groups-rings-fields

[vocal pebble]

on this note, do you know any good books you recommend for self studying ring theory?

[south patrol]

hm to be honest i remember being paranoid about varying definitions too and learnt some stuff from a uni course lol

but honestly i wouldn't really worry about the different definitions since there are quite standard things (all rings have units and maps preserve units)

i learnt most things through Atiyah-Macdonald's Commutative algebra, but, as its name suggests, it only treats commutative rings and does expect some ring theory already (I kinda learnt backwards using this which was good in some ways but ye)

There's also Lang's algebra I guess though perhaps that's not the most popular suggestion lol

[vocal pebble]

yeah, makes sense 👍 i would expect it to be pretty standard and here i should have just inferred it

[south patrol]

to be honest i think the ideal vs left/right ideal thing may be slightly less standard but i would generally assume ideal means two-sided

though i'm no expert on the terminology there (i basically work only with commutative rings)

[rocky cloak]

I'd say it's context dependant. If I'm in a context where I'm only interested in left ideals I'll just say ideal to refer to left ideal.

If I'm in a context where I'm talking about both left ideals and two-sided ideals I might use "ideal" to mean two-sided ideal.

[olive parrot]

i think i understand this now, prime powers are always coprime to each other, and the LCM of coprime numbers is their product which is maximal

correct me if i'm wrong

[bitter rover]

Yep! So computing Landau's function is about relating the sum and product of (powers of) primes.

The fact that RH comes in hopefully doesn't seem out of left field, at least.

[olive parrot]

that makes sense, thank you!

hm wait also

does this mean the cycle lengths of every partition of S_n whose LCM is L(n) consists of prime powers only

is there an easy enough proof for this that I can figure out

[rocky cloak]

I don't know if any example that achieves L(n) necessarily must break down into prime powers.

But it's fairly easy to prove that given any permutation you can find a permutation with the same order that is the product of cycles of prime power orders.

[olive parrot]

I care about if everything that achieves L(n) breaks down into prime powers

[rocky cloak]

Well then consider S6

[olive parrot]

the largest is legnth 5 and length 1 iirc

[rocky cloak]

? What is largest length 5?

[olive parrot]

L(6) = 5 is formed from LCM(5, 1) and 5 + 1 = 6

[rocky cloak]

So Sn always contains an n-cycle

So L(n) >= n

[olive parrot]

yeah

[rocky cloak]

... And 6 > 5

[olive parrot]

but that's valid because 5 is a prime power

[rocky cloak]

Okay, I have no idea what you're talking about

[icy bear]

just thought about it... should I do artin's exercises section by section or chapter by chapter?
I'm feeling kinda stupid rn cus just realised I've been doing it chapter by chapter even tho the exercises are separated into sections

[olive parrot]

the first sentence here

[icy bear]

damn that's why I've been feeling headaches reading this I've been reading the entire chapter before doing the sections' exercises I just have to acumulate so much information damn

[rocky cloak]

Yeah, so I'm trying to tell you that you should consider S6 because it will give a counter example

[bitter rover]

Take cycles of length 2 and 3 and 1

[rocky cloak]

Or a cycle of length 6

[olive parrot]

oh whoops sorry i meant to say L(6) = 6

but 2 3 and 1 is fine because the non trivial cycles have lengths of prime powers

[rocky cloak]

Yes, but 6 is not a prime power.

[olive parrot]

oh I think I see what you're saying

well, what if we ignore the N cycle

[rocky cloak]

I think L(11) can be achieved as a 6-cycle and a 5-cycle

[olive parrot]

ah, you're right i just wrote some code and 11 gave 6 and 5

[olive parrot]

then this is false, thanks for the help

[bitter rover]

If something has a 5 cycle and a 3 cycle, you'll find a permutation of the same order with a 15 cycle (for large enough n)

[olive parrot]

partitions of S_6, S_11, S_18, S_22, S_46 seem to be the only ones that violate this property for n < 80

[bitter rover]

L(15) is 105 and corresponds to 3+5+7

[olive parrot]

yes that is the only one of 105

and it consists of prime powers only so its valid

[olive parrot]

I have no idea what's special about the groups

the results from some short code i wrote[ [ 3, 2, 1 ], [ 6 ] ] 6 [ [ 5, 3, 2, 1 ], [ 6, 5 ] ] 11 [ [ 7, 5, 4, 3, 1, 1, 1 ], [ 7, 5, 4, 3, 2, 1 ], [ 7, 5, 4, 3, 3 ], [ 7, 6, 5, 4 ] ] 22 [ [ 13, 11, 7, 5, 4, 3, 1, 1, 1 ], [ 13, 11, 7, 5, 4, 3, 2, 1 ], [ 13, 11, 7, 5, 4, 3, 3 ], [ 13, 11, 7, 6, 5, 4 ] ] 46

[rocky cloak]

Well in all cases you have [3, 2, 1] that gets replaced by a [6]

[olive parrot]

i didn't see that, that's true!

so every partition with a sequence of primes whose sum is their product could potentially have a maximal LCM partition with a non-prime power

let me try to figure out if [3,2,1] and [6] is the only replacement

[bitter rover]

I'd guess it's either 2 and anything, or the fact that 2 and 3 are the only consecutive primes

[olive parrot]

maybe with larger N there become three trivial cycles in the maximal LCM partition

so then 5 2 1 1 1 becomes 10

[bitter rover]

Oh I see, well

If you had 1 1 1 that could become 3

[olive parrot]

3 is a prime power so 3 is fine

[crystal vale]

To prove, if n ≥ 5 then S_n has no subgroup of index t, 2< t < n.

Let H be a subgroup of index t, then there exists homomorphism S_n to S_t such that ker \subset H, now we know that only non-trivial proper normal subgroup of S_n is A_n if n≥5.

S_n cannot be imbedded into S_t so ker must be A_n but then it contradicts that H has index t > 2.

Is it correct?

[crystal vale]

Any hint? Given hint : use the fact {(12),(123..n)} generates S_n

Is there any way to do this one by group action?

[lusty marlin]

Yes

[lusty marlin]

Any p-cycle generates a subgroup of order p, where every non-identity element is a p-cycle

[crystal vale]

Sorry I get this point but I don't know how to go further

[cerulean swan]

Note that it suffices to generate every single transposition since everything can be written as the product of transpositions

[crystal vale]

Yes

But if transposition is (ij) and p - cycle is (a_1,...,i,..,j...,a_p) then if I can show there exists a p-cycle such that (i j,...,a_p) then I am done

[cerulean swan]

** transposition of neighboring elements

There are multiple ways to do this question, but working in the finite field $\mathbb F_p$ is probably the easiest way

[cloud walrus]

Sleepybear

[crystal vale]

How does the finite field help here?

[cerulean swan]

So if $k$ is the difference $j-i$ between the two elements in our transposition $(i\ j)$, then we can generate all elements of the form $(1\ 1+qk)$ for $q\in \mathbb N$

[cloud walrus]

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

[cerulean swan]

and then from here noting that $q$ has an inverse modulo $p$ finishes the problem

[cloud walrus]

Sleepybear

[cerulean swan]

does this make sense?

[crystal vale]

I am thinking about defining some action

[sonic coral]

i think you should look at conjugation

and what transpositions you can get from that

[crystal vale]

In ii) I proved that the intersection of all conjugation of H is the normal subgroup of G.

Since H has a finite index so it has finite conjugation, right?

If H has a finite index then xHx^-1 has a finite index, I think yes

Then I can do induction on i)

[crystal vale]

I don't get it why { H } is an orbit of size 1, G acts transitive on X by conjugation, right?

[next obsidian]

I don’t get this proof

It seems bunk

[crystal vale]

Me too

[next obsidian]

Maybe you let H act on X

Cuz H is also a p-group

But idk

And then {H} is an orbit

[crystal vale]

Yes

[serene dune]

is it normaliser lemma ?

to then prove sylow theorem ?

[crystal vale]

What is that?

[serene dune]

its index, normaliser, mod p, wait a min

https://www.math.clemson.edu/~macaule/classes/s24_math4120/slides/math4120_slides_chapter05_h.pdf

[crystal vale]

Got it yes H act on X

Thank you

[crystal vale]

122 pages

[serene dune]

search for key words

[grizzled spindle]

Kinda stuck on this last case for this problem

Like all I have is that the sum of the square is composite and I don’t see the line to produce the decomposition of a+bi

[next obsidian]

If p = 1 mod 4, then there exists c with c^2 =-1 mod p cuz (Z/pZ)^x = Z/p-1Z and you want a c of order 4.

So p divides (c-i)(c+i), but it doesn’t divide c-i or c+i so p isn’t prime.

This means you can write p = a•b for Gaussian integers an and b, and they aren’t units. But then

p^2 = N(p) = N(ab) = N(a)N(b), but then N(a) = p so a is a Gaussian prime. But N(a) = a•a-bar so when you write a in the form c + di you get it

This is kinda half of it, you need to do a little more to say any Gaussian prime is of the forms in the above

I’m not sure what work you’ve done so far so it’s hard to say how you can show any prime not of the other two types is that a + bi thing

[grizzled spindle]

oh i showed before this that a prime is the sum of 2 squares iff it is 4k+1 or 2

that was a bit hard but yeah this makes a few cases quite easy

im mostly stuc for when a^2+b^2 is composite

[next obsidian]

Oh well

Okay

Say x is a Gaussian prime

Then find a rational prime p dividing N(x)

Then by construction there is a Gaussian prime y dividing p

If p is 2 it’s 1 + i

If it’s 3 mod 4 then it’s p

And if it’s 1 mod 4 you do the other thingy to find that pair of primes

So now you have
y | p | N(x) = x•x-bar

So y divides x or x-bar

So y or y-bar divides x

And since those are all Gaussian primes, up to a unit you know x = y or y-bar

Where y or y-bar is either 1+i, 1-i, p for p = 3 mod 4, or a + bi where N(a+bi) = a^2 + b^2 = p

Where p is 3 mod 4

[grizzled spindle]

ah yeah i think i get it from that

didnt think of using conjugate

thx

[kind temple]

anything interesting you get from localizing a ring at its nilradical?

