groups-rings-fields

[tough raven]

I believe in the polynomial long division you use to reduce the S-polynomials, whichever of f and g you choose to write down first gets precedence, i.e., if you start with (f, g) you'll prefer x ↦ -z and if you started with (g, f) you'll prefer x ↦ -y. Of course, S(f, g) = y-z (or maybe z-y), so you'll also add a reduction y ↦ z that ensures it doesn't matter in the end. I believe that regardless of the starting order, you will end up with a Grobner basis; I'm not sure if it's the same Grobner basis, but any ideal has (wrt a given monomial ordering) a unique reduced Grobner basis, so if you simplify your Grobner basis to a reduced one at the end, the result will be invariant of the order you wrote the original basis in.

[barren sierra]

yes you may or may not get a different Grobner basis

[elfin wraith]

Yeah nice one, I felt that this would be the case, my sticking point was wondering if it gives you the same basis, which I doubt, but of course as you mention that doesnt really matter

[barren sierra]

if you want some sort of uniqueness yea look into reduced Grobner bases

Cox, Little, and O'Shea \emph{Ideals, Varieties, and Algorithms} chapter 2 Section 7

in fact Lemma 3 just above deals with what you want in your question

[elfin wraith]

Ah yeah nice one. I got the feeling that was the case but I didnt have the book with me to double check thanks!

[barren sierra]

I tried to plug in $y = a_i + a_{i, 1}x$ in for $y$ and expand and see what I got.
The result is that
\[
  f(x, a_1 + a_{i, 1}x) \equiv \sum_{j = 0}^n c_{n - j}(x) \left( a_i^j + j a_{i, 1} a_i^{j - 1} x \right) \pmod{x^2}
\]
but then I'm not sure where to go from here.
I don't have differentiation on my side, except maybe in a formal sense.

[cloud walrus]

Spamakin🎷

Spamakin🎷

[hidden wind]

anyone know some nice short reads i can look to if i’m interested in like the geometry of rings, like, the eisenstein integers

[rotund aurora]

What would you like to know? The geometry of numbers is a topic of algebraic number theory, treated in pretty much every text on algebraic number theory

[wraith cargo]

what does the geometry of rings mean here?

[rotund aurora]

I mean in a way the geometry of the Eisenstein integers is kind of trivial, it's just a lattice. It is interesting if you add arithmetic to it. You can ask for patterns on the Eisenstein primes, for instance. This gets very difficult really quick, of course

[tawny dune]

I don’t know much about this topic but maybe Hatcher’s book on the topology of numbers https://pi.math.cornell.edu/~hatcher/TN/TNpage.html would be interesting

[hidden wind]

thanks i def should have a look into algebraic nt seems like great fun mwahaha

[rotund aurora]

Tbh I don't understand the title of that book, the treatment seems standard. But I haven't read it

[hidden wind]

i am not quite sure tbh but i mean just looking at it makes it seem quite clear to me there should be something interesting to say here

[tawny dune]

I wouldn’t know haha I’m just putting it out there

[rotund aurora]

Do you know this? https://arxiv.org/pdf/math/0501314

idk if something like that holds for the Eisenstein primes

[hidden wind]

this seems like a very cool result but the article itself look like it’d take me at least a year to make sense of

[rotund aurora]

Yeah, I would like to learn it. You should first start with Szémeredi's theorem, which is already very much non-trivial

[hidden wind]

i was thinking simple things like for example one can ask something like what is its isometry group?

which should be pretty trivial yeah but probably can lead to more interesting questions

[barren sierra]

Dumb question but what actually is the inverse of 1 + x in R? I've shown it exists as I know from prior knowledge that the maximal ideal in R is the set of functions which vanish at (0, 0). But I would like to physically write out and check what the inverse is.

[rotund aurora]

Do you know about crystallographic groups?

[wraith cargo]

There isn't one

it's literally 1/(1+x)

like

[hidden wind]

just barely and just the two-dimensional case

[wraith cargo]

localization just means you take some elements and formally make inverses for them

[barren sierra]

Oh bruh

Yea you're right I was way overthinking it

Yea

[hidden wind]

i guess it should be something like a semidirect product of C6 and Z? wait no those two overlap

Z and C3

guh

[rotund aurora]

it should be Z^2 also

[hidden wind]

won’t it be the ring itself semidirect product with its units

[rotund aurora]

I mean you can write an isometry as a translation followed by a rotation, no? So you can write it in matrix form

[hidden wind]

yeah

after a couple of brainfarts i state with some confidence it should be $\mathbb Z[\omega] \ltimes C_3$

[cloud walrus]

rødbet-jens

[hidden wind]

where of course $C_3 \cong (\mathbb Z[\omega])^\ast = {1, \omega, \omega^2}$

[cloud walrus]

rødbet-jens

[rotund aurora]

it's C_6

[hidden wind]

whaaa

[rocky cloak]

I mean, you can think of R as a subring of k[[x, y]] if you want.

In which case the inverse is
1 - x + x^2 - x^3 + ...

[rotund aurora]

yeah like this ring is Z[primitive sixth root of unity]

[hidden wind]

where does the extra square root come from?

[rotund aurora]

Say zeta is an nth root of unity with n odd. Then -zeta is a 2n-th root of unity

[hidden wind]

ah right

[barren sierra]

Right right that's where I got to but I wanted to truncate that somehow lol

[hidden wind]

right so i'm missing the +- here

and the left side should be C6

i forgot -1 is a unit in Z

[rocky cloak]

Isn't 1+w also a unit?

[hidden wind]

yeah but that should be something like -w^2

[rocky cloak]

Oh right

That's true

[hidden wind]

i find it kinda unbelievable i should say anything true rn but i’ll take your word for it

i should probably eat something

[dull ginkgo]

I really chose a poor wording for that ngl sorry, i meant the matrices that are only a 1 at some part of the diagonal and 0 elsewhere

I.e the endomorphisms sending an element to it’s nth basis vector

[tough raven]

Still doesn't sound right - for ℂ/ℝ for example, with the basis (1,i), you'll get all matrices of the form [[a, -b], [b, a]].

[dull ginkgo]

I’ll describe what i meant when i get home

[tough raven]

To make it more notationally compact, you can write f(x, a_i + a_i1 x) ≡ f(x, a_i) + ∂f/∂y(x, a_i) (a_i1 x) ≡ f_1(x, a_i) + ∂f/∂y(0, a_i) (a_i1 x) = f(0, a_i) + ( C + ∂f/∂y(0, a_i) a_i1 ) x (mod x^2), where ∂f/∂y is a formal derivative and f_1(x, y) discards all terms divisible by x^2; and C is a polynomial expression of a_i and coefficients of f - in particular, does not involve x or a_i1. Since f(0, a_i) = 0, the desired requirement is met iff C + ∂f/∂y(0, a_i) a_i1 = 0. This can be done because ∂f/∂y(0, a_i) is invertible because a_i is a simple root of f(0, y).

[rapid junco]

Does anyone know the book that lists as a exercsise that if G/Z(G) is cyclic then G is abelian?

[coral spindle]

A lot of books list this as an exercise, I expect. Are you looking for a particular book or are you just wanting help with the question?

[slim kayak]

There are texts which don't leave it as an exercise?

I just figured it was tribute to the algebra gods just like the yoneda lemma

[coral spindle]

I'm frankly not sure, it's so standard I would have to check

[toxic zephyr]

can you explain this? S->G is that mean a bijection from S to G? not exactly what the bijection would be for a*b=(a-m)(b-m)+m bc it's a function from GxG?
i think i'm stupid

[mighty kiln]

Take S = R\{-1}, G = R\{0}

And the bijection f(x) = x+1

[rotund aurora]

For nxnxn Rubik's cube, God's number is of the same order as n^2/log n https://link.springer.com/chapter/10.1007/978-3-642-23719-5_58

[toxic zephyr]

okay, but like... how do we use the bijection to induce a group structure? i had actually tried to do this last night, trying to find some map from R{0}, the real multiplicative group, to R{-1} but i wasn't able to figure out how to finagle it.

is it like $a*_S b=f^{-1}(f(a)*_Gf(b))$ or something?

[cloud walrus]

eigentaylor (STfFGMOaPID)

[glad osprey]

yep, that works it should be straightforward to verify that this satisfies the group axioms

[toxic zephyr]

oh wow really? i thought i would have messed it up. okay imma try that rn thank you!

i think i'm struggling with associativity actually. it's not clear to me why

$f^{-1}(f(a)_Gf^{-1}(f(b)_Gf(c)))=f^{-1}(f^{-1}(f(a)_Gf(b))_Gf(c))$

[cloud walrus]

eigentaylor (STfFGMOaPID)

[tough raven]

I think you're missing two f's.

$a * (b * c) = f^{-1}(f(a) *_G \mathbf{f(}f^{-1}(f(b) *_G f(c))\mathbf{)}), (a * b) * c = f^{-1}(\mathbf{f(} f^{-1} (f(a) *_G f(b)) \mathbf{)} *_G f(c))$

[cloud walrus]

Raghuram

[toxic zephyr]

you're totally right

we get $f^{-1}(f(a)_Gf(f^{-1}(f(b)_Gf(c))))=f^{-1}(f(a)_G(f(b)_Gf(c)))$ which works bc of the associativity of $G$

[cloud walrus]

eigentaylor (STfFGMOaPID)

[toxic zephyr]

okay great tysm @tough raven @glad osprey @mighty kiln ❤️
that was actually a really exercise so thank you especially Arki

[toxic zephyr]

curious about this. i had an exercise where we show that G=Aut(H) acts on H via a group action, and we had to look at the G orbits of elements of H for small groups.
for H=Z4, we get the orbits {0},{2},{1,3}. for H=Zp i believe we get {0},{1,2,...,p-1}
i'm pretty sure because Zp is a field, that's why we get only two distinct orbits (not sure about that though). but is there like... a significance of the orbits of Z4 being {0},{2},{1,3}? like iirc G orbits partition the group H, but like... is there a special property of what the particular partitions are when the acting group is Aut(H)?

[tough raven]

For cyclic groups, the orbits of the automorphism group should just correspond to the possible orders.

Also for arbitrary groups, Aut(H) contains all conjugations, so the orbits are at least as coarse as the conjugacy classes (but some conjugacy classes might become the same orbit under the full automorphism group).

