groups-rings-fields

[south patrol]

Do you reckon it makes sense to say that as Z[x] is generated by x as a Z-algebra we'd have Z^n generated by the image of x (again as a Z algebra) which is impossible?

[hidden haven]

Yes

Works for n > 2 I think?

[south patrol]

Yeah that's what i mean

[hidden haven]

Wait actually I was imagining Z_2

[south patrol]

Ig the logic is reducing mod 2 means you don't need to worry about the fact we have polynomials in the image of x lol

[hidden haven]

ok nice lol

[south patrol]

but probably would work w/o reducing mod 2 anyway

[hidden haven]

[south patrol]

Hm

eh nah reduce mod 2 aha

[lavish nexus]

not sure how that last part works

different number of irreducibles on both sides?

like if f = b_1b_2
multiply both sides by a_1(y)a_2(y)
we get a_1a_2 * f = b'_1*b'_2 irreducibles

[chilly ocean]

Is my example wrong

Is Z3 cyclic

[lavish nexus]

you're right

[chilly ocean]

How do I know if z3 is cyclic lol

I’m kinda noob

[lavish nexus]

1 generates Z3

1= 1 1+1=2 1+1+1 = 0

[barren sierra]

your example is right but incomplete

@lavish nexus showed Z_3 is cyclic

but show that all subgroups of it are trivial or the whole group

[chilly ocean]

Yeah I’m kinda stuck

[barren sierra]

which you can do by looking at the subgroup generated by 0

and then the subgroup generated by 1

and then the subgroup generated by 2

and then you're done

[lavish nexus]

he already has an order argument

that proves it

[chilly ocean]

What would subgroups of z3 be

I know z3 is prime

[lavish nexus]

it's either order 3 or order 1

order 3 = Z3

order 1 = {0}

done

[chilly ocean]

Wouldn’t ]{0,1}.be a subgroup

[lavish nexus]

does 2|3

[chilly ocean]

No

[lavish nexus]

ok then it is not a subgroup

[chilly ocean]

Ahh ok

[proud bear]

another reason is that it is not closed under the group operation. like 1+1 isn't in {0,1}

[chilly ocean]

Thanks

[quick harness]

What does this notation mean ?

I know what it would mean without the square root of 2

[chilly ocean]

h is a polynomial in x with coefficients in Q(\sqrt 2). or are you asking what Q(\sqrt 2) is?

[quick harness]

So the polynomials look like a•√2 + b•√2•x + c•√2•x^2 ... ?

[lavish nexus]

the coefficients are in Q(sqrt(2))

so they can be anything in Q

and any linear combination of 1 and sqrt(2)

(which includes everything in Q)

[chilly ocean]

Why is the answer for c 1?

But I get 2

[maiden ocean]

How are you getting 2?

[chilly ocean]

i circled them @maiden ocean

[white jackal]

is it because of the uniqueness theory?

[maiden ocean]

I dont understand what the numbers you circled mean

[chilly ocean]

they have the same gcd

[maiden ocean]

What is that supposed to imply?

[chilly ocean]

it tells us how many subgroups of order 8 G has

[white jackal]

since 3 divides 9 that would just count as 1 right?

since they give the same subgroup

[maiden ocean]

The gcd tells you the order of the subgroup

Yes

[chilly ocean]

sooo

i got 2

[maiden ocean]

But if they have the same gcd then they correspond to the same subgroup

[chilly ocean]

okay

so

[maiden ocean]

Or rather like

[chilly ocean]

so is the answer for these type of problems is always 1 or 0?

[maiden ocean]

Yes

In the cyclic case specifically

[chilly ocean]

okayy

and for part e it would just be the number of generators?

[maiden ocean]

Remember the subgroup actually generated by 9 is more complicated than 3 because 9 doesnt divide 24

[maiden ocean]

Yeah

i mean thats what the problem asks

[chilly ocean]

wym

part e asks for normal subgroups

[maiden ocean]

Oh I misread and thought you meant (f), sorry

(e) is a little bit of a trick question because cyclic groups are abelian, so all their subgroups are normal

so its really just asking you to name the number of subgroups of G

[next obsidian]

wait is this under addition?

[maiden ocean]

Yes

[next obsidian]

aren't they the same

[maiden ocean]

Yes

[next obsidian]

oh

[maiden ocean]

Maybe that wording was bad

I meant that you cant just look at multiples of 9 until you hit 24

[next obsidian]

oh sure

[chilly ocean]

so

[next obsidian]

[chilly ocean]

for my case would it just be the subgroup: list i have?

for the normal subgroups

[maiden ocean]

Yes provided it is correct

[chilly ocean]

i think its correct

but the solutions say its 8

but i have 15

[maiden ocean]

Oh well your list does not account for duplicates. If i have a cyclic group G of order n then i have one unique subgroup of G for every divisor of n. 24 is divided by 1, 2, 3, 4, 6, 8, 12, 24

[chilly ocean]

im missing 1 also right?

[maiden ocean]

Im not really sure what your list means exactly

[chilly ocean]

those are just divisors of

24

[maiden ocean]

Uh, 9 doesnt divide 24? neither does 10

[chilly ocean]

oh yeah

i basically have the numbers that werent coprime

so in reality the number of normal subgroups would just be the number of divisors of 24?

[maiden ocean]

Yeah

Or rather theres one unique subgroup of a cyclic group for every divisor of its order

Since a cyclic group is abelian, those are all normal

[chilly ocean]

so this only applies for cyclic

[maiden ocean]

Yeah

In general the subgroups of a group might be quite complex, you can have multiple of the same order

But if a group is finite the order of the subgroup must divide the order of the larger group (lagranges theorem)

[chilly ocean]

yes

i think i get it now

thanks bro

[quick harness]

How is this any different from Z ?

[hidden haven]

Those aren't subsets of ℤ

[simple mulch]

like Z_2 = {0,1}

[quick harness]

oh, mb

[south patrol]

in what sense does it make sense to even take that union oof

[quick harness]

How is this any different from N then ?

[hidden haven]

Those aren't subsets of ℕ either

[south patrol]

Are you using Z_n to mean Z/nZ or smth else

[quick harness]

Zn as in class of Z mod n

[hidden haven]

What structures are you putting on these?

Are they groups?

