groups-rings-fields

[quiet pelican]

Two-cycles are transpositions (ie where a -> b -> a)
Disjoint means that it's a permutation that acts solely by swapping pairs of elements

(Disjoint means it's a product that looks like (a b)(c d)...(y z) where a, b, c, d, ..., y, z are distinct)

[knotty badger]

You’d also have to check that, if a permutation in S_4 has order 2, then it must either be a 2-cycle or a product of two disjoint 2-cycles

Which I don’t think the solution does

[quiet pelican]

Because (13)(24) and (24)(13) are the same

so the product you get from choosing two elements is the same as the product you get from choosing the other two elements

(in this example, the result of choosing 1, 3 is (13)(24) and the result from 2, 4 is (24)(13))

[urban geyser]

hamg on cat on my chest

[quiet pelican]

So for example you have the permutation that maps:
1 to 3
3 to 1
2 to 4
4 to 2
that is the product of two transpositions ( (13) and (24) ) and they contain disjoint sets of elements so are disjoint

[urban geyser]

I think i got it

thanks

[knotty badger]

Let $G$ be a group. What categorical properties does $[G, \mathbf{Set}]$ have?

[cloud walrus]

Pseudonium

[mighty kiln]

This category is called G-Set right

[knotty badger]

Yeah

[mighty kiln]

What properties are you looking for

[knotty badger]

For example, what limits/colimits exist

[mighty kiln]

Probably cocomplete cuz category of presheaves

[knotty badger]

Right, that makes sense

Ah, so then it’d have exponential objects too?

Actually yeah, this should be a topos right

[mighty kiln]

Yea cuz category of presheaves

[knotty badger]

Neat

Ok well that answers my Q lol

[chilly ocean]

The entire group can't be composed of order-35 elements. Suppose x has order 35. Then 35 is the smallest integer such that x^35 = 1. But then (x^7)^5=1 so x^7 has order at most 5

[mystic arrow]

no, because as its shown, if x has order 35, then x^7, which is in the group, has order 5

if x is a non-unit

[chilly ocean]

Whenever n is a composite number, if there's an element of order n, there's also elements of smaller order by the same argument

Do you see why it has order <= 5?

A product of more than 1 prime

What is the order of an element?

Do you agree that (x^7)^5 = 1?

We are assuming x is an element of order 35

From x we are obtaining a new element x^7

By multiplying x by itself 7 times

Maybe it will help to look at a concrete example

You know the group Z/35Z, right?

Here we will use additive notation so powers become multiplication

Anyway, the element 1 has order 35

We want to find an element of order different from 35

lets multiply 1 by 7

we get 7

and 7+7+7+7+7=35=0

so 7 has order 5

[urban geyser]

thsnk you

[chilly ocean]

Don't be discouraged, it happens to everyone sometimes to miss something that was evident in retrospect

[urban geyser]

thank u, still gonna clear my messages though lest i bring even greater dishonour to my ancestors

[delicate orchid]

Please refrain from doing that

[urban geyser]

[chilly ocean]

I don't think there's need to be ashamed of it. On the contrary, it shows courage to ask questions that might seem trivial to others. The real shame would be to not ask something and lose the opportunity to learn about it just because you're afraid to be judged imo

[urban geyser]

ur right, its sometimes tough out on these In-The-Same-Server-As-People-Who-Have-Been-Doing-This-For-10-Years streets

working on it though

[knotty frigate]

Dw about you get used to knowing less than someone in your areas of interest

Besides it’s not a bad thing to keep in mind unless it’s tied to a really unhealthy mindset

[mystic arrow]

by doing that, you deny others who might also be interested in understanding the problem you are having, the chance to learn from the responses to your question

you can also learn a lot by trying to solve other people's questions for yourself, and following the discussion around them

[arctic trail]

man

symmetric group is love

Galois theory is beautiful

[urban geyser]

sorry, i got it, just dealing w my own personal situation

wont do it again

[south patrol]

Hug

[urban geyser]

Okay gonna bite my pride, and ask for help on the same question. If |G|=35, and we assume no order-5 element exists, then |x|=35 implies |x^7|=5. x^7 is in G, and is order-5, so does this not immediately contradict the assumption we just made? Why the rest of the proof at all? Couldn’t we just end it there?

