groups-rings-fields

[minor wraith]

$x,y\in\langle x \rangle \leq \langle x \rangle \cap \langle y \rangle$

[pulsar dune]

we know that it’s a subgroup of both the group generated by a and the group generated by b

And this subgroup needs to divide the order of both

[cloud walrus]

Kakaka

[minor wraith]

right?

[pulsar dune]

the only numbers that divide both 15 and 24 are..

[minor wraith]

<x> is a subgroup by definition

[cobalt heath]

Uhh

[minor wraith]

wait

HAHAH

reverse inclusion

[cobalt heath]

Dw, I also often confuse (any kind of) direction

[minor wraith]

wait

so are you saying that

[cobalt heath]

(x might not be in <x> \cap <y>, consider 2Z \cap 3Z)

[minor wraith]

$|\langle a\rangle| =|a|$

we have two different defintions of order right

one is basically cardinality

[cloud walrus]

Kakaka

[pulsar dune]

yeah

[minor wraith]

other is multiplative order of an element

[cobalt heath]

Yeah so, how it coincides comes from structure of the cyclic group

[minor wraith]

so this is true?!?

[pulsar dune]

from my understanding yes

[minor wraith]

ohhhhhhh wait that's obvious

consider $\mathbb Z_7$

0,1,2,3,4,5,6

a generator is some primitive root

but everything can be written as a power of a primitive root

[pulsar dune]

pretty sure the answer should be that the group can have order 3 or 1 btw 🤔

[cloud walrus]

Kakaka

[minor wraith]

so multiplicative order = 6

yeah

Z_p for p prime

so order is always p-1

[pulsar dune]

yuh and all cyclic groups are isomorphic

[minor wraith]

exactly

[pulsar dune]

to z-p

[pulsar dune]

Uhh order should just be p?

[cobalt heath]

Imo understanding it through multiplicative group could be somewhat difficult

Z/pZ is an additive group, and it has order p

[minor wraith]

by FLT, it is p-1

[pulsar dune]

oh wait multiplicative group

I thought that was notation for z/pz mb

[minor wraith]

the primitive root generates everything except for zero

so the set generated also has p-1 elements

yeah

we can take exponents 0,...,p-2

so both multiplicative order and cardinality = p-1 = 6 in this example

[cobalt heath]

Hmm

[minor wraith]

wait

but how can we be sure that <a> <b> etc are cyclic

[cobalt heath]

@minor wraith what kind of definition for Z_p did you learn?

[minor wraith]

from number theory

literally complete system of residues modulo p

[cobalt heath]

Can you recall how it is defined?

Or like, Z_m for general case

[minor wraith]

Z_m won't necessarily have primitive roots/generators

you would need gcd(m,a)=1

[cobalt heath]

Was it denoted $\bZ_p$, not $\bZ_p^\times$?

[cloud walrus]

Absta

[minor wraith]

ah

I meant $\mathbb Z_p^{\times} \setminus {0}$

[cloud walrus]

Kakaka

[minor wraith]

you get the idea

because primitive root generates everything except zero

[cobalt heath]

Yeah, that was what confused us.

[minor wraith]

oh sorry

anyways

[cobalt heath]

It's okay

[minor wraith]

how do u know that <x> is cyclic

is it by definitoin

[cobalt heath]

<x> is defined to be cyclic group generated by x.

[minor wraith]

oh ok

[cobalt heath]

Yeah so, I think you should not regard <x> as a multiplicative group of integers

[minor wraith]

anyways

so how do I apply lagrange's theorem again?

because this is the def I have

[cobalt heath]

Yeah, so via subgroup relations

By identifying which is subgroup of which!

[minor wraith]

ohhhh

can $G$ be a subgroup too?

[cloud walrus]

Kakaka

[minor wraith]

not only a "group"

wait, a subgroup is basically just a group too

[cobalt heath]

Nah, in this settings, H is subgroup of G

[cobalt heath]

Indeed

[minor wraith]

so can I take <x> cap <y> = H

and <x>, <y> = G

so I try common divisors

of 15, 24

[minor wraith]

wait do I rule out the trivial divisor 1

[cobalt heath]

You gotta figure it out

Cannot just rule out the possibility

[minor wraith]

how?

think of an example?

[cobalt heath]

Yeah

[minor wraith]

the multiplicative group {1}

has order 1

additive group {0}

so the trivial groups

is always a subgroup any group

with order 1

so include it

wait

additive definition of order is min{k≥1 such that k(0) = 0} right

so yeah order 1

[minor wraith]

but this isn't a proof

I've shown it for two cases

is it because any group always has the trivial subgroup containing only the identity

which by definition has order 1

[cobalt heath]

Hmm, think of in this specific case

Namely, see if there is a case where <a> \cap <b> is trivial.

[minor wraith]

"triviail"

?

wdym

[cobalt heath]

Trivial group

That is, the group with 1 element (order 1)

[minor wraith]

still don't get it sorry

[cobalt heath]

I mean in the given case, see if H = <a> \cap <b> can have order 1.

This might not be easy, but imo doable.

[languid trellis]

I'm struggling with the second part of this exercise. In the finite case, this is banal, but I have been thinking about G infinite for a while and can't get anywhere. Can someone give a hint please?

pls ping 🙂

[minor wraith]

tips on filling in the rest of the entries?

really stuck

answers apparently

but don't get it

all I have is working from definition of inverses and identity

[languid trellis]

do you have a definition for x * y?

#1182793396776091708

[minor wraith]

nope...

main confusion

[languid trellis]

well you should write out some relations

c^2 = 1

that means c is its own inverse

[minor wraith]

wait, if identities exist, does that imply that inverses exist for every element of a group

oh by definiotn of gorup

[languid trellis]

no. a monoid is an object with an identity but maybe without inverse

an example of a "thing" with identity but not necessarily inverses is the set of all maps [n] -> [n]

anyways, we have c^2 = 1, a^2 = c

we also have ad = b

so d is the inverse of a

then you can start to combine these relations to get somewhere

[minor wraith]

I tried

and got nowheere

another thing I considered but also failed at is trying to maybe show associativity of a,b,c,d

so that I can construct a group and perhaps use group properities

[languid trellis]

associativity is inherent in a group

a group operation is associative by defin

[low wyvern]

u can use the property of a Latin square. That is every element should appear in a row/column no more than once.

[languid trellis]

it doesn't state that it has latin square property but I think it kinda has to lol

[low wyvern]

ye, I am pretty sure

[minor wraith]

wait by latin squarse u mean like this:

[low wyvern]

pretty sure that yz, zy ,xy etc should become one of the elements of the group and so u would have to find that out by looking at what works.

[minor wraith]

yeah so trial and error?

seems so inelegant

[languid trellis]

you are just computing stuff

[minor wraith]

computing what?

by property latin squares, I would need z,y,x,e on the last row right?

so that determines the rest of the elements

don't get what I am "computing"

[low wyvern]

the elements and u have done it

[dire siren]

yes

[glad osprey]

Aren't there many possible solutions to this? It doesn't say the inverses have to be unique or that the operation is associative

[languid trellis]

unique inverses and associativity are properties of groups

[glad osprey]

Yeah, but it doesn't say it's group table. Why would it specify that every element should have an inverse, instead of just saying it's a group?

[languid trellis]

it says that its a group table for this exercise, and given the solution he posted it only makes sense to assume its a group

[glad osprey]

I can't see where it says its a group table. I see he posted the solution of another exercise where it explicitly states to construct a group table. Either way, not a big deal, just the wording was a bit odd and ambiguous

[dull ginkgo]

(hint: each row must contain an element only once)

Same with column

No wait that is literally the fastest way to do the problem lmao

[elfin wraith]

This is more of a conceptual question but it is motivated by a problem I’m stuck on.

I really struggle to “picture” elements of quotient rings, it’s fine if I’m quotienting out something tangible, but in this problem I’ve been asked to find a monic, cubic in F_3[X] with no roots, then count the elements of F_3[X]/<X^3 +X^2+X+2> (more generally the ideal generated by any such cubic but that one felt like a decent choice)

But i really struggle to even picture what the elements of R/I are in this case. Like I see that I does not contain any polynomials of degree less than 3, so that’s probably a place to get started, since all of these will be representatives of equivalence classes and it’s not too hard to count how many of them there are.

I also know that my ideal is just the set of P*(X^3+X^2+X+2) for P in F_3[X], is it the case that each of these would be in one of the equivalence classes mentioned before?

I just find it really hard to actually work out what’s happening in situations like this, so does anyone have any advice as to how they tend to approach problems like this?

