groups-rings-fields

[tardy hedge]

(x-1/2)(x-1/3)

Nontrivial factoring over Q

I can factor this over Z?

I dunno why im confused

[south patrol]

Well it isn't even a poly over Z

A better example is like 2x^2 + 4 or 5x

[tardy hedge]

Ohh

[tardy hedge]

So i start with a poly over Z, if i can factor it over Q I can factor it over Z

Thats what Gauss lemma says?

Yeah tbh ill just go review the statement

[hollow fjord]

Does every finitely generated group can be presented by 2 generators ?

[barren sierra]

no, there are easy counterexamples

[south patrol]

Yes

(Nontrivially)

[south patrol]

Which do you have in mind?

I guess you can do smth silly like going to abelian groups e.g. if G can be generated by two elements then there's a surjection F2 -> G lol, and if you take G abelian then this map factors to give a surjection Z^2 -> G which is impossible for G = Z^n, n > 2, say

[barren sierra]

Z^3?

[south patrol]

I think nonabelian examples are a little trickier cause you can't just use this trick

[barren sierra]

Am I wrong?

[south patrol]

Nah I mean yours is a special case of what I just said aha

[barren sierra]

ok good thought I was going crazy lol

[south patrol]

I guess there are other trivial things like (Z/2)^3 lol

[barren sierra]

Sure sure

[south patrol]

Which may be the smallest example

[barren sierra]

It is yea

[south patrol]

I mean the only other candidate would be S_3 which is generated by two elements ofc

[thorn jay]

I often think I'm annoying for just talking here lmao, because I still get some things wrong

[frigid epoch]

Hey it's not like you confidently assert things that are incorrect haha...

[thorn jay]

Oh i do

Lmao

Thats the problem

But at least I accept my mistakes

Instead of the usual reaction of mathematicians it seems

[thorn jay]

(This was me in another server)

[tardy hedge]

Hahaha

what server is that

[thorn jay]

Redstone army

Its Mattbatwing's server

[tardy hedge]

ya Math Idiot

ur such a math idiot

[thorn jay]

Im an idiot who knows math

[tardy hedge]

complete buffoon ong

[thorn jay]

Its iconic ive been there for like 1.5 years

[tardy hedge]

I dunno what any of that is

[thorn jay]

He's a logical redstone youtuber

But he made a math vid once for SoME, and then i joined

I also met my current gf there

[tardy hedge]

Haha nice

So back to this q5, if I can assume here that the set of separable over F elements of E is a subfield of E then..

Write [F(a) : F] = [F(a) : F(a)s]*[F(a)s : F]

If a is inseparable over F, then min poly of a has degree p^n m , so p^n divides [F(a) : F(a)s]*F(a)s : F], but [F(a)s : F] has no factor of p because its a separable extension, so p^n divides [F(a) : F(a)s], so we can say for sure that the inseparable and separable degrees over F is the same as the inseparable and separable degrees of the min poly of a over F?

like, was that all necessary?

I dont know why its italicized

F is char p

[frigid epoch]

u were thinking of the generators of the homology of the torus

surely

[thorn jay]

Totally

[royal shadow]

you need to add \ before some special characters it is called an escape character to stop weird shit from happening to avoid bold letters or italics

[thorn jay]

Or just use latex

[royal shadow]

based

[tardy hedge]

For q6, the minimal polynomial of a in Es over Ks divides the minimum polynomial of a over F, and that one has distinct roots so the poly over Ks must have distinct roots too

Is that how that one goes?

[rocky cloak]

[F(a)s:F] can be a multiple of p

theres nothing stopping that

[tardy hedge]

I thought the derivative of the poly would be 0 though in char p

[rocky cloak]

the derivative of x^p + x is 1 for exmaple

[tardy hedge]

Oh shoot

[tardy hedge]

Can I have a hint for where im going wrong?

[rocky cloak]

I'm not sure what you mean.

Where you're going wrong is assuming [F(a)s:F] can't be a multiple of p when it can

[tardy hedge]

I mean like, am I approaching the problem the right way?

[rocky cloak]

Probably.

I'm not completely sure what kind of answer is expected here.

Like you can say something about ||the number of district roots||
or that ||the polynomial can be factored as f(x^p^k) for some seperable f.||

[tardy hedge]

But in the second thing u said, how do i know that the degree of the separable part is the same as [E : F]s

So yes the degree of p is degree of separable part times the p^k part

But how do i know thats the same as [F(a) : F(a)s] times the other

[rocky cloak]

So a^p^k is seperable hence
F(a^p^k) is a seperable extension.

Any element in F(a) has x^p^k in F(a^p^k), so can only be seperable if they're already in F(a^p^k). Hence those are all the seperable elements

[tardy hedge]

a^p^k is separable because its the root of a separable polynomial

We dont necessarily know that psep is the min poly of a^p^k do we?

[rocky cloak]

I mean, its not hard to show, but I guess it doesnt really matter

[frigid epoch]

Does anyone have an intuition for nonseparable polynomials

Like what they're there for and why they're interesting

[tough raven]

Just checking, but: if G is a group generated by a set S and k a commutative ring, then the kG-linear map kG^{(+)S} → kG, defined on the basis S by s ↦ s-1 for s in S, is injective iff G is freely generated by S, right?

And if G is not freely generated, say it has a presentation by some relations R, is the kernel generated by something indexed by R? (Maybe the relations need to generate ker(Free(S) → G) as a subgroup and not just as a normal subgroup?)

[crystal vale]

I don't know how characteristics help here?

I just know the basics of field theory so if it uses some more than it then I have to learn it

[tardy hedge]

I feel like I never see true/false questions like this anymore

[thorn jay]

Boolean rings mentioned

Coolest nontrivial isomorphism in the category of varieties

[tardy hedge]

So [E : F]s is the degree of psep?

[tardy hedge]

psep is an irreducible poly over F with a^p^k as a root so its the min poly

so [E : F]s is the same as [F(a^p^k) : F] which is deg psep

[tardy hedge]

Hm I don't actually understand that last line

[rocky cloak]

The only seperable polynomial that divides
x^p^n - b^p^n is (x-b)

[crystal vale]

Yes

[crystal vale]

You wanna?

[tardy hedge]

sure

[crystal vale]

Check DM

[rocky cloak]

It doesn't.

Only thing you need is that a degree n polynomial has at most n roots.

The way characteristic helps is that the characteristic 0 case and characteristic p case are very different, so you probably want to tackle them differently

[tardy hedge]

Man this is weird but ok

because thats purely inseparable

[rocky cloak]

The characteristic 0 case is very easy. Start by for example confirming for yourself that Q,+ is not isomorphic to Q^*

[crystal vale]

Yes it is not

[rocky cloak]

And why is it not?

[crystal vale]

Also R,+ and R^* is isomorphic

[rocky cloak]

They're not

R+ and R>0 are though

[crystal vale]

I mean yes they are not

[rocky cloak]

And why, what's the key difference?

[crystal vale]

Because 2x = -1 has a solution in Q,+ but x^2 = -1 has no solution

I mean I can prove, wait

[rocky cloak]

So that doesn't quite make sense, -1 is not like a group theory thing here

But maybe you can try to convince yourself that C,+ is not isomorphic to C^*

[crystal vale]

Let f : Q -> Q^* is an isomorphism then there is f(x) = -1, now we can write f(x/2 + x/2) = -1 and f(x/2 + x/2) = f(x/2)^2.

That means f(x/2)^2 = -1, which is not possible in Q^*

Is it correct?

Similarly for R and R^*

[rocky cloak]

Okay yes, this is correct

[crystal vale]

Yes that's what I meant

[crystal vale]

Here we can define the mapping x -> a^x, a>0

But key difference

[rocky cloak]

A very related idea is that if ||f(x) = -1, then f(2x) = 1||

[crystal vale]

So f(2x) = 1, then?

[rocky cloak]

Indeed, hence f is not injective

This even works in odd characteristic as well

[crystal vale]

Yes yes

[crystal vale]

You mean if F has odd characteristics then there is no field such that F, + is isomorphic to its multiplicative field structure

I am going to prove it

[rocky cloak]

Yeah, like that same argument works to prove that

[crystal vale]

Yes -1 and 1 are different in odd characteristics

Yes same exactly

Now we have to consider the case when characteristic p = 2.

[low cliff]

hello, need to prove that I2 is maximal

[crystal vale]

You mean 2A + xZ[x] is maximal in Z[x] ?

Try to find the quotient ring by this ideal

[low cliff]

i need to prove that the quotient ring is a field? how do i do that, is P is an element of A\I2, i need to prove that P has a multiplicative inverse?

