groups-rings-fields

[chilly radish]

You don't need a contrapositive here. Given an increasing chain of ideal, you want to somehow form a single ideal that you can use the f.g. assumption on

Then conclude from that the chain stabilises

[somber goblet]

can you somehow take union of this

or that's not always an ideal

[karmic moat]

prove this: if i have an increasing chain of ideals, then the union of all those ideals is again an ideal

[somber goblet]

oh wait a minute

do we need zorn for this 💀

[karmic moat]

no

[somber goblet]

i know this is true for finite chains obviously

im too tired for ts

[karmic moat]

it is fairly straightforward to prove for arbitrary chains

you just need to check that it satisfies the definition of an ideal

[somber goblet]

oh

im tired

[karmic moat]

If J_1 < J_2 < ... is an increasing chain of ideals, you could define J = \cup_i J_i. show that it's closed under sums and absorbs under multiplications

or you can do it whenever you're not tired

[somber goblet]

this might be enough rings for one day

im sorry

[karmic moat]

all good

[somber goblet]

i think im making progress

i think i understand why we care about notherian rings

[karmic moat]

yeah

one nice property is: every finitely generated module over a noetherian ring is a neotherian module (i.e., every submodule is finitely generated)

[somber goblet]

ooo i see

that is very nice

modules are getting more and more handy

[karmic moat]

if you care about geometry, Spec(A), the set of prime ideals of A, forms a topological space

if A is a Noetherian ring (every ascending chain of ideals terminates), then Spec A is a Noetherian topological space (every descending chain of subspaces terminates)

noetherian rings have finitely many minimal prime ideals, which correspond to irreducible components of Spec A

this is true of noetherian topological spaces in general

[somber goblet]

oh i see

since as you go up in ideals their closure shrinks? or something like that

[karmic moat]

yes, given an ideal I of A, you define V(I) = {p in Spec A such that I \subseteq p}

you give a topology on Spec A by defining closed sets to be of the form V(I)

V(-) is order reversing: if I < J, then V(J) < V(I)

[somber goblet]

i still have trouble wrapping my head around the points of Spec R being sets in R 😭😭

[karmic moat]

yeah i think a good way to first think about it is through the lens of classical algebraic geometry

[somber goblet]

makes sense within the roots of polynomials interpretation

[karmic moat]

for example, like C[x,y]

[somber goblet]

adding more polynomials = smaller set of shared roots

so I < J -> V(J) < V(I)

[karmic moat]

Hilbert's nullstellensatz says all maximal primes of C[x,y] are of the form (x-a, y-b) for a, b in C

so maximal ideals literally correspond to points in C^2

[somber goblet]

thats only for an ACF right?

[karmic moat]

yeah

if f(x,y) is an element of C[x,y], then f being in (x-a, y-b) is the same as saying f(a,b) = 0

[somber goblet]

yeah

prime ideals of k[x] correspond to elements of the algebraic closure of k

up to galois conjugation

which i think is beautiful

and is part of why i love commutative algebra

[chilly radish]

@somber goblet another useful condition is that R is noetherian iff any nonempty set of ideals has a maximal element.

[vocal pebble]

Also noetherian iff every prime idea is finitely generated

[swift root]

no because youve already constructed Xi and you want to construct Xi+1. If xi ∈ X \ X_i-1 then C(Xi ∪ { xi }) isnt necessarily different from Xi

[next obsidian]

I find proving the geometric version easier to do than showing directly finitely many minimal primes lol

[next obsidian]

When your ideas can’t be divided up any further

[next obsidian]

So prime ideals of k-bar[x] :^)

[azure cairn]

turnsintoa

[south patrol]

Ye lol

[balmy python]

hey sorry for bringing this back up but how would you sort smth like (124)(35)

as in how do you change the cycle shape when sorting into (1)(2)(3)(4)(5)

[tribal moss]

One convention would be to write each cycle with the lowest element first, then sort all the cycles according to their lowest elements.

In that case (124)(35) would already be in sorted form.

[balmy python]

i understand that but wouldn't I get something like (123)(45)?

wait i see what you mean

[tribal moss]

You don't renumber elements, that would make it a different permutation.

[balmy python]

yea

[tribal moss]

(wait a moment, that doesn't actually match the thing you're replying to. No, I have no idea there, sorry)

[balmy python]

😭

i'm a bit confused by the bubble sort method ngl i wan't to understand it a bit more

because if i had something like (124)(35) then by what he said i can take the inverse (142)(35), apply (24) and to obtain (124)(35) but it's not like i can really obtain (1)(2)(3)(4)(5) by applying transpositions?

ah i see

[tribal moss]

I think the most relevant point to make here is that bubblesort produces a proof the the neighbor-transpositions generate S_n.
But it is clearer if you work with two-line notation for the permutations, rather than cycle notation.

[balmy python]

wdym by neighbour-traspositions?

like (i,i+1)?

[tribal moss]

Yes.

[balmy python]

i see it's easier to understand with the two line notation

[tribal moss]

If you know how bubblesort works as a sorting algorithm, what it produces is, given any permutation sigma, a sequence of neighbor transpositions such that
[transpositions] o sigma = id

Thus the transpositions multiply together to sigma^-1, so the same transpositions multiplied together in the opposite order give sigma itself.

[balmy python]

yup

[copper kestrel]

i was talking with my professor yesterday about this and am still a bit confused

so since \alpha is algebraic over F(\beta), we know that there exists some polynomial f(x) \in F(\beta)[x] with coefficients in F(\beta) such that f(\alpha) = 0.

plugging in \alpha, we then can collect all terms that do not have beta in them, we will call that g_{\alpha}(x), and collect all terms that have beta in them, and factor it out. we then will call this term \beta • h_{\alpha}(x). Since \alpha is transcendental, we know that F(\alpha) is some field of quotients, so we can solve for beta to get

\beta = -g_{\alpha}(x) / h_{\alpha}

and then we have to somehow show that this is the root of some polynomial? does this sound like an okay method to do?

or at least im not sure how to reason that beta is the root of some polynomial

[rocky cloak]

I don't know about your suggested method, but if you just think about a polynomial in F[beta][x] evaluated at alpha as a polynomial in F[y, x] evaluated at y=beta, x=alpha you can just change the order in which you evaluate.

[copper kestrel]

oh

that seems much easier

[copper kestrel]

would you say that a polynomial in two variables is

$$\sum_{i,j}a_{ij}\alpha^i \beta^j$$ which is an element of $F(\beta)[x]$?

[cloud walrus]

tabby tabby

[copper kestrel]

or wait no i think beta and alpha should be flipped? i think?

[rocky cloak]

Well multiplication is commutative here

[copper kestrel]

bc we eventually want it to be in F(\alpha)

right

but we just want to show that when its an element of F(\alpha) we still have that it equals 0 making \beta algebraic over F(\alpha)

[balmy python]

can someone explain why the conjugacy class of 5 cycles in S5 splits into two conjugacy classes in A5?

[rocky cloak]

Think about the how S5 and A5 act on 5 cycles by conjugation.

Take a 5 cycle sigma and consider its orbit. Then you know the size of the orbit is |S5|/|C(sigma)|. Now for the same reason the size of the orbit by the action of A5 is
|A5|/|C(sigma) intersect A5|.

So if C(sigma) is contained in A5 the orbit would have exactly half the size

[balmy python]

is C the stabiliser?

[elfin wraith]

I think it’s best to just do this. Pick two 5 cycles in S_5 and work out how they’re conjugate.

You can also do some orbit stabiliser stuff to see this more abstractly

Ah Jagr beat me to it

[rocky cloak]

Centralizer, so yes stabilizer by conjugation

[elfin wraith]

But yeah like you can work this out very explicitly and this is maybe good to do (but what Jagr said is more generalisable, I just think there’s good intuition to be gained by working it out explicitly)

[tribal moss]

Yeah, there are only 5 different permutations p such that p^-1(12345)p = (12354) -- since the value of p(1) fixes everything else -- and you can check that all of them are odd.

[rocky cloak]

I guess if you know how conjugation works it's clear conjugating by something in A5 only gives you an even permutation of (12345), so it would be missing half the 5-cycles

[balmy python]

I mean what i worked out were the possible cycle shapes for the permutations in S5 to be even and obtained you should have 5, 3 + 1 + 1, 2 + 2 + 1 and the identity. Then i worked out the number of cycles with each cycle shape and kinda just said they're the conjugacy classes of A5. I think what i did wrong was assume that a conjugacy class in S5 would be one in A5 but that may not be true.

[balmy python]

It could split into further conjugacy classes as it's no longer for all x \in C, for all g \in S5, gxg^-1 \in C but rather x \in C, for all g \in A5, gxg^-1 \in C

[chilly radish]

Yes, that is your mistake

When you shrink the group conjugacy classes split up since the conjugating element might not exist anymore

[tribal moss]

In fact you don't need to work out the 5 possible choices for p -- it is enough that p=(45) works, and then (12345)^n(45) also works for n=1...4, which gives you 5 different p's that all works. Since I've just argued there are just 5 of them in total, that's all of them, and each (12345)^n(45) is clearly odd.

