Z/nZ such that a=b mod n...

[dull ginkgo]

Is the compactness of [0,1] needed for 14?

[cobalt heath]

Not read the problem but I am quite sure >.>

[dull ginkgo]

?

[cobalt heath]

Quite sure that anything involving [0,1] is abt compactness

[dull ginkgo]

Didn’t think I’d be using fucking HEINE BOREL

IN AN ALGEBRA TEXTBO

[cobalt heath]

Maybe you can see if this works for entire R if you wanna see closedness works

[dull ginkgo]

No like I can do it with compactness

[cobalt heath]

Compactness is /= heine borel tho

[dull ginkgo]

[0,1] being compact

[fading field]

you're not

you're using the fact that [0,1] is compact

heine borel relies on the fact that [0,1] is compact in R

[dull ginkgo]

Isn’t it proving that [0,1] is compact Heine Borel itself?

[fading field]

no

[dull ginkgo]

Actually paracompactness would work

[fading field]

the full Heine Borel statement is about R^n

[dull ginkgo]

Wait

[fading field]

interestingly, HB also not true for general metric spaces. It's not even true for complete metric spaces in general

[dull ginkgo]

HB basically asserts bounded -> compact for Euclidean metric spaces afaik from the proof I know. The converse is always true because compact image under the metric from X x X

anyway

[fading field]

bounded doesn't imply compact

of course

[dull ginkgo]

Bounded closed

[fading field]

i know you know

[dull ginkgo]

Bolzano Weierstrass shows that it’s sequentially compact though. Just takes more steps to go from sequentially compact to compact

[cobalt heath]

IIRC you need totally boundedness. Right? (And one more I forgot)

[dull ginkgo]

The FUCK

C^infty(R^n) HAS HEINE BOREL PROPERTY

[fading field]

[dull ginkgo]

Anyway

I think you can prove [0,1] is paracompact more easily

If you are a psychopath and needed to prove it for this problem without asserting [0,1] is compact a priori

[fading field]

usually with algebra books it's okay to assume that [0,1] is compact

i mean the point of algebra is to have a framework to talk about real math that's in the nitty-gritty of things

like algebra is really just machinery

so if it's not really relevant to the algebra, then you don't have to worry about it too much

[cobalt heath]

Don’t you need the metric here

[fading field]

you always need the metric

you need the metric if you're tlaking about R

even

[cobalt heath]

Yeah

[celest furnace]

Filthy animals take point set to #point-set-topology

[fading field]

lol

i think it's a useful exercise to talk about other contexts for which algebra helps us to think about

because like how else are we gonna use any of this algebra

[dull ginkgo]

(I am NOT latexing this shit, R is the reals)

Any ring morphism from C(X), X compact, to R is a valuation morphism. Also let i: x |-> 1 be the constant map of C(X)

Assume u isn’t a valuation morphism. Let v_x(f) = f(x) be the valuation morphism of x.

That means for each x in X there is a function f_x such that u(f_x) ≠ v_x(f_x). Let g_x = f_x - u(f_x)i.

Then v_x(g_x) ≠ 0, and u(f_x) = 0

Now for each x, the support of g_x exists, and the supports of g_x cover X. Thus by compactness, there is a finite number of supp(g_x_n) that cover X.

Therefore, the sum over g_x_n^2 is a function, but its support is over all of X. Therefore it admits an inverse in C(X), and is therefore a unit. This contradicts its image being 0, implying it lies in an ideal.

This beats the one problem that I had to use Urysohn’s Lemma for

[hidden wind]

whaaat really?

[cobalt heath]

I am still wondering what metric miz is talking abt

I heard C^infty does not come with natural metric

[hidden wind]

i’d expect the sup norm

[crystal vale]

In 15 question, they assumed R has unity?

[cobalt heath]

Hmmm. I mean, supremum of a smooth function could be infinity

Metric doesn't work well with infinity, does it

[hidden wind]

i mean the metric induced by it

sup(f - g)

[cobalt heath]

I'm not following

What if f(x) = x, g(x) = 0?

[lament bough]

then its 1

you bound the sup norm with something from above

[cobalt heath]

Hmmm

What was heine borel, again? I thought it was closed + bounded = compact

But if you bound the norm anyway, you will get closed = compact

[rocky cloak]

If you replace bounded by totally bounded, then Heine Borel should work in any complete metric space

[cobalt heath]

Ah, that makes sense.