[crystal vale]

Yes

you know if R is an integral domain and if M is maximal ideal then R/M is a field?

That is iff condition

[low cliff]

yes

but i dont know how to find the multiplicative inverse

[crystal vale]

You don't need to be find here

Can you map A to F, such that its kernel is I2?

And first guess what will be your F here?

Can you guess the structure of A/I2 ?

[crystal vale]

But isn't work in all fields which have characteristics is not 2 ?

[rocky cloak]

Yeah 0 or odd covers every characteristic expect 2

[thorn jay]

Its always char 2

[crystal vale]

So now we only need to consider 2

[thorn jay]

Its those boolean algebras i tell you

[crystal vale]

Oh

[low cliff]

F is f(A)?

bruh

[crystal vale]

No I mean first guess the structure of A/I2

[low cliff]

no

wdym its structure

[crystal vale]

What will it look like ?

Can you tell me what Z[x] /<x>, looks ?

[thorn jay]

Yeah, you can think if this as first quotienting by <x> and then quotienting by <m>

[crystal vale]

Yeah

[low cliff]

you mean its elements? Like all of the possible remainders are the integers

[thorn jay]

Yeah! Do you know what ring Z[x] / <x> is isomorphic to?

[crystal vale]

Char 2 implies that the ring is Boolean? I don't think so

Oh Boolean algebras

[thorn jay]

I was being overly stereotypical

[crystal vale]

What is that?

[thorn jay]

Like a true racist

[crystal vale]

I don't get it

[low cliff]

Z

[crystal vale]

Yes

[thorn jay]

Mhm, so then you quotient out by <m> and you get?

[low cliff]

Z/2?

[thorn jay]

Yes

So A/I_2 ≈ Z/2Z

(And in general A/I_m ≈ Z/mZ, shouldnt be too hard to see)

[low cliff]

Z/2Z is a field because 2 is prime

[thorn jay]

Yis

[low cliff]

ok thank you

[thorn jay]

Yw!

[crystal vale]

So if there is isomorphism between F and F^, char F = 2, then for every x in F^, x^2 = 1

[tough raven]

ie (x-1)^2 = 0

[acoustic igloo]

the number of irreducible cubic polynomials in Z_p[x] is p(p-1)^2(p+1)/3

[thorn jay]

Whats F^?

Oh

Lmao

[crystal vale]

That is, x = 1 because F is an integral domain, so it is not possible

Right?

Thanks, jagr, Raghuram, .enpeace_music

[thorn jay]

Dunno what i did but yw

[tardy hedge]

Lmao^^

[thorn jay]

I just got that aura ig

[rocky cloak]

Your presence alone is thankworthy

[tardy hedge]

Damn

[glad osprey]

thanks enpeace 🙏

[tardy hedge]

Jagr with the praise

Thank you enpeace

[thorn jay]

I am flattered

Then I would like to thank you, too

[tardy hedge]

What about me? Do I deserve a thanks? 🥰

[thorn jay]

Of course :3

[rocky cloak]

Thank you for posting questions and just being you

[thorn jay]

It keeps the channel having something to talk about

Even if tangentially related

[tardy hedge]

Wow this got randomly wholesome

[thorn jay]

[low cliff]

hello, question 2

[wicked patio]

^ resolved in a different channel

[barren sierra]

as a remark, you'll see this descending chain condition referred to as the domain being "Artinian"

similar to how for the ascending chain condition, which you've surely seen, is also referred to as "Noetherian"

[acoustic igloo]

let f(x)=x^3+17x+36. Mod the coefficients by 5 and let g(x)=x^3+2x+1. Since g(x) is of degree 3 with no zeros in the field Z_5, it's irreducible, and thus f(x) is irreducible over Q
is this right?

oh yeah i also need to mention the degrees of f(x) and g(x) are the same

[tough raven]

IG you should invoke Gauss's Lemma at the end to pass from irreducibility over ℤ to over ℚ. Otherwise, this looks correct.

[acoustic igloo]

sweet

thanks

[crystal vale]

May I ask which textbook are you using?

[chilly ocean]

Why a_n+b=(1)

[languid trellis]

a_n and b are coprime by construction

[chilly ocean]

Why coptime

[tardy hedge]

If irreducibles are the same as primes in a domain, is it a UFD?

[chilly ocean]

It's a ring

[languid trellis]

x_i + y_i = 1, xi in ai, yi in a_n, is the same as saying ai and an are coprime, which is the same as saying that xi = 1- yi, which is the same as saying xi = 1 mod (a_n). So, we're saying the product of the x_i (which is in b) is such that \prod xi = 1 mod (a_n), so there exists an element y' in a_n such that prod xi + y' = 1, hence these ideals are coprime

[tough raven]

I think not, because you still need (finite) factorisations into irreducibles to exist (although your hypothesis does guarantee uniqueness). A counterexample is given by the ring of holomorphic functions on a connected open subset U of ℂ. Any irreducible is (up to invertible elements) z-z0 for some z0 ∈ U and is prime because it generates the kernel of evaluation at z0. But a holomorphic function with infinitely many zeroes cannot be a finite product of (z-z0)'s and an invertible.

[south patrol]

Yes

Oh sorry yes lol

Given existence of factorisations this is true

Thanks Raghuram

[south patrol]

Note this is true under relatively mild hypotheses like the ring having no infinitely ascending chains of principal ideals

[tough raven]

TBF that's basically assuming what you want to prove.

[south patrol]

Lol I mean existence of some factorisation feels a lot weaker

Like for example this applies to all rings of integers

[tough raven]

Well, sure, they're Noetherian.

As long as you stay in the Noetherian plains, you can take these things for granted.

IDK why I said "plains", I just wanted to sound flowery.

[thorn jay]

Funny, I'd assume this to be true but I suppose thats true iff any join of a chain of principal ideals is again principal

[south patrol]

Yeah there are fun things like uh

if you take the direct limit of Z[t] -> Z[t^1/2] -> Z[t^1/4] -> ...

And consider the ideals like (t^1/2^n)

Lol

[languid trellis]

Very french of you.

shoutout the champ

[thorn jay]

Grothendiecken, even

[thorn jay]

This is why we consider mainly Noetherian rings...

[south patrol]

Down with noetherian hypotheses

[thorn jay]

Do you know how much niceness conditions it takes to get the theorem
R is noetherian iff R/I and I are
?

[south patrol]

Though rn stuff actually is noetherian which is a nice

[south patrol]

None lol

Though actually

[thorn jay]

I meant for general algebraic structures

[south patrol]

What do you mean by I being Noetherian

Oh lol

Idk what Noetherian means for other structures tbh

[thorn jay]

The same

Ascending chain condition is lattice-theoretic so naturally translates into the language of congruences

[south patrol]

I guess just idk this notion of algebraic structure well lmao

[thorn jay]

Nah thats okay

Just think of ideals/normal subgroups except take away all the niceness

And then you get congruence relations

[next obsidian]

Wow potato you don’t know something? NGMI, tbh

[languid trellis]

#notmypotato

[tardy hedge]

thanks

[thorn jay]

Anyhow, one sufficient condition is, a class of algebras containing its free algebras must satisfy:

• have a nullary operation 0
• congruences are fully determined by the 0 blocks
• congruences are 0-permutable, that is 0/R•S = 0/S•R where • denoted relational product
• congruences must 3-permute. That is, R • S • R = S • R • S

[tardy hedge]

i need to take a complex analysis course again

[tardy hedge]

lol plains

[thorn jay]

Im on a plain 🔥 🗣️

[tardy hedge]

yuhhh kurt

[languid trellis]

i can't complain

[thorn jay]

Glad to see we are all somewhat cultured here

[hidden wind]

not me i’ve no idea what y’all are talking about 💅🏼

[viral elk]

hello, I need help understanding the following expression.

[thorn jay]

Polynomials can have as many roots as their degree.
On the left, it's a degree |G| polynomial, and every element of G is a root, so that must be every root of that polynomial

On the right, it's the same except already factored

[wraith cargo]

Is G a ring?

[thorn jay]

Group probably

Because g^|G| = 1 for any g in a finite group G

Im assuming this is in some group ring

To do with representations perhaps?

[viral elk]

Yes, G it is a finite group.

[viral elk]

My confusion comes from:

I could see the left side has roots of unity, in the form of e^2πi/n , but the right side "g" it is supposed to be the elements of the finite multiplicative group. I do not see how they are the same, left seems to be in the complex numbers and the right not. What I am missing?