[quick harness]

yeah

[hidden haven]

Weird to take a union of groups but then they definitely don't sit inside ℤ

[quick harness]

No

It's a set

[south patrol]

Well Zn isn't a subset of Z though

[hidden haven]

But then this highly depends on what you mean by union

[quick harness]

I mean set union

[hidden haven]

Is this disjoint union

[next obsidian]

I think disjoint union

Or

[hidden haven]

If we do disjoint union we treat it as everything is formally disjoint

[next obsidian]

Maybe this is an inverse limit

Omegalol

[south patrol]

Lol

[next obsidian]

I mean Moldi

I don’t see how to put a group structure on the disjoint union

[south patrol]

aren't coproducts of groups just direct sums tho if im not being silly

[hidden haven]

Ye

[quick harness]

This isn't really helping

[next obsidian]

I feel like in group that’s a coproduct

Maybe

[hidden haven]

Yes

[next obsidian]

But like how do I multiply things from the two separate camps?

[hidden haven]

That's why I was saying it's weird to take a union of groups

[next obsidian]

Can you post the entire context for this?

[quick harness]

sure

[hidden haven]

It might be better to give more context

[quick harness]

wait a sec

[next obsidian]

Maybe with more context it’ll be obvious what it is

[south patrol]

Lol

[next obsidian]

https://tenor.com/view/sniper-monkey-funny-monke-funny-kill-gif-20219406

[hidden haven]

[quick harness]

I have to determine if given operation (let's say a * b) defines operation on set S =

such that (S, *) is group

[hidden haven]

[next obsidian]

So they have given you a specific operation?

[quick harness]

yeah

[next obsidian]

Or are you to come@up with one?

[hidden haven]

If it's a true or false problem then it could be literally anything though lmao

[next obsidian]

Okay, it’s probably supposed be the disjoint union then

What’s the given operation?

[quick harness]

wait a sec

have to english-ify it

[next obsidian]

Yeah that’s the disjoint union

[quick harness]

ok, so what does union look like ?

As in elements

[next obsidian]

It’s an element from any of those sets

But we keep them all distinct

So like an element looks like [n]_m for some n, some m

And we view [n]_m as an element of Z_m

[quick harness]

I think I get it

[next obsidian]

It’s a way to get a set that is the “union” of two sets which aren’t a subset of the same set

You just like artificially make it

https://en.m.wikipedia.org/wiki/Disjoint_union

There is probably a way to translate this page into your native language

But this should give you an idea of what it is

[quick harness]

ok, thanks

[delicate orchid]

me writing a proof that the free groups over two sets of the same cardinality must be isomorphic
https://tenor.com/view/spongebob-the-essay-hilarious-haha-gif-3565665

[next obsidian]

Just define a map and its inverse omegalol 4head

[delicate orchid]

there are a LOT of arrows

[next obsidian]

No there aren’t smfh

Take two generating sets

[delicate orchid]

there are at least three (3)

[next obsidian]

Pick a bijection

[delicate orchid]

oh yeah I've managed it

[next obsidian]

Then universal property handles it all

[delicate orchid]

yeah see that's the bit I'm still struggling with

I just drew a diagram and followed the arrows and then wrote it all as compositions

seems to work

[next obsidian]

It suffices to check to see if it’s the identity on generators

But a generator x would map to another one y in the other one

And you defined the map the other way to send y back to x

So the composition fixed all the generators

So it’s identity

[delicate orchid]

yeah I can see that it's obviously true

but it's proving it via the universal property that's hard

[next obsidian]

I don’t think so?

[delicate orchid]

for you, maybe

[next obsidian]

You know from the universal property where the generators go

And that’s all you need to keep track of

I mean okay there’s some truth to what you’re saying there in “for you, maybe”

But I’m just saying this because I think you’re over complicating it probably

And I just say that because I did that a lot too

[delicate orchid]

oh I absolutely am

[next obsidian]

So when i say “it’s easy” I really just mean there’s a way to simplicity it so that it becomes really simple

[delicate orchid]

I have to go through everything super explicitly and in detail to convince myself that it's right

[next obsidian]

Right so let’s try it out really quickly

And I’ll try to illustrate how you could do it totally rigorously

But still really simply

[pastel cliff]

and this is why i never get anything done

[delicate orchid]

like for instance, I know that a group homomorphism is uniquely determined by the image of the generators

but I have no clue how that applies in a universal property situation where it has to be to an arbitrary group

[next obsidian]

Right so

If you have a set S do you believe F(S) is generated by the singleton words on S?

It’s kinda defined so that’s true

[delicate orchid]

yeah that's kind of the definition

[next obsidian]

Okay so now suppose you had two isomorphic sets S and T

Or like same cardianlity lol

So you have a bijection f:S -> T

[delicate orchid]

yup

[next obsidian]

So you can define a map call it say, g:S -> F(T)

Sending s to the word f(s)

As a singleton word

[delicate orchid]

yeah that makes sense, it's just the composition of f with the inclusion map from T into F(T)

[next obsidian]

Yee

So now this induces a map g’:F(S) -> F(T)

And the commuting triangle says

g’(s) = f(s)

Where these are now both considered as singleton words

Sorry, f(s) is a singleton word

s is a singleton word

[delicate orchid]

ok yeah, and the triangle commutes because it satisfies the universal product right?

[next obsidian]

Yup

That’s the universal property

Okay so similarly

[delicate orchid]

sorry, I meant property not product

[next obsidian]

We have a function h:T -> F(S) sending t to f^-1(t) as a singleton word

This induces a map h’:F(T) -> F(S) with h’(t) = f^-1(t)

Same thing as before

[delicate orchid]

ah and simarly this introduces another map

[next obsidian]

Yeah

So now compose the two

[delicate orchid]

yes, this is sounding a lot like what I did just with fewer arrows

[next obsidian]

we see that h’•g’(s) = h’(f(s)) = f^-1(f(s)) = s

[delicate orchid]

compose the two and you get the identity map

[next obsidian]

And likewise

Yee

[delicate orchid]

yeah that's pretty much what I did

[next obsidian]

So you could do this with a few more arrows without using explicit elements

You have to use uniqueness

[delicate orchid]

that's what I did

[next obsidian]

Of the map in the universal property

But I think it’s fine to use explicit constructions when you have them

[delicate orchid]

I didn't realise I could get away with looking at specific elements

[next obsidian]

It’s nice to be able to do both, but you kind of need to know when you pick your battles

[delicate orchid]

especially since we haven't technically shown that the free group is generated by the singleton words

it obviously is (proof by "Look at it")

[next obsidian]

Well this follows like

By construction I feel idk hahaha

When you show the free group exists

I think you should show it’s generated by singleton words

You’re kinda missing the point without that lmfao

Actually the universal property shows it too

The uniqueness bit

Implies it

If they didn’t generate, there’s have to be some way to get non-unique maps because the map won’t be determined by the singletons

