groups-rings-fields

[tardy hedge]

But i dont really get it just at a more low level

If a^p = b^p we have a=b?

[hidden wind]

this is an identity in any integral domain

[glad osprey]

hmm, I found that the Frobenius endomorphism is injective more generally in a ring with no nilpotent elements. It's injective iff. there are no elements r with r^p = 0, and clearly there are no nilpotents in a field

[hidden wind]

well uh no not an identity, i mean, it holds true in any integral domain

[glad osprey]

(of prime characteristic p)

[hidden wind]

no you don’t even need that i’m pretty sure

[coral spindle]

C is an integral domain
Let a be a primitive pth root of unity, let b = 1.

[hidden wind]

oh

right i see my mistake, i should probably get out of bed and eat something it’s past 2 pm

[glad osprey]

without food and drink the hero is no good

(ancient norwegian proverb)

[south patrol]

Prove it

[rocky cloak]

It's all fun and games until someone whips out Håvamål

[frigid epoch]

Have a meal

[glad osprey]

Something that confused me for a sec: a^p =0 implies a = 0 is true in any integral domain, but you need Frobenius to be a homomorphism for this to imply injectivity. Dunno if this was your confusion too

[hidden wind]

i very embarasssingly must’ve conflated the exponent with a factor and somehow thought the cancellation rule (ax = ay implies x = y in an integral domain) applied, but of course we don’t have any common factor to cancel

[glad osprey]

Ah, easy mistake to make (especially when you're still in bed )

[tardy hedge]

Oh ya duh

Well not duh to u duh to me

This makes more sense now lol

[crystal vale]

I don't think we need N to be finite here

[coral spindle]

Correct

I don't know why they chose to do that

Oh, no wait I see. There is some fear that we would only prove that gNg^-1 \subseteq N for the case when N is infinite.

[crystal vale]

But is for every t so why do we consider this case?

[coral spindle]

Of course, this is sufficient, but perhaps this isn't known

[coral spindle]

What do you mean

[crystal vale]

I mean, I know gNg^-1 \subseteq N doesn't imply g in N_G(N), but here gNg^-1 \subseteq N for all g in G, right?

[coral spindle]

Yes

I mean, I know gNg^-1 \subseteq N doesn't imply g in N_G(N)
Does it not?

[crystal vale]

Not

Because we cannot say gNg^-1 = N

In a finite case it is true

[coral spindle]

No, we can. This is a nontrivial fact, but it is true.

Nonetheless yes, we can say it in the finite case.

It is easy in the finite case.

[crystal vale]

But yes here when it is given that gNg^-1 \subset N for all g in G then we can say gNg^-1 \subseteq N implies g in N_G(N)

[crystal vale]

How?

[rocky cloak]

It's only for every t in the generating set, not every t in G

[coral spindle]

Suppose $gNg^{-1} \subseteq N$. Then by applying the inverse map, $g^{-1}Ng \subseteq N$. Now conjugating by $g$ we obtain $N \subseteq gNg^{-1}$. So in fact we have the inclusion both ways. QED. [THIS PROOF IS INCORRECT]

[coral spindle]

Wait hold on, I'm wrong about this step applying the inverse map. I remember they prove this in hungerford

[crystal vale]

Yes

[rocky cloak]

Hence why you need finiteness, otherwise you cannot be sure that t^-1 N t is in N for example

[coral spindle]

Wow I didn't know about this, thank you for informing me

I was indeed mistaken

There is this theorem in Hungerford

,rotate

,rotate

[coral spindle]

Oh god this is hell

There we go

I see where this goes wrong now. It does suffice in general if we have a whole group, but I see for a single element it does not suffice, because we cannot assume that a^-1 also normalises N.

[rocky cloak]

You could modify the statement to be
tSt^-1 < N and t^-1 S t < N

Then you can drop the finiteness

[crystal vale]

So if it is for every t, tSt^-1 \subset N, cannot is it true for all g in G?

[coral spindle]

Your counterexample works for this too

Try choosing appropriate generating sets

[crystal vale]

Now I got it, thanks jagr

[tardy hedge]

Localization of Z popped up in my number theory homework

[coral spindle]

The worst possible notation choice 😭

[frigid epoch]

Heh

[tardy hedge]

Really why haha

Oh is it usually Z with a subscript 2 for localization at (2)?

[coral spindle]

The notation should be $\mathbb Z_{(2)}$. The notation $\mathbb Q_2$ is for the $2$-adics. This is especially annoying since $\mathbb Z_2$ often denotes the $2$-adic integers, and there's \emph{already} such an annoying clash of notation there!

[cloud walrus]

$\mathbf{Boytjie}$

[tardy hedge]

i see

[south patrol]

Yes Z_2 can mean a million things

1. 2-adics 2) the group/ring Z/2Z 3) the localisation of Z at the element 2 i.e. Z[2^-1] etc

3. is never used thankfully and 2) is something i would definitely avoid lol

[frigid epoch]

2-adics are so much less important than Z/2Z

I resent them for hogging that notation

[rocky cloak]

I mean, Z/2 is pretty sweet notation. Don't know why Z/2 would need both the notations for itself, kinda greedy ngl

[south patrol]

Ye Z/2 is very nice and matches what one says verbally lol

Well, I guess some people say it without the mod but then to me that refers to 2-adics lol

[south patrol]

Sad

[frigid epoch]

I don't like Z/2 cause that's not what's happening

[south patrol]

Wdym lol

You are modding out by the ideal generated by 2

[frigid epoch]

It feels incorrect

[south patrol]

I guess do you mean you prefer Z/2Z

[frigid epoch]

Like you just omitted a symbol

[south patrol]

or the ugly and pedantic Z/(2) lol

[frigid epoch]

Yeah

[frigid epoch]

And even that

[south patrol]

I mean ye i agree there like ye

but idk convenience here

[frigid epoch]

I mean I just try to use C_2 nowadays

[south patrol]

ig has the disadvantage that additive notation is often more convenient and fits in better ig

but ye for like "pure" group theory stuff it is v cool sure

[rocky cloak]

C_2 = group
Z/2 = Z-module or ring
Simple as that

[south patrol]

F_2 - field

[elfin wraith]

Still not over getting called out for using the “non standard” notation C_3 in my analytic number theory class

[south patrol]

Lmao

[elfin wraith]

Will upset me until the day I die

[sturdy spear]

i did (a) by computing each sigma^i for 1<=i<=12

found this online, i wonder if there is some initiative way to think why there are total #{n: gcd(12,n)=1} integers such that \sigma^i is 12 cycle

[barren sierra]

A cycle will generate a finite cyclic group. So then think about which elements of a finite cyclic group can also be generators of the cyclic group

for concreteness you can think about the possible generators of Z/nZ (it's all the same)

[sturdy spear]

iirc all m such that gcd(m,n)=1

wow

this is really motivative

[proud hedge]

can a zero divisor of a ring also have an left inverse?

[sturdy spear]

but a question, a cycle say of length 12 will generate a cyclic group but how do we know all the generators will have same cycle

[barren sierra]

wdym all the generators will have the same cycle?

I don't know what you mean

[sturdy spear]

sorry if it was not clear. Lemme think again

i think i am confused

If f is a permutation and let f is 12 cycle.
We wish to find i such that f^i is 12 cycle.
How those elements will work which can be generator for <f>
(if not talking about Z/nZ)

[barren sierra]

Oh oh I see what you mean I think. Ok one thing that would be good to think about is if sigma is the product of an n cycle and an m cycle which don't share any any values in the cycle (so they commute!) then what is the order of sigma?

So like if sigma = (1 2 3)(6 7 8 9)

and also specifically what is the order of an n-cycle

that should help you think about why 12-cycles correspond to generators of the cyclic group

[sturdy spear]

lcm of both

[barren sierra]

🔥 yup

[sturdy spear]

okay, i will think about it

https://tenor.com/view/epic-latex-fail-latex-latex-fail-bad-latex-bad-tex-gif-26615851

[candid patrol]

$\sigma^i$ is a $12$-cycle $\Rightarrow \langle \sigma^i \rangle = \langle \sigma \rangle \Rightarrow \frac{12}{pgcd(12,i)}= 12$, so you can just check if it works for $i \in {5,7,11}$

[glad osprey]

It definitely can't have both a left and right inverse. You can have a right zero divisor with a left inverse, for example the right shift operator (x1, x2, ...) |-> (0, x1, x2, ...) in the endomorphism ring of R^infty. I think you can't have a two-sided zero divisor with a left inverse though, not 100% sure

[cloud walrus]

UGOBEL

[sturdy spear]

maybe \gcd

[sturdy spear]

ah right right

i have seen this formula before as well

thank you guys for helping me
i will again think about it after having some tools

[proud hedge]

i also agree with you but i can't prove it

[rocky cloak]

Exercise: if an element has both a left and a right inverse, then the two inverses are equal and is in fact a two sided inverse

Possible hint: multiplication is associative

[proud hedge]

i've solved this before but i can't solve this one this is what i've got so far

if ab=0 and a is also a zero divisor then ax=0 (xa=0 cannot be true.) also ba is also a zero divisor

[rocky cloak]

Wait, what are you trying to prove here?