[urban geyser]

thank u

[south patrol]

x needn't have order 35

[dull ginkgo]

I LOVE CAUCHY THEOREM

[south patrol]

But yes this works if x does have order 35

So you should assume all elements except identity have order 7

[urban geyser]

Ah nvm I see—so the first part is if x has order 35, and the later part is the other case where it doesn’t

[south patrol]

Yes exactly

(In practice it is helpful to know what Miz has said, that actually if a prime p divides |G| then G has an element of order p)

[urban geyser]

this proof is basically separating into 4 cases—no order-5 element and |x|=35, no order-5 element and |x|=7, no order-7 element and |x|=35, and no order-7 element with |x|=5?

[south patrol]

Though ofc you shouldn't use that here, it at least tells you what to prove lool

[south patrol]

More precisely, it is like "there exists x with |x|=35", but otherwise yes

Another useful technique which doesn't seem to be used here is the following: Note that subgroups of order 5 intersect in either {1} or are the same, and every non-1 element of a subgroup of order 5 has order 5, so the number of order 5 elements is divisible by 4

(I can explain further if needed!)

But for example that shows that you can't have 34 elements of order 5 (or 34 elements of order 7 by a similar argument with 7 instead of 5)

[urban geyser]

I see, so… the first case we can prove just by separating out |x^7|=5 which is both in G and of order-5.
The second case we prove by showing that the cosets of H, which are order-7, paired with a g outside of H of order-7, “overcovers” the group as per Lagrange’s, and therefore that g or h cannot be of order 7 (and one must by exclusion be 5)
The third case we prove the same way, by finding |x^5|=7.
Fourth case we prove by taking a g outside of H of order-5, letting h be order-5, and thus generating 5 left cosets of order-5, which also “undercovers” the group.

[urban geyser]

Ah yeah I think i see

[arctic prawn]

One can also use the first Sylow Theorem. Since 35 = 5*7, there exists a Sylow 5-subgroup and a Sylow-7 subgroup. Now since each of these subgroups have prime order, they are cyclic. Hence, fixing generators for each, they are respectively order 5 and order 7 elements,

[urban geyser]

Ah, havent touched Sylow theorems yet

[arctic prawn]

I think one actually uses Cauchy's Theorem in the proof. So 🤷‍♂️

But in any case worth noting I suppose

[urban geyser]

omfg I just searched it up, that’s literally exactly what I needed what

Why am I given questions with stuff that i havent seen yet

[arctic prawn]

Oof

[dull ginkgo]

Interesting thing of note here

For cayley’s theorem, we use the action of G on itself G by conjugation to conclude there is an element of order p for p | |G|

For Sylow’s, we consider the action of G on its subsets of order p^k by left multiplication for k = v_p(|G|)

[knotty badger]

Yeah they’re both group action stuff

[dull ginkgo]

I wonder if you could use an action that proves both in one fell swoop

[mighty kiln]

There's an easier method

Suppose any element is order 1 or 5

Each element is in an order 5 subgroup and any two such subgroups trivially intersect

So 35 = 1 mod 4

[dull ginkgo]

The standard proof of Cayley’s theorem to me makes more intuitive sense than the Lovecraftian horror presented above

[mighty kiln]

Similarly if any element is order 1 or 7 then 35 = 1 mod 6

[rocky cloak]

Can you write out these proofs some more? I don't think I've seen them proved in this way

[dull ginkgo]

In a sec

[delicate orchid]

Oh my gosh just pair the elements up

[rocky cloak]

Like the proof I'm familiar with you take the cyclic group Cp acting on {(g1, ..., gp)| g1...gp = 1}

Then the fixed points are elements of order p.

