groups-rings-fields

[rotund aurora]

ye it's weird, in classical AG books I have read this is a preliminary result

[south patrol]

Yeah I'd call it Zariski's lemma

"a f.g. k-algebra which is a field is finite over k"

[rotund aurora]

ye, Vakil says that is also called Zariski's lemma

[south patrol]

Like i think uh

Zariski's lemma for a ring R is equivalent to R Jacobson

and of course fields are jacobson

then the alg closed assumption just lets you say finite => just the ground field

Thinking in terms of Jacobson ideals is kinda cool because then it's like, oh well in classical alg geo you can work with maximal ideals rather than Spec(R) etc because prime ideals can be approximated by maximals anyway

Also like lol, feel fre to ask about this stuff in #advanced-algebra

[cobalt heath]

Is this simple enough to prove?

[south patrol]

I'm not sure cause in my mind I'd prove Zariski's lemma using either the fact k is Noetherian (and hence also k[x] etc) or the fact it is a field (via Noether normalisation)

[rotund aurora]

Jacobson implies that the Jacobson radical is the same as the nilradical, correct?

[cobalt heath]

Hm, is finitely generated ring over a Noetherian ring a Noetherian?

[rotund aurora]

yes

[south patrol]

Hilbert basis theorem

[rotund aurora]

ye

[south patrol]

Yes

[rotund aurora]

and then quotient of noetherian is noetherian

[south patrol]

Though I would say "fintely generated algebra"

[rotund aurora]

but the converse doesn't hold, or does it?

[south patrol]

finitely generated ring to me means f.g. as a Z-algebra

[south patrol]

It does

WAIT

sorry

responding to the wrong thing

Do you mean "Jacobson = nilradical" is equivalent to Jacobson

[cobalt heath]

I thought Hilbert basis theorem is only about R into R[x]

[rotund aurora]

yes

[south patrol]

Then induction

[cobalt heath]

Yeah wait

[south patrol]

Quotient of a Noetherian ring is Noetherian

[cobalt heath]

Ah, so quotient is Noetherian as well

[south patrol]

ye easy via correspondence theorem

[south patrol]

Then I highly doubt it

Take a non-Jacobson domain

(i don't know any examples off the top of my head lol)

[cobalt heath]

But does Noetherian ring containing a field k have to be finite over k?

[rotund aurora]

mmh let me think

[south patrol]

No

C contains Q as a subring

[hollow mica]

double quotient is equal to single quotient for groups

[cobalt heath]

Hmm, then I don't see the connection to Zariski lemma here.

[south patrol]

I meant uh

i forget the name of the lemma

Artin-Tate lemma

Noetherian ring A, subrings A < B < C, C finite type over A and finite over B, then B is finite type over A

iiirc

youcan use that to prove Zariski's lemma

[cobalt heath]

Guess it is not immediate then, was wondering if there were easy way to do this

[rotund aurora]

Isn't $k[x,y]_{(x,y)}$ an example? The primes are $(0), (x,y), (x), (y)$, but this has just one maximal ideal

[cloud walrus]

croqueta3385

[south patrol]

Yes take anything local but not a field

I was about to type that 😭

But yes very nice

[delicate orchid]

holy smokes...

[south patrol]

Z_(p)

[delicate orchid]

society if Z_(p) was Z[1/p]

[south patrol]

etc

Lol

[rotund aurora]

right

[south patrol]

I don't like it cause uh

Like notation R_f for R[f^-1]

would mean Z_p = Z[1/p]

but then Z_p is p-adics to me

and some use it to mean Z/pZ

😭

so cursed

So personall I write R[f^-1], Z_p and Z/p

i guess some write $\mathbb Z_{p}^{\land}$

[cloud walrus]

Süßkartoffel

[south patrol]

forpadics

[delicate orchid]

yeah lunatics do lol

[rotund aurora]

xd

I have never seen that

[delicate orchid]

that's for topological p-completion

which, yes - is the same

but the vibes are off!

[rotund aurora]

how does a topological completion carry algebraic information like Z_p being a ring

[delicate orchid]

oh god I can't be bothered thinking about the massive simplicial complex based construction at 1am

ask potato

[south patrol]

What's the different lol

*difference

Lol

I guess do you mean it is Z_p as a topological ring vs as a ring

[delicate orchid]

well, standard algebraic completion via a limit (or colimit, who cares I can never remember - probably a limit though) is only really done on algebraic objects

[south patrol]

Limit

[delicate orchid]

but the type of p-completion I'm referring to can be done on any simplical complex

[south patrol]

Sure

[rotund aurora]

idk what topological completion means

[delicate orchid]

yeah it would need to be a limit for the p-adics to be unital lol

[rotund aurora]

do you mean like metric space completion?

[delicate orchid]

think p-complete, p-good, p-bad, spaces

these two notions of completion intersect for Z_p because... well the p-adics are pro-finite lol?

[dull ginkgo]

[south patrol]

Okay but isn't that a p different notion lol

[delicate orchid]

yes! that's my entire point!

RAAAAAAGHH

I only see the ^\wedge for that notion

[south patrol]

Okay I thought that was weird cause the people who write it for Z_p^ are the same people

Like topologists

[delicate orchid]

is something the matter, sir

[south patrol]

Hensel

[delicate orchid]

u lift me uppppp so I can stand on mountainnss

[south patrol]

Real

[rotund aurora]

what is $Spec,\prod_{n=1}^\infty \Z/2\Z$

[cloud walrus]

croqueta3385

[delicate orchid]

it's almost like there's an element in Spec for each subset of N

[rotund aurora]

idk if it's useful, but to describe the Spec as a set, you can think of a prime as a family F of subsets of N such that

• if X in F and Y subset X then Y in F (this is closure under multiplication by elements of the ring)
• if X cap Y in F then either X or Y in F (this is the prime property)
• and you should include that N is not in F, to avoid the unit ideal

[delicate orchid]

so F is closed under subsets and intersection, I swear I've seen this somewhere

[rotund aurora]

isn't this a filter

[delicate orchid]

oh on N?

[rotund aurora]

no

[delicate orchid]

yeah it is, I think

[rotund aurora]

nono

[delicate orchid]

oh

[rotund aurora]

for a filter, you would want X in F and X subset Y implies Y in F

but what we have goes the other way

and also N should be contained in the filter

[delicate orchid]

just order P(N) backwards then lol

[dull ginkgo]

$\rotatebox{180}{\text{why the fuck do we care about the prime spectrum}}$

[cloud walrus]

Dyfunction Executive

[rotund aurora]

ye that's what I was thinking

[delicate orchid]

does the intersecting condition corrispond to a filter condition?

yes of course

intersection of sets is their meet

[south patrol]

Hehe

[rotund aurora]

this is the usual definition for filter

[south patrol]

Boolean ring

Yeah

[delicate orchid]

right ok this is much easier than having to translate everything back and forth from poset land

[tough raven]

I believe in general, it's a compactification (Stone-Cech?) of the infinite disjoint union of Spec factors. There is a mathse/mathoverflow answer somewhere about Spec of infinite products.

In this case, a power of Z/2Z is a Boolean algebra — in fact a power set — and for those, prime ideals = maximal ideals = ultrafilters, which for power sets are ultrafilters in the set theory sense.

So Spec Π_{i in I} Z/2Z = {ultrafilters on the set I}.

[delicate orchid]

we were getting there... gradually

[rotund aurora]

it is true that the topology on the set of ultrafilters on N is the Stone-Cech compactification on N, but I just read that in a foot note, so I don't really know how it works lmao

[tough raven]

Yes, this is a prime filter in the poset of subsets of N; its dual is a “prime ideal”.
For Boolean algebras (in particular, power sets) you can show that this is equivalent to F being a maximal proper subset of 2^N with the properties except condition 2 (i.e., F is a maximal ideal basically).

There isn't really a better description of these things though.
There are exactly two kinds (using your kind of family F):

• for every n, all subsets of N{n} (the principal maximal ideals)
• maximal ideals which contain every finite subset (the free maximal ideals).
  The existence of the latter is a weaker form of the Axiom of Choice.

[tough raven]

Incidentally, if one is in a weird mood, one can define a “filter” in a ring as “1 + an ideal” and decide that possible quotients of a ring should be described by the preimage of 1 and write all the isomorphism theorems in terms of filters.
I think I started doing this once until I realised it was just 1 + ideal theory.

[rotund aurora]

so like in this way, you can prove that any filter is contained in an ultrafilter from the axiom that any proper ideal is contained in a maximal ideal right

[tough raven]

Yes, for filters of Boolean algebra (in particular, of sets).

In fact, IIRC, TFAE:

• filters of sets are contained in ultrafilters
• filters of Boolean algebras are contained in ultrafilters
• ideals of Boolean algebras are contained in prime ideal (Boolean Prime Ideal Theorem)
• products of compact Hausdorff spaces are compact
• product of discrete two-element spaces is compact
• the Compactness Theorem for first-order logic
• a complete and totally bounded uniform space is compact (converse true in ZF)
• a space is compact if every ultrafilter has a limit (converse true in ZF).