[glad osprey]

We're proving that a two-sided zero divisor can't have a left inverse. I think more specifically we can prove that a left zero divisor can't have left inverse

I think I found a proof, it wasn't that hard

[rocky cloak]

What does "two sided" zero divisor mean?

Just both a right and left zero divisor?

[glad osprey]

yeah

[rocky cloak]

Well isn't that obvious?

[glad osprey]

you mean this? Yeah, it was kinda obvious I guess

[rocky cloak]

Like ab = 0 -> a^-1 ab = 0 -> b=0

[proud hedge]

let's say a is an element of Ring such that for some x ax=0 and there exists element b such that ab=1.
can ba=1?

[proud hedge]

it's not two sided

[rocky cloak]

b*0 = bax = (ba)x = x

So if ba = 1, then x=0

[proud hedge]

in my text book x is a zero divisor if for some y xy=0 OR yx=0.

[glad osprey]

@rocky cloak is this example of a right zero divisor with a left inverse correct?

[rocky cloak]

Some non-zero y presumably

[proud hedge]

very good i actually just got to this

now i have another question

let's say a is an element of Ring such that for some x ax=0 and there exists element b such that ab=1.
is it possible that for some k ka=1?

Gosh i love ring theory so much 😆

[glad osprey]

isn't that what we just proved to be impossible?

[rocky cloak]

Yeah if ka = 1, then x=0

Also if ab=1 and ka=1, then b=k

[proud hedge]

no but we can use that

[proud hedge]

like this

@rocky cloak you are good with rings how did you learn it which text book did u study?

[languid trellis]

Have you watched the movie kung-fu panda?

[proud hedge]

yes why? 😆

[languid trellis]

Po going to traing with master shifu is the equivalent of what jagr has done

[proud hedge]

why what did he do?

[candid dove]

Jagr lore 🔥

[proud hedge]

Dude is he Serge Lang or something?

[languid trellis]

I don't understand why We are distinguishing case 1 from case 2. Can we not show that the maximal subgroup is unique so we have G_2 = H_2 always?

[rocky cloak]

I probably used Basic Abstract Algebra by Bhattacharya, but that would be close to ten years ago now.

Mostly learned by osmosis I guess

[rocky cloak]

A group can have several different maximal subgroups

[languid trellis]

I just realised, like (2) and (3) subgroups of (Z,+)

Thanks

[proud hedge]

you mean this site?
https://www.osmosis.org/

[rocky cloak]

No, like putting my head close to a smart person

[languid trellis]

And in the finite case, if we have G = HxK, H, K simple, then both H and K will be maximal normal?

[proud hedge]

Hmmm

[rocky cloak]

Knowledge diffuses

[proud hedge]

anyway thanks everyone for their help specially @rocky cloak
i am so glad that we have a community like this

[rocky cloak]

I'm sure there are people here with strong opinions/good recommendations for books about ring theory. Could always check #book-recommendations

[candid patrol]

^^

[rocky cloak]

My guess would be you get everything.

Is there any trick, besides just computing?

[acoustic igloo]

Does a normal subgroup have to be a union of conjugacy classes?

[candid patrol]

Of course

[acoustic igloo]

Thanks

I found that your subgroup contains the Klein 4-group btw

[candid patrol]

Let us set ( \varphi = \tau (\sigma \tau)^2 \tau = (1\ 5)(2\ 4) ) and ( K = \langle \sigma, \varphi \rangle ), which is a subgroup of H

[cloud walrus]

UGOBEL

[acoustic igloo]

But idk what to do next

[candid patrol]

The shows that [H : K] = 6

Then |H| = 6|K| = 60

[acoustic igloo]

How do you find [H : K] = 6?

[candid patrol]

I won't tell you lol

Try it

[acoustic igloo]

Ok

[thorn jay]

Any prime, in fact

[tough raven]

Because ||{1, 2, 3, 4, 5, 6} is a homogeneous space for H with Stab(6) = K||?

[candid patrol]

You can show that K = Stab(6) yes

But I don't really know what you mean by "homogeneous space"

[tardy hedge]

Are units in a ring considered irreducible ?

[south patrol]

Usually no (by convention)

e.g. if A is a UFD then irreducibles are equivalently primes

[tardy hedge]

And (unit) is not a prime ideal ?

[south patrol]

Yeah also by convention

[hidden wind]

i finally got around to write up my pages long elaboration on the step my textbook describes as “this is trivial”

[rocky cloak]

After careful consideration we have determined that it is indeed trivial.

[tardy hedge]

^hahahaha

[tough raven]

transitive H-set

[hidden wind]

i love homogeneous spaces

[tardy hedge]

Why

[potent condor]

Thanks ilyt

[tardy hedge]

Class clown mf

[tough raven]

https://math.stackexchange.com/a/1128937 so quotienting by homotopies is the same as inverting homotopy-equivalences (because universal pairs of homotopic maps are sections of a common retraction which is a homotopy-equivalence; hence making it an isomorphism makes them equal.

But inverting quasi-isomorphisms is not the same as equating maps which are the same on homology.

[wraith cargo]

well you already quotient out by chain homotopy right

the construction of the derived category is that

you quotient out by chain homotopy

and this already gives you that all homotopy equivalences are isomorphisms

but quasi-isomorphisms are a larger class of morphism

like much larger usually

so you also need to invert those to make them fully fledged isomorphisms

in the topological case you only care about homotopy equivalences

so you don't need to localize

[hidden wind]

i only know them from the very little diff geo i’ve done, in which case a the action on a homogeneous space lets you move local arguments wherever you want

[hidden wind]

^meant to reply to this

[rocky cloak]

The difference is that homotopy equivalences already have inverses, but quasi-isomorphisms do not, so you need to introduce new maps.

[tough raven]

Yes, but the point it's enough to just invert homotopy-equivalences; homotopic maps are automatically forced to be identified.

[tough raven]

Hmm.

[tough raven]

Well, they only "already" have inverses if you know that you are going to quotient out homotopies (so that gf is going to become id). If you were trying to invert homotopy-equivalences it wouldn't be clear that those are going to become inverses.

[wraith cargo]

I'm sort of confused
Why wouldn't it be clear?

you by definition have that fg is homotopic to the identity

if you quotient out by homotopy you're making homotopies into equalities

so it's like clear that these become isomorphisms?

[tough raven]

Can you prove (without using the explicit argument above) that if f is a homotopy-equivalence with homotopy-inverse g, then fg, gf have to become identities in the localisation at all homotopy-equivalences?

[tough raven]

I'm saying, if you invert he's (without intentionally trying to quotient anything), you end up with the exact same result. And this is unexpected (or at least not pure abstract nonsense).

[rocky cloak]

Yeah, it's not an immediate thing. But it should be less surprising.

[tardy hedge]

Today i learned jagr learned group theory 10 years ago

So I could maybe get somewhere in the ballpark to your level of knowledge in 10 years? I hope so, but i dont know if ill even have to time to properly review older material to get caught up to speed. Idk

I just feel like doing math in the school setting feels rushed vs when i was studying it as a hobby last year

I think u do learn more from a raw input sort of thing but i dont really have the time to properly understand everything compared to how i was trying to learn when i did it as a hobby

[sturdy spear]

For real?

[tardy hedge]

@rocky cloak we need ur input big bro

[glad osprey]

in 10 years you're gonna be big bro, kiand you're gonna be jagr2809

[thorn jay]

Jesus christ how many past jagr's were there and how long has it been going on for

[tardy hedge]

Omg ur right

[trail cave]

Can someone help with part (i) here? I know d1, d2 are associates iff they divide each other

I'm trying to use the first part of the definition but am confused

[tardy hedge]

Jagr is ong the goat though

Respect for boytjie too for sure but Jagr is goat status

[rose ridge]

seems like part (ii) of the definition would help you get to your 'iff' condition

[hidden wind]

it’s almost 2 am here in norway, i am not listening to the high one’s advice woops

[tardy hedge]

If K/F is algebraic, does the algebraic closure of F always contain K? Or is it that it contains a subfield isomorphic to K?

In what space does the algebraic closure lie in ..?

[trail cave]

Suppose all a_i | c, then d_1, d_2 | c. So c = d_1b, d_2e for some b, e. Hence d_1b = d_2e. Where do I go from there? I can't use inverses to get d_1b = d_2eb^-1 since this is only an integral domain right?

[hidden wind]

here’s a (very) loose translation of 23.

Foolish men
are awake all night
thoughts everywhere.
When morning comes
they are sleepy and tired
and all is tangled again.

ahem sry i’m way off topic

[rose ridge]

note that d_1, d_2 are both c's that satisfy the condition of (ii)

and yes, in general inverses dont exist in domains

[thorn jay]

In the (formally a class) lattice of field extensions

[thorn jay]

I mean K is a subfield of F no?