[rocky cloak]

Like a finite subgroup of the multiplicative group of a field? Are you trying to prove that G is cyclic or something?

[rocky cloak]

Like what is the context. Is G a subgroup of the multiplicative group of complex numbers?

And if not why are thinking of the polynomial as over the complex numbers?

[rain quest]

Hello, can anyone explain why two polynomials Q1 and Q2 in a quotient field F[X]/(P) where P is irreducible have the same roots if they are congruent to each others modulo P? this seems true and very obvious but i struggle to demonstrate it

[rocky cloak]

What do you mean exactly. Are they polynomials with coefficients in F[X]/P or polynomials in F[X], and are you considering roots in F or F[X]/P ?

[rain quest]

polynomials with coefficients in F[X] and roots in the field F[X]/(P), for context, i arrived to the final question of a lengthy homework to prove that Q(X)=X5-X+1 has no roots in K=F5/(P) and it seems like a good idea to simplify X5 using P(X)=X2+X+1

[rocky cloak]

Well if two polynomials are equal modulo P, then evaluating them at something will always give the same answer modulo P

Ifyou evaluate
f(x) + Pg(x)
at a you get
f(a) + Pg(a) = f(a) mod P

[viral elk]

Sorry I have not replied. I am driving. Give me a few minutes to reply to you fully.

[rain quest]

Hmm, i have not considered that, for some reason i reluctantly switch between treating the elements of K\F as constant values or as classes because of the X-inception, this is my first time tackling with such ideas, im still processing that P(X)=X2+X+1 has no roots in F5 but has two roots in K\F that are class of X and class of -X-1, its a bit alien lol

[delicate orchid]

we could be in Z[G][x]

that's the only option I see

[rocky cloak]

We could be, but then the statement would be false

[delicate orchid]

I agree lol

a group's order is not it's exponent

[rocky cloak]

Well the issue is that ZG is not an integral domain

[delicate orchid]

finding when the group algebra is an integral domain can't be that hard right :trollface:

it's just G is torsion free right

[rocky cloak]

It definitely isn't when G has torsion, but idk if being torsion free is enough...

Might be, idk

[delicate orchid]

you FELL for the troll https://en.wikipedia.org/wiki/Kaplansky's_conjectures

[rocky cloak]

Ah well, then I think "might be, idk" is the correct answer

[delicate orchid]

very interesting problem

[thorn jay]

of course such a simple fucking question is unsolved

[delicate orchid]

group algebras are LE HARD

notice how the field is NOT characteristic 0

[thorn jay]

"has been disproved in char 2"

of course it fucking has

[tardy hedge]

Le

Le Trollolooloololol

[glad osprey]

Are you on discord while driving? 💀

[tardy hedge]

Brah

[delicate orchid]

you aren't?

[tardy hedge]

😎

What are yall doin this Friday night

Imma be in toronto this summer

[languid trellis]

Why do we distinguish between a splitting field and a normal field?

Or rather, are there any relatively simple examples of an overfield that is one but not the other?

Also the "normal closure". These all feel like the same thing to me

[delicate orchid]

I'm gonna fuck around and find out

[tardy hedge]

yea basically i live in a small city rn and im so bored with it

im kinda hyped atm cause classes are almost done and i can be in toronto for a bit, a lot more going on there

[languid trellis]

wag wan shawty

[delicate orchid]

I'm marved.....

[tardy hedge]

some trono slang im assuming

[delicate orchid]

brother we live 5000 miles away how do WE know

and you don't

[tardy hedge]

no way marved is toronto slang too

[delicate orchid]

it means hungry

[tardy hedge]

bruh

[languid trellis]

bumping my question past the toronto slang discourse

[tardy hedge]

A normal field is the splitting field of a set of polynomials

is that the definition

This is Wew Lads Tbh^

that is HIS emote

like wtf is that guy

Ive always found it so funny

[frigid epoch]

A field extension F:G is normal iff no element in F is transcendent over G?

[languid trellis]

I guess normality is trying to mimic the conditions we get when working over finite fields - as soon as an extension of a finite field contains a root of a poly f, then it contains all the roots, whereas this isn't true in other cases. x^3-2 in Q[x] and Q(cbrt(2)) i guess

[frigid epoch]

Oh that

[languid trellis]

not necesssarily I don't think, if E/F has one root of f(x) in F[x] then it has all of them

[frigid epoch]

Yeah yeah

[distant summit]

Please would someone check my proof that $\Phi$ is injective?

Suppose $\Phi(A_1 \omega(z_1))=\Phi(A_2 \omega(z_2))$.

It follows that $\Phi(A_1 \omega(z_1)) \Phi(A_2 \omega(z_2))^{-1} = (1, I)$, which is equivalent to $\Phi(A_1 \omega(z_1)) \Phi((A_2 \omega(z_2))^{-1}) = (1,I)$.

$(A_2 \omega(z_2))^{-1} = \omega(z_2)^{-1} A_2 ^{-1}$.

Using the above fact and the fact that $\Phi$ is a homomorphism, we get $$ \Phi(A_1 \omega(z_1) \omega(z_2)^{-1} A_2 ^{-1})=(1,I)$$

Because $\omega(z)$ is diagonal, it is in the centre, i.e. it commutes with everything.

Thus $$\Phi(A_1 A_2 ^{-1} \omega(z_1) \omega(z_2)^{-1})=(1,I)$$

Since $\omega$ is a homomorphism, we can re-write this as $$\Phi(A_1 A_2 ^{-1} \omega(z_1 z_2 ^{-1}))=(1,I)$$

This is equivalent to $(z_1 z_2^{-1}, A_1 A_2 ^{-1})=(1,I)$, and so $z_1 z_2^{-1} = 1$ and $A_1 A_2^{-1}=I$.

Therefore $z_1=z_2$ and $A_1=A_2$, i.e. $\Phi$ is injective.

[cloud walrus]

Douglas

[rocky cloak]

Being a normal extension is equivalent to being the splitting field of some family of polynomials.

But the point is kinda different. Like the point of the splitting field is usually that some specific polynomials split

[rocky cloak]

It's levy aka GothamChess

He's a chess commentator. ...And I guess also a meme...

[frigid epoch]

I recall normal FEs being equivalent to normality of subgroups of the galois group, but I don't know if that can be formulated without a tower of FEs D:E:F

[languid trellis]

I guess another way to draw a clearer distinction is to look at the galois correspondence

hm

[languid trellis]

What is the "point" of a normal extension?

I suspect that it's "the theory works nicely"

[rocky cloak]

Corresponds to normal subgroups in the Galois correspondence.

Fixed by any automorphism of any field it embeds into

[languid trellis]

hmmm

[frigid epoch]

How to do cube roots in TeX

[rocky cloak]

$\sqrt[3]{x}$

[cloud walrus]

jagr2808

[languid trellis]

keith conrads notes are so fire

i love them somuch

[frigid epoch]

Consider $\bQ \subset \bQ(\sqrt[3]{2}) \subset \bQ(\sqrt[3]{2}, \sqrt[3]{2} \cdot \zeta_3, \sqrt[3]{2} \cdot \zeta_3^2)$. The first extension isn't normal, so $\mathrm{Gal}{\bQ}(\bQ(\sqrt[3]{2})) \ (\cong S_2)$ isn't a normal subgroup of $\mathrm{Gal}{\bQ}(\bQ(\sqrt[3]{2}, \sqrt[3]{2} \cdot \zeta_3, \sqrt[3]{2} \cdot \zeta_3^2)) \ (\cong S_3)$. And $\sqrt[3]{2} \mapsto \sqrt[3]{2} \cdot \zeta_3$ is an automorphism of the latter field that fixes $\bQ$ but not the middle field.

[delicate orchid]

Just show that the kernel is 0

[cloud walrus]

PKThoron

[languid trellis]

Thanks for the example, very neat. I will steal it

[frigid epoch]

Thanks I stole it myself

[glad osprey]

Gal(R/Q) is trivial, while Gal(Q-bar/Q) is uncountable, right?

[thorn jay]

And I mean, proving surjectivity is just proving the existence of a set theoretic section idk

[thorn jay]

Does R even have nontrivial automorphismsv

?*

[delicate orchid]

in this case it is legitimately easier

it's a diagonal matrix acting by conjugation

[glad osprey]

No, the only automorphism of R is the identity AFAIK

[delicate orchid]

oh yeah? watch THIS

[thorn jay]

Mm alright

[thorn jay]

Im watching

[delicate orchid]

[nervous gulp]

[sly crescent]

I’m watching too

Let R be the field with four elements. Then it has an automorphism of order 2.

