groups-rings-fields

[cobalt heath]

Then you already got associativity

[dull ginkgo]

it's just rote checking stuff which is rather obvious at a glance

convolution with the constant 1 function gives the sum over divisors then

and it's inverse must be the mobius function

so i guess it's a unit

[delicate bloom]

working out inverses is pretty easy fortunately

[dull ginkgo]

finding a g such that the sum over f(a)g(b) = 1 hm

[delicate bloom]

well

what's the identity first

it's not the function u(n)=1 for all n

[dull ginkgo]

the "dirac delta" in this case just \delta(n) = 1 if n = 1, else 0

[delicate bloom]

sure, I guess more like a kronecker delta lol

[dull ginkgo]

that's what i meant by 1 here, ring torture

i guess you can recursively define it

[delicate bloom]

yep

[dull ginkgo]

We would want f(1) to be a unit in whatever our image ring is tho

actually

hmm

[delicate bloom]

yes

although there is a stricter condition if you want f to be multiplicative which it doesn't need to be

[dull ginkgo]

We can consider the ring of characters from N under multiplication to an integral domain R, with multiplication being dirichlet convolution. Wouldn't the group of units of this new ring be the characters such that f(1) is a unit of R

i had to course correct that fast because probably zero divisors will fuck your day up

[delicate bloom]

you can just write out the recursive formula for the inverse of f explicitly

[dull ginkgo]

let g(n) be f(n)'s dirichlet inverse

then obviously g(1) = f(1)^-1

uhhh consider dirichlet convolution of g and f being 0 for n > 1

take out the pair where it's f(1)g(n)

then just, move the remaining sum to the other side, multiply by g(1)

boom g(n)

recursively lol or via strong induction

[delicate bloom]

no difference imo

[dull ginkgo]

yeh

[dull ginkgo]

actually I guess we won't have issues if R is not a domain

since f(1) is guaranteed to be a unit if f is a unit itself

welcome back to psychopathic generalization with miz at 2 am

[delicate bloom]

if you specifically want to focus just on the group of multiplicative functions the condition is a bit stronger

[dull ginkgo]

oh i was just considering the whole ring

[delicate bloom]

I know

[dull ginkgo]

multiplicative functions if f(1) = 0 then f(n) = 0 for all n

right?

[delicate bloom]

sure

[dull ginkgo]

the coprimality thing probably leaves things weird actually

1 is a unit so it's not coprime to anything

[delicate bloom]

f(n)=f(n*1)=f(n)f(1)

you better have f(1)=1

[dull ginkgo]

oh wait it's the kernel of the map f to f(1)

wait no

subgroup of that

[delicate bloom]

I'm talking about the group of multiplicative functions

the multiplicative functions don't form a ring

[dull ginkgo]

it's a subgroup of the group of units right?

excluding the trivial multiplicative function being 0

[delicate bloom]

yeah, excluding 0

[dull ginkgo]

neato

so the mobius inversion formula boils down to the mobius function being the dirichlet inverse of the constant 1

huh

i guess the def of it can be derived from the recursion too

thanks man, that's some neat shit

[delicate bloom]

actually seeing how this relates to dirichlet series would blow you away tbh

[dull ginkgo]

Isn't it used to prove infinitude of primes in arithmetic progressions via the characters to C

[delicate bloom]

$$D(f;s)=\sum_{n \ge 1} \frac{f(n)}{n^s}$$
so $\zeta(s)$ is the dirichlet generating function for all 1s.

[cloud walrus]

Merosity

[dull ginkgo]

1/zeta(s) must be the mobius above the n^s then

[delicate bloom]

exactly

and this is obvious from the euler product

$$\frac{1}{\zeta(s)}=D(\mu;s)=\prod_p \left(1-\frac{1}{p^s}\right)$$

[cloud walrus]

Merosity

[dull ginkgo]

oh true

that's for multiplicative functions right

[delicate bloom]

each prime only contributes once

[dull ginkgo]

same sort of idea

[delicate bloom]

every multiplicative function has an euler product yes

$$D(f;s) = \prod_p \sum_{k \ge 0} \frac{f(p^k)}{p^{ks}}$$

[cloud walrus]

Merosity

[dull ginkgo]

the analytic continuation crap probably isn't too important here because I assume it's just from analyticity and no lacunary bullshit

it's just there

[delicate bloom]

no analysis

[cobalt heath]

Are we doing analysis now

[delicate bloom]

just talking about formal series here as generating functions

[dull ginkgo]

no, also zeta is special because you can just do poisson summation on 1/(n + t)^s and get an immediate analytic continuation anyway so it doesn't work nicely for formal series anyway

analysis is boring

[cobalt heath]

Ah, generating function

[dull ginkgo]

unless there's cool algebra shit like harmonic analysis that shit rocks

Anyway thanks man lol

[delicate bloom]

yw

[dull ginkgo]

Here’s a little thing I wonder

Assume we have total multiplicative functions $h$ such that $\sum_{n = 1}^{\infty}{f(n)h(n)}$ and $\sum_{n = 1}^{\infty}{g(n)h(n)}$ converge, then would their product be $\sum_{n = 1}^{\infty}{(f * g)(n)h(n)}$?

[cloud walrus]

Miz

[dull ginkgo]

@delicate bloom would you need total multiplicativity for this, or would coprime be enough?

[delicate bloom]

I thought you said

[dull ginkgo]

convergence is only so it exists so I can ask in the first place about the convolution

[delicate bloom]

I don't really see a reason to be interested in this question

[cobalt heath]

I bet proving this is boring

[dull ginkgo]

fair, also it’s obvious now that I think about it

[delicate bloom]

you sure?

[dull ginkgo]

We’d need total multiplicativity of h, but if so then we’d have a double series over f(n)h(n)*g(m)h(m) = f(n)g(m)h(nm) so we can add up the terms with h(nm) together as a dirichlet convolution at nm

Thus proving it

[delicate bloom]

I don't buy it

[dull ginkgo]

and we know the product series converges if one of them is absolute which we can just append to the assumptions

The convergence itself or the equality if all three converge

Because tbf I don’t care about convergence I’m just assuming all 3 do

[delicate bloom]

not sure if I understand what you're saying now

[dull ginkgo]

If all three converge then the product is the sum over (f*g)(n)h(n)

[delicate bloom]

isnt the question asking if 2 converge, does the 3rd?

[dull ginkgo]

Moreso if all 3 converge then the product is the same as the third

I stated it poorly

[delicate bloom]

more seriously, if h(n)=1/n^3, this is completely multiplicative and then picking f(n)=g(n)=n makes the sums f(n)h(n)=1/n^2 converge, but the convolution (f * f)(n)= n*tau(n) but tau(n)/n^2 diverges

[dull ginkgo]

interesting

[delicate bloom]

while we're here since I just used it, try to prove that completely multiplicative functions distribute over the dirichlet convolution. Then as a lemma you get for $u(n)=1$ and $\delta$ being the identity, $u\star \mu = \delta$ take a completely multiplicative function C and multiply it through, $C \star \mu C = \delta$ so the dirichlet inverse of $C(n)$ is $\mu(n)C(n)$

[cloud walrus]

Merosity

[delicate bloom]

actually I didn't think about it, I just trusted wolfram alpha

,w sum from n=1 to infinity of tau(n)/n^2

[cloud walrus]

Results provided by WolframAlpha

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

Requested by merosity

[delicate bloom]

but it should just be zeta(2)^2 which converges smh

maybe WA is right and the analysis does shake out that way combining the two infinite series, whatever

[dull ginkgo]

Mero stop being gaslight by WA

[west nova]

how to show this

[dull ginkgo]

Show that (x^2 + 1, y^2 + 1) isn’t prime

Or that there’s a zero divisor

Want a hint?

[west nova]

does it boil down to showing that x²+1 and y²+1 aren't prime since ideals generated by irreducible elements are prime iff the elements are also prime elements

[dull ginkgo]

here’s a hint

Show x + y is a zero divisor

it involves a common identity :3

||x + y does not lie in (x^2 + 1, y^2 + 1) but (x + y)(x - y) = x^2 - y^2 = (x^2 + 1) - (y^2 + 1) = 0 in our ring thus it is a zero divisor||

[dull ginkgo]

uhhh

[delicate bloom]

yeah main thing I'd say is just play around with elements, the main red flag is we know C is a field where we've already added solutions to x^2+1, so adding more elements satisfying the same equation squeezes more roots than the degree of the polynomial can handle to call itself a field.

