groups-rings-fields

[young solstice]

so the thing is actually irrelevant

as in the exponenet

[astral ivy]

wait so we transformed 1996! into 1001!994! * 995?

then that's just 1001!994! = 995^-1

[young solstice]

ya and 1 more -ve sign

[chilly radish]

I misread the context

[dire wren]

i meant like you could consider equivalnece classes of elements such that right unit * a = right unit * b

then these equivalence classes form a associative structure with a unit

and that the if you compose two functions together, let us say f and g, then the answer depends only on the equivalence class of g, and f

so basically the only additional information is that given a element in an equivalence class, and a equivalence class, you have to choose a element in the equivalence class you get by composing these two

[swift root]

sure, but this has nothing to do with functions

you might as well do this with strings and identifying them

[dire wren]

it comes more naturally when considering it as functions, as you get this as a weird weaker form of injectivity

and injecivity is one of the first things you care about when looking at functions

[dire wren]

and anyway you do this, (where the second equivalence class is not the identity equivalence class, because then you already know the answer) gives you a associative structure with a right unit

[copper kestrel]

how does one get more comfortable with direct sums bc i feel like whenever i see them for groups or rings my brain shuts off

or like does anyone have any books that explain them nicely

i understand theyre just adding and their intersection is 0 but theyre still a bit confusing for me

[somber goblet]

for groups and rings they’re pretty much just direct product

[azure cairn]

did you do linalg

[copper kestrel]

i did yeah

[restive quail]

I think the nuance comes about for groups and rings when you look at infinite index families of stuff

Like i think for a direct sum its saying you can express any element uniquely as a finite combination of indices

Whereas a direct product might not necessarily have that condition so you could have elements you couldnt express as a finite combination (like an infinite sequence)

But im not too sure

[cursive spindle]

Are you confusing direct sums with direct products or just have an issue with something related to direct sums ?

[copper kestrel]

i was just looking at Z \oplus Z and seeing if theres an isomorphism between Z \oplus Z and Z[x]/(x^2 - 1) and i realized i didnt really know what Z \oplus Z is

[cursive spindle]

oh so you're trying to use CRT?

anyways direct sums in this context is pretty much what you might think it is

just your usual product

(a,b)

Note that if you try to use CRT in this case you have some problems. Do you see why?

[copper kestrel]

i wasnt trying anything i was just confused on what it was

[tulip otter]

another way to think of direct sums is as the subset of the direct product whose elements have finitely many non-zero entries

[cursive spindle]

Ah I see yeah I saw Z x Z and that ring and I thought you were trying to use CRT

[tulip otter]

you probably wanted to get number theory into the chat in one way or another while giving your answer

[cursive spindle]

sshhh

[copper kestrel]

hmmm

[tulip otter]

[cursive spindle]

no actually i wasnt

but I could

[tulip otter]

that would be the first time where something like this happens (and probably the last)

[vapid vale]

Just think of it as two copies of Z next to each other that don’t care about each other

An element is (a,b) with a, b in Z

[lunar wigeon]

who has you documents for probability ? you can send me

[somber goblet]

im trying to understand the proof that the ring of integers of a field extension is indeed a ring

i get that roots of integer monic polynomials correspond to finitely-generated Z-algebras

but idk how this gives us addition and multiplication

gemini is telling me Z[a, b] has the Z-algebras for (a + b) and ab as sub-algebras

but i want to make sure it's not making up some bullshit

this makes sense to me if it's correct

i really like algebraic nt

[flint timber]

Z[a,b] is finitely gen module so thats true

[somber goblet]

the proof it gave me sounds valid but im new to this so im not super certain

[flint timber]

it is standard theorem so dont doubt he will give wrong ans for this, just use finite generator arg to proove it

[chilly radish]

Ideally you should use a book written by humans

[flint timber]

Damn

[chilly radish]

That way you don't have to worry about it feeding you bs

[fading acorn]

coproducts in the category of (insert a structure)

[balmy sparrow]

what does k[x]_s mean? specifically, what is the s doing?

[quiet pelican]

It’s the ring k[x, y]/(sy - 1)
Ie, k[x] with an adjoined inverse for s

[errant pagoda]

I am just looking for help with this question from Artin Algebra. I really just don't understand the mapping that the phi is doing

[chilly radish]

Phi is understood here to be the identity on Z, so the only thing you need to define is where x gets mapped

x is allowed to be mapped to any element

[errant pagoda]

so right now we have x->1+sqrt(2) correct? but what is phi of a non unit constant?

[chilly radish]

It's just that constant

[errant pagoda]

like the homomorphism maps units to units correct? so 0-> 0 and 1->1

oh

[chilly radish]

phi is the identity/embedding on Z

[chilly radish]

0 is not a unit, unless by unit you mean identity

[errant pagoda]

ok that makes sense and phi(x^2)=phi(x)phi(x)=(1+sqrt(2)(1+sqrt(2))?

[errant pagoda]

yes sorry thats what i meant

[chilly radish]

For a general polynomial $$\sum_{i=1}^n a_i x^i$$

You have

$$\varphi\left(\sum_{i=1}^n a_i x^i\right)= \sum_{i=1}^n a_i \varphi(x)^i$$
i.e. evaluation at $\varphi(x)$

[cloud walrus]

ShiN

[errant pagoda]

thank u that makes sense

[coral steeple]

Is the indeterminate $x$ contained in $\mathbb C(x+1/x)$, the field obtained by adjoining the rational function $x+1/x$ to the complex numbers? I think it isn't, but I don't know how to prove this. I've found lots of elements in $\mathbb C(x+1/x)$ in the hopes of finding some property they share, but which $x$ doesn't have. This hasn't worked; I just have a bunch of rational functions with nothing I can find in common between them.

[cloud walrus]

person2709505

[velvet hull]

to give you the very loose argument, x + 1/x is invariant under the map that sends x to 1/x

while x is not

[marble hinge]

Say we have two groups G, H defined by the same number of generators and some relations. Then, if we map generators of G to generators of H and can derive relations in group H from relations in G, then we automatically have a surj homomorphism from G to H, right?

For example D_16 and D_8

[rocky cloak]

Yes, if you have a presentation of G then a homomorphism is exactly given by mapping the generators somewhere where the relations are satisfied.

If you map onto a generating set, then you map onto the whole group

[marble hinge]

Cool, I’m just trying to prove a relatively obvious thing that r -> r, s -> s is a homomorphism from D16 to D8 with the kernel isomorphic to C2

And trying to find the least verbose method to do so )

Other than saying “It’s obvious” :D

This would allow me to finish the proof that Aut(D8) =~ D8

[marble hinge]

Huh? How r^8 is not 1 in D8??

D8 is a dyhedral group of the square

I should have clarified the convention first, of course

But in both conventions r^8 = 1 in D8, so you probably made some typo in your message

[south patrol]

Oh yes I was eepy

Sorry lmao

[copper kestrel]

hello chat i am back for more (not my cat)

[errant pagoda]

Could anyone advice me on how to start the other direction. such that I can show that ideal is equivalent to the kernel

[quiet pelican]

Use the first isomorphism theorem

[errant pagoda]

sorry not completely sure how to apply this

I was trying to use the fact its a ufd and then everything has a unique factorisation?

[karmic dove]

[wraith cargo]

Prove that your polynomial is the minimal polynomial for 1+sqrt(2) so any polynomial that has it as a root has to be divisible by x^2 - 2x - 1

Specifically you'd wanna use the division algorithm for polynomials

[flint timber]

your own cat?

[copper kestrel]

no lol

[flint timber]

i have 3 cats all of them skinny, want a chubby cat, how to make a cat chubby

[somber goblet]

i am doing that as well lol

[azure cairn]

[somber goblet]

i just looked online since the proof in the book wasn’t making much sense to me

and i wanted to see if there was a proof that doesnt depend on choice of basis

[azure cairn]

Yeah I do that too

When a proof is hard to understand I sometimes just give it to an llm to explain in a less terse manner

Or to give better intuition behind it

[errant pagoda]

does anyone have any recommendations to tidy up this proof or any ways that something could be worded better. Just trying to improve my proof writing

[flint timber]

Dont write every word notations are enough at some places

[errant pagoda]

as in use the \exists instead of there exists and what not?

[flint timber]

yeah something like that

[errant pagoda]

oh really my professor left a comment on my previous work saying specifically not to

I guess its just his personal preference

thank you for the advice

[karmic moat]

im gonna disagree and say that in written solutions you should avoid using those kinds of symbols

[errant pagoda]

yeah that is what my professor said

[karmic moat]

in scratch work or whatever it's fine but when you submit solutions as homework, papers, whatever, try to avoid using that kind of notation

[glad osprey]

Trust your professor. What you wrote seems perfectly readable to me

[errant pagoda]

ok thats good to know. is there anything else in the proof you would recommend editing

[karmic moat]

i didn't check the work but i would suggest adding some more punctuation, you have some run-on sentences imo

not a huge issue but it just helps with the flow a bit

[errant pagoda]

thank you very much I was planning to anyway given he is the one assessing me after all and I am pretty fortunate to be studying somewhere where alot of the professors and lecturers have insane qualifications and education

[errant pagoda]

noted

[glad osprey]

You wrote x^2 - 2x - 1 = 0 is irreducible, but I think you meant just the left hand side

[karmic moat]

anyway you'll get better at writing just by doing it more and reading more proofs in textbooks and papers and whatnot

[errant pagoda]

as I dont need the 0 i wasnt sure but that makes sense

[glad osprey]

Yeah, it just doesn't make sense to say that an equation is irreducible

[swift root]

i concur, words are better than symbols

[flint timber]

i think professor should be treated in his own language

[swift root]

?