[next obsidian]

You can’t

Wait

Can xy be nilpotent when x and y aren’t?

Yes they can

k[x,y]/(xy)^2

[kind temple]

i thought you could localize at any multiplicative set

oh dang.

1

[next obsidian]

Yeah

But the complement of the nilradical isn’t multiplicative

Unless you mean set S = N(R)?

But then you localized at a set with 0 in it so it just becomes 0

[kind temple]

what is N(R)?

[next obsidian]

The nilradical

Of R

[kind temple]

okay. i totally forgot how localization works then. just running through some concepts on my own atm since im super out of practice

[next obsidian]

a/x = b/y if there is u in S where
u(bx - ay) = 0

If 0 is in S you set u = 0 and then everything is equal to each other, so you’re the 0 ring

[rocky cloak]

If it's prime, then you get a perfect ring.

If R is also Noetherian you get a local artinian ring.

[next obsidian]

How can you even localize a ring at the nilradical

Or is the second sentence contingent on the first

[rocky cloak]

Yeah if the nilradical is prime you can localize at it

[next obsidian]

Yeah okay

I guess that’s what the word “also” is doing

Tfw reading

[rocky cloak]

Actually I'm a bit unsure if you get a perfect ring, might just be semiperfect actually.

0-dimensional local ring at least

[next obsidian]

Yeh

[tardy hedge]

I have to work thru atiyah macdonald and im kinda not looking forward to it too much

Comm algebra does feel a bit boring sometimes

[next obsidian]

The start can be kinda meh I guess

Without understanding why it’s necessary or the geometric consequences

But Atiyah MacDonald can be done pretty quick

So think of it like taking medicine

[sonic coral]

my comm algebra class is using it next semester and i’m excited

[valid night]

A primitive polynomial with no integer roots is irreducible in $\mathbb{Z}[X]$ correct? The same goes for a polynomial with no rational roots in $\mathbb{Q}[X]$ and those with no real roots in $\mathbb{R}[X]$ right?

[cloud walrus]

Kroros

[rocky cloak]

Consider for example (x^2 + 1)^2

Is it irreducible? Does it have integer/rational/real roots?

[valid night]

Oh right of course

Thanks

[tardy hedge]

yeah true, i flipped through and noticed many of the chapters seem quite short

[rotund aurora]

Can every ideal of a one-dimensional noetherian integral domain be generated by at most 2 elements?

[next obsidian]

I want to say no, because you have to show this is true for a Dedekind domain via nontrivial methods

[rotund aurora]

For Dedekind domains it is true because the quotient by an ideal is a PIR (edit: need not be an integral domain lol)

[next obsidian]

That’s a really Monka way to prove it

But I just think it can’t be true for all 1-dim Noetherian rings

If u gotta prove it for Dedekind domains using some facts about Dedekind domains

[rocky cloak]

How about
Z[w] where w^3 = 16.

Can the ideal generated by 4w, 2w^2 and 16 be generated by just two elements?

It doesn't seem like it to me, but I could be wrong.

[rocky cloak]

Okay, so by this MSE
https://math.stackexchange.com/a/110734/306319

The ideal (8, 4x, 2x^2, x^3) in Z[x] cannot be generated by less than 4 elements.

This ideal contains x^3 - 16, so modding this out the image ideal cannot be generated by less than 3 elements. And the image ideal is (8, 4w, 2w^2)

And I guess this generalizes to give ideals with arbitrarily many generators by just considering
Z[w] with w^n = 2^n+1

[rotund aurora]

btw, just to sanity check, if I=(a1, a2,...,an) is an ideal and I is generated by m elements, say m<n, then it need not be true that these m elements are a subset of a1,..,an, right?

[next obsidian]

Yeah

[rocky cloak]

Yeah, it should be possible that {a1, ..., an} is minimal (meaning you can't remove any of the ai's) without n=m

[rocky cloak]

I guess (2, 3) in Z is a good example

[rotund aurora]

Is there a nice way to bound the number of generators of ideals? I'm only interested in orders of number fields really (aka a subring of O_K that contains a basis of K over Q)

in terms of the degree, for example

[wraith cargo]

I think the answer is no
I found at some point examples of noetherian rings where you have arbitrarily many generators for certain primes

I mean I found them online lol

[rotund aurora]

An order of a number field?

[wraith cargo]

https://projecteuclid.org/journals/journal-of-the-mathematical-society-of-japan/volume-26/issue-4/On-the-unboundedness-of-generators-of-prime-ideals-in-powerseries/10.2969/jmsj/02640722.full

no it's a different rings but still noetherian

idk what the situations is for orders

[stark gull]

help pls

[quaint fulcrum]

Hello! Could someone explain how the highlighted approach works to prove the theorem below

The theorem I am reffering to is "The sum of the cubes of any three consecutive integers is divisible by 9"

[next obsidian]

You’re asking a statement about some equation being equal to a fixed value (0 in this case) mod 9

The way modular arithmetic works, you can simply ask if
(n mod 9)^3 + (n + 1 mod 9)^3 + (n + 2 mod 9)^3 = 0 mod 9

If you modify n by a multiple of 9, whatever you have changed the equation by after doing all these powers and sums will still be a multiple of 9, so it doesn’t change the value when you’re already working mod 9

So you only need to handle the case where n = 0,1,…,8 because these are the 9 different values that exist mod 9

[night tartan]

Gal(L/sigma K) are precisely the elements of Gal(L/F) which fix sigmaK

[tardy hedge]

{2,3} is a basis for Z as a Z-module? But {1} is also a basis, and since Z is commutative I thought it had IRP

Omg nvm lol

{2,3} is not linearly independent 3x2+(-2)x3=0

But this does show an example of how modules differ from vector spaces tho right. In a vector space every generating set contains a basis

[keen badge]

How can I prove(if it's even true) that if a polynomial is irreducible in $\mathbb{C}(x)[y]$ then it's irreducible in $\mathbb{C}[x,y]$?

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[grave sedge]

C[x,y] is a subring of C(x)[y] so any factorization in the former would automatically be a factorization in the latter

[rocky cloak]

So you have to be a little careful, since something like xy is irreducible in C(x)[y], but reducible over C[x, y].

But if you require that the coefficients don't have any common factors (for example if it's monic in y) then a polynomial is irreducible in C[x, y] if and only if it is irreducible in C(x)[y].

This is Gauss lemma.

[grave sedge]

(this is indeed wrong, as i forgot that nonunits in C[x,y] can become units in C(x)[y])

[keen badge]

Isn't Gauss lemma is polynomial is irreducible in Z iff it's irreducible in Q? @rocky cloak

[rocky cloak]

Gauss' lemma is that a primitive polynomial is irreducible over a UFD iff it is irreducible over its field of fractions.

In particular Z is a UFD yes

[keen badge]

ohh

So a monic irreducible polynomial in C(x)[y] is also irreducible in C[x,y]

[rocky cloak]

Yes, though the direction you're asking about is the easy one. As Matteddy pointed out, and doesn't need any UFD stuff.

Any factorization in C[x, y] will give a factorization in C(x)[y] so you just have to exclude the case where one factor is a unit.

And for example monic would do that for you

[keen badge]

Ok. Ill try(to prove it)
Thank you both

[rocky cloak]

And of course you need your polynomial to actually be in C[x, y], otherwise the question doesn't make sense

[pearl fog]

https://kconrad.math.uconn.edu/blurbs/ringtheory/euclideanrk.pdf i have trouble making example of lemma 4.3

i consider 4=2*2=(1+sqrt(-3))(1-sqrt(-3))

but both 2 and 1+-sqrt(-3) are irred

[rocky cloak]

From the lemma you get a strict inclusion of ideals
(a) < (b)
applying the lemma again you get
(a) < (b) < (c)
and so on.

So whatever ring you consider would necessarily have to be non Noetherian

Since in a Noetherian ring a chain of ideals cannot increase forever

[pearl fog]

so this is under assumption non noetherian

[rocky cloak]

Well the theorem is also true for Noetherian rings. But a consequence is that every element is a product of irreducibles. So you can't find interesting examples

[pearl fog]

cuz this is from our proof PID->UFD, is this proof wrong?

a required to be non irred, but (1+sqrt(-3)) irred

[rocky cloak]

The proof isn't wrong, it's just pretty rare that you can't write an element as a product of irreducibles

One example could be
Q[x^2^-i : i in N]

Then x = x^1/2 * x^1/2
and x^1/2 = x^1/4 * x^1/4 etc

x cannot be written as a product of irreducibles

[pearl fog]

so such examples are usually "non discrete" enough for the chain of ideals to keep ascending

[rocky cloak]

Sure, I mean the example can still be fairly "discrete" it just has to be pretty infinite / unbounded

[potent condor]

I hate ambiguously worded questions in math

Does (2) mean "R_0 is a noetherian Z-module" or are we assuming R_0 is a field from the outset and (2) means "R_0 is a noetherian ring"?

The second interpretation makes it trivial but the first one sounds false

[tardy hedge]

M an A-module with Ann(M)=A, then M = 0?

Yea i was thinking smth about q

1

If xM = 0 x is in Ann(M), but if x is a unit then since Ann(M) is an ideal it should be all of A?

For context if it matters i am looking at nakayama lemma kind of things

[rocky cloak]

R0 should be a Noetherian ring. Your not assuming that it is a field, but if it is a field then it is Noetherian, so the condition would hold.

The point is that corollary 1.3 is about algebras over a field, so they're explaining the connection to that.

But what you're asked to prove is more general

And I guess in general when you have a ring and someone says it's Noetherian they mean it's Noetherian as a ring, not as a Z-module.

[potent condor]

How do we know R_0 is a ring?

[vivid kestrel]

if we have a polynomial of the form $y^2 + y \cdot s(z) + t(z) \in K[y, z]$ that is reducible, can we conclude that a factorization must be of the form $(y + p(z))(y + q(z))$

[potent condor]

Why must it contain 1?

