groups-rings-fields

[rocky cloak]

So here all ideals are of the form I x J, so one might guess that's true in general.

So say you have an ideal K in R x S, how might we produce an ideal in R and S to check if K is if the form IxJ?

[copper kestrel]

find the ideals of Z/2 and Z/4, set them up as a product I x J, then include the forms I x Z/4 and Z/2 x J

then compute the quotient of each

[copper kestrel]

i dont know

i've never done anything like that

[rocky cloak]

So an ideal is in particular a subset of RxS. Any way to take a subset of RxS and get a subset of R?

[copper kestrel]

i guess do the set theoretic way of K subset I x J and I x J subset K

[copper kestrel]

R x \varnothing maybe?

[rocky cloak]

Okay, say I have an element in RxS.

It looks like (r, s)

Is there any way to get an element in R from that?

[copper kestrel]

projection map?

[rocky cloak]

That would be a pretty natural way.

[copper kestrel]

oh and preimage of an ideal is an ideal

[rocky cloak]

So if K is an ideal in RxS and we apply the projection map, do we get an ideal in R?

[copper kestrel]

i believe the projection map is surjective so yes we do get an ideal in R

[rocky cloak]

Allright, so then we are able to create an I and J from K.

So now we need to check if K equals IxJ

[copper kestrel]

yep

[rocky cloak]

Maybe we first check if K is a subset of IxJ

[copper kestrel]

take a k in K, since its a subset of R x S its of the form k = (r,s), and since K is an ideal (a,b)K subset K, so (ar, bs) in K for all a,b in R x S

then somehow reason thats in I x J

like i know (ai, bj) in I x J

[rocky cloak]

Maybe recall to yourself how I and J were defined (how they're related to K)

[copper kestrel]

for all a,b in R x S

[copper kestrel]

like through the projection map?

[rocky cloak]

So you have (r, s) in K. Then what it means for K to be a subset is that this also is an element of IxJ

What does that mean? What does it mean for (r,s) to be a subset of IxJ?

[copper kestrel]

right

[copper kestrel]

every (r,s) is in I x J

[rocky cloak]

Okay, let me ask this, what is IxJ?

[copper kestrel]

an ideal

i think

[rocky cloak]

Maybe I'll go with an example:

Consider 2Z x 4Z the ideal in ZxZ. Is (3, 4) an element of 2Z x 4Z ?

[copper kestrel]

no bc 3 isnt in 2Z

[rocky cloak]

Right, so what is the condition that tells you whether
(r, s) is an element of IxJ?

[copper kestrel]

r has to be an element of I and s has to be an element of J

[rocky cloak]

Alright. So if (r, s) is in K. Would this be true?

[copper kestrel]

it depends what r and s are i think

[rocky cloak]

So, we wish to show that r is in I. What is the definition of I?

[copper kestrel]

its an ideal of R, so aI subset I for all a in R

(and Ia subset I)

[rocky cloak]

Well, it's not an arbitrary ideal. We defined it in a specific way

[copper kestrel]

oh wait but I is the R-projection of K

idk what its actually called

but J is the S-projection of K

[rocky cloak]

Yes, so what does this mean?

The projection should take (r, s) to... what?

[copper kestrel]

just r

[rocky cloak]

Right, so if (r, s) is in K, then...

[copper kestrel]

so every r element of K is in I

and every s element of K is in J

[rocky cloak]

So for an element (r, s) in K, we have r in I and s in J. Thus K is a subset of IxJ

[copper kestrel]

ah

so then every element of I xJ is an element of K by definition so boom K = I x J

[rocky cloak]

No, every element of K is an element of IxJ by definition, so K is a subset of IxJ

For the other inclusion you have to actually use that K is an ideal

[copper kestrel]

oh

[rocky cloak]

A hint could be that ||(r, s) = (r,0) + (0,s)||

[copper kestrel]

rats

[floral helm]

the reverse inclusion usually needs identities, do you have a 1? what happens next

[copper kestrel]

probably not, so theres something wrong with it

[rocky cloak]

I mean Z/n does have a 1

As all rings should

[elfin wraith]

As all rings do

[copper kestrel]

take an (r,s) in I x J, since I is the projection of an element r in K, we have that the preimage of r is (r, 0) i think?

do the same thing for J, get (0,s)

[rocky cloak]

Well it's (r, t) for some t at least

[copper kestrel]

ah

[rocky cloak]

I guess we have to come up with a way to turn this t into a 0

Something we can do to an element of an ideal that turns it 0.

Something like adding or maybe multiplying by something, as one does with ideals

[copper kestrel]

i have no idea, i want to add (0,-t)

but that doesnt make sense

[floral helm]

yeah if you got a 1 then multiplying by (1,0) is the move

[rocky cloak]

Yeah, I guess it's hard to add something without knowing more about K.

Multiplying might be easier, since we can always multiply by whatever and stay in the ideal

[copper kestrel]

oh true

my lord how did i not see that

and since K is an ideal its still in K

then do the same thing for J i guess

so i guess we can make it a whole proof

let K be an ideal of R x S and define I as the projection of R in R x S onto R and define J as the projection of S onto S. (idk how to word that better) we now want to show K = I x J. Take a k in K, k = (r,s). Clearly by definition every r in (r,s) is in I, and every s in (r,s) is in J, thus (r,s) in I x J and K subset I x J.

Conversely, take (r,s) in I x J. since I is the projection of an element r in K, we have the preimage of r is (r,t) for all t in J. Since K is an ideal, we have that (1,0)(r,t) = (r,0) in K. Similarly for an element s in J, we get (0,1)(t,s) = (0,s). Finally, we have (r,0) + (s,0) = (r,s), making I x J subset K. Thus K = I x J

very rough

[white oxide]

Okay now I'm confused, why are we even working in Zp in the first place if they're algebraic integers

Nvm we're working with congruences with algebraic integers lol

Wait how is this reasoning wrong? If w^3 = 1, then w^3 - 1 = 0 x p. Thus w^3 is congruent to 1 mod p

[wraith cargo]

Where does w^3 come into play

You're working with w^p

[white oxide]

I know I was just curious how that specific statement was wrong

[rocky cloak]

Is (-3/p) the legendre symbol?

[wraith cargo]

It isn't but that's not what you were originally claiming

You claimed that w^p is congruent to 1 mod p

[white oxide]

Yeah now I see

Anyways I'm just doing cases on if p is congruent to 0, 1, or 2 mod 3

1 mod 3 is easy, 2 mod 3 is giving me trouble

[white oxide]

yeah

[rocky cloak]

So then if Z/p has a third root it's 1 because (2w+1)^2 = -3

If Z/p does not have a third root, adjoining one you see that the square root of -3 is not in Z/p, so the symbol is -1

[white oxide]

Nvm, since 1 + w + w^2 = 0, we have w^2 = -w - 1. Thus if p is congruent to 2 mod 3, we have 2w^2 + 1 = 2(-w - 1) + 1 = -2w - 1 = -tau

[zealous sandal]

Can someone give me a hint on this problem? I've been trying to use the canonical projection from G to G/N, but I'm lost

[velvet hull]

yes, that is a very good idea

what do you have right now?

[zealous sandal]

Well my thought was to think of n_p (G) as [G: N_G (P)] for some P in Syl_p (G) and n_p (G/N) as [G/N : N__(G/N) (Q) ] for some Q in Syl_p (G/N), but idk how elements of Syl_p (G) and Syl_p (G/N) are related so it felt fruitless

But my hope was to do something with images or preimages

[velvet hull]

well at the very least you can get a well-defined map from Sylp(G) to Sylp(G/N)

do you see how?

[zealous sandal]

If P in Sylp(G), is PN in Sylp(G/N)? That's the only map I can think of, but I'm not sure how I would check this

[velvet hull]

what is the definition of a sylow p subgroup

[zealous sandal]

A subgroup of G which has order p^k where p^k divides |G|, but p^(k+1) doesn't

[rocky cloak]

And what is the order of PN/N?

[velvet hull]

there is a slightly better one you can use here; a sylow p subgroup is a maximal p subgroup of G

the 2nd Sylow theorem tells us that this statement is exactly equivalent to yours

[zealous sandal]

Sure

[velvet hull]

so you can think about the p subgroups of G and G/N

[zealous sandal]

oh wait so PN/N is isomorphic to P/(P intersect N), so PN/N has order p^k, so PN/N is a p-group

[rocky cloak]

Indeed it is.

Even more if you give some names to the largest power of p dividing G and N you might get something more quantative

[zealous sandal]

ok thanks guys hopefully this makes more sense once my meds have kicked in (:

[somber goblet]

sylow's proofs are hard for me to understand

[somber goblet]

oh wow thank you

[zealous sandal]

On this problem I have an idea but idk if it's a red herring. Can someone help me out?

So if I have a normal subgroup of order p then I'm done, otherwise let's suppose not. So I don't have a normal Sylow-p subgroup meaning that n_p > 1. By the third Sylow theorem, I know that n_p is congruent to 1 modulo p and also must divide p+1. Therefore my only choice is that n_p = p+1.

Now, I know that if two distinct Sylow-p subgroups have order p^k, that their intersection has at most p^(k-1) elements. Therefore, since my Sylow-p subgroups are of order p, their pairwise intersections are trivial. Hence, there are a total of (p-1)(p+1) distinct elements of G in my Sylow-p subgroups.