[errant pagoda]

it seems to be a trend that everyone agrees words a superior in this context

[flint timber]

for me lot of word make me loose petience

[swift root]

i dont know what you mean by "professor should be treated in his own language"

[swift root]

short attention span ahh

[errant pagoda]

Yeah but thats readibility for you specifically not a universally agreed upon proof

[karmic moat]

if i read something and it just used a ton of symbols in place of plain english i would lose patience

[swift root]

if you overuse symbols then your text becomes cluttered and non skimmable

[errant pagoda]

thats like saying because it makes sense to count in base 7 to you everyone should even though base 10 is the convention

[karmic moat]

i use symbols like \exists very rarely, like only when giving a board talk and commutative diagrams and stuff

[quiet pelican]

It’s a preference that basically every mathematician has
Unless you’re doing formal logic, don’t use those symbols
(Slightly hyperbolic, but they should rarely be used if you could use the words they represent instead)

[coral steeple]

I can see how to do it from there. Thanks

[glass sinew]

\exists is for the blackboard only

if you're typing anything out you have like no excuse to be obtuse lol

[ripe harbor]

I want to prove that the commutator $[S_n, S_n] = A_n$. So, we know that $$|S_n/A_n| = 2 \quad \implies \quad \text{abelian} \quad \implies \quad [S_n, S_n] \leq A_n.$$ How do we prove the other direction though?

[cloud walrus]

ILikeMathematics

[velvet hull]

are you allowed to use the fact that A_n is simple for n >= 5?

[ripe harbor]

Yes

[velvet hull]

then you are done for n >= 5

for the remaining few cases you can check them by hand

[ripe harbor]

Oh, thats good, right

[ripe harbor]

You mean actually write out [S_n, S_n]? That sounds like a a pain

[hollow finch]

Hello. I have been proving that S_4 = <(1 2 3 4), (1 2 4 3)> as exercise and I've come to conclusion that disprove it. So i think i might have mistake in my reasoning but i didn't found out. So here is my reasoning -> aa^(-1) = id, where a = (1 4)(1 3); aa^(-1) = e & order(1 2) = 2 => a(1 2)^2 * a^(-1)* (1 2)^2 = (a*(1 2) ) * (1 2) * (a^(-1) * (1 2) ) * (1 2) = e; I know that order( a*(1 2) ) = order( a^(-1) * (1 2) ) = 4, so i can compose this expression with inverse of a * (1 2) which equal (a * (1 2) )^3, so i get (1 2) * (a^(-1) * (1 2) ) * (1 2) = (a * (1 2))^3 , then (1 2) * a^(-1) = (a * (1 2))^3, and then a^(-1) * (1 2) = (1 2) * (a * (1 2))^3 * (1 2) = ( (1 2) * a * (1 2))^3. So it means that any element of generaor have form (a*(1 2))^n. But the order of a*(1 2) is 4 and order of S_4 is 24.

[velvet hull]

alternatively you can also show that [S_n,S_n] is a subset of A_n because it is generated by commutators, which are even permutations by construction

[ripe harbor]

Why is ghg^(-1)h^(-1) even?

[velvet hull]

apply the sign function to it

[ripe harbor]

Oh, sign is probably a homomorphism, right?

Then we can interchange and get it to sign(g)sign(g^(-1))sign(h)sign(h^(-1)) and then pull together again

[velvet hull]

there are a few equivalent ways to define what A_n is, one of them is that it is the kernel of the sign homomorphism

[ripe harbor]

Yeah thats 1

[ripe harbor]

Thats even better than this

Well much better lol

[velvet hull]

it's the same thing, you're just showing one sided inclusion

[ripe harbor]

You dont have to check cases by hand

[velvet hull]

no, both proofs only show that [S_n,S_n] is inside A_n

for the other containment you still need to work for it

[ripe harbor]

.

[velvet hull]

yes, the quotient being abelian implies that the commutator is contained inside A_n

not the other way around

[ripe harbor]

I can use that [S_n, S_n] is the smallest normal subgroup N so that S_n/N is abelian and that every other one for which S_n/H is abelian, we must have N c H

[ripe harbor]

Yes, and [S_n, S_n] in A_n is shown by what you said

So we are done

[velvet hull]

N is the commutator, H is A_n

so you have shown that [S_n] is in A_n twice

[ripe harbor]

crap

So, we need $A_n \leq [S_n, S_n]$

[cloud walrus]

ILikeMathematics

[velvet hull]

so this is still not too hard actually

A_n is generated by three cycles for n>= 3 (why?), so it is sufficient to show that the three cycles is in the commutator

[ripe harbor]

A_n is generated by 3-cycles, yes

Say N is normal in A_n and N contains a 3-cycle

Why does it follow that N = A_n?

[velvet hull]

not a 3-cycle

all the 3 cycles

[ripe harbor]

Oh

[velvet hull]

also the fact that [S_n] is normal in A_n is a nontrivial fact, so maybe don't use that

[ripe harbor]

Actually apparently it works with only one 3-cycle, just found a proof in my book

[velvet hull]

it works with one 3 cycle, if your subgroup is normal in S_n

because the conjugacy classes in S_n are exactly the cycle types

[velvet hull]

but if it is only normal in A_n then that is not guaranteed, because the conjugacy classes are no longer the cycle types

[ripe harbor]

This is for n >= 3 and N normal in A_n with N containing (a, b, c)

[velvet hull]

just prove the thing lol

[ripe harbor]

Oh, right haha

Ok so we want to prove all the 3-cycles are in [S_n, S_n], right?

That also seems non-trivial lol

[velvet hull]

it's not hard

[young solstice]

Hello

[hollow finch]

Why not ? S_4 is group, so it has identity element and associative op. So we can claim that a • e • a^-1 • e = e, then we can substitute (1 2)^2 and the as it associative we can change brackets like this (a • (1 2)) • (1 2) • (a^-1 • (1 2)) • (1 2) = e.

[velvet hull]

Oh wait, I misread

[young solstice]

I wanted to ask if someone can help me understand what a discrete group is. I understand the definition given in the book I am reading but I am not quite able to internalize what it means.

[velvet hull]

it is a group whose induced subspace topology, in the space of isometries, is discrete

[velvet hull]

what exactly is your proof trying to do?

what is a doing

So you have shown that a satisfies the relationship (12)a^{-1} = [a(12)]^2. So how do you conclude that any element of the generator has to have that form?

[limpid ferry]

What is a good approach for questions like: Is I=<5,x^2+4> a prime ideal in Z[x]?

[velvet hull]

so you want to check that the quotient is an integral domain. the third isomorphism theorem can help you out here

[warm badge]

you can also see if there are any obvious contradictions as u play with those generating elements a little bit. x²+4 - 5 = x²-1 = (x+1)(x-1). ask yourself whether it makes sense that any one of those factors already is in I.

[limpid ferry]

I tried this too, but I had a hard time figuring out how to show that both of two factors are in I or not, writing out polynomials multiplied together looks kinda messy

[warm badge]

is it?
if possible, let x + 1 = 5f(x) + (x²+4)g(x)

since lhs is a poly of degree 1, can g(x) be anything other than 0? otherwise obviously degrees of lhs and rhs don't match up.
but if g(x) = 0, then what about f(x)? can you finish from there?

[limpid ferry]

Yes, I will try this again. I was not confident that g(x) has to be 0, I was thinking about the product (x^2+4)g(x) can cancel some terms from 5f(x) and 5f(x) + (x^2+4)g(x) arrives at ax+b somehow

[hollow finch]

The proof trying to show that (1 2 4 3) = (1 2 3 4)^some_degree, where (1 2 3 4) = (1 4)•(1 3)•(1 2) = a • (1 2) and (1 2 4 3) = (1 3)•(1 4)•(1 2) = a^-1 • (1 2). The proof has shown that a^-1 • (1 2) = [(1 2) • a]^3 = [a • (1 2)]^-3. So we can substitute (1 2 3 4)^some_degree instead of (1 2 4 3). So any element of generator has form (1 2 3 4)^some_degree and the order of (1 2 3 4) is 4

[velvet hull]

okay, your original proof is very hard to read and will take me some time to parse through again, but the result you prove is straight up not true

you can compute the powers of (1234) by hand, and none of them are equal to (1243)

it's (1)(2)(3)(4), (1234), (13)(24) and (1432)

[onyx yoke]

2+2=?

I know it's a pretty complicated operation but could someone help me?

[wicked patio]

depends on the group

[onyx yoke]

what are you talking about...

[desert verge]

is this mod-pingable

[onyx yoke]

idk

[azure iron]

@onyx yoke the math topic channels are not for shitposting/trolling, take it to #chill

[wicked patio]

that is a sus profile banner

[onyx yoke]

oka

[azure cairn]

😌

[hollow finch]

Yeah ok. I clearly see that my reasoning definitely has some mistake . I’m kinda stuck on this exercise (like for 2 weeks). I don’t know maybe I should delay it

[velvet hull]

it's a difficult problem if you don't see the trick to it

[ripe harbor]

My book states that for $S \subseteq R$ multiplicatively closed, the map $$i_s: R \to S^{-1}R \quad r \mapsto \frac r1$$ is a ring homomorphism with $i_S(S) \subseteq (S^{-1}R)^\times$. But $S^{-1}R$ is just the set of the equivalence classes $r/s$ (the localization) with the equivalence relation you know, so isnt there a typo in this; shouldnt it be $0 \notin S$?