[thorn jay]

What do you mean "R the field of four elements"

Sorry

I had to

[sly crescent]

?

[glad osprey]

What's Gal(C/Q) btw? I know Gal(C/R) and Gal(R/Q), but I'm guessing there's no way to combine those to figure out Gal(C/Q)?

[delicate orchid]

it's AT LEAST C_2

[thorn jay]

No way

[delicate orchid]

way

[tough raven]

You don't usually say "Galois group" for transcendental extensions.

[glad osprey]

and at most?

[delicate orchid]

[nervous gulp]

[tough raven]

For various reasons, one of which is that AFAIK the correspondence doesn't work anymore.

[thorn jay]

No Galois correspondence?

[delicate orchid]

so what we really mean is Aut_Q(C)

[sly crescent]

Which is isomorphic to Aut(C)

[thorn jay]

She fix my Q till I Aut

[sly crescent]

Which is enormous

[tough raven]

Note that any automorphism of ℂ fixes ℚ, so this is just Aut(ℂ). This is "well-known" to be very uncountable (more precisely, its cardinality is equal to |ℂ|^|ℂ|); you can probably find a proof on mathse.

[glad osprey]

ah, right

[thorn jay]

I can imagine its such a group thats so big its not even fun anymore

[delicate orchid]

so, infinite?

[thorn jay]

Does it even admit some nice topological structure?

[thorn jay]

You and i agree

However, infinite groups with some topology on them are fine

[delicate orchid]

profinite at a stretch

[thorn jay]

Math is just us coping with the fact we're not finitists

[languid trellis]

pretty sure it is c2

[tough raven]

When algebraic groups

[thorn jay]

Those are cool

[languid trellis]

field and galois theory is still hard. does it get any cooler in the infinite case

[tough raven]

If you restrict to automorphisms which are continuous, yes.

[sly crescent]

It gets even harder

[thorn jay]

It gets harder, but im harderer

🙏

[glad osprey]

An element of Aut(Q-bar) is not necessarily an element of Aut(C), right? Is the converse true? You can restrict an automorphism of C to Q-bar, and since Q-bar is normal is the image of the restriction Q-bar?

[languid trellis]

this meme goes out to wew

[tough raven]

The normality implies that Aut(C) preserves Qbar, yes, so you get a group homomorphism Aut(C) → Aut(Qbar). Note that this group homomorphism is not injective, so you should be careful thinking of an element of Aut(C) as "being" an element of Aut(Qbar).

[glad osprey]

I see, thanks

[acoustic igloo]

i was just wondering if the quaternions are a group algebra ℝQ_8, but i guess not, because that would be like 8 dimensional

right?

[coral spindle]

Indeed

[acoustic igloo]

dang

[coral spindle]

But if you quotient identifying -1 in Q with -1 in R, you get H (iirc)

[tardy hedge]

Wow Boytjie makes an appearance

He comes out the shadows

[coral spindle]

In general if k is a subfield of C then kG is pretty much never a division ring

[coral spindle]

I have been Not Having Fun™️ on the server recently so have been avoiding it

[tardy hedge]

Oh dang really? Why?

[coral spindle]

Who's to say

[tardy hedge]

😭

[coral spindle]

I thinkkkkkk it mightttttt be when.... G is Abelian and k contains only those roots of unity coprime to |G|?

Something like this

[tardy hedge]

You missed the wholesome moment from yesterday

[coral spindle]

What happened?

[tardy hedge]

We had a moment of gratitude

[thorn jay]

pat pat make sure to take the breaks you need

[coral spindle]

Oh ok

*bites hand*

[tardy hedge]

We were thanking each other

[thorn jay]

AA
Okay, noted

[coral spindle]

This is just wrong isn't it

RC_3?

Yeah

Nah

But the example of RQ_8 is interesting because it contains the quaternions as a block

[thorn jay]

I have trouble imagining RG as a divison ring at all lmao

[coral spindle]

It's essentially never one, right

[thorn jay]

I cant imagine it being

I mean, take R = the real numbers

[coral spindle]

It will have as many blocks as it has irreps over k (in the case of char 0 at least)

So

there u go

Yeah ok actually I just explained it to myself

the only case is G=1

[thorn jay]

Rubber duckie method

[valid night]

Let $f \in \mathbb{Z}$ be a monic polynomial such that $f(0)$ is prime. Then $f$ has at most 3 different roots in $\mathbb{Q}$.

Because $f$ is prime at 0, its tail term must be prime. Since we can factor each root of $f$ out, the roots must divide the tail term, and the only roots that can do that are 1,-1 and -p.

Is this proof correct?

[cloud walrus]

Kroros

[tough raven]

For no finite non-trivial group, I think - (g-1)(g^{ord(g)-1} + ... + 1) = 0 and all that.

[coral spindle]

Oh that's a nice one

I was thinking in terms of rep theory

But that works more straightforwardly

[tough raven]

Huh, that's interesting. So there are always at least two irreps in the non-modular case.

[coral spindle]

I guess that is like, the central character of the trivial rep?

Yeah

Yeah this is actually not too hard to understand if you know a little bit about the Schur index lol

Like, all that happens in the case when k is not algebraically closed is that some reps get 'combined' in a certain way, according to Galois conjugacy

[tough raven]

Aaaaah when will I read about it

[coral spindle]

Oh dw it's not big deal

[tough raven]

You gave me Isaacs as a book reco right?

[coral spindle]

Yup! It's the go-to

[coral spindle]

But the trivial character will always be in its own Galois orbit... distinct from everything

And it's not hard at all to argue that a group always has at least two irreps unless it's trivial lol

[tough raven]

There's got to be more than that, right? Because real and quaternionic.

[coral spindle]

Basically just a cardinality argument

Yes that's right! The Schur index measures that difference

[tough raven]

Or IG if you just want to know how they combine, the distinction doesn't matter.

[coral spindle]

Yes exactly

You've understood it perfectly

[frigid epoch]

$\bZ\bN = \bZ[x]$?

[cloud walrus]

PKThoron

[frigid epoch]

And $\bZ\bZ = \bZ[x, x^{-1}]$

[cloud walrus]

PKThoron

[frigid epoch]

I assume furthermore that $\bZ M$ has zero divisors iff $M$ has torsion

[cloud walrus]

PKThoron

[coral spindle]

@delicate orchid got another one

[tough raven]

:omegalul:

[frigid epoch]

<_<

[acoustic igloo]

we view i in H as the same as i in C so that C is a subset of H?

because there's a problem that says this

well

nevermind

i think the statement must be right

[noble adder]

i finished the part of hungerford concerned with field extensions

and i also skimmed and read through the ruler and compass construction part

how important are ruler and compass constructions for like

galois theory and later field thoery

im guessing not a lot but just making sure

[rocky cloak]

Compass and ruler construction have basically no bearing on Galois theory, but it is a very famous application of Galois theory.

Many of the ancient Euclidean problems involving compass and ruler construction could only be solved after the advent of Galois theory.

[noble adder]

wait what kind of compass and ruler construction

hungerford covered the impossibility of trisecting 60 degrees and duplicating a unit cube without using galois theory at all

it was just field theory

[rocky cloak]

Well trisecting the angle, doubling the cube and squaring the circle where all solved at the time of the advent of Galois theory, but it is true that you don't have to build the full theory first to solve it.

However Galois theory gives a very comprehensible description of what sort of constructions are possible (c.f. constructable numbers)

[noble adder]

ok thank you so much !

[trail cave]

if R/I is a quotient ring, a,b in R, then does ab+I = 0+I imply ab = 0, or just that ab is in I?

I'm being asked whether R with ideal I having no zero divisors implies R/I has none

[bitter quiver]

the latter

take like C[x,y]/(xy) or smt

[coral spindle]

You should think of the ideal as defining what in R becomes the zero element in R/I

That is, a + I is the zero element iff a is in I.

C.f. the fact in a quotient group, we have gN = N iff g is in N.

[trail cave]

okay thank you

[delicate orchid]

https://tenor.com/view/me-and-the-boys-trolling-troll-me-and-the-boys-on-our-way-to-troll-a-bit-too-much-gif-26579516

I never intended for this....

[thorn jay]

Oppenheimer:

[next obsidian]

How tf is this not a theorem

[coral spindle]

Kaplansky would agree I suppose

Is Kaplansky still around?

Sadly not.

[south patrol]

Lol

Yeah I mean lots of friends in my office care about this

If I'm not mistaken lol

(Ggt gang)

[coral spindle]

Nerd underscore face

[rocky cloak]

Say you got a finite dimensional algebra A and A has finite injective dimension as a left module.