Well, that means that the G has a total of p(p+1)-(p-1)(p+1)=p+1 distinct elements not in a Sylow-p subgroup.

It feels too coincidental to just have p+1 elements lying around for them not to be the p+1 elements I'm looking for, but I'm not sure how I would proceed. Plus idk what my prof's hint is about. Am I on the right track? Or I am going crazy?

[velvet hull]

Yes, those are exactly the p+1 elements you are looking for

So, those p+1 are exactly all the non p-torsion elements of G

[zealous sandal]

Ok cool I'll keep thinking about it, I just wanted to make sure I wasn't completly off track

Thanks (:

[kind temple]

: )

[maiden crater]

So we need the norm here to have a way to compare elements in some sense , right

[rapid cave]

Compare in some sense

By more like to do division with reminder

[maiden crater]

so basically for the division lhrothm to make sense

[rapid cave]

Yeah

[desert verge]

curious that no triangle inequality is required for this type of norm

[wicked patio]

well its an extremely loose notion of a "norm"

just there to make sure the euclidean algorithm terminates

[thorn jay]

fairly weak seems like an understatement lol

[azure cairn]

yeah that's rly weak lol

is it normally called a norm?

id highkey call it something else

[elfin wraith]

I think my intro to rings course called it a Euclidean function

I don’t know how common that terminology is though

[floral helm]

yeah valuation is another big one, basically just needs to play nice with division

[elfin wraith]

Isn’t a valuation stronger than that?

[floral helm]

yeah valuations usually have way more structure like the additive property, euclidean functions are super barebones

[zealous sandal]

Suppose we have a subset X of a group G, and X is the kernel of some group homomorphism f:G->H. Does the first isomorphism theorem immediately tell us that X must in fact be a normal subgroup of G?

[rocky cloak]

Might depend on your definition of normal subgroup / your formulation of the first isomorphism theorem.

But it should at least be immediate either way

[astral ivy]

what to do to demonstrate closure of a group?

do I need to 'solve' it for x and y?

For example:

on an open interval $( \frac{-1}{2}, \frac{1}{2})$, operation $\ast$ is defined as: $x\ast y = \frac{x+y}{1+4xy}$

I'm latex illiterate when it comes to goofy brackets

[cloud walrus]

dan

[astral ivy]

so what do I do here?

I set it up like $\left| \frac{x+y}{1+4xy}\right| < \frac{1}{2}$

[cloud walrus]

dan

[rocky cloak]

You need to demonstrate that for any x and y in your group
x * y
is also an element of your group

[astral ivy]

I know but I'm not sure how

Am I supposed to use the inequality and show x and y for which it is true?

And what in case when boundaries aren't set like here?

[rocky cloak]

The goal would be to show that |x+y|/|1+4xy| < 1/2 whenever |x| and |y| are < 1/2.

So you would just have to use properties of the absolute value, for example |x + y| <= |x| + |y|

You can also split into cases of either x and y have different signs or whatever else would be useful

[limpid ferry]

Up to isomorphism there are two groups of order 14. My teacher just wrote down 2 groups Z_14 and D_7 without explaining, could someone give me an approach?
EDit:typo

[glass sinew]

D_4 does NOT have order 14

[limpid ferry]

my bad

[vapid vale]

do you know sylow's theorems?

[limpid ferry]

Yes, I do know all three

[vapid vale]

then you can conclude there is a unique subgroup of order 7 which is normal, and consider the semidirect product with some 2-subgroup

and there are two choices for that product

[limpid ferry]

I assume you mean internal semidirect product, all I know is its definition. I looked this up, they mentioned this as well but this where I don't know where to go next. Any theorem that I will use with this type of product?

[twilit wraith]

just from the definition + sylow theorems we can deduce all groups of order 14

note that all groups of order 14 have a normal subgroup of order 7 clearly

note that all semidirect products are uniquely determined (up to conjugacy classes of images of these actions) by what the action of a non-normal subgroup is on the normal one

i.e. a map from H to Aut(N)

where N is our normal subgroup and H is some subgroup such that H \cup N = G and H \cap N is trivial

but Aut(C7) = C6

and as theres only one element of order 2 in C7 theres only one nontrivial action and thus only one nonabelian group of order 14

but we already know D7 is nonabelian and order 14

so we must have that C14 and D7 are it

[somber goblet]

does this work for all groups of order 2p

[vapid vale]

yes

[proud vigil]

by a similar argument you can show that there's at most 2 groups of size pq for p < q primes, and you get the nontrivial one iff p divides q - 1

[candid patrol]

Let’s do pqr now

[proud vigil]

https://groupprops.subwiki.org/wiki/Hölder's_formula_for_the_number_of_groups_of_squarefree_order_up_to_isomorphism

wow that's elegant

i think you can do it inductively, repeatedly applying schur-zassenhaus and using the previous classifications

[zealous sandal]

On this problem, if we assume G doesn't have a normal subgroup of order p, then I've managed to show that the subset X = {x in G | ord(x) ≠ p} has p+1 elements. This should be the normal subgroup of order p+1 we are looking for, but I am struggling to show that it is in fact a subgroup. In particular, I can't figure out how to show closure. Can somebody help?

[wraith cargo]

idt that's a subgroup

also why not consider what's given in the hint?

[zealous sandal]

I haven't been able to figure out how to use the hint

My prof agreed that this is the set I should be looking at when I asked

We know that the number of Sylow-p subgroup of G is either 1 or p+1. So there are either p-1 or (p-1)(p+1) = p^2 -1 elements of order p in G, so I disagree with your first statement

[wraith cargo]

yeah I changed it lol

I realized I framed it incorrectly

hold on

let me retract everything I've said I think my argument was heavily flawed

[brisk tendon]

tdhis is the q

??

[zealous sandal]

yes?

[rocky cloak]

I don't think this is the intended solution given the hint: but something that works

||Note that if there are p+1 sylow p-subgroups, then every element of order p has (at least) p+1 conjugates, so cannot commute with any element not if order p||

||Take any non-identity element x of order not p. And let y be an element of order p.||

||As x and y cannot commute, yxy^-1 is a different element. So x will have (at least) p conjugates.||

||That means all the nonidentity elements not of order p are conjugates of x, hence have the same order.||

||Since this would include powers of x, its order must be prime, say q. And elements of G have either order p or q (or 1)||

||Then Cauchys theorem gives you that no other prime can divide the order of G, so p+1 is a power of q.||

||Then the p+1 elements not of order p must be the unique q-slow subgroup||

Okay, wait the hint actually simplifies this:

||Just consider the centralizer of x. It does not contain any element of order p, so x has at least p conjugates||

I mean, about the same but I guess that's what the hint was going for

[brisk tendon]

@rocky cloak ofc ya answered it lel

[rocky cloak]

So yeah to make a hint instead of a solution: ||an element x has |G/C(x)| conjugates||

[brisk tendon]

sleepy ? i dunno always figures yr frm europe

[rocky cloak]

,time

[cloud walrus]

The current time for jagr2808 is 11:15 PM (CET) on Thu, 19/02/2026.

[brisk tendon]

sorry deleted msgs to keeo chat clean

[zealous sandal]

Thank you 🙏 I had most of these pieces most figured out but I just couldn’t piece them together

Perhaps there is hope for me to finish my homework by tomorrow

[twilit wraith]

Alright I have to show that if $R$ is a PID, $B$ is a torsion $R$-module and $p \in R$ is prime, then if $pb = 0$ for some nonzero $b \in B$, then $\text{Ann}(B) \subseteq (p)$.

[cloud walrus]

hiidostuff

[twilit wraith]

I first considered the submodule Rb

so (p) is clearly a subideal of this submodule's annihilator

the annihilator itself has to be a principal idea tho, say (s)

it then follows that s divides p

so either s is a unit or is associate to p

if its associate to p the result follows

but why could s not be a unit?

if its a unit then all of R annihilates Rb

so then Ann(B) is a subset of R but thats trivial

[twilit wraith]

wow nvm im silly

1b = 0 but then b = 0 a contradiction

[ivory anvil]

if r_a is the ring of integers mod a, f(r_a) is the probability an element in r_a has a square root. it seems like the largest element in {f(r_b),f(r_b+1).....f(r_b+n)} is 0.5 as b and n get very large

[proud vigil]

is f(r_a) like the proportion of elements in r_a with a square root?

[ivory anvil]

yeah

[proud vigil]

your bio is so funny

[ivory anvil]

🎉

[proud vigil]

hm so we basically just wanna know the number of distinct quadratic residues mod a?

i thought that was always half the number of elements

wait no

[ivory anvil]

thats what my code seems to be converging to

[proud vigil]

oo

let me find the result im thinking of

[ivory anvil]

quadratic residue?

[proud vigil]

oh are you asking what they are or why i'm saying they're relevant

[ivory anvil]

i have never heard that term

[velvet hull]

It’s always half mod a prime

[proud vigil]

right

[velvet hull]

But if it’s a non prime that guarantees the existence of nilpotents and that drives down the %

[proud vigil]

they're basically exactly what you're finding, the perfect squares in a modulo ring

they're what's "left over" when you square a number

[ivory anvil]

by distinct do you mean only one?