Is C^\infty complete w.r.t. the uniform bounded metric

[rocky cloak]

I think so. Uniform convergence of smooth functions being smooth sounds quite reasonable

Certainly true if you replace smooth by continuous

[south patrol]

No. It's not a closed subspace of L^oo, say - indeed it is dense!

For example you can consider a sequence f_n of smooth functions which are 1 on [-1,-1] and 0 for |x| >= 1 + 1/n. Then (f_n) is Cauchy but not convergent (both because it converges to (the equivalence class of) the indicator of [-1,-1])

@cobalt heath

[cobalt heath]

Hmm

[south patrol]

For finite p, you can turn C^p(R) into a complete metric space in a natural way by putting the norm as the sum |f|_oo + |f'| + ... + |f^(p)|_oo i believe

[cobalt heath]

Yeah, on C^p you get natural complete metric

C^\infty tho

[dull ginkgo]

The completion is still W^p_infty(R) right

[delicate orchid]

the W....

[summer path]

the Wew..?

[still dew]

so using the hint if we consider the jordan matrix of A and look at the jordan blocks we are able to get a contradiction i think but i want to avoid matrices

is there a way to do this just by considering linear transformations

i.e my definition of diagonalizable is just there is a basis of eigenvectors and is it possible to do this just with that

[rocky cloak]

That doesn't seem Cauchy.

Like for any n you should be able to find x > 1, such that fn(x) > 1/2, then for N > 1/(x-1) you have |fn - fN| > 1/2

[south patrol]

Wait yeah I may have messed up lol

Uhhhh

There are things that work though lol

i think

Yeah so an easier example would be uh

Wait nvm

Ugh

[rocky cloak]

I guess you should be able to approximate any continuous function by smooth ones

So like |x| or whatever

[sonic coral]

if G is a transitive subgroup of S_n and S_m and i have a irreducible degree n polynomial for which G is the galois group, how can i construct a degree m polynomial from the degree n polynomial that has G as the galois group as well

i was thinking like multiplying by the second cyclotomic polynomial but then my new polynomial wouldn’t be irreducible

[winged void]

I have small question regarding proving isomorphism

I need to prove isomorphism between V_4 and (Z/12Z) under multiplication i drew two tables and i see they are isomorph

is that an acceptable prove

or is it more of intuition

prove

[lament bough]

it is acceptable but yeah kinda excessive

if i was asked to prove it i would say uhh i guess there is a homomorphism from Z_2 to (Z_12Z)* that sends 1 to 5 and also one that sends 1 to 7 therefore there is a product homomorphism from Z_2 x Z_2 to (Z_12Z)*

as one can see it is bijective and therefore an isomorphism

[winged void]

but is Z_2 * Z_2 similar to V_4

Klein four-group

[lament bough]

yeah

[winged void]

oh i did not know that

[lament bough]

i guess thats my definition of klein 4

[coral spindle]

One might even say they are equal

[lament bough]

whats your definition of K4?

[winged void]

literaly what is in the table that i drew

that is how its defined in my book

they did not say much about it

they say that if you multiply two elements

then you get the third elements

[lament bough]

i guess they really did want you to just draw the tables and compare

[winged void]

and every element the identity of it is the identity itself

[lament bough]

because its a beginners exercise

[winged void]

i was thinking for example take an arbitrair element from (Z/12Z)* for example a and then sends this elemnt to an arbitrair elemnt inV_4 for example 1 in a case f(1) = e f(5) = a f(7) = b f(11) = c and so on we see that each element is is aisgned to one element and as well injective as surjuctive

[winged void]

i was thinking more of that

maybe that is a better way to prove it or what do you think @coral spindle

[lament bough]

just drawing the tables and seeing that they are the same save for change of names is enough

its the same thing you just described but implicit instead of explicit

[winged void]

true

then i will do it the table way

it saves more time

[rocky cloak]

Like you want to explicitly construct it, or just show that it exists?

[languid trellis]

#1217259452621783154

[chilly ocean]

I am proving that every multiplicative Abelian group is a Z-module.