Then for Sylow you do some induction on the index

[delicate orchid]

Yeah that’s the standard method

[dull ginkgo]

If you aren’t a fan of strong induction and want a more intuitive way to understand Cayley imo

Here’s how I interpret it

Let’s say p | |G|, and there are no elements of order p in G

Thus there is obviously no elements of order p in any subgroup of G

So we can split up |G| into its orbit under the conjugation action on itself

[delicate orchid]

The conjugacy classes

[dull ginkgo]

One of the orbits, orbit of the identity, is the center and is abelian. p cannot divide the center since that would imply it has an element of order p since it’s abelian (that’s easy to verify)

[mighty kiln]

So class equation → p | order of center → order p element

[dull ginkgo]

Well the idea is to use the conjugacy classes to find centers of elements (subgroups) which p divides, and you can keep stripping down these sub groups (which p divides the order of) until we eventually have Z/pZ

But obviously, that has an element of order p lol

[mighty kiln]

Huh

[delicate orchid]

Centralisers

[rocky cloak]

Allright, so |G| = |Z(G)| + Sum_g |G/C(g)|

So if p doesn't divide |ZG| then there must also be a conjugacy class for which p doesn't divide G/C(g), hence p divides C(g), then you do induction

[delicate orchid]

It’s yummers

[dull ginkgo]

G contains no order p elements -> |Z(G)| can’t be divisible by p, p must not divide [G : C(x)] for some centralizer X(x) -> p | |G| = [G:C(x)] |C(x)| so p | |C(x)|, a smaller subgroup that doesn’t have an element of order p -> repeat to keep getting smaller centralizers which are always divisible by p in order which must eventually reach Z/pZ

If there is only one conjugacy class, then it’s Abelian which we have asserted cannot have order divisible by p

So our centralizers keep getting smaller and smaller until we reach Z/pZ or another abelian group

[rocky cloak]

Seems kinda complicated though. And you also have to prove it for abelian groups separately

[delicate orchid]

if there only one conjugacy class
My brother in heaven that is the trivial group

[grave sedge]

(it's cauchy, not cayley btw)

[dull ginkgo]

If it’s Abelian, and p divides the order, then :3

Which is the easier part imo

[delicate orchid]

Structure theorem

[mighty kiln]

Then :3

[delicate orchid]

Tensor A with Z/pZ

I won

[dull ginkgo]

In other words

If G is Abelian, and p | |G|, then we can actually compute an element of order p.

Starting from any element x

[delicate orchid]

u can do that in any group it’s called the p-part

Well, order p^k

[dull ginkgo]

Yeah lol

The contrapositive to using strong induction here is basically infinite descent lol

I wonder if you can do vieta jumping proofs using strong induction

[urban geyser]

Anything people are writing after this escapes me 😿

[mighty kiln]

You can ignore the Sylow discussion

[urban geyser]

I hope so, plus the Cayley and "just pair the elements up"

Sometimes i wonder if Im missing some type of prereq

too late now

[delicate orchid]

Pairing the elements only works for p = 2

Well it works for other primes but then they’re not pairs they’re p-tuples

[urban geyser]

Hopefully ill learn all this eventually

or itll be included in Artin

Probably gonna sign up for a second course

[dull ginkgo]

ylow

For Sylow:
Let p | |G|, and v_p(|G|) = N.
|G| = kp^N

Let G faithfully act on its subsets of size p^N (call this collection U) by left multiplication. For any set S of order p^N, let the stabilizer be denoted as N_S.

N_S acts faithfully on S, and the orbits are of the form (N_S)s, and are thus cosets, so S is a union of cosets of N_S. Thus. |N_S| | |S| = p^N, so |N_S| = p^n for some n.

For each subset S of size p^N, let f(S) = log_p(|N_S|). Since left multiplication by any g is bijective, f(gS) = f(S).