[rapid cave]

by CRT you only need to look at prime powers

[proud vigil]

like the number of distinct residues/squares

because we want the number of elements that have a square root, we don't care if an element has more than 1

typically multiple numbers will square to the same thing

[ivory anvil]

yeah thats what i was calculating

[proud vigil]

oh so you really just invented a prime number checker

[ivory anvil]

nah

because it can be larger than 0.5

im at 0.50000019999928 and generally every time i multiply b by ten another zero gets added between the 5 and 9

[proud vigil]

modulo odd prime p there's (p+1)/2 residues

so it'll be a little bigger than half but it'll converge to 1/2 when you find larger and larger primes

[ivory anvil]

ah

[proud vigil]

and it's easy to justify why n/2 + 1 is the upper bound on the number of residues mod n

[ivory anvil]

this number was from mod 5000018 which is composite

[ivory anvil]

worlds worst prime number checker seems like a fun challenge

[proud vigil]

oh sorry typo, fixed

[ivory anvil]

is mod 2 the only one where every element has a distinct square?

[proud vigil]

i guess you could do mod 1, and the only number is 0

that wouldn't be a ring though...

oh it can be a rng

or just a ring without unity

anyway this is boring; once you pass mod 2, 1 and -1 become distinct

[ivory anvil]

i thought mod n didnt have negative numbers

[noble nexus]

rings without unit are evil and bad

[proud vigil]

well first, mod n is just dividing the integers into equivalence classes, and you can identify every integer with the class it belongs to

usually we can write that like [x], where x is the integer and [x] is the class

mod n, [-1] = [n-1], they're both valid representatives of the same class

second, writing -1 indicates that we have the additive inverse of the 1 element, and we know additive inverses have to exist in a ring

[ivory anvil]

ah so just -1 in the sense of n-1 in mod n

[proud vigil]

also i should have added that odd primes don't uniquely achieve the floor(n/2) + 1 upper bound, so this is totally fine

maybe that makes it a pretty bad prime checker... but at least it gives you less numbers to check...

[ivory anvil]

whats the time complexity of checking if 2^prime-1 is prime?

is that the right form

i know theres something with 2^n that forms primes

[proud vigil]

yup those are called mersenne primes

https://en.wikipedia.org/wiki/Lucas–Lehmer_primality_test

[fossil beacon]

if a automorphism of Sl_2(R) s.t phi(x)=g^{-1) x g then phi is inner? for some no trivial x,

[somber goblet]

yeah

[somber goblet]

x doesnt have to be nontrivial

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

[fossil beacon]

understand my question

[rocky cloak]

The trivial ring is a ring (with 1) it just has 1=0.

[proud vigil]

yeah i realized that, i think in fields the convention is to not let 1 = 0 but i guess rings don't have issues with that

[rocky cloak]

I'm assuming u is a specific element here?

Since any automorphism would obviously take I and -I to themselves

[maiden crater]

what are the dots supposed to be here

like (2n)(3n-1)...( there's no obvious pattern here)

[quiet pelican]

The next one is (4, n-2)
Then (5, n-3)
And so on

What’s written is (2, n) and (3, n-1)
Which looks less clear because transpositions are usually written without the comma I’ve added

[maiden crater]

oh (2 n) (3 n-1)

lol

I read it was (2n) (3n-1) and was like huh?( which tbf are just the identity lol)

[maiden crater]

I suppose this can be proven with induction

[quiet pelican]

I’d prove it by “just look” /hj

[cursive spindle]

Almost all inductive proofs are just "look at it"

[maiden crater]

Here, theorm 2.36 is basically just description 2 , written formally right

[fossil beacon]

yes

[rocky cloak]

Okay, and you know what it is or...?

[fossil beacon]

say take unipotent

[rocky cloak]

I'm asking what the exercise says

Maybe they mean for all u

[fossil beacon]

no its not for all u

[rocky cloak]

Okay, then how are they defining u?

If it was for all u the exercise would make sense

[fossil beacon]

might be possible

[fossil beacon]

yes it might be for all u

[rocky cloak]

I guess you want to think about what alpha does to upper triangular and lower triangular matrices to see where the eigenvectors get moved.

Not sure if alpha is supposed to be an automorphism of lie groups or just groups or if it matters

[fossil beacon]

yeah please take lie group automorphism

this stumped me

some lie theory might need i think then

[cloud walrus]

Pom

[fossil beacon]

if it is irreducible we are done

[desert verge]

irreducible over what field

[fossil beacon]

perfect field

[desert verge]

huh

[rocky cloak]

Why no derivatives?

Anyway, you can treat this as a polynomial over Q. By the rational root theorem it has no rational roots. So the only possible ways it can factor into irreducibles is either being irreducible or the product of a degree 2 and degree 3 irreducible

Neither of which would give multiple roots

[tidal grail]

it isnt though

[fossil beacon]

char 0 field is perfect so here done, irreducible implies separable

[tidal grail]

ah yes

[tidal grail]

but proving that requires derivatives

or at least the proof i have in mind does

K char 0 and f irreducible over K and c some element in some field extension a root, then c has multiplicity 1.

Proof:
GCD(f',f)=1,therefore all roots have multiplicity 1

[desert verge]

"without using derivatives" in undergrad is crazyy

[maiden crater]

I mean technically derivatives require a concept of completeness no?

oh, in C

yeah ouch

[elfin wraith]

No? You can just take the formal derivative in any polynomial ring I think

[maiden crater]

I was thinking of the elements

[desert verge]

tbf to use derivatives to analyze polynomial roots u need stuff like ivt and real analysis so what i said is kinda dumb

[fossil beacon]

yesterday my prof did take some kahler differentials , in some module bla bla

[maiden crater]

fair

I'm still doing euclidian domains so

[desert verge]

how is that relevant

[fossil beacon]

like where is completeness related

[desert verge]

to what

[fossil beacon]

to derivative\

[desert verge]

it's not related

[next obsidian]

[delicate orchid]

it needs the Leibniz rule and not much else (those who know)

[thorn jay]

Ω_X/S

[cinder fox]

Is there a generalized form of the fundamental theorem of algebra for multivariable polynomials?

[wraith cargo]

this just follows from the fundamental theorem right

take f(x_1,...,x_n)

take any values for x_2,...,x_n

then you just have a single polynomial in x_1

and that's guaranteed to have a solution

[unkempt comet]

No

In fact differential algebra can be done over non-complete fields like Q(t)

[next obsidian]

Derivatives make sense for all polynomials and that’s all that matters

And okay, power series too as a treat

[tough raven]

Not quite: you need them to not make the resulting polynomial in x_1 constant. But this can be done.

[wraith cargo]

oh yes that's kind of an embarrasing oversight

[glad osprey]

I've heard the weak Nullstellensatz be described as a kind of fundamental theorem of algebra for multivariate polynomials: the polynomials f_1, ..., f_n in C[x_1, ..., x_n] have a common zero if the ideal they generate is not equal to all of C[x_1, ..., x_n]

[fossil beacon]

for better explaination just see daniel bump algebraic geometry, very friendly text

[next obsidian]

Not underrated among those who know 😍

Also I want to say the fact it works for power series is also hugely important

I think a fair few things that work in unramified mixed char we don’t know in ramified can be traced back to the fact you don’t have derivations acting on regular complete ramified local rings

[south patrol]

Lol I was gonna say like

de Rham cohomology

[next obsidian]

Rham de nuts in yo mouth

[halcyon sedge]

Can someone please suggest a good reference to study finitely generated groups and their properties in detail.

[fossil beacon]

looks fancy

[ivory anvil]

In mod 17, 3 isn’t a quadratic residue. If I define some a such that a^2 in mod 17 is 3, every number in mod 17 becomes a quadratic residue. In all the other prime mod n’s I’ve tested, I only needed to define a root for one unit to turn all other elements into a quadratic residue. Is that true for all prime mod n’s?

Also what’s the proper name of a?

[rapid cave]

Adjoining a root a of x^2 - 3 gives a quadratic extension F_17(a)/F_17 in some fixed algebraic closure. If instead we adjoin a root of x^2 - b to F_17 where b is a quadratic non-residue mod 17 we also get a quadratic extension of F_17, which is isomorphic to F_17(a) since two extensions of a finite field with the same number of elements are isomorphic.

Since F_17(b) contains a root of x^2-b then F_17(a) must also contain one.

[ivory anvil]

how do you know f_17(a) and f_17(b) have the same number of elements?

[rapid cave]

They are both quadratic extensions of F_17, so they have 17^2 elements

[ivory anvil]

3 is still not a quadratic residue after you add a root of 2 to mod 30

[rapid cave]

30 is not prime...

[maiden crater]

just wanted to share this as I found it kind of fun

It's easy tbj

[kind temple]

i would ||prove this by induction||

[maiden crater]

yea, I pretty much did the same thing

showed that 3 elements had a common genertaor

and we're done

(with the hardest part)

[candid patrol]

You’ve got an example of such a group ?

[novel star]

Q i think

[maiden crater]

(Q,+) yeah

[kind temple]

the cyclic groups are locally cyclic since subgroups of cyclic groups are cyclic

how do you start classifying the locally cyclic groups?

[candid patrol]

Of course but I was looking for something else

[candid patrol]

Nice thx

[proud vigil]

i think also if you take any subset of the prime numbers and restrict the rationals to denominators consisting of those primes as factors, then these are locally cyclic groups as well

mm i think

[kind temple]

so like the dyadic numbers?