$\$
Let (M,$\odot$) be a multiplicative Abelian group

Let $n \in \bZ$ and $a \in M$

($\bZ$ is the ring ($\bZ, +, \cdot$))

Also let $m \in \bZ$ and $b \in M$

$\$
I have defined the scalar multiplication as follows:

$na = $
$\left{ \begin{array}{cc}
a^n, n>0 \
1, n = 0 \
(a^{-1})^{|n|}, n<0
\end{array}
\right.$

$\$
I am having trouble in the case of $n,m<0$, trying to show:

$(n\cdot m)a= n(ma)$

[cloud walrus]

icebal²

[chilly ocean]

I am not sure how to use a^(-1) here. Here's my attempt:
(I am showing a^(-1) as a' here)

(n.m)a = (a')^(|n.m|) = (a^|m|)^|n| = |n|(|m|a) = (-n)((-m)a)

I do not know how to proceed from here, nor do I know if I'm on the right track

[sonic coral]

construct explicitly

i’m working with a more concrete example but that is the general situation i’m running into

i was thinking maybe composition with the correct degree irreducible polynomial

[south patrol]

Honestly easier just to use additive notation for your abelian group lol

[chilly ocean]

I'm kinda new to this topic so I'm trying to distinguish multiplicative/additive abelian groups via different notations of their binary operations, just to get a hang of it while proving stuff

[winged void]

I have small question regarding the following statment

i do not get an intuition behind this two statment

i thought first that a^k = e and k is the order and k must be the smallest integer to achieve that

[delicate orchid]

yup

that's the definition of order

[winged void]

sure but what do they mean wiht a^n = a^m

[chilly ocean]

~~If the order of the group is not finite, then you cannot have a k s.t. a^k = e because intuitively and roughly speaking, you won't be able to "cycle" around the elements, i.e., you cannot multiply a finite times to "cycle around" to e
So in this case, a^n = a^m iff n = m

But if |a| = k, then by definition a^k = e
So for all n,m≥k, a^n = a^m = e~~

Ignore

[delicate orchid]

this is completely wrong?

[winged void]

the last one is not true because lets say for example we take an element of mod 12

[delicate orchid]

a^k = e

a^k+1 = a

[winged void]

for example the order of 8

[delicate orchid]

not e

[winged void]

element 8 i mean here

and with operation addition

will be 8^3k = 0

but the smallest is 3

but multiple of that will as well lead to identity

[delicate orchid]

honestly I think there's just some text missing. The order is infinite then each power of a is different, otherwise a^n = a^m whenever n = m (mod |a|)

[winged void]

the other slide

says the following

[winged void]

but for me it does not make that sense as well

[delicate orchid]

how doesn't it make sense

assume the order is infinte and a^m = a^n for some n not equal to m, then a^(m-n) = e, so the order of a is m-n contradicting our assumption that it's infinite

k | n-m, then we have a^(n-m) = e so a^n = a^m

[winged void]

oh i see

[rocky cloak]

Well, I would be surprised if you can say anything at that level of generality, without using anything about the given polynomial. But let me know if you figure out something.

[coral steeple]

Can I define an endomorphism on Z/(m)[x] by sending x to whatever polynomial I want and fixing 1?

It seems to me you shouldn't be able to do this

[rocky cloak]

you can indeed

[delicate orchid]

cheeky evaluation map

[coral steeple]

How do you know that (x+1)(x+1) and x^2+2x+1 get sent to the same thing if you find the images of the factors before or after expanding, for example?

Looking at it I'm sort of expecting the answer to be "by definition" but it doesn't feel obvious lol

[delicate orchid]

without just saying "universal property"

assuming you're not doing anything stupid with higher powers of x : f(x^2+x+1) = f(x)^2+f(x)+f(1) = f(x)^2+f(x)+1 = (f(x)+1)^2 = f(x+1)^2 = f((x+1)^2)

this is just an evaluation map

you're evaluating a polynomial in R[x] on a different polynomial in the same ring - which seems a bit strange because we usually only evaluate R[x] -> R but you can evauluate in any ring R[x] -> S and it works just fine

[coral steeple]

Hmmmm I think that makes sense

[hollow mica]

assuming your ring is unital (and homomorphisms fix the unit), there’s a theorem that says that there is exactly one homeomorphism from a polynomial ring into another ring that sends x to any element. Some people call this the universal property of polynomial rings

[rapid junco]

This is in general false right?

You require S be an integral domain

[south patrol]

yeah i guess

since f(1)^2 = f(1)

[wraith cargo]

I believe it's enough to require 1 != 0 in S

[south patrol]

well for unital rings sure

hm

this is weird to me lol

[rapid junco]

can I see the argument for 1 != 0?