[west nova]

oh yeh there was this easy way

[rotund aurora]

This is wrong

It says it also diverges for sigma0(n)/n^6

sum sigma(n)/n^s=zeta(s)^2 for Re(s)>1 and at s=1 both sides diverge

[rotund aurora]

this is false for h(n)=1 already

I have been always interested in aspects of non-absolute convergence of Dirichlet series, but never really got the chance to explore it too much I must admit

I wonder what the function $(-1)^n*(-1)^n$ looks like

[cloud walrus]

croqueta3385

[delicate bloom]

I split it up into divisors ab=n where a=b mod 2 and a != b mod 2, something like tau(n)-2tau(n/2^v_2(n))

basically the power of 2 gets forced into one of the divisors when a != b mod 2, I think what I wrote is not right cause for odd numbers we just have tau(n) directly but heading to bed

I think trying to find examples of this form is ultimately doomed to fail since they look like products of convergent dirichlet series

[rotund aurora]

If n=2^k * m with m odd then the positive contribution comes from { d divides n such that d and n/d have the same parity } and the number of such pairs is d(m) · |k-1|

and the negative contribution will be 2·d(m) for k>=1 and 0 otherwise

you have the sum $\sum_{n\geq 1}\frac{d(m)(|k-1|-2\chi(n))}{n^s}$ where $\chi(n)=0$ if $n$ is odd and $1$ if it is even

[cloud walrus]

croqueta3385

[rotund aurora]

but $2\chi(n)=1+(-1)^n$ and it is possible that $\sum_{n\geq 1} \frac{d(m)(-1)^n}{n^s}$ converges

[cloud walrus]

croqueta3385

[rotund aurora]

so you will be left with $\sum_{n\geq 1}\frac{d(m)|k-1|-2d(m)}{n^s}$

[cloud walrus]

croqueta3385

[wraith swan]

Anyone a hint on how to proof there exists no element in (Z/pqZ)* of order (p-1)(q-1), the order of the group, for p /= q are odd primes? The exercise is to proof the group is not cyclic, by showing that element does not exist.

[glad osprey]

Maybe the Carmichael lambda function could help

[wraith swan]

Tak alfodr, I will read into it.

[rocky cloak]

Are you familiar with Chinese remainder theorem and how
(Z/pq)^* = (Z/p x Z/q)^* = (Z/p)^* x (Z/q)^*?

[wraith swan]

No, but I now get another hint they gave. Which was that if gcd(m, n) = 1 then C_m x C_n = C_{mn}. I thought that would only give me the order of (Z/pq)*, which I already got by just counting.

[hidden wind]

whaaat

why do i have to be so bad at elementary number theory

oh well summer is coming up and i plan to spend it learning ENT mwahaha

[wraith swan]

I will look into the Chinese remainder theorem, but where does the other come from? It feels like I missed something in the book, but it really isn't there...

I have proved 4.9, but couldn't see how I can use that

[rocky cloak]

I don't really see how 4.9 is relevant either

[wraith swan]

Well I guess (Z/pqZ)* ~= (Z/pZ x Z/qZ)*?

[rocky cloak]

Yeah, but without the ring structure it doesn't seem like you can see how ^* relates to it

[wraith swan]

Ok

[rocky cloak]

Anyway, the Chinese remainder theorem does say exactly that Z/pq = Z/p x Z/q as rings

So two integers are congruent modulo pq iff they're congruent modulo p and modulo q

From there you can use that x^(p-1) = 1 modulo p and x^(q-1) = 1 modulo q

[wraith swan]

Ok thank you

[tardy hedge]

So like a group action is kind of just realizing the permutations that G contains (cayleys theorem thing) in more concrete settings

does that make sense?

Probably better way of saying it is not "permutations that G contains" but the permutation groups isomorphic to normal subgroups of G?

in essence all of this is just different ways of writing out homomorphisms for G, right?

[coral spindle]

Yes

One way of thinking about group actions that I find somewhat helpful, conceptually, is that groups are an abstracted set of symmetries, and an action simply un-abstracts it and makes it an actual symmetry again.

Of course this is just how you conceptualise it

[tardy hedge]

Ah yeah

Yeah

Its a fun concept so far

Saying permutation groups isomorphic to normal subgroups of G was the correct way of phrasing it right

hm

[limpid ferry]

Hey I am struggling with proving this, maybe some hints will help!

[wraith swan]

What do you know about the order |[m]_n| in Z/nZ?

[wraith swan]

Z_n are the integers congruence modulo n, right?

[limpid ferry]

Z_n is the set {0,1,...,n-1} and a group under addition modulo n

[wraith swan]

Ok then can you do something with my hint? Can you find the order of an arbitrary element in Z_n?

[coral spindle]

Do you know Bézout's lemma?

[limpid ferry]

Supopse b = a^k, k being a generator, b has order n/d, n is the group's order and d is gcd(k,n)

[limpid ferry]

Yes

[coral spindle]

Use Bézout's lemma.

Gonna tell you this now

Lethe's hint requires you to do a much harder proof.

And besides, to prove it you will have to use Bézout's lemma anyway.

[wraith swan]

huh, does it?

[coral spindle]

Yes.

You are asking Mike to prove something far more complicated than what's necessary.

[wraith swan]

Ok I thought that the order |g^n| = |g|/gcd(|g|, n) was something basic to group theory. Was one of the first propositions i learned.

And only that is really needed, right?

[coral spindle]

So try proving it.

I would rather do this with tools that I know Mike has learned, and generally only with elementary number theory.

It is also completely unnecessary to derive a formula for the order of elements of Z/nZ here

[wraith swan]

[1]_1 has order n and thus generates Z_n. Any other element [m]_n = m [1]_1 and has order n/gcd(m, n) and generates Z_n if and only if gcd(m, n) = 1. ?

[coral spindle]

That's not what I was suggesting you do.

Try proving |g^n| = |g|/gcd(|g|, n).

[wraith swan]

But you can say the same about Bézout then, right?

[coral spindle]

Not really, it's much easier to prove, and in fact you will likely use Bézout to prove the above.

But more importantly

[wraith swan]

ok

[coral spindle]

I asked Mike if he knew Bézout.

[limpid ferry]

I have gcd(r,n) = ar + bn (a,b \in Z). Do I need to consider the order of r at all to continue?

[coral spindle]

No, you don't need the order of r at all.

Hint: r generates the whole group if and only if 1 is in <r>

[limpid ferry]

if Z_n = <r>, then I have <1> = <r>

[coral spindle]

Of course, yes.

It's a hint, so it's up to you to see why it's relevant.

[limpid ferry]

hey is the hint saying 1 = r^k or r= 1^k (k \in Z) ? I am not so sure about it

[quartz wind]

is this correct

[coral spindle]

Both.

But you are possibly making a mistake: we are looking at the additive group Z_n. No multiplication is happening, only repeated addition.

We would usually write 1 = k.r instead.

[quartz wind]

well its correct, but what i wanna know is, is this how you normally write these proofs? or do i need to clarify what im doing in some steps?

[dull ginkgo]

me when Abelian group sottue

[dull ginkgo]

as long as I understand it and I can describe it in English I tend to be lax when writing it down

Sometimes it can be a real pain in the dick to symbolically do, e.g. the shit with dirichlet convolutions I was doing last night

[quartz wind]

hmm okay thanks. the one proof based course i had so far i had to write down a table and say like, "for a,b in N, a+b in N is true by closure of N under addition" and every step had to be like that

[dull ginkgo]

[quartz wind]

but in books im seeing thats not the case

[delicate orchid]

just be careful, you can only conclude your assumption is correct from arriving at a tautology if each step of the proof is reversable

[dull ginkgo]

1:30 am

[delicate orchid]

I am not reading that

[quartz wind]

yeah you cant multiply both sides by 0 to get a tautology you mean im pretty sure i didnt do anything like that here

[dull ginkgo]

Never told you to britoid

[delicate orchid]

nah dw you didn't (or at least I'm not spotting anywhere where you did)

luckily group multiplication is an invertible operation!

[quartz wind]

alright ty

[dull ginkgo]

please specify wtf you mean here because either you mean left multiplication action on itself, or like the endomorphism ring on Z/nZ under addition

[limpid ferry]

Sorry for the wrong notation but I know we are using additive operation. I actually still struggles on how to move on from the given hint

[dull ginkgo]

What’s the group in context?

Multiplicative group (of units) mod n or just addition mod n (cyclic)

[limpid ferry]

addition mod n sir

[delicate orchid]

r u daft

[dull ginkgo]

yes

[delicate orchid]

what are the four properties a function GxG -> G needs to be a group operation

three, I suppose

closure always was stupid

[delicate orchid]

if r and n are coprime what does bezout's tell you

[limpid ferry]

ra+nb = 1(a,b \in Z)

[glad osprey]

It might help to know that r mod n is just a linear combination of r and n

[delicate orchid]

ok so if we're working mod n this is just ra = 1, agree?

[dull ginkgo]

What are we proving

[crystal vale]

This one

[south patrol]

Ah nice

[limpid ferry]

Yes, ra = 1 (mod n)

[delicate orchid]

can you see that 1 generates Z_n?

[limpid ferry]

I do

[delicate orchid]

so can you see how r therefore has to generate all of Z_n?

[coral spindle]

You are trying to find the r such that <r> = Z_n.

I have given you the hint that <r> = Z_n iff 1 \in <r>.

So perhaps you can use this to set a new goal that Bézout applies to.

[coral spindle]

And once you're happy with what Wew's helping you with, come back to this message here

[dull ginkgo]

So it must be |G|/k

[coral spindle]

Mizalign if you spoil another person's exercise again I will blow my top

[delicate orchid]

I'm mod abusing

FUCKLE 🗣️ 📢 🔥🔥🔥🔥🔥 FOOT THEY TALKIN BOUT?? 🗣️ 🔥

[south patrol]

Under what valuation

[dull ginkgo]

I thought they answered it already lmao and I was providing another way to think about it oof

Thanks for deleting it though I was gonna do that lol

[tardy hedge]

Somehow im confused on that simple proof that p-groups have nontrivial center. The one that uses class equation

[delicate orchid]

oh that's a good one

[coral spindle]

All time classic

[tardy hedge]

Hahahaha

[delicate orchid]

it's really just orbit stabilser

[coral spindle]

What's your confusion

[dull ginkgo]

Tldr let G (or Inn(G)) act on G by conjugation, then what is Z(G) in this context?