[balmy python]

So the number of conjugacy classes in a group are the total number of orbits under conjugation.

[tribal moss]

Yes, because "orbit under conjugation" is what "conjugacy class" means.

[balmy python]

yeah and in cycle notation g(1 2 3 4 5)g^-1 = (g(1) g(2) g(3) g(4) g(5)) which has 5 different representations if we want (g(1) g(2) g(3) g(4) g(5)) = (1 2 3 4 5)

so we have 60/5 = 12 by OS theorem

[copper kestrel]

i dont think i did this right but im not sure how else to do it (sorry for talking over yall feel free to ignore)

[rocky cloak]

Yeah, so one thing is you need to argue you can pick f in F[beta][x] instead of F(beta)[x].

And then lastly I guess you need to argue that the polynomial is not the 0 polynomial when viewed as a polynomial over F[alpha]

[copper kestrel]

i would have no idea how to argue that

[rocky cloak]

First one you mean?

[balmy python]

for the other cycle shapes, do i have to do the same thing?

[copper kestrel]

both 💀

[rocky cloak]

Think about what the elements of F(beta) are

[copper kestrel]

isnt it like f(beta)/g(beta)

[balmy python]

I don't really understand what the solution did

[copper kestrel]

but like f,g are polynomials

[rocky cloak]

Yeah, so if we had something like
f(beta)/g(beta) * x = 0
or similar.

How could we turn this into something without denominators

[copper kestrel]

multiply by g(beta)

[rocky cloak]

Yup, that's it

[copper kestrel]

ah

where does the transcendentality come in btw

[rocky cloak]

In the second part

I.e. showing that the polynomial is not the 0 polynomial

[copper kestrel]

ohhhh

[rocky cloak]

They're saying that C(sigma) is not contained in A5.

[copper kestrel]

does this look good so far? besides arguing that its not the zero polynomial?

[balmy python]

im sorry for being slow to understand but why does this imply you only have one conjugacy class?

[rocky cloak]

It's what I already wrote, but the size of the conjugacy class in S5 is
|S5|/|C(s)|, while the size in A5 is |A5|/|C(s) n A5|. Dividing these by eachother you get
2|C(s) n A5|/|C(s)|.

So if |C(s) n A5|/|C(s)| = 1 the conjugacy class gets split in two. Otherwise it stays the same size.

[balmy python]

I understand all up to: Otherwise it stays the same size.

if |C(s) n A5| < |C(s)| then why does it stay the same size?

am i missing something really simple

[rocky cloak]

We divide the size of the conjugacy class in S5 by the one for A5 and get 2|C(s) n A5|/|C(s)|, if this equals 1 they are the same size. If it equals 2 then one is twice as big as the other

[tribal moss]

The only sizes C(s) cap A5 can have are |C(s)| or half of that.

(This is general for subgroups of Sn, not just stabilizers in particular).

[balmy python]

Ah

[balmy python]

is there a name for this?

[tribal moss]

I don't think it has a fancy name.
The general principle is that if H is a subgroup of Sn, then H cap An is the kernel of the parity homomorphism H -> {±1}, so its index in H must be either 1 or 2.

[rocky cloak]

It's close to the second isomorphism theorem I guess.

Like for normal N in G you have HN/N = H/HnN
So the index of HnN in H divides the index of N in G

[copper kestrel]

i can't quite figure out why if the polynomial at the end is the zero polynomial then that means alpha is algebraic

is it easier than i think?

[rocky cloak]

The other way around.

If the polynomial is nonzero. Then beta is the root of a non-zero polynomial, so algebraic.

If the polynomial is 0, then beta being a root doesn't tell you anything.

[copper kestrel]

oh

then i am back to square 1

im sorry

[rocky cloak]

So what is the definition of a polynomial being the zero-polynomial?

[copper kestrel]

all coefficients are 0

[rocky cloak]

And what are the coefficients?

[copper kestrel]

a_ij \beta^j

[rocky cloak]

So you're mixing up alpha and beta for one

[copper kestrel]

sigh

[rocky cloak]

But the coefficients are some polynomials in alpha yes

And what do you know about evaluating a polynomial at alpha?

[copper kestrel]

it cant be zero bc its transcendental

[rocky cloak]

Boom boom

[copper kestrel]

so the coefficients of \sum_{i,j} a_{ij}\beta^j\alpha^i are polynomials in alpha, and since evaluating polynomials at alpha can be zero because alpha is transcendental, we have that beta is algebraic

[stone sky]

Quick question, the following is generally not true right? Let R be a ring with finite characteristic, show that if a,b in R with ab=0, then ba=0.

Because the set of 2x2 matrices with integer coefficients modulo 2 (i.e. only 1s and 0s) is a ring with characteristic 2, but there are examples where the multiplication is not commutative

[karmic moat]

Agreed

[rocky cloak]

Is there really a difference?

Like either you say take the smallest closed set not the union of irreducibles or you say take the largest ideal where R/I has infinitely many minimal primes

[novel star]

for the following theorem, is my proof correct? Let $A$ and $B$ be noetherian rings and $f:A\to C$, $g:B\to C$ be surjective ring homomorphisms. then the fiber product $A\times_CB$ is noetherian.\

Observe that the canonical projections are surjective, so we have an exact sequence of $A\times_CB$-modules
[
0\longrightarrow \ker f \longrightarrow A\times_CB \longrightarrow B \longrightarrow 0.
]
Furthermore, we have that $\ker f$ and $B$ are noetherian as $A\times_CB$-modules, so $A\times_CB$ is noetherian.

[cloud walrus]

qchs

[novel star]

i feel like theres some really basic error but im not sure what it is

[rocky cloak]

Looks good to me.

I would maybe add a little argument for why B and ker f are Noetherian

[novel star]

alr, ty!

[chilly radish]

The algebraic side is a lot less intuitive here though. On the algebra side I don't see how you can get as nice a proof. You need to break up the decomposition of I yourself (and also reduce to the case that I is radical)

Which is already less intuitive

[rocky cloak]

Pick I to be maximal with R/I infinitely many minimal primes.

Clearly I is not prime, so pick a and b with ab in I. Then any prime ideal containing I contains one of a or b. Say a is contained in infinitely many of the minimal primes, then R/(a + I) has infinitely many minimal primes contradiction

I feel like the things you're listing are what is needed for the geometric argument

Because you have to understand why spec R is Noetherian

[chilly radish]

Fair enough. I had a proof in mind using the fact that rad(I) must be a finite intersection of primes

Well, arguing that directly

[chilly radish]

Sure, but once you have all that thinking about the space is a lot easier

[rocky cloak]

Even still, I'm not sure I see which part is supposed to be easier.

Like the geometric argument I have in mind would just be exactly the same:

Let V(I) be minimal among closed sets that are not the union of finitely many closed irreducible sets. Then V(I) is not irreducible so it's a union V(a + I) u V(b + I). Then each of these are a finite union, contradiction.

In both cases the only non-obvios part is coming up with what to apply the Noetherian hypothesis to and that seems equally non-intuitive to me in both cases

I guess the advantage of the geometric argument is that there are much fewer things to do.

Like working with the ring you might be tempted to use localization or things being finitely generated. Whereas for geometry I don't know anything, so I'm only allowed to use the definition, which is all you need

[chilly radish]

I think you summed it up perfectly

The geometry in this case limits your options such that there is really only one path ahead (once you decide which defn of noetherianity you use)

[rocky cloak]

But that seems to be more about a function about being good at algebra and bad geometry. But maybe geometry is just limited for everyone

[round anvil]

for any group G, is G/Z(G) abelian?

[chilly radish]

What if Z(G)=1

[chilly radish]

That's possible

[round anvil]

ok thats enough

ty

[swift root]

if this were the case, then [G, G] ≤ Z(G), which would mean every group was both solvable and nilpotent

this would contradict the fact that there is no formula for the roots of a general quintic in terms of radicals

[round anvil]

ok

i'm solving a problem that ask to prove that G/Z(G) is not cyclic, since G isn't abelian

and i think: wtf but G/Z(G) cannot be abelian

[swift root]

it certainly can be abelian

just not cyclic

[round anvil]

why is G/Z(G) abelian?

with G not abelian

[swift root]

it can be

[round anvil]

how

[quiet pelican]

Consider say D_8
The centre is {1, r^2}
The quotient is C_2 x C_2

[swift root]

more generally the symmetries of a 2k-sided regular polygon, i believe

[round anvil]

if |G| = 2k and [G:Z(G)] = 2, so G/Z(G) is abelian

ok i see

[swift root]

that can never happen

because then G/Z(G) would be cyclic

[round anvil]

But I see that we need more hypotheses

[round anvil]

i agree

[swift root]

yes, G/Z(G) isnt always abelian

[round anvil]

ty

[rocky cloak]

I think D_2k modulo center is D_k. So only for k = 2 would this be abelian (D_4 being micois D_8)

[swift root]

oh okay that makes sense

[elfin wraith]

D_2n notation should be a criminal offence

Or you have to start writing S_n! In full, for the sake of consistency

[wraith cargo]

Uhhh can I have some S_720 with my S_120 please

Waiter waiter

More S_24 please!