[dull ginkgo]

I don’t think it’s usually easy to “picture” them or understand them conceptually outside of being a coset or “modulo I”

Unless they are isomorphic to something more well-behaved and describable

[elfin wraith]

It’s not necessarily about picturing it, but I just don’t know a good way to approach a problem like this, basically as soon as I’m generating an ideal I’m lost as to how to actually find the quoitent ring

I feel like in the example I gave, the fact that my polynomial is irreducible is probably some how relevant, but I’m just unsure about how to approach this

[dull ginkgo]

Finitely generated ideals are much, much nicer thank FUCK

[elfin wraith]

Ok sure, but I’m still struggling with a principle ideal and that doesn’t really help me

[dull ginkgo]

Do you mean like, interpreting the element of the quotient from it’s construction

Usually it’s based on context, for this case it’s actually a monic polynomial, which is nice

[elfin wraith]

I understand what an ideal is, and I understand what the elements of a quotient ring is, if you were to give me R/<x^2+1> or anything like that I’m perfectly happy because it’s quite concrete, but I struggle actually work with them in cases like that example that I gave

[dull ginkgo]

Usually you can work with them by proxy of a representative in the coset that they form

For polynomial rings usually there’s a minimal, I.e the remainder

so we have F_3 and the polynomial ideal generated by X^3 + X^2 + X + 2 = X^3 + X^2 + X - 1

[elfin wraith]

Like I see that there is at least 27 elements in the quotient ring since each polynomial with degree less than 3 has its own equivalence class, but I feel kinda stuck after that. For example, I’m not sure how to determine what the case is for all polynomials degree 3 or larger

[dull ginkgo]

Euclidean division should still work here

Each polynomial of degree geq three must be p(x) = d(x)(x^3 + x^2 + x - 1) + r(x) where r has degree 2

so p(x) under the quotient can basically be mapped to the remainder AX^2 + BX + C

So it’s a vector space as a module basically.

[elfin wraith]

Yeah no your right that’s what I was missing

[dull ginkgo]

it’s easy to show that if two polynomials mod q have the same remainder then they are equal to 0 under the quotient

So it’s isomorphic (forgetful gives bijection) as a vector space to F_3^3 so 27 :)

I don’t think characteristic here makes any difference

because of the “indeterminate” nature of X

[elfin wraith]

Yeah I was just struggling how to argue that all of the larger degrees would belong to the equivalence clssses we already had but it does just follow from the division algorithm

[dull ginkgo]

E.g. if p(x) has degree n in F_q then the quotient generated should be F_q^n

[dull ginkgo]

This also holds for general commutative integral rings if the polynomial is monic or it’s leading coefficient is a unit

It’ll be a finitely generated module

[chilly ocean]

Hey,
Can I get some help for this
I was initially attempting a induction on degree but couldn't do it
I also tried an induction on the number of non zero terms but couldn't do it
How does the general inductive algorithm work for this or
Is there a non constructive way to proceed here

[dull ginkgo]

for every i not in S hmm

[crystal vale]

How can I check the associativity for a group by the Cayley table?

[dull ginkgo]

Look at a,b,c elements. Assuming your table is row-major, you’re should end up at the same spot if you trace it on the table

[chilly ocean]

You could also use this iirc

Source : abstract Algebra by grillet

[dull ginkgo]

That is implying that S is bounded by above because the summation must be over the powers of t and there must be a maximum power unless you’re talking about the formal power series ring

[chilly ocean]

Uhh

But S is infinite

[dull ginkgo]

is k[t] the formal power series ring here

[chilly ocean]

Also we need to negate all the terms of the poly with powers not in S

No

Like

[dull ginkgo]

The subset of even numbers of N is infinite but it’s complement is not bounded

So it would make no sense here

[chilly ocean]

Huh

[dull ginkgo]

Because you can’t have an “infinite polynomial”

Unless it’s k[[t]] lol

[chilly ocean]

I mean I need to multiply something that if I have common power indices with S then only those power terms remain with maybe different coefficients

[dull ginkgo]

I think it’s asserting that S is cofinite

[chilly ocean]

If the poly has no common power indices with S then I would multiply 0

[dull ginkgo]

I.e it’s complement is finite

It’s essentially solving a linear system

[chilly ocean]

Wait let me type what I understand from the problem

[dull ginkgo]

NAH WAIT

I am an idiot

You have to prove the coefficients of g solve a linear system

[chilly ocean]

If S={even naturals}
If f(t)=x + x^3 + x^5
The I can take g(t) to be 0

And it satisfies the condition I believe

[dull ginkgo]

You can “pad out” the polynomials to be of the same degree

And use this: (once I type it out with LaTeX on mobile, pain)

[chilly ocean]

Elaborate

[dull ginkgo]

$\sum_{n = 0}^{N}{a_n X^n} \times \sum_{n = 0}^{N}{b_n X^n} = \sum_{n = 0}^{2N}{c_n X^n} \$ where $c_n = \sum_{i = 0}^{n}{a_{n - i}b_{i}}$

[cloud walrus]

Dyfunction Executive

[dull ginkgo]

Which the lower c_n terms give you a linear system of equations

Which the b_i are unknowns for

[crystal vale]

What if a number is repeated in the same column? Then any conclusion?

[dull ginkgo]

@chilly ocean if you are mental enough you can literally solve this linear system where c_n and a_n are known, b_n are unknown

Actually I have a better idea, but I need to drive home first

[chilly ocean]

Ping me when you re back

[crystal vale]

Thank you

[dull ginkgo]

Bro ngl I cannot process this fuckin question

[chilly ocean]

Same

[dull ginkgo]

It’s just the wording

[chilly ocean]

From what I understand it's this

[dull ginkgo]

But like, N\S need not be finite :(

So like,

What the fuck is it talking about

Like the sum UP TO some N?

[chilly ocean]

I'll repost the picture here just for convenience

[chilly ocean]

Why is it required to be finite

[dull ginkgo]

But that would imply that the polynomial f(x) divides every single one of those polynomials that comes from appending a_s(n) t*s(n) where s(n) is the enumeration of S

[dull ginkgo]

Brother you can’t have an infinite polynomial unless it’s a formal power series ring

[chilly ocean]

We don't need non zero on every s_i

[dull ginkgo]

No

[chilly ocean]

The question is from a recent exam I don't think it's wrong

[delicate orchid]

Ermmmm aktually….

Nothing in there implies you need power series

[chilly ocean]

Yes

Exactly

[delicate orchid]

It’s zero off of S doesn’t imply it’s non-zero on S

[chilly ocean]

Yes

[delicate orchid]

Anyway this question is craaazzzyyyy

[rocky cloak]

It doesn't say anything about what a should equal when i is in S, so you could just choose g=0.

Sort of a silly problem

[chilly ocean]

Yeah I thought of that

[dull ginkgo]

No it’s just the lack of conditions on S

Which means that a_n stops at some N

[chilly ocean]

But suppose I want to make a non zero poly because it was supposed to be of 4 points

[dull ginkgo]

But that fried my brain because that implies our f(x) divides some a_s(k) X^s(k)

[chilly ocean]

I don't think they will allow me to get away with just this

[dull ginkgo]

Let me try to restate this: let $s(k)$ be an enumeration of $S \subset \mathbb{N}$ and $a_n$ a sequence. The question asserts that for a polynomial $P(t) \in f(t)$, that there exists an $N$ such that $f(t) | \sum_{n = 0}^{N}{a_{s(n)} t^{s(n)}}$

[cloud walrus]

Dyfunction Executive

[rocky cloak]

If g is required to be nonzero, the problem becomes interesting. I'm thinking maybe you can do some sort of induction on the number of terms in f.

[chilly ocean]

Yes I was initially trying that I couldn't come up with a constructive algorithm

[dull ginkgo]

You want a constructive proof?

Not a contrapositive

[chilly ocean]

Anything works

As long as it's rigorous

[dull ginkgo]

I have an approach

[south patrol]

Cute

[dull ginkgo]

Let $S_N = \sum_{n = 0}^{N}{a_{s(n)} t^{s(n)}}$, then $S_{N + 1} - S_N = a_{s(n)} t^{s(n)}$.
By euclidean division $S_N = fd_N + r_N \Rightarrow a_{s(n)} t^{s(n)} = f (d_{N} - d_{N - 1}) + (r_{N} - r_{N - 1})$

[cloud walrus]

Dyfunction Executive

[spring horizon]

Does anyone have an intuitive approach as to why the neutral element of a Group is always part of the kernel (phi) (phi being an homomorphism)? -> I get the proof, but I dont really understand the concept behind it

[dull ginkgo]

what's the neutral element?

[rocky cloak]

Identity

[dull ginkgo]

If you are referring to the identity, the definition of a homomorphism of monoids (of which groups are) is that f(e_A) = e_B. the kernel is the preimage of the codomain's identity, so the domain's identity gotta be in it by def

[chilly ocean]

how is this intuitive

[ashen heron]

but it sounds so cool

[chilly ocean]

at least say something like "homomorphisms preserve structure and the neutral element is part of the structure" 😵‍💫

this is literally just "it's by definition"

maybe the definition does pin down what i'm getting at two messages above, but it's certainly not phrased in any intuitive terms like they asked for

[dull ginkgo]

fair, @spring horizon let me rephrase:

Saying that something is in the kernel is means that the homomorphism sends it to the neutral element in the image.