[coral spindle]

Let's hear the confusion first.

[tardy hedge]

It will have to wait until my break because my shift starts rn!

To be continued

[delicate orchid]

you know each orbit under the conjugation action has to divide the order of p, and thus is a power of p. The identity is definitely in the centre... so u got p^k-1 elements left!! That ain't a power of p!!! U need some other mfs to be in their own crib........ chill ass mfs................................ they ain't sharin this hizzouse....

only one dawg in this jawn...

[coral spindle]

U can't do this to me................ @ me

[crystal vale]

4th question

Let A be the set of all real numbers and B be the set of all natural numbers then let the permutation f(x) = 2x but it's inverse is g(x) = x/2 , B is invariant under f means f(B) is a subset of B but g(B) is not a subset of B

Is it correct?

[delicate orchid]

I used a similar technique in one of my results... yet to be published....

[crystal vale]

Do you manage any blog ?

[delicate orchid]

I am not tapped into the blogosphere, no

[delicate orchid]

yeah

nice counter example

[crystal vale]

Okay, thank you

[wraith swan]

Ok, back from dinner. Maybe it's a little handwavy but isn't it essentially
|g^m| is the smallest integer d such that (g^m)^d = g^(md) = e.
Then the order of g, |g|, divides md.
Doesn't it then follow that |g|/gcd(m, |g|) divides d and since d is the smallest such integer d = |g^m| = |g|/gcd(m, |g|)?

[coral spindle]

Doesn't it then follow that |g|/gcd(m, |g|) divides d
Yes
and since d is the smallest such integer d = |g^m| = |g|/gcd(m, |g|)?
No. It could be any multiple of |g|/gcd(m, |g|) at this point. You haven't justified why |g|/gcd(m, |g|) is the smallest.

[wraith swan]

hm ok

[coral spindle]

Hint: a useful lemma is to show that the smallest (by the usual order on {0, ..., n-1}) generator of <r> is equal to the index of <r> in Z_n.

There are other ways, but no matter.

[quartz wind]

is the writing here sufficient or nah

[wraith swan]

wait, I got that |g^m| is a multiple of |g|/gcd(m, |g|) and it is easy to show that (g^m)^(|g|/gcd(m, |g|))= e, wouldn't that show that |g|/gcd(m, |g|) is the smallest?

[coral spindle]

Indeed that does

[wraith swan]

But now to show |g| divides md... 😛

[coral spindle]

There are lots of small connecting steps, yes

[wraith swan]

Ok thank you

I didn't really do any number theory, so it is still pretty handwavy for me.

[languid trellis]

Suppose that f(x) is reducible in Q[x]. First of all, as the coefficients are in Z, we may take the gcd, d = gcd(a0, .... , an). So, we have f(x) = df_1(x), where f_1(x) is primitive. f(x) is reducible in Q[x] implies that f_1(x) is reducible, say f_1(x) = g_1(x)g_2(x). Call the coefficients of f_1, g_1, g_2, a'j, b'j, c'j, respectively, 1 <= j <= n. a'0 = b'0c'0, as p | a'0 but p^2 doesn't divide a'0, we have that p | b'0 exclusive or p | c'0. I'm slightly stuck from here, I would appreciate any hint

another idea I had was to show that f is irreducible in Z[x], which immediately implies that it is irreducible in Q[x], as Q is the field of fractions of Z

Or irred in Q[x] iff no rational roots

[rocky cloak]

So wlog we may assume p divides b0, and that p does not divide c0.

a1 = b1c0 + b0c1
means p divides b1c0, so p divides b1.

Continue inductively until you conclude that g1 is a multiple of p, reach a contradiction.

[languid trellis]

oh I literally wrote this down on my page, but I only realise now that g1, g2 are polynomials of degree < n, not n

That's clear now, thank you jagr

I did that then kept thinking. "oh but p doesn't need to divide b'n or c'n", but neither c'n nor b'n exist

[dull ginkgo]

Are the prime elements of D[X] for a gcf domain the primes of D and primitive, irreducible polynomials of D[X]

Actually yeah that seems to check out

[south patrol]

Also, a very pretty way is to consider this all modulo p

I believe this is probably the standard method nowadays

[dull ginkgo]

yeah that's what I did

[dull ginkgo]

From this, Z[X]/(p) ~= F_p[X] is obviously an integral domain (this is third iso theorem as X is disjoint from Z)

[limpid ferry]

Hey this is what I have found so far with your hint and still have not made any progress:
1 \in <r> => k.r = 1 (mod n) => k.r = k'.n + 1

[dull ginkgo]

tbh bezout here seems like mega overkill if it's for group orders

[coral spindle]

k.r = k'.n + 1 is equivalent to k.r - k'.n = 1.

[south patrol]

Idk how Bezout is mega overkill lol

Isn't that a special case of what is currently being proven

[coral spindle]

It is

[south patrol]

Sorry I mean, idk how to prove it withotu bezout

I don't think it actually follows but like yeah

[dull ginkgo]

are we proving multiplicative group mod n has inverses for coprime m

[coral spindle]

This theorem – that the units of Z/nZ are those numbers coprime to n — is equivalent to Bézout with the restriction that the GCD is 1.

[south patrol]

We're counting the number of generators of Z/nZ, which amounts to the same thing more or less

[dull ginkgo]

OHHHH

As a ring?

[coral spindle]

No

[dull ginkgo]

or just Z/nZ additive

[coral spindle]

I am rephrasing it.

[south patrol]

I think the two are equivalent tbf

[coral spindle]

The ring is a quotient of Z so it is generated by {} as a ring.

[dull ginkgo]

by (1) mod n

[coral spindle]

We have to talk about being generated as a group for it to be interesting.

[coral spindle]

No, I really mean that as a ring it is generated by {}.

The smallest subring of Z/nZ containing {} is Z/nZ.

Of course, unitality makes things work here.

[south patrol]

Oh in that sense sure lol

I was thinking like as a Z/nZ module lol I'm silly

[coral spindle]

I suppose it is also equivalent if we look at nonunital subrings

But only bc it's a quotient of Z!!!!!!!

[dull ginkgo]

sotrue

What are we starting out with

[coral spindle]

It is in fact true

[dull ginkgo]

We're trying to prove that Z/nZ (ADDITIVE) has one generator

[limpid ferry]

yes, if that's the answer then I have another question, is k.r the same as k multiplied by r?

[coral spindle]

mod n, k.r is kr

[dull ginkgo]

or just a cyclic group has one generator

[coral spindle]

But remember your definition of Z_n

[dull ginkgo]

I.e there is an element of order |G|

[coral spindle]

Annoyingly(!) your course defined it as Z_n = {0, 1, ... n-1}

In any case, you can just look modulo n.

You should know that if kr = 1 (mod n) then k.r = 1.

If you're not happy with that, then now is your time to say so

[dull ginkgo]

Trying to go about finding a general formula for this possibly

Last time posting this here because I always get stuck lol

Every monic linear polynomial is of the form X + a for a in F, so there are q many. All are irreducible

Now, every monic quadratic polynomial is of the form X + aX + b for (a,b) in F^2, thus there are q^2 many. We can remove quadratics that are squares of monics, i.e (X + a)^2 so there are q^2 - q remaining. Finally, the reducible monic quadratics are pairs (X - a)(X - b) where a neq b, so there are (N 2) many choices, or (q^2 - q)/2, half of the remaining. thus there are (q^2 - q)/2 irreducible monic quadratics

I am also trying to quantify another way that I am partitioning the sets of monic polynomials of degree n

through the divisors of n

[south patrol]

i imagine there is a way to interpret in terms of GL_n(F_q)

[rocky cloak]

An alternative approach could be to use that there is a unique field of any finite order.

And any monic irreducible polynomial is the minimal polynomial of some root

[south patrol]

like with companion matrices

[dull ginkgo]

That is corollary 2 which is almost 100 pages out

[south patrol]

Lol

Rippies

[dull ginkgo]

That gives $q^n = \sum_{d | n}{dN_F(d)}$

[cloud walrus]

Miz

[south patrol]

le moebius

[dull ginkgo]

and you can use Mobius inversion

Yeah

which the prior questions were about

[south patrol]

gotem

[dull ginkgo]

so I expect there is a connection that I am missing

[dull ginkgo]

At a glace it seems like partitioning the set of monic polynomials of degree N somehow

but I can't work it out

[rocky cloak]

So the approach I'm thinking:
g(n) = q^n
f(d) = number of elements that generate the field of order q^d
Then Möbius inversion gives
f(n) = sum_d mu(n/d) q^d
so there are
sum_d mu(n/d) q^d / n
irreducibles of degree n

[dull ginkgo]

once again

we haven't gotten there yet in the textbook

[rocky cloak]

I mean the exercise only asks you to do it for 2 and 3, which you can just do by hand

[dull ginkgo]

though this makes sense to me

[dim widget]

If F_q is any field with q elements then x \to x^p is an automorphism, so so is x \to x^q. Because any finite subgroup of the multiplicative group of a field is cyclic we know that x^q = x for all x \in F_q. This means that every x is a root of the polynomial x^{q-1} - 1 (so the field is unique, as it’s the splitting field of this polynomial over Fp), and as the multiplicative group is cyclic of order q-1 we see that there are Phi(q-1) generators of this field.

now you’ve gotten there!

so you can use Jagr’s formula

[rocky cloak]

The number of generators for the field and the number of generators for the multiplicative group isn't quite the same though.