For groups such that all automorphisms are conjugations, i.e., with no "outer automorphisms" (such as S_n for n ≥ 7), the orbits are exactly the conjugacy classes (so for S_n with n ≥ 7, each orbit is all permutations with a given cycle type).

For finite abelian groups, it should be possible but tedious to explicitly describe the orbits using the classification theorem (any such group is a product of unique cyclic groups ℤ/a1ℤ ⨯ ... ⨯ ℤ/akℤ, with 1 < a1 ∣ ... ∣ ak, or a unique product with all ai's being prime powers).

Beyond that, I'd guess that very little can really be said in general.

[toxic zephyr]

okay thank you! that is interesting

one more question: does the fact that a field has no nilpotent elements except 0 require proof beyond the justification that a field has no zero divisors? like i proved it as a lemma, but i feel like that's unnecessary.

i proved it as a lemma with induction. but like... do i really have to do that? lol

[tough raven]

IG you technically do need induction somewhere, to pinpoint the exact n such that x^n ≠ 0 but x^{n+1} = 0 so that you can apply the cancellative-ness of x.

But in practice, this is the kind of thing you prove once and then use instinctively forever, so exactly how you proved it isn't really important.

[dull ginkgo]

I actually remembered to respond.

We are considering End_R(B) as a right B-module, since B-scalings are nevertheless linear.

[tough raven]

Oh, sorry. I thought you were talking about a basis for End_B(B) over R instead of a basis for End_R(B) over End_B(B) = B^op.

[dull ginkgo]

yeah that's the mind bendy part about it

[tough raven]

I assume the module structure is given by (f⋅b)(x) := f(bx)?

[dull ginkgo]

ye

[glad osprey]

Do you really need induction for that? Can't you just use well-ordering to say there exists a smallest n s.t. x^n = 0?

[tough raven]

Same thing. 🤷

[glad osprey]

In the sense that well-ordering is the same as induction?

[tough raven]

Yes.

[tough raven]

This action is not well-defined if R is not central in B: (f⋅b)(r⋅x) = f(brx) is not necessarily equal to f(rbx) = r⋅(f⋅b)(x).

[glad osprey]

I would argue that explicitly using induction is unnecessary for that proof, unless your professor is really pedantic

[tough raven]

I think what works is to look at the subalgebra B^op = End_B(B) ⊆ End_R(B) = M_n(R^op) and observe that End_R(B) ≡ Hom_R(R^n, B) = Hom_R(R, B)^n ≡ B^n as left End_R(B)-modules and hence as End_B(B)-modules for any choice of R-basis of B, so that End_R(B) is free as a left End_B(B)-module, with the i^th basis element being the matrix with i^th row giving the coefficients of 1 (or your favourite unit of B) in the {e_i} basis and other rows 0.

[toxic zephyr]

how would you do the proof without it?
my thinking is that you need a^n=a a^(n-1)=0 implies a=0 or a^(n-1)=0. so if you can show a^m=0 implies a=0 for all 1<=m<n then that does it.
is there a simpler way?

or i guess that's more for an integral domain. there's probably a much faster way with multiplicative inverses.

or

idk

[glad osprey]

Assume a is nonzero and let n be the smallest integer s.t. a^n = 0, then a^(n-1) must be non-zero, so a a^(n-1) = 0 which is impossible

[dull ginkgo]

That’s what i meant to say originally, that’s what i did in the original observation i tried with ShiN

[tough raven]

Oh true, you can just note that (a^n)((a^{-1})^n) = 1 so that a^n is invertible, so it can't be 0.

Oh I see, that's why you took a basis consisting of units.

[dull ginkgo]

Are you taken to Eigentaylor

[tough raven]

No, the last message was to you.

[toxic zephyr]

oh lol that's even simpler

[tough raven]

If R is a ring (commutative if it helps) and M is a maximal ideal containing an ideal I, what further conditions are needed to guarantee that R/I is local iff I ⊇ M^n for some positive integer n?

I suppose (in the commutative case) it would be that for no prime ideal P other than M is M the unique maximal ideal containing P.

[tough raven]

Is a finitely generated Artinian module necessarily Noetherian? This is true over a commutative ring by https://math.stackexchange.com/a/2346330 (in short: it suffices to consider cyclic modules, i.e., R/I's; but I is two-sided, so we can switch from R-modules to R/I-modules, and Artinian ring ⇒ Noetherian ring).

[static mauve]

When tackling the below problem, I wasnt able to find groups that were Isomorphic to R_star, Q_star, C_star (all under multiplication) or Q under addition. Is that true? I feel like Im missing some isomorphisms here.

[abstract rock]

for R_star, what transformation would necessarily give you the other group on the real numbers (hint: how do you generalize multiplication as addition)
for Q_star perhaps you can find a similar arguement
for C* think about its relation to the real numbers

[next obsidian]

Damn, I was gonna prove it for commutative rings :(

[static mauve]

So I know that an ismorphism between <R, +> and <R+, mult> is e^x... So im trying to think of a function that does a bijective mapping between (-inf, inf) and (-inf, 0) U (0, inf)

[rocky cloak]

From Wikipedia:

but over noncommutative rings cyclic Artinian modules can have uncountable length as shown in the article of Hartley and summarized nicely in the Paul Cohn article dedicated to Hartley's memory.

https://doi.org/10.1112/S0024610797004912

The example in question is you take the noncommutative polynomial ring k<x1, x2, x3, ...> in countably many variables.

You define M to be the module generated by v1, v2, v3, ... with the relation
xi vj = v[j-1] (with v0 = 0)

Notice that all submodules of M are finite dimensional, so M is artinian not Noetherian.

Now adjoin an element u to M with
xi u = vi
and call this new module P.

Then P is cyclic, and P/M is one dimensional. So as P is an extension of M and P/M it is artinian.

[rocky cloak]

When comparing groups for isomorphism it can be smart to look at some invariants.
Do they have the same cardinality?
Do they have the same number of elements of a given order (how many elements of order 2 are there for example)?
Is the group divisible, i.e. can you for any x and n find y with y^n = x?

[coral steeple]

Thanks. Review is in order

[proven thunder]

Hey

if for all elements a, b of a group G, equations ax=b and ya=b have unique solutions in G, then G is a group?

I am supposed to prove this theorem.

The book I'm referring to has a comparatively lengthy solution

I solved it like this

We know that ax=b will always have unique solutions for x

Assume first solution to be e1, and the other solution to be e2

so ae1=ae2
By left cancellation law e1=e2

I proved the identity property of group like this but need to know if this method is correct as it seems very direct

[lone niche]

I don't think semigroups have cancellation law generally

[proven thunder]

Ah actually I made a typo

I have to prove it as a group

[lone niche]

I think the original question was correct. I recall someone asking a similar question last week.

Plus the new question is saying if G is a group satisfying... then G is a group.

[hidden wind]

my book states

The class of algebras is a natural extension of the class of rings

they’re talking about how every ring A has canonical structure as a Z-algebra given by mapping n to n1 where 1 is the unit in A

but in what sense is this an extension

?

[hidden wind]

this is pretty horrible notation so let me clarify… the homomorphism Z to A is given by $n \mapsto \sum_{k=1}^{|n|} \text{sgn}(n) 1_A$

[cloud walrus]

rødbet-jens

[hidden wind]

ok tbh that’s also horrible notation

[rocky cloak]

Just in the sense that a ring is a special case of a type of algebra. It's a Z-algebra.

[hidden wind]

right, so there will be more algebras than rings coming from this construction (would embedding be apropriate terminology here?)

[hidden wind]

i guess this works if sgn is a function from Z to {-1,1} in A

[rocky cloak]

Well, the map isn't necessarily injective, so maybe not an embedding.

Anyway, it depends a little bit. Not everyone defines algebras to be associative and unital, but even if you do you can consider algebras over rings that are not Z.

So that's how you'd get more algebras than just rings

[hidden wind]

ok thanks i think i’m following

but actually i should probabably rather be thinking about polynomial rings and ideals

[rocky cloak]

Polynomial rings are another good example of algebras

[hidden wind]

[elfin wraith]

Comalg exam took me 45 minutes out of 3 hours so either the easiest exam of my life or I’ve failed more drastically than ever before

[hidden wind]

i’m placing my money on the former mwahaha

[elfin wraith]

I am too, I feel incredibly good about everything I put down

Although it took me 3 passes to notice I claimed Z[x] was a PID lol

[rocky cloak]

Z is a PID, polynomial rings are PID, so surely Z[X] must be like a double PID

[slim kayak]

Z is my favourite field fr

[elfin wraith]

Unironically my thought process the first time around

Really just need to pray for the same energy going into non-com tomorrow (It absolutley will not happen)

[rocky cloak]

Those are some tight exam schedules

[elfin wraith]

Yeah 18 hours apart isn't too fun but it could be worst, my friend has diff geo friday morning, condesned matter physics friday afternoon and QFT Saturday morning

[rocky cloak]

Oof

[elfin wraith]

Condensed matter and diff geo are 80% exam and QFT is 100% so yeah ive seen happier men lol

[tardy hedge]

the (2,x) example not being principal is pretty much burned in my mind now, lol

[barren sierra]

Stuck on part b.
I've shown that
\[
  f(x, y_{i}^{(\ell + 1)}(x)) \equiv f(x, y_i^{(\ell)}(x) + (a_{i, \ell + 1} x^{\ell + 1}) \frac{\partial f}{\partial y}(x, y_{i}^{(\ell + 1)}(x)) \pmod{x^{\ell + 2}}.
\]
I know that I could isolate a term $C$ which is a polynomial in the $a_{i, j}$ and coeffients of $f$ from $(a_{i, \ell + 1} x^{\ell + 1}) \frac{\partial f}{\partial y}(x, y_{i}^{(\ell + 1)}(x))$ so my only real confusion is dealing with the first term.
I'm not sure how I can use the fact that $f(x, y_i^{(\ell)}(x) \equiv 0 \pmod{x^{\ell + 1}}$ to continue.

[cloud walrus]

Spamakin🎷

[fallen geyser]

What are the eigenvalues of matrices in SL_n(C)?

[barren sierra]

Well you're in SL_n(C) so the determinant is 1. What does that tell you about the eigenvalues?

[fallen geyser]

I know the product of the eigenvalues is then 1, but can more be said?

[barren sierra]

I mean specifically the eigenvalues all lie in one very specific region

But no I don't think you can say more

[fallen geyser]

Which region is that?