The structure of a group (which homomorphisms preserve) hinges upon any element multiplied by the neutral either way is just going to be that same element. A homomorphism will assure that wherever the neutral is sent to, it’s going to have that same property.

The only element with that property in the output is, of course, the image’s neutral.

If there were two neutrals, calling them a and b, then a * b = a , but also a * b = b. So they have to be equal

[dull ginkgo]

Wait is the a_s(n) sequence fixed or does it depend on f?

I give up, the problem’s worded weirdly

[stark helm]

decimal expansion of real and imaginary part is enough I guess

[dull ginkgo]

To actually compute the nth root usually uses a similar method to computing a root for a general polynomial so using the third or fourth order root equations probably just takes more effort

[chilly ocean]

this has been resolved
Someone in another server suggested to take a companion matrix for the given poly of f
The consider A^i such that i is in S

This set is LD

So consider the polynomial this obtained

As f is minimal poly of A

It divides this

Thank you for your patience help and suggestions everyone

[rocky cloak]

Alright, then I guess you want newtons method, or some other numerical analysis algorithm

Or just plug it into Wolfram alpha

[stark helm]

In fact, this is what the question given, so I might need a way that show some work.

[stark helm]

Find all the ideals of R[x] containing the ideal I = (x^2 + 1). Is it correct to argue that (x^2+1) is maximal in R[x], therefore the only ideal that contain ideal(x^2+1) is R[x] itself? Or we need to argue that why I is maximal??

[void cosmos]

yeah

you need to show that I is maximal

u can just consider the quotient

and use correspondonce isomoprhism theorem

same same ig

@stark helm

[dire siren]

showing the quotient is a field suffices

@stark helm alternatively, you can use the fact that if k is a field then k[x] is a PID

[stark helm]

If I show I is maximal, R[x]/I is a field, and there is a bijection from R to R/I, there does not exist intermediate ideal here, it seems to me that all is done

[delicate orchid]

A bijection from R to R/I? That’s not what the correspondence theorem says lol

[stark helm]

I think it should have a intermediate ideal J in correspondence theorem such that I subring J subring R. but in this question J=R.

[void cosmos]

there is a bijection between the ideals of R/I and the ideals that contain I

[stark helm]

question here, so if we can find the intermediate ideal J here , is it correct to wrtie a bijection between J and J/I?

[void cosmos]

wdym

by intermediate ideal

[stark helm]

like given I subring of J subring of R

[void cosmos]

ur assuming the negation or what

im lost sorry

[stark helm]

I mean the corresponodecne theorem itself

[void cosmos]

yeah you could say if we suppose there is an ideal that contains I then this would correspond to an ideal in R/I which is a contradiction as R/I has no ideals

[void cosmos]

the theorem says that there is a correspondence between ideals of R/I and ideals that contain I

for R ring and I ideal

[stark helm]

so since R/I is a field, but field only contains ideals 0 and itself. so there is no ideal that contains I to map to corresponding ideal in R/I?

[void cosmos]

yeah

[stark helm]

Do you think this one can describe correspondence theorem

[void cosmos]

yeah

by intermediate ideal you mean one that contais

contains the ideal

so yeah

[stark helm]

the map from J to J/I,I am wondering if the J in J/I mean the same J ? is it possible to mean a bijection from J to J'/I where J is subring of J'

[void cosmos]

huh

[stark helm]

Or it mean if J is subring of J', we have the bijection from J to J/I, and bijection from J' to J'/I?

[void cosmos]

the "map" in the text is a map from the set of ideals of R to the set of ideals of R/I

it is defined by sending the ideal "J" to the ideal "R/J"

that's it

this is the correspdoncone and u can prove it

[stark helm]

I might generalize it as if intermediate ideal J exist, then we have bijection between J and ideals that contain I . If J=R, then R/I can not have corresponding ideal, and that's it

[stark helm]

since I think it seems incorrect for this book to write a map from J to J/I, it should be from J to R/J

[void cosmos]

you are confused i think

this isnt a map from J to J/I

a map from J to J/I would involvle sending elements of the ideal J to elements of J/I

[stark helm]

so the book writes J->J/I, what does it want to express?

[void cosmos]

we want to find a map from a set of ideals to a set of ideals

so ur going to send an ideal to an ideal

this is how the map is defined

and this is a correspondence

[stark helm]

so it mean there are some ideals J contains I, and it maps to another set of ideals in R/I?

[void cosmos]

yea it maps to an ideal of R/I , which is J/I

this map is a correspondence

so for every ideal you have here you have exactly one there and vice versa

so with that an ideal that contains (x^2+1) would be an ideal of R[x]/(x^2+1)

the thing is idk if ur even allowed to use that

showing that R[x]/(x^2+1) is a field u might already assume that (x^2+1) is maximal

which is probably not the point of the problem

i think the point of the problem is to use a theorem someone mentioned here

that is, if R is a field then R[x] is a PID

so (x^2+1) is maximal iff it is prime

iff ired

[south patrol]

(as it is non-zero)

[void cosmos]

yeah a 0 ideal can be primee but not maximal u mean?

[stark helm]

I see, in PID, every nonzero prime ideal is maximal

[void cosmos]

yeah

you can do this manually or you can just see that since this is a quadratic then its enough to just say this has no roots and ur done

[stark helm]

that's the first method that I come up with for this question, so I can determine that it is maximal, actually the original question writes the hint: correspondece theorem for rings

[void cosmos]

weird ig unless you have R[x]/(x^2+1) is a field given for you

but thats just my stupid opinion who gives a fuck ig just use it then

[stark helm]

How will you justify this question? i know that E=F(a) and [F(a):F]=1, so [E:F]=1 and the root a in F is also in E. I don't know if it is complete to write E=F?

[celest furnace]

Show that if V is a vector space of dimension n and W is a subspace of V also of dimension n then V = W

Once you have this then you are done

[clear fiber]

Hi, I have written a proof attempt for part b of the problem. But I suspect that I have made a mistake. I never used the fact that B is the set of ideals in R that contain K. Nor did I use part (a) of the problem which the hint suggests to use. Can anyone see where I might have gone wrong?

The exercise 10 mentioned by the hint is this:

[toxic zephyr]

i think the hint is trying to get you to use the FHT

a possible issue with your proof is in assuming you can define phi by phi(f(I))=I. how do you know every element of A is the image of an ideal in B under f? i don't think this is necessarily true without using the fact that elements of B contain the kernel.

overall i think your proof has the right idea. but there are some gaps in how justified your definition is

if you prove that phi is well defined, i think this is a valid proof that avoids using part a

but the FHT might be a much quicker proof

the fact that part b uses K for ker(f) and part a uses K makes it seem like they really want you to leverage what R/K is isomorphic to

[stark helm]

ok，i get it. i actually have another idea now, to apply F[x]/(x-a) is isomorphic to F(a) and show that F is isomorphic to F[x]/(x-a), do you think this one is correct idea?

[celest furnace]

Yeah that works

Showing that F[x]/(x-a) is iso to F shouldn't be so bad and i think you defined E as F[x]/(x-a) so i think you're already done once you've done that?

[stark helm]

yes, I define E=F[x]/(x-a) and isomorphism to F(a). and proved F isomorphism to F[x]/(x-a), then F isomorphism to F(a)

[stark helm]

I am considering I have shown that [E:F]=1 and E is a vector space over field F, while F is a vector space over another field,so dimension of E is the same as F while F must be supspace of E( because E is extension field), do you think it is correct?

[clear fiber]

@toxic zephyr Thank you, your comments helped me write what I believe is a much more solid proof which I feel satisfied with. It ended up being incredibly verbose. There's probably a much cleaner way to write it.
The idea was

1. Let A be set of ideals in R which contain K.
   Let B be set of ideals in R/K.
   Let C be set of ideals in S
2. There is a bijection g:B -> A given by g(I/K) = I
3. FHT says S is isomorphic to R/K. Since isomorphisms map ideals to ideals, there is a bijection theta:C -> B between them.
4. So, phi:C -> A given by theta(g(J)) is a bijection because it is a composition of bijections

The actual verification of steps 2 and 3 was laborious

It involved both hints which the problem suggested

And the fact that A is the set of ideals in R which contain K, I think it only mattered in that the notation R/K usually assumes that

[toxic zephyr]

i think it's better to understand it as the set of ideals in R/K is just isomorphic to A, and that's what you're leveraging in your proof.

i.e. B is a stepping stone of an isomorphic ring. because that's what you showed: A, B, and C are all isomorphic.