Like for the field with 9 elements, there are 6 possible generators for the field, but only 4 that generate the multiplicative group.

Also, I guess generate the field should really be generate the extension, so actually depends on the base field.

[dim widget]

ah I didn’t really see your method

[dull ginkgo]

Idk where to use the PID condition in proving a PID is a UFD

or where I'm missing something

[dim widget]

but then it’s even easier to find the number of generators, so that’s good news

[dim widget]

the two places it’s useful is showing that gcds exist and showing every irreducible element is prime

[dull ginkgo]

every irreducible is prime?

yeah

is the main part here showing that principal ideals of irreducibles are maximal ideals

[rocky cloak]

That's a possible approach at least

[dull ginkgo]

i'd take that as a yes. Like for example in Z[X], 2 is irreducible but (2) is contained in (2, X + 1) i think

[dim widget]

I mean that doesn’t sound like a dumb thing to say

[dull ginkgo]

okay irreducible elements being maximal is what I was looking for

thanks

[barren sierra]

reviewing for quals in the fall

and it's been a while since I looked at some of this stuff

why do I care about

• composition series
• derived series
• lower central series

outside of classifying finite groups

[delicate orchid]

character theory

[barren sierra]

I haven't learned enough rep / character theory for these to show up I guess

[delicate orchid]

https://tenor.com/view/ash-ashh-baby-ia-gif-16729153814109504694

[rocky cloak]

Classifying finite groups is already a good reason tbh

[barren sierra]

yea fair

[dim widget]

Even if you don’t care about the classsification problem it’s a useful toolkit for analyzing how to build complicated groups out of smaller ones, and it shows up in nature

the quintessential example of the derived series is the natural filtration on unipotent nxn matrices over a field (which is the prototypical solvable group)

[barren sierra]

Trivial means there's a map r: B - A such that r o phi = id

I've gotten that A is abelian and then I'm stuck

[south patrol]

usually one says split exact

[barren sierra]

that'd be a map s: Z -> B such that psi o s = Z

as far as I can tell trivial isn't a standard term

[south patrol]

they are the same thing at least in abelian categories

groups aren't an abelian category so maybe the two don't coincide here hm

damn yeah they are different

[barren sierra]

they don't

[south patrol]

left split and right split etc

[vernal elk]

do you guys recommend that i study number theory before studying abstract algebra?

[barren sierra]

not a bad idea especially if you don't know proof writing that well

elementary number theory is relatively straightforward

I'd also look at a proof based linear algebra text as well since alot of those themes come up in abstract algebra

but if you feel comfortable with proof based lin alg then just starting with abstract algebra is also fine (I never took number theory before my first algebra course)

[summer path]

you can probably do elementary, or analytic (assuming you know little complex)

[vernal elk]

well i do understand proof writing etc pretty well but i really sucked my number theory classes back then. I know very little number theory. I now want to study abstract algebra, mostly because it pops up all the time in linear algebra and projective geometry, but from what I've see it in part generalizes many results of number theory, so I was wondering if I would get a better intuition of abstract algebra by first understanding number theory well

[barren sierra]

maybe but I mean you should be getting enough "motivation" (i.e. why we care about certain theorems and constructions) from lin alg and projective geometry

so I wouldn't say it's necessary

so really I think it's your call

it can't hurt

[vernal elk]

alright then, thanks!

[dull ginkgo]

would I have to use Eisenstein criterion to show that $\sum_{n = 0}^{p - 1}{x^{nk}}$ is irreducible

[cloud walrus]

Miz

[celest furnace]

No one is telling you to do anything. Do you mean could?

Also, what is k?

[dull ginkgo]

any integer

[celest furnace]

I don’t know if there is an easy proof of this

All the ones I know are just this is the minimal polynomial of some nth root of unity

[cobalt heath]

Is this irreducible?

I somehow doubt it

(I think 1+x+..+x^{p-1} is indeed irreducible)

[dull ginkgo]

like putting powers into polynomials I think is finnicky?

Like f(x) = x - 1 is irreducible right

but f(x^2) = x^2 - 1 = (x - 1)(x + 1)

[cobalt heath]

Yep, so you need to, like, verify individually

[dull ginkgo]

However $\sum_{n = 0}^{p - 1}{(x + 1)^{nk}} = \sum_{n = 0}^{p - 1}{\binom{p}{n + 1}x^{nk}}$ still satisfies Eisenstein's criterion

[cloud walrus]

Miz

[cobalt heath]

Oh, does it?

Then it is irreducible, yeah

[dull ginkgo]

the last coefficient is just p

so p^2 doesn't divide it

[cobalt heath]

How about other coe- ah

[dull ginkgo]

but p divides the rest

p doesn't divide n! for n < p

and the binomial coefficient has p! on the top and a denominator of two factorials with inputs less than p

[cobalt heath]

Interesting, so if Eisenstein holds for f(x), it holds for f(x^p) as well

[dull ginkgo]

so p will still divide it

[dull ginkgo]

yes

p | 0

[cobalt heath]

This seems off

[cobalt heath]

How does this hold? I think k shouls appear in the coefficients.

[dull ginkgo]

oh wait shit

[dull ginkgo]

no i don't think so,

it's just me doing x |-> x^k

on the basic sum

[cobalt heath]

Yeah then it should be

\sum (x^k + 1)^n = ...

Also an issue is that pk-th root of unity should be root of this polynomial, which covers almost all roots of unity.
But there are cyclotomic irreducible polynomials which does not have 1 as coefficient.

[dull ginkgo]

$\sum_{n = 0}^{p - 1}{x^n} = \frac{x^p - 1}{x - 1}$ so $x \mapsto x + 1$ gives $\sum_{n = 0}^{p - 1}{(x + 1)^n} = \frac{(x + 1)^p - 1}{x} = \frac{1}{x} \left(\sum_{n = 0}^{p}{\binom{p}{n}x^n} - 1\right) = \frac{1}{x} \sum_{n = 1}^{p}{\binom{p}{n}x^n} = \sum_{n = 1}^{p}{\binom{p}{n}x^{n - 1}} = \sum_{k = 0}^{p - 1}{\binom{p}{n + 1}x^{k}}$

jesus christ

there we go

FUCK LATEX ON MOB

[cloud walrus]

Miz

[dull ginkgo]

@cobalt heath this satisfies eisenstein's criterion

[cobalt heath]

Yeah, this proves irreducibility for 1 + x + .. + x^(p-1)

[dull ginkgo]

so when you put in x^k

your coeffs are just where c_(km) = a_m, else 0

so your leading is still 1, tailing is still p,

and your terms in between are the original (so they are divisible by p), or 0 (divisible by p always)

[cobalt heath]

Transformation x \mapsto x+1 does not work together with x \mapsto x^k, tho

Maybe try computing with small p and k

How about for p = 2, k = 2?

[dull ginkgo]

Look at the cyclotomic polynomials for p^n

wait wtf

you can just also have r

what

[dull ginkgo]

these LITERALLY ARE what I showed lmao

[cobalt heath]

Not really, this put in x^(p^k) instead of x^k

[dull ginkgo]

oh shit good point

[cobalt heath]

Notice pr is still there

[dull ginkgo]

yeah thanks

oh shit i guess that makes sense

mobius inversion with logarithms

[cobalt heath]

Ye-p

[dull ginkgo]

wait what the fuck

it seems to still hold????

[warped shadow]

typically the affine group is written as a semidirect product between the general linear group and the translation group, I don't get how $(t_2,M_2)(t_1,M_1)=(t_2+M_2t_1, M_2M_1)$ is consistent with $t_2g_2t_1g_1=t_2g_2t_1g_2^{-1}g_2g_1$. How is $g_2t_1g_2^{-1}$ equivalent to $M_2t_1$. Is it some kind of factorization of $M_2$?

[cloud walrus]

criver

[warped shadow]

I know how to make the above work for SE e.g. if I choose rotors to represent g2 and g1, since then the sandwich product g2 t1 g2^{-1} does what I want

i.e. a rotation matrix can be factorized into two matrices corresponding to the rotors

how does thst work out for the GL(n) case?

[dull ginkgo]

it

is agony to try to work out eisensteins

$\sum_{n = 0}^{p - 1}\sum_{m = 0}^{kn}{\binom{kn}{m}x^m}$

[cloud walrus]

Miz

[dull ginkgo]

i'd need to rearrange the sum so that m is on the outside

[cobalt heath]

Huh, for which is this

[cobalt heath]

Yeah this is it.