[barren sierra]

Well think about this

[wraith cargo]

I also have no idea what you mean lol

[rocky cloak]

They're nonzero I guess...

That is a region at least

[wraith cargo]

Yeah IG that's a fair conclusion lol

[barren sierra]

They have to lie on the unit circle no?

[wraith cargo]

No?

You can have x and 1/x

[barren sierra]

I thought they did that's what I'm getting at

Hmmm am I conflating this with something else

[rocky cloak]

Maybe you're thinking of SU(n)

[barren sierra]

Ah yea

My bad 💀

[barren sierra]

Anyone got any ideas on this 💀

I'm not sure what I can say about things mod x^l+2 knowing what they are mod x^l+1

This wasn't an issue in part a where I used the fact that f(0, a_i) = 0 because since it's 0 in k, it's zero mod x^2 because of quotient maps mapping 0 to 0

[fallen geyser]

Over any field, can you write V=H+N where H is a hyperplane?

[barren sierra]

What is V

[fallen geyser]

srry, a vector space

[cloud walrus]

Spamakin🎷

[fallen geyser]

Yes and yes

[barren sierra]

ok so then my next question is given what you know about the basis of a vector space, what do you think the answer is? Also this belongs in #linear-algebra not this channel so respond there and ping me

[hidden wind]

oh i am working my way through this one right now

well the problem as given to me is stated as

show the set I = {f in Z[x] : f(0) is even} is a nonprincipal ideal in Z[x]

so far i've only shown that I = (2,x)

bit stuck on how to show the ideal is not principal tbh

something something relation not possible

oh

nvm i had it i'm just stupid and forgot we're in the integers

[runic nimbus]

I can see all them implications from each definition, but how do I conclude this proof

[rocky cloak]

So
[G:H] = [G:KH] * [KH:H]
[KH:H] = [K:HnK] divides |K|, but it also divides [G:H].

So we must have [K:HnK] = 1.

[runic nimbus]

i see so if K is normal in G and H is any subgroup, them KH is a subgroup

[rocky cloak]

That's right

[runic nimbus]

Does K = H∩K

Therefore, K < H

[rocky cloak]

That's how the proof goes yeah

[toxic zephyr]

so i was tasked with finding GL2(Z2), and i did the painful determinant casework, and got 6 matrices. and then JUST NOW realized that there might be a much easier intuitive way to go about it. is this logic using possible linearly independent subsets the proper or better way to go about it? like can i just do this and use the fact that a matrix is invertible iff the columns form a linearly independent set?

[coral spindle]

That's right, and this can also be used to find a formula for |GL_n(F_q)| if you generalise appropriately

[toxic zephyr]

welp... i guess that can greatly shorten my proof
yeah i could have done an exercise where you do this for GL2(Z3) but i didn't wanna do the casework. i guess this way makes it much more managable.

[coral spindle]

It's still a bit of a pain to write down all the matrices, but yeah it does make it conceptually easier

[toxic zephyr]

yeah at least it's 2d vectors. so you just need to avoid scalar multiples

2 * 2 + 2 * 2 * 2 + 2 * 2 * 2 = 4+8+8=20
is that how many there are for GL2(Z3)?

[coral spindle]

Let me just calculate this on my own quickly

I think you've undercounted somewhat

[toxic zephyr]

oof yeah i forgot to double it for all possible column orders.
and it seems i also missed 4 more

[coral spindle]

Still a bit too few I'm afraid

The order of GL_2(F_3) should be ||8 x 6 = 48||

Oh wait yes that would work out I think?

Order of operations...

[toxic zephyr]

oh my other mistake i think was i didn't count any matrices with all nonzero entries. of which i'm sure there are a few

i was just trying to use the linearly independent sets i calculated for z2, and just multiply by 2 for all possible nonzero entries

rip

[coral spindle]

Here's a trick for making this calculation a bit easier.

You're really just counting ordered bases.

So how many choices are there for the first entry? You can choose anything but the zero vector, so there are 3^2 - 1 choices.

[rocky cloak]

So the first column can be anything nonzero, so that's 3^2 - 1 choices. Then the other column must just not be a multiple of that.

[coral spindle]

Then how many choices are there for the second entry? You just choose a vector outside of the span of the previous one, which is a 1d subspace. So there are 3^2 - 3 options

It should be clear how to make this into a general formula

[toxic zephyr]

yesss that makes so much more sense

that's a much smarter way to think about it

and that actually looks more familiar too

[rocky cloak]

And the with some generalizing you get
||(q^n - 1)(q^n - q)(q^n - q^2) ...(q^n - q^n-1)||

[toxic zephyr]

yeah yeah

nice

that's actually really cool

[elfin wraith]

While theres some discussion of couting stuff (something im horrifically bad at) there was a problem in my non com class a few weeks ago that im guessing has a nicer proof than the nonsense I did,

[rocky cloak]

What was the problem?

[elfin wraith]

"Let $S_s = k[x_1,\ldots,x_d]_{\leq s} = {f\in k[x_1,\ldots,x_d] | \deg f \leq s}$ and let $p_d\in \mathbb{Q}[s]$ be the unique polynomial such that $p_d(s) = \dim S_s$. Find a formula for $p_d(s)$ in two ways and prove it is correct."

[cloud walrus]

Nope

[elfin wraith]

So the first and realativley easy way is with the discrete derivative and induction, but then I tried a straight up counting argument for the second method, but the product of a sum im left with is so utterly unwieldy that im unsure if its useful

Ill go find my notes to see what I ended up with, but ive got no clue how youre ever supposed to see this horrific sum is actually just $\binom{d+s}{s}$

[cloud walrus]

Nope

[rocky cloak]

So it's the stars and bars thing right?

[vocal pebble]

okay so if |G| = 56, we'd have subgroups of order 2, and any group of order 2 isnt normal, so G cant be simple?

[elfin wraith]

Well my ipad is dead but its like $\prod_{j=1}^d\left(s - \left(\sum_{i=1}^d a_i\right) + 1\right)$ or something iirc

[rocky cloak]

Like if you think about the monomials in your ring, you just pick s elements from x1, ..., xd with repetition and multiply them together

[south patrol]

Not really sure what you mean

[rocky cloak]

And the formula for selection is (d+s choose s) per the stars and bars theorem

[south patrol]

The subgroups not being normal is the opposite of what you want

[cloud walrus]

Nope

[vocal pebble]

sorry i mean every group of order 2 is normal

[south patrol]

But it's also possible to have normal subgroups of order 2

[vocal pebble]

or wait no

this is a subgroup

[abstract rock]

probably bestnto start with sylow

[vocal pebble]

im dumb

[south patrol]

Being normal is a property of subgroups, not of groups absolutely

Yeah Sylow is probably best here

[elfin wraith]

See this may be realted to the thing about me being horrifc at counting arguments, im not sure I know about stars and bars

[rocky cloak]

Do you mean index 2 perhaps?

[vocal pebble]

indeed, i knew i was missing something simple

[south patrol]

Oh yeah btw jagr, I sorted out that Witt vectors thing, if you're interested lol

[rocky cloak]

https://en.m.wikipedia.org/wiki/Stars_and_bars_(combinatorics)

[elfin wraith]

My thinking was you have s choices for a_1, s-a_1 + 1 for a_2 etc and you multiply all of those, if theres a theorem that says thats the binomial them sick lol

[rocky cloak]

Yeah, I think you sendt a link, but I didn't read it.

It was true?

[south patrol]

oh i sent a link for the char 0 case

if A is artinian local with char 0 residue field k, then yes, there's a (ring-theoretic) section of A -> k

Uhh I didn't actually look into how this was constructed lol

But anyway, for a perfect char p field k it becomes true that A is an algebra over W(k)

There the idea is that you get a map W(A) -> W(k) by functoriality and this now admits a section

which i guess gives you a map W(k) -> W(A) -> A by composing etc

[elfin wraith]

Ah, yes this is nice and almost certainly something I shouldve learned by this point in my time doing maths

[south patrol]

here the idea is that given x in W(k) write x = F^n(y) where F is the frobenius on W(k),and then consider any preimage y' of y in W(R) and take F^n(y')

for large enough n this becomes a well defined ring homomorphism W(k) -> W(A) since the maximal ideal of A is a nilpotent ideal

[rocky cloak]

Hmmm

[south patrol]

Though tbf I don't see additivity without further thought

lol

Oh wait yeah this being a ring map is just cause the Frobenius is a ring map lol, once you show that this is all choice-independent

I'm not sure what happens with the char 0 situation though, but it should be easier

apparently this is basically Hensel's lemma, where the characteristic 0 probably helps derivatives not vanish

[rocky cloak]

So if k has characteristic 0, then at least Q embeds in A, that's clear.

Then I guess you can take a maximal subfield K of A. Then it's image will be a subfield of k. Say it's not everything, and pick a preimage x of an element not in the image.

Then there is a minimal polynomial such that p(x) is in the maximal ideal. Then you use Hensel's lemma to prove that p has a root mod m^n for all n. In particular when m^n = 0.

That seems like it should work.

[south patrol]

oh very nice

The issue is that the situation I was in is more general than this and now i don't see how it generalises lol

[vocal pebble]

so this should work, im posting here to confirm because it looks like im having brainfarts:
Number of 7-subgroups (say x) must divide 8, so x <=8, and x must be 1 mod 7. So, the only possibilities are: there are 1 7-subgroup(which means we are done) or eight 7-subgroups. If there are 8 7-subgroups, then consider number of subgroups in Syl2(G) ( call it y for now). y must be 1 mod 2(odd) and divide 7, so y<=7. So the only possibilities are y = 1(again done) or y=7. I want to claim x=8 and y=7 cannot happen together since then we will have 8(7-1)+7(8-1)>56 elements

[rocky cloak]

You can't quite count 7(8-1) as the 2-sylow subgroups can intersect in interesting ways.

[vocal pebble]

You are right i was sus when i was typing that. Even in the proof there is no gurantee they are free of intersection

[rocky cloak]

But you can make a similar counting argument

[elfin wraith]

Oh I had that as a homework question at the start of the semester, proof that I can do some counting arguments

[vocal pebble]

Ah wait the 7 subgroups intersect trivially so there can only be one 8 sub group lol (because 56-48=8)

I literally had that in my mind for the last few mins

Only the 8 subgroups can intersect non trivially and properly because they can intersect at elements of order 4 or 2...