[white oxide]

so I'm trying to prove that $G$ solvable implies $G/H$ solvable, but it doesn't make sense to let ${e} = G_0 \subset G_1 \subset \dots \subset G_n = G$ and consider or write $G_0/H \subset G_1/H \subset \dots \subset G/H$

[cloud walrus]

okeyokay

[white oxide]

because that would require H is a subgroup of the identity which doesn't make sense

[vivid tiger]

Well, if we have
G > H > e

Then you can refine

[white oxide]

are all subgroups of $G/H$ of the form $K/H$ where $K$ is a subgroup of $G$?

[cloud walrus]

okeyokay

[vivid tiger]

(using my arbitrarily chosen notation)
G > G1 > G2 ... > H > K1 > K2 > ... e

Now, this shows that H is solvable

To show that G/H is solvable, we need to write

[white oxide]

can u put spoilers pls if u do reveal solution

[vivid tiger]

My bad!

[white oxide]

ur good

i already read the proof that H is solvable

[vivid tiger]

Should I spoil this already, or should I leave you in suspense?

[white oxide]

he proved it so it's chill

lol if you could post the solution or some hint or something with spoilers that would be great

[vivid tiger]

The idea is to use what's between G and H.

You know that Gn/Gn+1 is abelian

So, you need to get something that starts with G/H, ends with e, uses the Gn somehow, and uses the fact that Gn/Gn+1 is abelian to get abelian quotients here.

It is a good idea to look over the isomorphism theorems.

Honestly I always found those screwey. Galois theory actually makes me understand them better.

So, you have the lattice theorem, which says that the opposite sides of the parallelogram in the lattice are isomorphic

[white oxide]

yeah the first one that comes to mind is the third isomorphism theorem

[vivid tiger]

You have the other theorem, which is what I think of as the tower theorem.

[white oxide]

or the collapsing theorem or whatever

(G/K)/(H/K) ~ G/H

[vivid tiger]

I think of G extending K and G extending H, where "extends" means "the latter is normal in the former". Tower theorem tells me that H extends K. It also tells me that if I take the big line from K to G and quotient out the top small line from H to G, I get the bottom small line from K to H.

If only I could draw pictures

This made no real sense to me until Galois theory, funnily enough

Anyways, this is indeed what you need

[vivid tiger]

So, this lets you take quotient of quotients.

If you make K in the G/K the same as K in H/K, then you get G/H after quotienting.

That is, if you "share a common denominator", you get the quotient of the numerators.

Hint hint nudge nudge

Still more dancing about the bush: ||One of those is nice. The other gives you sequence terms||

The answer: ||G/H > G1/H > ... Gn/H ... > e. This works since the quotients are (Gn/H)/(G{n+1}/H)=Gn/G{n+1} is abelian||

Once you get that,
Ex: Prove the converse: if H normal in G, and both H and G/H are solvable, then G is solvable.

[white oxide]

yeah i was just hoping that this would be true so that we could quotient out each of the groups in the last chain and apply that isomorphism theorem to show that they're abelian

but i don't think we can quotient out all of the groups in that chain for G, it wouldn't make sense

[vivid tiger]

I mean, this is basically it

Just don't do it for the entire chain

[white oxide]

oh

[vivid tiger]

The problem with doing it to, say, the chain obtained by refining G > e is that you can't quotient out H.

But, if you do it to the chain obtained by refining G > H > e, you now know when you need to stop.

[vivid tiger]

Sorry, I derped on reading this

[white oxide]

hmm okay i think i know what you're saying

wait you're not saying that we can insert H into some point in that tower for G right

[vivid tiger]

So, if we take any normal series that we desire, we can refine it

[white oxide]

what theorem is that again

[vivid tiger]

Schreier?

Refinement theorem.

[white oxide]

oh right

but what if we didn't assume schreier

like this is before the proof of schreier

to be annoying

[vivid tiger]

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

[vivid tiger]

Oh.

Umm

[white oxide]

LOL my b

[vivid tiger]

Actually, we don't need to prove uniqueness.

That is, we don't need unique refinements, which is what Schreier is for. We just need refinements

Suppose we have at any step G1 > G2.

Hm.

I think we do need uniqueness.

Uniqueness allows us to say that there's a refinement G > H > e has the same factor groups up to permutation as some refinement of the series that we know exists with abelian factor groups

That is, since G solvable, we know there's some G > G1 > ... > e with Gn/G{n+1} abelian. Schreier lets us refine this and refine G>H>e such that they have isomorphic factors up to permuting the factors, so that the refinement of G>H>e has abelian factors

And it really looks like proving this would basically just be Schreier.

Perhaps let us start with
G > [G,G] > [[G,G], [G,G]], ... which eventually terminates.

H is a subgroup of G, [H,H] subgroup of [G,G], ... so the derived series of H also terminates.

Now, we can cross our fingers and try to take quotients

G/H > [G/H, G/H] ...

Now, if [G,G]=e then [G/H, G/H] does too.

this seems fishy

so, there's definitely a problem with this reasoning, but since the terms of the derived series of G/H will be trivial when the terms of the derived series of G are trivial, we are done.

@white oxide where's the flaw with my logic? there must be one.

[white oxide]

i'm sorry my brain's not working right now and i kinda forgot schreier and everything so i'm probably not qualified

but the companion to Lang said that the proof of G solvable implies G/H solvable is essentially this paragraph

[vivid tiger]

Oh, I see

This is the "preimage of tower is tower, with image of factor groups as embedding" thing.

I like it.

I also remember it from reading Lang.

[white oxide]

wait don't spoil it i'll get it eventually 😭

i need to just come back i think

[vivid tiger]

I wasn't going to say anything else besides listing the three ways we have found

well, we have three proofs.

1. Schreier Refinement
2. The probably erroneous proof I gave that says that the D^n (G/H) [idr whether n is upstairs or downstairs for this one] is 0 if D^n(G) is.
3. The proof that uses the way towers work under homomorphisms G->G'. That is, T' tower gets preimaged to T tower, and factors F get imaged into factors F'.

3 is more general - Lang uses that idea for all the rest of the tower stuff

2 requires you to know that having a tower with abelian quotients is equivalent to the derived series
G > G1 > G2 > ...
with Gn = [G{n-1}, G{n-1}] where [G,H] = {gh(hg)^{-1} : g in G, h in H}, that is, the commutator subgroup. (I think group theorists are evil and write (hg)^{-1}gh)

The equivalence comes from the fact that a map G->A with A abelian factors through the abelianization of G, called G^{ab}, which is defined as G/[G,G].

This is just because the map must have [G,G] in the kernel, but in general you can't say more than that.

[tough raven]

gh(gh)^{-1} = 1… I think you meant ghg^{-1}h^{-1}.

[vivid tiger]

Yep, thanks 😆

[crystal vale]

I don't know this question makes sense or not.

I want to find Matrix corresponding to linear transformation.

T:P_2 -> P_2 where P_2 is a set of all real polynomials which has at most degree 2.

And T(p)= p', so what will be their matrix representation?

[dull marsh]

Start by picking a basis for P_2

Describe the images of basis vectors using the basis

Combine them into a matrix

[crystal vale]

Yeah I did but I didn't figure out how the matrix looks like

[dull marsh]

Show your work

[crystal vale]

I want a matrix A such that A[1 x x^2] = [1 2x 0] , right?

[celest furnace]

Have you seen that A applied to the first, second, and third basis vector are the first, second, third columns respectively?

[crystal vale]

Means A=[1 0 0, 0 2x 0, 0 0 0]?

[celest furnace]

You're ordering your basis as {x, x^2, 1}?

[crystal vale]

No

[celest furnace]

If youre ordering it as {1, x, x^2} then the [0,0,0] column should come first

[crystal vale]

Yes

[crystal vale]

So how this is right?

[celest furnace]

Yeah ur right thats wrong

[crystal vale]

It will be A[ 1 x x^2] = [0 1 2x]

[celest furnace]

Indeed

[crystal vale]

So matrix will be [0 0 0, 1 0 0, 0 2x 0]?

[celest furnace]

Yes

Interestingly enough you can see that this is not invertible--that's why we add the +C lol

[crystal vale]

• C ?

[celest furnace]

You know when do you an integral you add +C

[crystal vale]

Yes

So why here +C use?

[celest furnace]

Like if that matrix was invertible then there would be a unique anti-derivative is basically what i was getting at

[crystal vale]

I don't know what it means?

[celest furnace]

What part are you confused about

[crystal vale]

Why are we talking about the invertible matrix here?

If the matrix is invertible then the map is surjective?