Otherwise S^(-1)R = 0 and we cant have i_S(S) subseteq ...

[wraith cargo]

What's confusing you exactly?

Where does 0 ∈ S?

[ripe harbor]

Its not ruled out, and I thought it should be. Assume 0 in S. Then S^(-1)R = 0 and so (S^(-1)R)^x = emptyset. But i_S(S) != emptyset is possible, so the statement is wrong.

[wraith cargo]

Idk what you're talking about lol

Isn't 0 not in S by definition of a multiplicatively closed subset?

Oh well ig it isn't lol

Yes you need to assume that S doesn't contain any zero divisors you're right

[ripe harbor]

Thanks

[south patrol]

I would take the convention that 0^x is 0. This is standard in that it is needed for (-)^x to give an assignment from commutative rings to abelian groups

[solemn wadi]

I need help with proving that if R is a integral domain and P is a prime ideal,define D=R\P. To show D^{-1}R is a local ring

D is multiplicatively closed

[karmic dove]

What have you tried so far?

[solemn wadi]

Aahhh it's done

I couldnt think of making ideal of the units

[rocky cloak]

It is not the empty set. The unit group always contains 1, even if 1=0

[copper kestrel]

im still trying to find out if Z[x]/<x^{2} - 1> is isomorphic to Z (+) Z, i talked about it with a friend and they said that theyre not isomorphic sincec <x> is of order 3 but not in Z (+) Z, but that doesnt really make sense to me, im trying to figure it out a different way but i just dont get it

[young solstice]

wait Z[x]/<x^2 - 1> = {ax+b| a,b \in Z} \isomorphic {(a,b) | a,b \in Z} = Z^2 so it works right?

[copper kestrel]

i have no idea

[young solstice]

i am pretty sure it dose

[desert verge]

as rings?

[copper kestrel]

yes

[desert verge]

yeah idk

the way i think of approaching is, suppose there's an iso phi: Z[x]/<x^2-1> -> Z^2
phi(1) = (1,1) and say phi(x) = (a,b), etc

[desert verge]

this is true as sets at least

[copper kestrel]

thats what i was thinking, i know x is of order 2

so theoretically (a,b) should be of order 2

[desert verge]

is "order" the right word here

[copper kestrel]

maybe? i just know x^2 = 1

[desert verge]

yeah

[copper kestrel]

so that neuron activates in my head as order

[desert verge]

so (a,b)^2 = (a^2, b^2) = (1,1) i think

[copper kestrel]

which is only possible if a = 1 and b = -1 or a = -1 or b = 1 i think

bc (1,1) and (-1,-1) are already taken

[desert verge]

yep

[copper kestrel]

then either find something wrong or something good

[desert verge]

suppose wlog (a,b) = (1,-1)
now i think img(phi) = Z-span of (1,1) and (1,-1)

= {(p+q, p-q): p,q in Z}

ok let me stop there

[vapid vale]

Do you know Chinese remainder theorem

[desert verge]

kinda

if ur asking me

[copper kestrel]

if ur asking me i barely know it

my professors used it in previous proofs but i dont rly get it

[vapid vale]

So the ideal (x^2-1) can be written as the intersection of two ideals

[copper kestrel]

hm

[topaz solar]

Might be good to write out the statement and figure out what the ideals involved are

[copper kestrel]

what statement do you mean

[topaz solar]

Chinese remainder theorem

[vapid vale]

This might be an informative thing to work out logically, if it’s not clear how to express that ideal as the intersection of two ideals

[copper kestrel]

im gonna be totally honest i dont know what it is, its never defined in our book besides an exercise

[topaz solar]

This is why I am saying to do that

[ripe harbor]

Ah, youre right

[vapid vale]

Okay yoi would need a version a bit more general than that lol

[copper kestrel]

maybe this might be better?

[desert verge]

um wouldn't this just work

[copper kestrel]

but again we havent done this exercise

nor any exercises related to the chinese remainder theorem

by professor just pulled it out of nowhere

[topaz solar]

Yeah

[desert verge]

sharp 🙏

[copper kestrel]

are you trying to lead me to solve the problem or find a contradiction

general question

[desert verge]

i was trying to solve it myself tbh

[copper kestrel]

real

[desert verge]

/giving ideas

[vapid vale]

Ignore the CRT thing I’m sleep deprived and thought 1+1=1 but it doesn’t work

[topaz solar]

Try following that line of reasoning

[glad osprey]

The reason CRT doesn't work here is that (x + 1) and (x - 1) are not coprime, right?

[vapid vale]

Yeah

Another way you can do this is by showing Z[x]/(x^2-1) has no idempotents that are not 0 and 1

Whereas Z (+) Z has more

[next obsidian]

Fuck a Chmonk, I’m here for the funk

CRT is swag cuz u can just say injective is surjective isn’t that swaggy

[ripe harbor]

Let $f = c \ov g_1 \cdots \ov g_r$ with $f \in R[X]$ and $\ov g_i$ and thus the product primitive with $\ov g_i \in K[X] \supseteq R[X]$. My book says it right away follows that $c \in R$, but why?

[cloud walrus]

ILikeMathematics

[next obsidian]

wtf is g_i-bar

It seems like you’re missing a word

And what’s the assumptions on R

UFD?

[ripe harbor]

R[X] is a UFD, so we can write f as product with p_i irreducible, then you can write p_i = c_i p_i bar with p_i bar primitive

Thats how bar appears there

[next obsidian]

Your statement doesn’t say g_i-bar is primitive

[ripe harbor]

sorry yes, gi bar is primitive

[next obsidian]

I don’t understand what it means for it to be primitive but to be in K[x]

If it’s primitive then there shouldn’t be any denominators right?

Okay a lot of things don’t make sense to me

I don’t think you’ve written down all the relevant context

You should post the text directly

Like there’s nothing to stop c from being x based on what you’ve written

[ripe harbor]

Ok I made a mistake here, I meant to say K[X] is a UFD, but we dont know anything other than R UFD about R[X]. We dont know if R[X] is UFD yet, thats what we want to prove.

The idea is starting with f in R[X], seeing it as f in K[X] and since thats UFD, we can write it as product

f = eg_1 ... g_r with g_i irreducible in K[X] and e in K^x. Now write each gi as c_i bar(gi) with bar(gi) in R[X] primitive

[next obsidian]

What?

Is K the fraction field of R?

[ripe harbor]

Yep

K = Quot(R)

[next obsidian]

I see okay

[ripe harbor]

So we have f in R[X] and bar(g_1)bar(g_2)...bar(g_r) in R[X], and f = c * bar(g_1) ... bar(g_r), with c in K^x. Now apparently this allows us to conclude c in R

[next obsidian]

Uh you know gauss’s lemma?

[ripe harbor]

Yes, thats what allows us to say the product of the bar(gi) is still primitive

[next obsidian]

What is the statement you know

Word for word

Cuz the fact that R[x] is also a UFD is also kinda Gauss’s lemma

But that would obviously be circular lol

[ripe harbor]

Let R be a UFD, and let P,Q ∈ R[X] be primitive. Then PQ ∈ R[X] is primitive.

This is called Gauß lemma for me

[ripe harbor]

Yeah thats Gauß theorem for me lol

[next obsidian]

Okay

So this is really stupid but

Gimme a second

Write c = prod a_i/prod b_i

a_i and b_i are irreducible, no redundancies

Examine the constant of f

It looks like Prod a_i • Prod (g_i-bar)_0/Prod b_i

Where this horrendous _0 notation means it’s the 0th degree thing

then there’s some g_i-bar where b_1 divides (g_i-bar)_0, WLOG i = 1

Now look at the degree 1 term of f

Okay I can’t really write this down in an effective manner anymore but

It’s c•(Sum of coefficients from g_i-bar)•x

All but one of these have to contain (g_1-bar)_0 because its degree 1, which is divisible by b_1

Again because c contains b_1 in a denominator this means (Sum of coefficients from g_i-bar) must be divisible by b_1

Guh I can’t force it to always be g_1-bar

There must be a simpler answer lol

But it’s gonna boil down to the fact that for a UFD, to have something in R and not K, it can’t have denominators

Like it works exactly how you’d want it to

Anything that appears below must appear above with as much multiplicity

Okay sorry maybe this is too high brow

I think it’s easier lol

Okay yeah I’m being dumb

f is itself primitive right?

[ripe harbor]

Yes

[next obsidian]

Otherwise you can factor something out and it isn’t irreducible?

I guess actually that doesn’t even matter

Write g = Prod g_i-bar

And then f = c•g = cg_nx^n + … + cg_0

By the same argument as before, if c isn’t in R then some irreducible b divides g_n, g_n-1,…, g_0

So then g isn’t primitive, but that contradicts Gauss’s lemma

[ripe harbor]

Oh, thats nice

Thanks!