This carries over generally for order pq^3 because pq^3-q^3(p-1) = q^3 and there is only room for 1 q-subgroup

Nice

[south patrol]

Nice

[south patrol]

I feel morally obligated to point out (for cultural reasons) that every group of order p^a q^b is solvable (https://en.wikipedia.org/wiki/Burnside's_theorem), which implies the result you want is true even more generally

[tough raven]

Something seems off about a comm alg exam followed by a noncomm alg exam...

[tough raven]

Wait, didn't I answer part (a) somewhere above?

Oh this is part (b)

[charred iris]

I'm too tired to make a smart joke about exams commuting rn someone else do it

[tough raven]

Well, $f(x, y_i^{(\ell)}(x)) \equiv 0 \pmod{x^{l+1}} \implies f(x, y_i^{(\ell)}(x)) \equiv D x^{l+1} \pmod{x^{l+2}}$ for some $D$ which is a polynomial in the coefficients of $f$ and $y_i^{(\ell)}$. So you get $(\text{something}) x^{l+1} \pmod{x^{l+2}}$ which is the same as $\text{something} = 0$ (or at least $\equiv 0 \pmod{x}$).

[cloud walrus]

Raghuram

[tough raven]

I guess the second exam must be take-home, since it doesn't involve commuting...

[charred iris]

Outstanding.

[elfin wraith]

I would give my life for this exam to be take home

Instead it’s 1.5 hours to do 3 problems so either all the problems are trivial, I’ve done them before as homework, or I’m just unbelievably fucked

[late marsh]

i'm stuck on this problem i know to use burnside's lemma but it feels so daunting here

Exercise 12. Consider the colorings of vertices of a regular dodecahedron with 8 colors such that a face cannot have more than 3 vertices of the same color.

(a) How many such colorings exist?

(b) Consider two colorings equivalent if one can be obtained from the other through an action of an element of the group of symmetries of a regular dodecahedron. How many unique colorings exist?

(Hint: This group is isomorphic to the alternating group on 5 elements)

[late marsh]

this question was on my exam today lol

[noble belfry]

How do I show this

It’s trivial to show that if f(x) is irreducible then f(x)/a_n is irreducible

But I don’t think that’s what they mean

[abstract rock]

i mean a sketch will do you plenty i think, how far along have you gotten

[late marsh]

i feel like i could probably do (b) given the answer to (a)

ive done this type of problem with different numbers/groups but here i just have no clue how to count it for (a)

[abstract rock]

well start with one vertex colored, and walknyour way out remembering the constraint

you could also think about each face and mark the edges where they meet, see if that helps you

[late marsh]

it's not just a simple counting problem where i can do one vertex at a time independently because of the shared vertices between faces, even considering the constraint though is the problem

because then i have to consider so many different cases depending on which vertex is which color when i reach another face and that would be completely different depending on how the previous ones were chosen, so i can't just multiply by the choices for the next vertex

like try to think through it and youll see, i didn't just send this problem without trying anything

[abstract rock]

idk to me it sounds like 8 choose 5 or (8 choose 4 and 5 choose 2) for opposite faces, that leaves the necklace in the middle which i'll leave you to count

[tardy hedge]

Woah its nastydasty

Been a while

[abstract rock]

hey man, how ya been

[tardy hedge]

Im doing pretty alright thanks!

[late marsh]

i'm unsure why it would be 8 choose 5 for each of the opposite faces. wouldn't it be 8*8*8*7*6 for each?

then the necklace in the middle is still highly dependent on what those choices are

i haven't really had much experience with combinatorics/counting problems so i might be misinterpreting it

[abstract rock]

nonono, just fix a pair of faces, each one you can pick either a configuration with no common colors, a pair of bertices and oh sorry there is still 3 as well since it says no more than and not excluding three

[late marsh]

what?

[abstract rock]

if you fix a pair of opposite faces, that gives you two faces where you can count freely

[late marsh]

yes

(8*7*6*5*4) for just the no common case is already way more than 8 choose 5 though im confused

[abstract rock]

so if you count them as either having all unique colors, one or two pair of common colors, or one set of triplet common colors (sorry forgot to count the extra pairs) that gives you the count for each face

[late marsh]

so (8*7*6*5*4) (all unique) + (8*7*6*5) (one pair) + (8*7*6) (one triplet)?

for a single face

[abstract rock]

you'll need the addition of two pairs of different colors as well but yes that should be good since the permutation matters, sorry for my bad habits of combinatorial notation

[late marsh]

oh right yeah

that part makes sense because they're independent then but the necklace part just seems like i'd have to consider any case of the top/bottom faces being two pairs, one pair, unique, or triplet but it would probably depend on the permutation of them as well

as well as the rest of the necklace

[abstract rock]

so now that you have two faces considered, you can now look to the first triplet of points in the necklace, they correspond to three faces and you can count the cases discriminately amongst those three faces

[late marsh]

i'll give it a shot later

[abstract rock]

since they will have an opposite as well

oki

i hope this was helpful

im no combinatorist

[late marsh]

i think so it just feels like im going to have to do a lot of casework

[abstract rock]

yeah, thats jow they keep undergrads busy after all

[late marsh]

it isnt even a combinatorics class

[late marsh]

half of the class is grad students 😭 but they give them different problems on some assignments

[abstract rock]

sounds like a familiar setup to me

[late marsh]

like "undergrads do #1 and 3 others, grad students do #2 and 3 others"

[abstract rock]

they had us undergrads do the grad questions for bonus marks

[late marsh]

i would just get 0 for it

[abstract rock]

i liked when we had to prove how cantor bernstein fails for free groups

[abstract rock]

oh that stinks, they should acknowledge the effort if they're grading it anyway smh

[late marsh]

like if i did the #1 and 3 others and then also #2 it wouldnt be anything extra for #2

i usually did #1 and #2 because then the professor only grades those 2 and doesnt look at the others except for completion

and the first 2 seemed easier than the other problems

id do 2 of the others after 1 and 2

but one of the other 3 was usually too hard

this problem is the one i didnt do on the last hw

#2 for comparison was the same problem but for a triangle, 4 colors, D_3 and colorings with only 1 color not allowed

[abstract rock]

it seems more combinatorial in nature but i guess most finite group theory abstracted away from rep theory or dynamical systems do be that way

if i was to teach i'd prefer edging my students closer and closer into the lunacy of extensions but im biased what could i say

[trail cave]

This question just asks "Find all Abelian groups of orders 9, 16, 27, 32". Where do I start?

[abstract rock]

the fundamental theorem of finitely generated abelian groups probably

[trail cave]

ty

[trail cave]

So I have, for those of order 9, that if G is such a group then |G| = 3^2. So there's a sylow 3-subgroup which is unique, and hence G is isomorphic either to Z mod 9 or to Z mod 3 times Z mod 3. Hence there are two such abelian groups. Does that seem correct?

[abstract rock]

well your middle point about the sylow group should be clarified but otherwise yes there are two

[trail cave]

I got that from equivalency of the p-subgroup being unique with it being normal (and it's normal since the group is abelian)

thank you

[abstract rock]

yeah rhats why it doesnt really matter?

all subgroups of abelian groups are normal

[noble belfry]

Hey guys

Can someone explain like

Field extensions to me

And algebraic / transcendental ones

[abstract rock]

how much you know now?

[noble belfry]

Not very much about them

I know there’s always an extension field E of F in which f(x) has a zero

[abstract rock]

have you played with polynomial rings of fields yet?

[noble belfry]

Which is F[x]/<p(x)>

Where p is an irreducible factor

[noble belfry]

Yeah

[abstract rock]

well p(x) has to be an irreducible polynomial for the extension to just be well behaved

[noble belfry]

F[x]/<p(x)> is a field iff irreducible iirc

[abstract rock]

if it isnt irreducible its like when you adjoin an element already in your field

yes exactly

[noble belfry]

Cause <p(x)> is maximal if it’s irreducible

[tough raven]

Well, kind of. If a in F then adjoining a itself looks like F[X]/(X - a), and X-a is irreducible, so this is a field extension, it's just a boring one.

[noble belfry]

When we say F(a)

[tough raven]

I'd say taking F[X]/(f(x)) for f not irreducible is like refusing to make up your mind about which element you're adjoining.

[noble belfry]

Is that the smallest extension field that contains a and F

[rapid junco]

how does one go about this?

[noble belfry]

Or nah

[rapid junco]

this is what i have

how does one go further?

[tough raven]

Consider what [ℚ(a) : ℚ] can be.

[rapid junco]

Yeah I dont know how to deduce that though

oh wait it must divide 3

[noble belfry]

Wait guys

So pi isn’t transcendental over R right

But it’s transcendental over Q?

Are there any numbers that are transcendental over R

[delicate bloom]

x

[trail cave]

I'm having trouble with 5 here and questioning my solutions to 12 and 16. For 5, I know H, K are normal in G and that I can build isomorphisms from H to Hx{1_K} and likewise for K. But from there I don't know how to proceed.
For 12 I think it's sufficient to say that, since there's only one Sylow p-subgroup (of order 2), it's impossible. For 16 I think only the last two are isomorphic (same order), and that all are commutative since they're all cyclic. Is this correct?

Sorry for the big post I have an exam tomorrow and didn't want to spam them separately

[tight axle]

Have you tried the First Isomorhpism Theorem for #5?

[trail cave]

Yes but I don't know how to ensure that K, H are kernels in each case

I thought about using Caylays Theorem but I'm not sure its useful

@tight axle forgot to reply

[tight axle]

Consider the map H \times K -> H defined by (h,k) -> h. The kernel is then {(e_H, k) : k \in K} = {e_H} \times K, correct?

[trail cave]

Ok I see thank you

[tight axle]

Of course

[cobalt heath]

So say,
0 -> A -> B -> P -> 0 is exact
and P is projective.
Why is this sequence splitting, and how does one memorize this fact

[rocky cloak]

Well, the way to see it depends on how you define projective. But for example if you apply Hom(P, -) you get an exact sequence
Hom(P, B) -> Hom(P, P) -> 0
in particular there is a map P->B that maps to the identity, i.e. the composition P->B->P is the identity.

A way to remember is that projective modules are the ones that map onto things, but they can never be mapped onto themselves.