Does it have finite injective dimension as a right module? Surely yes....?

[next obsidian]

All algebras r commutative so obviously

🙂‍↕️

[rocky cloak]

Gorenstein symmetry conjecture solved 🎉

Okay okay, say you have a rigid module T of finite projective dimension, such that there is an exact sequence
0 -> A -> T^0 -> T^1 -> ...

Can the sequence be chosen to be finite?

(T^i in add T)

Say there exists an injective resolution
0 -> A -> I^0 -> I^1 -> ...
such that each I^i is projective-injective. Is A self-injective?

[worthy solar]

Anyone know some reference text or pdf about Chinese remainder theorem but in the context of rings and ideals? I have only seen it just for say over Z and dealing with x mod p1 ,.... x mod pn and getting some solution.

like a genearlization of the theorem

i guess i found something on the wiki but maybe there are some better resources

[sharp ice]

d&f must have one

[worthy solar]

🙏

Thanks

[chilly ocean]

Why the last sentence is true

[hidden wind]

CRT is a really nice theorem

[delicate orchid]

Because we’re assuming the result is true for all collections of n-1 prime ideals. So we can take an x_i that’s not in all of the p_j except for one

[chilly ocean]

What does the one mean

Can u understand this proof

[rocky cloak]

Like you have n primes. If you disregard one of them you have n-1 primes, hence can apply the induction hypothesis

So you do that for each i and get these elements xi

[distant summit]

.

[frigid epoch]

Let $K_n$ and $L_n$ be the set of abelian groups and general groups, respectively, of order $\leq n$. What's the behavior of $\frac{|K_n|}{|L_n|}$ as $n \to \infty$?

[cloud walrus]

PKThoron

[south patrol]

Well note if n is prime then this is 1

[frigid epoch]

Oh interesting

No wait

It's cumulative

So for n=11 you'd be counting the groups of order <11 as well

[next obsidian]

I’m p sure this becomes 0

[frigid epoch]

Yeah but how fast?

[thorn jay]

Fast enough

[next obsidian]

https://tenor.com/view/sonic-gif-25687409

[thorn jay]

Order less than or equal to

[south patrol]

Ah then it'll go to zero very quickly by considering groups of order 2^k I believe

[thorn jay]

Mhm

[south patrol]

You can show there are tons I think for nonabelian

[frigid epoch]

Well

I believe you

[thorn jay]

[next obsidian]

It’s funny too

I think most (not making precise statement) finite groups

Have order 2^n

[acoustic igloo]

if a group G is generated by S={a1 ... an} and we define a function f:S->G, then i think there is not necessarily a homomorphism phi:G->G with phi(s)=f(s) for all s in S

yeah because there could be too many generators

[south patrol]

Indeed that's why I mention these heh

[next obsidian]

That’s not the issue

the issue is that you don’t respect relations of G

[south patrol]

Lmao

[thorn jay]

Such a phi exists iff f(ai) • f(aj) = f(ai • aj) i believe

[next obsidian]

Take something like Z^2/{(a,-a)}

[acoustic igloo]

hmm

[next obsidian]

This is generated by (1,0) and (0,1)’s residue classes

[thorn jay]

Because then it factors through the natural map induced by the free group generated by S

[next obsidian]

But any group homomorphism phi:G -> G needs to send (1,0) and (0,1) to things which cancel each other out

Because (1,0) - (0,1) = e inside this group

[next obsidian]

And this condition means the f:{(1,0),(0,1)} -> G which induce a homomorphism are limited

[frigid epoch]

Now I wonder

[next obsidian]

This doesn’t make sense because f isn’t defined on ai•aj

f is only defined on {ai}

[thorn jay]

Right

[next obsidian]

The condition you need is to respect all the relations, but for non-abelian this is actually still not enough

[thorn jay]

f* from the free group then

[next obsidian]

Because when you write a presentation <S|R> you define it as the free product on S modded out by the normal closure of R

[acoustic igloo]

i thought they meant f(ai)f(aj)=f(ai aj) for all ai, aj in S

so it would be defined on aj

[next obsidian]

This doesn’t make sense, f’s domains doesn’t include ai aj

[acoustic igloo]

huh?

[next obsidian]

f is defined on S = {a1,a2,…,}

[acoustic igloo]

yeah

[thorn jay]

Yeah, i meant that the induced map
f* : F_Grp (S) -> G
Needs to preserve the relations defining G from S

[next obsidian]

This doesn’t need to include ai aj

[acoustic igloo]

i thought i and j are just indeces

[frigid epoch]

Take $N = \bN \setminus {2^n | n \in \bN}$ (so let's toss out the powers of 2). For each $n \in \bN$, what is the least $m_n \in N$ such that $|G_{2^n}| \leq |G_{m_n}|$, where $G_k$ is the set of groups exactly of order $k$?

[next obsidian]

What is ai aj?

Is that not the product of ai and aj?

[acoustic igloo]

oh

but if the group is written multiplicatively

[cloud walrus]

PKThoron

[acoustic igloo]

oh

it might not be in S

i see

[next obsidian]

Yes this is what I’ve been saying

[frigid epoch]

So when do the non-powers of 2 finally catch up to the powers of 2 in terms of group multitude?

[acoustic igloo]

so i need to know about presentations if i want to show that a homomorphism phi exists for some given f

[next obsidian]

Yes

[acoustic igloo]

thanks 🙏

[chilly ocean]

@lethal hatch

@lethal hatch

[charred bison]

how can i show that an irreducible polynomial of degree n exists over F_p^n[x]?

[acoustic igloo]

why do we need R to be commutative?

[frigid epoch]

Probably to tame elements of the form rgr'g'

[acoustic igloo]

(rg)(r'g') should be rr' gg' by definition right?

[delicate orchid]

yeah the elements of G should be in the centre of the group ring

honest answer is, it doesn't need to be commutative but it's difficult enough to deal with as is

[frigid epoch]

The elements of R

[delicate orchid]

no

[tardy hedge]

centre 🔥 not center 🤮

[delicate orchid]

the elements of G. If the elements of R were in the centre then your ring would just be commutative

[thorn jay]

If the elements of G lie in the center of the group ring then G must be commutative no?

[delicate orchid]

good point

you know what I mean though

elements of G commute with the elements of R

[thorn jay]

Centraliser, thenv

?

[delicate orchid]

as Axe said, by defnition of multiplication

or think of them as functions G -> R if you want

[frigid epoch]

Let's do group rings of a noncom ring over an abelian group

[thorn jay]

What if we remove the associativity condition though
Quasigroup rings

Sure, you wouldnt get a ring anymore

But you'd get a ringlike structure

Good enough 👍

[delicate orchid]

you can linearlise pretty much anything you want

[thorn jay]

:0 can I linearise my dad back into my life?

[delicate orchid]

no. absurd.

simply ridiculous

[tardy hedge]

there should be a childrens math tv show with superheros that go like "Lets Linearise It!"

[charred bison]

i couldnt think of any construction but i just need to prove existence

[frigid epoch]

So Wikipedia doesn't specify that the ring needs to be com

And you can indeed smush together a noncom ring and group

[acoustic igloo]

thanks

🙏

[frigid epoch]

Also you can do monoid rings

And probably rngs over other associative magmas

[acoustic igloo]

but i still want RG to be a ring

[thorn jay]

Thats called a semigroup 🙏

[thorn jay]

[acoustic igloo]

algebraic!

[thorn jay]

Linear, even!

[snow badger]

SUppose I have two systems of groups G_i, H_i, that I wish to take projective limits of. Is there a characterization in terms of the systems for when projlim G_i and projlim H_i are isomorphic?

[wispy light]

I don't know the answer for sure, but I'm pretty sure you could first consider the case where one of the inverse systems is a subsystem of the other. (For example Z/p^kZ where k ranges over all the naturals vs. only some of them.)

Let's call such things "good embeddings" of inverse systems if the induced map of inverse limits is an isomorphism.

And only then, for the general case, ask whether you can find "good" embeddings of G_i and H_j (notice the different índices now) into a common larger inverse system.

[snow badger]

Hmm that's a good idea

I'd bet that's an exact characterization in the profinite case

[tough raven]

Lemma 2.6 of https://ncatlab.org/nlab/show/pro-object#AsFormalCofilteredLimits should also be relevant.

[wispy light]

My ignorance of category theory in any real depth is going to be the end of me.