[cursive spindle]

My prof (which is the same one for numerical analysis) prefers this over D_n

Really annoying prof

[wraith cargo]

God I remember my algebra prof at UCSD

This was graduate algebra so the story makes more sense

I was learning category theory at the time since I was taking algtop

And during my algtop lectures I'd constantly ask if X was Y categorical construction

[quiet pelican]

D_{2n}

[wraith cargo]

Because I was a dumbasss

But then I started doing the same in my algebra class

And my professor said very sternly

"No more mentioning category theory in this room"

[cursive spindle]

Lmao

[quiet pelican]

Category theory is illegal 💯

[cursive spindle]

You sound like my friend

[wraith cargo]

This was about

3-4 years ago

I matured a lot since then but yeah

I was that person lol

[karmic moat]

i was too

[cursive spindle]

My friend is like 30 y.o and they still do this

All the time

[karmic moat]

2-3 years ago

[slim kayak]

The "undergrad category theorist" meme luckily entered my life very early and set me upon the path of salvation

[karmic moat]

HSCT

eek

[wraith cargo]

Me and my friends were all sort of young and stupid and eager to learn

I was especially like this because at the time I was super hard into homotopy theory

Tho I mean I say young

Younger than we are now

This wasn't 30 years ago

[slim kayak]

Aight unc

[karmic moat]

toniiiiiiiiiiiiight we are youuuuuuung

[wraith cargo]

I was a Lil' baby them

[balmy python]

if i was asked to factorise a polynomial in Fn, what would be the quickest way to do so? are there any tips?

e.g. x^4 + x^2 + 1 in F3

[somber goblet]

lucky for you, there are 3 possible roots in F3

so you can check those first tbh

[elfin wraith]

If I dont see an obvious factorisation, and its in something small like F_3 ill just look for roots

[chilly radish]

I remember when I asked if M_n(-) was a functor in my ug group theory class. The professor seemed really done but I honestly think it was better than the stuff the physicists were asking

[balmy python]

ok yeah true but what if there were no roots

my example was a bit shitty 😭

[somber goblet]

0 can’t be a root for instance

[balmy python]

take x^4 + 1 in F3

[chilly radish]

Then it's either a product of quadratics or irreducible

There are only 3 options for each coefficient, so you can check this quite directly

[balmy python]

true

what if you had some big degree

are there theorems which give more info?

[knotty badger]

What were the physicists asking

[chilly radish]

I actually don't remember, I just remembered it was embarrassing

[rocky cloak]

You can look for roots in F9 as well.

Or computing gcd with x^9 - x

[balmy python]

i guess you would be having a polynomial over a ring instead of a field right?

[balmy python]

what's the intuition behind this?

[rocky cloak]

The factors of x^q - x are exactly all the minimal polynomials of elements of Fq

Which comes from the elements of Fq being exactly the roots of x^q - x, which comes from Fq^* being cyclic

[rocky cloak]

Like are you asking about how to factor polynomials over arbitrary rings? That seems like it would be hard to do in general

[balmy python]

is this min polys in the linear algebra sense?

[balmy python]

nono i was just talking about fields but you said F9 which i thought was not a field

[elfin wraith]

F_9 is a field

[rocky cloak]

In general F_q refers to the field with q elements

[elfin wraith]

Theres a field for every p^n with p prime and n an natural

[rocky cloak]

I mean, it's the same sense I guess.

The minimal polynomial of z is the smallest polynomial with f(z) = 0

[balmy python]

oh sorry i mistook it for Z/nZ

[elfin wraith]

Exercise, prove that

[balmy python]

we've done that one in lectures lol

[elfin wraith]

Yeah its not so bad, a good thing to know if youre doing Galois though

[balmy python]

yeah

i might pick that next year

[balmy python]

our lecturer did this with field extensions

[elfin wraith]

Yeah you just take the splitting field of x^q-x. What is in your Galois course if youre doing this much field theory before Galois though lol

[balmy python]

oh no i don't do galois theory yet

this is our introductory algebra course

we do groups, rings and a bit on field extensions

So constructing fields from irreducible polynomials and showing their existence

[swift root]

we have a semester long (half year) Galois theory course separate from rings and fields

fields was basically everything you could do without Galois theory

[elfin wraith]

Whats left to be done in Galois theory?

[swift root]

Galois groups

would be funny if they also did some profinite group stuff

[elfin wraith]

The Galois theory course in my UG and MSc unis are both like 70% field theory or something

[swift root]

or rep theory

but that may be getting my hopes up

[wraith cargo]

Inverse Galois problem

,,,,

Dun dun dunnnnnn

[elfin wraith]

But yeah genuinely let me know if you can lol

Maybe the courses at my UG and MSc are just shit (I mean they are, theres a reason I didnt do it in UG and regret doing it here) but they must be really shit if thats the case

[swift root]

so im trying to get to the course but i recently changed my sim card and it isnt activated yet>.>

so i cant 2 factor authenticate into the place with all the courses

[elfin wraith]

Pain lmao

[swift root]

lemme see if there is another way in

[swift root]

yeS

[swift root]

thats slightly disappointing lol

how come that they refuse to give actually interesting math to undergrads 💔

genuinely no clue how they will split this over ~12 weeks

[elfin wraith]

Yeah so this is like maybe slightly than what my course does in 10 weeks, and does all of fields lol

And I already felt like if you know any ring theory a lot of the course is very straight forward

[swift root]

that has been my experiences with all the courses ive taken so far

straightforward

[swift root]

even the second year courses topology and rings and fields

[chilly radish]

I would hope so if it's a half year course

Maybe they'll do galois cohomology

[chilly radish]

This is so little

I am disgusted

[swift root]

modules is for some reason a masters course so i doubt it

[swift root]

when they slandering your goat but they lwk right

[chilly radish]

???????? (To it being a master's course)

You don't need modules for Galois cohomology though it's all vector spaces and tensors

[swift root]

true

[swift root]

i knOw

this uni is incredibly analysis pilled

[chilly radish]

I wish I could teach an undergrad or master's course on central simple algebras

That would be the dream

[swift root]

like almost 50% of 2nd and 3rd year electives is just analysis or diff geo

and a similar ratio for the mandatory subjects

[swift root]

and the diff geo doesnt even touch anything like de rham cohomology it seems

but i could be wrong

maybe ill just do some courses in other unis lol

[balmy python]

if m|fg and f,g are coprime but f|m and g|m is fg = m?

m, f and g are polynomials

[swift root]

f|m and g|m => fg = lcm(f, g) | m. But m | fg, so fg and m are associates

if all polynomials are monic then yes fg = m

[balmy python]

ah yeah okay

i'm wondering why my proof to a linear algebra problem doesn't work 😭

i guess it's right so far

[swift root]

:3

[balmy python]

this is what my lecturer said

so im guessing my proof is invalid somewhere

oh no

[swift root]

what is the statement?

[balmy python]

it's to prove that C(fg) is similar to the direct sum of C(f) and C(g) where f,g are coprime

[swift root]

whats C?

[balmy python]

companion matrix of the polynomials, sorry for not specifying.

[swift root]

i see a pretty nice proof

i believe

[balmy python]

same

what is yours?

[swift root]

For a polynomial P with real coefficients, let V_P be the ring R[x] / (P) as an R[x]-module, with the action of x being left-multiplication. I claim that this is is isomorphic to C(p) acting on R^n.

Indeed, let { e1, ..., en } be a basis of R^n, and define the map
φ : R^n -> V_P
ei -> x^i-1 + (P)
One can check that we have
φ( C(p) ⋅ ei) = x ⋅ φ(ei)
Furthermore, this map is an bijection simply by comparing dimensions.

Now, if f and g are coprime polynomials, then (fg) = (f) ∩ (g), so by the chinese remainder theorem, we have R[x]/(fg) ≅ R[x]/(f) × R[x]/(g) as rings, so V_fg ≅ V_g ⊕ V_f.

From the above we conclude that C(fg) acting on R^n+m is isomorphic to the direct sum of C(f) and C(g) acting on R^n+m. In other words, there is a basis change of R^n+m taking C(fg) to C(f) ⊕ C(g)

its not elementary or quick and easy but it is reasonably elegant

[balmy python]

what is wrong with mine?

[swift root]

what is m(x)?

[balmy python]

min pol of the sum

[swift root]

whats h?

[balmy python]

any polynomial

[swift root]

i dont get "f is the smallest degree polynomial such that h(C(f)) = 0"

[balmy python]

f is the min pol of c(f)

[swift root]

so?

i can think of plenty h such that h(C(f)) ≠ 0

[balmy python]

by def if h(c(f)) = 0 then f|h

[swift root]

that is true

but you wrote something different

You should write something like
"Let m be the minimal polynomial of C(f) ⊕ C(g). Then, as
0 = m(C(f) ⊕ C(g)) = m(C(f)) ⊕ m(C(g))
it follows that m(C(f)) = 0, and m(C(g)) = 0, so f | m and g |m, as f and g are the minimal polynomials of their companion matrices. [...]"