[cloud walrus]

eggman

[next obsidian]

It means R_0 is a Noetherian ring

R_0 is always a ring

I don’t really think this is ambiguous

[potent condor]

Why must R_0 always be a ring? I see nothing in the definition implying that

[next obsidian]

1 has to be of degree 0

Because it multiples everything to itself

If it has any other parts of any other degree it wouldn’t be able to do that because it would push some homogeneous element outside of the degree it’s supposed to live in

R_0 multiplies itself into itself

And is closed under additio

And has a 0

[next obsidian]

I think the answer is yes, and requires you to recognize that k[y,z] = k[z][y], and that reducible is a condition that can be stated purely ring theoretically. So to be reducible as a poly in y and z is the same as being reducible as a poly in y, over the ring k[z]. Then you notice that over k[z] this is a monic quadratic in y, and so has to factor as two linears in y

[potent condor]

oh i get it, thanks

[next obsidian]

Np

[vivid kestrel]

ah that figures, thanks

[potent condor]

Np

[cloud walrus]

person2709505

[coral steeple]

Correction: Lambda is the set of $P$-orbits of $\Sigma$. Also, can I write $\Lambda=^P\Sigma$?

[cloud walrus]

person2709505

[noble adder]

hey guys

can someone help me with a problem from gallian

i don't understand the solution and the final is in 45 minutes

[misty cosmos]

chat i need help:

[noble adder]

the problem asks to prove that a group of order 375 has a subgroup of order 15

[noble adder]

💀

[misty cosmos]

we gonna fail together

[noble adder]

🙏

[misty cosmos]

ok so there's only 1 sylow 5-subgroup

and there's 1 or 25 sylow 3-subgroups

if theres 1 its trivial

if there's 25 sylow 3-subgroups

[noble adder]

the solution says that if there are 25 sylow 3-subgroups, then the index of the normalizer is 25

why is that?

[misty cosmos]

hmm yeah fr i was confused on that too

this is not splendid

[vivid kestrel]

isn't that just one of the sylow theorems?

[noble adder]

really?

doesn't seem to be one

[noble adder]

OH I AM SO SORRY

[vivid kestrel]

its listed on the wikipedia page and in the lecture notes we got at least

yeah

[noble adder]

I DIDN'T SEE THAT

[misty cosmos]

embarassing!

[noble adder]

i agree 😔

[vivid kestrel]

good luck with your final

[misty cosmos]

its chill we both got bamboozled

[coral steeple]

Good luck with the final folks

[noble adder]

thank you eggman

thank you person2709505

😭 that actually isn't stated in our book

we have the weird version of sylow

[misty cosmos]

yeah gallian doesn't have it

[coral steeple]

What is the foregoing argument we are applying to P (the supposed element of Pi \setminus Sigma)? Surely not the one that starts "let P be in Sigma"? Does Sigma refer to the same thing on the second page as on the first?

Is it just me or is this proof written rather compactly

There is literally no other place in the proof where we have some chain of reasoning that starts with a Sylow p subgroup of our choosing

[rocky cloak]

I mean, the problem isn't asking for the subgroup to be normal.

So just take the product of your sylow 5-subgroup and one of the sylow 3-subgroups and that's a subgroup of order 15

[rocky cloak]

It's not so much a sylow thing as just in general the index of the normalizer is equal to the number of conjugates of a subgroup.

All the sylow subgroups are conjugate so then the number of conjugates is the number of sylow subgroups

[rocky cloak]

It's a little confusing that they reuse the symbol P.

But yeah Sigma is the same set as before, and the foregoing argument is the one where a sylow p-subgroup only fixes itself.

So if P is not in Sigma, then Sigma has no P-orbits of size 1, which means |Sigma| is 0 mod p.

But it's not 0 it's 1, hence contradiction.

[coral steeple]

Hm I think another way of saying this is that on the first page we let the new P (not in Sigma) partition Sigma into orbits right?

[rocky cloak]

Yes exactly. We partition Sigma into P-orbits each of which has size p^k for some k. But importantly not k=0

[crystal sundial]

I noticed that in the definition of prime ideals we recuiqre the ideal to live within a commutative ring. I'm a confused as to why we recquire this.

[coral steeple]

Ok, thanks. Your description is clearer I think

[rocky cloak]

You can define prime ideals for noncommutative rings (and people do), but they don't play any particularly interesting role in noncommutative algebra.

Their importance is for commutative algebra and algebraic geometry / algebraic number theory, so that's the setting most people define it in

[crystal sundial]

I see, thanks a lot for the insight!

[somber bluff]

where

is this wrong because when I do it I get that it is an antihomomorphism

[coral steeple]

Is there a minimally error-prone way to find the sign of a permutation? (Trying not to lose marks on an exam)

Finding a determinant seems finicky

[barren sierra]

why are you doing a determinant lol

What is your definition of sign @coral steeple

[coral steeple]

Determinant of the permutation matrix

[barren sierra]

ok yea there are a bunch of other definitions you should know

proving these are equivalent takes some work but at the least you should know some other characterizations

So one classic theorem is that every permutation can be expressed as a product of transpositions (I recommend you prove this at least, it's not hard)

[coral steeple]

Is that a good method, expanding everything out as in the proof that the transpositions generate Sn?

I'd need to check how many steps we used

[barren sierra]

For a permutation $\sigma$, the minimum number of transpositions needed to express $\sigma$ is called the length of $\sigma$. If $\sigma$ has length $\ell$, then it's sign is $(-1)^\ell$

[cloud walrus]

Spamakin🎷

[barren sierra]

But arguably the quickest way is to count the number of inversions

[coral steeple]

Oh hey I remember that from intro linear algebra (not this course though)

[barren sierra]

For a permutation $\sigma$, an inversion is a pair $(i, j)$ where $i < j$ but $\sigma(j) < \sigma(i)$. If the number of inversions of $\sigma$ is $k$, then it's sign is $(-1)^k$

these are all equivalent, but it isn't immediate that these are all equivalent

[cloud walrus]

Spamakin🎷

[coral steeple]

So we have 2^n pairs to check?

[barren sierra]

check your math again

simple counting

[coral steeple]

n^2 pairs

[barren sierra]

A little less, cause i < j

n choose 2

so like you don't need to check both (2, 3) and (3, 2)

[coral steeple]

Well that beats n!

Thanks

[somber bluff]

@tardy hedge ? am i correct in saying it as an antihomomorphism

[tardy hedge]

why did you tag me

Im assuming mistake?

[somber bluff]

i thought u meant something by hearting the message

im sorry

[tardy hedge]

no worries

[coral steeple]

Is G not {1,x,y,xy}?

I.e. C_4 or C_2 x C_2

Or is G not <x,y>

[coral steeple]

I suppose xy could be 1

[crystal vale]

In C4, there are no two distinct elements of order 2

[crystal vale]

Not necessarily, take D8 = < s, sr >

[coral steeple]

Well anyhow G is contained in {1,x,y,xy}, so it has order 3 or 4. With what you said in mind it is C3 or C_2^2. 2 doesn't divide 3 so it's C_2^2. No?

[crystal vale]

But what if xy≠yx ?

[abstract rock]

can you think of a presentation of the latter that would make sense as a dihedral group?

[coral steeple]

No, which is why I was confused...

[abstract rock]

think about the dihedral group

how do we usually construct it

or any example of a dihedral group i mean

[coral steeple]

I had convinced myself that xy=yx somehow

[abstract rock]

theres a way to do that

pretty easily from the givens

||whats the inverse of xy?||

[coral steeple]

yx

How does that help?

Figured it out

[abstract rock]

glad i could help

[coral steeple]

If A is a subset of a group G, and gAg^-1 is contained in A for all g in G, is <A> necessarily normal? This seems obvious but I can't prove it

My guess would be to look at the intersection of all normal subgroups containing A. I couldn't get that to work

Obviously if A is not finite then explicitly writing down elements of <A> is not possible

[mighty kiln]

Yes

[mighty kiln]

They are all of a certain form

[coral steeple]

$\langle A\rangle=\prod_{\substack{a\in I\subseteq A\I;\text{ finite}}}a^{\pm 1}$?? Really?

[cloud walrus]

person2709505

[coral steeple]

Well that might not be right but I mean finite products of powers of elements of A

Somehow I thought that you could only do this when A was finite, or otherwise G or A could be sufficiently big/unwieldy that you can't represent elements of <A>

[rotund aurora]

What is the simplest way to see Q(cbrt 2)!=Q(cbrt 3)?

[mighty kiln]

This always holds

You can also argue this abstractly

Conjugation by any element is an automorphism and preserves conjugacy classes

And A is a union of conjugacy classes

So the intersection <A> of subgroups containing A is preserved under this automorphism

[delicate bloom]

I was trying to keep it simple, but I don't think this approach can lead to something that works, but I'll post the attempt so far anyways.

for the sake of contradiction, suppose they're equal and call L= Q(cbrt 2) = Q(cbrt 3). Because it contains both cbrt(2) and cbrt(3) it contains their product, so L must contain K = Q(cbrt(6)) as a subfield.

[L:Q] = [L:K][K:Q]

But [L:Q]=3 and [K:Q]=3 means [L:K]=1 and so L=K.

I had thought it would be more obvious from this step now that all 3 fields are equal, but nothing stands out to me here.

[coral steeple]

Thanks

Hello all. If N is normal in G, a finite group and H is any subgroup of G, why does the order of pi(H) (pi the projection map G to G/N) divide the order of H?

[rocky cloak]

pi(H) is isomorphic to H/HcapN by the first isomorphism theorem.

So |H| = |HcapN| * |pi(H)|

[coral steeple]

Oh... of course I should have been thinking about ker pi|_H. Thanks!

[tardy hedge]

An A-module M can only be given a natural A/a-module structure when the ideal a is subset of Ann(M) right?

If you have an A-module M with maximal ideal m subset of Ann(M), M can be given an A/m-module strucutre so M is a vector space now, but if you take x in Ann(M)\m (so x is nonzero in A/m) and m in M, then xm = 0 for nonzero m, but i thought this cannot happen in a vector space?

What am I missing?