[clear fiber]

Yeah that is better

[rain grove]

I need help with some group theory:

Let (S, ∘) be a Semigroup. Prove that
S group
<=>
For every a,b the equations:
"a ∘ x = b"
"y ∘ a = b"
are solvable

[celest furnace]

is semigroup just group without inverses? In that case try specifying values of b to find left/right inverse

[ashen heron]

semigroup is just a set with an associative binary operation

[queen quarry]

first you would like to show that the neutral element exists, what does it need to satisfy?

[rocky cloak]

I guess semigroups must be defined to be nonempty for this to work.

[rain grove]

Okk so tryint to prove <=

• asociative is because it is a Semigroup
• Unit:
  There exists a unit element e so that for every a : a ∘ e = e ∘ a = a
  But the assumption is that For every a,b there exists x,y that those 2 formulas are true. If i fix a = b then i get
  a ∘ x = y ∘ a = a
  and idk what to do next

[queen quarry]

lets focus on right identity first

we know ax = a, but now we want to show that for all y in S: yx = y

for that we have 3 tools, namely
"the equation ax=b always has a solution"
"the equation ya=b always has a solution"
"ax=a"

What could we try to do?

[rain grove]

If we want to show yx = y then if we know ax = a for all a, we could say a = y right?

but why we want to show yx = y ?

[queen quarry]

ok now you're getting confused because of variable names so lets clear things up

lets fix an element g in S and we know that
g e = g
has a solution

Now we want to show that e is a right identity, so for an arbitrary s in S, we want to show that
s e = s

at the moment, e is only the solution of g e = g for that one specific g we chose, we don't know if it also solves s e = s

[rain grove]

oh okk, and how do we know its the only solution for that specific g?

[queen quarry]

we don't know that yet

[rain grove]

oh but we want to prove that that same e will solve s e = s for every other s in S

[queen quarry]

correct

the thing you're struggling with right now is that we don't know anything about s, right?

[rain grove]

yeah i mean i cant figure out how exactly that e (that was ok for g) is good for every s

[queen quarry]

well i kind of lied in my last sentence, we do know something about s. if we choose s as a constant for an equation, that equation always has a solution

[rain grove]

how would we do that?

[queen quarry]

we have 4 equations we could try, for fixed a and b in S:
sx = b
ax = s
ys = b
ya = s

and at some point we want to use that ge=g, so maybe we can choose a specific element of S

[rain grove]

I still dont see it

if g is a specific element, e is also a specific element, a and b are fixed, s is a constant and x and y are just some solutions. I dont see any variables i could exchange to use the equation ge = g

[queen quarry]

we can choose what we use for a or b

sometimes it helps to just try a few ideas, can we use the equation sx = b in any way?

the end result is supposed to be like
se = [...] = s

[rain grove]

Yeah we can

we could say b = g and then we have sx = ge

[queen quarry]

what's the next step?

[rain grove]

i guess i could use
b = g -> sx = g
a = g -> s = gx
and then if i insert s in the first equation
g x^2 = g = g e
so that means e = x^2

[queen quarry]

why is e=x²?

[rain grove]

but idk if i can say e = x^2 since S is not yet a group

and i cant just multiply the inverse element from the left

since there is no inverse element yet

yeah idk

[queen quarry]

i believe that here you mix up variables again

with multiple different variables being named x

[rain grove]

ok i dont understand the variables then

[queen quarry]

if you give me an a and b from S, I can give you an element x such that ax=b.
if you give me s and g, i give you an element x_1 such that s x_1 = g
give me g and s, and i give you an element x_2 with g x_2 = s
We don't know if x_1 = x_2 yet, these 2 are just some solutions of equations

[dire siren]

@rain grove there is some e such that ae=a
pick b arbitrary, and there is c such that ca=b
combine these to get be=b

[chilly ocean]

Is there an easy way to prove that theres not field extension between Q[x]/(x⁴-x-1) and Q?

[queen quarry]

is x^4-x-1 irreducible over Q?

[chilly ocean]

yeah (trust me)

if you want, consider the same question with x⁴-2x-2 instead, that one is irreducible by Eisenstein

[queen quarry]

is this true?
Q[x]/(x^4-x-1) | Q is a galois extension (i think) of degree 4, hence the galois group has 4 elements and by the main theorem of galois theory there are 4 in-between fields

[chilly ocean]

its not galois

[eager willow]

Its dimension is 4 over Q. So an intermediate extension must be a quadratic. If f(x) satisfies a quadratic equation modulo x^4-x-1 it will also satisfy the quadratic equation as a polynomial over F_2

So now look for elements in F_2[x]/x^4+x+1

[chilly ocean]

oh nice

[eager willow]

I'm not sure if this leads anywhere but just a thought

That polynomial x^p^n-x-1 has nice behavior mod p

It's saying the nth frobenius twist on x equals x+1

Yeah ok maybe this doesn't go anywhere because this finite field is just F_8, so it has F_4

[rocky cloak]

I'm not sure it qualifies as an easy way. But if you can show that the extension doesn't contain any other roots, then the Galois group must have order at least 12. Since the polynomial has 2 complex roots, conjugation is an element of the Galois group, so the Galois group cannot be A4. That leaves S4 as the only possibility.

The only index 4 subgroup is S3, and there are no subgroups between S3 and S4.

And if it's hard to show that it has two complex roots, I guess you can just repeat the argument for A4

[rain grove]

how do we know now that x_1 = x_2?

[queen quarry]

we can't know that, it's not true in general

[queen quarry]

the issue was that you declared both as x and used them as if they were the same

[rain grove]

Sure I understand so far

How do we continue from here and prove there is a unit which is the same for all?

[dire siren]

see my hint

[rain grove]

Hmm ok I think i figured out something:

g ∘ e_1 = g
e_2 ∘ g = g

for g, e_2 we can apply the original assumption and we have:

g^(-1) ∘ g = e_2
g ∘ g^(-1) = e_1

i just named the solution x,y to g^(-1) (its not necessarily the same yet)

but then using this into the first two equations I get e_1 = e_2

[dire siren]

that's right

but indeed, you might want to denote g^(-1) differently in those two instances, say g' and g'' for example

[rain grove]

ok idk what to do next

[dire siren]

you want to prove that e is a right identity

so for any b you want to prove that be=b

where I denoted the elements e_1 and e_2 by e

[winged void]

het i have a question regarding groups and its the following

[dull ginkgo]

*and nothing followed*

[winged void]

. Any permutation σ of X can be written as a product of a pairwise number disjoint

[winged void]

that follows it

like i do not understand it that well

can somoen give me an example small example

[dull ginkgo]

Pairwise disjoint transpositions ie two cycles

[dull ginkgo]

Are you asking about pairwise disjoint cycles

[vivid tiger]

pairwise disjoint transpositions?

[dull ginkgo]

I had a brain fart

Ignore that

[vivid tiger]

i thought the trick explicitly involved matching it up at one of the positions

oh, good, i hadn't forgotten the basics of permutations 🙂

[winged void]

not really sure because thats what the lemma in the book says

[dull ginkgo]

You have to consider what each disjoint cycle “touches”. Like for example we have (1, 2, 3, 4, 5). If we cycle like, 1 2 and 4, and 3, 5 independently, we don’t care what order we apply those permutations

Because they don’t touch the same elements

If they do, then it doesn’t, but otherwise if you apply one cycle then another, they commute because they don’t “affect” eachother

[winged void]

one minuit what do you mean with independently

like are they in the same cycle

[dull ginkgo]

No, we are considering their composition

So (1 3 5) (2 4) for instance

You cycle 2 and 4, then 1 3 5

[winged void]

sure or other way around does not matter

[dull ginkgo]

But if you do it in reverse, it’s going to be the same, because they don’t “affect” eachother

[winged void]

because they are like diffrent from each other

[dull ginkgo]

So if you have a permutation

You can look at each element, and see how it gets cycled

At some point it’s going to have to return to itself

[winged void]

sure agree on all that but (1 3 5)(2 4) is something else than (1 2 3 4 5)

right

[dull ginkgo]

Yes

(1 2 3 4 5) is a cycle

If you have a permutation. Each element is in a cycle (can be trivial) because it’s going to have to be permuted back to itself after being applied so many times

[winged void]

its like one cycle but can we write in a disjoint cycles not really right

[dull ginkgo]

No

It’s basically a “disjoint cycle” by itself, I.e it’s fully decomposed into cycles

Ben

*brb

[winged void]

i see i understand because like the lemma said . Any permutation σ of X can be written as a product of a pairwise number
disjoint and i did not get what it mean

so its like the thing that you explained about disjoint cycles

this is what i meant

one last question

[vivid tiger]

this construction is different

[winged void]

oeps

[vivid tiger]

this constructs e.g. (123) (45) by doing
(23)(12) and (45)

[winged void]

i see

but (23)(12) its not the same as (1 2 3 )

right ?

because here 2 -> 3 -> 2

[vivid tiger]

oh, shit, I fucked this up

here 2 goes to 1 and stays there

we don't want that

[dull ginkgo]

My favorite approach is the “bump” approach

Where you kind of propagate cycles up (like used for sorts) to permute elements

There’s also the other one where you just swap an element with 1, then swap 1 and the other

[vivid tiger]

(13)(12)

1 goes to 2, 2 goes to 3, 3 goes to 1, and I retain my math license

I remember figuring this out before seeing it and feeling very proud of myself

[dull ginkgo]

Can you still decompose permutations into disjoint cycles in the countable case?