[proud vigil]

oh yes

[twilit wraith]

i wonder if theres an uncountable locally cyclic group

R probably now that i think about it

the field structure of it as a ring makes it that way prob

nah nvm

taking a rational and an irrational as generator is a counterexample

shouldve been obvious from the get go ngl, oops

[kind temple]

Z^{\omega} should be locally cyclic?

[twilit wraith]

what about a copy of Z \oplus Z in it

[kind temple]

lmao right that isn't cyclic

[twilit wraith]

oh also just as a quick note

seems the prufer p group is also locally cyclic

which isnt surprising

seems that the classification of locally cyclic groups are direct sums of Q and the prufer p group

in which case its impossible for locally cyclic groups to be uncountable

ohhhh

this ends up just being a result of modules over PIDs

direct sum of Q is the free part and prufer p group is torsion

[velvet hull]

I suspect no, because my intution is telling me that locally cyclic groups should be limits of cyclic groups

and cofinite groups are always countable

[twilit wraith]

wait no no

if i took the subgroup generated by an element with torsion and an element without i wouldnt get a cyclic group

so its either free or torsion

[proud vigil]

https://groupprops.subwiki.org/wiki/Locally_cyclic_iff_subquotient_of_rationals

[twilit wraith]

Ghost ping :(

[indigo jetty]

Too abstract, how do solve this

What’s the essence of this strange definition

[delicate orchid]

twin... rotate yo damn phone 💔

[candid patrol]

,rotate

[delicate orchid]

u really are the goat...

[candid patrol]

that's my name 😎

[delicate orchid]

anyway doing cohomology of groups before simplicial sets/spaces is weird to me and it's no wonder this seems unmotivated

[delicate orchid]

the idea is you want to see the shape of your group G. And to see the shape you need a space: so you construct a space with a single loop for each element g in G, but you want gh to be related to g and h so for you add in a (degenerate) 2-simplex connecting each element to its product, which become 3-simplices for products of 3 elements and so on. Assuming A has trivial G-action for now, this map then sends each of these simplices to an alternating sum of their faces and squares to 0, making it perfect to form a chain complex with the objects being the sets of n-simpices and these maps forming the differentials. This turns out to be exactly the chain complex you want to compute cohomology (it yields the derived functors for the fixed point functor A -> A^G sending G-modules to their fixed points)

[delicate orchid]

these are trivial if you just write down what the statements mean. For example G acts transitively on X x X - the diagonal means, for instance, there is some g such that g(x, y) = (gx, gy) = (a, b), x != y, a != b. Solving part 1. You can see that the action is transitive on X by setting a = y, b = x; and so on

if you want a concrete example of a strictly 2-transitive group action I suggest thinking about GL_2(k) acting on k^2 for your favourite field k

or even simpler, I suppose, S_3 acting on 3 points

[fiery dirge]

can anybody help me understand the permutation group, especially things like multiplying from the left/right and conjugation.

Example: $H=⟨(12)⟩={e,(12)}$

Take $g=(13)$, $g\in G$.

Conjugate: $g(12)g^{-1}$
The result is $(13)(12)(13)=(23)$ but I didn't quite understand the process.

[cloud walrus]

BlaBla

[fiery dirge]

So what do I look at first? I look at 1. Right bracket, 1 -> 3, move to left bracket, there's no 3 there so I skip it. In the last one 3 -> 1 so it should be 1 -> 1?

Next, 2 is in the middle bracket first going from the right. 2->1 and then 1->3 so 2->3

I can't visualize it I need help understanding it.

[delicate orchid]

you're doing it correctly I'm not sure where the confusion is

[fiery dirge]

I don't understand this: (13)(12)(13)=(23)

Also, is $g^{-1} = (31) = (13) = g$?

[cloud walrus]

BlaBla

[fiery dirge]

well I'm confused how they got (23)

[delicate orchid]

if you swap something, then swapping it again gives you what you started with

[delicate orchid]

you just explained exactly how they got (23) perfectly lol

[fiery dirge]

lmfao

[delicate orchid]

let me draw a picture

[fiery dirge]

ahhh Im a degenerate

1 -> 3 and then 3 -> 1 so 1 is fixed

that's why it doesn't show up

[fiery dirge]

essentially these are like function compositions right? First one mapping, then it potentially maps to something else.

[delicate orchid]

I mean they are function compositions. Permutations are automorphisms of a set

[fiery dirge]

I don't see the automorphism though because the way I understood it, everything needs to change.

if we just have (13) then 2 remains fixed so it's an endomorphism?

I primarily used this illustration to understand it

but then again that endomorphgism mapping might just be kernel thing

[twilit wraith]

Note that endomorphisms of sets are just functions from a set to itself

Bc theres no structure for the morphism to preserve

[fiery dirge]

yeah I know. I just have trouble understanding the difference between endomorphism and automorphism

[twilit wraith]

An automorphism is a bijective endomorphism

Endomorphisms dont have to be invertible

Take {1,2,3} and map everything to 1

Thats an endomorphism with no inverse

The reason why we care about automorphisms of sets is that they form a group

Endos dont

[twilit wraith]

(For sets of cardinality n this is just the symmetric group of order n)

[fiery dirge]

ahh so about the case (12) where 3 -> 3. It's automorphism because 3 still gets mapped (or target set gets 'filled').

if 3 just dissapeared from the domain then it would be an endomorphism because it wouldn't be a bijection (3 in codomain would remain empty)

[twilit wraith]

The element (12) in S3 is the function sending 1 to 2, 2 to 1, and fixing everything else

Everything being 3 lol

But (12) is written the same way in S4, S5, etc

In which case 4 and 5 are fixed, etc

[fiery dirge]

so any Sn function will be an automorphism?

or there are cases where this isn't true?

[twilit wraith]

Every element in Sn is an automorphism of the set {1,2,...,n}

If it weren't an automorphism, it wouldn't have an inverse, which violates a group axiom

[indigo jetty]

Thanks very much

[knotty badger]

group cohomology of G = simplicial cohomology of BG?

[tall igloo]

its a beautiful result that these are equivalent

[south patrol]

i guess it's kinda funny if you view it as understanding $\varinjlim_{BG} M$ in two different ways

[cloud walrus]

Prismatic Potato

[azure cairn]

artin's book is so boring

hopefully it gets better im only a few chapters in

[velvet hull]

You can skip chapter 1 of artin

[azure cairn]

i did

[velvet hull]

You don’t think groups are cool? lol

[azure cairn]

.<

actually ch2 was good

i probably just dont like ch3 and ch4 cuz its linalg repeat lol

dyk if artin visits the minimal polynomial?

[velvet hull]

Okay yeah 3-6 is more linear algebra again

Just skip to chapter 7 lol

I forgot how LA pilled artin is

[fickle dirge]

Is the LA section in artin good

[velvet hull]

The definition and basic properties are explored but not an actual proof of the advanced theorems (Cayley Hamilton), that requires a lot of algebra that this book does the setup for

Oh wait it is covered

Much later, in chapter 14

[azure cairn]

wtf

Alright thx

[tulip otter]

in that case try lang

[feral slate]

If you want to torture yourself

[feral slate]

Safe to say that you probably won’t be bored

[torn shadow]

Hello, I was wondering if anyone recommends any books to read after Pinters Abstract Algebra. Im in 9th grade, so nothing too abstract, I’ve done calculus, but not linear algebra, Thank you for your help.

[vapid vale]

you've read pinter's book?

[torn shadow]

Yes

[azure cairn]

Linear Algebra Done Right.

[torn shadow]

Oh, good. I was thinking of reading that, but didn’t know if I should read a general applications based linear algebra book first.

[azure cairn]

Oh I was half trolling

But it's not a bad idea to read ladr because it'd go well with the algebra you already know

FIS (friedberg insel Spence) is good as well (wrt operator theoretic stuff) while still having a traditional approach to matrices, rref, and determinants

[azure cairn]

Also as a 9th grader you'll learn a first course in linalg anyways in school (assuming you go to college and do a major that will teach you linalg)

[torn shadow]

I know a small amount of all of those. Just from pinters

[torn shadow]

I self taught calculus so I won’t be doing it for awhile

Alright, thank you

[azure cairn]

Are there two pinter books or something I'm confused

[torn shadow]

I really liked the results of group theory and ring theory

Like symmetries of groups and ideal quotient rings

Integral domains etc

No, I haven’t

[azure cairn]

More ladr shilling but it has a section on quotient spaces and ch8 is useful when viewed through the lens of quotients

[torn shadow]

Oh, that sounds cool, thanks. Also do you know if John still wells number theory book is good? I’m asking because I already have it

[azure cairn]

My main point is that if you end up doing linalg I think ladr would be best with an abstract alg background

Of the books I know

[craggy oar]

In 9th grade?

[torn shadow]

Yeah, thats the main reason I started with pinters, to get more maturity with more abstract stuff

Ok, well thanks for all the advice.

Yeah, right

[craggy oar]

We indeed read Basic Number Theory in postgrad

[fossil beacon]

are you in india

[azure cairn]

Yea he's studying for the jee rn

[next obsidian]

I’m still on A Course In Arithmetic :/

[fossil beacon]

i think those courses are fine because in advanced courses , everything goes over head,

yes bro, for eg waterhouse

[fossil beacon]

it is so much dense , and prof following that , prof skips details also, so that became a nightmare

[fossil beacon]

Arithmatic theory is an advanced course , as i know

[indigo jetty]

Btw, is 6-transitive impossible for all groups?