[balmy python]

so what i mean is that since m(C(f) + C(g)) = m(C(f)) + m(C(g)), then we require that f | m

[balmy python]

yeah sorry it was more so rough working 😅

[swift root]

:>

i think its good practice to train yourself to be concise even at the start

[balmy python]

yeah that is true 😭

i am a bit tired so it is why i was lacking there

does the proof look sound though?

[swift root]

I think you should also clarify why m | fg instead of stating it

[balmy python]

ah that's because it was a problem sheet question so i am allowed to reference it 😭

i wrote it up to show a classmate

i shouldve specified

[swift root]

oh lol then its fine

i mean its like one or two lines lol

[balmy python]

what's a R[x]-module?

[balmy python]

yea lmao

[swift root]

an R-vector space V with a distinguished linear transformation T : V → V

a morphism is a linear transformation f : W -> V such that f ∘ T_W = T_V ∘ f

In particular, if youve got a vector space V with linear transformations T, T', then (V, T) and (V, T') are isomorphic iff T and T' are similar

this proof is probably very much outside of the scope of your course but i believe its nice and elegant

it also explains where the companion matrix comes from

[balmy python]

ive come across the chinese remaineder theorem in my algebra course

[balmy python]

what is an isomorphism in this context?

oh wait you said it lol

R[x]/(fg) ≅ R[x]/(f) × R[x]/(g) as rings, so V_fg ≅ V_g ⊕ V_f.
how do you get this implication?

[swift root]

idk why i said that tbh

the idea is that (f) and (g) are coprime submodules of V_fg with trivial intersection, so V_fg ≅ V_f ⊕ V_g

coprime meaning that (f) + (g) = V_fg

[twilit wraith]

alright galois theory question

Let f(x) in Z[x] be an irreducible quartic whose splitting field has Galois group S4 over Q. Let r be a root of f(x) and set K = Q(r). Show that K is a degree 4 extension of Q with no proper nontrivial subfields.

My buddy and I are having an argument as to if my argument is correct

[rocky cloak]

What's your argument

[twilit wraith]

I said this:

K is clearly degree 4. Suppose BWOC that L is a degree 2 extension of F contained in K. Then L is a degree 2 extension of F contained in the splitting field of f(x), and so by the fundamental theorem of Galois theory there is a subgroup of index 2 which fixes L. This subgroup must be isomorphic to A4, and so the subfield of degree 2 is fixed by even permutations of the roots of f(x), and thus Q(r) is fixed by such permutations as well. However, this is not the case, as a transposition of any of the other three roots fixes r and is an odd permutation, a contradiction.

[rocky cloak]

This is basically the canonical solution

[twilit wraith]

alright

i guess ill have to let him know

[quiet pelican]

An alternate way to note this is that Q <= L <= F <= splitting field of the given degrees would imply A_4 has a subgroup of size 6, which is like the canonical counterexample to that claim lol

[twilit wraith]

theres a second part asking if theres a galois extension of Q of degree 4 with no proper subfields

something about no normal subgroups of index 4 in S4 or smth

[rocky cloak]

If the Galois extension has degree 4 then the Galois group is of order...(?)

[twilit wraith]

i mean i thought it was like a galois subextension of the splitting field of f(x)

[rocky cloak]

Do you know any sort of relationship between the degree of a galois extension and the order of the Galois group?

[twilit wraith]

well i know theyre equal normally

[rocky cloak]

You say normally as though there are exceptions

[twilit wraith]

well yeah thats whats confusing to me

[rocky cloak]

What's the confusion?

[twilit wraith]

theres a portion of the fundamental theorem that says that if F < E < K with G = Gal(K/F) with H < G fixing E then [E : F] = |G : H|

[rocky cloak]

This is true yes.

[twilit wraith]

i think im just mixing up the notation or something like that

[rocky cloak]

E/F doesn't have to be a galois extension here if that's a point of confusion

[twilit wraith]

the main confusion is when it is

[velvet hull]

here is yet another alternate solution:
observe that the elements of the galois group is determined by their permutation on the roots of the polynomial that gives the splitting field.
f has 4 roots by definition, but the galois group is S4, so all the elements of the galois group are given by some permutation of the 4 roots exactly.
Hence, K(r) corresponds to the fixed field of the subgroup of S4 that fixes 1 exactly, which is isomorphic to S3.
In other words, by the fundamenal theorem, any subfield in between K and the splitting field corresponds to a subgroup between S3 and S4. But then this is not possible, as any element outside S3 has to move 1 by definition such that you would recover everything else in S4.

[twilit wraith]

i should try and explain in particular what im confused about

i know that |Gal(K/F)| = [K : F], but if i have a Galois extension E of F contained in K, then the subgroup H in Gal(K/F) fixing E satisfies [E : F] = |G : H|, but we also have that Gal(E/F) = [E:F]. However in the S4 case we have that F(r) is fixed by some subgroup of order 6 despite being an extension of degree 4

i mean i suppose the answer is that subgroups of a larger galois group and the galois group of a subextension have nothing to do with each other

[twilit wraith]

i think my friend gave this answer

[rocky cloak]

This is very much not true.

In the situation you had
F < E < K
then F/E is galois iff H is a normal subgroup, in which case Gal(E/F) = G/H

[rocky cloak]

In this example F(r)/F had degree 4 and [S4:S3] = 4. So they are equal as we should expect.

The fact that S3 has order 6 is not particularly relevant to anything (other than you can compute the index from it I guess)

[twilit wraith]

but we had that this copy of S3 in S4 is exactly the subgroup of S4 fixing F(r)

oh wow i just realized my mistake

Gal(K/F) is nothing even close to being a group that fixes K

that wouldnt even be an interesting object

[rocky cloak]

It's true that S3 is the galois group of K/E though

[twilit wraith]

yes

this does make sense though

i see my mistake now im not confused anymore

back to part 2 of this problem then i suppose

whether or not theres a degree 4 Galois extension of Q with no proper subfields

well this certainly cant be true

because theres always subgroups of order 2 in groups of order 4

so in fact part 2 didnt even have anything to do with the f(x) splitting field

in general its not even possible

thanks for the help jagr 🔥

[somber goblet]

let's say i have a galois extension $K/F$ and a group $H \leq \operatorname{Gal}(K/F)$. if my understanding of galois theory is correct, then $K^H$ extends $F$. is $[K^H : F] = [G : H]$?

[cloud walrus]

lexi

[somber goblet]

sanity check: $K^{\operatorname{Gal}(K/F) }= F$

[cloud walrus]

lexi

[somber goblet]

this is trivial...right?

by definition

[rocky cloak]

This is in fact equivalent to K/F being galois

[somber goblet]

oh

interesting

so if you take automorphisms fixing F of a non-galois extension, then the fixed field of that group is not the original field?

i guess it has extra junk

wait i dont see how this isnt always true by definition

[rocky cloak]

What definition are you thinking of?

[somber goblet]

Gal(K/F) = automorphisms of K that fix F

K^G = subfield of K fixed by G

[rocky cloak]

The classical example being
K = Q(cuberoot(2))

Then the only automorphism of K fixing Q is the trivial one

So K^Gal(K/Q) = K

[somber goblet]

oh

hmm

so maybe the right view is you don't have 'enough' group elements to make it work

since you're missing roots

and because you have a smaller group you have a bigger fixed field

as galois will tell you

[rocky cloak]

Yeah, if you have fewer roots then expected (either because the extension isn't normal so you're missing roots, or because the extension isn't seperable so some roots are counted twice) you won't get enough automorphisms to be Galois

Which is why Galois = normal+ seperable

[swift root]

okay infinite Galois theory was covered when they gave the course 4 years ago (it is a bi-annual course), and lecturer is one i do like

maybe they took it out or its dependent on the lecturer

[quiet pelican]

Profinite groups are cute

[somber goblet]

ok yeah

makes sense then

[swift root]

i love profinite groups rahhhh

[somber goblet]

ok why do we do all this stuff with automorphisms being linear independent

or something

idk the motivation

[rocky cloak]

Say you have a field extension K/F of degree n. Then you can think of the set of F linear maps K -> K as a vector space over K, and it will have dimension n.

The automorphism of K/F are elements of this space, so if you know they are linearly independent then you know there can be at most n of them.

In short |Gal(K/F)| <= [K:F]

[somber goblet]

wait why not n^2

[rocky cloak]

And when you have equality the automorphisms give a basis for this space

[rocky cloak]

Well it's n^2 dim over F and n^2 / n = n

[somber goblet]

oh wait over K

ok 👍 thats cool

so this is just for counting size of galois group

[somber goblet]

i should verify this

[chilly radish]

Hey for all the kiddies learning galois theory you should do this exercise:

Let t be transcendental over C, compute the Galois group of

C(t)/C(t^n+ 1/t^n)

C is the complex numbers

Ping for hints

[somber goblet]

does this mean $\mathbb C(t)/\mathbb C(t^n + t^{-n})$?