I don’t think so?

[winged void]

i guess

[vivid tiger]

you can still decompose permutations into disjoint cycles in the infinite case, i think

because we have a group action G on S

[winged void]

how 2 -> 3

[vivid tiger]

and the orbits must be disjoint

[vivid tiger]

follow 2

[winged void]

1 -> 3 and then

[dull ginkgo]

Yeah but do they need to be finite?

I guess you can consider infinite cycles

[winged void]

no

[vivid tiger]

2 goes to 1, then 1 to 3, so in total 2 to 3

[vivid tiger]

I was assuming infinite cycles. Otherwise of course you get a problem the moment you have an infinite orbit.

[dull ginkgo]

No in the countable case you might have to consider a “countable product” of cycles

Which of course you can’t do

[winged void]

i see

[vivid tiger]

Your theorem only talks about finite ones.

[dull ginkgo]

Has to be finite

[winged void]

i see

and one more last question maybe its a dump one

like they say in my book that $S_n$ has n! elements

[cloud walrus]

Mootje

[winged void]

i do not get what they mean exactly

[vivid tiger]

Do you know what a factorial is?

[winged void]

i know its a bijection

yes i know

n * n-1 * ... * 1

[vivid tiger]

Then, prove that the counting works.

[winged void]

but the thing do they mean like lets say for example

[vivid tiger]

I'll start you off: to every number in the input, you need to assign a number in the output

[winged void]

like the first elemnt lets say sigma(1) -> n elemnts and then sigma (2) -> n - 1 elements

before we specify

and at the end we have n * n -1 * ... 1

permutations

[vivid tiger]

Exactly

[dull ginkgo]

Hey so try to decompose the cycle on the Integers that sends n to n + 1 into transpositions

[winged void]

well why is that even important

[dull ginkgo]

The problem is that in groups you cannot consider an “infinite composition” I’m pretty sure

[vivid tiger]

...why is the number of permutations important?

That shows up everywhere in combinatorics

Comes up algebraically whenever the symbols you are multiplying can be permuted

[winged void]

i mean for a group

[vivid tiger]

Permutation groups are important. Thus, the number of them are

[dull ginkgo]

@vivid tiger consider S_Z and the successor permutation: You’d have to take the product of (n n + 1) over all n in Z but you can’t do that. Wouldn’t this be a contradiction

[winged void]

i know but why are we interested to know that

[vivid tiger]

Hmm.

But if you had infinitely many disjoint orbits, you could compose them.

[dull ginkgo]

Explain?

[vivid tiger]

Not directly using the group.

But, it seems clear to me what the permutation should be: if the cycles are C_i, then for x_i in the orbit of C_i, σ(x_i) should equal C_i(x_i)

[dull ginkgo]

excuse my lack of knowledge there was like only 2 pages in permutation groups in Jacobson only on the finite case

[vivid tiger]

And this is a permutation

I'm also finding this new.

[winged void]

i see

and one more laste question

[vivid tiger]

This isn't the group product, because you can't take infinite products in there.

[winged void]

there is an example that tells us why permutation are important

[vivid tiger]

why are we interested in permutation groups?

in groups?

[winged void]

or symmetry group

but do not get anything from

In the following example, we take for (X) the ordinary (Euclidean) flat plane (\mathbb{R}^2).
An isometry of (\mathbb{R}^2) is a mapping (\sigma: \mathbb{R}^2 \rightarrow \mathbb{R}^2) with the property that for all (P, Q \in \mathbb{R}^2) it holds
[ d(P, Q) = d(\sigma(P), \sigma(Q)), ]
where (d(P, Q)) indicates the distance between (P) and (Q); so if (P = (p_1, p_2)) and (Q = (q_1, q_2)) then
[ d(P, Q) = \sqrt{(p_1 - q_1)^2 + (p_2 - q_2)^2}. ]
We give some examples of isometries:
\begin{enumerate}
\item Translations. If (P = (p_1, p_2)), then the mapping (t_P: \mathbb{R}^2 \rightarrow \mathbb{R}^2) given by ((x_1, x_2) \mapsto (x_1 + p_1, x_2 + p_2)) is an isometry which we call translation over (P). Translations are bijective and the following rules apply:
[ t_O = \text{id}{\mathbb{R}^2}, \quad t_Q \circ t_P = t{P+Q}, \quad (t_P)^{-1} = t_{-P}, ]
where (O = (0, 0)) indicates the origin, and for (P = (p_1, p_2)) and (Q = (q_1, q_2)) we write
[ P + Q = (p_1 + q_1, p_2 + q_2) \quad \text{and} \quad -P = (-p_1, -p_2). ]
\end{enumerate}

[vivid tiger]

The permutation groups come up very often.

[cloud walrus]

Mootje

[vivid tiger]

An example with isometries cannot possibly be the simplest example showing you their importance

[dull ginkgo]

A lot of the finite case stuff depends explicitly on the pigeonhole principle to show order or finite orbits which don’t work nicely with even countably infinite sets. Furthermore it allows decomposition because there can only be finitely many cycle permutations composed

Infinite ones break down obviously

No “limit” structure

[vivid tiger]

But, I feel that there must be a way to do it.

After all, my construction that sends x_i to C_i(x_i) seems reasonable

[winged void]

but like in the example here above

do not really understand why they use d(sigma(P) , sigma(Q))

[vivid tiger]

As functions, if you have them such that the set on which f_i is not the identity is disjoint from the set on which any other f_j is not the identity, you can take a union of the functions as relations

[dull ginkgo]

In the finite case you can really lazily just take the product over (x \sigma(x))

[winged void]

sigma here is just a pair of function and does not has to do with anything with permutation or symmetric group

Am i right

[vivid tiger]

...?

the point is that σ preserves distance, that's what the definition says.

[winged void]

sure

but what has that to do with permutations

[vivid tiger]

Why does this have to have anything to do with permutations?

[winged void]

yes

because he introduced first symmetrical groups and then the permutation and then this example

so im not seeing the connection

[vivid tiger]

"symmetrical groups"?

[dull ginkgo]

I just like considering the orbits and the fact that you can swap em around because they don’t touch eachother

I prefer permutation groups instead of symmetry groups referring to S_n tbh

Symmetry group as a term is more general

Though synonymous with automorphism

[vivid tiger]

"the symmetric group" is the other name for S_n

terrible naming imo

"symmetric" being different from "symmetry"

[dull ginkgo]

I am at a red light sitting for a left turn so if I disappear that’s why

[vivid tiger]

is the implication that you'll crash and die and never come back or that you'll be busy driving

[dull ginkgo]

schrödinger’s driver

The observer in this case being the fucker riding my ass

[rain grove]

I dont how to prove that

[dire siren]

I basically told you how to prove it: #groups-rings-fields message

you can replace "a" by "g" if you want to stick with your notation

[frank cosmos]

can someone give me a hint to show its right noetherian

[rocky cloak]

Describe what all the right ideals are

One thing to notice is that an ideal will look like [I, X; 0, J] where I is an ideal of Z and J is an ideal of Q. So the question is just what X could be (it might depend on I and J).

[frank cosmos]

J must be trivial right

since Q is a field

if [x,y;0,z] is in a right ideal, so is [ax, bx+cy; 0, cz] for all a in Z, b,c in Q

[rocky cloak]

Yeah, so if x is nonzero, what does that tell you?

[frank cosmos]

bc+cy=Q?

[rocky cloak]

?

[frank cosmos]

in your notation, X=Q

[rocky cloak]

Yes, that's right

[frank cosmos]

so then we're good in this case since all ideals of Z are principal

now x=0

then still X=Q?

[rocky cloak]

Not necessarily

[frank cosmos]

if y!= 0

no

[rocky cloak]

Well, sort of

[frank cosmos]

im brainfarting

[rocky cloak]

Describing the ideal is actually a little more complicated than describing each coordinate here

[frank cosmos]

Okay, so we're assuming everything in the ideal is of the form [0, y; 0,z]. Then, for all c in Q, [0,cy; 0,cz] is in the ideal right?

[rocky cloak]

That's right

So you have elements that looks like pairs of rational numbers, and the action of the ring is multiplication by a rational number. This should feel familiar.

[frank cosmos]

so wouldnt this be generated by (at most) 2 elements, one with y!=0 and the other with z!=0?