Then gS = S iff g is in N_S. Thus the orbit of S is of size [G : N_S] = kp^N / p^f(S) = k p^(N - f(S)). Furthermore, this partitions f^-1(n) into orbits, so kp^(N - n) | |f^-1(n)|, and f^-1 partitions U, so |U| = |f^-1(N)| mod kp

Assume S is in f^-1(N), then N_S is a subset of size p^N, the same as S. Therefore S is just a single orbit, I.e a single coset of the form S =(N_S)x for some x. Likewise, if H is any subgroup of order p^N, then its orbits/cosets are all size p^N. Furthermore, H(Hx) = Hx so H is a subgroup of every N_(Hx), implying N_(Hx) = H. Thus the orbits of groups of order p^N partition f^-1(N), so the number of subgroups of order p^N = |f^-1(N)|/k

Abject horror of trying to type this on mobile from memory

Which is a ridiculous amount of shit that can be proven from a single group action

[urban geyser]

(Is this still stemming from the question i asked?)

[dull ginkgo]

No

And leaves open what happens when you consider |G| = m p^a q^b and instead consider pairs and subsets of size p^a q^b

Because the stabilizers will be of order p^n q^m

And we can partition that collection of sets of size p^a q^b by the pair (n,m) for their stabilizers

Actually there is an interesting bit here

Let’s say |G| = p^N q^M for prime p, q

G acts on its power set P(G) by left multiplication

If you consider a subset S where |S| | |G|

Then |Stab(S)| | |S| | |G|

So S can be matched with a pair (a,b) where |Stab(S)| = p^a q^b

But we know multiplication by g preserves this

So the sizes of these equivalence classes can be somewhat determined?

And we know the number of subsets of size N, it’s \binom{|G| + N - 1}{N}

Wait what the fuck

[delicate orchid]

Is this not just orb stab?

[dull ginkgo]

Well using orbit stab

[delicate orchid]

It’s clear that if your subset isn’t a subgroup nothing can stabilise it. And if it is a subgroup then it’s stabiliser is itself

[dull ginkgo]

I found a silly proof of Cauchy theorem using this technique lol

[delicate orchid]

Yes I saw your novel

[dull ginkgo]

That’s Sylow, dork

[dull ginkgo]

@delicate orchid I wonder if you can use this technique in a clever way to show it’s a solvable group constructively / recursively

without rep theory

and using the fact that any subgroup of G would be of similar form

[delicate orchid]

??? And cauchy’s is a trivial corollary of….?

[delicate orchid]

Even more than that, we can assume every subquotient is solvable using strong induction on order

I think the major obstacle is actually writing G as an extension in that case

[dull ginkgo]

Strong induction or infinite descent where we keep getting normal subgroups with abelian quotients which must terminate at the identity because the order is finite

[delicate orchid]

I feel like we’re missing something

Why must there be an initial abelian quotient

[dull ginkgo]

Also just for me to remember, a stabilizer is normal if the stabilizer is uniform across the orbit

I.e stab(y) = stab(x) if x = orb(y)

Let |G| = p^a q^b, p and q both prime.

Let P be the collection of subsets of G such that |S| | |P|. G acts faithfully on P by left multiplication.

For each S in P, Stab(S) acts on S and the orbits are cosets, so |Stab(S)| | |S|

If H is a subgroup of G, then by Lagrange theorem, H is in P, and so is any coset Hx. Furthermore H(Hx) so H is a subset of Stab(Hx), and |Stab(Hx)| | |Hx| = |H| so Stab(Hx) = H for any x

[chilly ocean]

how do i represent the group G of rotional symmetries of a cube as a group of pemutations of its edges? like how would i write that out?
i tried just labelling each edge with a number from 1-12, and so orbg(e) would be 12 yeah sense you can get to any edge from any edge though the group of rotational symmetries of a cube i thought is 24 and so that would mean that |StabG(e)| = 2? but im trying to imagine what rotation keeps an edge not moved after a rotation and cant think it

lol nvm i think you can just turn the cube upside down and it works

[delicate orchid]

consider an axis of rotation going through the midpoint of two opposite edges, then rotate by 180 degrees

I have no idea what you mean by turning a cube upside down, that doesn't fix any edges (naively)

[chilly ocean]

like this yeah?

[delicate orchid]

yurrrr

[chilly ocean]

lesss goo

yeah i imagined that as making it upside down

[delicate orchid]

I see lol

[chilly ocean]

lol ya and thanks though!