[cobalt heath]

Oh that's neat! Thanks

Interesting, writing Hom and the exact sequence makes it easier to think

I was trying to pull back a map by surjection, which my head refused to see

[vocal pebble]

Thank you! I had actually suspected this before going to sleep because the same proof also carries over to pq^n, so i was wondering how general it could be. Very interesting.!!

[elfin wraith]

Non com alg exam was actually extremely fair

I think I’ve actually done reasonably well, one of the questions was to show that the first Weyl algebra was simple, but I couldn’t exactly remember the proof, but I did remember proving the nth Weyl algebra is simple so just went over kill on it

My main concern now is that it’ll get scaled down which is always rough

[south patrol]

:)

[cloud lynx]

does every algebraically closed with char p=0 have the imaginary unit i?

[south patrol]

Well, what do you mean by i

[cloud lynx]

i=sqrt(-1)

[south patrol]

The equation x^2 + 1 = 0 has a root, so you can pick one of those roots

[cloud lynx]

I want to show that the circle and the hyperbola are isomorphic for k algebraically closed field with char p != 2

but its kinda hard

[rotund aurora]

They are also isomorphic for char p=2

[cloud lynx]

I know but its easy to show if its char p= 2 thats why my prof wrote char != 2

[cloud lynx]

so is this a yes?

[south patrol]

Well it's more that I wouldn't say "the imaginary unit"

But yes, there is a square root of -1

[cloud lynx]

ok now I have to think about char p != 2, 0

[rotund aurora]

x^2+1=0 has solutions in any alg closed field, regardless of characteristic

[cloud lynx]

ah okk I forgot the definition of algebraic closed field

true ups

tyy

[bitter rover]

An algebraically closed field is defined as a field where every non-constant polynomial has a root.

So by definition x^2 + 1 has a root over any algebraically closed field. There's nothing to prove. But you could ask other questions:

1. Is every field (isomorphic to) a subfield of some algebraically closed field?
2. Does every algebraically closed field have a subfield that isn't algebraically closed?

[hidden wind]

i have no personal relation to spectra, but i hate this notation

so ugly

[rotund aurora]

What do you suggest instead?

[south patrol]

Using the prime spectrum instead

[hidden wind]

i have nothing constructive to say i’m just bitching about a minor detail in what seems like an otherwise wonderful book

[rustic rapids]

I am a bit stuck with this. I know that ac are coprime. And i want to claim that somehow these 4 matrices are performing a form of euclidiean algothirm when row reducing. Then, we would eventually get RREF and wed be done. But im not entirely sure how to rigorously show it

[abstract rock]

whats the book?

[hidden wind]

smith, kahanpää, kekäläinen and traves’ an invitation to algebraic geometry

[tardy hedge]

ooh, an invitation, you say?

Am I invited too?

[rocky cloak]

So a change of coordinates turning
x^2 + y^2 = 1
into
x^2 - y^2 = 1
?

Then just replace y by iy, where i is a square root of -1.

[coral steeple]

Can someone let me know if this is not okay?

[abstract rock]

thanks

[cloud lynx]

Yes I did it thats why I asked about i= srqt(-1)

[lusty marlin]

This is correct

[coral steeple]

Thanks!

[hidden wind]

yes

you just need to but this way too expensive book

[abstract rock]

online library here i come

yarhar

[opaque finch]

I need a help in a problem

Let $L/K$ be a Galois extension and $\alpha \in L$ belongs to a normal basis of $L$ over $K$. Show that $L = K(\alpha)$.

[cloud walrus]

mycroftholmes1703

[opaque finch]

I understand the fact that there exists a $\beta$ such that $g_1(\beta),\dots,g_n(\beta)$ is a normal basis where $\text{Gal}(L:K) = {g_1,g_2,\dots,g_n}$. Then we have $g_i(\beta) = \alpha$ for some $i$. I just have to show that all over $g_j(\beta)$'s are just some powers of $\alpha$.

[cloud walrus]

mycroftholmes1703

[opaque finch]

I need help in that part

[abstract rock]

sounds like you need to show that basis is cyclic

[opaque finch]

exactly

[abstract rock]

what are some properties you know about cyclic groups?

[opaque finch]

they are generated by a single element

what properties are you referring exactly ?

@abstract rock

[rocky cloak]

Think about what the minimal polynomial of alpha could be.

How many conjugates does alpha have? What does that tell you about the minimal polynomial?

[opaque finch]

So at the top of my mind is that $g_j(\beta)$'s are basically the conjugates of $\alpha$. Then minimal polynomial is $m_{\alpha}(x) = \prod_{i = 1}^{n}(x - g_i(\beta))$. Then we have that $m_{\alpha}(x)$ is seperable. Then by primitive element theorem we get the result.

[cloud walrus]

mycroftholmes1703

[opaque finch]

Is this what you are trying to say @rocky cloak ?

[rocky cloak]

So I'm not sure what beta is, and you shouldn't really bring in the primitive element theorem for anything here.

Just think about the definition of normal basis. It's that the conjugates of alpha form a basis. So what does that say about the degree of the minimal polynomial?

[opaque finch]

is it of degree n ?

[rocky cloak]

If n is the degree of L/K, then yes

[opaque finch]

yes n is the degree of L/K

[rocky cloak]

A basis for an n-dimensional space must have n vectors

[opaque finch]

yes

[rocky cloak]

So then the minimal polynomial of alpha is at least of degree n.

So what does that say about K(alpha)?

[opaque finch]

So then $K(\alpha)$ is a degree $n$ extension over $K$.

[cloud walrus]

mycroftholmes1703

[opaque finch]

But $L$ is also a degree $n$ extension over $K$.

[cloud walrus]

mycroftholmes1703

[rocky cloak]

Yes, very good

So then you see the finish line?

[opaque finch]

Yes but is that even true ? That two degree $n$ Galois extensions are same ?

[cloud walrus]

mycroftholmes1703

[opaque finch]

Like is there always a unique Galois extension for a field of a specific degree ?

@rocky cloak

[rocky cloak]

No, but these aren't just two random extensions. They are related.

Is there for example an element they have in common?

[opaque finch]

ofc

$\alpha$

[cloud walrus]

mycroftholmes1703

[opaque finch]

is common to both extensions

[rocky cloak]

Right, so if alpha is in L, what can you then say about K(alpha)?

[opaque finch]

ahhh I am so stupid

yes now I get it finally

$\alpha \in L \implies K(\alpha) \subset L$ but both are degree $n$ and hence they must be the same

[cloud walrus]

mycroftholmes1703

[opaque finch]

I got so messed up

[azure wadi]

Under which math topic group, ring, fields comes under?

[abstract rock]

Abstract Algebra but the former two appear in multiple places, the latter is the basis for linear algebras

[azure wadi]

Fields come first then linear algebra?

[abstract rock]

more like you learn that cartesian products of fields with a scalar operation form vector spaces and alot of the consequences of that fact

[azure wadi]

Abstract Algebra is new to me

Never heard about it

I know it sounds bad

[coral spindle]

I would say that linear algebra occurs in multiple places too :)

[rocky cloak]

It's usually encountered in line 2nd or 3rd year of university. Commonly also just called "algebra"

(though non-math people call other stuff algebra, so the name "abstract algebra" was coined)

[azure wadi]

So it all comes under algebra? I thought calculus is algebra

[rocky cloak]

See this

[azure wadi]

I recently googled on Cryptography and apparently Groups Rings Field is very much involved in it

[rocky cloak]

Indeed it is

[abstract rock]

Question was just bout the topics of this channel, linear algebra would be out of scope for this conversation

[rapid junco]

If K is a Galois extension of a field F and H is a subgroup of Gal(K/F), show that there
is an element α in K such that H = {σ ∈ Gal(K/F) : σ(α) = α}, i.e. such that H is the
stabilizer of α

can I get an idea here?

just where should i choose alpha from

[rocky cloak]

Do you know the primitive element theorem?

[rapid junco]

yeah but K might be Galois but nessecarily finite

[rocky cloak]

But then it's not true right. Like just take K to be the algebraic closure of F and take H to be the trivial group.

[rapid junco]

is a subextension of a primitive extension primitive?

[rocky cloak]

And extension is primitive if and only if there are only finitely many intermediate subextensions. So yes

[rapid junco]

so the idea is that K^H is going to be primitive then

pretty much

[rocky cloak]

Yeah, by the Galois correspondence H should correspond to the field F(alpha)

[rapid junco]

wait im confused now

is this not what I want to show

also I think the last line H = Gal(K/K^H) is wrong right

[rocky cloak]

It says "finite Galois extension"

[rapid junco]

yeah

[rocky cloak]

So then there's no issue with the infinite stuff. You just pick alpha so K^H = F(alpha)

[rocky cloak]

No that's correct.

That's the Galois correspondence

H = Gal(K/K^H) and
E = K^Gal(K/E)

The line before that is fishy though. You should think about K^H/F not K/K^H

[rapid junco]

okay

but how do I use galois correspondence to recover H

[rocky cloak]

What do you mean?

[rapid junco]

hold up I think i got it

can you lmk what your thoughts are?

[rapid junco]

Show that every finite group is the Galois group of some polynomial over some field.

Again, little stuck on this.

I want to embed G into S_n, but from there I have no clue how to recover a polynomial.

[vocal pebble]

Give an example of a non abelian group such that every element in it has order 3?

The existence of such a group has an interesting implication: if we start with the binary string 01 and we are only allowed to insert and delete binary strings of the form XXX where X is a binary string, then the string 10 cannot be reached..

It turns out that the string 10 cannot be reached from 01, via a complstely different proof, so im womdering what exactly such a group would looks like

[coral spindle]

If you're asking for a hint on finding one: think about certain groups of matrices over the field with three elements and you will find one

If you just want a name, it's called the Heisenberg group

[vocal pebble]

Wait is this some well known problem?

[coral spindle]

I'd say it's relatively well known, yes

[sonic coral]

Lie groups go brrr

[vocal pebble]

I just saw it, how could i ever come up with this group

I mean looking at what it does to bases it seems obvious in hindsight...

[coral spindle]

It has to be of order at least 9, and it's not too hard to construct it as a semidirect product of C_3 x C_3 and C_3 after that, but no it's not obvious.

If someone were really determined they could find it

[vocal pebble]

Okay, il have to do some reading since idk what semidirect products are. Actually this was a question i found on a math competition and the proof that they provided used some bizzare invariant

[rapid junco]

bump, this is what i have but i feel its wrong

[alpine island]

This looks good to me.