(or if y=0 for all take the element with z!=0 etc)

[rocky cloak]

Unsure what you mean, but for example the ideal generated by
[0, 1; 0, 1] is certainly an ideal. And the ideal generated by [0, 1; 0, 0] is a different one.

[winged void]

small question what does it mean when we say $Q^{+}$

[cloud walrus]

Mootje

[winged void]

is it like the group where $Q_{>=0}$

[cloud walrus]

Mootje

[rocky cloak]

Could mean the group of rationals under addition. Could also mean positive rationals. Both are possible

[winged void]

well then in my case is it under addition

because first i thought that its positive rational but that cannot be true

[rocky cloak]

Well I don't know what your case is

[winged void]

because then its not a group

[rocky cloak]

Well the positive rationals is a group under multiplication, but it would be a little weird not to specify the operation in that case.

So yeah, addition seems most likely

[winged void]

exactly

[sour depot]

positive rationals under multiplication if you're talking about the set Q_{>0}

otherwise just rationals under addition

since positive or nonnegative rationals under addition isn't a group

[rocky cloak]

If you have the full context it might be easier to determine

[chilly ocean]

Whats a simple example of a number field with class number >1 that embeds into a number field with class number 1?

[frank cosmos]

i mean X=0 or X=Q

[rocky cloak]

Yes, but the structure of the ideal is a little more complicated than just specifying X.

[frank cosmos]

i guess i need an element [0,y;0,z] where y!=z

if there exist an element where z!=0!=y!=z the ideal is just [0,Q; 0,Q]

the rest of the cases just seem trivial, am i missing osmething

[rocky cloak]

So there are actually infinitely many such ideals.

[rocky cloak]

Think about this

[dire siren]

You should ask this in #advanced-number-theory, but I think I can answer your question: you cannot have a number field K of class number >1 inside a number field L of class number 1.

[frank cosmos]

i am not seeing this

[dire siren]

There is this result that says that a number field K has class number 1 if and only if O_K is a UFD.

[rocky cloak]

Alright, so what is the ideal generated by
[0, x; 0, y]?
For which values of (x, y) do you get the same ideal?

[chilly ocean]

i dont think thats true

[frank cosmos]

[0,cx;0,cy]

[chilly ocean]

a non-UFD can embed into an UFD

[frank cosmos]

and for the second question, if (x,y)=c*(x0,y0)

[rocky cloak]

So then you see that there are infinitely many

[frank cosmos]

i see

[dire siren]

yes, you are right, sorry for my confusion

[rocky cloak]

I believe Q(sqrt(-15)) < Q(sqrt(-3), sqrt(5)) should be an example.

At least sage says so

[dire siren]

I think Q(sqrt(-5)) < Q(sqrt(-1),sqrt(-5)) also works

[chilly ocean]

how did you find them so quickly

[rocky cloak]

Guessing and checking

[frank cosmos]

if every element in the ideal has x=0=y the ideal is 0.
if every element has x=0 then the ideal is generated by an element with y=0, if there is one, and 0 otherwise.
same thing for y=0
if every element has x=0 or y=0 but not both, the ideal is [0,Q ; 0,Q]
else, the ideal is either generated by 1 element or [0,Q;0,Q]

[chilly ocean]

According to this it works https://www.lmfdb.org/NumberField/4.0.400.1

[dire siren]

same here; I started with Q(sqrt(-5)), because it's standard, and then I just guessed

yes, I used LMFDB for checking

[rocky cloak]

Alright, so now you should have enough information to see that the ring is Noetherian.

[frank cosmos]

ye

[rocky cloak]

A similar example, that might even be easier to see, is that [Q, R; 0, R] is right artinian, but not left Noetherian.

[stark helm]

I feel very confused about how to describe this one? since we have already know it is a field that contains all rational numbers and pi, which aspect do we have to explain in depth?

[prime sundial]

describe a general element of Q(pi)

for example, Q(i) consists of elements of the form a + bi, where a,b in Q

what does Q(pi) consist of?

[stark helm]

do you mean describe the form a+b*pi, where a, b in Q?

[prime sundial]

well that isn't the form of elements in Q(pi)

consider what Q(x) is v.s. Q[x]

it then might help to think of whether pi is algebraic or not over Q

[stark helm]

it seems that Q(pi) over Q will have infinite basis?

[prime sundial]

remember that we can write Q(i) as a + bi because
a + bi + ci^2 + di^3 + ...
simplifies

[prime sundial]

yes

[stark helm]

since pi does not belong to Q

[prime sundial]

well that's not why

finite extensions are algebraic. if Q(pi) were to be a finite extension, it would suggest pi is algebraic

[prime sundial]

consider that Q(i) is a finite extension even though i doesn't belong to Q

[stark helm]

it has basis 1 and i only

[prime sundial]

to be pedantic, no, but that's beside the point

as in, there are many different bases, but it is dimension 2 yes

let's suppose the elements were of the form a + b * pi

now let a = 0 and b = 1

since we have a field, we should expect pi * (a + b * pi) to be in Q(pi)

then we'd want pi^2 in Q(pi), where pi^2 = a + b * pi, for a,b in Q

you may quickly realize this is impossible, since it would suggest a minimal polynomial for pi in the form
a + bx - x^2, which would then suggest pi is algebraic (which it isnt)

you can repeat this argument, to get to the point that we must at least include all polynomials on pi with coefficients in Q:
a + b*pi + c*pi^2 + d*pi^3 + ...

can you tell what elements are missing?

[stark helm]

missing elements? I think the pi should have infinite degree at last

[prime sundial]

yes we have at least all polynomials on pi

but we are still missing some elements

for example, pi^-1

[stark helm]

shouldn't the inverse be included in .... tend to be pi^n as n tends to infinity?

[prime sundial]

sort of, and that is a good question

but it isn't actually included

since we only consider finite polynomials

[stark helm]

I might think since i*i=i^2=-1. which is basically the same as1

[prime sundial]

we can go arbitrarily high in degree, but they will always be finite

if we had some finite polynomial p(pi) on pi with coefficients in Q such that
pi^-1 = p(pi), then it would suggest
1 = p(pi) * pi
0 = p(pi) * pi - 1, which would again suggest pi is algebraic over Q

so instead what we would want is to include all rational functions on pi with coefficients in Q

so we would like all of the expressions of the form
[\frac{a_0 + a_1\pi + a_2 \pi^2 + \dots}{b_0 + b_1\pi + b_2 \pi^2 + \dots}]
where (a_i,b_i\in \mathbb{Q}), and such that the denominator is nonzero

[cloud walrus]

maximo

[prime sundial]

this might not be enough, but it turns out that it is

Q(pi) is given by the elements of the above form

you can show this rigorously but i dont think i could explain that well. you could probably check MSE or wait for someone else to explain that if you need it

does that make sense karl?

[sour depot]

Q(\pi) is the same as the field of rational functions Q(t)

since Q is not algebraic over Q, it isn't the zero of any polynomial over Q, so adjoining \pi is the same as adjoining any indeterminate

[stark helm]

so you suppose a finite polynomial p(pi)=pi^-1 on Q(pi)? and also for p(pi)*pi-1, will this kind of irreducible appear on Q[x], i mean p(pi) is valid on Q[x]?

I am confused about p(pi)*pi-1 will appear on Q[x]?

[prime sundial]

the idea is to take a polynomial p(x) in Q[x]

and suppose that p(pi) = pi^-1

then p(pi) * pi = 1

and so p(pi) * pi - 1 = 0

this would be a minimum polynomial for pi:
p(x) * x - 1

but pi is transcendental

this tells you that you can't write pi^-1 as a polynomial on pi, with coefficients in Q

so [\pi^{-1}\ne a_0 + a_1\pi + a_2\pi^2+ \dots + a_n\pi^n,\ a_i\in\mathbb{Q}]

[cloud walrus]

maximo

[prime sundial]

there are maybe easier ways to say this, but this is the important result

[stark helm]

so because pi algebraic will also imply that pi^-1 algebraic, but in fact pi^-1 is transcendental, is that true?

[prime sundial]

pi is not algebraic

[stark helm]

so since we have Q(pi) over Q, pi should be adjoined in this field? can't it imply algebraic over Q?

[sour depot]

\pi isn't algebraic over Q, but you can adjoin it to Q just as you can adjoin anything else

[stark helm]

I see, since pi in Q(pi) can not figure out f(x) in Q that is algebriac over Q(not root of any f(x)

[sour depot]

right there is no polynomial f(x) over Q such that f(\pi)=0

so Q(\pi) is structurally the same as the field of rational functions Q(t) in any indeterminate t

[stark helm]

but it still seems confused to me why it happens? and also why pi^-1 can not appear here( is it really forces polynomial to be finite)?

[sour depot]

what

[celest furnace]

Yes elements of Q[x] have finite degree

If there was a relation you could multiply both sides by pi to get pi is algebraic

[stark helm]

because I see maximo justify why pi^-1 can not be written as a polynomial of pi, but still feel somewhat confused. also why Q(pi) is structured the same as Q(t)?