But like whatever this is a stupid stupid way to look at it

[delicate orchid]

maybe we did actually, I just don't think we called them singleton words

[next obsidian]

Yeah I didn’t know what to call them lol

[delicate orchid]

I think we just used a generating set

[next obsidian]

Lol

But anyway yeh

Arguing via universal property is good but

If you know what the maps do to elements

You should try to use that

This will become so important to prove things with the tensor product

[delicate orchid]

I probably will once I get more comfortable with the universal property as a concept - it's very good practice

[next obsidian]

Yeah I mean you should try to get used to using both

Really what’s going on is

The free group thing is a functor

From sets to groups

And for any functor, an invertible map gets sent to an invertible map

You could try to show that

Show that if you have a map of sets S -> T

It induces a map F(S) -> F(T)

We kinda did that

[delicate orchid]

yeah that sounds like what we did, and then also proved that if S -> T is bijective so is F(S) -> F(T)

[next obsidian]

And then show if you have a composition S -> T -> U

You get both this diagram

F(S) -> F(T) -> F(U)

That’s like doing the composition of the maps that are induced

But also you have a direct path

F(S) -> F(U)

By looking at the composition S -> U

And the map that induces

You wanna show that those end up being the same

[next obsidian]

This

[next obsidian]

And this

[delicate orchid]

oh so the functor preserves composition basically

[next obsidian]

That’s exactly what it is yeah

[delicate orchid]

nice

[next obsidian]

And uhhh

From this compose with composition thing

It ends up being derivable that the identity S -> S

Gets sent to the identity F(S) -> F(S)

But actually you can show that explicitly

But from here you can see why an invertible map S -> T

Gets sent to an invertible map F(S) -> F(T)

Yeah?

[delicate orchid]

yup

[next obsidian]

The inverse is given by the induced map from the inverse T -> S

[delicate orchid]

yeah exactly

[next obsidian]

Also the preserving composition bit

Requires the uniqueness

So you’ve actually pretty much already done this

Oh well actually there’s two ways lol

You can do it element wise

[delicate orchid]

who would've thought not looking at specific elements would lead to something more general 🤪

[next obsidian]

That’s like my proof of the fact you were doing

I manually showed that the inverse maps induced inverse maps on the free groups

But if you follow it, you see I really will have proved the composition thing

You could alternatively do it abstractly using arrows and diagrams like you did

[delicate orchid]

I like the arrows just follow the arrows they will never lead you astray

[next obsidian]

So true

[delicate orchid]

unless you draw one backwards and don't notice for an hour

[next obsidian]

Oof

[barren sierra]

Ok so suppose we have

P = |1 1|
    |0 1|

We have that B = PAP^-1 and C = PBP^-1 and A = PCP^-1. We also have A^2 = B^2 = C^2 = -I and AB = C. This is promising. However I'm not sure how to show this action by conjugation is well defined. Furthermore this set of {+-A, +-B, +-C} is almost the quaternion subgroup group, except you're missing +-I so I don't think I can use normality here to show this action is well defined. So I'm stuck.

[barren sierra]

P doesn't commute and neither does A so I can't get closure like that

[proud bear]

did you figure this out?

[barren sierra]

figure what out?

the question? @proud bear

no

gave up and started working at some other problem

[delicate orchid]

I'm not entirely sure why you're looking at relations between elements of the set - it doesn't have to be a group it just has to be closed under conjugation by elements in G'

[proud bear]

^

for any M in G', MAM^-1 has to +-A, +-B or +-C. if it is +-I, then you can multiply on the left by M^-1 and the right by M to get that A=+-I

[barren sierra]

well I was hoping to use the relations to see closure

[delicate orchid]

I see

just check all the elements there can't be that many

[barren sierra]

I want something more elegant tho

[delicate orchid]

seems to be 42 elements actually that is quite a lot

[proud bear]

i think you actually can use normality to show the action is well defined. when you conjugate elements of {+-A, +-B, +-C} by elements of G', you'll get some element of {+-I, +-A, +- B, +-C} (because {+-A, +-B, +-C} is a subset of the quaternion subgroup which you said was normal in G'). MXM^-1 does not equal +-I for any X in {+-A, +-B, +-C} (if it did that would imply that X= +-I, which can't happen). so {+-A, +-B, +-C} is closed under conjugation

[barren sierra]

ohhhhhhhhhhhhhhhh

I was close

[barren sierra]

48

[delicate orchid]

oh yeah that's what 6*8 is

have I mentioned how much I fucking hate 6 times 8?

it's like 7 times 8 if it was even more nonsensical and annoying

same goes for 6 times 7

[next obsidian]

Mf didn’t memorize his times tables in elementary school

🤣🤣

[delicate orchid]

we're in abstract algebra please call them elements of ideals in Z thanks

it makes me sound less stupid

also check pins in this channel for another "wew lads times tables" moment

[next obsidian]

Lmfao

[barren sierra]

idk man

7 * 8 is more annoying

[delicate orchid]

a controversial statement

[barren sierra]

I feel like your opinion is more controversial

[delicate orchid]

7*8 is just 49+7 but like 6*7 is 49 MINUS 7?!

hueyyyhhh whaaaa??

[lavish nexus]

feels like it should be deg(f)! or something like that

η permutes the roots

and there are deg(f) roots

[sturdy marsh]

yeah this is wrong

set F = F'

and let η = identity map

then the number of extensions of η is the size of the galois group

the size of the galois group doesnt agree with the degree of the polynomial in general

[lavish nexus]

Also if f is not irreducible like f = g*h both irreducible then it should be deg(g)!*deg(h)! right?

[sturdy marsh]

no you cant write down a formula involving just the degree

[lavish nexus]

Ok I can’t rely on these notes

ty

[hollow parrot]

Is dummit and foote a decent abstract algebra book

[sturdy marsh]

yes

[lavish nexus]

Yes

[sturdy marsh]

it's very good

[hollow parrot]

Thanks

[next obsidian]

What are these notes from?

[sturdy marsh]

EGA

elements de garbage algebrique

[next obsidian]

Oof

[prisma shuttle]

hey guys

for this problem

to show that I was a non-maximal ideal I used the counterexample as the ideal generated by I and (1,0,1,0,...). My mentor asked me bonus question, which is whether this ideal is maximal

i am pretty sure the answer is no and i think the proof involves smth with the axiom of choice, but i am not too sure about the details

[lavish nexus]

If you mod it out

doesn’t look like a field

[prisma shuttle]

wait can u explain a littel more

so u mod out by (1,0,1,0,...)?