[topaz solar]

Are you sure

[rocky cloak]

They're isomorphic as abelian groups, but not as rings

[topaz solar]

Maybe I’m just misthinking about something that seems like one though

Which is likely

[rocky cloak]

So for example Q[x]/(x^2 - 1) is isomorphic to Q^2. But the same argument doesn't work for Z because 2 is not invertible

[topaz solar]

Yeah

[rocky cloak]

And by reducing mod 2 you can see they are not isomorphic

[topaz solar]

I was probably sliding one of those /2 in without noticing 🥀

[ripe harbor]

To prove that something is not UFD, why is it enough to find two different prime factorizations of an element and leave it at that? UFD uniqueness only says that $$a = ep_1 \cdots p_m = f q_1 \dots q_n$$ implies $m = n$ and there exist $e_0, \dots, e_m \in R^\times$ with $e_0e = f$ and a permutation $\sigma$ with $e_ip_i = q_{\sigma(i)}$. So if we for example say $$6 = 2 \cdot 3 = (1 + i \sqrt 5)(1 - i \sqrt 5)$$ for $\mathbb Z[\sqrt{-5}]$, why does that prove anything without arguing why we cant have such elements?

[cloud walrus]

ILikeMathematics

[quiet pelican]

You need to argue that the elements you’ve got aren’t associates yeah

The way you typically do it here is by norm

N(2) = 4
N(3) = 9
N(1 +- isqrt(5)) = 6

[ripe harbor]

Is that the proper term for there existing invertible ei with eipi = qj?

[quiet pelican]

And every unit has norm 1 (and it’s multiplicative)

[quiet pelican]

Yeah
p_i and q_i are associates if they are unit multiples of each other

[rapid cave]

Associates mean that x = uy for some unit u

[quiet pelican]

(This might get slightly messier in the non-commutative setting idk what the correct defn is there)

[ripe harbor]

In the example I gave above, is there a way to see directly that they arent associates? After giving the definition, my book right away gives this example and concludes after giving the two factorizations that Z[sqrt(-5)] is not UFD

[topaz solar]

Try proving N(a) = N(ua) for a unit u and norm N

[quiet pelican]

Yeah I mean you can bash through some algebra to see it
(2 times anything will have every coef divisible by 2)
But norm is the more “natural” way to do it

[coral steeple]

A bit confused by the wording of this problem. Part iii is not so hard: you just need that the lower extension is normal. But this makes part ii kind of empty, unless we want to find some other description for the conjugate fields. This shouldn't be so easy to do, since I don't know how to think about $\operatorname{Gal}(\overline{\mathbb Q}/\mathbb Q)$. What might this question be asking?

[cloud walrus]

person2709505

[vocal pebble]

question is "empty" only if you can somehow show that Q(sqrt(2) + sqrt(5))/Q is normal without doing something like part i and ii. Otherwise, you can just follow the outline given to reach the conclusion in iii.

[coral steeple]

Well I think the lower extension is the splitting field of the minimal polynomial of sqrt2+sqrt5, hence normal

[vocal pebble]

i think so too, but theyre asking you to prove it, and as far as i can see in order to prove it you will have to basically do part ii in the process

[coral steeple]

Do we have the same definition of conjugate field? In this case, I'm looking at the set of all $\sigma(\mathbb Q(\sqrt2+\sqrt5))$, where $\sigma$ is in the Galois group of the larger extension (algebraic numbers extending the rationals)

[cloud walrus]

person2709505

[vocal pebble]

yes, and that set should just be Q(sqrt(2) + sqrt(5))

[coral steeple]

Right, but to show that I need to find every sigma

[vocal pebble]

which amounts to showing part ii. i dont know how else you will show that

[coral steeple]

Well I'm asking how it is possible to do part ii

[vocal pebble]

sigma preserves the minimal polynomial (so sqrt(2) + sqrt(5) gets mapped to some other root of the minimal polynomial)

so the conjugate fields will just be Q(a) where a is a root of the minimal polynomial

[coral steeple]

Okay, how do we know that every such Q(a) will be attained?

[vocal pebble]

because for every a which is a root of the minimal polynomial, sqrt(2) + sqrt(5) -> a is an embedding of Q(sqrt(2) + sqrt(5)) in Q bar fixing Q

[coral steeple]

An automorphism of Q bar?

hm

I suppose you want to show first that sqrt2+sqrt5 -> a is an automorphism of the splitting field of the minimal polynomial of sqrt2+sqrt5 over Q?

And if I understand you, this extends to an automorphism of Q bar

...Don't answer that, I'll try and figure out

[sacred minnow]

Question ; What does * means in group theory ?

[coral steeple]

Context?

[sacred minnow]

(Sorry for bad translation), if you have E (non empty) and a law *. What does that mean.

[coral steeple]

It's a function $E\times E\to E$

[cloud walrus]

person2709505

[sacred minnow]

oh ok

[coral steeple]

For example, if you take E to be the real numbers, you can take * to be (a,b) -> a+b. This is the group of real numbers under addition

[vocal pebble]

i think it will be easier for you to use the definition of normal field that is: a field is normal if it is the splitting field of some polynomial. In this case we want to show that Q(sqrt(2) + sqrt(5)) is the splitting field of the min poly of sqrt(2) + sqrt(5) which is x^4 - 7x^2 + 9. In order to do that, note that the other roots of this polynomial are -(sqrt(2) + sqrt(5)), -sqrt(2) + sqrt(5), sqrt(2) - sqrt(5). So it suffices to show that Q(any of those roots) = Q(sqrt(2) + sqrt(5)). (which is part ii)

[sacred minnow]

ok ok thx

[vocal pebble]

from here on you just need to do some algebra to get your result

[coral steeple]

Yes, that was my original idea

But once I've done that, I think it suffices to just cite this

Here K is Q and M is Q(sqrt2+sqrt5)

(and L is Q bar)

[coral steeple]

Rather, you can show that all the roots of the minpoly lie in Q(sqrt2+sqrt5); they are just obtained by taking inverses and negatives of sqrt2+sqrt5

[vocal pebble]

+-3*inverse and negative of sqrt2 + sqrt 5 yeah

[coral steeple]

Yeah

[coral steeple]

I mean, part ii is asking us to exhibit these other simple extensions as automorphisms of Q bar, right? So there's no way to get around your step where you extend these automorphisms to that larger field

[raw yarrow]

just starting out with group theory

[coral steeple]

Anyhow, I think I can see how to go about either way of doing the problem now. Thanks @vocal pebble

[raw yarrow]

kernels killed me

[coral steeple]

But once you understand kernels and the isomorphism theorem, you understand homomorphisms, so it's good to think about them

[raw yarrow]

tbh rn i have an exam on my head so i cant go too deep into groups but i'm loving it so far

[coral steeple]

Does an automorphism of Q(sqrt2+sqrt5) fixing Q extend to an automorphism of Q bar fixing Q? In general, for a tower of extensions K\subseteq L\subseteq M, does an element of Gal(L/K) extend to some element of Gal(M/K)? I understand that the answer to at least the first question is yes, but it is very unclear to me how you'd define such an extension.

[rocky cloak]

Yes, you would define this using Zorn's lemma / transfinite induction

[coral steeple]

For the second question?

[rocky cloak]

For both, though for the second you would just need normal induction (assuming M/K finite)

[coral steeple]

I don't know about transfinite induction, and unfortunately I think I need it since Q bar/Q(sqrt2+sqrt5) is infinite

What partial order would you use to apply Zorn?

Do you want to look at the largest intermediate field with an extension, then show that it's the whole field?

[rocky cloak]

Consider pairs (M, f) with M containing Q(sq2, sq5) and f an automorphism extending the automorphism in question

Then (M, f) < (N, g) if g extends f

[rocky cloak]

Pretty much, yeah

[coral steeple]

Yes I see it in your hint

[coral steeple]

What if there are no fields which admit an extension? Doesn't Zorn just say that Q(sqrt2,sqrt5) is maximal?

I might try to fix that by considering nontrivial field extensions, but then it is necessary to show that the poset of pairs (M,f) is nonempty

[rocky cloak]

Q(sq2, sq5) admits an extension (trivially), so the set is not empty.

Zorn gives you a maximal element and you just need to show it's the whole alg closure.

[coral steeple]

Although according to your other hint, you can show that a finite extension admits a basis

[coral steeple]

How is this for a start: following your hint, we can use induction to produce an extension in any finite extension of Q(sqrt2+sqrt5). Hence, Zorn's lemma shows that there is a maximal element (M,f), and M/Q(sqrt2+sqrt5) is not finite

And I guess then I'd try to show that Q bar is the smallest transcendental infinite extension of Q (edit: this is false of course)

[rocky cloak]

So Qbar is not a transcendental extension.

If you have an induction proof ready you should just be done. Just take any (M, f) where M is not all of Qbar and use your induction step to make a larger M

[coral steeple]

Excuse me, we know that M/Q(sqrt2+sqrt5) is not finite

[limpid ferry]

ngl, I have been figuring out how g(x) has to be 0

[coral steeple]

Ah, ok yes I see how you're doing this

[delicate orchid]

Theres an x^2g(x) on the right and 0x^2 on the left

[limpid ferry]

How about x+1=5(x^2+...)+(x^2+4)(-5+...)?

[delicate orchid]

What

[karmic moat]

if x + 1 = 5f(x) + (x^2+4)g(x), then when you pass to Z/5Z you get x + 1 = (x^2 + 4)g(x) mod 5, and consider a degree argument in here

[limpid ferry]

you mean something like deg(fg)=degf+degg?