[cloud walrus]

lexi

[chilly radish]

Yea

[somber goblet]

hmm first i need to prove that's actually an extension

[swift root]

thats by definition

[somber goblet]

that's not obvious

[swift root]

C(t) is the field of rational polynomials

t^n + t^-n is a rational polynomial

(t^2n +1)/t^n, if you want to be explicit

[somber goblet]

oh it is

yeah

ok i might not be cut out for this exercise 😭

[rocky cloak]

Any magic geometry motivating this example, or just something neat to compute?

[chilly radish]

I'm sure there's some sort of magic geometry underlying these by passing to riemann surfaces but I just think it's an instructive exercise and fun to compute

[somber goblet]

also, is this for all n or one specific n

[chilly radish]

Any n

[rocky cloak]

Pick your favorite

[somber goblet]

🔥

[rocky cloak]

Might start with n=1 if you need a warm up

[somber goblet]

yeah fs

[frail shoal]

||hm, i know it's nontrivial because t |-> t^-1 fixes the subfield. ok close enough||

✍️

[chilly radish]

Good start

[frail shoal]

ok maybe i should try harder

[swift root]

||For warm up take n = 1
t is a root of X^2 - (t + t^-1) X + 1, so the extension is finite and algebraic. Furthermore, it is nontrivial, as it has a nontrivial automorphism as keith said. Thus the degree is 2, and the automorphism group is C2, the nontrivial element being t |-> t^-1||

[frail shoal]

oh that first part is clever

[frail shoal]

||we also have t |-> zeta_n t||

||so it's probably D_n cuz rule of "cute answer is the correct one"||

ok back to not trying cuz i dont like galois theory

[quiet pelican]

You’ve basically already solved it lol

[frail shoal]

ok but like "it's cute" isn't a good proof

[quiet pelican]

Namely, ||you’ve found 2n automorphisms
Now prove the extension is of degree at most 2n||

[frail shoal]

oh good point

[chilly radish]

Well you first need ||to verify that these subgroups actually satisfy the correct action and that the intersection is trivial||

[swift root]

||More generally, t is a root of X^2n - (t^n + t^-n)X^n + 1, and the degree of the extension is bounded by 2n.

Keith has already exhibited 2n automorphisms, so the extension is Galois of degree 2n with Galois group Dn||

[chilly radish]

You need to check that it's the correct action!

[somber goblet]

||i guess i want to check to make sure this is algebraic?||

[chilly radish]

And actually splits the way you expect it to

[somber goblet]

||it probably is, because they have the same transcendence degree (i think this is how that works)||

[chilly radish]

Yea, you should start by identifying a potential minimal polynomial

[elfin wraith]

I got this far with just my noggin and decided that working out that this is actually the answer is much to hard for the head and at 2am

[frail shoal]

whao that's clever (again)

i needa put that in my mental toolbox

[elfin wraith]

But micois principle of cute answer is the right one saves the day

[frail shoal]

for the future

[somber goblet]

yea, for sure

[quiet pelican]

Wdym by ||correct action||?

Do you mean ||well defined?||

[frail shoal]

wait that was my principle 😭

[chilly radish]

||To be D_n, C_2 has to act on C_n via sending a generator of C_n to its inverse||

[elfin wraith]

Damn sorry, misattributing it right in front of you 😭

[quiet pelican]

Ok yeah but I’m assuming you’ve done the obvious thing of noting you have a|| rotation and a reflection lol||

[somber goblet]

or i can consider ||the vector space generated by the adjoin and finding a finite basis? but polynomial might be easier.||

[chilly radish]

||Also that these intersect trivially so that this is actually a semidirect product||

[velvet hull]

Good question ty
Galois group is ||S_2n||, is that correct

[chilly radish]

That's way too bug

Big

[somber goblet]

||since we're adjoining a single element it's going to be algebraic over base field iff the adjoint is||

[velvet hull]

Oh wait it has to be C linear as well

[swift root]

This t^n + t^-n business reminds me a lot of a factor that comes up in quantum knot invariants

[small yacht]

i would make a joke but i'm not sure if it falls under the forbidden category

[cursive spindle]

tfw I recognize this

[quiet pelican]

||Tbh that’s kinda overkill for stuff that’s routine calculation that those two elements satisfy the relations of D_{2n}
And given that they obviously have the right orders, it can’t be a proper quotient||

[chilly radish]

You're overthinking it

[elfin wraith]

It does feel like this should be some geometry nonsense

[swift root]

its frustrating because i kinda get the derivations of the HOMFLY polynomial from some lie algebra representation but the "q deformation" stuff is just kinda thrown out there in the books ive read without any proper explanation or derivation besides "it works out nice"

[chilly radish]

||I mean even if they have that relation a priori I don't see why that should imply trivial intersection on its own||

[velvet hull]

Okay right now I’m thinking it is ||Aut Z/nZ^2||, I think this is correct no but I got it, it is the automorphism group of “that set” but i described it incorrectly

[cursive spindle]

wait am i schizo thinking rn

||is it dihedral?||

okay yeah then it is geometric

[swift root]

||The C2 and Cn actions should have trivial intersection because else that would induce an algebraicity relation on t||

[quiet pelican]

||If they have the correct relations we get a map from D_{2n} to the group they generate
Every quotient, by trivial calculation is of the form D_{2k} for k dividing n (including some small trivial cases eg D_2)
But we know our group has an element of order n, so it must be D_{2n}||

[swift root]

thats very nice

[cursive spindle]

||tfw you can realize it as a semidirect product||

meow good exercise

[chilly radish]

Ok yea, that's one way to do it, but in this case appealing to the field theory is way nicer

[quiet pelican]

where “trivial calculation” means ||“I’m 99% sure this is on our 1st year groups sheets”||

[chilly radish]

And again I think it's instructive

The way enpeace did, which is immediate

[cursive spindle]

now this chat looks like the epstein files

[swift root]

speaking of, does anyone know a good book on quantum knot invariants? that doesnt read like its written by a mathematical physicist (/s)

[small yacht]

that's the joke i wanted to make ugh
i was afraid i'd get smitten under the changelog

[quiet pelican]

Groups are cuter :3

[swift root]

valid fear lol

[somber goblet]

i ||found a polynomial||

i think

[swift root]

in particular the connection with lie algebra representations im interested in

[somber goblet]

||idk if its irreducible tho :(||

how do you spoiler texit

can you do that

[quiet pelican]

||this isn’t easy to do directly||

,texsp

[cloud walrus]

Please give me something to compile, for example latex ,tex The solutions to \(x^2 = 1\) are \(x = \pm 1\).See ,help and ,help tex for detailed usage and further examples!

Apr 8, 2026, 1:07 AM

[quiet pelican]

This is how you spoiler tex

[small yacht]

also uhm how do i go about proving this
I know that it at the very least G/H has to contain H, and aH for some a. a also has to have second order otherwise a^-1H would be in G/H. But im kind of lost how to go from there,,

[somber goblet]

,texsp g

[cloud walrus]

lexi

[somber goblet]

🔥

[quiet pelican]

You want to prove that aH = Ha right?
What’s another description of aH?
(What elements aren’t in aH?)

[rocky cloak]

What's the definition of normal you're working with?

[somber goblet]

||i let a = t^n + t^{-n} and found a polynomial via t^2n - at^n - 1||

[quiet pelican]

One of your signs is wrong

[small yacht]

elements of H?

[quiet pelican]

Yes
So what’s another description of Ha (do the same thing)

[small yacht]

elements that aren't in H...
wowzers!

[somber goblet]

really? i got ||t^n - a = t^-n, so (t^n - a)t^n = 1 = t^2n - at^n, so t^2n - at^n - 1 = 0||

[quiet pelican]

Shouldn’t t^n - a = t^n - (t^n + t^{-n}) = -t^{-n}?

[somber goblet]

...

idk how i missed that

that's embarrasing

[chilly radish]

Remember that cosets partition the space

That is, H u aH = G, and H, aH are disjoint

Oh ok you got it

[somber goblet]

ok, ||t^2n - at^n + 1||

[small yacht]

Mhm
Though i am looking through the chapter to find the probably more rigorous way they'd want us to handle this using this intuition

[somber goblet]

how tf do i tell that this is actually the min poly

[chilly radish]

There's a clever trick here

The hint is to use Gauss' lemma (general version for UFDs). But also you don't actually need to show this for the galois group computation

[rocky cloak]

If you ||prove the extension has degree 2n|| then ||the minimal polynomial must be degree 2n, so is this one||

[chilly radish]

Do you have another idea in mind or do you just mean ||by finding 2n automorphisms and noting the degree is at most 2n||

[somber goblet]

oh really?

wait do i just ||immediately start thinking about automorphisms||

[rocky cloak]

Yeah that

[somber goblet]

||did i really overthink it that hard||

[chilly radish]

Well

Now you know the degree is at most ||2n||

[somber goblet]

||i think i still need to know degree||

[chilly radish]

So just find at least ||2n|| automorphisms

Then the degree will be forced to be ||2n||

[somber goblet]

||but it only meets that bound if it's truly minimal?||

[chilly radish]

Yea

[somber goblet]

||oh i see||

[chilly radish]

It's a polynomial with t as a root, so the min poly divides it

[somber goblet]

yep

||so now i just start hunting for automorphisms||

[chilly radish]

This is already rigorous, since normality is equivalent to aH=Ha for any a

I don't think there's any way to do this element-wise that doesn't go through a similar kind of assertion

[small yacht]

Ik it’s just idk why he would put that exercise in this chapter then rather then putting it somewhere earlier right after learning about normal groups

Eh I’m probably overthinking it

[small yacht]

Is there a way to do this non element wise?