[lavish nexus]

D/I is (Z/2)^n n < inf?

wait no

[next obsidian]

It shouldn’t be

[lavish nexus]

no I read I wrong

[next obsidian]

Consider the following

Oh hmmmm

Okay what about this one then

Add in this vector

(0,0,0,1,0,0,0,1,0,0,0,1,…)

Then you shouldn’t be able to get the identity element (1,1,…) inside of that ideal

Because you basically have to fill in the k-th spot over all k which are 2 mod 4

[prisma shuttle]

hm i don't think i understand

how does this prove that the new ideal is nonmaximal

[next obsidian]

I created a proper ideal containing it

The same way you took your ideal plus the element (1,0,1,0,…) to show it wasn’t maximal

[prisma shuttle]

how do u know that this type of element isn't in the ideal generated though

[next obsidian]

Well it’s almost the same argument you’d use to show that the ideal generated by I and (1,0,1,0,…) is a proper ideal

Write out the finite sum, and you’ll see that in order to sum to (1,1,…) you’d need to produce something which has infinitely many 1s inside of the k-th spots for k = 2 mod 4

All you can use are

(1,0,1,0,…)

Which doesn’t touch the spots where k = 0,2 mod 4

The element

(0,0,0,1,0,0,0,1,…)

Which doesn’t touch the spots where k = 2 mod 4

And then a finite number of elements in I, but each of those only has finite many non-zero entires

And we’d them to sum to something with infinitely many non-zero entries

[prisma shuttle]

oh i see

thats v big brain

[next obsidian]

[prisma shuttle]

do u know if there is a way to like continue this

using the axiom of choice

[next obsidian]

Yeah you could just keep going

I don’t think you need choice

[prisma shuttle]

to show that we can never generate a maximal ideal

[next obsidian]

Just do it inductively

You could do it so like you add in

(0,0,0,0,0,0,0,1,…)

So like just kinda cutting in half

Then you have an issue for k = 4 mod 8

Then go again and it’s for k = 8 mod 16

Blah blah blah

Just take the idea I used and just keep going further and further

[prisma shuttle]

ok thx so much

[next obsidian]

[next obsidian]

So true brofibration

[south temple]

Hey guys, I am trying to determine for which coefficients the polynomial
$$
f = x^2 + ay^2 + 2by + c
$$
is irreducible in $\mathbb{C}[x,y]$. I was wondering if I'm on the right track...

What I've tried doing so far is using Gauss' Lemma: We have a UFD $\mathbb{C}[y]$, its fraction field $\mathbb{C}(y)$, and the rings $\mathbb{C}(y)[x]$ and $\mathbb{C}[y][x] = \mathbb{C}[x,y]$. The lemma tells us that if $f$ is reducible in $\mathbb{C}(y)[x]$, then $f$ is reducible in $\mathbb{C}[x,y]$. We can see from our polynomial that $f$ is a difference of squares in $\mathbb{C}(y)[x]$, hence $ay^2 + 2by + c$ must be a perfect square.

[cloud walrus]

vov&sons

[south temple]

from there I was just gonna complete the square and solve for when the term outside the square is zero...

[prisma shuttle]

can someone explain to me why my sol for this problem is incorrect

my professor said its because i assumed that all chains are countable (that they can be indexed by a subset of the natural numbers), which is not always true

i understand this but now idk how to repair my proof (or if its even repariable)

[next obsidian]

You do exactly the same thing but with arbitrary chains

You just have to change how your index it

Just say you have a chain {I_alpha}_alpha in A an arbitrary chain with I_alpha < I_beta for all alpha < beta

[prisma shuttle]

why does this work and not my proof

isn't it basically the same thing

[next obsidian]

Yes

They’re splitting hairs

[prisma shuttle]

the example he gave me for which it didnt work was a polynomial ring with infinitely many variables

[next obsidian]

It’s because you could have uncountable chains

And your proof is indexing using the natural numbers

If you just changed it to be an arbitrary indexing set this works

The idea works totally fine, it’s just that as written you couldn’t apply Zorn’s lemma because Zorn’s lemma requires you to show even uncountable chains have an upper bound

[sturdy marsh]

also, make sure that your indexing set is actually a set (this isnt a problem here, but you will probably use zorn a lot later)

[next obsidian]

Well that’s implicit in the definition of a chain right?

[sturdy marsh]

yes

[prisma shuttle]

how do i make sure of this

[next obsidian]

It isn’t “make sure”

As much as it will always be

[prisma shuttle]

is it ok if I just write let the indexing be {I_k} in A such that for all \alpha, \beta, I_{\alpha} < I_{\beta} iff \alpha < \beta

[sturdy marsh]

in this case your indexing set is a subset of a power set of a set, so youre okay

[next obsidian]

Brofib what are you talking about

Zorn’s if you’re applying it to sets

You’d never have an issue with the indexing set not being a set

You have to show every totally ordered subset has an upper bound

So you can always index it with a set

If you’re trying to do global choice level of Zorn’s it becomes an issue sure

[sturdy marsh]

alright here's a fake "proof"

Look at the poset (this is where it fails ig) of all sets. Every chain has an upper bound (union). So there's a maximal element.

[next obsidian]

Yeah but that’s not about the indexing set

The issue is you’re trying to apply Zorn’s to something that isn’t a set

[sturdy marsh]

oh see what you mean

yeah that's not what I meant

[next obsidian]

If you’re trying to use global Zorn’s then you have to handle class sized chains

Okay

You made it sound like you have to provide an argument for why you only have to handle set-sized chains

When using Zorn’s lemma

[sturdy marsh]

yeah I worded it incorrectly

[next obsidian]

Okay

[next obsidian]

Yeah

Sorry for the confusing stuff we were writing

Don’t worry about that, basically what Brofib ration is saying is that to apply Zorn’s you need a partially ordered SET

So you can’t use Zorn’s on the class of all sets

That’s like a partially ordered class

But in this case your poset is the set of ideals of a ring which is obviously a set

[barren sierra]

protip

8 does not divide 36

this would have been good for me to realize

when trying to find 2-Sylow subgroups of S3 x S3

💀

[proud bear]

[barren sierra]

it really felt like 8 should divide 36

[sturdy marsh]

you know what's funny

4999 is prime

[next obsidian]

,w factorize 4999

[cloud walrus]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by Chmonkey 2.0#9767

[next obsidian]

No

https://tenor.com/view/lunadials-topi-dude-from-breaking-gif-21214320

[barren sierra]

[sturdy marsh]

[rustic crown]

woah

4999 prime!!?