[trail cave]

I see, ty

[tardy hedge]

Right yeah

[thorn jay]

So if F is a subfield of it's algebraic closure, then by how set inclusion works, K must be a subfield of F's algebraic closure

[tardy hedge]

F is subfield of K in my example right

But alg closure of F should include K because an element in K is algebraic over F so its in F bar by definition?

i guess who cares about talking about "isomorphic" because its just relabelling so who cares i guess?

[rose ridge]

i think 'contains a field isomorphic to K' is the right way to go in general

after all the algebraic closure itself is only unique up to iso

[rose ridge]

the main issue here is that there are multiple ways to add roots of an algebraic equation

but yea in practice like you said its just a matter of relabeling roots so theres not much harm in assuming that the closure of F contains K

[glad osprey]

The notation K/F means that K is a field extension of F, so F is a subset of K. I prefer the notation F⊆K

[rose ridge]

a simple way to do this is to take the algebraic closure over K in the first place

[coral shale]

I've seen and like the notation K : F

[royal shadow]

if a field K embeds in L and L embeds in K are they isomorphic?

[lusty marlin]

Yes

[royal shadow]

holy crap thats based

lets go field theory

how difficult is it to prove ?

seems like its not true: https://mathoverflow.net/questions/89718/classes-of-fields-and-cantor-schröder-bernstein

[tough raven]

me

[tough raven]

It's dangerous to call this a lattice.

The join isn't well-defined (invariant under isomorphisms of field extensions).

[tough raven]

For any algebraic extension K/F and any algebraic closure L/F, there exists an embedding K → L over F. So the latter: L includes a subextension which is isomorphic to K (as an extension of F).

[tough raven]

Probably not. If thes embeddings are as algebraic extensions of a common subfield E then yes, I think so.

[lusty marlin]

Oh damn
Sorry about that

[royal shadow]

that is no problem my friend

I stumbled down a nice rabbit hole

[mental plaza]

Hello, I need a recommendation for abstract álgebra resources.

I left uni several years ago, but I used to have a solid mathematics undergrad background. Can someone point me to the right resources ?

[tough raven]

What topic within abstract algebra?

[royal shadow]

Dummit & Foote is a book I see a lot of people recommend

covers advanced topics aswell but starts from the basics, its like 900 pages so quite a lot of stuff in there

[pearl fog]

Uhm consider extension e^i(2pi/7) over Q, x^7-1=0 but is reducible, so min poly is x^6+x^5+...+x+1 but this says extension degree is 6?

[royal shadow]

why do you say its reducible?

[mental plaza]

I need to brush up on groups Fields and rings, then I need something that pushes further. I’m pretty solid in linear algebra.

[mental plaza]

Thanks I’ll check it

[tough raven]

So the minimal polynomial and the extension both have degree 6. Sounds correct.

[tough raven]

I would second the recommendation of Dummit & Foote then. For field theory I would also suggest Milne's notes on field and Galois theory.

[royal shadow]

oh right yeah im dumb maybe I need to go study more

[tough raven]

Well, I don't think studying more fixes not noticing 6 ≠ 6, but taking a break might.

[pearl fog]

But the basis are 1, e^i2pi/7, e^i2pi2/7, ..., e^i2pi6/7 7 in total?

[mental plaza]

Thank you so much

[royal shadow]

from your minimal polynomial you get e^(i2pi*6/7) as a linear combination of all your other powers

[pearl fog]

x^7-1=(x-1)(x^6+x^5+...+x+1)?

[tough raven]

That's the set of 7^th roots of unity, but (perhaps surprisingly) it is not a basis.

[pearl fog]

Oh sure

Bruh

[royal shadow]

💀

[tough raven]

The sum of all n^th roots of unity is 0, which gives you one linear relation.

In general, the degree can be less than n-1, so there are even more relations.

[pearl fog]

Well I'm only considering prime power here so I think I'm good

[tough raven]

Even for prime powers, it's not n-1.

The degree for p^k is p^{k-1}(p-1).

[pearl fog]

I see by cutting x^p^k-1+...+1 apart into sections of length p then it's irred from eisenstein?

[royal shadow]

https://math.stackexchange.com/questions/1257071/splitting-fields-and-their-degrees

[static mauve]

Been trying to understand the line that is underlined in red

[crystal vale]

what's your doubt here?

[static mauve]

that xa^-1 is an element of (aH)(a^-1H)

I feel like that is sorta what the book is getting at, but I havent been able to show this out with individual elements

[crystal vale]

you know what does it mean by when operation is well defined?

[static mauve]

It means you can "trust is" it to true, right?

Maybe you are getting at something more specific?

[crystal vale]

trust? How can you show some operation is well - defined?

if you have to show some operation is well defined you have to show if a = c and b = d then ab = cd, where ab defined as a operates with b

[static mauve]

Ah, I had no idea

[crystal vale]

so here they assumed that given operation in G/H is well defined

[crystal vale]

so if x in aH that's means xH = aH, by definition of well defined, (xH)(a^-1H) = (aH)(a^-1H) => xa^-1H = H

[static mauve]

Ok, so not realizing that xH = aH is where I specifically was going wrong, I think.

This seems so obvious now 🙃
@crystal vale thanks for the help!

[crystal vale]

you can also at Dummit and foote, it is exactly same but the idea is why do we need N to be normal to make G/N group with given condition, so both are equivalent

[royal shadow]

if you define a function that takes as input a group of people and returns a name by: pick a random person in that group and return their name, it is not a well defined function because you dont get the same name every time you use your function, same with multiplication of cosets, you define multiplication of cosets by picking some random elements from coset 1 and coset 2 and multiply them together and then go back to cosets right but theres not reason to believe it is well defined because it could depend on which elements you pick. it turns out it doesnt matter if the subgroup is normal

its like if you feed this name function a group of people that all have the same name it becomes well defined, doesnt matter which random person is picked

[static mauve]

Yeah, that totally makes sense. Thanks for the clarification!

[thorn jay]

Oh cant read 💀

[lusty marlin]

Yes it is nice

Makes one think about which objects in general do not satisfy a CSB type property

[rocky cloak]

It can be tricky to understand everything up front. Often it's easier to understand something when you have a wider context from things you learned later.

Then you just need an excuse to think about things you "already know". Teaching or hanging out at these sort of math forums works.

I guess it's also worth it to spend some time trying to connect new ideas to old ones. Does what you're learning look like anything from e.g. linear algebra? If so, is the result expected unexpected. Helps with remember the new ideas, and also gives deeper understanding of the old ones.

[tardy hedge]

The last sentence there

I know by fundamental theorem that quotient group is the galois group for the subextension over Fp but why is it generated by image of frob p

I guess it makes sense actually

[hidden wind]

which book is this?

[sturdy spear]

looks like D&F

[tardy hedge]

Dummit and foote

Is a finite degree separable extension Galois?

In dummit foote it says "finite, separable and normal extensions (are galois)"

Oh so the normal extension part is for the splitting field condition sort of thing

You could have an example of a finite separable extension that is not a splitting field, and so not Galois?

Like Q(cube root 2) over Q ?

That is finite so algebraic and separable because every polynomial over Q is separable

But not splitting field

[glad osprey]

Yep exactly, Q <= Q(cbrt(2)) is finite and separable, but not Galois

[frigid epoch]

S_5 not solvable implies Abel-Ruffini

Does S_6 having a unique automorphism imply anything for polynomials?

[tardy hedge]

if f is irreducible it must be coprime to its derivative Df so f is separable, except for the one case where Df is identically 0?

[rocky cloak]

It appears to say something
https://math.stackexchange.com/a/1630600/306319

[hollow tartan]

couldnt find the dfn of a G set anywhere else, but according to nlab its a continuous group action on a discrete topological space, but didnt understand the part that says its a free coproduct completion of G-orbits, from what i read its like a disjoint union of cosets? to put it intuitively?

[kind temple]

a G set is just a set X which affords a permutation representation

i.e., a group homomorphism G —> Sym(X), or in other words, a group G acting on a set X

[hollow tartan]

yes this is from the discrete topology kinda makes sense i guess

this group action stuff is very interesting, how rich is something like geometric invariant theory in studying this? (kinda got recd this)

[kind temple]

i would presume quite rich as wikipedia describes geometric invariant theory as the study of group actions on algebraic varieties or schemes to construct quotients

[hollow tartan]

yeah it sounds too good to be true haha

since very large finite groups are considered a complete mistery

[flint raven]

I'm just starting group theory and learned that the additive group Z is a cyclic group? I'm not sure I understand how that is possible as there seems to be 2 generators needed, 1 and -1. Are inverses considered as a single element in this context?