[karmic moat]

Yeah

All I really mean is g(x) has to be 0 mod 5

[wraith cargo]

You know that you can just use Euclid's algorithm to find these

Calculate the gcd of 5 and x^2 + 4 and from this you can get polynomials f and g s.t. gcd = 5f(x) + (x^2+4)g(x) and if the gcd divides x+1 you just multiply by the quotient

[limpid ferry]

My prof just taught this like this morning. Prior to this I only know the division algorithm for F[x]

[karmic moat]

So if g(x) is 0 mod 5 then g(x) = 5h(x) for some h(x) in Z[x]

Then you go back and see that x+1=5f(x) + (x^2+4)g(x) = 5[f(x) + (x^2+4)h(x)]

Now use the fact that f(x) and h(x) are in Z[x]. Emphasis on the Z

[coral steeple]

How can I show that x^8-x^4+1 is irreducible over Q? I know it's the 24th cyclotomic polynomial, but I don't want to use that. By reducing the coefficients mod 3, I've shown that it can only reduce as a product of two integer quartics whose coefficients reduce mod 3 the same as the polynomial x^4+1. This leads to an involved brute force calculation to try and show that this cannot happen, and I'm wondering if there's some other method

[velvet hull]

in what sense are you unwilling to use the fact that it is the 24th cyclotomic polynomial?

[coral steeple]

I guess I was just anticipating someone saying "the 24th cyclotomic polynomial is irreducible"

Unless I mentioned it

[velvet hull]

you can show that Q[x]/(x^8-x^4+1) is a Galois extension, but that does require you to implicitly know that it is the 24th cyclotomic polynomial to be able to construct the galois group

[coral steeple]

Is there a lot of work that goes into these definitions? We haven't covered cyclotomic polynomials in class (but will in a few weeks, so definitely should not be using lots of the theory)

[velvet hull]

the definition itself is not too difficult, but showing that the cyclotomic polynomials are always irreducible is a bit involved

[coral steeple]

Yes, I expect we'll cover that. I'm also trying to use the information I found by reducing mod 3 to rule out a reduction after reducing mod 2

Should I expect that reducing by (a possibly very big) prime can show the irreducibility of some or all polynomials irreducible over Z?

[velvet hull]

yeah, I don't expect that strategy to work

[vocal pebble]

Another way could be to factorise it in C, and then show that any partition won't give you a product of polynomials in Q

[coral steeple]

Oh that's true isn't it

Although I only know one root over C, so I'd still have to factor a septic?

[velvet hull]

but being able to factorize it in C also basically requires you to already know it is cyclotomic

one you know it is the 24th cyclo the factorization is immediate

[vocal pebble]

Regard it as a polynomial in x^4, so you have t^2-t+1 which factorizes as (t-something)(t-something')=(x^4-something)(x^4-something')

[velvet hull]

oh, ah that is nice

[vocal pebble]

And those parts are easy to factorise

[coral steeple]

That is nice

[raw yarrow]

trying to prove that the kernel is a normal subgroup

[lusty marlin]

Recall the definition of normal subgroup

It should be easy to see

[raw yarrow]

i do know that

i just need to prove that f(axa^-1) = e'

oh wait

obviously the homomorphism is extendible

just do it recursively

f(abc) = f(a) f(bc) = f(a) f(b) f(c)

but what does it mean for non associative operations

like subtraction

wait if the operation is non associative is the set even a group in the first place

[lusty marlin]

A group operation is associative by definition

[raw yarrow]

yep

got it

tyy

[foggy tartan]

in this case, it's a loop

[raw yarrow]

beyond my scope rn lowk

$$
\text{Find the generator of the group } G(\mathbb{Z}, +)
$$

[cloud walrus]

fluX

[raw yarrow]

I don't think this group is cyclic

how do you cover negative numbers with a positive generator under addition

and vice versa

[foggy tartan]

existence of inverses

[raw yarrow]

ah, if a is the generator then a^-1 is also a generator

so the inverse element of the group G(Z, +) is such that aa^-1 = e or a + a^-1 = 0
or a^-1 = -a

cool

now is the identity element, 0, obtained by setting n = 0

like, what it means is

don't do anything to this element, hence it is 0?

don't add to it, don't subtract from it

[foggy tartan]

e + a = a for each a; i.e. e = 0, the empty action

[raw yarrow]

thinking of the group like an array of integers with all bits set to 0

[foggy tartan]

i guess you could see Z as the rotational symmetries of a regular apeirogon

[raw yarrow]

anyways

$$
1^0 = 1^{(1-1)} = 1^1 + 1^{-1} = 1^1 + (1^{-1})^1 = 1^1 + (-1)^1 = 1 + (-1) = 0
$$

[cloud walrus]

fluX

[raw yarrow]

@foggy tartan

[foggy tartan]

exponentiation?

[raw yarrow]

under addition

[foggy tartan]

for an additive group, we usually write $g^n$ like $ng$ in the "multiplying" sense. but yeah this is it

[raw yarrow]

$$ a^n = a * a * \cdots n\text{ times }\cdots * a \forall a \in G \text{ with operator } *$$

[cloud walrus]

fluX

velverette

[raw yarrow]

@foggy tartan i have a huge exam tomorrow and i opened my book today 😭

[foggy tartan]

oh damn!

what are the topics?

[raw yarrow]

its ok im having lots of fun

[raw yarrow]

groups, morphisms, kernel, codes, cosets, posets, lattices

[foggy tartan]

damn

not so bad though

[raw yarrow]

oh wait look at this proof i just did

$$
\text{Prove that } O(a) = O(a^{-1})
$$

[cloud walrus]

fluX

[foggy tartan]

nice!

(you can actually simplify this by realizing that because $e=e^{-1}$, then we have $a^p=(a^p)^{-1}=a^{-p}=(a^{-1})^p$)

[cloud walrus]

velverette

[ripe harbor]

I want to prove that $[S_n, S_n]$ for all $n \geq 2$ contains all 3-cycles $(a, b, k)$. How would we do that?

Wait

[cloud walrus]

ILikeMathematics

[ripe harbor]

(this is to follow [S_n, S_n] = A_n since we know A_n is generated by 3-cycles, and we can show [S_n, S_n] c A_n)

[rocky cloak]

I think the famous example should be
x^4 + 1
Which is irreducible (it is in fact the 8th cyclotomic polynomial). But it's reducible modulo every prime because p^2 - 1 is divisible by 8

In fact similar argument applies to your polynomial. Since the roots are 24th roots of unity and (Z/24)^* = (C2)^3 this splits over any field of order p^2, so it factors into quadratics (or linear terms) over every prime

[raw yarrow]

im having a lot of trouble understanding cosets

and computing them

[ripe harbor]

Is there any useful lemma we have here?

[foggy tartan]

for n >=2 [S_n,S_n] = A_n, which contains all even permutations, thus all 3 cycles

[foggy tartan]

cosets of a subgroup are "shifts" of the subgroup by some element

consider the subgroup 3Z of multiples of 3 in Z. what are the cosets of 3Z in Z?

[raw yarrow]

Z is {..., -3, -2, -1, 0, 1, 2, 3 ...}

3Z is. {..., -9, -6, -3, 0, 3, 6, 9 ...}

cosets of 3Z in Z are just 0 + 3Z, 1 + 3Z, 2 + 3Z, 4 + 3Z (not 3 + 3Z, it's already covered) and so on

[rocky cloak]

Try to compute the commutator of two 3-cycles

[foggy tartan]

what can you say about 1 + 3Z and 4 + 3Z?

[raw yarrow]

oh it's the same

okay so there are only 3 unique cosets

[foggy tartan]

yes

[raw yarrow]

okay and what's so special abt these

[foggy tartan]

the number of cosets divides the order of the group

furthermore, the order of the group G is the order of any subgroup H times the number of cosets of H in G. in symbols $|G|=|H|\cdot[G:H]$ where $[G:H]$ is the number of (left) cosets of $H$ in $G$

[cloud walrus]

velverette

[foggy tartan]

this is lagrange's theorem, and characterizes the possible orders of subgroups for a group among many other things

[raw yarrow]

head hurts rn i think i need a breajk

but i'll come back to this

ty

[foggy tartan]

sorry to infodump lol

rest well

[raw yarrow]

No worries, you've been helping me a lot today
Thank you

[ripe harbor]

In case theyre disjunct, we will have (a, b, c)(d, e, f)(c, b, a)(f, e, d) = id, the terms commute

[rocky cloak]

Yeah, and what is there is some overlap?

[ripe harbor]

Uh, should I try through all the different cases or can we somehow visualize this?

[rocky cloak]

Yeah, I mean shouldn't be that many cases.

If the overlap is all 3 then you just get the identity again. So the two cases should be an overlap of 1 or an overlap of 2

I'm realizing now you asked for [Sn, Sn] = An, and my hints are actually for the stronger-ish [An, An] = An (n>=5)

[ripe harbor]

(a, b, c)(b, a, c)(c, b , a)(c, a, b)

oh right

Thats id

[rocky cloak]

If you just want [Sn, Sn] = An, then prove it for n=3 and notice that the proof generalizes

[ripe harbor]

Actually, my book already has that A_n is simple for n >= 5 so we only need n <= 4 since we can quickly get [S_n, S_n] <= A_n but yeah maybe just generalizing right away is more elegant

[ripe harbor]

Whats the idea behind the proof? Can we somehow do some neat dimensional argument or anything

Without having to mess around too much

[rocky cloak]

To prove that [S3, S3] = A3, you mean? I was suggesting just computing the commutators

[ripe harbor]

Alright.