[chilly radish]

What I mean is aH=Ha instead of proving ghg^-1 in H for arbitrary H

[small yacht]

Ah I see, then

[somber goblet]

||i don't really know what function field automorphisms look like||

||maybe i can compute the galois groups of both relative to C? but neither of them are algebraic extensions of C||

[chilly radish]

So, think of it this way

Any map C(t)->C(t) is determined by where you choose t. You can choose to send t to any rational function and this will map you into a subfield

You want this subfield to be all of C(t) for this to be an automorphism

In practice that means you send t to something of the form

(at+b)/(ct+d)

But you don't really need to know this

[somber goblet]

yes that was my thought

[chilly radish]

Just that you send t to a rational function

[somber goblet]

t is the only thing that can be moved

so this automorphism structure is isomorphic to C(t) somehow?

[chilly radish]

The way to justify this is that for any element f\in C(t) non-constant,

You have a map C[t]->C(t) given by t->f(t). Since any nonzero element gets mapped to a nonzero element (If some polynomial mapped to 0, it would mean f(t) is algebraic, but this is impossible)

this extends to a map of fields

C(t)->C(t)

[chilly radish]

Well no because most choices will not give an automorphism

[somber goblet]

oh true

we need another element generating the field

oh i feel so slow taking so long on this

[chilly radish]

Why

[somber goblet]

like t needs to map to another generator

[chilly radish]

Oh

Yea, like I said those would be the "degree 1" rational functions

So you're limited to things of the form (at+b)/(ct+d)

Try to think what you could possibly map t to, such that t^n+1/t^n would be fixed

Cuz we want this to be an element of the Galois group

There aren't many options

[somber goblet]

oh i see

ok right

hmm i feel like i almost have it but still nothing

[somber goblet]

im going to think about this later

[small yacht]

Is GLn(R)/SLn(R) isomorphic to R?

As groups

[crystal vale]

I think it is isomorphic to R\{0}

And R\{0} is not isomorphic to R as group

[elfin wraith]

Yeah, GL_n is invertible matrices, so det(M)\neq 0

Then the kernel of that map is just SL_n and it should be first iso

[wraith cargo]

me when GL_n is the semidirect product of R^x and SL_n :woag:

[rocky cloak]

R{0} (with multiplication) is isomorphic to R x Z/2

[small yacht]

Yeah you’re right I forgot to add the non zero part

[small yacht]

that’s how I found it
I just wanted to make sure

[delicate orchid]

that's basically R. Let them have it, it is the season for giving and forgiveness after all

[crystal vale]

How is this basically R?

[fickle dirge]

What does it mean "in the contrary case, our process chose f_m+1"

[tardy hedge]

I think just setting up a contradiction

What book is that brw

Btw

[fickle dirge]

so, assuming I ≠ <f_1,...,f_m> ?

[fickle dirge]

eisenbud commutative algebra with a view towards algebraic geometry

[tardy hedge]

Yes so pick f_m+1 now and continue

[fickle dirge]

but what is m+1

[tardy hedge]

Its just the next index

[fickle dirge]

oh right haha

[tardy hedge]

Just a label yeah

[fickle dirge]

got it

thanks

[rocky cloak]

The difference is only Z/2

[tulip otter]

Let J=(4+3i)Z[i] and f:Z->Z[i]/J be the homomorphism defined by f(n)=n+J. Prove that f is surjective.

any hints?

[tardy hedge]

Maybe you can break it up into different projections or something

[tulip otter]

I tried to break it up into a composition of surjective maps but I didnt find a thing like that, tho i was fixated on Z->Z[i]->Z[i]/J so idk

I didnt find a surjective homomorphism Z->Z[i]

[tardy hedge]

Ik i was just thinking that too

If it was (4+i)Z[i] would be done right cause sure you dont map to i in Z->Z[i] but i is -4 in the quotient

[tulip otter]

the prof said that I should use the fact that 25 in (4+3i)Z[i], fix a+bi in Z[i] and then find the desired n by starting from a+bi+J=n+J or in other words from
a-n+bi\in J

[tardy hedge]

Im not sure why “desired n”

[tulip otter]

yea, and oh you mean that since both 1 and i are in the image then you can find an integer because {1,i} generates Z[i]?

[tardy hedge]

Yeah

[tulip otter]

I am not sure what you mean, I just meant the n whose image is a+bi+J

[tardy hedge]

Oh ok

Ok yeah doing that is probably just some algebra with complex numbers i guess

[tulip otter]

yea but I still couldnt find such an n

what I did something like this: fix a+bi in Z[i]. we want to find n such that a-n+bi+J=J, so a-n+bi=(4+3i)(c+di) and then expanding gives
a-n+bi=4c-3d+i(4d+3c)

[cursive spindle]

Any map from R to R/I is surjective

I guess if you want to do it explicitly it boils down to explicit computations in the complex numbers

[tulip otter]

any? But also this isnt a map R->R/I unless you mean to decompose Z->Z[i]/J into Z->Z[i]->Z[i]/J or something like that

[cursive spindle]

Ah sorry I missed that lmao

[tulip otter]

haha np

otherwise this would just be the canonical homomorphism Z[i]->Z[i]/J which is surjective as you said

[tardy hedge]

System of equations time? Lol

[tulip otter]

lol

[cursive spindle]

If you do Z -> Z[i] -> Z[i]/J then the second part is surjective automatically but the first part is injective

[tulip otter]

I mean sure but what should i do with the system, find c and d in terms of a,b and n?

[tulip otter]

right, I tried to think about a surjective map Z->Z[i] but such a map shouldnt exist lol

[cursive spindle]

Okay here's a NT observation

How many elements are there in the quotient?

[tulip otter]

25 because the kernel is 25Z so this quotient should be isomorphic to Z/25Z

[cursive spindle]

okay then you basically have it now

[tulip otter]

(assuming the map is surjective of course)

[tulip otter]

do you mean because the image and the quotient have the same order?

or does 25\cong 1 mod 3 have something to do with this?

[cursive spindle]

Surjectivity here works since Re(4+3i)=4 and Im(4+3i)=3 are coprime

[tulip otter]

why does (Re(4+3i),Im(4+3i))=1 imply surjectivity tho

[cursive spindle]

Noggin

[cursive spindle]

huh

okay let's backtrack

Did you compute kernel?

[tulip otter]

did the message you are replying to reach you now?

[cursive spindle]

yeah weird

[tulip otter]

oh i see

[tulip otter]

yea 25Z

[tulip otter]

sorry noggin isnt working rn

[cursive spindle]

right so Z/ker f = im f

[tulip otter]

tho I am trying

[tulip otter]

sure

[cursive spindle]

but you know Z/im f has size 25

The quotient you want also has size 25

[tulip otter]

yea

[tulip otter]

I have done this by cheating lol

[cursive spindle]

[tulip otter]

I should count it properly

[cursive spindle]

Monkey not for u but the fact they didn't do norm maps

[tulip otter]

wait whats norm maps

[cursive spindle]

N(a+bi) = a^2+b^2

[tulip otter]

is it literally the norm on Z[i] or something

[cursive spindle]

Yup

[tulip otter]

oh its literally that

but wait where did you use that

[cursive spindle]

Z[i]/(alpha) has size N(alpha)

[tulip otter]

ohh

[tulip otter]

ok so for now I will use this

so the image and the whole quotient have the same order

and then we are done no?