[celest furnace]

No particular reason just pointing out a cool observation

[crystal vale]

Nice

[crystal vale]

I think it is wrong matrix will be [0 0 0, 1 0 0, 0 2 0]

[plush pulsar]

if a group and its subgroup both have the same order, does that mean they are equal?

[delicate orchid]

if your group is finite, yes

otherwise no

consider Z and 2Z

[plush pulsar]

I agree with this

[plush pulsar]

but this seems falsi

false*

[delicate orchid]

1 is in Z

1 is not in 2Z

they are not equal

[plush pulsar]

order is concerned with the total number of elements

not with specific elements i think

[delicate orchid]

yes? and they're the same order?

[coral spindle]

Yes, and Z and 2Z have the same order

[delicate orchid]

but they're not equal as sets because they have different elements

if you want non-isomorphic then the free group on two elements contains the free group on n elements for all n as a proper subgroup

[coral spindle]

You are aware of the standard results about infinity I hope?

[dull ginkgo]

My brain fried on what order was for a second

Cardinality not the maximum element order

[plush pulsar]

yeah

[dull ginkgo]

[delicate orchid]

doesn't matter for Z and 2Z

infinite exponent irregardless

[coral spindle]

So you should know that Z and 2Z are in bijection

[plush pulsar]

yes

[coral spindle]

So they have the same order

Tadah

[plush pulsar]

this seems intuitive

[coral spindle]

You don’t say

[plush pulsar]

🙏

[coral spindle]

Baruch atah adonoi etc

Sice we’re praying

[chilly ocean]

https://media.discordapp.net/attachments/811392071117701120/1215235621895938098/1709804766340.jpg?ex=65fc035e&is=65e98e5e&hm=d0b0729e9a35f4aebed1c38bab08ce020935ac02a09f3a31f1882eb33367610a&

Can I get some help with this

So in the case that z lies on the unit circle with irrational argument

I think the subgroup generated by them will be dense

In S_1

In case when z is not in S_1 it's a bunch of points in the closure along with zero?

[tough raven]

Irrational multiple of 2π, technically.

[chilly ocean]

Yes

My bad

[tough raven]

OK, I thought I remembered the outline of a solution, but I don't 💀.

OK right.

First you show that the cyclic subgroup contains points on the unit circle arbitrarily close to 1 (other than 1 itself).

Then this implies that there can be no arcs of non-zero length in the unit circle that contain no points from the cyclic subgroup.

From there, it's straightforward.

[chilly ocean]

Huh not necessarily?

Which cyclic subgroup are you referring to

[tough raven]

Generated by z

[young plover]

I've been discussing the following with a friend, but i'm not too happy with his conclusion. help would be aprreciated

let f:X->Y and A∩X={}

is f(A)={} or undefined?

Also, is this the most appropriate channel to ask this question?

[chilly ocean]

Why does it have points arbitrary close to 1?

[tough raven]

In the case that z lies on the unit circle with irrational-multiple-of-2π argument.

[chilly ocean]

Oh

Right👍

The irrational multiples of two pi case in the closure we get the entire S_1?

[tough raven]

I'd say it's empty personally, but also that this “shouldn't” matter practically.

[tough raven]

As for the channel, probably not this, but I don't know the correct one. Maybe #discussion?

[tough raven]

Yep, and thus quotient R.

[chilly ocean]

It's clear in that case to me

[chilly ocean]

This one is causing trouble

[tough raven]

BTW leaving a fun problem for others to try later:

Find all finite groups G such that G acts by even permutations in the left-regular action.

[tough raven]

Suppose the cyclic subgroup contains an α on the unit circle (and not 1) with argument less than 2π/n. What does this imply?

[chilly ocean]

It's not possible

[young plover]

oh thanks, i'll bring the question there. i agree that in practice it isn't that useful a question. he was trying to prove that f(A∩X) = f(A) for an arbitrary A and X still the domain and i just wasn't comfortable with giving meaning to f(A) i fA had elements outside the domain

[chilly ocean]

If A is modulus of z then if A is not 1 then cyclic subgroup of A will never be on the unit circle?

[tough raven]

You mean if z doesn't lie on the unit circle?

[chilly ocean]

Yes if it's generator is outside s_1 iff any of its group elements are outside it?

Which makes everything outside it?

[delicate orchid]

Yeah, cause |z^n| = A^n which diverges away from 1 unless A = 1

[chilly ocean]

Yes also I think all those kind of groups are similar in quotienting

[tough raven]

Is there a category which contains groups but also contains e.g. sets with a transitive group action?

[hardy scaffold]

I mean you can consider the category of pairs (G,X) where G acts on X and the morphisms from (G,X) to (H,Y) pairs (a,b) where a is a group hom and b is a set map such that b(gx) = a(g)b(x)

[tough raven]

Sometimes it seems like it would be nice to consider both on an equal footing. For example, Haar measure (is there Haar measure on a coset space of a locally compact group?) or something.

[tough raven]

Hmm

I guess I was hoping for a more “direct” definition.

A free transitive (say, left) action on a set X can be specified by an operation (x,y,z) |---> (yx^{-1})z (specifying the “parallelograms”) if I'm not mistaken.

[hardy scaffold]

you can pushforward the haar measure

if what you quotient by isn't compact though this can cause issues

[tough raven]

Oh.

Can the condition for the Haar measure be expressed in terms of properties of the quotient space?

[tough raven]

That would fit with this idea.

[hardy scaffold]

This has a much better answer than I gave: https://mathoverflow.net/questions/21704/haar-measure-on-a-quotient

[chilly ocean]

Also what's up with the idea that closure of a group is a group

Is it true in general

[hardy scaffold]

yes

the closure of a subgroup of a topological group is a group

[south patrol]

multiplication is continuous

[glad osprey]

2Z is isomorphic to 3Z, right? But Z/2Z is not isomorphic to Z/3Z?

[delicate orchid]

correct

[glad osprey]

That really freaks me out. I thought two isomorphic groups have the exact same structure. What structure is ignored by isomorphism that makes Z/2Z and Z/3Z not isomorphic?

[paper flint]

2Z and 3Z do have the same structure

[south patrol]

They "sit inside" Z in different ways

Baiscally you want like

[delicate orchid]

trying to think of a finite group example

[south patrol]

An isomorphic of maps 2Z -> Z and 3Z -> Z

Rather than just isomorphisms of domains and codomains

to conclude that the quotients are iso

[rocky cloak]

C4xC2

[delicate orchid]

yeah there we go

[south patrol]

C4 = C2 x C2 by CRT

:)

[delicate orchid]

I was thinking non-abelian for some reason

[rocky cloak]

S3xC3

[delicate orchid]

another example, R[x]/(x) and R[x]/(x^2)

[delicate orchid]

ok then. NOT a product. I ain't BOUT THAT

[south patrol]

$\Sigma_3 \times \mathbf Z/3$

[cloud walrus]

Süßkartoffel

[south patrol]

Just to annoy wew

[glad osprey]

This is somewhat over my head, but is an isomorphism of maps a kind of natural transformation? Whereas an isomorphism is a functor?

[south patrol]

Yes, I mean in the sense of isomorphism of objects in the arrow category of a category C i.e. functors {* -> *} -> C

so concretely, that means like

A-> B and C -> D being iso if we have a square

A -> B
^ ^
| |
C -> D

where the vertical maps are isomorphisms

So the problem here is okay, youhave the vertical maps, but the diagram doesn't commute

[chilly ocean]

So what's the inverse of zero in closure of subgroup generated by say 2i

[south patrol]

What do you mean by zero when you are talking about closure in C^x

[chilly ocean]

Ah we are talking closure in C*

[south patrol]

Yes and C isn't a topological group under multiplication

at least

the typical multiplication doesn't turn it into a topological group

[chilly ocean]

Then what is the closure of this in C*

[south patrol]

It's already closed

[chilly ocean]

Okay

[south patrol]

I think

[hardy scaffold]

yes

[glad osprey]

I'll need some time to digest this, but thanks 🙏

[south patrol]

only limit point in C is 0, which isn'tin C^x

[chilly ocean]

It's just that I don't see how the quotient of C*/<2i>

Is iso to s_1 x s_1

[hardy scaffold]

C* is just R x S^1

[chilly ocean]

I agree

[south patrol]

Brain hurty

(0,infty) x S^1

[chilly ocean]

It should be R+

Yes

[dull ginkgo]

Shategory theory?

[hardy scaffold]

then <2i> is just Z. even if it sits inside of R x S^1 in a different way, their quotients are still the same

think about the group automorphism that shears the S^1 factor

[chilly ocean]

I have no clue

[dull ginkgo]

[chilly ocean]

Would you elaborate pls

Is that levy?

[hardy scaffold]

sorry, so R x S^1/Z (where Z is Z x {id_{S_1}}) is S^1 x S^1

[chilly ocean]

Hmmmmmm

[delicate orchid]

R/Z is S^1

but we need to specify where this Z subgroup is tbh

[hardy scaffold]

but now think about the automorphism of R x S^1 -> R x S^1 sending (x,e^{i t}) to (x, e^{i(t+ax)}).