[sour depot]

an indeterminate t doesn't satisfy any polynomial relations over Q, and \pi doesn't satisfy any polynomial relations over Q

[stark helm]

so t can be any irratonals right? such as Q(sqrt(2)), Q(pi)? does that refer to the same structure?

[sour depot]

no

something like \sqrt{2} satisfies algebraic relations over Q

t can be any transcendental over Q

[stark helm]

like x^2-2?

[sour depot]

yes

any transcendental behaves like an indeterminate algebraically speaking

[stark helm]

Q(t) represent any rational functions with coefficient Q, and t is transcendental. pi is also transcendental over Q, so it structure similar?

[sour depot]

t is an indeterminate here but it behaves the same way if it's some fixed transcendental yes

and yes you're looking at the field of rational functions over Q

[stark helm]

I think I get your argument now, since pi is not algebraic, but p(x)*x-1 behvaes like pi is algebriac but it doesn't. we write p(pi) in form of pi^-1, and it is impossible to write out.

[prime sundial]

yes, so now you have that pi^-1 cannot be written as
a + bpi + cpi^2 + dpi^3 + ... + kpi^n

so you need to bring in rational functions

[prime sundial]

like these

[stark helm]

I see, really appreciate your response

[dull ginkgo]

Hey chat in general is $A, B \subset G \bigwedge |A| + |B| > |G| \Rightarrow AB = G$ for finite group $G$ a hard exercise?

i.e is it worth my time because it looks awful

[coral spindle]

Well it's false so I think it would be pretty tricky

[dull ginkgo]

Really?

I should specify G is finite

[coral spindle]

I assumed G was finite.

[dull ginkgo]

It's an exercise in jacobson

[coral spindle]

You also meant |G| not G

[dull ginkgo]

Yeah sorry

[cloud walrus]

Dyfunction Executive

[coral spindle]

Oh wait > |G| not >= |G|

[dull ginkgo]

Yep, that's the kicker

[celest furnace]

Yeah u we’re thinking two copies of alternating group right

[dull ginkgo]

I was going to use Inclusion-Exclusion

[coral spindle]

No

But it was a similar idea

[celest furnace]

Good idea

[coral spindle]

Yeah, that's a good strategy.

[dire siren]

{e} and G-{e} is a counterexample for >=

[coral spindle]

We know

It's worth checking, is this just subsets or do A, B have to be subgroups?

[dull ginkgo]

actually idk how to use inclusion-exclusion here

[dull ginkgo]

subsets, subgroups you could use the prior exercise

[coral spindle]

OK

[celest furnace]

Maybe pigeonhole

[dull ginkgo]

i.e |A| |B| = |AB| |A cap B|

[dull ginkgo]

yeah, thinking about using that

If they are disjoint subsets it's easy

gonna ponder for a bit

[celest furnace]

Hint:: ||think about normality||

[dull ginkgo]

might hijack the commutator/normalzier of A or B and use pigeonhole in a second if the idea I have doesn't work rq

[celest furnace]

Wait it’s just subsets?

[dull ginkgo]

Yep

It's the second to last exercise of this section

torture

I could try to do some shit with inverses

[dire siren]

yes, inverses should work

take some arbitrary element g

[dull ginkgo]

no

[dire siren]

and consider gB^(-1)

[dull ginkgo]

don't give an immediate answer a

AA

true

[dull ginkgo]

Thank you.

Just got it

[dire siren]

no problem

if you want to try a similar problem, here is one: Prove that in a finite field any element is the sum of two squares.

actually I think it's just a corollary of that problem

[dull ginkgo]

Assume $|A| + |B| > |G|, \exists g \in G \setminus AB$. therefore $A^-1g \subseteq G \setminus B \Rightarrow |A^{-1}g| \leq |G \setminus B|$ However, the map $x \mapsto x^{-1}g$ is a bijection between $A$ and $A^{-1}g$ by group cancellativity, thus $|A| = |A^{-1}g|$. This implies $|A| \leq |G \setminus B| \ $. By Inclusion-Exclusion: $|B| + |G \setminus B| = |G| < |A| + |B| \Rightarrow |G \setminus B| < |A| = |A^{-1}g|$, contradicting $|A^{-1}g| \leq |G \setminus B|$

[cloud walrus]

Dyfunction Executive

[dull ginkgo]

WHEW

That SUCKED

[dire siren]

the solution is fine

I had a slightly different idea, but essentially the same

my thought process was that we want to prove the reversed inclusion, so that for any g we want to prove that there are a in A and b in B such that ab=g

equivalent to a in A and b in B such that a=gb^(-1)

[dull ginkgo]

Pigeonhole I assume

smart

[dire siren]

that's why I chose the set gB^(-1)

[dull ginkgo]

The contradiction in mine is literally pigeonhole's contrapositive in retrospect lmao

The next exercise is to find a subgroup of index 2 in a group of even order

I think I have an idea via cayley's theorem that will work to find an element of order 2, but it might actually work for any prime dividing the order of the group

[celest furnace]

What extra conditions on the group are there

A_5 doesn’t have an index 2 subgroup for example

[dull ginkgo]

a = (1 2)(3 4). Order of a is 2.
<a> = {e, a}
[A_5 : <a>] = 2

no wait

messed that up

brain fart

order of the group is 2

I misread the problem

[dull ginkgo]

This exercise requires me to find Aut(S_3):
let g_1 = (1 2), g_2 = (2 3), g_3 = (3 1). These are generators of S_3, and the only elements of order 2. Any automorphism is uniquely determined by the images of these generators, which also must be order 2, therefore gives a permutation of the set of generators. This provides a homomorphism from Aut(S_3) -> S_3. It is mono because automorphisms are determined by generators uniquely, and it is epi because every element in the image can be used to describe an automorphism of S_3. Therefore S_3 is iso to Aut(S_3)

In general, if a group G is generated by a set of elements K, then is there an isomorphism between G and S_|K| (even if G, K are infinite)?

[prime sundial]

this only holds supposing that all the generators have the same relations

[dull ginkgo]

ah

how else would I prove Aut(S_3) ~= S_3

[prime sundial]

im not saying your proof is incorrect

your generators all behave the same

[dull ginkgo]

No i'd have to prove the relations

what would be a better proof?

I'm going off of Jacobson and this section doesn't really go over how to find automorphism groups explicitly

I'll leave it be for now, it's just self study. I'm mostly worried that I am going to not understand if I come across another problem like it

[prime sundial]

the fact that S_3 is so small might be hinting at just cranking it out

has jacobson gone over presentations yet

[dull ginkgo]

no lol

[prime sundial]

did they just give the table for S_3 or

[dull ginkgo]

no

It's just "Determine Aut(S_3)"

literally that's all the question says

[prime sundial]

yeah i was asking silly questions nvm

[dull ginkgo]

I am confused by this next problem in mainly the way it's worded

Like isn't G_L literally a subgroup in Aut(G)

[prime sundial]

ok this doesnt even hold in this case i think. you may have 2 distinct generating sets, e.g. (123) and (12)

[prime sundial]

is a_L(g) = ag

[dull ginkgo]

yes

NO WAIT

SET G

brain fart

[dull ginkgo]

I just realized I mixed it up with Inn(G)

$\mathrm{Aut}(G)$ and $G_L$ I assume have trivial intersection, so $|\mathrm{Hol}(G)| = |\mathrm{Aut}(G)| |G_L|$

[cloud walrus]

Dyfunction Executive

[dull ginkgo]

yeah because that would imply a(e) = e => a = e

[prime sundial]

ah nice yeah

[dull ginkgo]

@prime sundial Is it a good exercise to prove that Out(S_6) is nontrivial

or is it actually difficult

I haven't tried it, all I know is that it isn't

[prime sundial]

no idea

im sure it'd be interesting, not sure how difficult it is

[dull ginkgo]

proving SL(2,Z)'s abelianization is Z_12 is fun because of the use of the euclidean algorithm but it's also like

a pain in the balls in terms of presentations

So fun in fact that I am tired of working on jacobson and I am going to try generating the relations again

I know Euclidean Algorithm implies $\begin{bmatrix} a & b \ c & d \end{bmatrix} \in \mathrm{SL}(2,\mathbb{Z})$ can be reduced to the identity (the gcd of a and c) via repeated use of $\begin{bmatrix} 1 & q \ 0 & 1 \end{bmatrix}$ or it's transpose where $q$ is the euclidean divisor of $a$ by $b$ or vice versa

[cloud walrus]

Dyfunction Executive

[dull ginkgo]

The problem is i'm pretty sure the relations are fucking awful

Actually it isn't so bad because of the inverse map and the transpose map both being antiautomorphisms

wait what the fuck

The inverse of the transpose is an automorphism on SL(2,Z)

[dull ginkgo]

Holy fuck this makes it real easy

$\mathrm{SL}(2,\mathbb{Z}) \cong \langle x, y | x^y y^x = (xy^{-1})^6 = e \rangle$

[cloud walrus]

Dyfunction Executive

[dull ginkgo]

x^y = x is the relation for the abelianization, therefore xy = e, so xy^-1 = x^2, so x^12 = e

therefore H_1(SL(2,Z),Z) = Z/12Z

[chilly radish]

This is very non obvious, it can be a good guided exercises but there's several steps you're unlikely to come up with on your own

[dull ginkgo]

non-obvious

Let me guess, Sylow/Conjugate shit

[chilly radish]

It involves it yes

[cursive edge]

for this proof that the identity in G' is phi(e), we show that phi(a)*phi(e) = phi(a). But since phi isnt necessarily surjective this doesnt show that g'*phi(e) = g' for all g' in G' (it just shows it for all g' in the image phi(G)). So how is the proof complete?