[coral steeple]

If I have a decomposition of a permutation into disjoint cycles, then I can write each cycle (i1 i2 ... ik) as (j ik) ... (j i1) for j not among the i_l. Hence sgn(i1 ... ik) is (-1)^k. Doesn't that mean that the sign of a permutation is (-1)^(length of the cycle decomposition)?

[mighty kiln]

The first two assertions are slightly wrong

The third depends on how you define length

[tardy hedge]

My supervisor told me I got A+ on the module theory course and exam. Pretty much i owe it to u guys 😂

Im not sure where i would be knowledge wise if it werent for this server

[velvet hull]

1st small detail: sgn(i_1 ... i_k) is equal to (-1)^(k-1), not (-1)^k. Otherwise you'd be saying that any 2-swap is an even permutation

[rocky cloak]

If m is maximal and contained in Ann(M), then m=Ann(M).

So there is no such x

[velvet hull]

but yes, if the cycle type decomposition of of some permutation sigma is rho_1 rho_2 ... rho_k, then sign(sigma) = sign(rho_1) * sign(rho_2) * .... * sign(rho_k) = (-1)^[(n_1-1) + (n_2 - 1) + ... + (n_k-1)] as you'd expect

[velvet hull]

it's not the length of the cycle decomposition. unfortunately just knowing that tells you nothing about whether if a permutation is even or odd. Specifically, you need to know the cycle type of the permutation, i.e. the length of each disjoint cycle, which is what you need

[rocky cloak]

Well, sometimes you have canonical ring homomorphisms from A/a to A.

For example Q[x]/(x) is canonically a subring of Q[x]. So I guess in those cases you could make M into an A/a module without a being in the annihilator.

This is sort of a rare case though

[coral steeple]

Ah right of course. Well that claim was right up to constants at least

[coral steeple]

Nice! This seems easier than counting inversions assuming you have the cycle decomposition as well...

[cloud walrus]

UGOBEL

[rustic crown]

A4 is the standard counterexample for "converse of lagrange's theorem"

6 divides 12, but A4 doesn't have a subgroup of order 6

[candid patrol]

Fact

But proove it

[quiet pelican]

There are two non-commuting elements of order two in S_3

[wraith cargo]

(12) and (23)?

[quiet pelican]

Yes

[wraith cargo]

Ah wait

I THOUGHT YOU WERE ASKING IF THERE EXIST LOL

Uoohhhh

[quiet pelican]

It’s a proof that S_3 isn’t a subgroup of A_4

As all the order 2 elements of A_4 commute

[tardy hedge]

Thank you!

[candid patrol]

Yeah but prooving that S3 isn’t a subgroup of A4 dosn’t mean that A4 doesn’t have a subgroup of order 3!

My english is so bad lmao

[quiet pelican]

That C_6 isn’t a subgroup either is easy

[candid patrol]

C ?

[quiet pelican]

Cyclic

(The cyclic group of order 6)

[candid patrol]

Z6 ?

[quiet pelican]

Yes, that’s another way of denoting the same thing

[candid patrol]

We don’t have the same notation lol

[quiet pelican]

C_6, Z/6Z and (in an irritating clash of notation) Z_6 can all denote the same thing

[candid patrol]

Ok ok it’s good to know

But explain me why A4 doesn’t have a subgroup of order 6

[quiet pelican]

I generally use C_6 for the group, Z/6Z for the ring (if you don’t know what that is, it’s just a different type of structure) and don’t use Z_6

[quiet pelican]

The groups of order 6 are C_6 and S_3 (prove this if you haven’t seen it)

Don’t call me bro

[candid patrol]

Oh ok ok i forgot it

[quiet pelican]

A_4 doesn’t have an element of order 6 (prove this) so can’t have a subgroup isomorphic to C_6
A_4 doesn’t have two non-commuting elements of order 2 (prove this) so can’t have a subgroup isomorphic to S_3

[candid patrol]

Why ? Sorry but it’s not bad

[quiet pelican]

And I’d prefer not to be called bro

[candid patrol]

Ok ok mb

[candid patrol]

I got it thx

[noble adder]

is D_3 isomorphic to S_3 then?

[quiet pelican]

Yes (assuming you use the convention where D_3 is the symmetry group of the triangle)
although the other convention doesn't make sense here so lol

[noble adder]

Lol yes

[warm dove]

Another neat argument - a subgroup of order 6 would be index 2 and therefore normal. Now, in A_4 there are two distinct conjugacy classes of the 3-cycles (work out what these are yourself). In particular, a 3-cycle in A_4 is never conjugate to its inverse. But, if such a subgroup H is normal and contains a 3-cycle x, x must necessarily be conjugate to x^-1 since H is normal, a contradiction.

I like this a lot because it doesn’t invoke the fact that the only groups of order 6 are S_3 and Z_6

[candid patrol]

Yeah i had this version

A4 ~ A4/S3 = {+1,-1}

[quiet pelican]

Why does H normal => x, x^-1 conjugate?

[candid patrol]

Because of [G : H] = 2

[warm dove]

This actually isn’t correct I think off the top of my head

My bad

A better way to do it is to consider a 3-cycle a not in H

Then H, aH partition A_4

So a^2 must be in H since a is in aH but then a^2 a^2 = a a^3 = a in H

A contradiction

Thanks for pointing out the error

Either way the argument is still nice bc it doesn’t invoke anything specific about groups of order 6

[warm dove]

Hmm I guess it’s true

Because SOME 3-cycle in H must be conjugate to x^-1

And then since the 3-cycles in H are conjugate to each other (exactly one conjugacy class is contained in H) we’re done

By composition of conjugation

[quiet pelican]

Why?

[warm dove]

gHg^-1 = H by normality

But H contains x^-1 if it contains x

And 3-cycles can ONLY be conjugate to 3-cycles (since conjugation preserves cycle decomposition in S_4 it does so in A_4)

So only the 3-cycles in H can be possibly conjugate to the 3-cycle x^-1

And we know it is conjugate by normality

So at least one of the 3-cycles is conjugate and we’re done

(I’m not very good at group theory if you can’t tell)

[quiet pelican]

One of the three cycles is conjugate to x^-1, but we don't know that it's x

[warm dove]

Yeah, but all the 3-cycles are conjugate to each other in H

Since H contains the conjugacy class

So then we can compose the conjugations

[quiet pelican]

We don't know that

[warm dove]

Hmm

[quiet pelican]

The 3-cycles could be a union of two (or more) conjugacy classes

[warm dove]

Then we have too many elements

Each conjugacy class has 3 elements and H is of order 6

So if we have two classes we automatically have at least 6 elements not including the identity

[quiet pelican]

conjugacy class has 3 elements
How?

[warm dove]

Just computing it

[quiet pelican]

How are you computing it?

[warm dove]

Just by writing out the conjugations

I have an abstract proof in my abstract algebra homework somewhere

For the fact that x isn’t conjugate to x^-1

But idr the argument off the top of my head

[quiet pelican]

Each ccl of 3-cycles in A_4 has size 4
But these could split in H

[warm dove]

Direct computation works though

[warm dove]

H is normal, so gHg^-1 = H and so conjugation by anything fixes H

So if we have anything in the conjugacy class

We have the whole conjugacy class

[quiet pelican]

Yes
My point is that while |ccl_{A_4}(x)| = 4, we could have |ccl_{H}(x)| = 2 (as the element conjugating x to half of the conjugacy class is in A_4 but not H)

[warm dove]

It doesnt matter if it’s in H

By normality

Right?

Normality means gHg^-1 = H for all g in G = A_4

[quiet pelican]

I mean yeah the whole A_4 ccl will be in H
But that won't necessarily be one ccl in H

[warm dove]

Yeah but it doesn’t need to be conjugate in H

Being conjugate in A_4 is enough

Because if x and y are conjugate to each other and y is our element such that y is conjugate to x^-1

We compose the conjugations in A_4

Which gives the contradiction

[cloud walrus]

UGOBEL

[tardy hedge]

Oh dang ok it was confusing me looking at the rings of fractions construction in d/f versus in atiyah

In d/f they assume the multiplicatively closed subset ALSO has no zero divisors

they dont do this in a/m

[rocky cloak]

That's weird. So d/f never defines localization at primes?

[next obsidian]

Eh?

[tardy hedge]

They dont in section 7.5 which is what i was looking through

[next obsidian]

Surely that can’t be true…

[tardy hedge]

I think they revisit the topic in chapter 15

I feel like every time I learn about the ring of fractions thing, there is some added detail that I learn that I didnt know about before lol

the first time i learned about it, it was only for an ID so it was the field of fractions

[rocky cloak]

Well I guess it makes sense to define it first for non-zero divisors, before jumping in to the added complexity

[tardy hedge]

then i saw that oh the ring doesnt have to be a domain and u can just add in some fractions etc

yea

[rocky cloak]

But either way the constructions are pretty similar.

You just need to be careful that
a/b = sa/sb
in the zero divisor case

[coral steeple]

I think I aced my group theory final. Many thanks to everyone who's helped me on here.

[tardy hedge]

wdym by this?

When you are putting the relation on A x S where S is just a mult closed subset (could have zero divisors), the ring of fractions still has inverses for any pair (a,b) where a and b are in S right?

[rocky cloak]

Like if sa = 0 for some s, then a/b should be 0, because
a/b = sa/sb = 0/sb

[tardy hedge]

oh ok so you cant say that every (a,b) has an inverse just when a,b are in S

[rocky cloak]

I'm not sure I totally follow what you're saying, but to construct the localization you still look at AxS, except you mod out by the ideal of things annihilated by something in S.

[tardy hedge]

my question is, in the ring of fractions on A x S, is every element (a,b) where a and b are in S a unit?

No, right?

[tardy hedge]

because of this problem, some (a,b) pairs are actually just 0

[rocky cloak]

They're units yes. The inverse is (b, a)

[tardy hedge]

ay yai yai

But (a,b) is 0?

[rocky cloak]

If a unit is 0 then every element of the ring is 0.