[wicked patio]

S_n on {1,2,...,n} should be n-transitive

[indigo jetty]

except Sn and An？

[wicked patio]

Any group with S_6 (or A_6 ig) as a quotient can be 6-transitive

There's probably others

[delicate orchid]

there are no 6-transitive groups other than {S_n, A_{n+2} : n >= 6}

[delicate orchid]

A_n is n-2 transitive so A_8 is the smallest one that works

[frigid epoch]

Crazy fact

[indigo jetty]

I want to know why? what is the essence of this fact?

even group of infinite order can be 6-transitive?

[quiet pelican]

S_6 x C_2 be like

[delicate orchid]

Only proof I know of is the classification of finite simple groups and all groups are finite

[delicate orchid]

This isn’t 6-transitive

How do you map (7,8) to (1,2)

[quiet pelican]

I’m not having it act as a subgroup of S_8

[delicate orchid]

Then how are you having it act

[quiet pelican]

Via S_6 x C_2 -> S_6

[next obsidian]

Noob wew

[delicate orchid]

That’s essentially the regular group action which is sharply 1-transitive. I will need to be convinced that this is 6-transitive

Because I literally have a paper open from 1967 saying that the only two known (at that time, later proven by CFSG) 6-transitive groups are precisely what I said they were

[next obsidian]

Hmmm

[quiet pelican]

How do they define 6-transitive?

[tough raven]

Which action?

[next obsidian]

I looked it up and n-transitive requires faithfulness of the action?

[quiet pelican]

Because I do remember this statement holding with some extra conditions but icr them

[tough raven]

Ah OK k-transitive subgroup of Sn

[next obsidian]

I didn’t know this was a thing

[delicate orchid]

Hmm perhaps we are working with different definitions

[rocky cloak]

I mean obviously if you define a group to be something transitive you mean a subgroup of Sn.

Otherwise you would define the action to be as such

[delicate orchid]

Yeah this

[next obsidian]

Hmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

Truge…

I mean I guess

If you don’t enforce that every group is n-transitive for all n

By acting on {*}

[quiet pelican]

Via which embedding into which S_n?
(Because this isn’t well defined for groups not given as a subgroup of S_n)

[rocky cloak]

An and Sn are both given as subgroups of Sn

So nothing weird about saying
Sn and An+2 are the only 6-transitive subgroups

[next obsidian]

What does An+2 act on to be n-transitive

[tough raven]

I think the classification of k-transitive actions is supposed to classify the possible images of such an action (presumably on a finite set).

[rocky cloak]

{1, ..., n+2}

[next obsidian]

Oh

Okay, so it isn’t a subset of it or anything lol

Wait

Ofc it isn’t

[tough raven]

Which is equivalent to classifying k-transitive subgroups of Sn (implicitly acting on {1,...,n}) for each k and n.

[next obsidian]

It wouldn’t be closed

lol

[tough raven]

Let A be a finite-dimensionaa algebra over a field k. For any a in A, are the characteristic polynomials of left and right multiplication by a on A the same?

unital associative

No, consider the monoid algebra of the monoid {1, a1, ..., an} with the multiplication xy = x if y = 1 else y; then right multiplication by any of a1, ..., an has rank 1 but left multiplication has rank n.

However, it is true that lmul has det non-zero iff rmul does. In this case, are the determinants necessarily equal? Is there an "explanation" for this, e.g., the eigenvalue of the two mul maps must differ only by 0's or something?

[next obsidian]

Me when I answer my own question on MSE

[rocky cloak]

Consider the ring of 2x2 upper triangular matrices. Let e1, e2, f be the standard basis with f nilpotent.

For x = ae1 + be2 you have
xe1 = e1x = ae1
xe2 = e2x = be2
xf = af
fx = bf

So left multiplication has determinant a^2 b and right multiplication has a b^2

[tough raven]

Right. The determinants are still polynomials on A though. So they must have the same irreducible factors. The multiplicities must be interesting.

For upper-triangular, we get a1^n ... an^1 on the left and a1^1 ... an^n on the right, where a1, ..., an are the diagonal entries.

I think those are the multiplicities of composition factors in the regular representation.

Oh duh

[rocky cloak]

They are the idempotents for the indecomposable projectives, so it makes sense

[tough raven]

Is the result true for division rings (fd over our base field) at least?

[rocky cloak]

I would still think no, but worth thinking about maybe

[fossil beacon]

can i know your google scholar profile

[ripe harbor]

#help-2 message might fit here

[tough raven]

OK so take any (fd) A. A/J(A) is semisimple. There is a canonical correspondence between left and right irreps of the latter, so the same is true for A. (Maybe this more conceptually follows from duality between A-mod and mod-A?) In fact a left and right irrep correspond iff they have the same kernel. Call a fd division ring D and natural number n good if the left and right irrep of M_n(D) have the same determinant/characteristic polynomial (as a polynomial on the vector space M_n(D)). I also believe that everything is good, but I've only proved it for D = k the base field so far. Anyway the det/char poly on any left or right rep is the product of the same for the composition factors. In particular if all the simple ring factors of A/J(A) are good, then for the regular rep of A the left and right det/char polys are the product of the irreducible ones with exponents the multiplicities, and they differ exactly in how the composition factors of A as a left and right module differ.

[tough raven]

Doesn't one need to have papers for that?

[fossil beacon]

u arent in phd yet?

[rocky cloak]

Okay, so I guess we may assume it's central in which case tensoring with the algebraic closure reduces to the ring of nxn-matrices.

So then I guess it's true

[tough raven]

Why can we assume it's central?

[rocky cloak]

I was thinking that if you have K/k and a K-linear map then the k-determinant is determined by the K-determinant

Haven't thought that thought all the way through

[tough raven]

Oh. I think I actually proved this for characteristic polynomials at some point. You get an interesting answer.

Nice so this gives the exponents are multiplicities of left/right composition factors and nothing else differs always.

IG it would complete things to show the determinant on an irrep is an irreducible polynomial.

We can assume we are working with M_n(D) acting on D^n.

[rocky cloak]

Then the determinant is just the determinant

[tough raven]

Ah, if D = k yes.

Although TBF IDK how to prove that det is irreducible I just know it's .

[fossil beacon]

by eisestine

[tough raven]

Really? For which prime?

Also isn't Eisenstein for univariate polynomials? You'd have to pick which n^2-1 of the n^2 variables to treat as scalars.

[fossil beacon]

try for 2 x 2

[tough raven]

Sure, det(a, b, c, d) = ad - bc. So which variable and which prime do I use Eisenstein for?

OK I forgot that det is multilinear

[fossil beacon]

xw-zy treat any one variable as prime

[rocky cloak]

Doesn't Eisenstein just guarantee degree irreducibility? (Which is obvious from being degree 1)

[rocky cloak]

Slick proof:

Mn(k) x Mn(k) -> V(det)
(P, Q) |-> P (some fixed rank n-1 matrix) Q
is a surjective map of varieties.

As the former is irreducible, so is the latter.

[tough raven]

I've heard this but it doesn't rule out a power of a prime ideal as far as I know.

[rocky cloak]

That's true, but I guess that's an easier case to rule out

[tough raven]

OK yeah write det = M1 a1 - ... ± Mn an as say first column expansion. Then because this is linear in a1 the only way to factor it is a common factor of M1 and C = M2 a2 - ... ± Mn (in polynomials over the other n^2-1 variables). Inductively M1 is irreducible, so the only issue is if M1 divides C. But because M1 doesn't use any of a2, ..., an it would have to divide their coefficients, i.e. M2, ..., Mn. But these are irreducibles which are no associate (they have different variables) so we're done.

Using det or equivalently the characteristic polynomial seems kind of cleaner than using the trace for character theory TBH... IG the latter is more obvious for the OG setting of groups since it is linear so can be specified on just the group elements.

[crystal vale]

this is an assigment problem, so i discarded first two, so how do i solve later one?

hint only

[rocky cloak]

What do you mean you discarded the first two?

[crystal vale]

i showed they can't be galois group

for given f

oh no

sorry

[quiet pelican]

What about ||x^4 + x^3 + x^2 + x + 1||?

[crystal vale]

so it is hit and trial?

[rocky cloak]

I guess a hint then is that it's an unsolved problem whether every finite group can be a galois group over Q

[tough raven]

I think a reasonable hint is that ||they are all possible (except I'm slightly uncertain about A_4) and you should try to find examples of each, because that will be instructive||.

[quiet pelican]

Is “Can every transitive subgroup of S_n be a Galois group of an irreducible poly over Q” unsolved?

[rustic crown]

all shouldn't be possible ||as it needs to be a transitive subgroup of S_4||

[tough raven]

||These all embed as transitive subgroups.||

[rocky cloak]

Yes
https://en.wikipedia.org/wiki/Inverse_Galois_problem

[rustic crown]

not ||Z/2 × Z/2|| right?