[snow badger]

Thanks! In my case the index posets are not necessarily the same so I can't use functoriality

[snow badger]

We'd need a denseness type assuption for the morphisms to gurintee one isn't a quotient of the other

[tough raven]

ie Hom_pro(G, H) = \varprojlim_j Hom_pro(G, H_j) and every homomorphism from G to H_j (regarded as a "discrete" pro-group) factors through some G_i, so Hom_pro(G, H_j) = \varinjlim_i Hom(G_i, H_j).

[thorn jay]

Category theory is the ultimate

[tough raven]

They don't need to be the same there either.

[tough raven]

IG this should be close to a correct concrete condition: there exist monotone functions j: I → J and i: J → I (where G_i for i ∈ I and H_j for j ∈ J) and homomorphisms f_j: G_i(j) → H_j ∀ j ∈ J, g_i: H_j(i) → G_i ∀ i ∈ I, such that ∀ j ∈ J, f_j ∘ g_i(j): H_j(i(j)) → H_j is the map of the inverse system and similarly for g_i ∘ f_j(i) for all i ∈ I.

[tardy hedge]

Yooo algebros

Who up who up

Enpeace where u at

You guys are familiar faces I hope one day i dont get on here and everyone is gone

[trail rose]

A group G acts on a set X. A set S containing exactly one element from each orbit is a

1️⃣ System of representatives
2️⃣ Complete system of representatives
3️⃣ Set of representatives
4️⃣ Complete set of representatives
5️⃣ Transversal
6️⃣ Cross-section
7️⃣ other

[tardy hedge]

Bruh these quizzes are funny

[trail rose]

it's not a quiz, it's a poll, it's me trying to figure out what the convention is

cuz all (most?) of these are correct answers

[tardy hedge]

Phew

Ya cuz i was confused

[wispy light]

A right inverse of the quotient map X -> X/G.

Because the data of S is actually a map X/G -> X.

[sly crescent]

What do you call the group of isometries of a circle?

1️⃣ O(2)
2️⃣ ΓU(1)
3️⃣ something else

[thorn jay]

Sleeping

Haha

[thorn jay]

I genuinely read 2 as closure system

My brain is rotting

[candid tulip]

I'm confused on why is d. considered false

[trail rose]

it is true, wdym

this is literally Cayley's theorem

[candid tulip]

My book marks it as false

Maybe it's a typo then on the author's part

[thorn jay]

Yeah lmao

[wispy light]

I think Cayley's theorem is often stated in a highly misleading way.

One shouldn't say that G is a subgroup of Aut(G), but rather of Aut(U(G)), where U : Grp -> Set is the forgetful functor.

[south patrol]

I assume that is what is meant by S_G though right?

Otherwise that is bad notation

But it seems perhaps it just is bad notation lol

Indeed

[thorn jay]

People say Aut(G)?
Just use Sym(G)

[thorn jay]

YeH

Yeah

[south patrol]

Aut to me is way more common and I've only seen Sym for like sets tbh lol

[rocky cloak]

Ive never seen anyone use Aut for a permuation group

[wispy light]

Unfortunately, when I took my first algebra course, my professor (not an algebraist) said Aut(G), and I spent like three days trying to find the embedding of V_4 into Aut(V_4).

[tardy hedge]

"discriminant is a square in K"

ok, that means the square root of the discriminant is in K?

ty

[lusty marlin]

Who even states it as G being a subgroup of Aut(G) lol

[thorn jay]

Well, we are taking G here to be a set

Burris and Sankappanavar's notation is just always superior tbh

Bold for the algebra, regular for the underlying set

G < Sym(G)

[thorn jay]

Lmao

I mean, if your group is centerless then its true ;3

Hell, you also have an embedding
Aut G -> Aut Aut G

[late matrix]

whats wrong with the first way?

[lusty marlin]

Because Aut(G) usually denotes the group of group automorphisms of G

And it is not true in general that G is isomorphic to a subgroup of this group

[late matrix]

left multiplication is a group automorphism tho right? so you get a homomorphism G -> Aut(G) by left multiplication

[thorn jay]

conugation is a group automorphism

[lusty marlin]

left multiplication is not a group automorphism

[thorn jay]

agah =/= agh

that would in fact require a to be the identity

[late matrix]

oh right

i got defn wrong

ic ic

[tough raven]

a gah ≠ agh hmm

singular vs uncountable expression of frustration

[thorn jay]

me when I have to explain the nuances of expressions of frustration

[merry coyote]

this seems to be higly unlikely

this is more likely in new zealand

i see

[wispy light]

That joke is unfair to mathematicians. You know, mathematicians would axiomatically define a class of objects that are black on one side, then label them with Greek or Latin (or, more recently, German or French) adjectives depending on whether the black-on-one-side object in question bleats, has wool, etc., and study how these properties relate to one another.

[acoustic igloo]

the book gives an answer for (d)?

my edition only gives answers for a,c,e,g,i

[south patrol]

lol

[glad osprey]

Two questions: the elements of Q(ζ) can be written as a + bζ + cζ^2 + dζ^3 where a, b, c, d in Q, right? I feel like ζ^4 should also be involved, but Q(ζ) has only degree 4 over Q

Secondly, for gamma_2 for example, the automorphism is uniquely determined by sending ζ to ζ^2, but it's not literally the map x |-> x^2, correct? I think it would send a + bζ + cζ^2 + dζ^3 to a + bζ^2 + cζ^4 + dζ^6 = a + dζ + bζ^2 + cζ^4, but not 100% sure

[rocky cloak]

z^4 = -(1 + z + z^2 + z^3)

And yes, not every x is mapped to x^2, that would not be a homomorphism

[glad osprey]

Ah, I was confused by how to get rid of the ζ^4, but I didn't use noggin long enough thanks

[trail cave]

in Z adjoint isqrt(3), are the units 1 and -1? I got this from the norm a^2 + 3b^2 = 1 if and only if a+bisqrt(3) is a unit, and the only time this occurs is if a is 1 or -1

[glad osprey]

Yep, that seems correct 👍

[trail cave]

ty

[empty perch]

Hey folks, just a quick reminder that the Abstract Algebra Lecture series is meeting regularly again ^_^
We will meet to discuss bilinear forms in about 15 minutes over in the ⁠⁠#1055201711679082516. More information can be found in our thread: ⁠⁠#1317307081535000606

[grizzled crow]

Is it more common to see * or $\circ$ for an arbitrary operation? I prefer $\circ$ since my brain thinks multiplication when it sees *

[cloud walrus]

babonk

[rocky cloak]

Juxtaposition or $\cdot$ is pretty common.

Thinking multiplication is probably a good mental image for an arbitrary operation anyway

[cloud walrus]

jagr2808

[grizzled crow]

Juxtaposition?

Like ab?

[frigid epoch]

$a \times b \otimes c \oplus d \odot e \bullet f$

[cloud walrus]

PKThoron

[frigid epoch]

Oh or good old $*$, probably a top choice

[cloud walrus]

PKThoron

[rocky cloak]

$a\star b$

[cloud walrus]

jagr2808

[rocky cloak]

$a\boxtimes b$

[cloud walrus]

jagr2808

[south patrol]

Exterior tensor product

[charred iris]

just write groups additively (yes even the nonabelian ones) 🗿

[boreal inlet]

real

[shrewd token]

Is the definition of irreducibility for monomial ideals equivalent to the general definition of irreducibility? By this, I mean if I is a monomial ideal, then the definition of irreducibility for monomial ideals requires that I can not be written as the intersection of two strictly larger monomial ideals while the general definition of irreducibility requires that I can not be written as the intersection of two strictly larger ideals. Are these two notions equivalent?

[rocky cloak]

In Z[sqrt(-5)]
(3) = (3, 1+sqrt(-5))\cap(3, 1-sqrt(-5))

[acoustic igloo]

(true or false):

any subfield of R can be ordered in only one way
I think it's false because i think {a+b sqrt(2) | a,b in Q} is a field with the automorphism a + b sqrt(2) -> a - b sqrt(2) which induces a different ordering

[shrewd token]

I am not sure if I follow, this shows $(3)$ is not irreducible in $Z[\sqrt{-5}]$ but $Z[\sqrt{-5}]$ is not a polynomial ring. Let say we are working over the ring $S = \mathbb{K}[x_1, \dots, x_n]$ where $\mathbb{K}$ is a field.

[cloud walrus]

sheaf

[rocky cloak]

The polynomial ring is a ufd, so there it should be true

...I think at least

Yes. If an element is irreducible it is prime, so then if it contains an intersection it contains one of the ideals

[shrewd token]

do you mind writing what the analogue for the example is in a polynomial ring?

[rocky cloak]

What do you mean?

The analogue of what?