But that could happen sure

You'll notice this only happens if 0 is in S though

[tardy hedge]

I think i got hung up on ur example with the zero divisor in S

in that case we have a unit being 0

(a,b) ~ 0

but that came from simply just having a zero divisor in S?

[rocky cloak]

(a, b) is only a unit if a is in S

[tardy hedge]

Oh

[rocky cloak]

Otherwise there is no (b, a)

[tardy hedge]

yea

jumpscare

[velvet hull]

yes

I think what you're confused on is the construction of the field of fractions. if S is a ring, then we "want" to say that its field of fractions is the set of all elements of the form a/b, but you're discovering right now that it doesn't work out when S has zero divisors

so, in practise, the field of fractions of S is constructed as S^2 / <R>

where R is the ideal generated by the equivalence relation on the ordered pairs

[rocky cloak]

I guess a nice example might be

A = k[x, y]/(xy) and S = {1, x, x^2, ...}

Then for example y/x = 0 because y/x = yx/x^2 = 0/x^2 = 0

The localization is isomorphic to k[x, x^-]

You can always think of localization in two steps:
killing of everything annihilated by S. Then in this new quotient ring the elements of S are no longer zero-divisors.

Then you do localization as you would without zero divisors.

[tardy hedge]

so if a is in S and it is a zero divisor then (a,b) is a unit but its also 0 so the ring of fractions is 0?

oh wait

[rocky cloak]

Not all zero-divisors become 0.

Just those annihilated by some s in S

You may note in the example above that x is a zero divisor in S.

[tardy hedge]

Thanks, these are unexpectedly subtle details to me that i didnt notice at first

[tardy hedge]

so that bs is still in S so u can write (a,b) = (as, bs)?

[rocky cloak]

Yup that's the idea

[tropic spade]

I'm trying to show x^4+x^2 y^2 +y^3 is irreducible over C. I don't see a nice way to apply eisenstein. I was figuring maybe I can prove it has no root in C[x] since it is degree 3 over C[x] but I don't see why this couldn't be true either.

I've also considered trying to reduce mod some convenient ideal but I have gotten nowhere with that either.

Any tips or ideas?

[rocky cloak]

I mean looking for roots is not so bad.

It can't have degree 3 or higher, so you just have to plug in a + bx + cx^2

[tropic spade]

That feels kinda rough having to cube that square that and so on

[tropic spade]

The end result would just be to try and get an inconsistent system of equations in a,b and c right?

[rocky cloak]

Yeah, I mean it's kinda immediate that a=0, then from there it shouldn't be much work

[delicate bloom]

I could have sworn you said later that you didn't care if it required more high tech machinery but was too busy to respond at the time. Example I was gonna give was in the 31-adics you have cbrt(2) but not cbrt(3) which can be checked by modular arithmetic. So the even larger field Q_31 has Q(cbrt(2)) as a subfield but not Q(cbrt(3)), so they can't be equal.

[tropic spade]

Okay, pretty sure I see it. I was making this way too complicated lmao. Thank you for the help!

[tropic spade]

Okay, I have spoken too soon. Why can't it have degree 3 or higher?

[rocky cloak]

Well what happens if you plug in something of degree 3 or higher?

[tropic spade]

Leading term degree too big to cancel

[rocky cloak]

Bingo

[tropic spade]

I feel so dumb lmao

Thanks again

[warm dove]

Oh doot congrats on pink mod (or apologies? Both are kind of appropiate tbh)

[wicked patio]

gemini gave an explanation which roughly works out to this, I'm curious if it's right or I'm missing something

x^3-2 is irreducible over Z[cbrt(3)] by Eisenstein's and thus over Q(cbrt(3)) by Gauss's

[delicate bloom]

I guess you now have to prove Z[cbrt(3)] is an integral domain

[wicked patio]

well its a subring of R

[delicate bloom]

nice

[wicked patio]

nvm you need it to be a UFD to apply gauss'

which, maybe it is but good luck proving

[cobalt heath]

Wdym, can you elaborate?

Ah nvm, I confused what was the polynomial ring and what was the coefficient ring

[delicate bloom]

nice, then the 31-adic way is undefeated 😌

[rotund aurora]

Yeah I did realize

[rotund aurora]

This is fine as long as you pass to a completion. But some care is needed, because this doesn't work if you do this the other way around (ie, try to apply Eisenstein's criterion to x^3-3 from Z[cbrt 2]). So in the end you have to compute ramifications, but you might just notice that the ramifications are different in the first place (I said this but I deleted idk why)

[woeful sage]

oh you can construct a field of fractions even when S is not an integral domain?

I never knew that 👀

[rotund aurora]

Your method is somewhat computational tho, a priori you don't know which prime will work (I haven't checked if 31 is the smallest). Let n>1 an integer, a, b distinct positive integers free of nth powers. Then I think Q(a^(1/n)) and Q(b^(1/n)) should be distinct. Can you find a prime p such that x^n-a is irreducible mod p while x^n-b splits completely mod p, both being separable?

[candid patrol]

Bruhh... the dual of A4 isn't trivial at all :///

it is for n >= 5 lmao i'm unlucky

[tropic spade]

No. If you had a non-integral domain R you can pick an x,y in R-{0} where xy=0. If R had a field of fractions F, R embeds into F, so xy=0 in F and this is impossible in a field because every field is also an integral domain.

The general trick is called localization. It's similar but because of the issue I mentioned it won't always give you a field.

You do it relative to a multiplicatively closed subset of your commutative ring. In the special case where your ring is an integral domain and your subset is R-{0} you get the field of fractions.

The elements of the multiplicatively closed subsets become units when you do the construction.

The equivalence relation is sorta different but I think by cancellation in the integral domain case they end up being the same relation.

I could be mixing some of this stuff up tho

[tardy hedge]

If you allow 0 as a denominator in ring of fractions then you get the zero ring right

[wicked patio]

yeah

[surreal swan]

Going through basics of group theory
Assume N is a normal subgroup of G
Then there is a correspondence of subgroups of G containing N and subgroups of G/N
This correspondence also preserves quite a lot of structure
A < B iff mapping(A) < mapping(B)
A normal in B iff mapping(A) normal in mapping(B)

Anyone got intuition on the properties being preserved?
This feels pretty important to grok, and it hasn't really clicked for me yet
Given a property, I wouldn't know how to easily check if it's being preserved by the correspondence

Is it easier to think about normal/quotient groups through homomorphisms?
As in, ker Gamma instead of N, Image of Gamma instead of G/N

[abstract rock]

Well the first Isomorphism theorem gives you the latter about why homomorphisms are important when discussing normal and quotient groups

the correspondence theorem is further in that if you have a subgroup H that containsN normal in G and a homomorphism Phi: G to G/N the image of the subgroup is also a subgroup

theres more to it than that

we get a nice bit of lattice theory for free from it, and its like the trees by inclusion of subgroups (disregarding the converging node of the whole group) with a root that is the normal subgroup are (almost) perserved across homomorphisms with the normal group as its kernel

[tardy hedge]

How important is the material on tensor products in atiyah macdonald? In my reading course, it doesnt look like my professor is assigning me questions about tensor product stuff so id rather skip some stuff for now if i can

I dont mean important generally, i imagine tensor products are important (although atm i havent seen really how yet lol), i just want to be able to read through a/m and if I can safely ignore some tensor product stuff then i would like to know

It kinda seems like they just pop up in a/m to illustrate how tensor product interacts with whatever construction the chapter is introducing

[abstract rock]

Personally if the content is well written I see no reason to pass it up but if you feel its kind of flagrantly put there and its not critical to your course dont pay it much mind

[tardy hedge]

Yeah, honesty its mostly because i need to get this reading course done asap

Normally i would like to spend ample time understanding everything

Flagrantly is a good word

[abstract rock]

Tensor products are worth a visit, mostly because tensors show up everywhere when you want to multilinearize complicated systems

[tardy hedge]

I see

Ya they seem powerful in that way

@abstract rock wow ur pfp

[abstract rock]

,av

[cloud walrus]

nastydastywillbreakyournecksee's Avatar

Click here to view the image.

[tardy hedge]

Wow

Jay Cutler evolution

When you localize at p, the ring Ap has exactly one maximal ideal, but it can have other proper ideals right?

[surreal swan]

@abstract rock TY for the writeup, atm when I have questions like "is something preserved under the correspondence", I'll start off asking "is it preserved under preimage of a homomorphism", feels pretty natural from some examples I worked through

[abstract rock]

the unfortunate thing anout the correspondence theorem, and why i said almost, is that the correspondence is simply a bijevtion

it may not preserve the edges of the tree i mentioned

this makes it act like a dual sword, you cant make too explicit a statement without using more information

[next obsidian]

Yah

[molten viper]

In particular, I believe the ideals contained in your ideal you localize remain ideals of your localization

[next obsidian]

Yes, every ideal of a localization A_S looks like I_S = IA_S

But you can have non equal ideals I and J with IA_S = JA_S

This is not true however if you assert that I and J are prime (or primary)

[delicate bloom]

yeah it's a bit computational, but it's sorta simple compared to the other diophantine stuff I think you were suggesting earlier. I just tried finding the smallest p where only one of 2 or 3 was a cubic residue.

I'm thinking it might be possible to prove for every pair a,b with gcd(a,b)=1 and fixed n that there exists a prime p where exactly one is an nth power residue and the other isn't. At least if that ends up being relatively easy to prove we get the more general statement about field extensions from it

maybe the gcd condition isn't necessary lol idk I just made this up

[surreal swan]

Do solvable finite groups with a trivial center exist?

[rocky cloak]

Like S3 for example

[sly crescent]

Trivial group

[glad osprey]

I'm doing some exercises in Naive Lie Theory, is $$\begin{bmatrix} \cos(\theta) & -\sin(\theta) & 0 & 0 \ \sin(\theta) & \cos(\theta) & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \end{bmatrix}$$ correct for 2.6.1?

[cloud walrus]

sheddow

[glad osprey]

(and 2.6.2 would be the same but swapping rows/columns 1, 2 with 3, 4, right?)