[tough raven]

||Consider all (ab)(cd)'s.||

[rustic crown]

so it has ||2|| orbits

[tough raven]

||Gal(ℚ(sqrt(2), sqrt(3))/ℚ)||

[rocky cloak]

I guess you're asking a slightly stronger question, but I would be surprised if it wasn't also true (and hence unsolved)

[crystal vale]

why do you guys hiding?

what does it mean by transitive subgroups? i am not sure

[tough raven]

||No, not {1, (12), (34), (12)(34)}. {1, (12)(34), (13)(24), (14)(23)}.||

[rustic crown]

yea but we want Gal of splitting field of an irred degree 4 poly so if we take a primitive element in that extention then the splitting field will strictly contain the degree 4 extension

[quiet pelican]

Same
But yeah my point is that tje question being asked at least seems stronger

[rustic crown]

so yea this group doesn't ||act transitively on the 4 roots, it has 2 orbits||

or am i confused

[tough raven]

I think you should double-check the orbits.

[rocky cloak]

I don't think these groups can be transitive groups on any other number than 4 elements though [okay A4 and S4 can of course act on much bigger things]

[crystal vale]

can anyone tell me what is transitive subgroups?

[rustic crown]

ah oops

[rocky cloak]

For any pair of elements x and y there is a group element g with
g(x) = y

[rustic crown]

okie this was wrong lol

[crystal vale]

so here G acting on some sets? i think G acting on roots of minimal polynomial

[candid patrol]

Galois group of P is transitive iff P is irreductible

[crystal vale]

i see

one direction seems trivial

[tough raven]

Actually funnily enough it's always wrong in a sense. Even if the degree N of the Galois extension is greater than that n of your starting f, the extension has a primitive element and the minimal polynomial g of that has degree N. So your Galois group G acts transitively on roots of f but freely ane transitively on roots of g.

[rustic crown]

yea i realized 🙈 det rusty

sorry for confusing you @crystal vale >.<

[crystal vale]

i didn't get confused, just i don't understand what's going on?

let me understand

so here f is irreducible so G acting transitively on roots of f

so G acting on set which has cardinality 4 and G acting transitively

so how does this information help me?

[candid patrol]

is Z/2 x Z/2 a transitive group for exemple ?

identified as V4 in S4

[crystal vale]

is there something that some groups can't act transitively on set whose cardinality 4?

[rustic crown]

yee S_3 sitting inside S_4 for instance can't

[tough raven]

G also acts faithfully, because the roots generate the extension so G is determined by what it maps the roots to.

So the action must embed G as a subgroup of S_4 which acts transitively on {1, 2, 3, 4}.

[tough raven]

By the orbit-stabiliser equation, |G| would have to be divisible by 4. In fact a group G has a transitive action on a set of cardinality 4 iff G has a subgroup H of index 4 (you can take the action of G on G/H).

[crystal vale]

yes, because stab is a subgroup and by orbit stabiliser theorem |G| / |stab| = 4

[crystal vale]

i don't get it

[crystal vale]

i understand when say G acts transitively on set but what does it mean by G is transitive?

[tough raven]

Yes. So in terms of using the transitive action idea to rule out a Galois group, if G has no subgroup of index 4 (in particular, if |G| is not divisible by 4), it can't be a Galois group of a degree-4 irreducible polynomial. If it has, then the test is inconclusive and you have to try something else.

[rustic crown]

like @tough raven said, by orbit stabilizer a transitive subgroup will have order divisible by 4. but S_3 here has order = 6 which isn't divisible by 4 so it can't possibly be a galois group of an irred degree 4 poly

[tough raven]

It means some action has been fixed and it's asking about that action.

[crystal vale]

but in a given problem every group is divisible by 4

[crystal vale]

yes

i think it is stronger condition that we need subgroup of G such that it has index 4, does order of G is divisible by 4 implies it has subgroup of index 4?

[tough raven]

Actually you can make it stronger: G needs to have H of index 4 such that the intersection of all conjugates by G of H (= the largest normal subgroup of G contained in H) is trivial.

[candid patrol]

In Sn, a subgroup H is transitive if it acts transitively on {1,2, ..., n}

[tough raven]

This corresponds to the action of G on G/H being faithful.

[crystal vale]

i don't know how to check Z/2Z \times Z/2Z transitive group or not?

[candid patrol]

Mabye start with S3 in S4

[tough raven]

You can't without picking an action.

[ripe harbor]

Is there a general method for finding an explicit construction for a (constructible) regular n-gon? I.e. a pentagon?

[candid patrol]

i mean, Galois group of P is transitive (identified as a subgroup of S_deg(P) with the natural action) iff P is irreductible

[ripe harbor]

You can memorize the constructions to n = 3, 5, 17, ..., (Fermat primes) then is there a way to add them up to construct e.g. a 3 * 5 = 15-gon?

[tough raven]

Oh yes multiplication is possible (though I don't think it's by any means short).

Constructing a regular n-gon is equivalent to constructing cos(2pi/n) (or equivalently sin(2pi/n)). If you can construct cos(a), sin(a), cos(b), sin(b), you can construct cos(a+b), sin(a+b) by trig formulas. Now if you got the regular m-gon and regular n-gon, you got cos, sin of 2pi/m = n 2pi/mn, 2pi/n = m 2pi/mn. By adding up a bunch of these you can get to gcd(m, n) 2pi/mn = 2pi/lcm(m, n). So you can make a regular lcm(m, n)-gon.

The really difficult part would be Fermat primes. They're constructible by a bunch of Galois theory but working it out would involve finding a formula for the primitive F_n = (2^2^n + 1)th root of unity using 2^n nested square roots.

[ripe harbor]

Hm, ok. I found an exercise that asks you to find 3 different explicit constructions of a pentagon, are these just something you might see along the way and remember, or is there some general way to find other constructions given one of them?

[crystal vale]

Oh Z/2Z \times Z/2Z can act transitively on set of cardinality 4

[tough raven]

This IDK; I never studied or thought about explicit constructions of regular n-gons and I think at this point it gets too specialised to figure out just from the general theory.

[ripe harbor]

Thanks

[tough raven]

Bah I can't stop thinking about this

Specifically if there's a larger class than semisimple and commutative algebras where the left and right composition factors agree.

[tough raven]

And the irreducibility of the determinant here.

Also it's pity this doesn't work for non-associative algebras, esp. Lie algebras where I think it would be very useful.

[tough raven]

Suppose D is central with dimension d = k^2 and base-change to the algebraic closure (or any splitting field). Then the algebra becomes isomorphic to M_{nk}, the module to M_k^n (with M_{nk} = M_n(M_k) acting, which is isomorphic to M_{nk⨯nk} acting on M_{nk⨯k}), so the determinant becomes det_{nk}^k under a linear change of variables (not necessarily defined over the base field).

The change of variables phi depends only on D, not n, but the relevant question is: "which powers of phi(det_{nk}) are defined over the base field?". And this could perhaps depend on n.

[zealous sandal]

So my prof just introduced Holomorphs and I’m curious why we care. It feels somewhat random to consider the group G ⋊ Aut(G)

[candid patrol]

It is not

The action is tautological

[noble nexus]

holomorph is a cool name

[thorn jay]

the main reason to consider it

[astral ivy]

I feel like I'm very close to understanding how to count elements of certain order but I'm not there yet, can anybody help me?

I know that if x | n (x being the wanted element order and n being the order of the group), then phi(n) is the result. However, I'm completely lost when it comes to group products.

For example, what is the number of elements of order 9 in Z27 x Z3? I know there's the lcm(a, b) = 9, but I don't know what a and b are (I know they're from Z27 and Z3 respectively).

[coral spindle]

It's worth noting that there's not a uniform way to count elements of a given order in any old group.You seem to have gotten a bit muddled about which groups the results you're talking about apply to.

[lusty marlin]

What you said in the second paragraph holds only for cyclic groups

[coral spindle]

The result about phi(n) is applicable only to cyclic groups, like KnightWatch said, which means it only holds for groups like Z_m

If you want to think about the group Z_27 x Z_3, you need to think differently, since that's not cyclic (maybe try proving why?)

Here's a way to get started. Let G and H be any old groups. Remember that the group G x H consists of elements (g, h) where g and h are elements of G and H respectively, and the product is just (g, h)(g', h') = (gg', hh')

Now notice therefore that (g, h)^n = (g^n, h^n). So what's an element of order n in G x H? It must be one where g^n = 1 and h^n = 1

Maybe you can use this to figure out the answer for the group Z_27 x Z_3 now

[candid patrol]

it si for abelian finite groups at least

[coral spindle]

Oh whoops @astral ivy forgot to ping you with what I wrote above, sorry

[coral spindle]

This is true in principle but in practise it's harder. When we have an f.g. Abelian group in the form of a product of cyclic groups, it's easy. But we often aren't given them in that form. It's not a priori easy to calculate the number of elements of order two in a group like (Z/123Z)^\times, at least not without knowing some number theory

[candid patrol]

Why (Z/123)^times lol ? I mean, by the p-Sylow decomposition, computing lcm is esay

[azure cairn]

FINE

i admit artin is better now that ive read sections of his book that arent things ik already.

[granite galleon]

Artin or Waerden? 🤔

[granite galleon]

Just needing someone to confirm this (I've done the math but I'm a little tired and can't stop feeling a little anxious about it)... If \lambda : R \to End_R (R_R) (R-endomorphisms as right regular module) by left multiplication, it is an isomorphism, but if \rho : R \to End_R( _R R) (R-endomorphisms as left regular module) by right multiplication, it is an anti-isomorphism. Are those statements true?

[elfin wraith]

Yes, End(R_R) is isomorphic to R as a right module, and R^op as a left module

[granite galleon]

Oh, lord. It really feels like something is wrong when working with duality... Thank you so much.