[rotund aurora]

It should be L/L^G

[tough raven]

Must a von Neumann regular ring have trivial Jacobson radical?

[rocky cloak]

This is one of the listed "facts" on Wikipedia at least

[tough raven]

Huh. I really should have checked that first. Thanks.

[chilly ocean]

Integral domain have unity or not

[rocky cloak]

All rings have unity.

There solved it for you

[chilly ocean]

My book doesn't say ring must have unity in defination

Lol

[cobalt heath]

Thinking about it, a non-associative algebra without unit just a vectorspace with canonical bilinear map?

[elfin wraith]

There are some people who say rings don’t require unity but that’s weird (weird as in definitely not the usual choice)

People tend to refer to rings without unity as rngs these days

[wooden rain]

Random number generators

[abstract rock]

should be rings mod i

[south rain]

Is sending full proofs (medium to short proofs), for someone to check, if anyone has time of course, allowed here, or frowned upon?

[abstract rock]

No not a problem at all as long as its ontopic

just make sure its in a pdf if its big enough

[south rain]

I hope this image suffices.

[rapid junco]

What is rhe difference between a radical extension and a purely inseperable extension?

[south patrol]

They are very different objects - what makes you think they're the same/similar

[rapid junco]

Radical extension is one that the primitive element has some poee that is in the base field right?

But a purely inseperable extension is one such that for every a in K a to the p power is in F for some p power

[rotund aurora]

This tells you that every purely inseparable extension is a radical extension, but not all radical extensions are purely inseparable

A radical extension is an extension generated by elements x such that for some k x^k is in the base field. So it need not be finite

[south patrol]

This is a property of them but not a definition

[rapid junco]

Ah I see

[white oxide]

Can somebody help me see why the first highlighted sentence is true?

[placid shale]

Artin's Abstract Algebra reading group

[crystal vale]

I am missing something here, I don't get it why do we need f(R) in the centre of A ?

When I verify the axiom for R-module structure of A, I didn't use this property

So it makes A to R-module in both left and right R-module structure

[crystal vale]

Let M be a finite Abelian group then we know that it can be Z-module. Is it Q-module?

No, because if it is then there exists a ring homomorphism f: Q -> End(M) such that f(1) = 1 then Q is isomorphic to f(Q) which is subfield contained in End(M), but here Q is infinite countable set and since M is finite so End(M) must be finite so it cannot contain infinite countable set.

Is it correct?

[void cosmos]

how can you have a finite Q-module

Q is a field

a finite abelian group has torsion

[crystal vale]

Is my argument correct?

[void cosmos]

clarify first step

there exists a ring homo

[crystal vale]

If it has Q-module then we map r to f(r) such that f(r) in End(M), f(r)•a = ra

[void cosmos]

whats f

its late so ig im not following sorry

if it has a Q-module then that means you have an action by rationals

what now

[crystal vale]

I mean f: R -> End(M), and we define f(r) in End(M) such that f(r)•m = rm, right side rm define by Q-module structure

[void cosmos]

u mean send the rational q --> f_q(r) = q*r

right?

[crystal vale]

Yes

q to f_q

[void cosmos]

yes

[crystal vale]

Where f_q(r) = qr

[void cosmos]

now how can u prove the claim that Q is isomorphic to f(Q)

and what kind of isomoprhism is this

ring isomorphism?

[crystal vale]

Because it maps 1 to 1 and Q is a field so f is injective

Because the kernel is an ideal

So the kernel can be {0} or Q

[void cosmos]

i mean

[crystal vale]

Yes ring homomorphism

[void cosmos]

okay but then ur arguing that Q is infinite countable and f(Q) isn't?

and hence cant have a bijection?

[crystal vale]

Yes

[void cosmos]

yeah ig its cool

[crystal vale]

Okay thank you

[cobalt heath]

Why does one say "non-associative algebra" instead of that?

Like, it is quite far from being an algebra.

[rocky cloak]

Well, "vector space with bilinear form" is a little longer and clunkier.

It's a also somewhat contextual, like people who study non associative algebras do care about varies restrictions one can place on the operation and how they relate, associativity among them.

It isn't really that far from being an algebra. Even though associative is a big deal.

[cobalt heath]

Hmm, interesting

[languid trellis]

Is it true that $G/N \rtimes G \cong G$? I do recall seeing this somewhere but I'm now not entirely convinced it's true.

[cloud walrus]

swifteeee

[languid trellis]

N Is a normal subgroup of G, a group

I feel like there's more hypotheses

or maybe its N \rtimes G/N

[languid trellis]

In fact I'm quite sure this can't be true as maps G/N x G -> G will struggle to be injective

[rocky cloak]

You mean
$N \rtimes G/N$?

This is true sometimes, but not in general

[cloud walrus]

jagr2808

[languid trellis]

I realised soon after posting that this is what I mean

[rocky cloak]

If N isn't abelian, then you don't even have an action of G/N on N

But even if N is abelian, G/N isn't necessarily a subgroup of G

[languid trellis]

Yes that was my struggle. how do i identify G/N with a subgroup of G

Oh well

It was worth a think. Thank you jagr

[tough cairn]

is this post uni?

no it shsould be in uni, but like u getting ur masters

probably

correct me if I am wrong

[rocky cloak]

You mean the level of the topics in this channel?

Around 2nd or 3rd year uni

[tough cairn]

yeah

sorry if I interjected on a grand discovery

I was doing some calc 3 and just needed a side note cause I have no clue what the heck is going on

[tough cairn]

I see

[coral spindle]

After putting this the right way round, try G=Z/4 and N being its unique nontrivial normal subgroup

[languid trellis]

C4/C2 is the cyclic group of order 2, and everything is abelian, C2 \rtimes C2 = C2 x C2 which, if what I was trying to prove is true, would show that C4 is isomorphic to the klein 4group, which it is not

C2 im identifying with the subgrou of order 2 in C4, {1,x^2}

which is its only nontrivial subgroup, and as C4 is abelian it is also normal

[abstract rock]

your first dip into the nature of the group extension problem

which has been resolved for abelian kernels (for the most part i believe, something in the back of my mind is trying to remind me ofna case but i cant be assed)

[rocky cloak]

I guess it depends on what resolved means. When the kernel is abelian it's classified by the second group cohomology.

You also have a very similar construction to group cohomology classifying the extension for non-abelian kernels as well.

But computing group cohomology isn't necessarily easy.

[abstract rock]

I thought central extensions required the kernels to have a non trivial centre

or are younreferring to another construction for non abelian extensions?

[rocky cloak]

Central extensions correspond to group cohomology with trivial coefficients. Extensions with abelian kernels correspond to group cohomology, and then you have a generalized construction where instead of an abelian group with an action G -> Aut(A) you consider "actions" of the form G -> Out(A) and allow A to be non-abelian.

[abstract rock]

Oh yes I think i saw mention of this (actions on G-> Out(A)) on groupprop wiki, i dont remember finding a source, do you have one you suggestion?

[rocky cloak]

This MSE suggests Ken Brown's Cohomology of groups

https://mathoverflow.net/a/129731/157483

[abstract rock]

thanks

[rocky cloak]

But anyway I guess, the main point is that for group cohomology you at least have the tools of homological algebra to help you compute it, so in that sense the problem is "solved".

Not sure how well those tools translate to the non-abelian case.

[crystal vale]

It is given that M is left R-module, I am not getting the 4) last part, if I define the ring homomorphism f: R -> Hom(M, M) where f(r) : M -> M such that f(r)(m) = rm, then how f(r) in the centre of Hom(M,M) ?

In solution they make Hom(M, M) is a left R-module and defining r•phi = phi • r for all phi in Hom(M,M) and r in R.

[rocky cloak]

Well, it's in the center by definition.

Like take g in Hom_R(M, M).

Then
g o f(r) = g(rm)
and
f(r) o g = rg(m)

The definition of Hom_R(M, M) are exactly those gs such that these expressions are equal.

The thing that breaks in the non-commutative case is that f(r) isn't contained in Hom_R(M, M)

But otherwise the above equations hold true

[coral spindle]

There you go! That’s a counterexample

[crystal vale]

I got it I thought it is Hom(M, M) not Hom_R(M, M)

[crystal vale]

I see why f(r) is not contained in Hom_R(M, M)

Because f(r) does not follow the scalar property of transformation

[rocky cloak]

No it's not. It's Hom_R(M, M)

[crystal vale]

Yes

[crystal vale]

Here, there exists r_1 and r_2 such that r_1r_2 ≠ r_2r_1, but isn't possible r_1r_2m = r_2r_1 m for all m in M ?

[languid trellis]

I think Q_8 is another one if we consider the ses 1 -> Z/nZ -> G -> Z/2Z -> 0?

To say G = Z/nZ \rtimes Z/2Z would imply that D8 and Q8 are iso

Thanks boyt!

[rocky cloak]

Yes, it could happen. But it's not true in general

[crystal vale]

Okay so there can be possibility

I think when M = 0 it can be possible

[coral spindle]

Damn that was fast! Yes, that’s another example

The general version of this is the dicyclic group of order 4n

It’s the nonsplit version of the dihedral group

[crystal vale]

If I remove the necessity of f(R) in centre of Hom_R(M,M) and replace Hom_R(M, M) with Hom(M, M), is it still ring homomorphism R to hom(M, M), I think yes

[rocky cloak]

I'm not sure what you're asking.

The action of R defineres a ring homomorphism R -> End(M), yes

[crystal vale]

Yes I was talking about the action of R on M

[abstract rock]

I havent found anything convincing yet in my research but I've been chipping away at it with my spare time

[languid trellis]

Gotcha, I'll have a Google and a think 👍

[tough raven]

I just realised that this can be used to construct a module with an arbitrary totally well ordered set I as its lattice of non-zero cyclic submodules (and therefore an arbitrary successor ordinal as its submodule lattice). Indeed, let k be any division ring and consider R = k<X_i : i ∈ J>, where |J| ≥ sup_{x ∈ I} |{y ∈ I : y < x}| (for example you could take J = I) and fix a surjection f_i: J → {y ∈ I : y < x} for each i ∈ I. Let M be the free k-vector space on I with basis {m_i : i ∈ I}, and let x_j ⋅ m_i := m_{f_i(j)}, defining an R-module structure on M. Then for any v ∈ M \ {0}, if i = "deg"(v) := max {i ∈ I : v_i ≠ 0}, Rv = span_k {m_j : j ≤ i} (prove this by well-founded induction on i).