This comes from a linear transformation on R x R sending (x,y) to (x, y+ax), then you quotient the second factor to get S^1

so if you have a group generated by (x, e^{i t}) for x non zero, then the quotient R x S^1/<x> is S^1 x S^1

[dull ginkgo]

Lie group more like cap group amirite

[chilly ocean]

Sorry i think i need to write this down

I still have difficulties understanding it

I will let you know after sometime

[chilly ocean]

Is the 'a' here sth specific?

[hardy scaffold]

no

[chilly ocean]

Ok what map do u take here

log x?

[hardy scaffold]

yes

[chilly ocean]

For the modulus?

[chilly ocean]

Would you mind telling me again why this step is being done

Just briefly

[hardy scaffold]

to show that when we quotient by the subgroup generated by (x, e^{it}) for x nonzero, it is isomorphic to the quotient by (x,1)

[chilly ocean]

Oh

Okay i think it's clear to me now

Thanks a lot V5_

[tough raven]

C^× is isomorphic to R × S^1 by the isomorphism (t, θ) -> θ exp((ln 2 + 2πi/4) t).
Under this isomorphism, Z × {1} corresponds to ‹2i›.

To understand this isomorphism a bit better, look at the image of R × {1}.

[rocky cloak]

Semidirect product of C4 with C4
<x, y | x^4 = y^4 = 1, xy = yx^- >

C2 is a subgroup as (x^2) and as (y^2)

First has quotient C2xC4, second has quotient D4.

[delicate orchid]

kino cinema

whoops just realised this post is over 3 hours old, sorry for the ping haha

[toxic zephyr]

so i've been continuing to explore F^n as an F[x] module via a matrix A: p(x)v=p(A)v, and i think i've decomposed it, but i'm not totally sure. i basically used the jordan decomp: breaking it up into the regular eigenvectors and eigenchains.

i think it makes sense any regular eigenvector corresponds to a one dimensional subspace F[x]/<x-k> since (x-k)v will be zero if v has eigenvalue k.
and we can leverage an eigenchain by multiplying by multiples of (x-k) to get us the rest of the basis for each generalized eigenspace.
can someone just sanity check me that this vibes well?

[delicate orchid]

oh my golly gosh

[glossy crag]

This doesn't make sense, m_i and k_i shouldn't be different.

[toxic zephyr]

mi is the geometric multiplicity, and ki is the length of an eigenchain.

(if there are multiple eigenchains then we just let other lambda_i's to be the same and choose those mi=0

this is a matrix direct sum, not just a regular sum of matrices

[glossy crag]

Still doesn't make sense, if I and J are diagonal/Jordan blocks.

[toxic zephyr]

why not

I is the identity and J is a jordan block

[glossy crag]

Geometric multiplicity of an eigenvalue is the size of the largest Jordan block with that value

[toxic zephyr]

uhhhhhhhhh

[glossy crag]

An identity block is not a Jordan block, it's a sum of Jordan blocks of length one, so what you say would mean that each eigenvalue has geomult 1, making the matrix diagonalisable.

[toxic zephyr]

geo mult is the dimension of the eigenspace

[toxic zephyr]

um... if you have a scalar matrix direct sum a jordan block, that's not diagonalizable

[glossy crag]

Yes, but then you can't say that the size of the identity block is the geomult.

[glossy crag]

Anyway, I may be wrong, but what you wrote here looks like complete porridge to me.

You should look into the decomposition for modules over PIDs and how that relates to Jordan form.

[toxic zephyr]

oh then it's the geo mult-1 or -the number of eigenchains

[rocky cloak]

So the point is it shouldn't be one identity matrix and one Jordan block for each eigenvalue. You should have a sum of Jordan blocks of varying sizes for each eigenvalue.

After that you are correct in how a Jordan block of size k corresponds to F[x]/(x - lambda)^k

[toxic zephyr]

i was just doing this for simplicity

[rocky cloak]

So the geometric and and algebraic multiplicity is not enough information

[toxic zephyr]

what

no i said it corresponds to (F[x]/<x-lambda>)^m

i guess the main summary of what i'm doing is saying that an eigenvector corresponds to F[x]/<x-lambda> and an eigenchain of length k corresponds to F[x]/<(x-lambda)^k>

[rocky cloak]

The geometric multiplicity tells you the number of Jordan blocks, but the distribution of their sizes can still vary quite a bit (even if you know the biggest one)

[glossy crag]

Does it? Isn't it the size of the largest block?

[toxic zephyr]

its the nullity of (A-lambdaI). so its the number of independent eigenvectors/jordan blocks

[rocky cloak]

No, the geometric multiplicity is the size of the eigenspace. Each block gives you an eigenvector, so the dimension is the number of blocks

[toxic zephyr]

i dont think you can get the exact size of the jordan blocks from just the alg and geo mult unless you know theres only one eigenchain

if theres only one eigenchain, then the size is alg mult-geo mult+1 or something

[glossy crag]

Ah, I'm confusing geomult with the prime decomposition of the minimal polynomial, I think.

[rocky cloak]

Yeah, the minimal polynomial will tell you the size of the biggest block.

[toxic zephyr]

i guess we can make it even more compact by just only using jordan blocks and allowing duplicate eigenvalues (i.e. just summing along the eigenchains, including those of length 1, regular eigenvectors)

and then the iso comes from the start of the eigenchain and then we can just multiply by (x-1)^k to get the other basis vectors from the chain. and doing that for all such jordan blocks gives us the full jordan basis

[rocky cloak]

I guess people often go the other way. Using the structure of F[x] modules to prove Jordan decomposition of matrices 🙂

[toxic zephyr]

oh yeah huh I guess thats a similar problem

so does this mean the generators for the jordan chains are a basis for the module or just that they generate the module? not sure if we need to mod out by the minimal polynomial to really get that

[toxic zephyr]

so i think i can sorta see how this would go. but the key seems to be that it's a direct sum of F[x]/<(x-k)^l>. how do we know that if a matrix has eigenvalues 1 and 3 (for example) that we won't have F[x]/<(x-1)(x-3)>?

[rocky cloak]

F[x]/(x-1)(x-3) = F[x]/(x-1) (+) F[x]/(x-3)

[toxic zephyr]

is it the linear independence of eigenvectors with differing eigenvalues?

oh

[rocky cloak]

You can use the Chinese remainder theorem for this for example

[toxic zephyr]

im gonna try to prove this

[rocky cloak]

It's very much the same flavor to that yeah

[toxic zephyr]

so the isomorphism p(x)+<(x-1)(x-3)> to (p(1)+<x-1>,p(3)+<x-3>) clearly works. so it seems like it would be possible to combine (at least part of) the decomp into F[x]/<p_A(x)> where p_A is the minimal polynomial. i would expect it would be the parts corresponding to the largest eigenchains?

i haven't thought about it too much, but i'm guessing this principle extends to F[x]/<(x-a)^k1(x-b)^k2>=F[x]/<(x-a)^k1>(+)F[x]/<(x-b)^k2>? something something taylor series
iirc the fg module over PID decomp theorem says that what you mod out by in the direct sum can have a division chain. so we should be able to combine the terms to get a decomp
F[x]/<p_A(x)>(+)F[x]/<p1(x)>(+)...(+)F[x]/<pl(x)>
where p_l | ... | p1 | p_A(x)?
i wonder if we can say anything about what the p_i's must be

is that true actually? if we have some decomp R/(b1)(+)...(+)R/(bn), then we can rearrange/combine them into the form R/(d1)(+)...(+)R/(dn) where d1|d2|...|dn?
i would guess yes, but im not sure

[cobalt heath]

Yes, I think you can follow the same reasoning

[dull ginkgo]

Trying to process something

Assume $A$ and $B$ are subgroups of abelian group $G$, then does a homomorphism $\mu$ induce a map from $\frac{A}{B} \rightarrow \frac{\mu(A)}{\mu(B)}$, where $\mu(A)$ is the image of A

[cloud walrus]

A Fibrous Powder

[cobalt heath]

Yeah, I think so

The map A -> mu(A) sends B to mu(B)

So you should be able to quotient like that

[dull ginkgo]

I drew out a diagram and realized I'm a moron

[cobalt heath]

Uh, wdym moron

[dull ginkgo]

I am a big goober brain

brain only booger

[cobalt heath]

Imposter syndrome taking another victim

[dull ginkgo]

sus

[cobalt heath]

What kind of diagram did you draw

Hm does this follow from exact sequences

[dull ginkgo]

no just traced along it to get a map

also isn't it true that $0 \hookrightarrow A \hookrightarrow B \twoheadrightarrow C \twoheadrightarrow 0$ imply $C \cong B/A$

[cloud walrus]

A Fibrous Powder

[dull ginkgo]

i'm pretty sure

what makes me wonder is if we have an morphism backwards from B

[cobalt heath]

Yes

[cobalt heath]

If you have exact sequence 0 -> A -> B -> C -> 0 with one projection B -> A, it is called splitting condition, I think

[dull ginkgo]

Huh

might try and prove that

[dull ginkgo]

well I mean A into B must be an injection, B into A that identifies that injection is a hmmm

[cobalt heath]

B -> A is a projection

So that A -> B -> A is identity, but not B -> A -> B

[dull ginkgo]

ah I see

[dull ginkgo]

Huh so that specifies when $A \times \frac{B}{A} \cong A \otimes \frac{B}{A} \cong C$?