[rain grove]

In the definition of Algebra $\mathbb{H}$ we can replace the role of real numbers with elements of any field F. Elements are still in the form $h = \lambda_0 + \lambda_1i+\labda_2j +\lambda_3k$ except $\lambda_i \in F$

We get a 4-dimensional algebra over field F: $\mathbb{H}_{\mathbb{F}}

[cloud walrus]

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

[rain grove]

Then I have a exercise where I have to show that $\mathbb{H}_{\mathbb{C}}$ has divisors of zero. But im confused cause If Im using the field $\mathbb{C}$ for the Quaternions, doesn't that mean its not an Algebra anymore because $\lambda h_1 h_2 \neq h_1 \lambda h_2$ for $\lambda \in \mathbb{C}$

[cloud walrus]

OHHELLNAH

[dull ginkgo]

You can replace it with any ring actually

[dull ginkgo]

Doesn’t necessarily need to be an algebra, it is still a ring which Q_8 injects into as a group, and so does C as a ring (to R * identity of Q_8)

[dull ginkgo]

Actually this is really interesting

Because the imaginary in C is DISTINCT from the imaginaries of H

[rain grove]

But it clearly says we get a 4dimensional algebra over field F. And the "algebra" first of all its not 4 dimensional but 2dimensional and its not even an algebra... but im sure im misunderstanding the definition somehow

[dull ginkgo]

Remember

The imaginary of C is distinct from H

And we DEFINE

[rain grove]

Ohhh

[dull ginkgo]

C[Q_8] such that any element of C commutes with the i, j, k of H

So yes it still forms an associative algebra

If it helps, write the Q_8 “imaginaries” with a little karet to keep them distinct

But like, they should’ve introduced something like

[rain grove]

Yeah I think i get it, we have i_C and i_H

[dull ginkgo]

The “complex algebra over the complex numbers”

Where we have i_C and i

Which I’m pretty sure has zero divisors

And is a subalgebra of the bigger one

we just have that the two “imaginaries” commute

take (i + i_c)(i - i_c) = i^2 - i_c^2 = 0

You have to be careful, because really it’s

i*e + i_c

Because the “identity” is a basis element

[dull ginkgo]

Actually it’s a really weird construction to deal with

Because it’s like (-1)_c is distinct from -1

[rain grove]

hmm yeah

[dull ginkgo]

Nah it still works

They way they define it is as a quotient of the construction i thought it was, there is no issue

[rain grove]

What information in the definition makes it distinct?

[dull ginkgo]

If we have a ring R, and a group G, we can create a ring called R[G]. The group ring of R and G

It’s simple, we take the elements of G to be a basis of R^|G| (all but finitely many are 0 though). And we define multiplication naturally as (ag)(bh) = (ab)(gh) and extend it. So therefore we have elements of G in it, where they behave as they would through multiplication as the basis, and R embeds into it too as Re, where e is the identity of the group :3

I naively thought that the quaternion algebra is R[Q_8] but the problem is that “-1” is an element of the group, but we want r(-1) = -r e (e the identity).

So what we can do is “set” e - (-1) = 0 in our algebra/ring, so we consider the principal ideal (e - (-1)) and quotient out by it

And now observing what we had, that element we took the product of is in that ideal, so is 0 in the quotient

[rain grove]

ty for writing this but idk what ideal is yet and I don't understand lots of things you said here

[dull ginkgo]

No biggie,

Depends on what you took prior, some textbooks do algebras after big sections on rings

[rain grove]

Do you maybe know why or from where you know its distinct?

[dull ginkgo]

It wasn’t distinct

My construction was wrong

It is not distinct in H_C

(i + i_H)(i - i_H) = i^2 - (i_H)^2 = -1 - -1 = 0 as expected

[rain grove]

So its not a 4 dimensional algebra?

[dull ginkgo]

The “scalar” i still squares to “scalar” -1, and so does “basis” i

[dull ginkgo]

The quaternion i, j, k still are basis elements, along with an “e” we omit for “purely complex” scalars

So like how we have like a + bi + cj + dk

Really we have a “e” behind the a like ae

[rain grove]

why is (1,j) not a basis?