[shrewd token]

I don't understand what this example is showing

[rocky cloak]

Ah wait, I was thinking principal ideal, not monomial ideal

Got some wires crossed there

Anyway, it should be there yeah
https://mathoverflow.net/q/162739/157483

[wispy light]

What's an irreducible ideal?

[rocky cloak]

One that is not the intersection of bigger ideals

[candid tulip]

Yep at the back of the book

All odd numbered exercises that don't ask for definitions or proofs have answers

[shrewd token]

I am not too familiar with graded rings, the second answer says the the result for N^n-grading (which corresponds to monomial ideals) is well known but the reference uses some module theory I don't know. Is there a more elementary proof for this case.

[rocky cloak]

I mean, the first answer is an elementary proof

Oops that's only N-graded

[shrewd token]

The proof shows (0) is graded-irreducible implies it is irreducible. Perhaps I am missing something but I don't see how this immediately generalizes to arbitrary graded ideals.

[rocky cloak]

If you have some ideal you can always take the quotient to reduce to the case of the (0) ideal

[sly crescent]

Is there a field whose additive group and multiplicative group are isomorphic

[wispy light]

Can't have many roots of 1, right?

[thorn jay]

Has to be nonzero char at least

Because a 0 char field, has an additive group without torsion (else you'd either have divisors)

[wispy light]

I mean, if it's of characteristic p, then its multiplicative group can't have elements of order p, right?

But additively every nonzero element has order p.

[thorn jay]

Then given that -1 * -1 = 1, there would be a + a = 0
=> 2 • a = 0
=> 2 = 0 as a cannot be 0

So the field must have char 2

[wispy light]

But if F has characteristic 2, then no element of F has order 2 multiplicatively.

Because the only square root of 1 is 1, in characteristic 2.

[thorn jay]

Yeah

So
1 * 1 = 1
Gets turned into
0 + 0 = 0

[wispy light]

The best thing we'll get is that there's a group homomorphism from F to F^*, namely exp, when F is like R or C, me thinks.

[thorn jay]

Mhm

[wispy light]

Local isomorphisms, even, when you consider them with their usual Lie group structure.

Alas, impossible in characteristic p.

[thorn jay]

[shrewd token]

Ah of course, thanks

[cloud walrus]

James

[trail cave]

For showing I is an ideal in T, I take a/b, c/d in I and get:
$a = 5a', b=5b'$ so $\frac{5a'}{b} - \frac{5c'}{d} = \frac{5(a'd - c'b)}{bd}$

[cloud walrus]

Bottlecap Desu~(Bottlecap Gang)

[trail cave]

But how do I show that the gcd of the numerator and denominator is 1 here?

[velvet hull]

you could argue that if you kept dividing out gcd's the process will eventually stop

but imo it's "kind of obvious"

in the same way that "Q formally consists of simplified fractions" is sometimes omitted in saying that it just consists of fractions

[trail cave]

I guess I don't really see how it's obvious but I realize it might be hard to explain

[worthy solar]

This must be a typo right? How is Z_2[x]/<x^2+1> a field if (x+1)(x+1) = x^2+2x+1 = x^2+1 = 0 mod x^2+1 so it cant be a field

[glad osprey]

You don't have to show that 5(a'd - c'b) and bd are coprime - I don't think that's true in general. You only need to reduce it to a fraction where the numerator and denominator are coprime, you can do that by dividing by the GCD

[knotty pilot]

Yeah its a typo

[worthy solar]

Guess I'll have to figure out what the prof meant. I wonder if it was x^2+x+1

[knotty pilot]

Maybe the question setter meant Z2 or Z2[x]/<x² + x + 1>

[velvet hull]

likely that

[worthy solar]

[trail cave]

oh ofc this makes more sense

[glad osprey]

But tbh, I don't know why they specify that the numerator and denominator must be coprime, it doesn't really matter since we can always reduce fractions anyways

[trail cave]

thank you both I was kind of missing the point of the condition as just reducing the fractions

[glad osprey]

Hmm, I guess requiring gcd of 1 is to make the set condition unambiguous, without it you would have 1/2 in T but 5/10 not in T

Gotta go to bed now tho, it's 4 AM 💀

[static mauve]

Does this question actually make sense? I think there is a typo in here but Im kinda uncertain:

Show that the set of all g ∈ G such that i_g : G → G is the identity inner automorphism i_e is a normal subgroup of a group G.

[sharp ice]

if i_g means conjugation under g then it makes sense. the set of all g such that gxg^-1 = x for all x in G. that subgroup is the center of G

[static mauve]

Ah i see! Thanks for the clarification

[grizzled siren]

Heya, how can i show that $x \in A$ is invertible iff $x+J(A)$ is invertible in $A/J(A)$, with $J(A)$ the jacobson radical?

[cloud walrus]

mak601

[rocky cloak]

Say x + J is invertible. That means there exists a y with 1 - yx in J.

One characterization of J gives that 1 - (1-yx) = yx is invertible say with inverse z. So zyx = 1

[grizzled siren]

ohhh I should have been able to do that

thanks ! 🙂

[tardy hedge]

If I define a map of Fp(a) (a algebraic) to itself by sending a to some other root of a polynomial over F that a satisfies, is that map actually an automorphism?

I am asking because I didn't send a to another root of the minimal polynomial, just to some other polynomial so it might be a different irreducible factor

like u need to send a to other roots of the irreducible factors of polynomials that a satisfy right

[next obsidian]

Let a not in Fp, and consider its min polynomial f. Now 1 is a root of f(x)(x-1) so this can’t be true

[tardy hedge]

I ask because of (b)

We have a, and we know a+1 also satisfies that polynomial. From that alone, can we really conclude that it extends to an automorphism?

I was thinking, what if a+1 is not a root of the same irreducible factor

[next obsidian]

Okay so

I think this is more nuanced

K = Fp[x]/(x^p - x - a) under the isomorphism sending x to alpha

You can define a map Fp[x] -> Fp[x]/(x^p - x - a) sending x to x+1, and this kills x^p - x - a

So it extends to a map on the quotient

So you get the map K -> K sending alpha to alpha + 1

It’s an automorphism because either you can see that the map alpha -> alpha - 1 is also defined the same way

Or just see that doing this map p-times in a row results in id

[tardy hedge]

Hmm this is weird

[tardy hedge]

Isnt this true assuming x^p-x-a is irreducible

[next obsidian]

Huh

Oh yah

[tardy hedge]

But we can't yet assume that right

[next obsidian]

Okay well here’s a really roundabout way

I just came up with this now

[tardy hedge]

Ok

[next obsidian]

Call this polynomial we mod out by f

Then f(1), f(2),… can’t be zero

Until f(p)

Cuz otherwise a = -n^p + n which is in F

Err

Okay this is more complicated than it needs to be

We just assumed f has no roots in F

[tardy hedge]

Yea i dont really think my prof was intending it to be complicated here

I dunno

[next obsidian]

So fixing alpha, alpha + 1, alpha + 2,… enumerates the roots

So let alpha’s irreducible factor in f be called g

Clearly g isn’t linear

Else f has a root in F

[tardy hedge]

Yeah

[next obsidian]

So g has at least two roots, which need to be of the form alpha, alpha+k

[tardy hedge]

why

alpha +k?

[next obsidian]

Cuz I wrote down every root of f

[tardy hedge]

Oh because f has distinct p roots a+i

[next obsidian]

So alpha -> alpha+k is an automorphism of K

But because p is prime, we can obtain via composition every map alpha -> alpha + n for n in {0,1,…,p-1}

Remembering that we’re in char p

And also obviously each of these automorphisms are different cuz if not you get alpha + n = alpha + m, so m = n which is a contradiction

So I just cooked up p distinct automorphisms of K

Which would for example, force deg g = p

[tardy hedge]

I think I get the gist of it but I will need a moment to think it through

[next obsidian]

The crux of it is this

You wanted to show alpha -> alpha + 1 is an automropjism

To then generate every automorphism alpha -> alpha + k

(This is saying 1 is a generator of Z/pZ*)

[tardy hedge]

right

[next obsidian]

But actually because p is prime, any alpha -> alpha + n will generate all those automorphisms

Because everything generates (Z/pZ)*

[tardy hedge]

yeah

[tardy hedge]

Why do we need this last part rn

[next obsidian]

Cuz it demonstrates p different automorpjsims of K

[tardy hedge]

Oh cause this

[next obsidian]

Forcing the degree of the min poly of alpha to be at least p

[tardy hedge]

yeah thats for the later part of the Q right

[next obsidian]

Sure

I mean this lets you cook up alpha -> alpha + 1

[tardy hedge]

Yeah

[next obsidian]

But then like, who cares, cuz any of these automorphisms handles b)

I worded it that way cuz I wanted to drive home the fact that all of these automorphsims are functionally the same

[tardy hedge]

ok this is an important point then right

[next obsidian]

Yeah, I don’t see easily how to immediately guarantee alpha + 1 is another root

But I could get it for some alpha + k which was enough

[tardy hedge]

Yeah, I wonder if my prof is expecting this kind of answer or not. I mean maybe

at first glance it seemed like it was just a simple thing

Thanks though that was informative

[tardy hedge]

So: Given F(a) with a algebraic over F, a map sending a to another root of the minimal polynomial of a over F necessarily induces an automorphism of F(a)

just want to cement it

[tardy hedge]

(B) immediately mentions the Galois group of K. A priori, we dont know that K is Galois extension of F do we?