[cloud walrus]

A Fibrous Powder

[cobalt heath]

It’s direct sum instead of tensor product, but yes

[dull ginkgo]

oplus lol

[dull ginkgo]

I was fucking around with a cochain complex and idk why my brain decided to short circuit on this lmao

[cobalt heath]

Happens always for me

[dull ginkgo]

I am kind of being drawn into category theory aaaaa

the notion of (co)limits are kinda sick

[cobalt heath]

It’s okay to give in

Yeah, limits are quite universal

[dull ginkgo]

in a weird way they're like the join and meet for subsets in lattices

suprema and infima?

i.e if A gets mapped commutatively into our diagram, it has to pass through the product, and if everything maps into B, then it has to go through the coproduct

[cobalt heath]

Oh, this is also an intuition I bumped into

For me, limit is like inf, colimit is like sup

[dull ginkgo]

direct sum works out nicely for abelian groups since the finite support means you can add up an elements component images

[dull ginkgo]

yep

Sheafs are like fucking entire categories associated with topologies though I keep seeking to find a "categorical" way to describe the gluing and locality axioms. I keep getting way ahead of myself though

[cobalt heath]

Yeah

[dull ginkgo]

someone described it here once and I kinda just stared at the def and thought it was sick

[cobalt heath]

I heard in #algebraic-geometry that they are equalizers

[cobalt heath]

Indeed

[dull ginkgo]

anyway with the cochain complexes I keep trying to expand this subgroup lattice system into an almost homologic "system"?

hard to describe

Using the fact that $A \subset B \subset C \Rightarrow \frac{B}{A} \hookrightarrow \frac{C}{A}$ and $\frac{C}{A} \twoheadrightarrow \frac{C}{B}$

[cloud walrus]

A Fibrous Powder

[dull ginkgo]

Which... is an exact sequence :D

$0 \hookrightarrow \frac{B}{A} \hookrightarrow \frac{C}{A} \twoheadrightarrow \frac{C}{B} \twoheadrightarrow 0$

[cloud walrus]

A Fibrous Powder

[dull ginkgo]

so that's kinda neat

@cobalt heath does this have any strong applications?

[cobalt heath]

Dunno about this one

Seems like (C/B) ~= (C/A) / (B/A)

[dull ginkgo]

That’s how I got the last epi

[cobalt heath]

Which.. what was the name of the isomorphism theorem

[dull ginkgo]

3rd I think?

[cobalt heath]

Yeah, seems like it would have the same applications

[dull ginkgo]

a homie's gonna ponder

for regular chain complexes it's just im(f) in ker(f)

for double ones it's nuts

lmao

i.e v(ker(d)) is in ker(h) and im(v)

v(ker(h)) is in ker(v) and im(h)

[dull ginkgo]

gonna see if I can do some homological sillyness using this goofball exact sequence applied a lot

[cobalt heath]

Huhh

[dull ginkgo]

it's neato because the image of the whole thing under h or v connects at the bottom to the next one

creating this massive network

exactness at h or v just makes the left or right inclusion equal

[cobalt heath]

Is this a commuting diagram?

[dull ginkgo]

it's really an inclusion lattice lol

a line going up implies the lower is in the upper

[cobalt heath]

Where is h and v from?

[dull ginkgo]

horizontal morphism or vertical morphism

it's a double complex

Really it's im(h_n - 1,m)

but I'm just not using the indicies because the domains are implied

[cobalt heath]

Ahh, like chain complex but two directions

[dull ginkgo]

yeppers

d is the diagonal, or h o v

or v o h

commutes so it doesn't matter lol

[cobalt heath]

Nice

[dull ginkgo]

I think there's a proof of the salamander lemma like this using spectral sequences maybe?

someone told me this approach looks similar to those but whenever I look them up I get jumpscared by mathematical machinations beyond my raisin brain

Anyway if h is exact at our X

then ker(h) = im(h) = H for convenience

this time h(ker(d)) is in ker(h) cap ker(v)

and v(ker(d)) is in im(h) + im(v)

[chilly ocean]

here there's no data on the dimension of V when doing 1 -> 2 suppose eigenvctors span v but what if the dimension of V divides n and not actually n

wont we still have all the conditions satisfied and cardinality lesser

i undertand the root of the poly wont repeat for x^n - 1 as its derivative only zero is 0

as char of field is 0

ah wait

its more like its sufficient for some generator of the set in 2 to be a eigevalue of T?

[chilly ocean]

what about T as identity transformation on R^3

T^3=I

V is spanned by eigenvectors of T

the cardinality of the set {x in R,x^3=1} is 1

also -I in R^4 works

as a counterexample

[chilly radish]

Presumably n is minimal

[chilly ocean]

Ah then 1->2 is very easy?

Cause all eigenvalues are distinct roots

[chilly radish]

Yea

That's the converse too

Do you see why?

[chilly ocean]

Yeah

It becomes minimal poly

[chilly radish]

Ye

[chilly ocean]

And it factors distinctly

Well they never wrote n is minimal

[chilly radish]

No but otherwise as you noted it's false

[chilly ocean]

I spent hours without any progress

Right I should have looked for counter first

Thank you for the suggestions

[chilly radish]

No problem

Don't blame yourself

It's the problem writer's fault

[chilly ocean]

It's from a quals exam

💀

[chilly radish]

Like otherwise, n is a multiple of the minimal number k, say for n=2 we get
T^2k-I=0 =(T^k-I)(T^k+I)
And then we actually need k-th roots of unity, and maybe there are all k-th roots but not all n-th roots (take k=Q(xi_k) for xi_k a primitive root of unity).
2->1 is true still tho

Since if there's nth roots there's k-th roots for k|n

[rocky cloak]

The statement is that it holds for arbitrary T. So finding one example of T is not enough for the statement to be true.

[chilly ocean]

Hmm?

T be a linear transformation satisfying...

Oh wait

No but

Ok this is confusing

[chilly ocean]

I was actually giving a counterexample here tho

[rocky cloak]

No, you said that T = id satisfies (i), then tried to conclude that (i) was true

[chilly ocean]

A=>B if both are true then this is true

[chilly ocean]

No I was saying 2nd statement is not satisfied

It's true value should also be true

[rocky cloak]

Yes, but that's only a contradiction if (i) is true

Showing that (ii) is false without determining (i) doesn't tell you anything

[chilly ocean]

Nah but we can say they aren't equivalent

[rocky cloak]

But they are equivalent

[chilly ocean]

How

[rocky cloak]

Because you can always construct a T that has minimal polynomial x^n - 1

[chilly ocean]

Yes companion matrix

[rocky cloak]

So if you don't have enough roots of unity, then (i) can't be true

[chilly ocean]

But

Huh

Isnt this of the form (A=>B) <=> C

[rocky cloak]

The statement is
If T:V -> V does A, then B.

If there exists a T such that A is true, but B is false then the statement isn't true.

The fact that there exists T for which B is true or that A isn't, doesn't tell you anything conclusive about the statement

[chilly ocean]

Okay but how do u know a implication is right

If A is true and B is true in A=>B then it's true right?

[rocky cloak]

Yeah so the implication would have to hold for any choice of T.

[chilly ocean]

Right I see now there's a 'forall' T satisfying T^k=id there

Hidden

[rocky cloak]

For example if I say

x^2 = 1 implies x = 1

You can't just say, plugging in 1, we see that the statement is true

[chilly ocean]

I agree

So how is this done

Then

[rocky cloak]

Well, pretty much like twisted said. You just construct a specific T, that you use to prove (ii). And conversely you figure out what the minimal polynomial of T could be.

[cobalt heath]

TIL Eigenspace of identity spans entire space

[rocky cloak]

Finally, all the eigenvectors are mine! Muhahaha!
-Identity transformation, probably

[chilly ocean]

So basically this power of T has to be minimal by taking a specific T (like the companion matrix) and the other side ((2)->(1) holds regardless)?

[chilly ocean]

Any scalar multiple of identity transformation also

Thank you for the clarification

[pliant forge]

Suppose F=GF(p^n) such that |F^*| = p^n - 1. How can we then identify F with a group so that we can calculate primitive elements of the multiplicative group corresponding to F^*?

[delicate orchid]

I'm not following what you mean. Do you just want the result that F^* is cyclic or?