[wraith cargo]

no?

it's generated by 1
That means you can add and subtract it since inverses exist

[velvet hull]

generation includes for the operation of taking inverses as well

it's not just the "forward" direction

[acoustic igloo]

it inclues the identity too

[flint raven]

oh okok thanks guys

[hollow tartan]

think of it as a ring, quite literal, no pun intended

[south patrol]

Ye the subgroup generated by an element g is best defined as the smallest subgroup containing g

And indeed that reflects the terminology

[velvet hull]

when you phrase it that way, you can probably recharacterise that statement in terms of a universal property

[south patrol]

For finite groups you can get away with just taking powers as each element has finite order anyway

[velvet hull]

idk why I never made that connection

it is just the free group on the generating set mod out the presentation of the ambient group anyways

[south patrol]

Ye

[frigid epoch]

As a monoid, Z is generated by 2 elements :>

[south patrol]

I guess here viewing it as a universal property is slightly awkward but it is definitely a more universal or invariant characterisation

[velvet hull]

its slightly more awkward since there are existing relations but it sounds doable

[south patrol]

Ig here I would be pedantic and say that "G is generated by n elements" doesn't define n well lol

As in you can say Z is generated by n elements for any n >= 1 lol

[hollow tartan]

sounds like someone like cat theory (cool)

[south patrol]

But ye

[frigid epoch]

Minimum number of generators

[south patrol]

Did you know that Z \ 0 is the free monoid on countably many generators

[velvet hull]

anyways, could someone sanity check my proof here?
I can't see where I went wrong with it, but I seem to have proven that if aM = M as (finitely generated) modules for any single ideal a in the base ring R, then M is trivial:

https://cdn.discordapp.com/attachments/1281763418952699975/1348338901886046268/image.png?ex=67d265b5&is=67d11435&hm=b6e4b9d52402a165afe2a13e28d2e03b826ad94e17f9890b58dba4508921a3e0&

https://cdn.discordapp.com/attachments/1281763418952699975/1348338956340695131/image.png?ex=67d265c2&is=67d11442&hm=f5cfaddda041e61712bbdd1ef0a73ca5de718df6560960e22324a72febb35155&

the only place where the proof seems sus to me is the "A in M_n(R) is invertible iff det(A) is a unit", but that sounds like a reasonable assumption to make without giving it more thought

tl;dr my proof somehow avoids using the local assumption but still looks correct to me so im confused

[rocky cloak]

So what you're proving is that the determinant of I-aij is 1 modulo m.

But this doesn't imply it's a unit if m is just an arbitrary maximal ideal.

For example 3 is 1 modulo (2) but is not a unit in Z.

So what you have to use is that because A is local, being a unit modulo m means being a unit (or more generally, not being in m means being a unit)

[hollow tartan]

isnt this just proving a_ij not being I, if it were then is the whole thing

or is it more subtle

[rocky cloak]

I mean, the 0-matrix is not the only matrix that isn't invertible

I - aij can be noninvertible without aij being equal to I

[frigid epoch]

N \ 0 is free, -1 makes it unfree 🙂

[thorn jay]

Z \ 0 ≈ Q^(Q) x C_2

[frigid epoch]

Yep

[thorn jay]

Or well

[frigid epoch]

Well N^(N)

[thorn jay]

With the right free monoid structure on Q^(Q)

[tough raven]

Is Q^Q a new emoticon?

[south patrol]

Ah ye I'm dumb. I was gonna say N \ 0 and then was like eh -1 is f8ne but obviously squares to 1 lmao

[tough raven]

Horn-rimmed glasses with the horns at the bottom and pointing to the same side...

[thorn jay]

Youre right

Free product and all that

[thorn jay]

Just enlarge your type to include the constant -1 which squares to 1

Then, given that the monoids are commutative im inclined to believe Z \ 0 is free

[velvet hull]

Wait, so you’re saying that a unique maximal ideal, if it exists, is exactly just the set of all nonunits?

I guess it makes sense

[hollow tartan]

Oh that's a good point

[next obsidian]

Yes, being local is equivalent to having the nonunits form an ideal

[velvet hull]

it feels weird, lmao like this is too simple of a characterisation

[hollow tartan]

Just by dfn kinda makes sense bc of 1

[velvet hull]

so to finish the proof, is just to say that 1 = det(...) can't be in m, therefore det(...) is a unit and the matrix is invertible (by locality)

[rocky cloak]

A ring if local if any of the following equivalent conditions are true:
It has a unique maximal left ideal
It has a unique maximal right ideal
It has a unique maximal two-sided ideal, and the quotient by that ideal is a division ring
The sum of non-units is again a nonunit.
For every x either x is a unit or 1-x is.

[velvet hull]

ok that tracks

[tardy hedge]

Yuh

HChan approves ✅

[hollow tartan]

I'd love to see your ref bc hungerford pg 147 has fewer equivalent statements

[rocky cloak]

Which of them are in hungerford?

[chilly ocean]

$\bZ$ is also the pushout(in Grpd) of ${0\leftrightarrows1}\leftarrow{0, 1}\rightarrow{•}$ , pretty interesting

[next obsidian]

jagr doesn’t subscribe to [x,y] = 0 gang which means jagr is different from Chmonkey

[cloud walrus]

NAT Enthusiast

[rocky cloak]

We may be different, but do we commute?

[next obsidian]

I think so

❤️

[rocky cloak]

radical

[hollow tartan]

the gtm version i think

[rocky cloak]

gtm?

[tardy hedge]

Graduate text version

[hollow tartan]

yeah the yellow books

[tardy hedge]

GRADUATE TEXTS IN MATHEMATICS

[rocky cloak]

Okay, but which statements?

[hollow tartan]

it's almost no difference, but i like your version it's a bit more precise, thats why i asked

[rocky cloak]

Alright, I might try to look up a reference tomorrow

[tardy hedge]

Jagr the 🐐

[tardy hedge]

THE GOAT THE GOAT I TELL YOU

I will now stop because words are meaningless if they are often repeated

[trail cave]

Could someone help on this? I've tried some arguments but they're not going anywhere I think. My professor gave an argument for when we have a^2 + b^2 = p a prime but 145 isn't prime. I was thinking that maybe the fact 145 isn't prime implies a+bi isn't prime (where it's norm is 145), hence not irreducible, so we can factor then look at the norms of the factors as 29 and 5. But that seems wrong

[south patrol]

Use unique factorisation ye

[trail cave]

Following the argument my prof. gave and using unique factorization I have something like:
Let a+bi be a Gaussian integer with norm 145. 145 is not prime in the integers, so since the Gaussian integers are a PID, a+bi is not prime and thus not irreducible. Hence a+bi = (c+di)(e+fi) for some Gaussian integers as shown. This is unique, since the Gaussian integers are a UFD. WLOG, let the norm N(c+di) = 29. Then N(e+fi) = 5.
(from here my professor gave an argument like this, but I don't really know what quadratic residues are since I've never taken number theory):
c^2 + d^2 = 29, so c^+d^2 is congruent to 0 mod 29. So there exist q such that cq is congruent to 1 mod 29. Hence q^2c^2 + q^2d^2 is congruent to 0 mod 29, and q^2d^2 is congruent to -1 mod 29. So -1 is a quadratic residue of mod 29.

[south patrol]

The last bit is a little overcomplicated: if c^2 + d^2 = 29 then wlog c is not divisible by 29 and then (d/c)^2 = -1 mod 29. But also like idk what proving that -1 is a quadratic residue has to do w the conclusion

I would start by factorising 145 within the Gaussians

And then you can use uniqueness there

To get a complete (finite) list of possibilities

[trail cave]

I'll try that

thank you

[keen badge]

If $[x,y]$ is the commutator of $x,y \in G$($G$ is a group) then I need to prove that:
$$
[ab,c]=(b^{-1}[a,c]b)[b,c]
$$
I tried to expand it, but it looks wrong... any help?

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[delicate orchid]

b^-1a^-1c^-1acbb^-1c^-1bc = b^-1a^-1c^-1abc = [ab, c]

think this works

if you do the commutator the other way I agree it breaks

[keen badge]

But the commutator defined as [x,y]=xyx^(-1)y^(-1)

[delicate orchid]

not my problem

[wraith cargo]

yeah then it isn't true

[keen badge]

OHH WELL, thanks

[noble adder]

hi all
i made it to a summer program (yay!) and will probably be doing a research paper on algebraic number theory

i dont have any idea what the specific topic is, but i think i need to build some prerequisites: specifically, field theory and galois theory

i've just done ring theory and the second semester is field and galois theory in the class im taking right now but i want to get ahead of the game so how unrealistic is it to study chapter 5 in hungerford's abstract algebra in ~6-7 weeks if i work hard

kinda a soft question but i didnt know where to ask it

[tardy hedge]

What is chapter 5?

[violet spade]

Hi, I've already done part (a) of this question. I'm struggling with (b), trying to prove that g is in K[X]. If g is in K[X], then we can say that #X(L/K)=[L:K], hence g has correct degree, and we can choose varphi(alpha)=alpha so g(alpha)=0 for some alpha. Could anyone please give me a himt to prove this is in K[X]?