[tulip otter]

but then i should use this somehow

[cursive spindle]

Yup

[cursive spindle]

Dw about it

If you don't have the technology I said then you can compute the elements of Z[i]/(4+3i) explicitly or realize them as a sublattice whose fundamental cells has area N(4+3i)=25

[tulip otter]

I mean I dont think that I can even use the fact that a subring have the same order as the whole ring implies that the 2 rings are equal

[tulip otter]

what do i search for to find the relation between the gcd of the real and imaginary parts and the surjectivity

I want to find more about this, but I really cant think properly rn

[cursive spindle]

huh

You are overthinking this

[tulip otter]

no, I am not even thinking

[cursive spindle]

but yeah if the codomain was infinite sure I guess there are counterexamples

The quotient is finite

[tulip otter]

no what I meant by cant use is that I am not allowed to use it

[cursive spindle]

But it's just a set theory fact lmao

[tulip otter]

I mean yea true

[cursive spindle]

1984

[tulip otter]

whats that lol

[tulip otter]

it makes sense if you are implying that the technology we are using rn is from 1984

[cursive spindle]

I don't believe you are not allowed to use that

[tulip otter]

yea I think we probably can use it

[south patrol]

67

[tulip otter]

so whats the relation between the gcd and surjectivity, where can I read about that if its some special case of something that holds in a more general setting or something like that

[summer surge]

Tough an argument with norms more elegant imo, I suppose, you actually solved your problem here? You just need to find c and d such that imaginary part of rhs is b. It is a linear Diophantine equation

[cursive spindle]

Think about 2+2i and 4+3i

[tulip otter]

oh yea its just this lol, I was fixated on using both equations simultaneously but yea that makes things clear. tysm

[tulip otter]

so you want me to think about the surjectivity of f:Z->Z[i]/J when J=(2+2i)Z[i] and when J=(4+3i)Z[i]?

[rocky cloak]

If you've identified that it should probably be Z/25.

Then think about which element of Z/25 should correspond to i. This will then tell you which n you should use

[tulip otter]

(4+3i) should be mapped to (25) so (i) is mapped to (7)?

[summer surge]

Maybe its usefull to compare sizes of Ker f and Z[i] / J in both cases

In a setting of solution with Diophantine equations it is also clear why gcd is matter

[rocky cloak]

So then if i-7 is in the ideal. Then
a + bi + J = a + 7b + J

[tulip otter]

and n=a+7b should work in that case

[cursive spindle]

yup

[rocky cloak]

So all you have to do then is determine if i-7 is in the ideal

[tulip otter]

so now I just solve the system of equations given by i-7=(4+3i)(c+di)

[rocky cloak]

That's technically true, but it's probably easier to think about how you would invert 3 modulo 25 and go from there

But yeah I guess it's just a linear system of equations to solve, so not too bad

[tulip otter]

but doesnt all this implicitly assume that Z[i]/J is isomorphic to Z/25?

[rocky cloak]

No, just that 25 is in the ideal

(that you have presumably shown already)

[tulip otter]

ah right yea, the fact that i corresponds to 7 only depends on this

[rocky cloak]

Another way to think about it, avoiding computation even is to factor the map as
Z -> Z[i] -> Z/25[i] -> (Z/25[i])/(4+3i)

Then as 3 is a unit modulo 25 this is the same as (Z/25[i])/(i - a) for some a. So the original ideal is (25, i-a) for some a.

(that a happens to be 7, but you don't need to compute it)

[coral steeple]

What does $\alpha=\frac{2\beta-4}{12-\beta}$ have to do with finding the minimal polynomial?

[cloud walrus]

person2709505

[rocky cloak]

You can throw the expression into the minimal polynomial of alpha to get the polynomial below. Then you just have to check that it's irreducible / argue that the minimal polynomial of beta most have degree 3

[coral steeple]

Hmm. The only method I know for problems like this would involve writing the powers of beta as linear combinations of the first few powers of alpha, then solving a linear equation to find the minpoly

[tulip otter]

ohhh nice. Why did you think about factoring this way and then going back to Z[i]/(4+3i) after that?

[coral steeple]

Ok, I see that

[tulip otter]

is it just because 25 is in (4+3i)?

[rocky cloak]

Yeah, it's just about reducing the problem as much as possible.

We knew 25 was in the ideal and we're comfortable working modulo 25, then get 25 out of here!

[tulip otter]

yea I see why the gcd matters using diophantine equations, because b is arbitrary so if the gcd is not 1 then it wont divide b for some b and then there wont be a preimage there

[tulip otter]

based

[tulip otter]

@cursive spindle you probably didnt want me to view in this way did you?

anyways tysm everyone, have a great day/night!

[cursive spindle]

yes it all boils down to the diophantine equation

[tulip otter]

lol, its kinda funny how I was too close to the answer in the exam but then I kept moving in circles

[cursive spindle]

many such cases

[tulip otter]

I knew that you would say this, its like the standard response you use in situations like this

[cursive spindle]

idk exams do be like that what can I say

[tulip otter]

real

[cursive spindle]

tfw one time I proved a much harder statement cuz I was unsure if my proof was correct

[tulip otter]

anyways whats important is that I learned something new

[wraith cargo]

anyways yeah in the future for these types of questions it's good to imagine this kind of diagram

I had this in mind but then brainfarted everywhere

[cursive spindle]

tfw sublattice of Z^2

[wraith cargo]

you basically count the number of elements inside that grid

and it's easy to show this number is exactly the area of your fundamental cell in the lattice

or even writing the down the elements explicity isn't a horrible idea

[tulip otter]

I see, the idea of thinking about things geometrically and picturing things actually never crosses my mind tbh

I dont really like it that much (well at least until now), thats why I never think about things in this way

[wraith cargo]

born to be the antithesis of a geometer

[tulip otter]

like its not intentional, it just never crosses my mind

[tulip otter]

lol

but well I have to change this because interesting stuff seem to rely on geometry too

[wraith cargo]

like here specifically

this works

[cursive spindle]

tfw you don't read what I write

[wraith cargo]

I wouldn't recommend this is every possible situation

[tulip otter]

no vro I read it when you said that i can either compute the elements explicitly or think of it geometrically as a sublattice

but I didnt want to get into a geometric argument thats why I didnt talk about that

and at the same time i wanted to know what the relation between the gcd and surjectivity was so I asked you about that instead

[cursive spindle]

I tend to yawn when there is no geometric argument

[wraith cargo]

https://tenor.com/view/fake-larp-larp-larp-sahur-sahur-larp-larping-gif-18307483096237429166

[cursive spindle]

it's true tho lmao

[frail shoal]

what did i miss

ah, bro is a yawner

[cursive spindle]

https://tenor.com/view/woahm-anime-girl-anime-girl-anime-girl-sleepy-gif-1859288045321179131

[swift root]

still no monke

[frail shoal]

sometimes a fella wanna rabbit

[swift root]

yk thats valid

[cursive spindle]

https://klipy.com/gifs/bunny-too-cute

[glossy warren]

cant wait to start learning algebra very soon very soon

[next obsidian]

Girl started HRT and was like “I need to start algebra too”

[quiet pelican]

💯

[swift root]

many such cases

[somber goblet]

many such cases

[quiet pelican]

many such cases

[somber goblet]

i’m sure theres a natural transformation joke somewhere here

endo(crine)morphism

[next obsidian]

True

[trail cave]

If I want to show $Q(\sqrt{3}-\sqrt(5))=Q(\sqrt{3},\sqrt{5})$, how do I show that the first is contained in the second? I'm struggling with it though I feel like it should be simple. Is there something to do after squaring?

[cloud walrus]

Bottlecap Desu~(Bottlecap Gang)

[next obsidian]

Surely you mean that the second is contained in the first?

You just gotta play with it vro

This isn’t really a thing you can give a hint on without just saying what equation does it

[twilit wraith]

sqrt(3) - sqrt(5) is clearly an element of the latter

[trail cave]

Yea, sorry. I guess a hint doesn't make sense but I'm asking because this is an old homework I got points taken off on and it may be on my exam tomorrow

[trail cave]

I meant the other direction, sorry

If I square it then I can find $\sqrt{3}\sqrt{5} \in Q(\sqrt{3}-\sqrt{5})$ but I don't think that's the right direction

[cloud walrus]

Bottlecap Desu~(Bottlecap Gang)

[next obsidian]

Okay

Multiply that into sqrt(3) - sqrt(5)

Clearly you need to use an element not in Q to get further and adding doesn’t work well with radicals

You get 3sqrt(5) - 5sqrt(3)

Now you can subtract away some multiple of sqrt(3) - sqrt(5) to get a multiple of just sqrt(3) or sqrt(5)

Now divide and you win

[trail cave]

Thank you 🙂

[twilit wraith]

galois theory likely

oh nvm its not necessary

[next obsidian]

What the f are you talking about vro

😭

[trail cave]

lol

this is for my algebraic number theory senior seminar

[twilit wraith]

i mean the galois theory way is simple too

[trail cave]

but we've spent 3 months on just abstract algebra field theory stuff and then 2 days on some random theorems that he said were the number theory part and now an exam

[twilit wraith]

sqrt 3 - sqrt 5 is only fixed by the identity automorphism

so its the whole thing

[next obsidian]

You’d need to show the extension is normal or whatever

I don’t think it’s clear that the RHS is a splitting field

Without this equality

Idk

[twilit wraith]

i mean Q(sqrt 3, sqrt 5) is the splitting field of something

so it works out

oh actually maybe ts is not clear

nvm it is

(x^2 - 3)(x^2 - 5) is the polynomial this field is the splitting field of

and its separable

so we have a galois extension

[pastel vortex]

Can someone explain the euler’s totient function to me. It seems very unintuitive