[rocky cloak]

I guess you can start with the polynomial x^(p^n - 1) - 1. Then divide out x^d - 1 for all proper divisors d of p^n - 1. Then you should be left with the polynomial whose roots are all the primitive roots.

[pliant forge]

no i need to find all generators of F^*

[delicate orchid]

sure cyclotomic polynomials would work but they want something to do with "identifying with a group"?

[rocky cloak]

Once you have a primitive root, then you get the isomorphism between F^* and Z/(p^n - 1)

[delicate orchid]

yeah but they said F lol

if it was F^* I wouldn't be confused

[pliant forge]

yeah my professor does something like this: «identify F with Z_16» where in this case F was GF(17)

[rocky cloak]

Yeah that should be
Identify F^* with Z/16

[pliant forge]

i see

[rocky cloak]

So you need to "find" them, meaning you want to describe their minimal polynomial or something? Or you just want to count how many there are?

[pliant forge]

Counting is simple isnt it? its just phi(p^n-1) no?

[rocky cloak]

Indeed

Well, assuming you know at least one exists anyway

[pliant forge]

this was the question given «Find generators for the multiplicative groups of fields with 8,
13, and 17 elements.»

[rocky cloak]

With these sizes, I think the guess and check method is the most efficient

[delicate orchid]

once you have one of the generators it's a lot easier to get the rest of them as well

[pliant forge]

yeah. just was stuck with representing it so i could check

[rocky cloak]

For GF(8) you mean, or...?

[delicate orchid]

well for F_13 and F_17 you can just use the fact that they're isomorphic to Z/13Z and Z/17Z

and for F_8 I'd just go with F_2[x]/(x^3+x+1)

[pliant forge]

well for that one all of them are so no problem

all except 0 and 1*

[pliant forge]

right. not sure why i was stuck on that

Since number of elements of F_p is prime, p, and the characteristic of F_p is a prime it must be p and contain a subfield that is iso to Z/pZ, so the entire set

[delicate orchid]

sure, and then a simple cardina- yup you got it

[chilly radish]

Ok I think I misunderstood what the question asker wanted

In this case @chilly ocean I agree with jagr's interpretation. I was taking T to be fixed and then it asking us to prove the equivalence for a given T, which would require n to be minimal, but it says that an operator satisfying T^n=I being diagonalisable implies that there are n n-th roots of unity

So for 2=>1 you really do just need to find one operator who satisfies both of those

And n is fixed from the start

[rocky cloak]

If T is fixed then it wouldn't make sense even if we assume what was meant was that n is minimal.

Because if T doesn't satisfy T^n = id, then (i) is automatically satisfied.

[chilly radish]

Ok fine then "fix T with T^n=I" is what I was really assuming

Just cuz the wording on this problem is pretty bad

[rocky cloak]

They certainly could have afforded an "for all" somewhere

Or just replaced "V is a vector space" with "for any vector space V"

[teal vessel]

the group of sets S_a={x in R | x - flr(x) = a} under addition is literally just R/Z, right? the quotient group of the additive group of reals mod integers. Sometimes I feel like my tendency to just start constructing things without reference to the literal section of the text I'm in (in this case the chapter on quotient groups) really comes back to bite me.

[coral spindle]

x - flr(x) = a is equivalent to x + Z = a + Z

[long obsidian]

Suppose I had two commutative rings with identity R and S

Is it true that $(R/nil(R))\times (S/nil(S))= (R\times S) /nil(R\times S)$?

I think the canonical projection maps are $f:(R/nil(R))\times (S/nil(S))\to (R/nil(R)$ defined by $f((r,s)+nil(R\times S))= r+nil(R)$ and similarly $g:(R/nil(R))\times (S/nil(S))\to (S/nil(S)$ defined by $g((r,s)+nil(R\times S))= s+nil(S)$. Does this sound right?

[cloud walrus]

HausdorffT1

[dull ginkgo]

is $\mathrm{nil}$ the nilradical?

[cloud walrus]

A Fibrous Powder

[dull ginkgo]

This is for exercise 1.10 of Jacobson:

[languid trellis]

where r u looking

[dull ginkgo]

Prove $\frac{\bigcap H_a}{K} = \bigcap \frac{H_a}{K}$ for $K \trianglelefteq H_a \subseteq G$

[languid trellis]

ah

can't help them lols

[dull ginkgo]

I was going to see if this is a good proof so uh if you want to do this one don't read it

[celest furnace]

Isn't this equality?

[dull ginkgo]

yeah

[cloud walrus]

A Fibrous Powder

[dull ginkgo]

w a i t

technically this can be done set theoretically because the quotient map is surjective

[delicate orchid]

let H denote the intersection of all those mfs cause I can't be bothered typing it
let kH be something on the left, now think about what the things on the right have to be

[dull ginkgo]

I got derailed because I had an idea of an alternative way to prove it

[delicate orchid]

go on?

[dull ginkgo]

typing it out lol

[delicate orchid]

also, just when you're done, is this a finite collection of subgroups or an arbitary collection

[dull ginkgo]

Let $\pi$ be the projection map, then what matters here is that $S \subseteq \frac{G}{K} \Rightarrow \pi(\pi^{-1}(S)) = S$. By definition $\pi(H_a) = \frac{H_a}{K}$. Preimages preserve intersections, so $\pi^{-1} \left( \bigcap \frac{H_a}{K} \right) = \bigcap \pi^{-1} \left( \frac{H_a}{K} \right)$. Because of the congruence relation, $\pi^{-1}(xK) = y \LeftrightArrow \pi(y) = \pi(x)$, so $\pi \left( \frac{H_a}{K} \right) = H_a$

ALWAYS THE -1 POWERS

So applying $\pi$ and using the first property implies $\bigcap \frac{H_a}{K} = \frac{\bigcap H_a}{K}$

[cloud walrus]

A Fibrous Powder

[dull ginkgo]

god I hate latex

[delicate orchid]

there isn't a single command that's camel case

this ain't minecraft 1.20

oh brother this guy STINKS

[dull ginkgo]

I worked with java for like 3 hours this morning

[delicate orchid]

Let $\pi$ be the projection map, then what matters here is that $S \subseteq \frac{G}{K} \Rightarrow \pi(\pi^{-1}(S)) = S$. By definition $\pi(H_a) = \frac{H_a}{K}$. Preimages preserve intersections, so $\pi^{-1} \left( \bigcap \frac{H_a}{K} \right) = \bigcap \pi^{-1} \left( \frac{H_a}{K} \right)$. Because of the congruence relation, $\pi^{-1}(xK) = y \leftrightArrow \pi(y) = \pi(x)$, so $\pi \left( \frac{H_a}{K} \right) = H_a$
So applying $\pi$ and using the first property implies $\bigcap \frac{H_a}{K} = \frac{\bigcap H_a}{K}$

[cloud walrus]

Wew Lads Tbh
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[dull ginkgo]

The camelcase is there

[delicate orchid]

tada!

anyway I like this idea

[dull ginkgo]

It's a bit excessive, but the one I wrote down did it via an element approach

and showing two way \subseteq

[delicate orchid]

yes and that's where I was trying to nudge you towards

the proof for finite collections of subgroups is u literally just show that aH_1 \cap bH_2 = c(H_1 \cap H_2) for some c in G and then induct

[stark helm]

a quick question: how to use RRT(rational root theorem) to show that 2^(1/3)+sqrt(3) is irrational?

[dull ginkgo]

in Q(2^1/3, 3^1/2)

[delicate orchid]

finding the minimal polynomial might be a good start

[delicate orchid]

or show that Q(2^1/3+sqrt(3)) isn't Q via the tower law I guess

[dull ginkgo]

[vivid tiger]

...it's a degree 6 extension?

so,6≠1?

QED?

[delicate orchid]

ok thanks??

anyway

[vivid tiger]

am I missing something?

I'm asking to see if I understood you correctly

[dull ginkgo]

prove it's a degree 6 extension
because Q(2^1/3 + sqrt(3)) isn't Q

[vivid tiger]

Other way around

[delicate orchid]

if you want to use the RRT, try applying to whatever minimal polynomial you find

[delicate orchid]

you're completely missing the entire point irregardless

you're saying "it is because it is"

[vivid tiger]

I am?

[dull ginkgo]

The element approach I used was just flat out quantification over all of the a and the two way inequality by def (inclusion "infimality") of the intersection lmao

[delicate orchid]

and unless fibrous was going to actually use the RRT at some point then it was irrelevant to the question asked too

[crystal turtle]

****** regardless 🤓 🤓

[delicate orchid]

infmality is a word u just made up

[dull ginkgo]

that's why I used quotes

[delicate orchid]

go back to #category-theory we're doing things that actually mean something here

[crystal turtle]

[stark helm]

do you mean x=2^(1/3)+sqrt(3) and becomes x^6=(2^(1/3)+sqrt(3))^6??