But, we know there at least p distinct automorphisms of K, and the degree of K over F is at most p, so from that we know the extension is Galois and can also conclude degree of K over F is p

[hidden cairn]

what is the purpose of verifying (a) and (b) in this proof?

this is from dummit and foote

the theorem^

v is the R-module homomorphism which produces the maximal element (v(N)) of all images of N under homomorphisms

a_1 is the generator of v(N) in R. R is a PID.

[tardy hedge]

Well can we see the rest of the proof?

What was said before?

[tardy hedge]

Ty

I looked at this proof last term so its nice to review

[hidden cairn]

i think i get the purpose now

its not that we have to check a and b any time we want to put an element v in a basis for some module M but rather that these verifications are needed only to justify the induction step

which comes after this part of the proof

so i just had to read a bit more

[worthy solar]

How do I show my field extension is the minimal field extension so that it forms a splitting field

Like I showed I wanted some polynomial to split over my field

So I verified it has all the roots of the polynima of degree 3 and contains the field coefficients but I'm not sure how to show it is necessary minimal. I mean it is a degree 3 extension so I must be?

[charred bison]

if F is a field and F^x denotes the multiplicative group, what could this notation mean

$F^{\times 2}$

[cloud walrus]

CoolShot

[charred bison]

i really have no idea

[thorn jay]

The multiplicative group disjoint union'd with itself

Haha

Get it, "times two"

Ill see myself out

[kind temple]

i think i remember seeing this as the multiplicative subgroup of squares

[charred bison]

ohh

i guess that could make sense

[tardy hedge]

The splitting field is the field extension generated by all the roots of the polynomial

So if your extension was generated by the roots of your polynomial then its the splitting field

[worthy solar]

Well I was given a field which was some quotient field which I know to be isomorphic to F(a) where a is a root of the polynomial. It happens to also contain all 3 roots. So I can't say like like I am unsure if I can or can't say it was by definition generated so as to be minimal

[tardy hedge]

F(a) is by definition the smallest field containing F and a

[worthy solar]

Because there was some theorem about quotienting by an irreducible polynomial then that quotient field is isomorphic to the field extension of your original field with any root of that polynomial

[tardy hedge]

The splitting field must necessarily contain F(a)

[worthy solar]

Yes

[tardy hedge]

So if F(a) also has the other roots

Then its the splitting field

[tardy hedge]

Yeah

Thats not really relevant here though

[worthy solar]

Okay.. it was just weird it happened to contain all the doors even though it was generated by 1

Well I guess maybe because of closer

Yeah that was why

[tardy hedge]

Hm?

[worthy solar]

Closure***

Because my field was z_2 coefficients and I had x as a root, so I also got x^2 and x2+x (which happened to be the other roots). I guess it was because it's a degree 3 polynomial which was irreducible so it cant have an extension smaller than 3? I'll a little fuzzy on this idea/topic

[tardy hedge]

Yeah adjoining a root of that degree 3 irreducible polynomial will generate a degree 3 extension over the base field

That extension doesn’t always contain the other roots of the polynomial but it could

The degree of the splitting field of a polynomial of degree n over the base field is at most n!

[worthy solar]

Only n!

[tardy hedge]

Yeah its worth thinking about why its that

[gusty thistle]

Suppose $f$ is a polynomial in one variable over a field $K$. Is the minimal polynomial of $x$ over $K(f)$ the polyomial $m(y) = f(y)-f$ ? If so how do you see this? If not how do you compute the degree $[K(x):K(f)]$ ?

[cloud walrus]

stable compass needle

[tardy hedge]

K(f)?

[gusty thistle]

Yea i mean the subfield of K(x) generated by f lol

i should have been more clear lole

[tardy hedge]

I think the question as stated is unclear

[gusty thistle]

i see should have phrased this question in a more repsonsible manner

[tardy hedge]

Or not precise or smth

Yeah

[gusty thistle]

lol

Let $K$ be a field and $f \in K[x]$. Then $f$ generates a subfield $K(f)$ of $K(x) = \operatorname{Frac} K[x]$. I want to determine the degree $[K(x):K(f)]$ and I would like to know the minimal polynomial of $x$ over $K(f)$. I think its supposed to be the polynomial $m=f(X)-f \in K(f)[X]$.

[cloud walrus]

stable compass needle

[rocky cloak]

Imagine a polynomial
an x^n + an-1 x^n-1 + ... + a0 = 0
with ai in K(f).

By clearing denominators you can assume ai in K[f], so this is an equation in K[x].

Let d be the degree of f, and assume n < d.

The degree of ai x^i is i modulo d (unless ai = 0), so all the terms have distinct degrees. Hence the degree of the equation is just the max of the degree of each term. But then the equation can't be 0, contradiction. So the degree of x is at least d.

Since m has degree d it must be the minimal polynomial of x

[gusty thistle]

That is great I will go over it tomorrow

Thank you !

Im glad its true

[glad osprey]

Does a faithful transitive group action of G on A induce an isomorphism between G and Aut(A)? The faithfulness ensures injectivity atleast, but I'm not sure transitivity implies that it's surjective

[rocky cloak]

Cn acts faithfully and transitively on {1, ..., n} but is not Sn

[glad osprey]

I see, thanks Is there an adjective that implies a group action is surjective?

[thorn jay]

boring
(/j)

often group actions are interesting because they preserve a certain structure of the objects

if your group action is surjective then it preserves no structure besides being a bijection

[hidden wind]

on {1,…,n}, that the action is n-transitive i guess is the term you’re looking for

or i might be misunderstanding something

[glad osprey]

I think that is what I'm looking for, thanks 👍 atleast on for set automorphisms on {1, ..., n} that works, but I for automorphisms in another category I think there might not be enough automorphisms for a group action to be n-transitive, even though it is surjective

[hidden wind]

i’m not sure if i understand what you’re asking for in asking for the action to be surjective

[glad osprey]

hmm, but take a group G where all automorphisms are inner, then isn't the conjugation action of G on G surjective? Might be misunderstanding you, not sure if this counts as preserving no structure

I mean I want the group action to be surjective as a map G -> Aut(A)

[hidden wind]

right

[glad osprey]

these might be stupid question btw, I just realized I didn't 100% understand group actions, so I had to brush up on them

[hidden wind]

in the context of A just being a set, so Aut A = S_A is just the bijevtio bijections of A, the action of G on A being free is equivalent to the map G to Aut A being injective, so asking for surjectivity as well is the same as simple asking that G is Aut A

I should sleep

if A is in some more specific category than Set there’s probably more interesting things to say

[glad osprey]

Yep, if the group action is both free and surjective then that's pretty boring, but I could imagine a case where it's surjective but not injective, for example if there's a big group acting on a smaller object

[thorn jay]

Aut(A) in what category?

[glad osprey]

any I guess, I'm just talking about group actions in general

[thorn jay]

If it is in the category of sets then a surjective group action is rather boring, in my opinion

[thorn jay]

A group action is a homomorphism f : G -> Aut_C(X) where X is an object in a (locally small) category C

So a surjective group action is one where f is surjective

[glad osprey]

yep, we're on the same page

[thorn jay]

Right

[thorn jay]

But in the category of groups, whether or not the group action by conjugation is surjective is interesting

Although I must say that we care more about it being injective than surjective

So i suppose free/faithful group actions are simply more interesting :3

[glad osprey]

I see I haven't used group actions that much, so I can't think of an example where it being surjective is important

[hidden wind]

i guess a stupid example is Aut A x Z/2 -> Aut A given by simply forgetting the Z/2 part

[glad osprey]

but I can imagine free/faithful ones are more interesting

[glad osprey]

lol, good example

[rocky cloak]

Or both: https://en.m.wikipedia.org/wiki/Complete_group