Like, I know I'm supposed to use part (a)

but yeah i'ts not entirely obvious that phi applied to the coefficients is g again

[rocky cloak]

What's the relationship between the roots of a polynomial and its coefficients?

Remember that phi preserves addition and multiplication

[void cosmos]

i understand how to finish the proof if i show that the image is discrete.
i don't understand how we get that |g+mg_1| < |g_1|/sqrt(2)
if we have a g in B so that p(g) < |g_1|/2, then the element cg_1+g (c in R) projects to g in W. they are of the same norm. now im stuck and i cant follow

do i get to pick any m i want? it doens't matter which m do i pick cuz it still projects into g in R.g_1 right

i think i can show that the norm of g+mg_1 is less than the norm of g if this m has norm less than 1/2

does this work

that is, consider g in B with p(g) (the orthgonal projection onto the orthgonal complement of Rg_1) has norm less than |g_1|/2. consider the element x = g+mg_1. this is in B and has norm <= |g+mg_1| <= |g_1|/2 + |g_1|/2 ?

[rocky cloak]

It's worded kinda confusingly, but I think the point is that
g = rho(g) + xg1 for some real number x.
By adding some integer multiple mg1 we can make |x| <= 1/2.

Then
|g + mg1|^2 = |rho(g)|^2 + x^2 |g1|^2 < |g1|^2 / 4 + |g1|^2 / 4 = |g1|^2 / 2

So g + mg1 has norm less than g1, contradiction

[void cosmos]

g=rho(g)+xg_1 where rho(g) is the orthgonal projecton onto the orthgonal complement

we may wlog (probably similar to what i said above?) assume |x| <1/2

and then what u typed out follows

yeah got it

tysm king

[noble adder]

fields and galois theory

[velvet hull]

Honestly, 6-7 weeks for Galois theory sounds decent

[noble adder]

ok, thank you!

[elfin prairie]

Hello. I need some rudimentary intuition of Borel Subgroups and perhaps from there work towards understanding the definition.

[void cosmos]

im sorry but where did we use the fact that the graph is a closed (and hence a submanifold)?

ig if i dont have that it is a submanifold it doens't make sense to talk about Lp? (Lp is the induced map in the lie algebra/tangent space at e)

[tough raven]

I think that's it.

[void cosmos]

yeah

so to recap the proof, the idea is that the projection induces a bijection on the tangent spaces and hence by the inverse function theorem they are local diffeos?

but if htye are inveritble local diffeos then they are diffeos?

and hence f is just a copmosition of diffeos which is a diffeo?

@tough raven

[tough raven]

I think so. To be really precise, the inverse function theorem tells you that the inverse of Gamma_f → G, namely G → Gamma_f (which is a priori only continuous) is differentiable.

[void cosmos]

i see

thank you so much

[tardy hedge]

(b), it shouldve said irreducible over F right

[tough raven]

Yes, it splits over K since it has p distinct roots a, a+1, ..., a+p-1 = a-1.

[tardy hedge]

How do I know that a to a+1 extends to an automorphism of K? I saw examples that for K fg over F, you need to always send generators to another root of the min poly over F , but you may not just choose it arbitrarily cause it may not be an automorphism

I have been trying to understand when this works or not

The example was like for the splitting field of x^3-2 over Q. Q(cuberoot(2), p(cuberoot(2)))

[tardy hedge]

Also, since [K = F(a) : Fp] = p, we only can say K is Fp^p because we know there is exactly one finite field of order p^p?

[acoustic igloo]

what is an internal direct product?

i think it's a way of saying G is isomorphic to HxK

but it's not actually a set of definite objects

but idk if that's right, i'm guessing

[tardy hedge]

That polynomial x^(p^n)-x over Fp is separable but its not irreducible right

because x = 1 is always a root

[tardy hedge]

For (e), how many elements are Artin-Schreier? Does that just mean how many elements of Fp^p are roots ? Isnt it just p many because its degree p

[tardy hedge]

Yeah, im not totally sure what question e) is asking for

[next obsidian]

It’s a silly thing

It’s when you find two normal subgroups H,K such that H\cap K = 1 and HK = G

If this is true you get that G ≈ H x K

[acoustic igloo]

that makes sense

thanks

[west sinew]

If we look at (x) (the ideal generated by x) and k[x] as modules over k[x], are they isomorphic k[x]-modules? k is some field

[next obsidian]

Yes

[sturdy spear]

I wanna prove that: Let $\sigma = (12\cdots m)$ be $m$ cycle. Then $\sigma^i$ is $m$ cycle as well if and only if $\gcd(m,i)=1$.\
Assume $\gcd(i,m)=1$ then notice $\sigma^i(1)=i+1$, $\sigma^i(2)=2i+1$ ans so on upto $\sigma^i(m)=mi+1\equiv 1$. Wlog assume $i\in {1,2,\ldots, m}$. Since gcd is 1 so $i\neq m$. Now $1\leq i\leq m-1$ implies that $ai+1 \not\equiv_m 0$ so $ai+1 \in {1,\ldots, m-1}$. Finally it suffices to $i+1,\ldots, 1$ a4e distinct. If $ai=bi$ then $(a-b)i=0$. Since $i\neq 0$, $a=b$.

[cloud walrus]

Abstract Afzal

[sturdy spear]

Does it make some sense?

ans and

[vagrant zinc]

Name of book ?

[west sinew]

How? The map $x\hookrightarrow k[x]$ doesn't even seem to be surjective

[cloud walrus]

adi

[vagrant zinc]

x to which algebraic structure does it belong?

[next obsidian]

That’s not the map you use

[west sinew]

Oh. Do you do it the other way? $k[x]\rightarrow (x)$, with the map being multiplication by $x$? This seems to be surjective with trivial kernel.

[cloud walrus]

adi

[next obsidian]

Yes

[west sinew]

Cool thanks. It's a bit odd that no map from $(x)\rightarrow k[x]$ is forthcoming, though...

[cloud walrus]

adi

[next obsidian]

Yeah there is, divide by x, it’s just the inverse

[west sinew]

Oh! Yeah my bad

[sturdy spear]

I have seen some people writing

,, \sigma^i(a_k)=a_{(k+i)\mod m}

[cloud walrus]

Abstract Afzal

[sturdy spear]

(sigma is m cycle), but in the book author says if k+i>m then take mod m

[vagrant zinc]

Dummit

[sturdy spear]

Yes

[static yew]

for the field of 4 elements using {0, 1, a, b}, the valid bases are:

{1,a}; {1,b}; {a,b}*

and no others, right?

*

[bitter vigil]

Why are these the same field?

I don’t see how root 5 + root 7 is in the second field

Like in general when is it true that like

R[alpha]R[beta] = R[alpha,beta]?

[tardy hedge]

What is Q(sqrt5)Q(sqrt7)

Elements there are products ab with a in first b in second..?

[tardy hedge]

In dummit and foote KH for K and H fields is the smallest field containing both K and H

[tough raven]

Always. For additive subgroups H, K of a ring R, HK almost always means the additive subgroup generated by {hk : h ∈ H, k ∈ K}, i.e., {h1 k1 + ... + hn kn : hi ∈ H, ki ∈ K}.

[tough raven]

In field theory with subfields it's sometimes extended even further like this (though this only makes a difference when both H and K are transcendental over H ∩ K).

[woeful sage]

kinda feels like the product of ideals

[languid trellis]

Can anyone provide any intuition for the definition of a solvable group?

[coral spindle]

Think of the composition series. A solvable (finite) group has a composition series consisting entirely of cyclic groups.

[languid trellis]

I agree : )

I think I asked my question poorly, I would like to know why we defined "solvable" to begin with

[coral spindle]

You are aware of what a solvable extension of fields is, I hope?

We can split up a solvable extension of fields into cyclic extensions

In turn, the Galois group is split up into cyclic subquotients

This correspondence defines solvability on the two sides

[languid trellis]

Oh, I haven't gotten there yet. I think that's in the next chapter

I'll ask again soon

[coral spindle]

Spoilers!

Well yeah there you go. The point is that you get cyclic extensions when your field is solvable.

field extension* is solvable

[languid trellis]

Gotcha.

thank u pookie bear

[coral spindle]

May u always be swift, eeeeeeeeeeee

[delicate orchid]

they're like metacyclic groups but you can make a big sequence instead of a short exact sequence. This means that they are nice.

[tardy hedge]

Why is every algebraic extension a union of finite extensions?

[coral spindle]

Let E:F be algebraic. Let a in E. You agree that F(a) is a finite extension?

[tardy hedge]

Yeah

[coral spindle]

Take the union of all such things.

[tardy hedge]

Yea fair enough lol

[thorn jay]

:0 does this mean that the lattice of algebraic extensions of F is algebraic?

[tardy hedge]

the composite of all quadratic fields is Galois over Q because its the splitting field of a set of separable polynomials over Q?

those being some x^2-a things?