[dull ginkgo]

just temporarily set your category to Grp like how people try to visualize 4d manifolds by setting N = 4

Category theorists are just mathcord spy mains

[stark helm]

I am wondering what does minimal polynomial refer to here?

[delicate orchid]

hold on lemme actually compute the minimal polynomial lol

[opens magma like a boss]

[stark helm]

what is hold on lemma?

[delicate orchid]

am I being trolled

[stark helm]

I mean if it is correct to start on x=2^(1/3)+sqrt(3), but I don't know how to proceed here since I tried square or cube on both sides, but it seems not work

[delicate orchid]

ok the minimal polynomial is grosssss

[stark helm]

then if there is any method using RRT?

[delicate orchid]

well I have the minimal polynomial, and the RRT does indeed give us that there are no rational solutions

it's just finding that minimal polynomial without just throwing a computer at it

[stark helm]

in fact, i found that it will have the stuffs like x^3-2= the side that appears stuffs like 2^(2/3) and sqrt(3)

[stark helm]

then i actually can directly say since left sides is rational while right one is irrational, then done

but I don't know if that satisfies RRT?

[dull ginkgo]

doesn't rational root theorem ONLY need the leading and trailing coeffient

[delicate orchid]

do you need to use RRT or can we use literally anything else

[dull ginkgo]

I.E the product of all the conjugates

[delicate orchid]

if you say so!

[dull ginkgo]

and our leading coefficient is just 1

[delicate orchid]

yeah cause it's minimal

[stark helm]

so can I just say that the last coefficient is irrational, so no rational numbers can divide it?

[delicate orchid]

the last coefficient of the minimal polynomial is -23

[dull ginkgo]

The rational root theorem states that if the root is (coprime) p/q, then p divides the leading coeffient and is thus 1, and p divides the trailing coefficient, i.e the product of all the conjugates.

-23 is prime

[delicate orchid]

yeah

good luck multiplying all the conjugates together nerds

[dull ginkgo]

true

@delicate orchid wait doesn't x in H_a iff xK in H_a/K like, make it immediate lmao

[delicate orchid]

finish the train of thought

I wanna see if you'll figure out where I was hinting towards with this

or if there's a faster way I missed lol

[dull ginkgo]

It's just using that two way:

For all a, x in H implies x is in H_a, thus xK is in all H_a/K thus is in the intersection of all H_a/K
For all a, xK in all H_a/K implies x is in all H_a, and is thus in the intersection of all H_a

then it's just defining the image as (intersection H_a)/K

[delicate orchid]

yup that is faster

[dull ginkgo]

My brain went "oh you gotta prove the image is (intersection H_a)/K" but that's the goddamn def lmao

[dull ginkgo]

it all hinges around that x in H iff xK in H/K property which frankly like

all of the group iso theorems pull from

[severe tide]

what is a surjective ring homo $f:{p/q\in\mathbb Q:q\neq 47k}\to\mathbb Z_{47}$ such that $\ker f={p/q\in\mathbb Q: q\neq 47k, p=101\ell}$

[cloud walrus]

kevinhardy2

[severe tide]

k,l in Z

[dull ginkgo]

[dull ginkgo]

I mean I would look at how this hypothetical map would affect Z

Actually the domain itself is interesting

[teal lily]

Hey guys,sorry to interrupt, am fairly new to groups and proof writing and was just wondering if someone would check over this proof i have written , not sure if this is the right place to do that.

[severe tide]

?

[dull ginkgo]

I haven't done much ring theory but here's my take on it

I'm gonna call your domain Q_47. We know Q_47 contains the ring Z generated by 1

The only real difference between Q and Q_47 is that 47 and any multiple of it are no longer a unit

This makes (47) an ideal in Q_47, and a maximal one at that.

Q_47/(47) is a field

wait

[terse crystal]

Send it

[dull ginkgo]

@severe tide I don't think it's even possible. Here's my two cents. Your domain ring, I'll call $\mathbb{Q}{47}$, is not a field, as $47$ is not a unit. Therefore, it's generated set $(47)$ is an \textbf{ideal} in $\mathbb{Q}{47}$. We are looking for an epimorphism $f : \mathbb{Q}{47} \twoheadrightarrow \mathbb{F}{47}$ (a finite field of \textbf{characteristic 47}). Therefore, automatically $(47)$ is in the kernel of $f$. It's a field because $47$ is prime and thus irreducible, meaning $(47)$ is a \textbf{maximal ideal}. The special kernel you want is not possible, as it would have to strictly contain a maximal ideal for it to be onto.

[cloud walrus]

A Fibrous Powder

[terse crystal]

If Z_p he means Z/pZ yeah impossible

There exists k such that f(a/b)=kab^-1, then ker f is either Q_p (when k=0 mod p) or 0 (otherwise)

[dull ginkgo]

Also your kernel is the whole of your domain, because $1$ would have to be in your kernel, since $\frac{1}{101}$ is in there too, so $101 * \frac{1}{101} = 1 \in \mathrm{ker}(f)$. Since multiplying anything by an element of your kernel will be inside your kernel, everything's in the kernel

[severe tide]

Ok what about instead

[cloud walrus]

A Fibrous Powder

[dull ginkgo]

@severe tide are you talking about the 47-adic integers?

[severe tide]

No youre rivt

[dull ginkgo]

okay just wanted to make sure because p-adics are way out of my league

[cobalt heath]

Rivt?

[severe tide]

How about this

[cobalt heath]

p-adic integers are just directed limit of Z/p^k Z

[severe tide]

Is a+bx a unit in R[x] if b is nilpotent and a is a unit?

Not shre hownto show this

[dull ginkgo]

I haven't done this before.

[severe tide]

I'm confuse

[dull ginkgo]

It might help to assume there's another polynomial such that (a + bx)p(x) = 1

a has to be a unit and we know the constant coeff of the other polynomial has to be a^-1

[cobalt heath]

Hmm, it divides a^n + b^n x^n = a^n

So it is a unit

[severe tide]

?

[dull ginkgo]

I don't know off the top of my head how to show that a polynomial is a unit iff all but it's constant coefficients are nilpotent, and it's constant is a unit

[cobalt heath]

x + y divides x^m + y^m when m is odd

Yknow, polynomial factorization

[dull ginkgo]

OH SHIT YEAH

It's the alternating sum of the powers shifting from one to the other

[cobalt heath]

Yeah, tbh I don't like when one has to do computation to prove

[dull ginkgo]

same

[cobalt heath]

Maybe there is a way to involve advanced technique to avoid factorization? But then it's harder..

[severe tide]

So you saying the solution is

[dull ginkgo]

$X^{n+1} - a^{n + 1} = (X - a) \sum_{k = 0}^{n}{a^{n - k}X^k}$ is kinda by induction

[cloud walrus]

A Fibrous Powder

[dull ginkgo]

I also think it's true in any commutative ring

[severe tide]

What is that telling me

[severe tide]

'

[dull ginkgo]

Assume you have $p(X) = bX + a$ where $b^N = 0$. Then $b^N X^N +a^N = (bX + a)\sum_{k = 0}^{N - 1}{(-a)^{N - k - 1} (bX)^{k}}$

Therefore

[cloud walrus]

A Fibrous Powder

[dull ginkgo]

$a^N = p(x) \sum_{k = 0}^{N - 1}{(-a)^{N - k - 1} (bX)^{k}} \Rightarrow 1 = p(x) \left( a^{-N} \sum_{k = 0}^{N - 1}{(-a)^{N - k - 1} (bX)^{k}} \right)$

[cloud walrus]

A Fibrous Powder

[dull ginkgo]

i.e p(x) is a unit

[dull ginkgo]

@severe tide That sum in parenthesis is literally your inverse lmao

so there is an accidental calculation of the unit's inverse lol

[cobalt heath]

Notable thing to recall:
If some v divides a unit, then v is itself a unit

[dull ginkgo]

that actually works to prove that p(x) is a unit iff p(x) = u + v(X)X, where u is a unit and v(X) is a nilpotent polynomial

[cobalt heath]

Hmm, does v have to be nilpotent

[dull ginkgo]

yes

It's an inductive proof

Basically using a + bx + cx^2 + dx^3 = a + x(b + x(c + x(...)))

which you have to show all but the last constant coefficient is nilpotent

at least that makes sense to me

The whole gist is that if f(X) is nilpotent, and so is a, then a + f(X)X is also nilpotent

[cobalt heath]

I don't think so.
On the ring R[x]/x^2,
1 + x is a unit with inverse 1 - x.
But 1 is not nilpotent

[dull ginkgo]

good point

[dull ginkgo]

no wait this is purely over R[X], not any of its quotient rings

[cobalt heath]

But then, how can v(X) be nilpotent?

[dull ginkgo]

v(X) is nilpotent as an element of R[X}

i might be misinterpreting what you're saying though

[cobalt heath]

If R is domain, R[X] is also a domain

I guess something happens if R is not a domain?