[sturdy marsh]

wrong react det

[rustic crown]

but its a cute prime

123456789011 is also prime

[next obsidian]

,w factorize 123456789011

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Chmonkey 2.0#9767

[next obsidian]

No…

https://tenor.com/view/lunadials-topi-dude-from-breaking-gif-21214320

[rustic crown]

[next obsidian]

Did you know that ,w 987654321/123456789

,w 987654321/123456789

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Chmonkey 2.0#9767

[next obsidian]

Good approximation for 8

[rustic crown]

[tough raven]

Abstractest of algebra

[spice whale]

are finite fields algebraically closed?

[proud bear]

no

if a_1,...,a_q are all the elements of the finite field, then (x-a_1)...(x-a_q)+1 has no roots

[spice whale]

ah

[lethal dune]

@next obsidian u didn't record the talk right?

[next obsidian]

I did not

[lethal dune]

sad

[waxen hedge]

Rip

I wanted to watch it

[lethal dune]

ya, bad timezone

[hidden haven]

Imagine not staying up till 3 am just to watch

[lethal dune]

had to wake up early for class, otherwise I would have

[hidden haven]

Imagine attending class over

[lethal dune]

is this new? don't remember seeing this before

[hidden haven]

This is the shit the pepe slots are being used for

[lethal dune]

is this simonkey

[rustic crown]

this is eevee

[delicate orchid]

prove it

[waxen hedge]

What did you talk about ? In a few sentences, don't bother

[hidden haven]

see #events

Direct limits pretty much

[rigid cave]

basically showed us their chmonkey plushie collection

[runic hemlock]

did you know 30,000,000 plus or minus 1 are both prime

[lethal dune]

nice twin

reminds me of this

[rustic crown]

🙈

[rustic crown]

,w factor (3*10^7)^2 - 1

[cloud walrus]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by det#7067

[lethal cipher]

Hey y'all, not too sure where to put this. Since Galois theory has quite a bit of abstract algebra involved, I figured this was good enough. Anyways, I am having a really hard time figuring out how to approach this and I can really use some help.

[rustic crown]

have you proven that S_n acts on F[x1, ..., xn]?

[lethal cipher]

what do you mean by acts?

[rustic crown]

oh in the sense of a group action.

what we need here is that tau(sigma(f)) = (tau * sigma)(f)

pick some arbitrary tau in S_n and just see what would happen when you act it on those two expressions (that sum and product i mean)

[hidden haven]

Galois theory is abstract algebra though

[lethal cipher]

How would I do that on a general polynomial f?

[rustic crown]

that's the thing, you don't need to actually know anything about f. just compare what happens to the expressions before and after the acting by tau

(recall that a polynomial g in F[x1, ..., xn] is called symmetric if and only if tau(g) = g for every tau in S_n)

[lethal cipher]

Well, it seems like what we are trying to get to is that tau of the product of sigma f is equal to the product of tau*sigma f.

[rustic crown]

yep, that would be the first nice thing to expect. do you see why this is the case?

so abstracting out what you said. we want to show that
tau(g * h) = tau(g) * tau(h)

for arbitrary g, h in F[x1, ..., xn]

[lethal cipher]

I'm having a hard time putting it into words, but I do believe I understand why this is, yeah.

[rustic crown]

okie, so let's now worry about manipulating the product of tau*sigma f

we want to show that, the product expression in the problem is symmetric

so this should turn out to be product of sigma f

if you're unsure at any point of time, try out an example!

pick some value of n, say n = 3 and pick some specific f

[lethal cipher]

Well, it sounds like what we are gathering is that tau*sigma_i is itself a permutation, and we want to show that no permutation is deleted by composing tau.

[rustic crown]

putting it into a statement, we need to show that for permutation alpha and beta
tau * alpha = tau * beta <=> alpha = beta

(but we've skipped a step btw...
we should make sure that tau(sigma(f)) = (tau * sigma)(f))

[lethal cipher]

Well that one honestly isn't too hard

[rustic crown]

yep!

[lethal cipher]

Actually isn't the group operation of Sn function composition?

I am not sure why we need to do that extra step

[rustic crown]

yea but recall that f here is something in F[x1, ..., xn]

it's not something in {1, 2, ..., n} that we can directly appeal to the definition of function composition

sigma(f) is defined in a different way.... by permuting indeterminates in f

[lethal cipher]

Well, it kinda is right? because {1,2,...,n} has an easy bijection to the variables {x1,...,xn}

[rustic crown]

it sure is... (there is a thing called universal property of polynomial rings, and with that it's exactly that!)

(but ignore that for now)

[lethal cipher]

Respectfully ignored

[rustic crown]

so do you see what exactly happens when we apply tau to
sigma_1(f) * .... * sigma_N(f)?

(N here is n!)

(sorry about my sentences, i often change what i'm trying to say in the middle >.<)

[lethal cipher]

Ummm, well tau is acting on the whole structure, but that's the same as tau acting on each sigma_i(f) individually.

Because really you are just acting on n! different objects

[rustic crown]

yep, so by composing tau with {sigma_1, ..., sigma_n!} we just permute these... no permutation will be deleted, and no two will collide and become a single permutation

[lethal cipher]

Oh duh! tau^-1* tau*alpha=tau^-1* tau* beta

Got to love not thinking about the basics of group theory xp

Okay, so here's my stab at a proof. Let $f\in F[x_1,\dots,x_n]$ and $\sigma \in S_n$. Now apply a permutation $\tau$ onto the product and sum of $\sigma\star f$, this is equivalent to acting on each $\sigma\star f$ individually. Since $\tau\circ \sigma_1=\tau\circ \sigma_2 \Rightarrow \sigma_1=\sigma_2$, for every $\sigma\in S_n$, $\tau\circ \sigma$ is a unique permutation. So the product and sum collects every permutation of f.

[cloud walrus]

dackid

[rustic crown]

rest is commutativity under multiplication

[lethal cipher]

And that comes naturally since F[x_1,...,x_n] is a field

[rustic crown]

(it's not a field, but sure is a commutative ring!)

[lethal cipher]

F is a field, so shouldn't F[x1,...,xn]?

oh, inverses

Nevermind

[rustic crown]

yea exactly >.<

[lethal cipher]

we'd need the rational functions for that

Okay cool! That wasn't nearly as bad as I thought!

I am going to have another question soon, because the next one looks a bit tough to grasp

If you want, feel free to take a look at it while I finish number 7

Oh, this is easy!