[dull ginkgo]

that's the x -> x + 1 substitution

We are summing over pairs $(n,m) \in \mathbb{N}^2$ such that $n \leq p - 1$ and $m \leq kn$

[cloud walrus]

Miz

[cobalt heath]

Yes, you see, pain

[dull ginkgo]

WAIT

WE CAN WORK WITH THIS

The leading coefficient is 1 right, and the trailing is the number of occurances of m = 0

of course there's one per 0 to p - 1

so p many times

p^2 doesn't divide the trailing

[cobalt heath]

In between?

[dull ginkgo]

what do you mean

eh too painful

[cobalt heath]

The terms in between leading and trailing

Need to be divisible by p, as well.

[dull ginkgo]

yeah

which is some difficult binomial shit tbh

@cobalt heath if $\binom{n}{m}$ divides $\binom{nk}{m}$ then we're golden

[cloud walrus]

Miz

[dull ginkgo]

idk if that's true

[warped shadow]

t2 g2 t1 g2^{-1}(v) = t2 g2 (g2^{-1}(v) + t1) = v + t2 + g2 t_1 actually which is fine

[dull ginkgo]

Seems to be true for every example I can think of

[dull ginkgo]

oh wait lol it's asking if $\frac{n!}{(n - m)!m!}$ divides $\frac{(nk)!}{(nk - m)!m!}$

[cloud walrus]

Miz

[dull ginkgo]

$\frac{(nk)!(n - m)!}{(nk - m)!(n)!}$

[cloud walrus]

Miz

[dull ginkgo]

$\frac{\prod_{i = 0}^{m - 1}{(pk - i)}}{\prod_{i = 0}^{m - 1}{(p - i)}}$

[cloud walrus]

Miz

[dull ginkgo]

anyway

Hmmm

Assume there exists nonzero nonunit c

If D[X] is a PID then every principal ideal generated by an irreducible element is maximal

So (x + c) would be maximal, but is contained in ideal (x, c)

Every element of (x, c) is of the form wx + (ux + v)c = (w + uc)x + vc
every element of (x + c) is of the form ax + ac

So if those ideals were equal, then (w + uc)x + vc = ax + ac

[tribal moss]

Isn't it simpler just to show directly that <x,c> cannot be principal?

[dull ginkgo]

oh i guess lol

[tribal moss]

Or perhaps not simpler but at least more enlightening.

But the minimal degree of <x,c> is 0, since the ideal contains c.

[dull ginkgo]

hmm ye

@tribal moss Assume $(d) = (c, x) \neq D$, therefore $d$ is a \textbf{non-unit}. Thus $x + c = d(ax + b) = dax + db$. However, the linear coefficient implies that $da = 1$ which implies that $d$ is a unit.

[cloud walrus]

Miz

[tribal moss]

Yes, but you can also just try to make x instead of x+c.
The underlying point is that in <d>, every coefficient of every element is divisible by d, but that is not the case for x = 1x+0.

[dull ginkgo]

that also works lmao

thanks

tbh idk what this is asking. Jacobson hasn't brought up the field of fractions of F[X], denoted horribly as F(X), yet in the textbook

[tribal moss]

Also, since <x,c> consists of exactly the polynomials whose constant term is a multiple of c, it is not all of D[x], so d has to be non-unit.

(Oh, you already said that and I misread it).

[tribal moss]

Nevertheless F(t) can hardly mean anything but that field of fractions.

[dull ginkgo]

My proof that i wrote down is as follows:
Assume D is a non-division domain, then there exists a nonzero nonunit a. Assume D[X] is a PID.
Let J = (X, a).

J is proper, as if 1 was in (X, a) then there would have to be some A(X)X + B(X)a = 1, but setting X = 0, B(0)a = 1, contradicting a being a unit.

Assume J = (f(X)). Then there must exist some A(X) : f(X)A(X) = c, but deg(f) + deg(q) = 0, thus f is a constant, so f(X) = d != 0. But likewise, there must exist some A(X) : dA(X) = X. deg(A) = 1, thus, A(x) = aX+ b. However, then da = 1, a contradiction again

I actually think I can skip the proper ideal check

[tribal moss]

Isn't the proper ideal check how you get a contradiction out of da=1?

[dull ginkgo]

a is a nonunit either way lol

[tribal moss]

Why?

[dull ginkgo]

oh shit i accidentally wrote c

(X, a), a is the nonunit from it being a non-division domain lmao

[tribal moss]

Ah, okay.

But then how do you get that a is also the leading coefficient of A?

[dull ginkgo]

brain fart, on my paper I used c as a nonunit, and A(X) = ax + b and jumping between what I used here and on paper fried my brain a bit

oh wait shit yeah

idk where I fucked up here gimme a hot second

oh i know yeah you're right I need the proper ideal check

[cobalt heath]

This is usual notation tho

[dull ginkgo]

no but like if you haven't done any abstract algebra before you wouldn't know what to do

Anyway actually showing that, hm

[cobalt heath]

Ah true

F(x)
"What is this, a function?"

[dull ginkgo]

kinda not sure what to do here

I guess i could start by assuming f(x) is reducible over F(T)

there we go

[cobalt heath]

Yeah

Proof by contradiction would be good approach

Maybe if you are familiar with categorical approach, you might use universal property? Idk

[dull ginkgo]

that's what I've been trying to do

Starting with F[X,Y] as a ring of origin basically

[cobalt heath]

Ahh right

Once you show it is irreducible in the ring, then you are done

[tribal moss]

If you start by clearing denominators, you get p(x,t)q(x,t) = f(x)g(t) in F[x,t].

[dull ginkgo]

i should've tried a routine like that, i kept going a categorical route

[tribal moss]

And then you can look at only the terms that have maximal degree in t, in each of p,q,g.

[dull ginkgo]

I wonder if there is a way to do it categorically

oh

[cobalt heath]

Idk categorical but

[dull ginkgo]

f(x) embeds into (F[t])[X] from F[X]

and is irreducible still

[cobalt heath]

You can consider F[x, t] / f(x) F[x, t]

[cobalt heath]

Are you sure about this?

[dull ginkgo]

If f(x) is irreducible in F[x] then it is irreducible in F[x,y] as it would factor over the onto-map that sends y to 0 into F[X] no?

[cobalt heath]

Oh

That's a great one!

[dull ginkgo]

Then i can use Lemma 3

[cobalt heath]

Btw can you identify why the same logic does not apply to Q[x, i] ?

[dull ginkgo]

Which was from the text lol

[dull ginkgo]

Q[X,Y]/(Y^2 + 1) you mean right

[cobalt heath]

Yeah it is equivalent to that

[dull ginkgo]

i isn't just some random indeterminate i hate when textbooks do that

[cobalt heath]

Yeaa

[tribal moss]

What's the relation between F and D there? F is only mentioned once in the statement of the lemma ...

[dull ginkgo]

F[X] = D

[cobalt heath]

?

[dull ginkgo]

Okay let me be consistent with my variables

D = F[t]

so Frac(D) = F(t)

D[X] is iso to F[T, X] (i forget if it's canonical lmao)

[tribal moss]

By why doesn't the statement of the lemma simply say F[x] instead of D?

[cobalt heath]

I think tropo mean what they are in this lemma.

[dull ginkgo]

This is from the text,

it's a lemma for integral domains

[cobalt heath]

Isn’t F field of fractions for D

[tribal moss]

There must be some hidden assumptions that relate D and F.

[dull ginkgo]

yes

[tribal moss]

Oookay, it didn't say that.

[dull ginkgo]

Sorry

it's using the D being GCD, but our F[T] is pid so lol

[tribal moss]

Does "factorial" mean UFD?

[cobalt heath]

They should have added name like “Gauss lemma”

[dull ginkgo]

yeah Unique factor domain sorry

[cobalt heath]

It is Bourbakian nomenclature (IIRC)

Who also calls compactness “quasi-compact”, idk

[dull ginkgo]

@cobalt heath Here's the whole proof:
Assume F is a field, then F[T] is a PID and is therefore a UFD. Let f(X) in F[X] be irreducible.
The ring of polynomials over F[T], F[T][X] is iso to F[X][T].

Due to this, F[X] embeds into F[T][X]. If f(X) is reducible in F[T][X], then let h(T, X) | f(X).

However, there is the onto map from F[T][X] to F[X] that sends T to t in F, which fixes f(X) and sends h(T, X) to h(t, X) in F[X].

However that would imply h(t, X) | f(X), and since f(X) is irreducible, h(t, X) must be a unit in F[X], i.e some nonzero constant c in F for each t.

Therefore h(T, X) = k(T) as it is independent of X. However, that would imply k(T) | f(X) in f[X][T]. Then, k(T) would be dividing f(X), essentially a polynomial that is just a constant F[X] coefficient in F[X][T], so k must be ultimately just a nonzero constant in F, thus a unit.

Therefore f(X) is irreducible over F[T][X], and by lemma 3, F(T)[X]

Juggling X and T is awful

The funny thing is that this isn't a proof by contradiction at this level

[south patrol]

What is this a proof of

[south patrol]

Yeah I believe so, Lang uses this too

And I've seen it basically only in Lang or algebraic geometry

[south patrol]

Oh this

[crystal vale]

What is meant by the statement, unitary modules over Z are simply abelian groups ?

Modules are always Abelian group, right?