[dull marsh]

If you have a group and elements a, x in it such that ax = a, it is sufficient to deduce that x is the identity

Just multiply by the inverse on the left

[cursive edge]

ah okay so it doesnt really matter that phi(a) cant be any element in G'

[dull marsh]

Yeah

[fluid compass]

I am working on a homework problem where I need to show for a ring homomorphism f: R --> S that the preimage of f (denoted A) is A = I + ker(f) for some ideal I of R. The way I see it, I can let x be in A, suppose x is not in I, and show it is in ker(f) (and vice versa), but I am struggling to come up with anything useful properties by excluding a from I.

[fickle brook]

im having a bit of trouble proving W is unique for this thm

ive only established thus far that they have the same dimension

er

namely that if there exist two different FG-submodules W and W' of V such that V = U oplus W = U oplus W' then their dimensions are the same

but idk how to proceed from here

[cobalt heath]

The theorem statement does not seem to contain uniqueness

But uniqueness under isomorphism might be proved using exact sequences, I think

[fickle brook]

yeah my prof told me to prove uniqueness tho

[cobalt heath]

I see

[fickle brook]

and... i have to admit i have no idea how to use exact sequences here 🥲

like i've proved existence alright.

[cobalt heath]

You can use short five lemma

[fickle brook]

what is that

[cobalt heath]

Ah, sorry, maybe you cannot use that one then.

[fickle brook]

the textbook is James & Liebeck

i have a .djvu copy on hand if you want one

[cobalt heath]

Do you know of a way to express W with inclusion U -> V?

(Also since you have V = U + W, you can see projection V -> U instead - W also plays some role here)

[rocky cloak]

Say V = U + W = U + W'

We can think of the identity on V as an isomorphism
U + W -> U + W'

Write this map as a matrix and note that it restricts to the identity on U. Conclude that the induced map W -> W' is an isomorphism.

[cobalt heath]

Is matrix form valid for modules

[rocky cloak]

Yes, a matrix is just a convenient way to describe a map between direct sums.

But here everything is a vector space, so you can just think of it like linear algebra if you prefer

[cobalt heath]

I mean, being an isomorphism in F-vectorspace is entirely different beast than isomorphism in FG-module.
But it does give a good intuition!

[rustic crown]

a lil more generally, in any category with finite products and coproducts, a matrix is just a map from a coprod ⊔ A_i to a product ∏ B_j. By the universal property you just need to specify maps φ_ij : A_i --> B_j.

[rocky cloak]

Not really. If you have a homomorphism of FG modules, then it's in isomorphism iff it is an isomorphism of vector spaces.

[cobalt heath]

Ah, nice that we have biproduct

[cobalt heath]

How- ah wait

Sometimes even bijective homomorphism isomorphism, I forgor

[cobalt heath]

This reminds me, one specific important case is the "matrix over integers". (Maybe this is unimodular matrices)

[rotund aurora]

If A/nil(A) is a nontrivial product of rings, does it follow that A is also a nontrivial product of rings?

I think I have an example where it fails

mmh like I believe it's false

essentially, I'm asking if from e^2=e+n where n is nilpotent I can find an e' such that (e')^2=e'

I'm thinking about Z[x,y]/(x^2, y^2-y-x)

[rustic crown]

geometrically this is obvious. A nonzero ring A is a product iff Spec A is disconnected.

And Spec A and (Spec A)_red have the same underlying space

[rotund aurora]

I'm pretty sure this is circular reasoning, but I'd like to be proven wrong

[rustic crown]

Algebraically I think you can do some newton's method type argument to get a root of x² = x, starting from a root mod nil(A)

[rotund aurora]

ohh nice

[rustic crown]

Nah it isn't circular. You just need to show that the structure sheaf is a sheaf.

If Spec A is disjoint union of two opens U and V, we can define an element in A that restricts to 1 on U and 0 on V

That's gives an idempotent

[rocky cloak]

Is A commutative?

[rotund aurora]

wait how you do that, I need to divide in Newton's method

[rustic crown]

(I assumed this )

[rotund aurora]

yeah

[rocky cloak]

Then it's true

Just let e' be an idempotent of A/nil. And let e be a preimage.

Then e(1-e) is in the nilradical hence nilpotent. Say e^t(1-e)^t = 0.

Then expand
1^2t = (e + (1-e))^2t

With some fiddling you get that sum[k=0 to t] 2tCk e^2t-k (1-e)^k is idempotent lying over e'

Then if the ring is commutative all idempotents are central, so correspond to factors in a product.

[rotund aurora]

why can you do that? If you assume Spec is disconnected, then you just have two ideals R and S such that R+S=(1) and RS subseteq nil(A), how do you proceed?

[rustic crown]

Doing it with closed subsets is harder.

[rustic crown]

Doing with open is immediate if you know the structure sheaf is a sheaf

[rotund aurora]

whatever you can say about opens you can just take the complement and it works for closeds lol

[rotund aurora]

I'm pretty sure your argument doesn't work. The nilpotent elements are invisible to the structure sheaf

or at least I'm not getting it

[rotund aurora]

This just tells you that you have an x such that x=1 mod p for all p in U and x=0 mod q for all q in V. Now, x^2=x mod p for all p in U and x^2=x mod q for all q in V, i.e., x^2=x mod the nilpotents

[rocky cloak]

The nilpotent elements are seen by the structure sheaf, but it's not clear to me whether one would have to prove something analogous when proving that the structure sheaf is a sheaf in the first place

[rotund aurora]

How are they seen? they are zero everywhere

I suppose they do make a difference in some situations, but I'm not seeing how det's argument works even assuming the structure sheaf is a sheaf

[rocky cloak]

Theyre not 0, the rings in the sheaf are localizations, not quotients.

[rotund aurora]

they are 0 mod p for all p, as functions they evaluate to zero everywhere

[next obsidian]

Det’s argument is simple

[rocky cloak]

But the sheaf contains the information of the global sections, so has all information about the ring

[next obsidian]

You can glue the function 0 on V and 1 on U to a global element which becomes an element of the ring, call it x

To check x = x^2 you do this at stalks

For a point in U the germ is 1^2 = 1

On V it’s 0^2 = 0

So x = x^2 but x ≠ 0 or 1 because those elements are 0 on U and V, or 1 on U and V

[rotund aurora]

ok, let n be nilpotent and consider x+n. Your argument also works for x+n, no?

[next obsidian]

No

The stalks are localizations

[rotund aurora]

ah

[rocky cloak]

But this does sort of just push the question to proving that the structure sheaf is a sheaf at all.

[next obsidian]

Yeah

[rocky cloak]

But I guess with the appropriate machinery it's enough to define a sheaf on a base, then just check the conditions on D(f)

A lot of machinary for something that can be done with the binomial theorem.

[delicate orchid]

glue
not in my channel you don't #point-set-topology #algebraic-geometry #diff-geo-diff-top

[rocky cloak]

Also, this proof also works for noncommutative rings, so is even better.

[summer path]

is this the first time someone has pinged to some other advanced channel from grf

[stark helm]

I want to prove that Q(sqrt(2)) is not isomorphic to Q(sqrt(3)), do you think this is a correct solution?

[rocky cloak]

You seem to imply that the mapping
a + bsqrt2 |-> a' + b'sqrt3
maps a and b completely independently of one another. While the value of a' could depend on both a and b.

I think it would be easier to prove that one has some invariant property the other doesn't have, like that only one of them has a square root of 2 for example.

[rotund aurora]

Say $x+y=1$, $(xy)^2=0$. Is your claim that $\sum_{k=0}^2 \binom{4}{k}x^{4-k}y^k=x^2(x^2+4xy+6y^2)$ is idempotent?

[cloud walrus]

croqueta3385

[stark helm]

You mean like suppose phi(sqrt(2))=a+bsqrt(3)? and we first need to prove isomorphism of phi right? I am doubting that if we have 1+2sqrt(3) and 3+4sqrt(3), will its domian always remain sqrt(2)?