Interestingly, this requires "sufficiently many variables" to go throuh. Over a Noetherian ring, finitely generated Artinian (or non-Artinian, for that matter) modules are Noetherian. Perhaps this generalises to notions of Noetherian and Artinian with respect to ordinals/cardinals (e.g. a-Noetherian iff every submodule is inaccessible by a-directed joins for a a cardinal, or a-Noetherian iff every increasing a-chain of submodules stabilises for a a limit ordinal).

[arctic trail]

Is there any injective (additive) morphism from Zp to Z hat which preserves integers?

that is to say

f: Z_p -> Z
f(n) = (n mod 1, n mod 2, ...) if n is an integer

[mighty kiln]

Yes, the profinite integers is the direct product of Zp over primes p

[arctic trail]

the trivial one does not preserve integers

[mighty kiln]

Oops

Probably none, since profinite integers don't contain rationals

[pearl radish]

Could someone please help me write a formal solution for

1. Z(G) of group G = {(2x2 matrix: alpha beta 0 alpha), alpha > 0, Beta in R}
2. What subgroups does a cyclic group of ord7 have?

[pearl radish]

I know hwo to solve these problems btw

I just don't know how to write a formal solution

Like if you want just tell me each thing I ought to mention, so e.g., the definition of the group, subgroup, lagrange theorem, Z(G), cyclic group

idk if i'm missing anything

@slim stag

<@&286206848099549185>

[arctic trail]

this group is abelian innit

cyclic groups of prime order only have 2 subgroups

[pearl radish]

i know due to lagrnage but how to write it formally

@arctic trail

[stone sky]

Just say by Lagrange’s theorem that since the cyclic group of order 7 has prime order, the order of the subgroups must divide the order of the group. Indeed, the only divisors are 1 and 7. Clearly, a subgroup with order 1 is possible being the identity, and then a subgroup with order 7 is also possible, being the entire group itself.

you can formalize it further but that’s the idea

and you could also write it better but it’s an outline

[pearl radish]

what about the other question

[molten viper]

I've fallen down a rabbit hole and discovered that there are exactly 2 groups of order 18515 = 5.7.23^2 up to isomorphism. One of them is the cyclic group of order 18515. My current problem is that I don't know what group the second one is and I'm not sure how I would figure out what it is

any ideas?

nvm someone jsut came up with it

It's apparently Z_35 x Z_23 x Z_23

[coral spindle]

You could also see this with ease if you use the classification of finitely generated Abelian groups :)

[molten viper]

Well, up until this moment I did not even know if the 2nd one was abelian

[coral spindle]

The classification tells you at a glance at the factorisation that there are two of this order which are Abelian

[molten viper]

You're referring to the one that goes like

$G \cong \Z^n \times \Z/p_1 \times \cdots \times \Z/p_n$

like every finitely generated abelian group (in particular, the finite ones) are of this form?

[coral spindle]

Yes

The theorem also tells you a standard form which tells you exactly when they are distinct

[molten viper]

Oh interestinmg

It hasn['t really come up in class recently so I guess I didn't think of it

so how does one prove these 2 are all of them? Sylow theory plus some other facts?

[coral spindle]

I assume you just show that there are unique Sylow subgroups for all primes, but hey, I only claimed you could find the other Abelian one easily

Idk how the argument would go

[molten viper]

So, this all got started because an old qual problem had "prove that every group of this order with an element of order 35 is abelian"

[abstract rock]

Well if you're allowed to state theorems you havent proved this group has order of the form p^n*q^m

woops sorry misread earlier messages!

[molten viper]

and the assumption of having an element of order 35 seemed unnecessary

(you can instead prove that the group has a normal subgroup of order 35 and this is all you need to use a recognition theorem)

ig the question then becomes "are there any non-abelian groups of order 35"

One would imagine not

[abstract rock]

well first you state that there is no simple group of order 35=7*5 by showing there are no simple groups of order p*q then you can use schur zassenhaus to show that the remaining possibilities must be semidirect products

[molten viper]

I think we actually proved the "no simple groups of order pq"

and we've proved that there are no nontrivial semidirect products of order pq for distinct pq

interesting!

[abstract rock]

aren't groups great

[molten viper]

I love em

My next reading might be how to construct a topological space with arbitrary fundamental group

I'm told there's 2 ways to do it

one with group actions, another with presentations

[serene dune]

what is the question?

[pearl radish]

too late

[serene dune]

there's a little twist about the divisibility but yeah you are right

if that's your attitude, then never post and certainly never tag any role

[molten viper]

what's the twist? I mean p,q prime btw

[serene dune]

p divides q-1

take 2,3 as example

[abstract rock]

that follows from the semidirect product being non trivial i suppose

[molten viper]

right, 2,3. But I think there's no non-trivial C_2 \rightarrow Aut(C_3)

since 0 must map to the identity

[sonic coral]

consider D_6

[serene dune]

in that case you get the possibility of acting non trivially

[molten viper]

wait am I being goofy

[abstract rock]

maybe a lil

[coral spindle]

Aut(C_3) is isomorphic to C_2

[abstract rock]

thats ok man

[serene dune]

u want me to post the construction?

wait

[molten viper]

No I've definitely seen this before and I should probably construct it myself

So, Aut(C3) is C2 yes, namely these isomorphisms are 1 \mapsto 1 and 1 \mapsto 2

[serene dune]

right!

[molten viper]

Oh wait

Yeah there's 2 homomorphisms C_2 -> C_2

the zero map and the identity map

zero map gives Z_6

identity map is a nontrivial homomorphism (a fact I seem to have forgotten, lol)

and you get D_3 from this

[abstract rock]

bingo

[molten viper]

I guess I need to modify the set of things I call "trivial" in my head

[serene dune]

triviality is when the brain wants comfort

[molten viper]

true

I guess the reason I got confused is because I tend to say things like "G is a trivial normal subgroup of G"

which isn't technically correct

trivial is when the fact is obviously true and not useful

[serene dune]

its so hard to get what's going on in someone's head

[molten viper]

especially your own

but yeah, I'm now doing some like

"true/false" questions that our prof gave us

[serene dune]

than lord, we have theorems and lemmas as shinning stars

[molten viper]

we've done a lot of ring theory too, so I'm kind of forgetting the group theory

not forgetting

just rusty

[serene dune]

hey, i feel im so weak in rings

any problem set/exercise you have in mind

that will make me a lil bit mature

[molten viper]

To steal some problems I've done recently \

1. Prove that $\mathbb{Z}[x]$ is not a PID \
2. Prove that for a finite ring $R$, $R$ is a domain if and only if it is a field \
3. Show that every proper ideal of a ring $R$ is contained in some maximal ideal of $R$ \
4. Show that a ring $R$ is noetherian if and only if all of its ideals are finitely generated \
   For one of these problems, you'll need to use Zorn's Lemma

[cloud walrus]

Galstaff, Sorcerer of Light

[molten viper]

if these seem daunting, perhaps open up a text like Dummit and Foot and do some of the early ring exercises

[serene dune]

hmm, thanks i want to extensively solve a book ig

[tardy hedge]

Ive always liked q2

[grizzled spindle]

Just making sure I’m not crazy, this action also works if I just take gf(x) = f(gx) right

[rocky cloak]

Well it's no longer an action unless G is abelian (it doesn't preserve multiplication), but otherwise sure

[coral steeple]

I'm trying to prove that groups of order $p^2$ are abelian using the class equation and contradiction. I've been able to show that if such a $G$ is not abelian, then [\underbrace{|G|}{p^2\text{ elements}}=\underbrace{|C|}{p\text{ elements}}+\sum_{i=1}^{p-1}\underbrace{[G:C(x_i)]}_{p\text{ elements}}]

[cloud walrus]

person2709505

[serene dune]

yes, on the track

[coral steeple]

From roughly where will the contradiction come? Of course the counts match up

[serene dune]

the quotient group G/Z can't be nontrivially cyclic

[coral steeple]

What's Z here?

[serene dune]

Z(G) is the centre, which contains those element(s) which commute(s) with the whole G

[coral steeple]

Ah, I see. I will think about that, thank you.

[south patrol]

Well there is an easier solution in my view but ye

Consider ||the centraliser of an element not in the centre||

[coral steeple]

Ok, got it.

[coral steeple]

Is there some kind of contradiction from the fact that all the xi have the same centraliser, namely the centre?

[south patrol]

What elements are in the centraliser of g

[coral steeple]

The ones fixed by conjugation by g

I can't figure it out at this hour but I will think about that more in the morning, thank you @south patrol

[arctic trail]

what do the Sylow theorems say in this case

wait

[molten viper]

So you get that there is a unique subgroup of order 7 and 23^2

[arctic trail]

how many groups of order 5.7 are there

I think only 1 right

which is the cyclic one

[molten viper]

It will end up being that way, but it requires some work

[arctic trail]

I can use sylow innit

n7 = 1 (mod 7)
If there were 8 I'd have more than 35 elements

[molten viper]

Yeah counting elements gets you to direct product recognition I believe

[arctic trail]

don't p groups have non-trivial center

it would follow G/Z(G) is either isomorphic to C_p or {1} (either case is cyclic) so G itself must be abelian

[arctic trail]

take gf(x) = f(g^{-1}x) or gf(x) = f(xg)

[arctic trail]

2. is hard

easiest solution is the class equation

[molten viper]

It has an easier solution that is similar to a basic proof in group theory

[arctic trail]

3. is true???

[arctic trail]

how?

domains don't have to be abelian

[molten viper]

As far as I’ve ever heard, a domain must be commutative

[arctic trail]

that's an integral domain

I've heard domain for just

ab = 0 => a = 0 or b = 0

[molten viper]

That’s bizarre

[arctic trail]

anyway, now you know that if you have a zero divisor in a finite ring

cool ass theorem

[molten viper]

Yeah sorry wikipedia is wrong here

But yeah what you’re proving is that no zero divisors —> division ring

[arctic trail]

wdym?

[molten viper]

But you HINT: ||consider a nonzero r and count all the unique powers of r||

[arctic trail]

I know the proof haha

no worries

[molten viper]

Gotcha

[molten viper]

Then prove it :P

[coral spindle]

No, you can have noncommutative domains

I'm surprised you haven't seen this in books. Wikipedia is in fact correct here.