[mighty kiln]

There is no extra structure from being a Z-module

i.e. homomorphisms of Z-modules = homomorphisms of Abelian groups

[south patrol]

A similar statement is: for commutative rings, an A-algebra is a ring B with a map A -> B. But every ring admits a unique map from Z, so there is no extra data in saying something is a Z-algebra vs just a ring

This often happens w Z ig

[mighty kiln]

Initial object moment

[south patrol]

Ye

Another thing is this. A module M over a ring R is an abelian group M with a ring map R -> End_Z(M). By the argument above, if R = Z then this map always exists uniquely

[vivid tiger]

Have you heard of Lang and his entire rings?

[fresh gate]

I have a question

in terms of symmetric groups

what is A_n?

[south patrol]

A_n is the "alternating group" on n letters.

have you heard of the sign of a permutation?

you can assign each permutation (i.e. element of S_n) a parity ("even" or "odd"), and A_n is the subgroup of even permutations

[crystal vale]

What does it mean? I don't get it

[delicate orchid]

Not only are Z-modules abelian groups, abelian groups are Z-modules

[mighty kiln]

Do you know what a homomorphism of modules / a homomorphism of groups is

[delicate orchid]

They’re isomorphic as categories

[crystal vale]

I know about Homomorphism of groups

[mighty kiln]

There's a definition for homomorphisms between two R-modules

It turns out for Z-modules a map is a module homomorphism iff it is a group homomorphism

[crystal vale]

Oh in terms of linear transformation?

[mighty kiln]

R as in any ring

[crystal vale]

Yes

[mighty kiln]

But yea module homomorphism is basically linear map

Preserves addition and scaling by elements of R

[south patrol]

Often one says "R-linear map". And being Z-linear is the same as being additive i.e. homomorphism of abelian groups

[crystal vale]

Okay

Let f: M ->N be a module and M and N are Z-modules.

Then f(x+y) = f(x) +f(y) and f(nx) = nf(x).
But how Z-modules help to show that module homomorphism is a group homomorphism because, it is always true on R-module, right?

Yes group homomorphism is not always module homomorphism

[fresh gate]

how does that work?

[south patrol]

So one thing is that you can write any permutation as a product of transpositions. if you write it as a product of an even # of transpositions, then we say the permutation is even. (simialrly for odd ofc). Then you need to check this is well-defined

[south patrol]

module homomorphisms are group homomorphisms by definition

[crystal vale]

Yes

To show group homomorphism is module homomorphism we need Module over Z ?

[fresh gate]

interesting

[crystal vale]

Yes, but every module is an abelian group, right? What is meant by Z-module is abelian group

[coral spindle]

Here's a fun one: write the permutation as a permutation matrix. Then the permutation is even if the determinant is 1, and odd if the determinant is -1.

[loud merlin]

Reread the post. Yes, every module is an abelian group
But every abelian group is a Z-module
They're equivalent

[crystal vale]

Got it

[tribal moss]

No?

[glad osprey]

Let R be the ring of 3x3 matrices of the form [a 0 0; b a 0; c d a], where a, b, c, d is in Z/2Z. Is there a way to find all the proper ideals without guessing and checking?

I know atleast that a must be 0, since a proper ideal cannot contain a unit

[south patrol]

$\begin{pmatrix} a & \ b & a & \ c & d & a \end{pmatrix}$

[cloud walrus]

Elliptic Potato

[south patrol]

Ah, lower triangular

[coral spindle]

Lower triangular with constant diagonal*

[south patrol]

well "ah, they are lower triangular"

jk yeah i should've said that thanks

but yes, agreed, we need a = 0 for all things in there

Well that only leaves like a few matrices and if we look at how they multiply it should be ok

[glad osprey]

Yeah, there's 8 (?) combinations after setting a=0, so maybe it's not so bad

[coral spindle]

So there are only 8 elements where a = 0 -- in the first place I would say that guessing and checking is not too hard -- but let's think about this anyway.

Let B, D, C be the matrices where b, d, c are the only nonzero entries. I can see that DB = C and BD = 0. So perhaps this helps you out?

In fact {1, B, D, C} forms a basis for this ring as a F_2-vector space. Kinda cute.

Note also that C multiplied by any other non-unit element is 0.

And of course you've already noted that all ideals are subsets of {0, C, D, B, C+D, C+B, B+D, C+B+D}

[glad osprey]

Ah, so there's actually 2^8 candidates. But since DB=C, any ideal containing D or B must contain C

[coral spindle]

That's right

So one way of continuing from here is to classify the ideals generated by a single element and consider if there are any new ones which arise as the sum of these (as all ideals do.)

Since you're working with a finite ring, you only need to consider finite sums too!

[glad osprey]

That's a nice approach 👍 I've found that {0, C}, {0, B, C} and {0, D, C} are ideals, then taking sums of those yields 3 new ideals

[coral spindle]

That seems about right to me

[glad osprey]

I think that's all of them, either way the question asked for only 5 ideals, so it'll do 👍 thanks for the help 🙏

[coral spindle]

No worries

[noble lynx]

how is the last statement true? I thought x^p - x = 0 in F_p.

[coral spindle]

You are confusing the polynomial x^p - x with the function x^p - x.

Remember that polynomials are formal: they are symbols, not functions.

When you've worked previously in rings such as Z, there was no need to distinguish between the polynomials Z[x] and the polynomial functions, because they were isomorphic.

However in finite rings, there is an enormous difference between the two as you have discovered.

The ring R[x] is always an infinite ring when R is not the zero ring.

[coral spindle]

So to summarise: your observation that x^p - x = 0 is actually saying that when you evaluate the polynomial x^p - x in F_p[x] at any value in F_p, you do get the value 0. But the polynomial itself is not 0; you have just sent it through a ring homomorphism whose kernel it lands in.

[noble lynx]

is there notation for specifying whether someone is talking about the ring of polynomial functions or ring of formal polynomials?

or maybe the distinsion doesn't come up as often as I am thinking

[celest furnace]

Haven’t personally seen one

[rocky cloak]

If your field is infinite, then there is no difference.

And if you're field is finite, then every function is polynomial, so there's no need to distinguish between functions and polynomial functions.

[celest furnace]

Oh wow that second fact is very cool never thought of it like that

[rocky cloak]

So I guess the only time this would relevant is if you're talking about the polynomial functions over varying fields. Don't think there's a standard notation for that, so would just introduce it inline

[coral spindle]

R[x] means polynomials. I don't know of any standard notation for the set of polynomial functions, but frankly you likely will not need it in your textbook.

[dire siren]

in some countries you would use capital letters for the indeterminate when you are referring to polynomials

[coral spindle]

But this doesn't mean that using a lowercase letter refers to the functions.

It simply means that it refers to an element in a larger ring instead.

[dire siren]

maybe it depends? In Romania we really write polynomials with capital letters and functions with lowercase letters

[coral spindle]

R[x] typically does not refer to polynomial functions on R.

[dire siren]

oh I see, I was a bit off-topic

that would still be R[X] lol

[coral spindle]

Yeah, what I was saying is that typically that would mean that there is some previously mentioned ring L containing both x and R

So it's just very frustrating notation overall isn't it

Well. For beginners.

[noble lynx]

when they refer to the ring R[u_1, ..., u_n] do they consider R \subset S and consider image of some mapping phi: R[x_1, ..., x_n] -> S where phi is the identity on R and maps x_1 to u_1 in S, ..., x_n to u_n in S? Or do they literally mean the polynomial ring R[u_1, ..., u_n]?

since otherwise the mapping is always a isomorphism

this is probably better context actually, ignore the earlier picture

[coral spindle]

Yes well here we are

when they refer to the ring R[u_1, ..., u_n] do they consider R \subset S and consider image of some mapping phi: R[x_1, ..., x_n] -> S where phi is the identity on R and maps x_1 to u_1 in S, ..., x_n to u_n in S?
Yes.

We use the same notation for two different things. Isn't that annoying huh

[summer path]

Wait it's literally the same?

Huh

[coral spindle]

The universal property of the polynomial ring is that it's the initial thing amongst things that could be referred to with that notation

[warped shadow]

What is the notation used for the Lie algebra $${z\in\mathbb{C},:, z=-\overline{z}} = {i\theta\in\mathbb{C},:,\theta\in\mathbb{R}}$$ of the circle group $S^1 = {z\in\mathbb{C},:, \overline{z}z = 1}$? Surely it is not $\mathfrak{so}(2)$ since then I wouldn't be able to differentiate it from the matrix group?

[cloud walrus]

criver

[warped shadow]

should I just use u(1) for it?

[rotund aurora]

Couldnt you quotient the polynomials by x^q-x?

[rocky cloak]

Well, you would quotient by x^q - x, but that of course only works if you know q before hand

(or know that q is finite at least)

[rotund aurora]

Well right if you want to treat this for all fields uniformly its going to be annoying

[rotund aurora]

It's funny this maybe suggests that "q" should be 1

I mean it doesnt make any sense lol but q=p^f where f is the dimension over the prime field. In char 0 the prime fielf is Q... but maybe some other argument suggests that f=infty for infinite fields and then 0^infty=1 sounds plausible

Im just saying nonsense, nvm me

[barren sierra]

https://media.discordapp.net/attachments/496784958430380033/1241899262682337280/image.png?ex=664c8911&is=664b3791&hm=5e1398a65b6d17ec41f27b9861cafb1f2117997cfa04ca55337c78b6d4029aac&=&format=webp&quality=lossless&width=1921&height=333

ok I'm back

I've shown that A is abelian

and B/A is isomorphic to Z by exactness

(trivial means left split)

Ok so in theory I want to show that B/A x A is isomorphic to B

not sure how to do that

[rocky cloak]

Step 1: let x be a perimage of 1 in B, and let C be the subgroup generated by x.
Step 2: show that A\cap C = 1
Step 3: show that B is the internal direct product of A and C
Step 4: C = Z

[vivid tiger]

he uses "entire" to mean an integral domain

[tribal moss]

Wonderful.