Since C collects all non-identity permutations of B, BC is the product of all permutations of B, which we just proved is symmetric

for (b) A/B(BC)=AC is a product of symmetric polynomials, which is symmetric

And since AC and BC can be represented as elementary symmetric polynomials, then so do AC/BC

[rustic crown]

just one tiiinnyyy error

A/B isn't a symmetric polynomial

it's a symmetric rational function

[lethal cipher]

you are right, it is a symmetric rational function

And the product of symmetric rational functions is of course symmetric.

This one looked wayy scarier than it actually was

[lethal cipher]

do any of you understand what that sum is? I am a bit confused on it

[waxen hedge]

Maybe $x_1^2x_2 + x_2^2x_3 + x_3^2x_1 + x_1^2x_3 + x_2^2x_1$ ?

If you are indeed reading about symmetric polynomials

[cloud walrus]

Adrien

[sinful mirage]

I have a confusion related to characters. how can one deduce this?

I thought a character is by definition the trace of the representative matrix

i.e. tr(pi). why tr e(pi(H))?

[fickle tapir]

abstract algebra

[delicate orchid]

abstract algebra

damn he only got as far as real complex analysis

rip king

[next obsidian]

It’s your job to finish what he started

[spice whale]

what are some algebraically closed fields other than C?

[rustic crown]

field of algebraic numbers (over Q)

[chilly ocean]

{0}

[spice whale]

oh...

[hidden haven]

Algebraic closure of any field

[spice whale]

wait

[chilly ocean]

my answer is a meme

[spice whale]

i know

[hidden haven]

What's the characteristic of the 0 field?

[spice whale]

it's not a field

[rustic crown]

1

[hidden haven]

based

[chilly ocean]

b-but 1 = 0!! it's a field!!!

[delicate orchid]

it is 0

it's called F_0 for a reason you silly billies

[hidden haven]

Every non zero element has an inverse

What's the issue

[chilly ocean]

you mean F_un?

[delicate orchid]

france detected ⚠️ france detected 🚨

[rustic crown]

Fun Things are Fun >.<

[hidden haven]

Big disagree

[rustic crown]

watch K-On! lol

[hidden haven]

No

Big disagree

[delicate orchid]

I'm going to agree actually

[hidden haven]

Is it a cartoon

[delicate orchid]

_ _

[rustic crown]

no a weeb anime for degenerates

[hidden haven]

Isn't that stuff for kids det

[chilly ocean]

[rustic crown]

i'm a kid

(i saw that)

[chilly ocean]

[hidden haven]

🪚

[spice whale]

so

[delicate orchid]

det is actually a primary schooler (I ardly knew 'er)

[hidden haven]

Having seen det irl

[chilly ocean]

i am learn noether normalization

#groups-rings-fields

[hidden haven]

That just seems weird

[delicate orchid]

wow abstract alg meet up and I wasn't invited

[hidden haven]

[delicate orchid]

I see how it is

[hidden haven]

F

[spice whale]

would the algebraic closure of a finite field be finite?

wait no

[chilly ocean]

no

[hidden haven]

No

[rustic crown]

nope

[spice whale]

nvm

[chilly ocean]

no finite field is algebraically closed

[hidden haven]

[chilly ocean]

its countable

[rustic crown]

{0}

[chilly ocean]

...except for that one

[spice whale]

I'm looking for test cases for linearity of the arithmetic derivative

[hidden haven]

Stop moving goalposts

[chilly ocean]

whats the arithmetic derivative

i don't feel like doing my homework right now

[spice whale]

here

D_K(z) is the arithmetic derivative

for l(p) = 1/p, it has the property D(p) = 1 for prime p

prime p referring to a prime in the base UFD

[rustic crown]

it's defined all multiplicatively... why would one expect it to be linear?

[spice whale]

it's linear over C[X]

in which it acts as a normal derivative over polynomials

which is why I'm investigating whether it's linear for polynomial rings over other algebraically closed fields

[rustic crown]

so you define l(x-a) = 1/(x-a)?

[spice whale]

yes

[rustic crown]

wait so D(x^n) = x^n * 1/x = x^(n-1)

[spice whale]

no

[rustic crown]

we probably count primes with multiplicity then

[spice whale]

n*x^(n-1)

like the normal derivative

if you define l(p) = 1 then L_K(z) is a prime factor counter

[rustic crown]

with multiplicity right?

[spice whale]

Ω(z)

wdym with multiplicity

[rustic crown]

like L(p^2) = 2?

[spice whale]

yes

wait

no

wait

yes

[rustic crown]

[spice whale]

sorry

getting confused

yes

[rustic crown]

lol issokie >.<

[spice whale]

for l(p) = 1

L(p²) = 2

in that case L(z) is just the big omega function

[rustic crown]

right, so for alg closed field k, the thing for k[x] would be the normal derivative

[spice whale]

yes

[rustic crown]

assuming again l(x-a) = 1/(x-a)

[spice whale]

yeah

i mean you can just say l(p) = 1/p
because if k isn't algebraically closed, then not all irreducibles will be linear polynomials

[rustic crown]

right

but over Q and stuff, it will be bad

[spice whale]

wait actually i need to think abt that one
it definitely is over C[X]

[rustic crown]

D(x^2 + 1) = 1 as well then

[spice whale]

yep

[rustic crown]

the proof is same actually

[spice whale]

wait

so it will be linear over K[X] where K is algebraically closed?

[rustic crown]

yea looks like it

just one thing tho

your definition is a bit weird

l is a function from the set of primes in the UFD to its fraction field

not to K

[spice whale]

oh yeah

well

uh

[rustic crown]

oh wait

[spice whale]

just define it over the fraction field

then

it can always be extended (i think)

[rustic crown]

so yea

[spice whale]

yeah

[cloud walrus]

det

[spice whale]

yeah

[rustic crown]

but linearity might fail over say Q

D(x^2) = 2x
D(1) = 0

[spice whale]

it does

[rustic crown]

but D(x^2+1) = 1

[spice whale]

I'm wondering

whether it's linear on any other UFDs

[rustic crown]

[spice whale]

definitely not on integers or gaussian integers

because it's a weird property

[rustic crown]

yea

D(n) = nD(1) = 0

[spice whale]

only works for the trivial derivative

l(p) = 0

[rustic crown]

btw one more thing... it will work for K only if you choose your primes to be monic linear

[spice whale]

yes

[rustic crown]

2x is also a prime in C[x]

[spice whale]

no

well

i guess

but

it's all the same up to isomorphism

[rustic crown]

[spice whale]

sorry

nvm

[rustic crown]

D(x) = 1 would mean D(2x) = 2

but it will also be 1 since it's a prime on its own

[spice whale]

yeah

this also applies to

any other UFD with units other than 1

[rustic crown]

-1 is going to be a unit unless char = 2 (where it's just 1)

[spice whale]

for naturals

the only unit is 1

idk what a characteristic is yet

[delicate orchid]

1+1+1+1+1+... = 0

characteristic is the lowest number of 1s required

[hidden haven]

It's the non negative generator of the kernel of the characteristic map 😌

[spice whale]

wait

[hidden haven]

Characteristic map for a ring R is the unique map ℤ → R

[spice whale]

so like C, Z, Q[X,Y] etc have characteristic infinity?