[arctic trail]

It was a question

[molten viper]

I have commutative algebra brainrot I guess

[molten viper]

Its true yeah

Well its equivalent to the axiom of choice

[arctic trail]

Ah lol

Zorns lemma

I was thinking the union of a chain could be the whole ring

but no, cause then 1 is in one of the proper ideals

makes sense

[crystal vale]

Yes

[cloud lynx]

how does looking at the degree tell me that's injective?

[wraith cargo]

I believe that's meant to show it's not surjective

basically show that there's some degree of polynomial that's never reached

[cloud lynx]

ahh oh

ty

[warm dove]

The first example that comes to mind is the quaternions but that seems awfully complex - is there a simpler example?

[coral spindle]

k<x,y>, the ring of non-commuting polynomials in two variables

over some field k

[south patrol]

Lol I realise now the issue that to me domain usually is an abbreviation for integral domains (which are commutative)

[rocky cloak]

The Weyl algebra (ring of operators in k[x] generated by multiplication by x and d/dx)

[south patrol]

But I suppose if you already have the assumption of commutativity then this is true

[south patrol]

Tbf this is a nice example though as it is a division algebra

Probably the simplest / most familiar example thereof

[grizzled spindle]

But don’t you have g(h(f(x)) = f(ghx) = gh(f(x))

Like this direction still preserves multiplication like this

If we take g(f(x)) = f(gx) as the action

Isn’t the action commutative in either case

[dull ginkgo]

Iirc integrality is where there is no zero divisors, being a domain is commutativity

[rocky cloak]

g(hf(x)) = hf(gx) = f(hgx) =/= f(ghx)

[grizzled spindle]

Oh you can’t do h first?

That makes sense

[rocky cloak]

So it's important to note that g and h act on the function f.

So you can act by h first, which gives you the function hf(x) = f(hx)

Then when you act on it by g you get
g(hf)(x) = hf(gx) = f(hgx)

[grizzled spindle]

Ah yeah I see

g needs to act before h on x

[rocky cloak]

If instead X was a right module you could define
gf(x) = f(xg) though

And that would be fine.

Or you can define a right action
fg(x) = f(gx)

[arctic trail]

yeah that's what I wrote

[arctic trail]

.

these actions show up a lot in a thing I studied

[rocky cloak]

Yeah same

[grizzled spindle]

I think I’ve seen them for vector spaces too but not much else for now

[arctic trail]

If you have like two grupos R,S and an action on R this defines an action on Hom(R,S)

[south rain]

Is this correct?

[south rain]

Thanks!

[woven panther]

Is this true?

$\mathbb{Q}(\sqrt{3}, \sqrt{5}, \sqrt{12})/\mathbb{Q} = \mathbb{Q}(\sqrt{3}, \sqrt{5})/\mathbb{Q}$?

[cloud walrus]

Bean Man

[woven panther]

As $\sqrt{12} \in Q(\sqrt{3})$?

[cloud walrus]

Bean Man

[coral spindle]

Yes. Note the bit that matters here is just $\mathbb Q(\sqrt 3, \sqrt 5, \sqrt{12}) = \mathbb Q(\sqrt 3, \sqrt 5)$.

[cloud walrus]

$\mathbf{Boytjie}$

[woven panther]

I'm a little confused as to why

$\mathbb{Q}(\sqrt[3]{2}, \sqrt{3}, i)$ over $\mathbb{Q}$ isn't Galois

[cloud walrus]

Bean Man

[rocky cloak]

What makes you say it's not Galois?

[woven panther]

I thought it was but I'm told all the roots of $\sqrt[3]{2}$ aren't in the extension

[cloud walrus]

Bean Man

[rocky cloak]

Well, they're in there. Which roots are they claiming are not there?

So for clarity
Q(cbrt(2), sqrt(3)), Q(cbrt(2), i) or Q(cbrt(2)) are not Galois.

But Q(cbrt(2), sqrt(3), i) is.

[minor fulcrum]

If I have a field K with a topology that makes it a topological ring, does that imply that it's a topological field i.e. the map a -> a^-1 is continuous on K{0}?

I strongly suspect the answer is no but I'm having a hard time getting a counterexample

[warm dove]

any tips for locking in for my absalg final (groups and rings)

😭

things to do or whatever

I know everything well but I'm just concerned abt blanking on computations/whatever

[kind temple]

do practice problems from textbooks

[crystal vale]

If H is a proper subgroup of a finite group G, then G is not the union of all
the conjugates of H.

Any hint?

I was thinking about using some action

And yes all its conjugates are proper

[crystal vale]

To prove, g and g^-1 are conjugate in S_n.

Since g and g^-1 have the same cycle structure therefore it can be conjugated to each other, is it correct?

Then I can say in any finite group G, g and g^-1 are conjugate

Because the number of conjugates of H is |G | / | N(H) | and since |H | ≤ | N(H) | so there are at most [G:H] conjugates

Done, thank you TTerra

[lusty marlin]

Does this hold for infinite groups too or is there a counterexample?

[crystal vale]

If G is a finite group with conjugacy classes C 1, ... , Cm, and if gi \in Ci , then
G=<g1,···,gm>·

This is the second part of that question, so now I think we can take H = < g1,..,gm> and but G can be written as a union of conjugation classes of H therefore H cannot be a proper subgroup of G, correct?

[crystal vale]

I think yes

[cobalt heath]

This is interesting
https://math.stackexchange.com/questions/121526/union-of-the-conjugates-of-a-proper-subgroup#121533

Basically it generalizes to finite index case, but not infinite index case.

[crystal vale]

I understand a little bit, how this reduces the finite case, I mean let H be the subgroup of finite index n, and then take an action on the set of left Cosets, it gives G -> S_n with kernel K, such that K\subset H. Now how does it reduce the finite case?

[crystal vale]

To show the number of conjugacy classes in S_n are the same as the number of partitions of S_n.

For every partition we will take the element of that cycle structure, I mean if partition such that a+b+c = n, then we can take the element of xyz such that x has length a, y has length b and z has length c and these are disjoint, correct?

[crystal vale]

Is there any better source for studying A_n group?

[rocky cloak]

Consider Q with a basis of open sets given by
q + nZ for q in Q and n in Z.

[coral steeple]

Oh yeah I think that's more or less what yeshua said

[chilly swallow]

I am looking for a concise book recommendation on abstract algebra, specifically covering groups, rings, and fields, including finite fields.

[lusty marlin]

'Abstract Algebra' by Dummit and Foote

[olive parrot]

To determine the maximal order of an element in a symmetric group (the value of Landau's Function) would solve the Riemann Hypothesis, so it must be very hard to list all the possible orders of the elements of 𝑆𝑛
in general.

Can someone elaborate on this? What's the exact connection between finding all orders of elements in a symmetric group and the Riemann hypothesis?

[bitter rover]

What is the maximum value of Landau's function in terms of the cycle structure of the permutations?

[olive parrot]

the lcm of their transpositions

[bitter rover]

Was expecting you to talk about partitions of n.

But you look at all the partitions of n and take the one with the largest lcm.

Did you mean partition?

[rocky cloak]

From Wikipedia, if g(n) is the maximal order of an element in Sn then
log(g(n)) < sqrt(Li^-1(n)) for sufficiently large n
is equivalent to the Riemann hypothesis.

https://en.m.wikipedia.org/wiki/Landau's_function

[bitter rover]

Anyhow, if you think about how to make that as big as possible, you'll see you want cycles whose lengths are prime powers whose sum doesn't exceed n.

At which point the fact that RH could be involved should be less surprising

[rocky cloak]

And it seems a proof of equivalence is here
https://arxiv.org/abs/1907.07664

Though I didn't read it

[olive parrot]

yes sorry

[olive parrot]

do you mean removing orders that divide higher orders

wdym by sum doesnt exceed n

[bitter rover]

I should've said "You'll want non-trivial cycles whose lengths are prime powers and whose sum doesn't exceed n." The sum of the cycle lengths in the cycle decomposition of an element of S_n always add up to exactly n, but some of those cycles might be trivial cycles (i.e., fixed points).

Think about why both L(5) = 6 and L(6) = 6, for example.

I'd just suggest you sit down and think about how you'd try to make the order of an element as larger as possible by changing the cycle structure. It'll be clear after a bit why you care about cycles w/ prime-power length.

[eager cape]

im confused with this part

how can that ever not be 0

[wraith cargo]

Because those are generalized eigenvectors

[languid trellis]

I don't see how we justify "The centre of G/C1 has the form C2/C1 where C2 is normal in G". Can someone help?

Something is telling me this is related to the correspondence theorem but I can't figure out any details

[south patrol]

Yes, this is a variant of the correspondence theorem

Under the correspondence, normal subgroups correspond to normal subgroups

And the centre is always a normal subgroup

[languid trellis]

We have the canonical surjection pi : G -> G/C1. G/C1 is a p-group, so has non-trivial centre. Then by the correspondence theorem, this is the image of a normal subgroup of G containing ker pi = C1.

?

So we have the identification of C2/C1 with the centre of G/C1, C2 normal in G containing ker pi = C1

[vocal pebble]

this shouldnt be possible if the ideal isnt two sided ?

i could come up w/ a proof for two sided ideals but not for 1 sided ideals by considering what you do in groups by taking cosets, but i dont think that proof will carry over to any ideal in general because it need not be two sided which i needed because i had to show that the multiplication of these cosets is well defined

[south patrol]

Sure, this should be about two-sided ideals

Any kernel is clearly a two-sided ideal, so the statement can't be true for merely 1-sided ideals

[vocal pebble]

this too, yeah its for two sided ideals

[south patrol]

Indeed that then becomes the correct statement

:)

Idk why they said I should be a proper ideal though

it is still true for I = R since that is the kernel of the unique map R -> 0

[vocal pebble]

yeah i can just take the 0 homomorphism

oops bad typo lol

[south patrol]

(warning, this is the only "0 homomorphism" provided you assume ring maps preserve 1 (and that rings have 1))

you likely already know this but the way you said it left some room for that interpretation, so i thought it may be worth mentioning lol

[vocal pebble]

yeah i was just reading through random sources online so im having troubles with definitions since they keep changing (i should just be reading from one place)