[barren sierra]

thanks! But is there any way to do it the way I was thinking?

i.e. find such a isomorphism

oh I guess this proof (if I unravel things) gives me that morphism in a way

[rocky cloak]

I mean, that is basically what's happening here. C is isomorphic to B/A and B is isomorphic to AxC

[barren sierra]

yea

How do I show that B is an internal direct product of A and C?

in particular how do I show that AC = B and C = <x> is normal in B?

is it valid to say that for $b \in B$, $\psi(bxb^{-1}) = 1 = \psi(x)$ so $bxb^{-1}x^{-1} \in \ker(\psi) = \text{im}(\phi) = A \subseteq Z(B)$ so that $bxb^{-1}x^{-1} = e$?

[cloud walrus]

Spamakin🎷

[barren sierra]

nvm no that doesn't follow

xy in center doesn't mean xy = yx

[long geyser]

when it says F = {1, a, a^2, a^3}, should the 1 be changed to 0?

doesn't make sense otherwise

[rocky cloak]

So you need to show two things:
That AC = B
And that ac = ca for a in A and c in C.

The second one you get because A is in the center.

For the first one, think about a b in B, and it's image in B/A

[delicate orchid]

yeah

[barren sierra]

don't you need that A, C are normal in B as well

[rocky cloak]

It follows from B being the direct product of them

[barren sierra]

oh wait it's sufficent to show AC = CA = B and A cap C = e for internal direct product?

[long geyser]

so many typos/mistakes in this book

[barren sierra]

I thought you also needed A, C normal in B

[rocky cloak]

No, you need ac = ca

[long geyser]

it's third edition too...

odd

[rocky cloak]

I.e. A and C commute

[barren sierra]

hm ok I should probably prove that equivalence of conditions since I don't remember that one

ok then that makes sense assuming that equivalence

[long geyser]

this is obvious with the hint, right? any polynomial in R[x] can be regarded as a poly in Frac(R)[x], where it has at most n roots

[delicate orchid]

if you indeed have that result about polynomials in fields, then yes

[long geyser]

yes

[dull ginkgo]

do you need to use field of fractions?

Couldn’t you use degree inequality and remainder theorem?

[long geyser]

what theorems are you referring to

[dull ginkgo]

deg(fg) = deg(f) + deg(g)
deg(f + g) <= max(def(f),deg(g))
For domains

and for even noncommutative rings f(x) = q(x)(x - a) + f(a) for some polynomial q(x)

If you’re a sociopath you can calculate said q(x) in a noncommutative ring

[long geyser]

f(x) = q(x)(x - a) + f(a) for some polynomial q(x)

we only proved this in the case of a field

[dull ginkgo]

[long geyser]

proved in general only for monic f

[dull ginkgo]

Well…

$x^n = 1 + (x - 1) \sum_{k= 0}^{n - 1}{x^k}$

[delicate bloom]

need a - in there

[dull ginkgo]

Oh shit thanks

I am on mobile best with me

[cloud walrus]

Miz

[rocky cloak]

The relevant part is really that (x-a) is monic

[long geyser]

oh wait fuck

yes

[dull ginkgo]

But either way

[long geyser]

the f in the theorem was the divider

[dull ginkgo]

If there was more than n roots, then by repeated use of the degree (in)equalities and the remainder theorem you’d run out of degrees to work with

[rocky cloak]

But I guess there is some subtelty with R[x] not necessarily being a UFD

[dull ginkgo]

It holds for noncommutative polynomial rings, once again

[rocky cloak]

What holds?

[dull ginkgo]

Remainder theorem

[rocky cloak]

Yes, but the remainder theorem is not the question. It's whatever a polynomial can have more than n roots

[dull ginkgo]

You need a domain to have deg(fg) = deg(f) + deg(g)

[long geyser]

the thing is that division algorithm doesn't guarantee uniqueness, so what if we had 2 different q's you could go down and they give different roots

not possible?

[dull ginkgo]

I think this was an exercise in Jacobson I did let me check

[rocky cloak]

Yeah, I guess once you have that you're pretty much good

Like if (x - a)(x - b) = 0, then x is either a or b because you're in a domain

[long geyser]

oh wait, you do have uniqueness

in domain

ok

I prefer the solution in the book though

[dull ginkgo]

You can still do something like the Euclidean algorithm for general domains

You just need to introduce a power of the leading coefficient to the right side

Which domains don’t have nilpotent elements so idk if that’s a problem tbh

[tribal moss]

OTOH if ab=0 with a,b both nonzero, then (x-a)(x-b) = x² - (a+b)x has too many roots, namely at least a, b, and 0.

[dull ginkgo]

Domains

[long geyser]

cool

[dull ginkgo]

@long geyser

However the at-most-degree-many-roots thing doesn't hold for general rings. The ring of quaternions has infinitely many solutions to x^2 = -1

[tribal moss]

Hmm, this doesn't deal with Z/4Z, though -- only one zero divisor to play with. Is there a polynomial over Z/4Z that has too many roots for its degree?

[dull ginkgo]

hmm

[long geyser]

lol, maybe rings with only one zero divisor still enjoy this property

[tribal moss]

Formal polynomials over noncommutative rings are still somewhat well-behaved -- the problem with them is that they don't have well-defined evaluation maps -- or, if you try to define such a map in the obvious way you may not get a ring homomorphism out of it. That ruins a lot of properties that we otherwise tend to take for granted.

[dull ginkgo]

My take is that it's more formulated for central elements

because commutivity

Or R-algebras

[tribal moss]

Ah! 2x²=2x holds for every element of Z/4Z.

[dull ginkgo]

omg so true!!

[tribal moss]

There don't seem to be any monic examples, though.

[barren sierra]

How do I classify all groups $G$ such that $0 \to \mathbb{Z}/2\mathbb{Z} \to G \to \mathbb{Z}/2\mathbb{Z} \to 0$ is exact? I think it's just $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/4\mathbb{Z}$ but how do I show that's all of them?

[cloud walrus]

Spamakin🎷

[tribal moss]

It has to have four elements, right?

And those are the only two groups of order 4.

[rocky cloak]

Imagine a perimage of 1 in G, what can it's order be? Consider the cases

[barren sierra]

Why would it have to have 4? How do you come to that conclusion?

[dull ginkgo]

The morphism into G must be injective, the morphism out of G must be epi

but the image must be the kernel of the next

use legrange's

:3

[long geyser]

Is there anything special about F_2 here? my proof didn't involve the fact

[tribal moss]

Beause you can quotient out the injective image of the first Z/2Z and get a quotient of order 2.

[long geyser]

btw F_2(x) = Frac(F_2[x]), is this standard terminology?

[rocky cloak]

|A| = |A/B|x|B|

[barren sierra]

Yea

[tribal moss]

Yes.

[barren sierra]

Ah

[tribal moss]

R(x) = field of fractions of R[x] for any domain R.

[long geyser]

👍

[rocky cloak]

Would people actually use that if R is not a field?

Like saying Z(x) = Q(x) seems weird to me

[dull ginkgo]

Awful

[tribal moss]

Hmm, now that you mention it, I'm not sure if I have seen that.

[dull ginkgo]

torture

[long geyser]

?

[barren sierra]

Z(x)

[dull ginkgo]

I actually don’t think it’s necessary

I guess it’s like nudging towards Frobenius endomorphism but that doesn’t cause any contradiction since -f(x) = f(x) anyway cuz char 2 lmao

[tawny magnet]

maybe they want to hint at how to coook up an inseparable extension

[long geyser]

ah, you're right

I shouldn't have cropped the stuff in parentheses

[delicate bloom]

fun way to see that is f'(t)=0

[long geyser]

lol

[south patrol]

nice

another would be to apply Gauss' lemma and say it'd be irredcible over F_2[x]

which is obvious by degree considerations

or by Eisenstein's criterion lol

Lol dumb question

is there standard notation for the non-unital ring generated by some elements

Like a polynomial ring minus the 0th degree bit lol

I've never seen this so probably nothing too standard lol

[dull ginkgo]

I literally just used this lmao

[south patrol]

Lol

Yeah but idk at least in algebra i don't often come across non-unital things

[dull ginkgo]

For the psycho x^r(x) = x for every x in ring problem

[south patrol]

i am doing it more though via homotopy theory

ah lol

yeah

[dull ginkgo]

It sucks, and it's named after jacobson, wtf not

[south patrol]

lol

the important thing is that like cohomology of a space forms a graded ring, but often you want to consider what happens when you chuck away zeroth degree lol

[dull ginkgo]

I imposed a partial order on R that x | y if y = f(x) for some integer polynomial without a constant (because there is no gaurunteed identity)

[south patrol]

oof

[dull ginkgo]

Well if x | y then xy = yx

but it turns out I think it's a bit easier because I forgot that if nx = 0, and my = 0, then nm(xy -yx) = nx my - my nx = 0 lol

which I think may actually be a route to prove it more easily

[south patrol]

why the | notation lol

idk would confuse me but fair

[dull ginkgo]

lazy

[noble lynx]

what are the minimal pre reqs to read something like matsumura or atiyah? I am trying to speed run to it to hopefully get ready for ag in the fall

I think I should be familiar with basic ring and galois theory in a week or two, but do I need to know a lot about tensor products, cat theory, or advanced module stuff?