I am looking at Galois theory for infinite field extensions bc i have to do a presentation on it

Can someone help me with this example? Im not totally clear why we can say its the countable direct product of Z2

and what does it mean by "not direct sum!"

I'm also not sure why Gal(L/Q) is uncountable

[next obsidian]

Because it’s a direct product of counts my many Z/2Z

This is a cardinal arithmetic thing

[tardy hedge]

Yeah im quite unfamiliar with that stuff unfortunately

[next obsidian]

If it was a direct sum only finitely many of the spots can be nonzero, so in this case it means that you can flip the sign of infinitlely many of the sqrt(p) in a single automorphism

[next obsidian]

Look at what happens when you take sqrt(p) for p < N, and then take the limit as N -> infinity. You end up taking the limit of (Z/pZ)^{p(N) + 1} where p(N) is the number of primes <= N

[tardy hedge]

Hmm tbh im kind of confused by both of that ngl

[tardy hedge]

In the finite case, are direct sum and direct products the same? But for infinite collections, direct sum means finitely many are nonzero?

[next obsidian]

Yes

[tardy hedge]

I know that each automorphism must take sqrt(d) to some root of the min poly of sqrt(d) over Q

So there are only two choices for each, positive or negative square root

But im not sure why we know each choice extends to an automorphism

[next obsidian]

It sounds like u have math to do

[tardy hedge]

Is every element there a finite Q combo of sqrt(d) for some d's?

[tough raven]

Within a fixed algebraic extension, yes. This is because an intermediate field is the same as a subalgebra over the base field.

[thorn jay]

Yes thats what I meant haha

[acoustic igloo]

is this an error?

[thorn jay]

No

[tough raven]

This does use the relatively easy but still non-trivial fact that intermediate rings are automatically fields.

[thorn jay]

Theyre identifying T with the corresponding elements in Q(R, T)

[tough raven]

Mfw non-zero element of T even though 0 ∉ T

[acoustic igloo]

yeah T was already defined not to contain 0

[thorn jay]

So a ring R with F < R < E and F, E fields and E/F algebraic is also a field?

[tough raven]

It has to be an algebraic extension.

E/F

yes

[thorn jay]

Huh.. I'll think of this a bit

[acoustic igloo]

yeah it makes sense actually because some elements of R won't have inverses

some elements of Q(R,T) won't have inverses

you're right

[thorn jay]

Ohh of course, because let a be an element in R \ {0}
Then F(a) ≈ F[x] / I, so a^-1 is some polynomial of a with coefficients in F, in particular, a^-1 in R, meaning that R is a field

[thorn jay]

In fact, you can see it as just adding the inverses of the elements in T

(And then making everything well defined as a ring)

Thats why T cant contain 0, you'd just collapse everything to the zero ring

And T being multiplicative too, as if a and b have inverses, then ab must have one too

[tardy hedge]

What the heck kinda book is this

Ive never seen a text tell me “you know what, think about it first for about 15 minutes or so, “

[acoustic igloo]

it's kind of funny, there was a true/false question: "the theorem of lagrange is a nice result"

[tardy hedge]

Hahaha

[acoustic igloo]

lol

[tardy hedge]

Ya i saw a final exam from my undergrad have a question saying “state the most important theorem learned in class” (points given for good taste)

Something like that lol

[thorn jay]

True

But orbit stabiliser theorem is better

[thorn jay]

What class was this lol

[tardy hedge]

Dont remember, i think it was an analysis class though (Gross!!)

(Stinky!!)

(PEEEYEWWW)

😎

[thorn jay]

Ewwwww

Do you recall what you answered?

[thorn jay]

They reeeaalllyyy wanna make sure that your given element isnt 0

[tough raven]

The book is my spirit animal

[tough raven]

True. They've also required it not to contain zero divisors (which makes not containing 0 redundant unless R = 0 to begin with), so that the map R → Q(R, S) is injective.

[thorn jay]

I personally think we should just let T be any subset and refrain from this constructive behaviour

Let the universal properties reign

[tough raven]

It's the opposite of constructive

But avoiding zero-divisors simplifies the definition of Q(R, S), which is good the first time one sees localisations.

[thorn jay]

This is constructive

[thorn jay]

Yeah thats fair

[tough raven]

Assuming 0 ∉ T makes it less constructive.

Anyway

[thorn jay]

They literally say "following the construction in this section"

:P

[rocky cloak]

Classy exercise. But I'm guessing they give full marks to everyone who states any theorem from class(?)

[thorn jay]

Imagine someone accidentally states a lemma

Then it's up to the personal definition of "theorem" for the teacher

[tardy hedge]

Later ill try to find the exam and show u guys

[thorn jay]

Okay!!

[rocky cloak]

Lemma is a one-off theorem

Proposition is a theorem we didn't bother proving

Theorem is the reason we wrote this paper

[acoustic igloo]

can i assume that "a" is nonzero?

[delicate orchid]

Lemma is a result used in later results
Proposition is a result not used in later results, but is still worth mentioning (e.g. for an example, or context)
Theorem is the reason we wrote this paper

this is my heuristic

[tardy hedge]

In a proposition u basically just propose things 🔥

Lol^

[frigid epoch]

imo proposition is stuff like "this construction has all the desired properties"

like quotient ring is associative and distributive and everything

[thorn jay]

I generally use propositions for fairly basic facts that still need to be proved but aren't necessarily worth calling a theorem, but it's also not for a specific theorem so I wouldn't call it a lemma
e.g. "so and so defined maps are morphisms of closure systems"

[thorn jay]

yes exactly

[thorn jay]

homework paper ahh

[frigid epoch]

proposition: L is an inner model of ZFC

[thorn jay]

false, L is an algebraically closed field

[delicate orchid]

that's fair

[thorn jay]

unless proving it is an actually worthwhile result and takes more ingenuity than a would be homework problem

[tardy hedge]

For a Galois group acting on the roots of a separable polynomial, each orbit necessarily corresponds to permutations of roots of one irreducible factor right?

So the # of orbits is also like the # of irreducible factors of that polynomial

Also not sure if i needed to say separable there^

[tough raven]

Hmm. The automorphism group of any normal algebraic extension permutes conjugates transitively, yes.

[tardy hedge]

A normal extension over F is an extension that is a splitting field for a set of polynomial over F?

Is it redundant to say normal “algebraic” extension then? If you have K/F where K is the splitting field of some set of polynomials over F, would K not already be algebraic?

[tardy hedge]

Nice emote lol

[tough raven]

: mathgtmheart:

[tough raven]

Probably, but I said algebraic just to be sure.

[tardy hedge]

I feel u

Galois theory has so many terms

[frigid epoch]

better than AG

[tardy hedge]

AG just seems so hard

[frigid epoch]

flat étale unramified proper regular

[frigid epoch]

it's not even about anything good tbh tbh

[tardy hedge]

When ur learning that sort of stuff is there even a geometry picture associated to it or like

[frigid epoch]

i think étale is the same as covering map in topology

[tardy hedge]

I have a book called the geometry of schemes

One day ill read it

[frigid epoch]

not to be confused with étalé

[tardy hedge]

I see switch friend code

R u excited for new console?

[frigid epoch]

yeah a bit

[tardy hedge]

Im excited for new games

Im a nintendo fan but havent cared for so long

[frigid epoch]

hoping pokemon ZA will be good

[tardy hedge]

Ong i liked Wii U era better

[frigid epoch]

hoping MK9 will be great

[south patrol]

Good book

[tardy hedge]

Me too

[south patrol]

Yeah you are a good book

[frigid epoch]

switch is a failure after like the first two years

[tardy hedge]

Haha thats such an unpopular opinion but i agree

[frigid epoch]

and even then
SMO < SM3DW
SSBU < SSB4

[tardy hedge]

True

And totk was disappointing

[frigid epoch]

absolutely

[tardy hedge]

Classic zelda moment where it got such praise the moment it released then everyone turned on it later lol

Like skyward sword

[thorn jay]

I mean in a sense it's the algebra most close to highschool algebra, as it is at least loosely related to solving equations

Perhaps not post-Grothendiecken AG

[tardy hedge]

Is Galois theory about solving equations

[frigid epoch]

yeah but tbh algebraic curves just dont excite me

honestly silly things like Spec(Z) are kinda more fun to think about

[thorn jay]

I mean you can always do universal algebraic geometry

[tardy hedge]

Whyd i even ask that

[frigid epoch]

but even then there's such a flood of terms

[frigid epoch]

it's about NOT solving equations

[thorn jay]

Galois theory is solving equations when everything is all nice and happy

Imagine trying to solve equations over sets

Or magmas

[frigid epoch]

who even INVENTED magmas

where do they even show up...!

[thorn jay]

Universal algebra

[south patrol]

Galois theory is making strong assumptions on fields so everything works

Jk

[thorn jay]

👍

[south patrol]

Now the question is where universal algebra turns up

[frigid epoch]

lol

[thorn jay]

I actually stumbled upon it through Latin squares, which has to be the genuinely most "me" thing to ever happen

But I'd argue any time you wanna do something with solving and solvability of equations, lattices and universal algebra show up

[frigid epoch]

lol

and when do you wanna solve equations?