[molten viper]

That seems right to me

[glad osprey]

For 2.6.3, is it enough to just observe that $$\begin{bmatrix} \cos(\theta) & -\sin(\theta) & 0 & 0 \ \sin(\theta) & \cos(\theta) & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \end{bmatrix}$$ and $$\begin{bmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & \cos(\theta) & -\sin(\theta) \ 0 & 0 & \sin(\theta) & \cos(\theta)\end{bmatrix}$$ are both subgroups isomorphic to $S^1$ with trivial intersection?

[cloud walrus]

sheddow

[quiet pelican]

No
You need to observe that they commute as well

Then you’re essentially done

[glad osprey]

I see, thanks

[glad osprey]

I know that topologically, S^3 is clearly different from S1 x S1 x S1 (the latter is not simply connected for one), but I'm not seeing a group theoretic argument. Any hints?

Figured it out, S^3 is not abelian

[rotund aurora]

(Answering the second paragraph) Take a=1 and b=2^n lol. Recall also that 16 is an 8th power mod every p. Anyway, there are some edge cases that need to be excluded.

Even for n=2 this seems not entirely obvious to me. Sure, you can use quadratic reciprocity. You can also use reciprocity for nth powers for the general case, although some more care would be needed here (for n an odd prime Eisenstein reciprocity would suffice). Ideally tho, we would want something simpler

[delicate bloom]

I'm not surprised those conditions don't work

was just brainstorming

personally I'm satisfied with this proof being simple enough, at this point I'd only be interested in trying to generalize the idea to come up with a lemma to show "there exists a prime p" without having to actually cook it up, but that'd definitely be more work

like I guess the annoying generalization to go for would be, given f,g polynomials irreducible over Q, does there exist a Q_p where one stays irreducible and the other splits completely?

I am pretty sure I can cook up counterexamples to that so don't waste your time lol

[rotund aurora]

I think the Grunwald-Wang theorem might be a possible generalization? I have never payed attention to it until now, the statement in Neukirch was kind of cryptic when I first read it lol. https://en.wikipedia.org/wiki/Grunwald–Wang_theorem

It's also in Artin-Tate, chapter X (the screenshot is from Neukirch's cohomology of number fields)

[delicate bloom]

interesting I'll think about it, better than wasting my time trying to put a condition on say, divisors of the resultant of f, g or some random garbage lol

[rotund aurora]

16 being an 8th power mod all primes is pretty much the only exception (this is accounted for in the GW theorem). In Theorems 0.6 and 0.7 here https://ishanina.github.io/PROMYS/Local-Global Principle for nth Powers/Local-Global_Principle_for_nth_Powers.html the result is stated carefully.

Morally a similar statement to what you are suggesting is the one that follow from Chebotarev, that says that if Spl(K)=Spl(L) for L,K Galois number fields then L=K. There are examples where Spl(K)=Spl(L) but L!=K, but in that case L and K are not both Galois over Q. So probably this kind of problems are harder in the non-abelian case, or even the non-solvable case. But dunno, I have to do other things now

[delicate bloom]

cool, thanks. Yeah I'm tired so won't be thinking about this tonight either

[smoky coral]

R(+,•)
a+b=a+b-1
a•b=a+b-1
Can this be a field cuz it is following all the axioms?

+and- on the rhs in both is std ones

[delicate bloom]

is your + and * the same operation?

[smoky coral]

They are addition and multiplication respectively but yes I have defined them so as to give the same result
a+b=a•b

[rocky cloak]

The only ring for which + = * is the trivial ring

[delicate bloom]

every element under + has an inverse, but not every element under * has an inverse, so they can't be the same in a field.

[rocky cloak]

But
a+b = a + b - 1
a*b = ab - b - a + 2
is a field

[smoky coral]

Yes the additive identity shouldn't be having a multiplicative inverse .here also the additive identity 1 is not having an multiplicative inverse

[smoky coral]

I see

[rocky cloak]

As for the axioms, distributivity fails in your example

[smoky coral]

Oh shit yeah missed it...thanks

[whole jetty]

Hey. How can i show that the induced representation of the identity (Ind^G_H 1) is reducible?

[rocky cloak]

What do you mean when you say "the identity" here?

Like the trivial representation?

It's not so hard to see that the induced representation contains the trivial representation as a subrepresentation, so unless H=G this would give a proper subrepresentation

(assuming H has finite index relatively prime to the characteristic of your field)

[whole jetty]

I still dont understant very well induced representations,

I am strugling a bit to get the concept.

I am a master student on theoretical physics and thought it would be fun to take a group class. But, i didn't thought how hard it would be since my mathmatical background is a little bit different. Ahahahhahahah

[rocky cloak]

So if V is a representation of H, then one construction of the induced representation is to pick representatives g1, ..., gm for the cosets gH.

Then the induced representation is the direct sum of m copies of V (one for each coset). Then g in G acts on (gi, v) by first picking gj such that ggiH = gjH, then h in H so that ggi = gjh. Then
g(gi, v) = (gj, hv).

If V is the trivial representation, then we can ignore the action of h. So you just have G acting on a vector space with basis G/H

I would recommend looking at some examples if you find it difficult

[whole jetty]

Okok thank you 🙂

[candid patrol]

What’s the fastest way to find the dual of Sn ?

Knowing the dual of Sn is isomorph to the dual of Sn/An which contains 2 elements, we can find it quickly

But I guess that we have to know the signature morphism

[serene dune]

$$\frac{1}{1+x}= \sum_{k=0}^{\infty}(-1)^{k}x^{k}.$$

[cloud walrus]

yeshua

[minor fulcrum]

Is there a name for the elements of U(1) with finite order

[wraith cargo]

roots of unity

[serene dune]

why does this make sense when x is nilpotent, or x^n = 0

good album, btw

[minor fulcrum]

Is there a name for the group of all roots of unity

[wraith cargo]

it's just called the group of roots of unity I don't believe there's a special name

[wraith cargo]

well it makes sense because the RHS is a finite sum

so there aren't any convergence problems

[serene dune]

alr, so my guess was true, now what should i write for the proof

if x is nilpotent, 1+x is a unit, so is 1-x

[serene dune]

mentioning this will suffice ?

[wraith cargo]

you can just write 1 = 1 - x^n = (1-x)(1+x+x^2+... + x^{n-1})
So 1-x is a unit

[serene dune]

right!

that a very slick way to use the backdoor

[tulip otter]

is this closed under *

0<1+xy<2 so 1/(1+xy)>1/2 with no upper bound

and -2<x+y<2 so x*y need not be an element of (-1;1) right ?

[velvet hull]

in fact, it doesn't necessary need to make sense just when x is nilpotent - but of course it has to make sense that way if we're restricted to only taking about polynomial rings

[velvet hull]

from the algebraic point of view, power series can be considered independently whether if the sum strictly "converges" or not

specifically what I'm talking about right now is the ring of formal power series, denoted R[[x]]

and in that ring, (1+x)^-1 = sum (-1)^k x^k

(note that I'm choosing to write (1+x)^-1 instead of 1/(1+x) here - because even though they functionally behave the exact same, I want you to think about it as the inverse function of 1+x, and not as the fraction)

another way of approaching this, is that you can view sum (-1)^k x^k as the result you get by solving for the generating function of the inverse of 1+x

[velvet hull]

one slick way to make this final step, by the way, is to notice that f(x) = -x is an automorphism of any polynomial ring

and thus since 1+x is a unit, f(1+x) = 1-x must also be a unit :)

[serene dune]

very slick

[serene dune]

hmm i see, that would be overkilling of the problem hypothesis here

[velvet hull]

a neat way of thinking about this, and yeah it's probably very much outside the scope of whatever the original problem was, is that the inverse of 1+x always exists inside of R[[x]]

[serene dune]

i have heard just a lil bit about that ring, but it never made sense to me

[velvet hull]

But as R[x] is contained within R[[x]] as a subring, if you substitute your x to be nilpotent then that inverse "collapses" into the lower polynomial ring

[velvet hull]

lol, you just throw any notion of convergence out the window

that's it

[serene dune]

may i ask for a good text and problemset for rings

[velvet hull]

you just purely think of power series as objects satisfying some algebraic properties

[serene dune]

i wanna be better at it

[velvet hull]

hmm

[serene dune]

im trying to solve undergraduate algebra by lang rn

[velvet hull]

let me check it out first real quick I didn't even know there was an undergrad version lmaoo

[serene dune]

haha, me neither then i found it in my folder

[velvet hull]

I used artin alegrba, which is free online
the content is roughly the same, but lang split the chapter on rings into a chaper on rings + a chapter on polynomial (..rings, lol)
as far as I could tell the content is roughly the same, so you can just focus on the problems I think
http://www.alefenu.com/libri/artin.pdf

[serene dune]

oh yeah, im aware about that one, ok i will solve these two then

[serene dune]

well, G after being an abelian group with +

how can it mess up to not form a ring

[next obsidian]

There’s lots of answers but here’s just one example

Say every element in G has finite order, but this is unbounded

Then it can’t be a ring, because 1 would have some finite order k but then
0 = 0•x = k•x = x + x + … + x

So the order of every element x in R is <= k

[next obsidian]

But by this assumption you have some x with order > k cuz it’s unbounded

An example of an abelian group with this property is like, Q/Z

[serene dune]

Q/Z type beat ?

[next obsidian]

Hahaha

[serene dune]

well a tangent, but can we generalise the construction of Q/Z via free groups ?

[rocky cloak]

Most would just call it Q/Z if you just care up to isomorphism

[tulip otter]

?

[crystal sundial]

Hi, I am confused as to why the gcd assumption is necessary for the converse of this result:

More specifically, I don't see why I cannot argue it in this manner (which doesn't use the gcd assumption): Suppose p(x) in R[x] is reducible. Then there exists a(x), b(x) both non units in R[x], such that p(x) = a(x)b(x). Now, since a(x), b(x) are non units in R[x], they are non constant (since R[x] is an integral domain) so the same factorization shows that p(x) is reducible in F[x]

here is our definition of irreducible and reducible:

[surreal swan]

^ i misread as "the god assumption", was confused and intrigued

[wraith nexus]

2x is reducible in Z[x] but not Q[x]

[wraith nexus]

I don't see how R[x] integral domain implies a(x),b(x) nonconstant

without assuming gcd requirement

[crystal sundial]

Here is my argument: In an integral domain, deg ( pq) = deg p + deg q as long both p and q are nonzero. If x is a unit, then there exists x^-1 such that deg(xx^-1) = deg(1) = 0. Thus both x and x^-1 must be of degree 0, and thus constant.