[delicate orchid]

you'd think so but we say they have char 0

[spice whale]

makes sense

[delicate orchid]

mainly due to moldi's definition

which is why moldi's defition is better

but cmon who thinks about it like that

[hidden haven]

Yes that is moldi's definition I came up with it myself

[delicate orchid]

I will cite you

like I will put a screenshot of that discord post if I need to cite what a characteristic is

[delicate orchid]

yeah I saw the words "unique map"

[spice whale]

iso theorem 😳

[gusty halo]

Z is initial in the category of rings

[hidden haven]

Universal property of ℤ

[spice whale]

I'm lost in a sea of words
maybe I'll understand it all in a year's time

[gusty halo]

just read riehl then everything will make sense

[spice whale]

actually i probably will depending on how fast i go through pugh analysis

[hidden haven]

Imagine reading Riehl instead of the OG mac lane

[spice whale]

is riehl alg topology?

or just algebra

[hidden haven]

All the words we said amount to "for any ring R, there exists a unique map ℤ → R"

[hidden haven]

Cat theory

[gusty halo]

tbh the reason i preferred riehl was because its shorter lol

[hidden haven]

It skips so much

[gusty halo]

also im always a bit conflicted about reading old things

like maybe there were new developments that should definitely be mentioned

i guess most of the time it doesnt matter

but it still bugs me

[hidden haven]

Riehl is like mac lane but rearranged and shortened with added examples

[gusty halo]

i do feel like she could have put abelian categories in there

[chilly ocean]

Written so well though

[gusty halo]

but other than that i think the content is good

[hidden haven]

Real men read badly written books

[gusty halo]

i also think that riehl is like the second ever book i really worked through

[delicate orchid]

real men write their own books and then read them

[gusty halo]

usually i dont find the time to properly and carefully read everything

and just look for the things i need

[chilly ocean]

Same

[spice whale]

I've got a pdf of knapp basic algebra

[hidden haven]

Mac lane took my first time

[gusty halo]

there are way too many interesting books to read each of them completely and carefully

[spice whale]

i think it introduces linalg independently

[gusty halo]

you already finished mac lane?

im still on the kan extensions chapter in riehl

but rn i gotta prepare for exams

and kan extensions arent exactly needed there

[hidden haven]

Still have to finish Kan extensions lol

I have read till the coend formula for Kan extensions

Riehl doesn't even do ends and coends does it

[gusty halo]

after i finish riehl i planned to skim through mac lane and look at things that werent in riehl and that might be useful

[gusty halo]

no she doesnt

[hidden haven]

Does Riehl have horizontal composition of natural transformations?

[gusty halo]

yes ofc

[spice whale]

would it be more useful to look at alg topology or general cat theory (meow) after algebra

[gusty halo]

depends on what you want to do

[hidden haven]

Does it compose nat trans with only functors or does it horizontally compose any 2

Because I couldn't find it when I went through it lul

[gusty halo]

its on page 45

[hidden haven]

I only saw like ηF and Fη

Oh

[spice whale]

both seem like very interesting fields of maths
foundations and topology

[hidden haven]

I see

[gusty halo]

if you havent seen a lot of maths yet then i doubt category theory will make much sense or even be useful

[spice whale]

ok fair

so

[hidden haven]

Have you seen point set topology though?

[gusty halo]

topology is pretty basic though, doesnt need many requirements

point set that is

like munkres

[spice whale]

is point set required for algebraic

or like

[gusty halo]

well the basics yes

[hidden haven]

Basics are

[gusty halo]

you need to be comfortable with continuous maps, product spaces, quotient spaces, compactness, connectedness, etc

[spice whale]

ok

I'll go through pugh, the two knapp books, rudin 2, then probably I'll check out some more general topology

[gusty halo]

i mean on the other hand you could literally just dive in and everytime somethings comes up you havent seen you look that up somewhere and try to understand it

top down as opposed to bottom up

[hidden haven]

I'd suggest point set topology then cat theory then alg top

[gusty halo]

i heard scholze learned AG that way before he heard of linear algebra lel

[spice whale]

hm

[hidden haven]

But cat thy can be done any time

[spice whale]

yeah

i have a plan
pugh, knapp 1, complex analysis, knapp 2, point set, cat theory, logic/model theory, alg top

probably

in the summer I'll probably do like 2 at the same time

interleave

[gusty halo]

as i discussed above, depending on how thorough you plan to be in working through this, that might take some years

[spice whale]

i have plenty of time

[gusty halo]

ok then nice

[spice whale]

besides I'm gonna have to revise it all in university anyway

[gusty halo]

i wish i had started in highschool

but oh well

it is what it is

[hidden haven]

I wish I had started in my previous life 🤧

[gusty halo]

i was actually really interested in numbers in kindergarten but when i went to school that kind of faded away. Only in uni i rediscovered my interest for maths

[delicate orchid]

I LOVE NUMBERS!

[spice whale]

NT

[hidden haven]

They should make nom theory 2 with less numbers

[spice whale]

modular arithmetic moment

[gusty halo]

number theory in F_2

[hidden haven]

😄

[prisma shuttle]

can someone explain how this shows that $\text{char} \mathbb{Q} = 0$?

[cloud walrus]

JustKeepRunning

[prisma shuttle]

this problem part b(p26 sec 3 chapter 7 d & f)

how does this show B

[sharp sonnet]

this looks like a typo

is this your issue? or what is?