[dull ginkgo]

zero divisors

Do you want an ideal theoretic proof I suppose

[cobalt heath]

Let me think

[dull ginkgo]

We have R[X], and we also have the nilradical K = Nil(R[X]). It's an ideal, which we can verify using binomial theorem or the route shown above. Therefore if p(X) is in K, then Xp(X) is in K

[cobalt heath]

Ah perhaps this is adjacent to the Jacobian ideal

Jacobson*

[dull ginkgo]

Lol it's as simple as all of our coefficients being in our nilradical

Nil(R[X]) = X * Nil(R) lmao

because it's an ideal, it's going to "aborb" all of our X powers

[cobalt heath]

IIRC, in commutative case maximal nilradical is the Jacobson ideal

[dull ginkgo]

we just have to SHOW it's an ideal

[dull ginkgo]

jacobson radical i thought is the intersection of all maximal ideals

Actually I think i know why specifically this occurs

There's actually a weird ideal-theoretic thing behind this that eludes me

I wonder when it is the case that, for an ideal $V \subset R[X]$ that $p(X)$ is in $V$ if and only if all of it's coefficients lie in $V \cap R$

[cloud walrus]

A Fibrous Powder

[cobalt heath]

I was sloppy

J(R) contains every nil ideal of R. If R is left or right Artinian, then J(R) is a nilpotent ideal.

Also nilpotent ideal /= nil ideal. Sorry..

[dull ginkgo]

no worries lol

[cobalt heath]

But I recall in the proof of characterization of Jacobian radical, something being a nilpotent element appeared a lot

[dull ginkgo]

Jacobian or jacobson?

For commutative rings the Jacobson radical is the intersection of all maximal ideals

There's a rediculous characterization that uses the fact that if x lies in an ideal A, then 1 + x cannot lie in A

I think it's that forall r in R that 1 + rx is a unit for x in J(R)

[cobalt heath]

Jacobson

Lol, my brain is not working today

[cobalt heath]

Yeah, indeed

[dull ginkgo]

The set of units in any commutative ideal is literally $R \setminus \bigcup M$

[cloud walrus]

A Fibrous Powder

[dull ginkgo]

the set of units is the complement of the union of all maximal ideals

i.e the union of ALL strict ideals

If an element doesn't lie in an ideal, then it's a unit. It seems a bit weird at first, but if x doesn't lie in any strict ideal then (x) must be R, implying there is a y such that xy = 1

[cobalt heath]

But on R[X], if 1 + a is an ideal, a has to be nilpotent, right

[dull ginkgo]

Which is a funny way to say that a field is a ring with no nontrivial ideals

[dull ginkgo]

?

(1 + a)?

[cobalt heath]

I am not functioning

[dull ginkgo]

i.e the subring generated by 1 + a

i feel ya tbh

[cobalt heath]

1 + a being a unit

[dull ginkgo]

yeppers

[cobalt heath]

Might be a consequence of R[X] being artinian

[dull ginkgo]

?

i.e it cannot be artinian unless it's "trivial" in terms of ideals

[cobalt heath]

Wait it isn't

[dull ginkgo]

No

[cobalt heath]

Hmmm, this is strange.

[dull ginkgo]

(X^n) is an infinitely decreasing sequence of ideals

it's NEVER artinian

[cobalt heath]

Yeah

So, note that as you said, if 1 + a is unit, a is nilpotent, so for all r, 1 + ra is a unit.

[dull ginkgo]

wait let me think

[summer path]

I don't know why but I picture the sequence of ideals (x^n) as a duck waddling infinitely far away

[cobalt heath]

That means if 1 + a is unit, a is in J(R[X])

[dull ginkgo]

you have just inserted a distracting cognitohazard in my brain thanks

[dull ginkgo]

1 + ra would have to be a unit for ALL r

which actually

[cobalt heath]

That holds if a is nilpotent

Because ra is also nilpotent

[dull ginkgo]

is Jac(R[X})) = X*Jac(R) too?

I know Jac(R) contains Nil(R) by def

Wait no

it's vice versa

[cobalt heath]

Well, we showed earlier that if a is nilpotent, then 1 + a is unit

ra is also nilpotent, so 1 + ra is also unit

[dull ginkgo]

I don't think that's true in general then?

let me ponder for a sec

Speaking of jacobson i need to do exercises from basic alg tonight

[cobalt heath]

Since when was Jacobson a basic alg

[dull ginkgo]

I am going off of Jacobson Basic Algebra I

I'm still on the group section

[dull ginkgo]

Actually you're right, Nil(R[X]) = Jac(R[X]) = Nil(R)[X}

The ideal structure of R[X] in general is interesting

And how it extends R's

I guess for each ideal J in R, J[X] is an ideal in R

[cobalt heath]

Ah, the actually more difficult part

[dull ginkgo]

mf on the group section partaking in ring theory (mf needs to stay in his lane)

Let $R$ be a commutative ring with unity, $R[X]$ be its ring of polynomials. Then every ideal, $J \subset R$, can be extended to the ideal $J[X] \subset R$. However, for any ideal $K \subset R[X]$, there's a restriction to an ideal of $R$, $K \cap R$. mf Galois connection

[cloud walrus]

A Fibrous Powder

[dull ginkgo]

Also that $(K \cap R)[X] \subseteq K$ and $J[X] \cap R = J$

[cloud walrus]

A Fibrous Powder

[dull ginkgo]

Now I ponder, when is $(K \cap R)[X] = K$

[cloud walrus]

A Fibrous Powder

[dull ginkgo]

K \cap R is the ideal of constant coefficients in K

[cobalt heath]

Think It is when K = J[X]

[dull ginkgo]

J[X] is the ideal of polynomials who's coefficients are all in J

so yeah, when K = J[X] for some ideal J in K

if K is entirely disjoint from R, then the only constant coefficients are 0

Therefore K = X*V for some other ideal V?

[cobalt heath]

https://math.stackexchange.com/questions/329236/jacobson-radical-equal-to-nilradical-in-rx
Oh, so it is actually equal

Seems jacobson is so often coincides with nilradical

[dull ginkgo]

But yeah lol

I wonder

Assume R is noetherian, then every sequence of ideals I_n must stabilize

R[X]/(X) is iso to R

so any sequence of ideals strictly within (X) must stabilize

So if we have a strictly increasing sequence of ideals in R[X], they must eventually intersect R

However it's not gaurunteed that if ideal A strictly lies in ideal B

[cobalt heath]

Vacuous truth

[dull ginkgo]

implies their intersections preserve the strict inclusion

[dull ginkgo]

?

[cobalt heath]

(R[X] does not have strictly increasing seq of ideals)

[dull ginkgo]

I am trying to think of another way to prove hilbert basis

[cobalt heath]

Ah

[tough raven]

Ideals within X =/= ideals of R[X]/(X).

[boreal inlet]

Uhm so I'm working with the primitive nth roots of unity over a field of characteristic 0 or a prime which doesn't divide n. The nth primitive root we picked (let's say zeta_n) now has a minimal polynomial. Then there are a few conditions we need to show to be equivalent.

1. degree of the minimal poly is Totient of n

2. the minimal polynomial is basically the cyclotomic polynomial, that is, all primitive roots of unity are roots of this

3. Aut(K(zeta_n)/K) is isomorphic to the group of units of the cyclic group of order n under the map [a] -> sigma^a

Where sigma^a takes zeta_n to (zeta_n)^a

What I was thinking is -

If sigma is an automorphism, then sigma(zeta_n) must be a root of the minimal polynomial also right?

[cobalt heath]

Indeed since the automorphism preserves K

You can exchange roots "without noticing" on the K-level

[boreal inlet]

So if I just show the polynomial is separable

I'm done

Because

The polynomial has at most phi(n) roots

If all roots are distinct

Then all the primitive roots must be a root of it

The group action already is well-defined

[cobalt heath]

Yeah, I think so

[boreal inlet]

Wait I'm having a hard time even convincing myself

https://cdn.discordapp.com/emojis/722022152328445985.png?quality=lossless&size=48

Uhm so

zeta_n is a root, so sigma(zeta_n) is a root and so is sigma^k(zeta_n) for all k.

If order of sigma is n, we are done

[cobalt heath]

The condition 3 says that sigma is of order n

[boreal inlet]

Because what I am trying to achieve is to cycle through all the roots

[boreal inlet]

Exactly, but currently I only have assumed condition 1

I have to prove condition 1 -> 2 -> 3

[cobalt heath]

Ahh

[boreal inlet]

Then back to 1

[cobalt heath]

For 1 -> 2, I think you have to use some argument by exhaustion

That is, like, totient of n is "big enough"

[boreal inlet]

For that if the minimal polynomial is separable we are definitely done

For char 0 we are done

If char p but p doesn't divide n

How do we say that

[cobalt heath]

Ah, non-separability