[thorn jay]

Also as a general framework and toolbox to do algebraic structures on

[thorn jay]

Idfk man why does galois theory exist

[frigid epoch]

lmao

[thorn jay]

I think something as rudimentary as solving equations but operators aren't necessarily not horrible to work with is a fairly good motivation

-# regardless of whether or not thats why you got interested in the first place

[tardy hedge]

Woah cool text effect

[thorn jay]

Yeah

-#

Just like # enlarges it

I AM BIG

[south patrol]

That is quite cool

[thorn jay]

I wanted to prove some impossibility of a certain algebraic structure of a given cardinality existing (Euler's conjecture for order 6) and needed a framework for studying more universal algebraic structures

[frigid epoch]

can you prove the impossibility of a field of order 1 existing

[thorn jay]

I did, in fact, not end up proving it
I was too dumb

[thorn jay]

Suppose field of order 1 exists

cry
Lets not suppose field of order 1 exists

Well the most harmful thing about the zero ring being a field is that 1 would be prime

Qwq

[tardy hedge]

I was thinking about this for a while. I feel like im close , but i havent yet used rhe fact that l is prime, and im not sure where that comes in

I know that any irreducible degree l poly over Fp must be a factor of x^(p^l)-x

[sturdy spear]

Can I ask, what book is this?

[tardy hedge]

Its my professors homework he wrote

Sometimes he gets stuff from dummit and foote though

[sturdy spear]

Oh got it. Thank you!

[tardy hedge]

this is one of those problems where im like daaamn i feel so close so i been locked in for hours

but i should take a break

[acoustic igloo]

The product of two units in Z_n is always a unit, right?
and the product of a unit and a non-unit is never a unit?

[tardy hedge]

The units of a ring form a group under multiplication

So that is true in any ring

[acoustic igloo]

thanks 🙏

[tough raven]

#「graduate-lounge」 moment

[tough raven]

Don't need to, since you will actually do it.

[acoustic igloo]

are these answers right?

i wasn't sure if it should be n in Z+ or n in Z

[tough raven]

You can write down an answer to this for any degree n. I think the answer simplifies for n a prime. So that might be all it is.

[tough raven]

It doesn't matter, since m/2^{n-1} = 2m/2^n i.e., you can always increase the exponent in the denominator by changing the numerator.

[tough raven]

Yes, but can you simplify the second one?

[acoustic igloo]

hmm

m/(2^n 3^(n-1))

[tough raven]

That's true. But you can make it simpler if you change variables to m' = 2m and n' = n+1.

[acoustic igloo]

m'/(2^n' 3^(n'-2))

[acoustic igloo]

if it doesn't matter that means we should have an isomorphism from the field with Z+ to the field with Z
i wonder how to make this isomorphism

[tough raven]

... OK, I'm doing something wrong.

Anyway, what I meant is that the set is also {m/6^n : m ∈ ℤ, n ∈ ℤ^+}.

Because m/6^n = 3(2m)/6^(n+1).

[acoustic igloo]

ok i see what you mean

[tough raven]

You can construct the isomorphism using the very fraction rewriting procedure. For example, map m/2^n to m/2^n for n ∈ ℤ^+ and map m/2^n to 2^{max(n,0)-n} m / 2^{max(n,0)} for n ∈ ℤ.

[acoustic igloo]

oh they are actually equal as sets

i see

thanks 🙏

[thorn jay]

I mean, neither the algebraic geometries nor coherent conditions on Set are particularly interesting

As the free object on n generators in Set is just { 1, ..., n}, the coordinate "algebras" of a set X are just the finite sets of cardinality smaller than |X|, and the affine varieties are all isomorphic to X^n for some n

and the coherent conditions on Set are all principal, so basically the same as the algebraic geometries

i.e., nothing much to be solved here

🔥

[delicate orchid]

why? it's a beautiful encapsulation of schur-weyl duality. See further: https://ncatlab.org/nlab/show/field+with+one+element

[thorn jay]

Im still somewhat confused as to what properties and phenomena are being generalised as expressions over q with q=1

Unless im misunderstanding

[delicate orchid]

Well for one, the sylow structure. Which is what I care about

Number 2, it turns linear algebra into combinatorics. I.e. GL_n(1) is just S_n, F_1-modules are pointed sets, the F_1 representation ring is the Burnside ring

So on and so forth

The funnest fact is that they’re studying geometry over this object to try and prove the Riemann hypothesis

[thorn jay]

Thats really cool, alright
I'll let it sink in for a bit

[delicate orchid]

Let that sink in

[thorn jay]

Might be worth looking further into when I've finally bothered to learn rep theory

[thorn jay]

Yeah, cmon, it's cold outside

That poor sink

[tardy hedge]

there are (p^l-p)/l irreducible polys of degree l over Fp?

alright well googling shows that was indeed the answer, ok cool

I dont know what is different if l was not prime though

[tardy hedge]

Ok, I get why now

[tardy hedge]

So Gal has uncountably many subgroups of order 2, and if the correspondence worked how it normally does, each subgroup of order 2 should correspond to a field L > K > Q with [L : K] = 2?

But [L : K] = 2 would need K to be the composite of all quadratic fields but take away one sqrt(p)?

and there are only countably many ways to do that?

[next obsidian]

Yeah

[glad osprey]

I guess p^l - p wouldn't be divisible by l? Fermat moment

[rocky cloak]

This is true, but I think an easier justification is that

Gal(L/Q) = (Z/2)^N is an F2 uncountable F2 vector space, hence has an uncountable number of index 2 subgroups (just picking a basis an projecting onto a basis vector).

But there are only countable many degree 2 subextension of L because L is countable.

[tardy hedge]

Eeeh ok wait

The index of a subgroup should correspond to [E : Q] for a subfield K > E > Q right

Tbh i think there is a lot in that that i do not understand

[exotic anchor]

Hey guys, i have a quick question: I have to get all the possible tuples of (s2,s3,s5) such that these subgroups are a valid sylow-subgroup combination of a group of order 30. So 30=532, that is clear and i also know that these are the possible solutions: s2 is in (1,3,5,15); s3 in (1,2,5,10); s5 in (1,2,3,6). When i consider the equation s2+2s3+4s5+1 (for identity), does this have to be <= than 30 or =?

[keen badge]

If $H,K\le G$ and $H\lhd G$ with $H \cap K={e}$. Does it mean that $HK\cong H\times K$/

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[thorn jay]

No

Consider G = S_3, H = A_3, K = { 1, (1 2) }

S_3 is noncommutative, while H ≈ C_3 and K ≈ C_2, meaning that H x K is commutative

What you do have though, is that HK is a so-called semidirect product of K by H

[keen badge]

Say $H, K\lhd G$. When does $HK\cong H\times K$?

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[thorn jay]

Always

(If their intersection is trivial)

[keen badge]

ohh lol

How can I prove it?

I thought maybe doing something with $\phi:H\times K\to G$ defined by $\phi(h,k)=hk$ and then first isomorphism theorem, but I wasn't sure if it would work.

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[thorn jay]

I think the reverse direction is actually easier, somewhat

May be a little convoluted

[candid patrol]

Canonic isomorphism

[candid patrol]

You need |HK| = |G|

[candid patrol]

Bro what ?

[thorn jay]

Take the homomorphism
HK -> HK/H x HK/K
x -> (xH, xK)
Show that it must be injective and surjective (use the fact that conjugation is an automorphism)
Then use the third isomorphism theorem to see that
HK/H ≈ K/(H \cap K) ≈ K
HK/K ≈ H/(H \cap K) ≈ H
Thus finally HK ≈ H x K

[candid patrol]

Why would H being normal in HK ?

[thorn jay]

Hypothesis is that both H and K are normal subgroups

[thorn jay]

.

[candid patrol]

Ok but we don’t need this hypothesis

[thorn jay]

Only that H is in the normaliser of K and K in the normaliser of H ig

[rocky cloak]

One thing you can do is to consider the commutator of an element h in H and k in K. Using normality you can show that the commutator is contained in both H and K, hence is trivial.

So in fact HK = HxK iff both H and K are normal in HK

[acoustic igloo]

do polynomials over C always have solutions in C?

[tardy hedge]

yeah

C is algebraically closed

which means every polynomial over C has a root in C, which means it has all its roots in C

the fact that C is algebraically closed is called the fundamental theorem of algebra

[rocky cloak]

I guess
1 = 0
Technically is a polynomial equation without solutions in C

[chilly ocean]

fundamental theorem of algebra is for non-constant polynomials

[acoustic igloo]

oh yeah degree has to be at least 1

[tardy hedge]

holy moly new person

wassup Ricci ma boi

[chilly ocean]

hi boy

[tardy hedge]

Im here all the time asking about groups rings and fields

this place is my home

[chilly ocean]

i want to ask a ring theorey question

[tardy hedge]

go ahead

[rocky cloak]

[chilly ocean]

Let $R$ and $ S$ be distinct subrings of $\mathbb{Q}$, each with exactly two prime ideals. Show that $1/ 2$ belongs to at least one of $R$ and $S$.