(there should be an equals sign before the minus iota)

[prisma shuttle]

no i mean like why do you need the homomorphism and everything

can't you just note that adding 1 repeatedly will never get you to 0 in Q

[sharp sonnet]

sure

which is equivalent to the map Z -> Q being injective, which is shown here

[barren sierra]

https://proofwiki.org/wiki/Group_of_Order_105_has_Normal_Sylow_5-Subgroup_or_Normal_Sylow_7-Subgroup

I am lost at one part

why

I don't get that

[next obsidian]

Apply Lagrange

Only divisors of a prime are itself and 1

[barren sierra]

ah

[spice whale]

(and other units in UFDs other than the natural numbers)

[next obsidian]

This is N

I don’t care about that

[spice whale]

oh

yeah fine

[barren sierra]

this seems too easy lmao

(the identity comes from the fact that prime subgroups have trivial intersection)

[prisma shuttle]

hey guys for the set $\overline{E} = \frac{\mathbb{F}_2[x]}{(x^2+x+1)}$ i am tyring to prove that $\overline{E}^{\times}$ is isomorphic to a cyclic group of order 3

[cloud walrus]

JustKeepRunning

[prisma shuttle]

a solution i found online is below

can someone explain why x(x+1)=1 and (x+1)(x+1)=x

because i thought x(x+1)=xx+x=x+x=0 and (x+1)(x+1)=xx+2x+1=3x+1=x+1,$ but thats not what it says in the solution

[sharp sonnet]

what makes you think x(x+1) = x + x

its x^2 + x

(which is -1 = 1)

[pastel cliff]

is there such thing as "intervertible" or did this guy mean invertible

[next obsidian]

This is supposed to be invertible

[pastel cliff]

ok just double checking, ya never know

[south temple]

Hey guys, if I have a polynomial $f = ax^2 + by^2 + cxy + dx + ey$, is it possible to somehow "force" the $x^2$ term to have a nonzero coefficient?

[cloud walrus]

vov&sons

[south temple]

using an affine transformation

I'm trying to transform this polynomial to the polynomial $x^2 + by^2 + ey$ using affine transformations, but this requires me to cancel out the $a$ coefficient

[cloud walrus]

vov&sons

[south temple]

my prof said that this is possible, but I'm just super confused on how one can guarantee a nonzero coeff for x^2 at ALL times

[prisma shuttle]

i think you can use a rotation?

[south temple]

can you elaborate?

[prisma shuttle]

oh wait no thats for xy

[tribal moss]

If b != 0, then interchange x and y. If a=b=0, then hopefully c!=0. In that case switch coordinates to u=x+y, v=x-y, and then u² and v² will both have nonzero coefficients.

[south temple]

right, but what if a=b=c=0?

oh maybe if a=b=c=0, then we don't have a conic section anymore so we don't really care about that case

I should have mentioned the whole purpose of this is to find all reducible conic sections in C[x,y]

[tribal moss]

Yes, in the a=b=c=0 case you can't get it to the form you wanted, because with x^2+by^2+ey there's a straight line (namely the y-axis) whose image is [0,infty) whereas there is no such straight line for dx+ey+f.

[wary ginkgo]

weird question, but would Algebraic Graph Theory be under Abstract Algebra? if not, then what field of mathematics would it be defined under?

[hidden haven]

#discrete-math or #combinatorial-structures perhaps

[rigid cave]

Riehl wants to construct the orbit category associated to a group G. She says "its objects are subgroups H of G, which we identify with the left G-set G/H of left cosets of H." I don't understand the G-set stuff here. On one hand, the category is supposed to have subrgoups as objects but on the other hand it seems like the objects are sets G/H

I don't really know what this means sorry

so H is isomorphic to G/H?

oh okay I see lmao, thank you so much!

[woeful flint]

I'm trying to show that $\mathbb{Z}/2\mathbb{Z}$ is not a flat $\mathbb{Z}/36\mathbb{Z}$-module. In fact I'm not even entirely sure if this is the case. Any hints as to how I can go about doing this?

[cloud walrus]

Kraft Macaroni

[hidden haven]

Try to exploit the torsion

I think it is generally true that modules with torsion can't be flat

||consider the map R → R which is multiplication by a non zero divisor annihilating the given module||

[barren sierra]

So I've shown that this action is well defined but I'm not sure how to show that it describes a surjective homomorphism. I know that G' acting on X does give a homomorphism but showing that it is surjective I'm not sure how to do

[hidden haven]

Wait this doesn't work here because 2 is a zero divisor torsion doesn't make sense for this

[runic hemlock]

Consider multiplication by 2: Z/18-> Z/36, as a map of Z/36-modules. When tensored by Z/2 it stops being injective

[tough raven]

%\DeclareMathOperator{\ch}{ch}

For a field extension $L | K$ and $α$ in $L$, we can generalise the trace $Tr_{L|K}$ and norm $N_{L|K}$ of $α$ to the characteristic polynomial $p_{ch}(α)$ again by defining it to be that of the K-linear multiplication map. $p_{ch}(α)$ is in fact $p(X)^{ [L:K(α)] }$ where $p$ is the minimal polynomial of $α$.

Does the result of transitivity of trace and norm for field extensions $E | L | K$ generalise to the characteristic polynomial?

[cloud walrus]

Raghuram

[proud bear]

@barren sierra so if we label {+-A} as 1, {+-B} as 2 and {+-C} as 3 then the matrix P you found here gets mapped to (1 2 3). If you can find some other matrix Q such that B=QAQ^-1 and A=QBQ^-1 then this will get mapped to (1 2) (C=QCQ^-1 because AB=C). Then since every element of Sym(X)=S_3 is a product of (1 2 3) and (1 2), phi is surjective. For example (1 3 2)=(1 2 3)(1 2 3) so P^2 is mapped to (1 3 2) by phi and (1 3)=(1 2 3)(1 2) so PQ is mapped to (1 3)

[tough raven]

Please tell me if I should take this to #advanced-number-theory or somewhere else.

[barren sierra]

@proud bear ok that makes sense.

[chilly ocean]

What is the point of ideal quotients

I would not mind a geometric interpretation but I want an algebraic one, along with some places its used in

[next obsidian]

Wdym by ideal quotients

[tough raven]

\newcommand*{\chp}[1]{p^C_{#1}}

Call the characteristic polynomial $p^C$ actually. For $α$ in $L$, we get $\chp{E|L}(α) = (X - α)^n$ where $n = [E : L]$, and $\chp{E|K}(α) = p_{\min}(α)^n$, which suggests $(X - α)$ is getting mapped to $p_{\min}(α)$, or irreducible polynomials over $L$ get mapped to the irreducible polynomial over $K$ they divide.

… which I guess I can prove.

[cloud walrus]

Raghuram

[chilly ocean]

for ideals a,b in ring R, (a:b):={r in R| rb subset of a

[next obsidian]

Ohhh okay

Give me a second I can give an example that’s useful

So these show up in a certain exact sequence