[south patrol]

are they assuming rings have unit but homomorphisms don't respect the unit? that's definitely non-standard

bruh

[wraith cargo]

I might have had a brainfart hmhmhmh

[south patrol]

i'm pretty sure for example that Z/2 -> Z/2 x Z/2 sending 1 -> (1,0) and 0 -> 0 is a ring homomorphism for example

under this definition

so that would be a counterexample to the exercise more generally

[rapid junco]

right

ill assume they mean integral domain then

[south patrol]

basically you need the ring to have no idempotents besides 0,1

otherwise you can do stuff like this

but yeah i've never seen someone define homomorphisms of unital rings in a way which doesn't have f(1)=1

I guess this exercise shows it makes little difference

[wraith cargo]

here we go

okay yeah I had a brainfart you have to assume that S is integral

[south patrol]

hm is this d&f

or artin

[wraith cargo]

d&f yeah

[south patrol]

ok sure

yeah this is why i didn't learn about rings from there lol

[wraith cargo]

I did 😭

[woeful comet]

I'm trying to do this and I'm a bit confused, there is no rotation in the Klein-4 group sooo........ what do I do? Maybe just finding the left and right cosets would be easier but then it'd be longer?

WAIT

for example, for a vertical flip in V4, is like doing a flip in D4

and then horizontal is the same as doing two rotations then a flip

Maybe I cooked.....

thoughts?

[last spoke]

@woeful comet V is a subgroup of D4

Altho I’m confused by the notation. Conjugation by r produces the empty set?

[woeful comet]

no, I haven't filled it out yet

bc I was confused what the elements would be

[woeful comet]

yes, ik this

[last spoke]

oooh

well u just need 2 elements of order 2 that commute with each other right

[woeful comet]

why only 2?

[last spoke]

that suffices to generate V. a and b with order 2, so a, b, a^2 = e, and ab

[woeful comet]

ahh

Hmm

yes you can generate V with just two elements of order 2 however, idk how that plays into finding the normalizer for V4 in D4

[last spoke]

once you find the elements that look like V then you can compute the conjugates and see which ones fix V as a subset

[woeful comet]

what does it mean to "look like" V

like, a similar structure when looking at a Cayley graph of it?

[last spoke]

just casual language for isomorphic

so whichever subset of D4 is isomorphic to V, you find the conjugates of that set

[woeful comet]

hmm

Alright

and I assume, if there are multiple subgroups that are isomorphic, then the same would apply to those

well I think there would be at most 2 actually

[last spoke]

good thinking

you only need to find one tho

[woeful comet]

How come

oH, since they're isomorphic, they're just the same?

Hmm, well wait

that doesn't immediately follow (in my head) actually, I wonder if that's actually accurate

[last spoke]

the task is to find V as a subgroup of D4. so if you find a subgroup of D4 that's isomorphic to V, then you've successfully completed the task

[woeful comet]

Fair enough

true

Thanks

[coral spindle]

5x^2 + 5 is divisible by x^2 + 1

Typo lol

Oh wait wrong way around. My mistake.

x+1 divides x^2-1.

[winged void]

I have question regarding conjugate

conjugate is another word for similar right

but they call it just conjugate in group theory

but in linear algebra its similarity

like two transformation are conjugate similar is the same

[rocky cloak]

Well, similarity is also used to refer to non-invertible matrices, so it covers a bit more than the word conjugate.

But yes, the meaning is essentially the same.

[winged void]

i see but why its actually important is it only usefull in linear algebra

[winged void]

because i do not really see why would it be true for something else

[coral spindle]

How do I show an elephant is african? I might struggle to show it for an indian elephant.

Irreducibility is the condition you're looking for.

[cloud walrus]

AlleluiaAlleluia

[coral spindle]

Typically not, no

In the first place, take k_i all squares.

[delicate orchid]

what if they were all prime? what then, wokerati?

[coral spindle]

Oooh spooky triangle

Illuminati etc

[summer path]

What if they're all off by a square

[delicate orchid]

you could also have like, n m nm

[south patrol]

then yes right

iirc this is slightly annoying but uh lemme thonk

Hm isn't this like. Suppose $\sqrt{p_{k+1}} = a_0 + a_1 \sqrt{p_1} + \dots + a_k \sqrt{p_k}$. Consider the element of $\mathrm{Gal}(\mathbf Q(\sqrt{p_1},\dots,\sqrt{p_k})/\mathbf Q)$ sending $\sqrt{p_1} \mapsto -\sqrt{p_1}$ and fixing $\sqrt{p_2},..,\sqrt{p_k}$. Then $a_1 \sqrt{p_1} = 0$ or $= a_1 \sqrt{p_1} = \sqrt{p_k}$. Second scenario is impossible unless $a_1 = 0$