Now since, a(x) and b(x) are both non unit they must be nonconstant polys (i think)

[tardy hedge]

Not sure what a meets S means

[whole jetty]

Can we show this with Frobecius reciprocity?

[wraith nexus]

why

your proof does not show that

[crystal sundial]

oh wait

I see my mistake now

constants can be non unit too

thanks for the time and help

[wraith nexus]

np

[tardy hedge]

why are extension and contractions of ideals make my brain stop

so annoying

Not sure about proof of iii)

Dont know what implication is coming from where

too short

[whole jetty]

Can we decompose a inner porduct between two characters into to inner products?

[next obsidian]

Being a non-unit doesn’t mean non-constant

Consider 2 in Z[x]

But the problem is if you factored as a•b(x) with a constant, then the gcd of the coefficients wasn’t 1

So this forces the factorization to actually use two non-constants

[next obsidian]

Oops

[dim wagon]

does anyone recognise the minus symbol at the botton right?

[rocky cloak]

$\ominus$

[cloud walrus]

jagr2808

[dim wagon]

i mean the meaning of the symbol

[rocky cloak]

Presumably it's the inverse with respect to oplus

And indeed that checks out with the formulas

[dim wagon]

hm im trying it out now but why would $(x,y)+(x,y+x^2)=(x+x,y+y+x^2+x^2)=(0,0)$

[cloud walrus]

somethingwrong

[rocky cloak]

So x+x = 0, and similarly y+y and x^2 + x^2 so it becomes (0, 0) which is the identity

[dim wagon]

oh cause GF(q) is just C_2*C_2...C_2 a times so everything is an involution

okay thank you i got it

[barren sierra]

For a monomial $\overline{x}^\alpha$, let $|\alpha| = \sum_{i = 1}^n \alpha_i$ for $\alpha \in \mathbb{Z}_{\geq 0}^n$.
An ordering $>$ on monomials is \emph{degree-anticompatible*} if $|\alpha| \leq |\beta| \implies \overline{x}^\alpha \geq \overline{x}^\beta$.
Exercise 4.4.3 of Cox, Little, and O'Shea's \emph{Using Algebraic Geometry} is the following:

Let $>$ be a degree-anticompatible ordering on either of the local rings $k[\overline{x}]_{\langle x_1, \ldots, x_n\rangle}$ or $k[[\overline{x}]]$.
Show that any nonempty set of monomials $\mathcal{S}$ is bounded below.

This should be straightforward but after mulling over it for some time it's not coming to me. It also isn't clear why we need to consider just these local rings when we can have degree-anticompatible orderings over any polynomial ring.

[cloud walrus]

Spamakin🎷

[serene dune]

is there any canonical construction of a two sided ideal in a noncommutative ring ?

[crystal vale]

I think yes, there is no reason for no

[rocky cloak]

The ideal generated by a single element x consists of sums of things of the form rxs with r and s in your ring.

But note that these sums can't necessarily be simplified down to one term as in the one sided case, as in for example
rxs + txu
might be the simplest form of this element.

[grizzled spindle]

Is there a notion of ring presentation like there is for groups?

I was thinking about the integers and it seemed like the way a ring would naturally arise without extra relations

In the same way that for groups, the free group on one element is isomorphic to the integers

[south patrol]

Yes there is

[grizzled spindle]

There’s also a “natural map” I saw in class from the integers to any ring and it was pretty clearly unique and I’m not sure if there’s another ring that has something similar to this

When I say map I mean ring homomorphism here

[south patrol]

Usually one says a map of commutative rings A -> B is of finite presentation if there exists some n and a surjection A[x1,...,xn] -> B with finitely generated kernel

[rocky cloak]

The free ring is the ring of noncommutative polynomials over Z (or just polynomials if you're rings are commutative).

So in particular Z is the free ring on 0 generators

[south patrol]

I'm sure there is a noncommutative variant

with <> instead of []

[grizzled spindle]

Are the x’s the generating elements of the kernel

[grizzled spindle]

I feel like this is kinda cheating cause Z is already in the definition but that does make sense

[rocky cloak]

I mean that's not the definition of free ring, that's just what it happens to be

The definition is the adjoint to the forgetful functor

[grizzled spindle]

Oh so the definition comes from category theory?

I mean this definition of free ring does seem natural

I’m not sure how it’s formulated

[serene dune]

L+M might not be the union of the subsets i.e. L\cupM, right ?

then whats is the technical distinction bw the cases when it is vs when it is not, is that distinction recognised in the literaure as anything ?

[south patrol]

Pretty sure that the sum is the union iff one is contained in the other

[serene dune]

seeing one direction was not that hard

[rocky cloak]

So in general for some algebraic object, say rings

The free ring on a set X is the ring F(X) such that homomorphisms F(X) -> R are exactly given by functions (between sets) X -> R.

So for example a map Z[x] -> R is exactly given by choosing one element in R that x maps to, and you can choose this element arbitrarily.

The definition is the same for groups, abelian groups, monoids, etc.

[tardy hedge]

For 3, if x is in a^ec then xs is in a for some s right

[tardy hedge]

.

[whole jetty]

Can I say that the permutation representation is the S_n representation?

[rocky cloak]

I mean Sn has many representations. Though I guess that one stands out in particular

[whole jetty]

We could show that, if something is a permutation representatio, it's always reducible making as an argument the Sn group being always reducible for n>1? This makes sense?

[rocky cloak]

I don't know that you mean by "the Sn group being always reducible".

And if you're just talking about permutation representations of arbitrary groups, then they don't have that much to do with Sn anymore. Although you can involve Sn if you like.

[whole jetty]

Oh okok

[serene dune]

are we talking abuot representation theory ?

[whole jetty]

Yea

I am trying to show that the Induced representation of the trivial representation is reducible.

[serene dune]

oh ok then, higher echelon stuffs

[whole jetty]

And since i found that the induced representation of the trivial representation is the permutation representation, maybe i could argue that since its that I could conclude that is reducible.

No sure if this is true tho.

[tardy hedge]

Can someone please tell me what “meets” means

[rocky cloak]

So G acts on G/H by multiplication on the left. And inducing the trivial permutation from H gives you the permutation representation of G/H.

One way to show that this is reducible is to show that the trivial representation is a subrepresentation.

I.e. you have to find an element in the span of G/H that is fixed by the action of G. Are you able to do that?

[rocky cloak]

Has non-empty intersection with

[serene dune]

M is an ideal of a C.R. ( M^2 \subsetneq M ),
equality holds when M is idempotent.

[cloud walrus]

yeshua

[serene dune]

true ?

[candid patrol]

If we know that Sn have only An as a normal none trivial subgroup for n =3 and n >= 5, we can get quickly D(Sn) because D(Sn) isn't trivial, and Sn/An is abelian.

[whole jetty]

Hmmmmm... not sure.

[candid patrol]

n = 4 is borring but it's ok

[rocky cloak]

What if we take an example.

Say G = {g, 1} is a cyclic group of order 2, and say H is the trivial group.

Then we're just looking at CG = {ag + b} where G acts by left multiplication. Can you find an element that is fixed by the action of g?

[whole jetty]

Ok, I may sound dumb but I am struggling a bit here. Ahhahahah can you say the element and give me another example? Sometimes I need a concrete example.

[rocky cloak]

Well, maybe I can give you one more hint first.

g * (ag + b) = ag*g + bg = bg + a

So when will
ag + b = bg + a

[whole jetty]

Okkk, so... every element when a=b?

[rocky cloak]

Indeed, and maybe you can make a guess at the general pattern here as well

[whole jetty]

A general pattern for a cyclic group larger than 2 or?

[rocky cloak]

Or for any group and permutation module

[whole jetty]

Ok, so here... let me think...

[whole jetty]

Ok, first thing first. Using this definition I can show that the induced representation of the trivial representation gives the permutation representation because when acting g on φ(x) = will give φ(xg), since φ(g,h⁻1)=1(h)φ(g)=φ(g)?

[grizzled spindle]

Ahh I see, thx a lot

[whole jetty]

Is it the sum?

It makes sense, since the permutation dont change the sum.

[rocky cloak]

That's right, the sum of all the basis vectors remains the same if you permute them

[whole jetty]

To be fair I was researching and saw that. Then i thought, ohhhhhhhhhhhhh of course.

[whole jetty]

There is still one thing bugging me... that is, where the induction representation of the trivial representation takes place.

I am probalby confusing things.

[rocky cloak]

The induced representation of the trivial representation of H, is the permutation module of G/H.

[whole jetty]

But how can i show that?

I think you now understand that i dont understand induced representations. Ahahahhah

[rocky cloak]

Depends on your definition/construction of the induced representation

[whole jetty]

hmmmm...i use the defenition of the serre's book.

[rocky cloak]

I'm afraid I don't have his book memorized

[whole jetty]

Ahahha sorry sorry my bad.

[barren sierra]

Part e. is giving me more trouble than I'd like to admit 💀. Any thoughts would be appreciated on how to show that any set with 3 or more vectors must have a linear dependance in R

[tardy hedge]

Is S^-1A/S^-1p = Sbar^-1(A/p) just a straightforward application of first iso thm?

Nvm its in a prev prop. i will look

well actually that was for modules

Actually, like I was thinking about this earlier

The text showed that formation of modules of fractions is exact

does that apply to exact sequences of rings? I also never even seen exact sequence for rings

I know a ring R can just be viewed as an R-module , so probably right?