[next obsidian]

For Matsumura, yeah it helps a lot

For AM, no

If you don’t know much about tensor products Matsumura is above your pay grade atm. AM covers tensor products in like chapter 2 or 3, so that is a better fit

[noble lynx]

I see, for matsumura, what do you think beyond basic ring/galois and module theory(I only know like structure theorem and nothing else pretty much) is required?

Ill def add tensor products to my list

[next obsidian]

There’s technically a treatment of derived functors as well as direct / inverse limits in an appendix. I find it hard to believe that someone who doesn’t know them going in however will be able to follow all the arguments

I think a working knowledge of homological algebra (a first course) is probably the main thing, but besides that one just has to have a certain degree of maturity

The book won’t spell out every detail in every argument

[noble lynx]

I see, it seems a bit far away then, I will try to do jacobson until I know what those mean, thanks for the help

[lilac mango]

How would you calculate the Galois group of x⁴+4x²+2?

The roots are +-isqrt(2+sqrt(2)) and +-isqrt(2-sqrt(2))

I know the Galois group is Z4 I just have to prove it

[cobalt heath]

Maybe explicitly find the elements of it

[lilac mango]

So my idea is that since the galois group G is transitive on the roots, there exists an automorphism that sends isqrt(2+sqrt(2)) to isqrt(2-sqrt(2)) and want to prove that it is forced to send isqrt(2-sqrt(2)) to -isqrt(2+sqrt(2))

[lilac mango]

How would you go about that?

[cobalt heath]

Yeah likely you can use field axioms to show how automorphism behaves

[lilac mango]

Yes I tried that but it didn't quite work

Probably I'm missing sthing

[cobalt heath]

The field extension in between might help as well

Can you identify subfield extensions?

[lilac mango]

The only one shoukd be sqrt(2)

Q(sqrt(2)) that is

[cobalt heath]

Yep, so you can think of how the automorphism would act on sqrt 2

And infer stuffs from that.

[lilac mango]

It would take sqrt(2) to -sqrt(2) I have that too

[cobalt heath]

Yeah so

How did you get sqrt(2) btw?

Think about that. How it appears in this picture

[lilac mango]

I mean I got it by raising phi(isqrt(2+sqrt(2))) and isqrt(2-sqrt(2)) to the power of 2 and

And got phi(-2-sqrt(2))=-2+sqrt(2)

So phi(sqrt(2))=-sqrt(2) since Q is fixed by phi

I was trying to do something like that to prove that isqrt(2-sqrt(2)) has to fall in -isqrt(2+sqrt(2)) but I don't know how to do it as isqrt(2+sqrt(2)) and -isqrt(2+sqrt(2)) feel basically as if they were algebraically the same

[cobalt heath]

What is phi?

Ah, the automorphism?

[lilac mango]

The automorphism that takes isqrt(2+sqrt(2)) to isqrt(2-sqrt(2))

[lilac mango]

Yes

I mean it is as of now AN automorphism that takes...

But it should be the only one

[cobalt heath]

Hmm, then let me ask how you got isqrt(2+sqrt(2)) instead. Like how did you compute it?

[lilac mango]

If the galois group is in fact cyclic of order 4

[lilac mango]

Well I calculated the roots of x⁴+4x²+2 and well x² has to be -2+-sqrt(2)

So the posibilities are the roots of that

[cobalt heath]

That comes from factorization, right?

[lilac mango]

Yes

[cobalt heath]

Let me be more direct. What is factorization of the poly on Q(sqrt(2))

[lilac mango]

Let me do that

[cobalt heath]

Ah wait, now I get your frustration - this does not work easily. Hmm just a sec

I thought I knew where my intuition came from, but now I am not sure.
Anyway, @lilac mango you can try multiplying the two

Namely, i sqrt(2 + sqrt(2)) * i sqrt(2 - sqrt(2)) .

[lilac mango]

I have tried that already but I couldn't see how it helped really

I thought of something maybe it could be useful

The automorphisms fixing Q(sqrt(2)) are only the identity and conjugation

[cobalt heath]

Did you compute the product? It is simple, isnt it?

[lilac mango]

That's not too hard to proof I think

[lilac mango]

Yes

-sqrt(2)

[cobalt heath]

We already know where phi sends sqrt(2).

[lilac mango]

Oh yeah so phi can't fix the product

Yeah that makes sense

Since it doesn't fix the product, we are done

phi has to send isqrt(2-sqrt(2)) to -isqrt(2+sqrt(2)) since if it sends it to isqrt(2+sqrt(2)) then it would fix -sqrt(2) but it doesn't

[cobalt heath]

Indeed.

[lilac mango]

And a similar argument should prove that there are only 4 automorphisms

Like I would have to prove the automorphism is completely determined by what it does to, say, isqrt(2+sqrt(2))

So the order is 4 and I'm done

Or is there some smarter way of doing it?

[tribal moss]

You already have the ingredients. Give names to your four roots:
A = isqrt(2+sqrt2)
B = -isqrt(2+sqrt2)
C = isqrt(2-sqrt2)
D = -isqrt(2-sqrt2)
If you know f(A), then obviously f(B) = f(-A) = -f(A).
Since AB = 2+sqrt2, we have f(sqrt2) = f(A)f(B) - 2.
Since AD = sqrt2 we have f(D) = f(sqrt2)/f(A)
And f(C) must be the one left over.

[lilac mango]

Thanks Absta, you were very helpful 🙂

[cobalt heath]

No problem! Good that you got it despite all my blunders and my mistakes
Good luck with your algebra endeavour!

[lilac mango]

Oh wow yeah you're right

I was doing it case by case but yeah this is much more efficient

Thanks

[tribal moss]

I think you need to check explicitly that it's not V4, though, unless I missed an argument for that.

[delicate bloom]

I think since Q(sqrt(2)) is a subfield, each is a quadratic extension over the previous only gives you +- so you're stuck with klein 4 group

[lilac mango]

Yeah you missed it. The one that takes A to C is (A C B D) since sigma takes sqrt(2) to -sqrt(2) and sigma(AC)=sigma(-sqrt(2))=sqrt(2) so sigma doesn't fix AC so must take C to B

[delicate bloom]

Q(sqrt(2))/Q has conjugation that fixes Q, Q(sqrt(-2+sqrt(2))/Q(sqrt(2)) has conjugation that fixes Q(sqrt(2)), and then you could conjugate both for the third (non identity element)

[tribal moss]

I'm not too strong in Galois theory. but does a tower of extensions necessarily give us a direct product? I'd have thought the best we could hope for in general was a normal subgroup and quotient.

[next obsidian]

That gives you the size by Lagrange and then you can conclude by just knowing groups of order 4

The fact that Q(isqrt(2)) is also a sub field says it isn’t C_4

[lilac mango]

Except it is C4

[next obsidian]

Wat

[lilac mango]

Q(isqrt(2)) isn't a subfield

[lilac mango]

.

[next obsidian]

Oh teehee so true

[summer path]

teehee

[next obsidian]

But you can just start analyzing sub fields

Or I guess just get the order 4 automorphism by observation

[lilac mango]

Yeah since I don't know how I would discard that there were some other subfield of degree 2 over Q otherwise

[delicate bloom]

ah just came back to this, I see my mistake was thinking since it fixed the subfield it fixed the elements cool makes sense

[cobalt heath]

Ahh, Galois is so confusion

[summer path]

Why is this mass spammed

[tardy nimbus]

No i think your intuition is right if what you mean is "does tower -> direct product of cyclic groups". That is probably false since there are non-abelian solvable groups

[delicate bloom]

it's definitely false since that example had its galois group a cyclic group with 4 elements

[tribal moss]

The point I had in mind is that it looked like Merosity was assuming the Galois group would be the product of the Galois groups of each layer in the tower, in the message immediately above mine.

[delicate bloom]

yeah, I was haha

[chilly ocean]

If the permutation groups of two sets are isomorphic does that mean that there is a bijection between them

I think the answer is yes and I was trying to construct such a bijection, does taking x in one set and y in other and if f is the isomorphism then considering mapping n in first set to f((x,n))(y) work?

[tardy nimbus]

what do you mean by "permutation groups of two sets". I think generally speaking two objects can have the same automorphism groups but not be isomorphic. That might even mean they can have different numbers of elements

[tardy nimbus]

also are these sets finite?

[chilly ocean]

No

[chilly ocean]

I meant of A and B are two sets and S_A ~= S_B then is |A|=|B|?

|X| denotes the cardinality of X

[tardy nimbus]

this might get hard then

thats beyond my knowledge, is there a set theory of formal logic subchat here somewhere? they would be quick to create weird counterexamples if some existed

[loud merlin]

#foundations