[cloud walrus]

Süßkartoffel

[south patrol]

but then that'd give you that sqrt(p_k+1) is rational lol

[summer path]

Square root of three plus one is rational, so by induction all square root of primes plus one are rational

Oh wait it was subscript

[cloud walrus]

AlleluiaAlleluia

[south patrol]

lol

Depends a lot on the k_i

yes i mean still depends

it's either 1 or 2 of course

well sqrt(k_m) satisfies x^2 - k_m = 0

so its iminimal poly is of degree 1 or 2

[summer path]

iniminomininal

[last spoke]

what's your understanding of what a field extension is?

[south patrol]

I suggest you review degrees of extensions ig

but the important thing here is that the degree of a simple extension is the degree of the minimal polynomial of the generator

[rocky cloak]

I'm not sure what you mean. Conjugation comes up all the time. Very useful when talking about group actions for example.

[vapid vale]

little unsatisfied in my lie theory knowledge – any reading recs?

[amber wraith]

Naive Lie Theory by Stillwell is a good read, not too heavy at all

[woeful comet]

Guys, rate my diagram I made

ah shit, svg no work in discord

BOOM

yeaaaaaaaaaa that's clean

[delicate orchid]

it's wrong but looks nice

[woeful comet]

FUCK

[woeful comet]

wait how

[delicate orchid]

integral domains are not always division rings and fields are commutative

[woeful comet]

uH

[delicate orchid]

fields should be the intersection and integral domains should be a subset of commutative rings

[woeful comet]

wait can u give me an example of when an integral domain isn't a division ring

[delicate orchid]

the integers?

[woeful comet]

I mixed up my definition of fields and integral domains

but also I misdefined one of them

on top of that

so gg

okay I think I fixed it

wait but I think there are some IDs that are div rings

[coral steeple]

All fields are integral domains no?

[woeful comet]

i'm so cooked

😭

OKAY, NOW I COOKED

YEAHHHHHHHHH

[dull ginkgo]

Quaternions

Integral Domains are commutative domains

• Domains are rings without zero divisors, i.e. the {0} ideal is prime.
• Integral domains are commutative domains
• Division rings are domains.
• Fields are commutative division rings thus are integral domains or are commutative domains @woeful comet

[woeful comet]

HUGE math lore just dropped

ok lemme study this

[dull ginkgo]

Simple rings are a thing too

[woeful comet]

what is an ideal? "{0} ideal "

[dull ginkgo]

you'll get to it

[woeful comet]

[dull ginkgo]

An ideal is an additive subgroup of a ring that absorbs multiplication on both sides. I.e if we have an ideal A, then multiplying any element of a to the left or right by any other element of the ring is still in the ideal A

For example, the integers are a ring, and the even integers form an ideal

[amber wraith]

kernels of ring homomorphisms being the prototypical example (kindof how kernels of group homomorphisms are the prototypical normal subgroups)

[dull ginkgo]

first iso theorem trollface

[coral steeple]

Aren't the kernels of ring homomorphisms exactly the ideals of a ring?

[dull ginkgo]

The two-sided ideals

[coral steeple]

Oh yeah

What good is a one sided ideal though

[dull ginkgo]

technically there are left and right ideals which are basically R-modules inside of itself

[coral steeple]

Ah

[amber wraith]

yeah, the idea of (f(ab) = f(a)f(b)) => ( f(b) = 0 => f(ab) = 0), it's *the * motivating example in a way

[dull ginkgo]

Let D be an integral domain and let D embed into fields A and B such that both are generated by D, I am supposed to find unique isomorphism between A and B

In other words the image of D into A and B, a(D) and b(D) has the property that any ring between a(D) and A (respectively for B) is A

The embedding gives an isomorphism between a(D) and b(D)

Now I need to give an isomorphism between A and B from that

[south patrol]

This isn't true

Well

Like what do you mean by generated by D lol

[dull ginkgo]

Isn't the standard definition that the generator of a ring (or structre) is the intersection of all rings containing it

[south patrol]

But a(D) is already a subring

Unless this is some weird nonunital ring thing

Lol

[dull ginkgo]

if a ring lies between a(D) and A then it is either a(D) or A

let me be more precise with that

[south patrol]