[elfin wraith]

What do you mean?

[granite galleon]

Somehow, easy things turn odd or confusing when I have to start to think about their dual concept or statement that I only feel relieve just after I'd done all the work around it instead of accepting it

[elfin wraith]

I guess the key point here is just that right R modules are the same thing as left R^op modules

Which is just kinda a case of unwrapping what those words mean

[granite galleon]

That's true. Ty!

[fossil beacon]

can we classify continuous automorphism of Sl_2(R)

[fossil beacon]

i think we can do for lie group automorphism

i want to show any lie group automorphism induces Adjoint lie algebra iso

[sly crescent]

Yes

The continuous automorphism group of SL_2(R) is PSL_2(R)

[fossil beacon]

hints and steps

[sly crescent]

Figure out the inner automorphism group and the outer automorphism group

[fossil beacon]

thanks for your valuable answer

[fossil beacon]

https://math.stackexchange.com/a/1632734/1166895 this tells about it, of course not vary obvious

[fossil beacon]

this shown that every induced lie algebra automorphism will be Adjoint so by lie theorem every automorphism is adjoint also SL_2(R) is simply connected we get it is just inner

[crystal vale]

again this is an assignment problem, so if it has infinitely many roots, then L\{0} has infinite subgroup, {x \in K | x^n = 1 for some n in N }, but it is proper subgroup

i don't know where i have to use finitness of extension

[wicked patio]

I'm confused on what you're saying

[crystal vale]

me too

[wicked patio]

We need finiteness of extension because, e.g. C is a field extension of Q with infinitely many roots of unity

[crystal vale]

ah

i see

so any hint how do i prove it?

[wicked patio]

The only way I can think to do it rn relies on every finite field extension being finite dimensional, idk if you're allowed to use that

[astral jasper]

Out of curiosity, how else are finite field extensions defined?

[crystal vale]

yeah

[wicked patio]

Oh, I thought it was like finitely generated as a ring or something

[fossil beacon]

just see that you can irreducible polynomials of every degree, like cyclotomics, if L has infinite root of unity of course it has to be infnite dim,

[wicked patio]

I think I'm just being dumb

[crystal vale]

sorry i don't get it

[wicked patio]

All finitely generated field extensions are finite is the theorem

[crystal vale]

no

Q(\pie) is finitely generated but not finite extension

[wicked patio]

Finitely generated as a ring

[astral jasper]

Try to find the degree of the minimal polynomial of a primitive n-th root of unity

[crystal vale]

yes it euler function at n

[astral jasper]

Yeah, its degree is n-1

[crystal vale]

?

[astral jasper]

The degree of the polynomial

[wicked patio]

For prime n only

[azure cairn]

shouldnt it be for every positive integer n

[astral jasper]

Oop my bad

[wicked patio]

But yeah this is on the right track

[fossil beacon]

infinitely many roots will force arbitrary large irreducible poly, as cyclotomics

[crystal vale]

how?

[fossil beacon]

u have cyctomics

[velvet hull]

you mean as an algebra?

[wicked patio]

That's probably what it was yeah

makes more sense

Oh, weird, is there a counterexample for R then

as in the reals

[velvet hull]

okay I misremembered the exact statement

if R is finitely generated as an F-algebra such that R itself is also a field, then that implies that R is finite dimensional as an F vector space

but that requires quite the nontrivial proof

[wicked patio]

Yeah I think the F finite case is weird or something

well idk

[desert verge]

perhaps it's useful to note that if there infinitely roots of unity then there are infinitely many primitive units of unity

[wicked patio]

anyways I forgot that finite was defined as finite dimensional

and the other thing was the theorem

[desert verge]

wait maybe this is dumb because every root of unity is a primitive nth root of unity for some n

like obviously true

i suppose "primitive root of unity" in of itself doesn't mean much

[wicked patio]

Maybe it's easier to think about it containing primitive nth roots for arbitrarily large n

Then, you can show the extension has degree at least phi(n)

[desert verge]

mm ok

because there are phi(n) primitive n-th roots of unity

[crystal vale]

How?

[desert verge]

idk

[wicked patio]

Well every root of unity is a primitive root of unity

[crystal vale]

How?

[wicked patio]

Can you find one that isn't?

Every root of unity is primitive for some n

[desert verge]

every root of unity can be written as exp(i 2pi*n/m)
reduce the fraction n/m to n'/m' and ur root of unity is a primitive m'-th root of unity

[crystal vale]

I see

Yes

Exactly

Got it

Thank you @desert verge @wicked patio

[wicked patio]

No problem

[desert verge]

ur welcome.. but doesn't seem like we've solved the original problem

[fossil beacon]

that was quite trivial,

[desert verge]

thanks

for ur very useful comment /s

[crystal vale]

Yes

[desert verge]

maybe if u take a single primitive n-th of unity for infinitely many n, you get a Q-independent set

no idea

[wicked patio]

You just have to show that the degree is arbitrarily large

so, phi(n) goes to infinity as n goes to infinity

[crystal vale]

How do I show, so for given n, there are finitely many d such that f(d) ≤ n, where f is Euler function

[candid patrol]

With integer factorization

[astral jasper]

You can show that the totient function is multiplicative, that it goes to infinity on large primes, and also goes to infinity on large powers of "small" primes

[crystal vale]

I know it is multiplicative

I know f(p) = p-1

So it is for large n we have m such that f(m) > n

[lusty marlin]

You can prove, using the expression for the totient function ϕ in terms of prime factorisation, that ϕ(n)≥√n for all n other than 2 and 6.

This is stronger than what you need but it isn't much work to prove.

[crystal vale]

so i think it is enough to show for only p^n

[crystal vale]

[crystal vale]

i got it, thank you, mse and aops

[crystal vale]

so i find out the Galois group is D4( order 8 ), but i have to find the lattice of subfields? is there any easy way rather than doing computations?

[rocky cloak]

Do you know the fundamental theorem of galois theory (aka the Galois correspondence)?

[crystal vale]

yes

but i don't want to find fixed fields

[rocky cloak]

What do you mean, why would you do that?

FTGT tells you the lattice of subfields is exactly (the opposite of) the lattice of subgroups

[crystal vale]

yes

[rocky cloak]

So you just need to describe the lattice of subgroups

[crystal vale]

so i have to find the fixed fields of subgroups right?

[rocky cloak]

I mean you can, but the exercise isn't asking you to do that

[crystal vale]

then what it is asking?

[rocky cloak]

To describe that lattice of subfields

[crystal vale]

but i have to write explicitly subfields?

[rocky cloak]

Say the galois group was Z/4 then we would have the lattice

Q < K^2/4 < K

[rocky cloak]

Why?

What does that even mean exactly?

[crystal vale]

sorry i don't know what does it mean by K^2/4?

[rocky cloak]

The fixed field of the subgroup 2Z/4Z

[crystal vale]

so is it fine to write it without explicitly find the subfields?

[rocky cloak]

I mean your describing what the fields are and which contain eachother, I don't see how you would need anything more to describe the lattice of subfields

[crystal vale]

so do i need to find subfields explicitly or not?

[rocky cloak]

If you don't feel this describes them explicitly enough then feel free to describe them as specific as you like

[crystal vale]

Okay let me ask this to my professor

[rocky cloak]

All I'm saying is that the exercise is asking you about the lattice of subfields. It does not say "describe what each subfield is as specific as possible"

[crystal vale]

Yes I understand

[crystal vale]

so i am not sure this extension is normal or not, how do i find it?

how do i show \sqrt{2-\sqrt{2} } is not in K?

do i have to write explicitly in terms of basis and arrive at contradiction?

[candid patrol]

sqrt(2 + sqrt 2) * sqrt(2 - sqrt 2) = sqrt 2

[crystal vale]

Yes

[candid patrol]

So sqrt(2-sqrt 2) = sqrt 2 * (2 + sqrt 2)^{-1} € K

because sqrt 2 € K

[crystal vale]

I see thanks

[crystal vale]

Jagr, is it correct?

[fossil beacon]

fun fact for bi quadratic extension galois group of irreducible is D_4

[crystal vale]

no

you mean if K is splitting field of polynomial f which degree 4 and f is irreducible then Gal(K/Q) is D4? then no

[fossil beacon]

i meant bi quadratic

[crystal vale]

what does it mean by bi quadratic?

i am not sure what to do here, just adjoin the roots?

[rapid cave]

this does give you a splitting field. Just simply into a nice description of K

[crystal vale]

i don't know what nice description it can have?

[quiet pelican]

There are two “canonical” generators
I don’t quite mean “canonical” and more like
“The answers that any two humans who know the area well will give you are the same”

[sonic coral]

a quadratic in y evaluated at y = x^2

[candid patrol]

Don’t forget the normal subgroups

[crystal vale]

so do i have to mention corresponding extension is Galois extension?

[crystal vale]

i don't know i just know it is Q(\sqrt 2, (10)^1/3, w), where w is a primitive 3rd root of unity

[quiet pelican]

Yeah that’s how I’d write it

Now just prove that’s the splitting field

[tulip otter]

x^n=x implies x(x^{n-1}-1)=0 so an element of A is either a unit or a zero divisor. Now let p be any prime ideal in A and consider the ring A/p, we know that A/p is an integral domain and the goal is to prove that it is in fact a field. Now 0 in p so any zero divisor is also in A/p, but the zero element of A/p is p so any non-zero element in A/p is of the form x+p where x is a unit not in p and hence it has an inverse x^{-1}+p so that p is maximal

is this correct?