If the overlap is all 3 then you just get the identity again
How did you know? I "verified" it with one case, (a, b, c)(b, a, c)(c, b , a)(c, a, b), but that doesnt say anything, does it? Stuff could be shuffled some other way, how do we know its still id

[rocky cloak]

There are just two 3-cycles on 3 elements and they are inverses of eachother

[ripe harbor]

The two middle ones arent

[rocky cloak]

What do you mean?

[ripe harbor]

(b, a, c) and (c, b, a) are not inverses of each other

[rocky cloak]

Because they're the same

[ripe harbor]

Oh, right haha

[ripe harbor]

If I shuffle some other way, why can we be sure it will behave the same way?

[broken tulip]

It's easier to see if you write down them in two rows, with input being in the first row and the corresponding outputs being in the second row

[ripe harbor]

this time g = (a, c, b) and h = (a, b, c) or something

[rocky cloak]

I mean, just count the 3-cycles:

a maps to either b or c, that's two choices. If a maps to b then b must map to c and similarly interchanging b and c

So (a, b, c) and (a, c, b) are the only 3-cycles

[ripe harbor]

Oh, alright, so really we are checking
g = (a, b, c), h = (a, c, b)
g = h = (a, b, c)
and the first case the other way around

And after confirming they are all id, we conclude

If the overlap is all 3 then you just get the identity again

[raw yarrow]

what's a boolean algebra

[copper kestrel]

still trying to prove Z x Z / <(a,b)> is cyclic if and only if gcd(a,b) = 1 and still have made 0 progress other than putting down the definitions

i just dont get how i'm supposed to get from the generator to the gcd of a and b

bc the generator of Z x Z/ <(a,b)> is some element (c,d) in the group such that n(c,d) = Z x Z/<(a,b)>

but then somehow i have to get bi + aj = 1 for i,j in Z

[rocky cloak]

So, assuming b is nonzero (1, 0) gives you an element of infinite order. So if you can find an element that has finite order then it can't be cyclic

[quiet pelican]

Bezout

[raw yarrow]

Can we appreciate Hasse diagrams for a second

[copper kestrel]

arent all elements in a cyclic group of finite order though?

[raw yarrow]

The way the Hasse diagram of the poset (P(A), subset operator) draws out the card(A) dimensional square/cube/tesseract wtv

[rocky cloak]

Z is a cyclic group

[copper kestrel]

oh right

[ripe harbor]

Let H <= S_5 with a 2 and a 5-cycle. Then I want to show H = S_5.

We know 2 | |H| and 5 | |H|, and the composition of 2 and 5-cycles give 4-cycles, so 4 | |H|. Thus |H| in {20, 40, 60, 120}. How do we rule out the first 3?

[rocky cloak]

And in fact
Z^2 /(a, b) = Z (+) Z/gcd(a, b)

[copper kestrel]

i didnt know that

[rocky cloak]

No, but that's basically what you're showing now

Just a little spoiler for you

[copper kestrel]

oh

how do you even figure that out

also isnt Z^2 and Z (+) Z the same

at least my book says theyre the same since Z is abelian

but how do you figure out you want to show that Z^2/<(a,b)> = Z^2/gcd(a,b)

bc that seems to come out of nowhere

oh i see there is exactly one example in by book that does something even slightly related to this

[copper kestrel]

im genuinely so frustrated bc how was i supposed to figure this out??

[frigid epoch]

picture Z^2 in your head

[foggy tartan]

set endowed with the operations of boolean logic ("and", "or", "not") and two constants "true" and "false"

e.g. the power set of a nonempty set X has a boolean algebra structure on it

[frigid epoch]

tbh I'm not sure what Z^2/gcd(a,b) means

[foggy tartan]

gcd is just a number, what does the quotient here mean

[frigid epoch]

Z^2/<(gcd(a,b),0)> ?
Z^2/<(0,gcd(a,b))> ?
Z^2/<(gcd(a,b),gcd(a,b))> ?

oh, looking at jagr's statement, they seemed to refer to Z (+) (Z/gcd(a,b))

[copper kestrel]

actually im now a bit more confused, arent Z x Z and Z (+) Z not cyclic?

[raw yarrow]

like this?

[copper kestrel]

so how can it be cyclic if gcd(a,b) = 1?

[raw yarrow]

also, so if i know how to work with groups and logic, i will be able to solve problems on boolean algebra, right/

[next obsidian]

Z x Z/1Z = Z

[copper kestrel]

oh i didnt realize if was Z/1Z, that makes sense

[foggy tartan]

boolean algebra is a lot of comp sci adjacent stuff, it's close to arithmetic

[next obsidian]

I mean that’s if the GCD is 1

Otherwise no

[copper kestrel]

yeyeye

[next obsidian]

Cake

[raw yarrow]

i am a cs major

[foggy tartan]

the cake is a lie algebra something something

[next obsidian]

I had some cake this morning

It was leftover from a bday party yesterday

Burrito plus chocolate cake breakfast oh yeah

[foggy tartan]

what a vibe

rate my proof

[rocky cloak]

I mean, just try to determine what the order of the elements in Z^2 /(a, b) is and it just pops out

[copper kestrel]

yeah, it makes sense. i just dont understand why my brain cant compute it

i know its fundamental but jeez

[noble nexus]

better proof: think about a subset as a function from X to Z/2 (i.e. the characteristic function) then addition of functions is symmetric difference and multiplication of functions is intersection.

clearly the set of functions from any set to a field form a ring under pointwise operations

[foggy tartan]

doing this exercise right now funnily

Prove that the map P(X) -> R defined by A -> \chi_A is a ring isomorphism.

[broken tulip]

This one is from Aluffi, right?

[foggy tartan]

d&f actually

wouldn't be surprised if that overlapped

[broken tulip]

Yeah, textbook authors borrow problems all the time

[broken tulip]

This is how I did it, IIRC

[next obsidian]

Better proof is to cite the exercise from Aluffi

See [reference]

[foggy tartan]

proof by outsourcing

[swift root]

outsource it to a paper you will write in the future, and then in that paper prove it by citing your original paper

[maiden crater]

But if we consider (K:HK). Won't it just be 1

as K is entirely contained in HK

and similarly isn't (HnK:H)=1

or am I spewing nonsense

[swift root]

were considering (HK : K), (K : HK) isnt defined, as HK is not a subgroup of K

[raw yarrow]

how would the hasse diagram for the poset {X, |} look like
X = {2, 3, 5, 7, 9, 10, 20}

[maiden crater]

oh right

[raw yarrow]

should be this, no?

[swift root]

you can draw jt

[maiden crater]

okay that is neater than I expected

[swift root]

i would draw it differently, but yes

[raw yarrow]

how would u draw it

in a neater way

[swift root]

not have crossing lines

[raw yarrow]

oh alright, yeah

i try to follow the order of ascending numbers from LTR

on any given level

[swift root]

always draw hass diagrams how it looks the cleanest, it is good for your soul

[ripe harbor]

We could try saying WLOG let the cycles be (1, 2, 3, 4, 5) and (1, 2). And then get every transposition. But why can we say that WLOG?

[raw yarrow]

for this poset, there is no least / greatest member, right?

[foggy tartan]

nope

[foggy tartan]

lattice of subgroups of D_16

[raw yarrow]

oh hn i have to cover ts rn

cover

hahaha

im going crazy

[swift root]

this is actively harmful

[maiden crater]

I thought we'd begin by establishing $\phi: G \to G /H$ as $x \to x+H$ with $Ker( \phi)=H$

[cloud walrus]

wai

[maiden crater]

Then $G/H \cong (G \setminus H)/H$

[cloud walrus]

wai

[swift root]

?

no

[maiden crater]

isn't that the image

[swift root]

G\H is not a group

[maiden crater]

that union with e

[lusty marlin]

G ∖ H is not even a group. How do you quotient it by a subgroup?

[maiden crater]

this is fine, right

[swift root]

sure i guess

[maiden crater]

Then first iso theorm

[swift root]

but why

[foggy tartan]

if you're applying 1st isomorphism thm, G/H \cong G/ker(\phi) = G/H

no info gained

[swift root]

the first isomorphism theorem says here that G/H ≅ G/H

[maiden crater]

oh right

[swift root]

lol

[maiden crater]

😔

lemme think

[next obsidian]

Big if true.

[foggy tartan]

@maiden crater suppose there exist x,y in G such that xy = yx

[maiden crater]

yea, that just struck me

[raw yarrow]

anyone know anything about group codes 😭

[ripe harbor]

Is there any motivation on why you would consider sts^(-1)?

[foggy tartan]

by hypothesis x,y not in H thus form cosets of H. what can you say then?

[ripe harbor]

If you want to get transpositions

Why not st or ts or st^2 or s^2t^2 or tst^(-1)?

[maiden crater]

Let $x \in c(G)$. Let $y$ be arbitrary. Then consider $xy=yx$. This would mean $xy+H=yx+H$. But $G/H$ is not commutative

[foggy tartan]

G/H, not G\H

[cloud walrus]

wai

[maiden crater]

and we're done

[foggy tartan]

this works you can be explicit about xyH = xHyH is not yHxH = yxH

so then you have the actual noncommuting elements xH, yH in G/H

[ripe harbor]

Or is it just trying out something "random"?

[foggy tartan]

conjugation is a pretty natural first choice. it preserves cycle parity

[ripe harbor]

What is cycle parity?