No but I mean that like the ring generated by D should just be the image of D already at least under the usual definitions

[dull ginkgo]

oh, i mean the field generated sorry

[south patrol]

I imagine what you want is that A is the smallest field in which D embeds

Yes sure

[dull ginkgo]

so any field between a(D) and A is either a(D) [if it's a field] or A

brain fart

[south patrol]

Well I claim you can write esch element of A in a specific form

[dull ginkgo]

I was going to try to do it through this definition instead of proving each element is a sum over ab^-1's

[south patrol]

Well you canjuzt work out that that is the case and it is kinda clear

But yeah tbf there is a universal property but it basically involves this lol

[dull ginkgo]

If I define product of sets AB to be finite sums of ab where a in A, b in B

Then it's trivial that D(D^-1) is a subring in their respective fields, and due to it being generated, D(D^-1) = A and likewise B due to the embedding

which you can therefore define an isomorphism I guess through mapping the inverses

[south patrol]

(D\0)^-1

But yeah

[dull ginkgo]

yeah thanks

[sly rain]

The way I understand is we need commutativity (or image of x_i to commute with every element in S) in S in general so that the homomorphism can respect multiplication.
ι : Z -> S ι(n)=n*1_s
f: {s} -> S s maps to f(s)

So e.g. during multiplicating the images we have ι(2) ι(1) f(s) + ι(3) f(s) ι(4), so we want f(s) to commute with the elements so that we can distribute.
But apparently this is automatic in this situation.
I think i dont get why this is automatic in this case, maybe give a hint or ask a question that leads to the answer as nothing major is going on here i think

[rocky cloak]

Any element s commutes with 1, and commutes with itself.

It will therefore also commute with any sums or products of such.

So the image will be a commutative subring, no matter if S is commutative or not.

[sly rain]

oh well, that was it. Thanks

[rocky cloak]

Sorry I couldn't give a better hint, but I'm not sure there is really that much going on

[sly rain]

No dont be sorry lol, I guess it was indeed a technical detail and I didnt make that connection in my head. Imo its fine giving away the solution in a situation like this

[swift prawn]

Hello! suppose you have some irreducible (depressed) cubic p(x) = x^3 + px + q over some field F. Why does the discriminant being non-square ensure that the extension of the splitting field E of p have degree 6?

[rustic crown]

p needs to be irreducible

[swift prawn]

I know that the square root of the discriminant is in the splitting field, but why does that mean that the splitting field is (1) degree 6 and 2 of the form Q(sqrt(D), alpha) for some root alpha of p

[swift prawn]

yea sorry I'll add that in

[rustic crown]

because [F(alpha):F] = 3 and [F(sqrt(D)):F] = 2

so [F(alpha, sqrt(D)):F]=6

and this must be the whole thing as you might have proven that degree of splitting field is at most (deg p)!

[swift prawn]

right

ok yea I think that makes sense

[rustic crown]

[swift prawn]

why does this have all the roots of the polynomial? and why does the [F(alpha): F] = 3 and [F(sqrt(D)): F] = 2 give you this? I guess since sqrt(D) wouldn't be in F(alpha) probably

or I guess why does adjoining the discriminant give you everything intuitively

[rustic crown]

say K is the splitting field. Then we know that [K:F] <= 6

and we know sqrt(D), alpha are in K

this means that we have a tower of field extensions F --> F(alpha) --> F(alpha, sqrt(D)) and F --> F(sqrt(D)) --> F(alpha, sqrt(D))

this means both [F(alpha):F] and [F(sqrt(D)):F] divide [F(alpha, sqrt(D)):F]

so [F(alpha, sqrt(D)):F] is divisible by both 2 and 3.

[swift prawn]

right

and 2 and 3 are coprime so that means that [F(alpha, sqrt(D)):F] = 6

[rustic crown]

right so F(alpha, sqrt(D)) --> K is an extension with degree 1

[swift prawn]

and this is because alpha is in K by definition and sqrt(D) in K by some other fact

[rustic crown]

right

you have an expression for sqrt(D)

[swift prawn]

yea

in terms of the product of roots

or well you know what I mean

[rustic crown]

product of differences yea

[swift prawn]

yea

right so sqrt(D) is in K as well

and so since its a degree one extension

they're actually the same

very nice.. thank you!

[rustic crown]

[rapid junco]

If I is an ideal then I^2 = I right?

I^2 \subset I \cap I = I.