[next obsidian]

Wdym

I think it’s quite intuitive, it has nice properties which make it easy to compute (assuming you can factor), and the definition is very simple

[pastel vortex]

Not to me 😅

[frail shoal]

do you feel you understand divisibility well

[pastel vortex]

Kind of

[next obsidian]

I think this probably is better in #elementary-number-theory just fyi

[twilit wraith]

do you know the integers mod n

[cursive spindle]

👁️

[pastel vortex]

Oh okay i asked this here because it seems its used to get the order of a multiplicative group

That i understand though

[frail shoal]

it's funny you say that cuz i was just about to sorta use that to explain how i intuit it

well ok i was still working out the details of my explanation

i think there's sorta a chain of concepts here

divisibility, gcd, and then euler totient

[pastel vortex]

I understand divisibility and gcd or at least i think. I might be missing something

[next obsidian]

Like

You need to say what you’re confused with

[vocal pebble]

it is defined so

[next obsidian]

This is too vague which is why I said I think this is simple

[pastel vortex]

The fraction inside the function seems unintuitive

[next obsidian]

The definition is “# of divisors” which is simple

The properties make it eas to compute

[vocal pebble]

the point is that a residue class [a] is invertible in Z/(n) iff a is coprime to n (so counting the order of this multiplicative group is the same as counting the number of coprime integers)

[next obsidian]

Understanding “why” the properties hold could not be simple and if that’s where you’re confused someone can explain that

[frail shoal]

gcd(m, n) is the largest k such that k | m and k | n. the way i was gonna explain it is, imagine you keep adding m in mod n arithmetic. so you look at 0, m, 2m, ... mod n. then gcd(m, n) can be thought of as the smallest nonzero number you can reach this way

[next obsidian]

Wait

Am I smoking crack rn

[frail shoal]

for example, mod 6, let's add 4 repeatedly
0 4 2 0 4 2 0 ...

notice how it only ever gets as small as 2

[next obsidian]

Lmfao

[frail shoal]

so gcd(4, 6) = 2

this is how i think of it when i visualize it in my head

[next obsidian]

Okay, yeah totient is coprime stuff, but the point still stands basically

[frail shoal]

even if this isn't the formal definition

[pastel vortex]

Yes i understand this

[frail shoal]

so essentially, totient function counts how many numbers have gcd 1

that is, they hit everything

[pastel vortex]

Like the function itself

[next obsidian]

Put the n inside

[frail shoal]

oh yea that's a shit definition to start with

it's true but a theorem (as opposed to a starting point)

[pastel vortex]

Wait its diving over all such prime divisors right?

[frail shoal]

yea

so for instance, φ(24) = 24 * (1 - 1/2) * (1 - 1/3)

because 2 and 3 are the prime divisors of 24

there is a way to think about this intuitively i think

[next obsidian]

Okay sorry, I am being confusing

[frail shoal]

suppose that gcd(n, 24) ≠ 1. then there must be some common prime p such that p | n and p | 24

[next obsidian]

Prove that the totient is multiplicative across coprime things

[novel star]

as the order of the group of units in Z/nZ?

[next obsidian]

Literally just write out everything as a product of primes

[frail shoal]

so first we delete 1/2 of the numbers 0 through 23, because 2 divides them. then we delete 1/3 of that, because 3 divides them

[next obsidian]

And this becomes clear

[twilit wraith]

the best definition of phi(n) is the number of positive integers less than n that are coprime to n

we care about this because the elements of Z/nZ which have multiplicative inverses are exactly the ones that are coprime to n

[next obsidian]

Like if m and n share no factors right

[frail shoal]

that's also true yea

[pastel vortex]

I should probably do that yea

[next obsidian]

But try to do that

And then this formula turns into this

For a power of a prime p^n

Show that phi(p^n) = p^n(1-1/p)

Which is p^n - p^n-1

[frail shoal]

oop i forgot to say this

[next obsidian]

This is hella clear, you just remove one in every p things, the multiples of p

Now you write n as a product of powers of primes p_i

Apply the multiplicativity and the nProd(1-1/p) pops out

[pastel vortex]

Sorry just reading through yalls explaination

[next obsidian]

Btw I think I can prove it like this.

phi(n) = n - # of things sharing factors with n that are less than or equal to n.

Assume m,n are coprime then
phi(n)•phi(m) = mn - n(share factors with m) - m(share factors with n) + (share factors with m)(share factors with n)

[fleet cairn]

A function $f: A \rightarrow B$ is said to be a monomorphism iff for all set Z and for all function $a', a'' : Z \rightarrow A$. It is true that $f \circ a'$ = $f \circ a''$ implies $a' = a''$.

Can someone give an example of a monomorphic function, and prove why it fits the definition? (Don't say it's because it's injective)

And I can't talk back and forth, so I'd appreciate if it said all at once.

[cloud walrus]

JŦК

[pastel vortex]

Wait this was a good explanation

I think i get it more

[frail shoal]

thanks

[pastel vortex]

Okay got it

Thank guys i need to look into it more but i think i understand better

[next obsidian]

So we need to show that the number of things which share a factor with mn is given by the latter part of this expression

Take the things which share a factor with m, and then take all multiples of it up to multiplying by n, this is n(shares factor with m) many. Do similar with things sharing a factor with n to get the n(shares factor with m) + m(shares factor with n)

You however overcounted by exactly (shares factor with m)(shares factor with n)

[next obsidian]

Getting this to be correct is where the m,n coprime is needed I believe

Cuz it’s literally like, how many things of the form x(shares a factor with m) will be a multiple of something sharing a factor with n? Well that’s exactly when x shares a factor with n cuz m and n share no factors

So there’s exactly # of shares a factor with n many choices for x

Anyway, my notation is so unbelievably ass so I hope it makes sense

I typed this on phone

[coral steeple]

How is this: f = x^5-6x+3 is not solvable by radicals over Q. Since (f,f')=1, f is separable. By Eisenstein's criterion with the prime 3, f is irreducible. Hence, its Galois group contains every automorphism which interchanges any two distinct roots and leaves the other three fixed, so (since the transpositions of S5 generate the whole group) the Galois group is S5. This is not solvable, so the splitting field of f is not solvable over Q, meaning that f is not solvable by radicals over Q.

Actually I don't even need to show that f is separable; this follows from irreducibility in char 0

[coral steeple]

This is not correct

I guess the Galois group has a 5-cycle by Cauchy's theorem, and it also contains complex conjugation, so these are enough to generate S5

(and the accompanying statement is that any irreducible separable quintic with a complex root is not solvable by radicals)

[rocky cloak]

It's also key that there are exactly two complex roots. Otherwise complex conjugation is not a transposition

For example x^5 - 2 is solvable

[coral steeple]

Well if there is one complex root then there is its conjugate no?

Oh hmm

Oh I see what you mean

Shoot

I suppose the way to do it is to graph it and find three (and only three) roots in R

[coral steeple]

Is $\mathbb Q(\sqrt3,\sqrt7)\subseteq\mathbb Q(\sqrt3,\sqrt7,\exp(2\pi i/5))$ a degree 4 extension?

[cloud walrus]

person2709505

[coral steeple]

The question is to find the Galois group of (x^2-3)(x^2-7)(x^5-1).

Ah, maybe I can think about the extension as $\mathbb Q(\zeta_5)\subseteq\mathbb Q(\zeta_5,\sqrt 3+\lambda\sqrt 7)$, where lambda is a primitive element of the adjunction of the two square roots

[cloud walrus]

person2709505

[coral steeple]

Nope

(meant to write "where lambda is chosen such that sqrt3+lambda sqrt7 is a primitive element", anyhow)

[vocal pebble]

the galois group acts transitively on the irreducible factors, so one way might be to try to figure out what that (if I can count, 16 element) subgroup of S_9 is

[desert verge]

surely there are many such subgroups

[vocal pebble]

Yeah maybe, but the ones in question are all isomorphic to the galois group of that polynomial

The automorphisms act in a certain way. If you labelled the roots 1 through 9 and looked at the action that should correspond to one concrete subgroup

[coral steeple]

If I had to guess, it would be C_2^2 times the group of units of Z/5Z. It's only a guess but idk how to justify recovering the group from its action on the irreducible factors any more than I can justify the guess

[vocal pebble]

Well different automorphisms must permute the roots of atleast one irreducible factor differently

[coral steeple]

brb

[vocal pebble]

So you have a injective map from the galois group to the guess group, and you can show its surjective, because any element in C2xC2xC4 will correspond to an automorphism in the galois group

[coral steeple]

It's the case that for a tower K \subset L \subset M, elements of the lower Galois group can be extended to elements of the upper right? At least if we have some assumptions on K \subset M? I think that's what I want to use here, but I don't think we've proved it...

[vocal pebble]

But you can just directly provide an isomorphism between the galois group and your guess

[coral steeple]

I don't have very many results at my disposal to say what the elements of my galois group are. I know that if there are nontrivial elements then they permute the roots of the irreducible factors...