$Q$ is PID these prime ideals are generated by some prime elements

[rocky cloak]

So 0 is a prime ideal, so these rings have exactly one maximal ideal.

So can look at the intersection of this ideal with Z.

[chilly ocean]

subrings are local rings

[rocky cloak]

These will be yeah

[chilly ocean]

can we take for example localization of Z at 5 and 3 are the local rings as subrings of Q

[rocky cloak]

I mean, those will be an example sure

[crystal vale]

I can show the upper bound, how to show the lower bound?

[chilly ocean]

lower bound is easy to show

just use the definition of lcm

m divides the quotient so is n it means lcm will also divide

[crystal vale]

Oh you mean m divides | G : H \cap K | ?

[chilly ocean]

yes

if you solved the first part it will be immediate from that

[crystal vale]

But why will it be true?

[chilly ocean]

how did you solve first part

it will be obvious from there

[crystal vale]

I showed that every right coset of H\cap K is the intersection of the right coset of H and the right coset of K

So it will be finite and upper bound by mn

[chilly ocean]

you cant do that with cosets

[crystal vale]

Why?

[acoustic igloo]

does this work?
[G : H cap K] = [G : H] [H : H cap K] = a multiple of m
[G : H cap K] = [G : K] [K : H cap K] = a multiple of n

[crystal vale]

But we didn't prove it yet

[acoustic igloo]

which part?

[crystal vale]

[G:H ] = [G: K ] [K : H ]

It is next exercise

I done

[acoustic igloo]

oh

[crystal vale]

But I don't want to use that one

[crystal vale]

We can because that's the same idea used in Basic Algebra by Nathan Jacobson

[chilly ocean]

let me remind it , the solution was using some group action

[crystal vale]

Okay

Can I ask how? Because K is not normal in G, or you mean just mapping without any algebraic structure?

[frigid epoch]

I forgot if I already asked this but

[dim widget]

Everything you want to know about is just about the size of those sets

So just a surjection of sets is enough

[frigid epoch]

What's an example of a ring where you can't construct a maximal ideal and actually need to resort to Zorn?

[crystal vale]

Yes

[crystal vale]

I think Q will work with modify multiplication ( you mean the ring which has no maximal ideal ) , right?

[crystal vale]

So we don't need K to be normal, thanks

[chilly ocean]

sorry unable to remind that

[crystal vale]

Np

[chilly ocean]

you just use next excersise

[crystal vale]

But it is in Dummit and foote so I don't think the author wants to use group action here

[crystal vale]

Okay

[chilly ocean]

for next excersise define a map from $[G:H \cap K] \mapsto [G:H]$ by $h(H \cap K)\mapsto hH$

[cloud walrus]

Ricci 10

[chilly ocean]

CHECK this map is well defined you will your result

[crystal vale]

Next exercise is [ G : H ] = [ G : K ] [ K : H ]

And I already did it

[chilly ocean]

okay

now show $H/H\cap K$ IS finite order

[cloud walrus]

Ricci 10

[crystal vale]

It comes from [ G : H \cap K ] = [ G : H ] [ H : H \cap K ]

[chilly ocean]

i was avoiding your jacobson solution

[crystal vale]

That's not Jacobson's solution

I ses

I see

[crystal vale]

But when we have a tower, why do we have to do something different?

[candid patrol]

Are you still struggling ?

[crystal vale]

Not exactly

[crystal vale]

But can I get the lower bound without using tower law?

[tough raven]

If you really want to know, I guess one can look up why this implies AoC. Then take a family of sets whose product is not provably non-empty without choice and use that to make a ring.

[chilly ocean]

My whole aim was to solve this problem without using any other result from here

[chilly ocean]

If you show this explicitly you are done

[crystal vale]

Okay then cannot we just prove that tower law?

[chilly ocean]

okay you can

[chilly ocean]

what intersection with Z will give me

[tough raven]

You can show that the possible R's or S's are precisely ℤ_(p) for every prime number p.

[errant portal]

How do I show that H is a subgroup of G if G = <RxR,+> and H = {(x,y) : y=2x}?

[elfin wraith]

Do you know of the subgroup test?

Of course you can directly just check the group axioms hold (and it is clearly a subset), but if you know the subgroup test you can do this slightly faster

[errant portal]

yeah, but I can't really figure out how to use it for that example

[elfin wraith]

Well where are you getting stuck?

[errant portal]

I'm not exactly sure what my end result should look like when proving that it is closed to w/r to addition. I define (x_1,y_1),(x_2,y_2) to be inside H and add them and get (x_1+x_2,y_1,y_2) but I'm not even sure im thats where im supposed to be going

[elfin wraith]

Well you have a way to relate x and y in H (im also guessing you mean y_1+y_2 there)

[errant portal]

should I then make it into (x_1+x_2,2x_1+2x_2)?

I could then turn it into (x_1+x_2,x_1+x_2)(1,2) but idk if im allowed to do that since multiplication is not the operation

or perhaps I want to turn it into (x_1+x_2,2(x_1+x_2)) then define some new variable (which I guess can be n) then I have (n,2n) but im not sure how to conclude that part if that is the correct step

[acoustic igloo]

If F is a field, then the units in F[x] are precisely the units in F
this is true right?

thanks 🙏

[acoustic igloo]

wait

wouldn't that make it false?

like 2 is nilpotent in Z_4 right?

oh but i have a field

i see

[chilly ocean]

how to show projective simple linear group over Z_5 is simple without iwasava theorem on doubly trasitive group action

[crystal vale]

You can try to show by using deg(f•g) = deg(f) + deg (g) in Euclidean domain

So can you see any polynomial of degree more than 0 can't be unit

In F[x]

[acoustic igloo]

that makes sense

[chilly ocean]

i think i have seen this in commutative algebra by balwant singh

[crystal vale]

Yeah this is a basic result

[chilly ocean]

is this question should be asked her or in other server

[sturdy spear]

symmetry group of some n-gon (in 2-D) is called Dihedral group right?

[subtle crypt]

If you can flip it over, yes.

Wait.

Yeah, if you can flip the n-gon over, the dihedral group will represent rotations and flips.

[sturdy spear]

oh wait i think i stated incorrectly

symmetry gorup = dihedral only if n=3

S_3 \iso D_6

[spice whale]

yes

because 3 is the only n such that n! = 2n

[sturdy spear]

true, for other cases dihedral group is like a subspace of symmetry group (precisely, D_2n is isomoprhic to some subgroup of S_n)

[spice whale]

yes

well

[sturdy spear]

its really awesome, if two subgroups are isomorphic they possesses almost same properties. Then only thing that differs both is the representation of elements.

[spice whale]

for n>3

[sturdy spear]

ah yes yes

[spice whale]

for n<3 this isn't true

because D_n doesn't act faithfully on the vertices anymore

[sturdy spear]

sorry, wdym by faithfully

[spice whale]

no non-identity element has the same action as the identity

that's what a faithful action is

[sturdy spear]

oh

i haven't read actions yet, so can i consider it as D2n (for n<3) would not be well defined (i mean would not exists)?

[spice whale]

i mean they're perfectly well defined

[spice whale]

⟨x,y | x^n = y^2 = 1, xy = yx^-1⟩ is perfectly understandable for n = 1 or 2

D_1 is just C_2, and D_2 is just K_4

(i will be referring to them as D_n because i think D_2n is silly notation)

[sturdy spear]

yeah indeed it make sense. (i was thinking about 2-gon and 1-gon geometric figures thats why said it isn't defined lol)

Thank you for this

[spice whale]

if you think of them bulging out in the right way you recover these groups

[spice whale]

you're welcome :)

[vocal cliff]

no D_2n is goated

[spice whale]

why

why would you index a sequence of groups with their orders, rather than with the obvious number each one is associated to

[vocal cliff]

the obvious number associated to it IS its order

dont tell me youre a tau enjoyer too 🤦‍♂️

[spice whale]

it's literally the symmetries of the n-gon

defined with a generator of order 2, and a generator of order n

[vocal cliff]

and theres 2n such symmetries

https://tenor.com/view/lbj-gif-13596181858079823880

[lusty marlin]

The D_n notation is better at conveying the natural embedding of the group into S_n

Among other things

[spice whale]

there are n! symmetries of a set of size n, but we don't say S_n! for the symmetric group on n elements

[vocal cliff]

https://tenor.com/view/lebron-james-lebron-screaming-lebron-lebron-james-screaming-gif-5629213437781461656

[late matrix]

every group is a subgroup of a symmetry group

[sturdy spear]

fr?

is (Z,+) a subgroup fo some symmetric group?

if p is odd then a group of order p^3 is isomorphic to some symmetric group?