NVM I \subset I^2 doesnt work.

FWIW I am trying to understand this set of inclusions

P is a maximal ideal.

so P^2 \subset P \cap P = P.

a and b are not in P.

so where does the containment (a) P lie in?

[south patrol]

No

An easy counterexample is that if I is the ideal pZ of Z then (pZ)^2 = p^2Z

[south patrol]

P

Since P is an ideal

Like note (a). P = aP

[rapid junco]

ah yeah (a) P = aP but aP is contained in P since P is closed under mult by a

[south patrol]

yes

Well you don't even need the first bit like

IJ is always contained in I and J

[cobalt heath]

Why is Etingof exercises harder than Hartshorne’s to me

[still dew]

Wait i should have posted this in lin alg channel

My bad

[pseudo sierra]

Hi, I need help with a concept that I’ve forgotten and cant seem to remember what its called

Let R be a ring with identity and A a set. Then R^A is an R-Module

What is R^A?
I cant remember what this is called so I also cant search for it in google

[delicate orchid]

It’s just the set of functions from A into R

More generally it’s called an exponential object

[pseudo sierra]

Ah thank you very much

[shrewd orbit]

it's motivated by the fact that if $A$ and $B$ are finite sets, there are exactly $|A|^{|B|}$ maps from $B$ to $A$.

[cloud walrus]

Orange

[hidden haven]

Silly potato, it is I am, not I is.

[crystal vale]

How do we define maximal 2 sided ideal in ring R? M is maximal 2 sided ideal if there is no proper 2 sided ideal which is not M such that M contains in it, right?

In a local ring they define that if a set of all non-units is an ideal then it is a local ring, so it is a two sided ideal, right?

If I want to prove that for all x belongs to R either x is unit or 1-x is unit then R is local ring. And R has unity.

So let I be the set of all non -units elements of R. Then it is non - empty.

Let a and b belong to R then if a-b is unit then there exists c such that (a-b)c= 1.
So ac = 1 + bc but since ac and bc are both non- units so it implies that 1+ bc is unit but then ac becomes unit, it contradiction.
Hence a-b is in I. And if a is in I then ar and ra both are in I. Thus I is an ideal.

Is it correct?

[rocky cloak]

You have to be a little careful. It's not in general true that the product of non-units is a non-unit.

The problem being that an element can have a one sided inverse, without being a unit. So you have to show that that can't happen in your case.

Possible hint: ||product of unit and non-unit is a non-unit||

[hollow mica]

Are there any proofs of Newton's identities (the one concerning symmetric polynomials and power sums) using the fundamental theorem of elementary symmetric polynomials?

The proofs I find online are muy complicated

[languid trellis]

Jacobson, Basic Algebra I, pg138-140. fyi the proof of Newton's identites are in an exercise

[hollow mica]

Thanks

[dim widget]

the one I really like is seeing the identities as a consequence of the Cayley-Hamilton identity

If $X$ is a generic matrix then $Ch_X(X) = 0$, if $x_1, \dots, x_n$ are the eigenvalues of $X$ then this means that if $s_i$ are the elementary symmetric polynomials, then $\sum_i (-1)^i s_i(x_1, \dots, x_n)X^{n-i} = 0$ taking traces we get that $\sum_i (-1)^i s_ip_{n-i} = 0$ where $p_{j}$ is the power sum polynomial of degree $j$.

[cloud walrus]

kålrot

[dim widget]

this gives the newton identity for $p_n$ and the other ones are obtained by letting $X$ have some zero eigenvalues

[cloud walrus]

kålrot

[south patrol]

nice

hadn't sen that

[hollow mica]

Wow this is really nice

Did you see it somewhere or just come up with it

[dim widget]

I came up with it some years ago, but many people have had the same idea

[hollow mica]

I see

I found an article about it now

[fading field]

i am not clicking that

[vivid tiger]

it's not a rickroll

it's an article giving a matrix proof of Newton's identities

This totally isn't an elaborate trick.

[celest furnace]

You liar! How could you!

[vivid tiger]

i am totally telling the truth

[hollow mica]

you can right click copy link

[crystal vale]

So for closed under additive inverse it is correct but not for ar and ra in I

let a and b are non-units , so 1+b is unit then a(1+b) is non- unit , by closed under additive inverse ab will be in I.

And if a is non - unit and b is unit then ab is non- unit.

Is it correct?