[rocky cloak]

Not sure what exact sequence of rings should mean, as the kernel of a ring homomorphism is not a ring

[tardy hedge]

yeah

[ripe crest]

We have

1. group acting on a set: G-set
2. ring acting on an abelian group: R-module
3. field acting on an abelian group: k-vector space

what are some other examples of actions

[abstract rock]

Semigroup S acting on an object: S-act

Group linearly acting on a vector space: Representation

Group acting on a ring: G-ring

Group acting on a topological space: G-space

and so on

[ripe crest]

ooh yes, that's a good one

the other ones I haven't heard of

[abstract rock]

the collection of actions from a group on an object form an action groupoid (ala group with a partial operation)

[ripe crest]

do you have an example for this

[abstract rock]

a field lol

[ripe crest]

hmm

[abstract rock]

to be fair im leaving out some adjectives

And Im not exactly an expert on those kinds of objects

[cloud walrus]

yeshua

[serene dune]

i think this is true ?

can we generalise this for any number of finite ideals in an arbitrary CR ?

then can we lose the hypothesis of PI and finitely generated as well ?

[velvet hull]

it's true, because L \cap M = <lcm(a,b)> and LM = <ab>

so if LM = L \cap M, lcm(a,b) = ab and so gcd(a,b) = 1

I'm pretty sure the argument used can be generalised to PIDs to show that if $\prod_{j=1}^k I_j = \bigcap_{j=1}^k I_j$, then $\gcd(n_1, \cdots, n_k) = 1$

[cloud walrus]

HChan

[velvet hull]

what do you mean by "lose the hypothesis of PI"? because if L and M are no longer PIs the question kind of loses its meaning, no?

[serene dune]

oh i meant what if instead of generated by one element they are general ideals

but i wanted to take one step at once to see what is going on

[empty kernel]

Let G,+ be a group and H a subgroup off G such that [G:H] = 3 and such that there exists an x in G\H with x+H=H+x. Prove that H is a normal subgroup

i have no clue how to start on this

also i used + as my operation but it isnt like the normal + it is just a random operation. If i used * discord would make my text go italic

[empty kernel]

<@&286206848099549185>

[bitter rover]

How many cosets are there in G\H? Are there any you can name specifically? How many are left after you do that?

[empty kernel]

a coset is x+H right? im dutch so we use a completely different word lol

oh yeah so that would mean 3 cosets

one of which has to be e+H where e is the neutral element

oh wait for one of the three x+H= H+x

and that would obv be the one where x=e cuz e+H=H=H+e

OMG WAIT IM SO DUM

x is part of G\H so like x cant be equal to e obv

so e+H=H=H+e
x+H=H=H+x
so there is only one left that we need to prove it for

but

bc the cosets form a partition must y+H = G(e+H U x+H) = G(H+e U H+x) = H+y

so it goes for all three cosets which means H is a normal subgroup

@bitter rover epic that was exactly that small little push i was looking for

thanks

[bitter rover]

🙂

[barren sierra]

Bumping this

Thought about it some more and I don't have anything

[rocky cloak]

Just pass to the field of fractions.

Then you see the its a submodule of a 2d vector space

[potent condor]

Suppose $U \subset R$ is a multiplicatively closed subset of a ring. My book asks me to show that for any ideal $J \subset R$ we have $R \cap J R\left[U^{-1}\right]=\sum_{f \in U}\left(J: f^{\infty}\right)$ where $\left(J: f^{\infty}\right):=\cup_{n=1}^{\infty}\left(J: f^n\right)$.

How does the intersection $R \cap J R\left[U^{-1}\right]$ make sense? If the localization map is not one-to-one, what does it mean?

[cloud walrus]

EdgarAlnGrow

[rocky cloak]

The preimage by the localization map would be the natural thing to consider, but I'd agree it's a fairly big abuse of notation if it's not injective

[next obsidian]

Nah

This is defined in Matsumura for example

Given any algebra S you defined I\cap R to be the preimage

I like the notation so it isn’t an abuse of notation

[rocky cloak]

Chmonkey has spoken! The abuse is no more

[rustic crown]

[potent condor]

Im still reporting all of ur accounts for abusive behavior

[next obsidian]

What

[rustic crown]

why does chuchu like it?

[next obsidian]

Simple

Often times you say S is an R algebra without naming the map

So if u wanna do preimage you need to then go back and say f:R -> S

To write f^-1(I)

[rustic crown]

<

what about Ann_R(S/I)

[next obsidian]

RAHHHH

[potent condor]

wdym? is I a submodule of S over R and you're taking its preimage under the scalar multiplication map R -> S?

[next obsidian]

Yeah exactly

[potent condor]

so what are $S$ and $I$ in $R\cap JR[U^{-1}]$? I would assume $JR[U^{-1}]$ is $S$, but it's not an algebra if $J$ doesn't contain $1$

[cloud walrus]

EdgarAlnGrow

[potent condor]

oh nvmnd i get it

I is JR[U^{-1}]

and S is R[U^{-1}]

[next obsidian]

Yeah

It’s push pull

You’ve extended the ideal into the localization

And then contracted it back

You fatten up the ideal as a result

[keen terrace]

Guys it may be a dumb question, I'm still new to this.. if I have a ring, how can I verify if it's a principal ideals ring without checking every single ideal?
My guess is, if it's a field, it's a PID, then it's a principal ideal ring..

[bitter rover]

How can you verify every number is even or odd without checking every single number?

[keen terrace]

If it's divisible by 2 I guess

[bitter rover]

Well, it's not like you check every number. You prove it using induction and the fact that the natural numbers are well-ordered, or the like.

In other words, you can't prove it without relying on something else you know about the natural numbers.

Same with a PIR. You'll need to be given something else about a ring (beyond it merely being a ring) to say anything definitive.

[barren sierra]

This highly depends on the ring

For example you can prove ℤ is a PID with little work because you know things about integers

You can prove polynomial rings in general aren't PIDs, such as ℤ[x], because you know things about polynomials

But there is no general algorithm to do this

[next obsidian]

Just apply the most basic dimension theory

[barren sierra]

so true that's totally what I was implying they should do

[next obsidian]

Also realistically speaking every ring I know being a PID

[barren sierra]

where's that thread on complicated proofs of basic facts

[next obsidian]

Is because you showed it’s a Euclidean domain

I think the only exception is a Dedekind domain where every prime is prinicipap is a PID

Cuz u got ideal factorization

[barren sierra]

https://mathoverflow.net/questions/42512/awfully-sophisticated-proof-for-simple-facts this one lol

[next obsidian]

My dimension theory is that dimension go up by at least 1 when u do [x] and PIDs r dimension 1

[unkempt stream]

I am completely stuck on this stupid idiot problem

The idea of "such that PA" basically implies I am only alotted row operations

and it's rather easy to understand how you can go down the line, and for each column vector, get the gcd of the column terms below it to the diagonal, and wipe out the lower terms. However i cannot think of an easy way to get the condition that the value of the diagonal terms are the largest in the row

oh wait no it's naturally like that using as I mentioned, misread on the indicies

[tough raven]

Use row operations R_j → R_j - c R_i to change the ji^th entry for j < i?

[slim wren]

Why do people even willingly chose to do this

Like what is there to gain here

[ripe crest]

do we have cases where the group operation distributes over the moniod operation
as opposed to a ring

[rocky cloak]

If you loosen monoid to semigroup, then you have something like the positive real numbers (with the roles of addition and multiplication reversed).

As for the monoid case, if + is the monoid and * is the group, then
x*0 = x*(0+0) = x*0 + x*0
so your monoid cannot be cancelative.

Another property you would need if 0^- is the inverse of 0, then
x = 0*0^-*x = (0+0)*0^-*x = 1*x + 1*x = x+x

A similar idea is a distributive lattice, where you will have two monoids that both distribute over eachother.

[slim wren]

??

[tardy hedge]

what

[candid patrol]

Hello guys

[tardy hedge]

Greetings

[slim wren]

Jobs bro jobs 😭😭

[candid patrol]

I wanna proove that there is no surjective morphism from Sn to Sn-1 for n >= 5

[cloud walrus]

UGOBEL

[tardy hedge]

Ye

[slim wren]

Torture

Math gets to a point bro

[tardy hedge]

@slim wren you are loved

[candid patrol]

what does that mean ?

[candid patrol]

I don't get why An is simple => Ker phi = An

[vocal pebble]

because ker phi intersection A_n is normal subgroup of An and nontrivial

[candid patrol]

Why of An ? The intersection of ker phi and An should be a normal subgroup of Sn no ?

[vocal pebble]

well its a subset of An

and its the intersection of two subgroups so a subgroup of An (taking An as a group in its own right)

[coral spindle]

Try proving that it is also a normal subgroup of A_n

[vocal pebble]

so taking An as a group in its own right you have ker phi interseciton An a subgroup of An (moreover normal)

[candid patrol]

Imagine G = Z/27, H = 9Z/27 and N = 3Z/27, then H is simple, but N intersection H = H so it's a subset of H and N, but we don't have H = N even if H is simple

[coral spindle]

Indeed, there is a further fact you need.

You need to show that the kernel is not bigger than An

[keen terrace]

Is it correct? (I don't think so but oh well)In the b), can I say something like "since fi is the canonical projection (or at least it seems so(?)), it is surjective, and since I is a two-sided (bilateral? Idk in english) ideal of Z[x], it is the ker of the
canonical projection. From this, and since the image of fi is Z (because it's surjective), using the first homomorphism theorem, I can say that Z[x]/I is isomorphic to Z"? Still new to the subject
Just a sec and I'll send the exercise

[coral spindle]

Hint: there is only one subgroup of S_n that is bigger than A_n

[candid patrol]

Sn but you mean normal ?

[coral spindle]

No

I do not mean normal

Recall the theorem of Lagrange and think about the order of A_n. If you have some group between A_n and S_n, then what?

[candid patrol]

pgcd(n!/2,n!) = 2 ?

omg I'm lost

[vocal pebble]

The question is can there be a subgroup of Sn that contains An (other than An and Sn)

[candid patrol]

why contains An ?

[vocal pebble]

So this subgroup should have more than n!/2 elements . Now think about the consequence that has wrt lagranges theorem