[foggy tartan]

odd/even permutations?

[next obsidian]

When you conjugate an element sts^-1, what you do is if writing t down in cycle notation, you just apply s to each thing in the cycle notation. So (123) gets taken to (s(1)s(2)s(3))

[ripe harbor]

Ah, the sign, alright

[next obsidian]

This shows that you can just apply whatever elements of S_n to things in a very controlled and specific manner, letting you produce the types of elements you want

[ripe harbor]

Oh

This is nice!

[ripe harbor]

Btw, here, we have that H has an arbitrary 2 and an arbitrary 5 cycle and WLOG, they assumed they are (1, 2, 3, 4, 5) and (1, 2). Why can we say that WLOG without any justification? How do we know nothing goes differently for everything else

[next obsidian]

Just rename shit

[ripe harbor]

oh

[next obsidian]

Labels are arbitrary

You can also rotate a cycle like Mario and it doesn’t change. (12) = (21)

I think what you do need to reason about tho is that (ab) shows up in that order somewhere in the 5-cycle

Idk what you’re trying to do here actually but you might be able to replace the 5-cycle with a power of itself which might also help

To get to this situation

So like after relabeling you can always get to the situation the 2-cycle is (12) and the 5-cycle is (1abcd)

If 2 = a or d then you can relabel to get (12345) or you rotate to (21abc) and then rewrite (12) = (21), and then relabel and you’re back to (12) (12345)

If 2 = b or c you can replace (1abcd) with either its square or its cube to have 1 go to 2 so then you can relabel to get (12345)

[coral steeple]

I didn't expect to get an example, thanks!

[ripe harbor]

Is there an easier way to show that [S_2, S_2] = A_2, [S_3, S_3] = A_3 and [S_4, S_4] = A_4 (without knowing that [S_n, S_n] = A_n) or some quicker way to go through all the possibilities? Writing out [g, h] for every possible g and h gives a ton of possibilities

[rocky cloak]

(ab)(bc)(ab)(bc) = (a c b)

So if you know An is generated by 3-cycles that shows [Sn, Sn] = An

So immediate for S2 and S3 and not too hard to check for S4

[ripe harbor]

Are you suggesting showing by hand that all the 3-cycles are in S_4 or show this for S_n in general?

[rocky cloak]

Show that A4 is generated by 3-cycles

[ripe harbor]

I can use that A_n is generated by 3-cycles

[rocky cloak]

Okay, then there's nothing more to show

[ripe harbor]

I still need to show these 3 cycles are in [S_n, S_n]?

[rocky cloak]

I don't know what you mean by that

[ripe harbor]

Ah

You did that at the beginning and a, c, b were arbitrary

Alright, yeah that shows it

[ripe harbor]

How did you think of this? Why (ab) and (bc)? Is this just something random you picked?

[rocky cloak]

Like start with considering S3.

Then you just have to compute the commutator of two things, so (1 2) and (2 3) would make sense, and then that's what you get

You can also do
(a b c) (a b) (a c b) (a b) = (a c b)
or whichever nontrivial commutator in S3

[ripe harbor]

So really, we dont need anything else, right? With just this one line you prove it for all n

That [S_n, S_n] = A_n

[rocky cloak]

Yeah, or I guess that An is generated by 3-cycles requires a little bit as well

[ripe harbor]

Well luckily my book did that already

But yeah, thanks a lot

[coral steeple]

An update to the un-explained step in this problem: the professor has revised this argument to look more like what you suggested

[foggy tartan]

struggling to see how \phi(1_R) is a zero divisor

[sudden edge]

use 1_R ^2=1_R

[foggy tartan]

oh wait it was really just that lol

thanks

[unborn helm]

I wrote some code that finds roots of unity in terms of radical expressions and as a little bit of an add-on, finds roots of general solvable polynomials in terms of radical expressions for solvable polynomials of degree < 24. It may interest people as it involves some amount of Galois theory. I still need to push the database entries for higher-degree than quintic solvable polynomials though:
https://github.com/NadiaYvette/surd

[ripe harbor]

Why is $f^{-1}(f(J)) = J$ for $\ker(f) \subseteq J$?

Whats a way to see this?

[swift root]

correspondence theorem, essentially

[ripe harbor]

Is there a more elementary way? I think that might be a stronger statement of what Im proving

[swift root]

its essentially exactly that statement

(along with f(f^-1(J)) = J for all J whenever f is surjective but that is elementary set theory)

[chilly radish]

Explicitly, J is contained in f^-1(f(J)) always. Show that if x has f(x) \in f(J) then x is actually an element of J

If you can't do this, go back to the definitions

[unborn helm]

It might be slightly nicer in terms of output quality if it dealt with ring extensions starting from the integers and only dealt with fields of fractions at the very end, but most of the algorithms assume that they're operating over the rationals. There are also some issues round the branch cuts for root extraction where the radicand most often differs by a sign from what would otherwise be collectible in like terms etc. (for which I use very basic Gröbner code), meaning I need to pull out (-1)^(1/k) for a k-th root, which in turn has to be expressed in radicals with cyclotomic affairs. There's not really a natural place that fits into algorithms that avoid analytic issues.

[swift root]

(and if youve done that, congrats! you proved the correspondence theorem)

[somber goblet]

thats cool

[unborn helm]

It seems like whenever you're working over Q(α) you should almost always be able to work over Z(α) until the very end and only divide by integer denominators then.

[somber goblet]

this reminds me of algebraic integers

i am curious, how do you encode numbers in ring extensions

[unborn helm]

Yes, I'm trying to think of ways to work in something akin to them. I'm not sure the concept I'm describing completely lines up, though.

[somber goblet]

ive been wanting to do algebra/algebraic nt stuff with a computer program so i am very curious about how your project works

[unborn helm]

What things are using right now for real extensions are rational bracketing intervals paired with minimal polynomials. For complex extensions, the uniqueness issue with the branch cut pops up and needs cleaning up after the fact.

[somber goblet]

i see, that's clever

[unborn helm]

Ideally I'd be able to only ever take roots of positive reals. Cardano's casus irreducibilis means that I have to consider the argument of the radicand and pull out factors (hopefully -1 suffices) so the result is always in the right half plane.

[somber goblet]

i might try to do this with 100% precision in a finite field

[unborn helm]

Choosing the right root of -1 is troublesome even after all that.

[unborn helm]

I learned a lot doing it.

[somber goblet]

have you tried in a finite field?

[unborn helm]

No, it's a lot easier in them as I understand it.

[somber goblet]

seems like a computer would have an easier time working in one

[cloud walrus]

ILikeMathematics

[rocky cloak]

Subfields of alg closed fields aren't usually alg closed

For example Q is a subfield of C

[ripe harbor]

oh

[unborn helm]

@eager lark I'm willing to be corrected if that's incorrect.

[ripe harbor]

Why is phi(K_1) alg closed? Why does K_1 alg closed imply that

[rocky cloak]

K1 is isomorphic to phi(K1)

[ripe harbor]

oh

[frigid epoch]

all field morphisms are injective

[ripe harbor]

Wait why, we just consider K_1 -> phi(K_1) with a -> phi(a), right? And thats surjective and a ring homomorphism. Why is it injective?

[frigid epoch]

if $a$ is in the kernel, then so is $aa\ii$

[cloud walrus]

PKThoron

[ripe harbor]

So?

[frigid epoch]

that would mean 1 is in the kernel

ring morphisms always send 1 to 1 though

of course there's a ring where 1 is 0, the zero ring, but that isn't a field

[ripe harbor]

Oh

Thanks!

[frigid epoch]

so all ring morphisms from a field to a non-zero ring are injective

[knotty badger]

Took me a while to understand but it makes sense now

[copper kestrel]

wowie

[swift root]

the joy of simple objects in a variety of algebras where the top congruence is generated by nullary operations

[elfin wraith]

That’s just what I was thinking

[swift root]

should be your first thought always

[coral whale]

In a field F must the additive inverse of any element of F also be in F?

[tall igloo]

yes

[coral whale]

and if so
I am confused on what makes a field algebraically closed. What would an “open” field look like?

[tall igloo]

part of the data of a field is an abelian group structure on the set that defines addition, and groups are closed under inverses

[tall igloo]

can you elaborate on your confusion? I don't see what closure under additive inverses has to do with algebraically closed

[next obsidian]

Algebraically closed is about being closed under algebraic extensions

[coral whale]

I see. im misunderstanding the notion of algebraic closure It was a misguided question

Thank you

[tall igloo]

also just to clarify the terminology - closed here means closed in the sense of, if i apply an operation then the result of the operation is still inside the field. here the operation is taking roots of polynomials

so there isnt really a notion of an algebraically open field. closed sets in metric spaces are called closed because they are closed under evaluating limits of sequences inside a closed set, and so it makes sense to call the dual thing open. but there isn't really some dual notion of algebraic closure

[coral whale]

thank you

[tall igloo]

yup nw. if you need help with understanding the notion of algebraic closure feel free to ask!

[left grail]

R is ring with no zero divisors and every subring of R is an ideal then R is commutative
The proof plz

[coral whale]

Eisenbud is a wonderful expositor

[next obsidian]

Lmfao

[coral whale]

i am very excited to read his commutative algebra book

[left grail]

Any one can help ?