[quiet pelican]

Yes

Streamlined proof: if p is prime, A/p is an integral domain where every element satisfies x(x^{n-1} - 1) = 0. If x =/= 0 we can therefore cancel it to get x^{n-1} = 1, so x is a unit with inverse x^{n-2}

[tulip otter]

I also have another question, is there anything interesting that happens if the nilradical and the jacobson radical are equal?

[next obsidian]

Look up what a Jacobson ring is

[delicate orchid]

maybe I will. but NOT because you told me to

[next obsidian]

I don’t care about you

[tulip otter]

alright

tysm mico and Chmonkey, have a great day/night

[azure cairn]

the first condition requires closure right

[quiet pelican]

“Together with a law of composition” requires closure

[azure cairn]

if "equipped" were used there instead of "together", that wouldnt imply closure

[quiet pelican]

It would

[azure cairn]

how?

[quiet pelican]

Because a law of composition on a set G is a function G x G -> G

[azure cairn]

yes

so the law of composition is the thing that guarantees closure

oh wait i misread the first bullet point lmao

yeah so by definition of a law of composition, G is guaranteed closure when equipped with one

[quiet pelican]

Yes

When closure is stated, it’s usually stated as a redundant axiom to help ease the transition into the defn of a subgroup

[glad osprey]

Just nitpicking, but closure is not a property of G, it's the law of composition that is closed

[quiet pelican]

It’s a property of G (the group) not G (the set)

[glad osprey]

Or G under the law of composition I guess

[azure cairn]

yea

when denoting the multiplicative group of $\bR$, is it fine to write $(\bR \setminus {0}, \times)$?

[cloud walrus]

Altanis

[azure cairn]

it feels iffy to write it as (R, *) since 0 is obviously not in the group

[lusty marlin]

Yes

[azure cairn]

my book writes it as R^+ and R^(\times) which i dont like 😢

[quiet pelican]

R^\times or R^* are common for the multiplicative group

[azure cairn]

im fine with R^\times and R^* but i dont like R^+

cuz i use that for positive reals

and i dont want to mix and match notation so if i do (R, +) ill do (R, *)

but then (R, *) feels wrong to me because 0 isnt in the group

[coral spindle]

Well it's definitely technically wrong too

For what it's worth, I've seen books mix (R, +) and R^\times

[rocky cloak]

The easy solution would probably be to write R and R^\times

[azure cairn]

Eh idm verbosity that much

[vapid vale]

you can also just say (as an additive group) or something

under addition

why use symbol when word do trick

[azure cairn]

i dont mind verbosity in symbols

but itd be annoying to write out (as an additive/multiplicative group) every time

[vapid vale]

you don't need to every time, but fixing it for clarity at the start of something might be useful if there is ambiguity, in any case R^\times is fine for multiplicative

[azure cairn]

hm

ig ur right

[vapid vale]

anyways i think its decent practice to be as clear as possible, and if you have to puzzle over a notational standard to stick to, its probably not ideal to everyone

[azure cairn]

now another thing is that i like how (R, +) and (R \ {0}, \times) look

but that might be solely due to its novelty

like how i used to like the notation f: X -> Y in 10th grade but now idc

[rocky cloak]

What? That notation is great. Is there even an alternative worth considering?

[azure cairn]

no i still like it, but ive gotten used to it

[rocky cloak]

I guess good notation is like good special effects. They're best when you don't notice it

[alpine island]

should I show in detail that the natural maps here are well-defined & homomorphisms? or should I just do the computation that they are inverses to each other?

[velvet hull]

yeah you should show everything

on the level of modules, showing well-definedness usually nontrivial

[noble nexus]

well-definedness should be automatic if you prove it in the right way (just verify both satisfy the universal property)

ok though I suppose you need maps from M\otimes B and N\otimes B into the tensor product on the left

but even then the maps you get should come from the universal property

[novel star]

is there a way to prove the fact that projective modules over local rings are free without using full axiom of choice

[karmic moat]

Wouldnt this give an axiom of choice-free proof that every vector space has a basis? Just take the local ring to be a field, then every module over it is free (and thus projective)

[next obsidian]

This is trivial using tensor rules

[novel star]

🙁

[next obsidian]

You need to prove commutativitity, associativity, and that tensoring by B over B does nothing

[novel star]

logic why must u hurt me so

[next obsidian]

Each of these isomorphisms exist in a very simple way which you can write down explicitly

And then you can compose them

[next obsidian]

What?

[alpine island]

How would associativity work here when we're tensoring over different rings?

[karmic moat]

Maybe i misread the question actually

[next obsidian]

Okay I tried to type it out but it was annoying

But you have to prove associativity on the level of bi-modules

If you have a right A-module, an A-B bimodule, and a left B-module you can show that

M (x)_A (N (x)_B L)

This comes down to showing that the universal property plays nice with bimodules

From there you do

(M x_A B) x_B (N x_A B) =
(M x_A B) x_B (B x_A N) =
((M x_A B) x_B B) x_A N =
(M x_A B) x_A N =
M x_A (B x_A N) =
M x_A (N x_A B) =
(M x_A N) x_A B

[karmic moat]

I feel like my logic is fine

• suppose you can prove the statement “every projective module over a local ring is free” without choice (or any equivalents)
• then let your local ring be a field and M a projective module over that field
• then it follows from the claim that M is a k-vector space with a basis
  So it’s a corollary that every vector space admits a basis, and we didnt use choice
  But the claim that vector spaces admit bases is equivalent to choice

[next obsidian]

Why is M automatically projective lol

[karmic moat]

Yeah my original message was probably wrong, but the new one is fine, just take M to be a projective module

[next obsidian]

But you haven’t shown all vector spaces have a basis

You showed all projective vector spaces have a basis

[karmic moat]

Yes i just realized that

:p

Whoops

[alpine island]

okay this makes sense

[next obsidian]

I wonder if maybe you can get away with just dependent choice or something

[alpine island]

we haven't done bimodules at all, so I think I'm just typing out the slop and maybe hand waving part of it

but this is a nice way to do it that's less element-wise grinding

[next obsidian]

I mean

It’s just a universal property thing

Just show that both of them factor the right kind of B-bilinear maps

[alpine island]

that's true

[next obsidian]

And use the fact that maps out of M (x)_A B -> N for a B-module N is exactly A-linear maps M -> N

This is the adjunction of push-pull

Again, a lemma

Really really simple

[alpine island]

I mean I'm using the fact that elements of the form n (x) m form a basis, and I can just work on them individually, which is easy

a lot of this can be copy-pasted in the tex

[next obsidian]

It’ll end up turning into “maps from this into a B-module L is an A-bilinear map from M x N -> L”

[alpine island]

true

[next obsidian]

How do you show the map exists?

[alpine island]

$$\begin{array}{{2}{@{}c}{15}{@{,}c}}
f & {}:{} & (M&\otimes_A&N)&\otimes_A&B& {}\longrightarrow{} & (M&\otimes_A&B)&\otimes_B&(N&\otimes_A&B) \
&& (m&\otimes&n)&\otimes&b & {}\longmapsto{} & (m&\otimes&b)&\otimes&(n&\otimes&1_B)\
\end{array}\qquad\text{and}\qquad\begin{array}{{2}{@{}c}{15}{@{,}c}}
g & {}:{} & (M&\otimes_A&B)&\otimes_B&(N&\otimes_A&B) & {}\longrightarrow{} & (M&\otimes_A&N)&\otimes_A&B\
&& (m&\otimes&b)&\otimes&(n&\otimes&b') & {}\longmapsto{} &(m&\otimes&n)&\otimes&bb'\
\end{array}$$

[cloud walrus]

nHail

[next obsidian]

And I think it’s gonna be virtually impossible to show it’s injective manually

[alpine island]

just take these two maps, fg = id, gf = id

[next obsidian]

You can almost never define a map out of a tensor product like that

I mean what you would need to do is show that the map out of the similar product is bilinear

But the relations inside the tensor product are so fucked trying to define on simple tensors won’t work

You can’t show it’s well-defined

Or well, doing so is equivalent to saying the map out of the product is bilinear

[alpine island]

can't you? we have this from class

[next obsidian]

Okay yeah yeah this is showing the map out of the product is bilinear

[alpine island]

yeah exactly

[next obsidian]

I think it’s kind of backwards to do it this way though tbh but I’m just a hater

Tensor products should be worked with universally

And then you unwind to see what the map actually is

[alpine island]

yeah I think we're just kinda brushing over the bilinear map thing and working with linear maps from the tensor

which as you said is equivalent

maybe \otimes is one character longer than \times

[next obsidian]

I mean it’s the same thing lol

The thing says “above universal property” which is gonna be that one

[alpine island]

it's like a page up, there are some other remarks in between saying that {a (x) b} is a generating set, etc.

[next obsidian]

No I mean I know what that was gonna say

This is saying it factors bilinear maps and is universal for it

[alpine island]

oops, swapped the word order of this and thought you said "which is that one gonna be"

yeah this is the only relevant universal property

[next obsidian]

Wellllllll relevant here sure

Hom-tensor is… well it’s the same thing, but in a more useful form

Theoretically

[alpine island]

sure

[next obsidian]

Anyway whatevs