[long obsidian]

Is there a special name for rings that embed into themselves? The best example I have is the polynomial ring k[x]. An example of such an embedding is f:k[x]->k[x] given by f(x)=x^k for some non negative k integer

[barren sierra]

I mean every ring embeds into themselves if you wanna be a smartass about it: just take the identity

[barren sierra]

but x -> x^p isn't a homomorphism unless k is a field of characteristic p

[long obsidian]

Hmm this feels like your right but I can't tell why lol. Do you have an example of you can't distribute right ?

[long obsidian]

I guess I originally thought k[x] had a proper subset that was an isomorphic ring to itself

[cobalt heath]

Yeah you have the homomorphism induced by x -> g(x) for any polynomial g.

Because k[x] is a free commutative algebra

[barren sierra]

(x + y)² is not x² + y²

Unless you're in characteristic 2

[long obsidian]

What is k[x^2] then? I thought this expression made sense as a ring and you had to make sense of it as being k[x] embedded into itself

[lapis latch]

I think the map you'd want there is p(x) -> p(x^k)

[cobalt heath]

Yeah, it makes sense

Unfortunately, understanding this fully might require algebraic geometry, methinks

[long obsidian]

Oh wait yeah that's what I mean. To substitute not to square

[cobalt heath]

(Note k[x^2] is indeed isomorphic to k[x])

[cobalt heath]

It makes sense both ways

[long obsidian]

Okay so is there a cool name for these rings that have proper subsets that are isomorphic as rings to them? Maybe it's cheap I guess the integers have this with the evens

[topaz solar]

They meant x -> x^k as in literally the element named x, not as a function rule. k[x] -> k[x^k] < k[x]

[barren sierra]

Ohhhhh

My bad

[topaz solar]

[cobalt heath]

Woah, that's concise and clear description

[topaz solar]

Uh so I don’t know such a term for plain embeddings

But if you ask for infinitary elementary embeddings and R is countable, then this occurs iff Aut(R) as a Polish space (it’s isomorphic to a subset of Sym(N) under some bijection R ~ N) has a left invariant complete metric I believe, and there’s a paper by Gao about this

Note this is a DST/infinitary logic concern, of course, but this is equivalent to having an infinitary elementary extension that’s uncountable

So if there’s a proper (infinitary) elementary extension, there’s a big one

Maybe not quite what you were looking for, but it’s related

https://www.cambridge.org/core/journals/journal-of-symbolic-logic/article/abs/on-automorphism-groups-of-countable-structures/4AE1E8B57AA6D7F30AD8A2206E685CDC

[cobalt heath]

It's model theory all along

[topaz solar]

So, if you have any uncountable ring R with a countable infinitary elementary subring P, then this is the same as R’s Scott sentence being equivalent to an L_omega_1,omega sentence (the Scott sentence of P), and P therefore has this property

I do not know anything about embeddings in general or related statements about larger cardinalities (I.e. without smaller Scott sentences)

[topaz solar]

Well I mean, this stuff is not too far from “Galois away from fields”

[topaz solar]

https://arxiv.org/abs/1211.7145 Ok so, there is something on that, but it was withdrawn it seems. I'll keep this on tabs to check and see about salvagable partial results buried in it later maybe though oop

That’s been enough hijacking w/ logic notions for one goofy lil question though

[rocky cloak]

No, you're assuming that ac can't be a unit, in your proof for additive inverses as well.

[crystal vale]

Because a is non-unit and c is unit so ac is non-unit, right?

Because if ac is a unit then there exists d such that acd =1 then it shows that cda =1 then cd will be multiplicative inverse of a which is contradiction

[rocky cloak]

Ah, yes. You're right. Then you're all set

[crystal vale]

Thank you

I want to prove that for every subgroup H of S_n for n>=2, either all the permutations in H are even or exactly half of them are even.
First H must have an order of multiple of 2.

If I let there be π permutation of odd numbers of transposition in H then if let mapping A - > B where A is set of all even permutation in H and B is set of all odd permutations in H and a->πa then it is injective and well defined mapping. And for onto let p be in B then π^(-1)p be in A(H is subgroup )then ππ(-1)p = p , thus it is onto. Hence |A| = |B|

It is the same as when we prove that A_n has exactly n!/2 elements.

Is it correct?

[south patrol]

That sounds great to me

Another way to rephrase this is to note the even permutations are the kernel of the map sgn: H -> Z/2 and the kernel either has index 1 or 2. But it's essentially the same proof.