[limpid ferry]

how to prove <1+i> is a maximal (or prime) ideal in Z[i] using the minimum of theorems

[vapid vale]

show the quotient is a field

[limpid ferry]

do I need to use that in conjunction with Third isomorphism theorem?

[vapid vale]

the correspondence theorem

[glass sinew]

what have you tried?

[next obsidian]

Upside down pic nice

[glass sinew]

in the second one you haven't proven R commutative

just that aR = Ra, which is guaranteed by hypothesis

in the first picture what's the reason for defining <a>?

[azure cairn]

did you tilt ur head 90deg

[indigo vale]

Why does |GL2(F2)| = 6? I am not understanding something

[rapid cave]

You have 3 nonzero vectors (1,0),(0,1),(1,1)

Any choice of two of them (with order) gives an invertible 2x2 matrix over F2

[indigo vale]

Does F2 != Z/2Z?

[rapid cave]

They are isomorphic, which pretty much means they are the same

[indigo vale]

ok, thank you

[foggy tartan]

x^3-x^2-4 has splitting field basis {1,i*sqrt7} and galois group Z/2 right

[indigo vale]

Oh, i think I was misunderstanding Z/2Z.... this is simply the integers mod 2?

[rapid cave]

Yes

[rapid cave]

x^3-x^2-4=(x^2+x+2)(x-2). x^2+x+2 is irreducible over Q with discriminant -7, so the degree of the splitting field is of degree 2, with basis 1,sqrt(-7) and galois group Z/2Z as its the only group of order 2.

So you are correct

[indigo vale]

Am i correct that four of these elements will have infinite order?

[rapid cave]

No. Its a finite group so every element has finite order.

[indigo vale]

ok i see

[indigo vale]

Do all the elements have unique order? is there a faster way to do this than multiplying each matrix by itself until i return to identity?

[foggy tartan]

thanks!

[quick laurel]

By Cayley's Theorem, every group is isomorphic to a subgroup of the symmetric group - for a group $G$, is there a name for the minimum $n$ such that $G$ is isomorphic to a copy of a subgroup of $S_n$?

[cloud walrus]

rubixcyouber

[candid patrol]

Yes there is

But i don’t remember lol

Maybe minimal permutation representation

[rocky cloak]

It appears to be called the minimal faithful permutation degree

Even has a GAP function

[next obsidian]

I wonder if you can say anything about its growth rate as a function of |G|

Like obviously it’ll get minimized for G = S_n, so I’m thinking of something like “for most G the size is…”

Inentionally very open ended

[rocky cloak]

I'm curious for how many groups it equals |G|.

I know it does for Q8, but does it work for any bigger groups

[next obsidian]

I have no idea what sort of thing could exist as an obstruction so it seems hard to analyze

[frigid epoch]

is there an equivalent name for the minimal representation dimension $n$ in $G \to \operatorname{GL}_n(k)$?

[cloud walrus]

PKThoron

[next obsidian]

Your mom

[rocky cloak]

Hmm, any G-set is the union of things of the form G/H with kernel given by the core of H. So if G has a unique minimal subgroup then a faithful permutations rep would need to contain G/1

[next obsidian]

[rocky cloak]

I think this goes for all the generalized quaternion groups

[next obsidian]

Uh, sure

[frigid epoch]

oh yeah faithful

because, meme as it is, $S_n$ is just $\operatorname{GL}_n(\mathbb{F}_1)$

[cloud walrus]

PKThoron

[rocky cloak]

People turning facts into memes now?

[frigid epoch]

well I'm of two minds

the meme is more me talking about F_1 😛

[sudden edge]

what if F_1

[rapid cave]

If there is an element of order 3, the subgroup it generates has 3 elements, where one of them is the identity and every other element is also of order 3.

[frigid epoch]

a strange hypothesized object that behaves like a field of characteristic 1, but isn't actually a field

by "hypothesized" I mean that there are a lot of other objects in maths that behave like vector spaces, algebraic groups, curves etc. over this object, but the object itself cannot be found in a field or ring-theoretic sense

pointed finite sets are the finite-dimensional "vector spaces" over it (note that these "vector spaces" don't actually have a notion of addition) and their general linear group is $S_n$

[cloud walrus]

PKThoron

[frigid epoch]

there's a bit of hype around it because if this "field with one element" were better understood, it could lead to advances for the riemann hypothesis, which to my knowledge has been solved over other finite fields

[sudden edge]

why we need it

[copper kestrel]

problem 2 and 4 my professor said to use smith normal form but im not quite sure how that would help :/ i was thinking about focusing on problem 4

i know we want to show that Z^10/N is isomorphic to some infinite group

i think

if it was nice i could just say let <g_1,...,g_7> = N, then we can show an isomorphism of

Z/g_1 x Z/g_2 x ... x Z/g_7 x Z x Z x Z

but i don't think that is correct

[quiet pelican]

We want to change from the usual basis of Z^10 to a basis where we can do this

[copper kestrel]

ooh

is that where smith normal form comes in?

[quiet pelican]

Ie, a basis in which g_i = a_i e_i, where e_i are the basis vectors and a_i are integers

[copper kestrel]

omg i like smith normal form

im not quite sure how to compute it though..

[quiet pelican]

You don’t need to compute such a basis
Just argue that it exists
Then it’s fairly clear (once you’ve done that) that Z^10/N = Z/a_1 Z x … x Z/a_7 Z x Z x Z x Z

[copper kestrel]

hmmm

i mean we would need to show that Z^10/N is a free abelian group then

bc then it has a basis and then by smith normal form properties we have that smith normal form's new basis is still a basis for Z^10/N

[rocky cloak]

I think you might be confused.

Z^10 / N should not be free.

As N is generated by 7 elements you can think of it as the image of a map
Z^7 -> Z^10

This would correspond to a 10x7-matrix that you can apply Smith normal form to

[copper kestrel]

blinks

[copper kestrel]

hmmmmm

i guess that makes sense

how does that give us the basis such that we have the nice isomorphism?

[rocky cloak]

Maybe I don't know what you're asking, but for a map Z^n -> Z^m, the Smith normal form is a choice of basis for Z^n and Z^m so that the matrix has a very nice form

[copper kestrel]

that makes sense, i was asking how the smith normal form basis "exists" such that if N = <g1,...,g7> then smith normal form gives us g_i = a_i e_i so that Z^10/N = Z/a1 x Z/a2 x ... x Z/a7 x Z x Z x Z

bc i dont think that 10 x 7 matrix always exists for every 10 dim by 7 dim factor group

unless it does and i dont know what im saying

[rocky cloak]

Like you just let the columns of the matrix be the generators of your group

[copper kestrel]

ohhhhhhh

so then it always exists?

[copper kestrel]

actually now that i think about it i dont really get this

why can we think of N as the image of a map of Z^7 to Z^10

[next obsidian]

The definition of generated

[copper kestrel]

huh

[next obsidian]

A map Z^n -> G is a choice of n elements in G

You send e_i to g_i

[copper kestrel]

right

[next obsidian]

And then extend linearly so Sum a_ie_i goes to Sum a_ig_i

To say that S generates G is exactly saying the map Z^S -> G picking out the elements of S is surjective

[copper kestrel]

interesting

i will try to remember that

[next obsidian]

Wut de incineroar doing

[copper kestrel]

LMAO truth

[next obsidian]

This holds exactly for modules btw

Where G is now a module over a ring R

And you replace Z with R

For when you do module theory

[copper kestrel]

module theory shudders

we did a little bit of that which is where i had to learn smith normal form

[next obsidian]

Yeh

[copper kestrel]

and we let the columns generate the Z/a1 x .... x Z x Z x Z thing?

[rocky cloak]

After changing to smith normal form your columns will be
(a1, 0, 0, ...)
(0, a2, 0, ...)
etc.

So then the quotient is
Z/a1 x Z/a2 x ...

[copper kestrel]

ah i see

smith normal form is so strange

i dont really get it

i have one day to get comfortable with it bc the final is on tuesday 💀

[elfin wraith]

what you do is open up a sage terminal

[copper kestrel]

and smith normal form exists bc we have integer entries in the matrix right?

[copper kestrel]

idk what that means :(

[south patrol]

"Use a computer"

Sage is a computer program

[copper kestrel]

ohhhhhhh

sorry i am not cultured

[elfin wraith]

How dare you not know about a random computer algebra software while you’re learning intro to algebra

Banable offence

[copper kestrel]

sobs

modulators this is mod aboos!!!

[rocky cloak]

Yes, or more generally for entries in a PID

[copper kestrel]

ahhhhh

[karmic moat]

Real cultured people use Macaulay2

[copper kestrel]

truth nuke

[copper kestrel]

why does the number of reducible polynomials matter? im guessing its like. find number of reducible polynomials of that form and subtract it from something

also i think 5b is just euler phi(1000) maybe?

[south patrol]

yes, the total number

[copper kestrel]

that makes sense

[south patrol]

I was really confused at first though (read Z_p as the p-adics and was like huh so answer is just "infinity")

[copper kestrel]

LOLLL

i'm always bad at these total number of polynomials sort of thing

i know theres p^2 polynomials of the form x^2 + ax + b i'm pretty sure

[south patrol]

yup

[copper kestrel]

so then since its of degree 2 if its reducible we can find the solutions such that (c)^2 + a(c) + b = 0

or i guess it would be quadratic formula

x^2 + ax + b = 0 mod p

-a \pm sqrt(a^2 -4b) / 2 mod p

[cursive spindle]

tfw combinatorics problem