[dull ginkgo]

That is the basis vector

[dull ginkgo]

The dimension of an algebra considers it as a vector space and disregards the algebra

(1,i,j) would form a generating set for the algebra, not a basis for the underlying vector space

One would say “spanning set” if it generated the vector space

[rain grove]

ok this is my argument: (1,j) is the basis of the vector space because scalars are in the form \lambda_0+\lambda_1 i. so then I could with any linear combination make all elements from H

why is this wrong if the identitys "i" are NOT distinct?

[rain grove]

Yeah I think I understand this

[dull ginkgo]

Like here’s an example. The space of real polynomials forms an algebra under multiplication, and is generated by (1,X), but is NOT a finite dimensional basis, which would be the powers of X

[dull ginkgo]

I see your point.

The problem is if we are viewing the elements of C as “scalars” and the i,j,k as basis vectors

Because with what you mentioned, if not, we can just split it up and make it into a normal quaternion,

since the (a + bi)j = aj + bk for instance

So we need the “scalar i” to be distinct from the “basis i, j, k”

[rain grove]

so they are distinct??

[dull ginkgo]

Yes

otherwise we would have the same as the real one

[rain grove]

Oh omg ok I thought you said they arent and got confused

sry

[dull ginkgo]

I misread what you asked so it’s on me too

[rain grove]

npnp and ty

[dull ginkgo]

It’s confusing because like. You could like, take H_C right

And define H_H_C without problem and then have a whole different set of “i,j,k”

So really it’s about the “i, j, k” being basis vectors

And squaring to the -1 scalar

And since there’s basically 4 you’re introducing, including the “identity” one, it’s a 4 dim algebra over H_C

algebras can be defined over more general stuff

[arctic trail]

and 1

[rain grove]

So definig it this way, the identities has to be distinct unless you specifically say we make "C" as a set of quaternions in the form a+bi

[arctic trail]

1,i,j,k are basis vectors

don't forget 1 😛

[dull ginkgo]

That’s what i called “e”

[arctic trail]

uhhh why

[dull ginkgo]

Because it’s a bit fucky because 1 is seen as a “scalar”, and not a vector

So it’s nice to say we just omit an e

[arctic trail]

they're the same thing

[dull ginkgo]

It just helped me a lot put it into intuition

[arctic trail]

the field of scalars is naturally embedded in the algebra

r -> r.1

[dull ginkgo]

Well it’s like how you can embed R into a monoid algebra R[M] and just omit writing out the identity of M

That sorta business

[coral spindle]

Note that it is also generated by X alone - the smallest k-subalgebra of k[X] containing X is the whole thing, since every subalgebra contains k also!

[dull ginkgo]

Oh yeah good point thanku :3

[arctic trail]

people care about monoid algebras? 😭

[coral spindle]

Yeah well kinda

[arctic trail]

(I'm not talking about group algebras)

[dull ginkgo]

They appear a lot

[delicate orchid]

do they

[dull ginkgo]

Also like for instance

[coral spindle]

The rep theory of monoids is less uninteresting than it might seem

[dull ginkgo]

The polynomial rings of any number of variables are the monoid algebras of direct sums of N

[arctic trail]

(Obviously studying R[G] is literally representation theory when G is a group)

[arctic trail]

oh alright

[delicate orchid]

it's pretty universal that "module over [...]-algebra" is a "[...]-rep"

[dull ginkgo]

Also the free monoids in finite number of words have an ordering that respects multiplication so you can do silly shit with that on the monoid algebra (lexicographic)

[delicate orchid]

like category reps

[coral spindle]

Plus you can get some facts about nice Markov chains by looking at the reps of associated monoids

[dull ginkgo]

Like consider “leading coefficients” in that regard

[coral spindle]

But I cannot remember details I’m afraid

[arctic trail]

man

$\begin{pmatrix}a & b\-b& a\end{pmatrix}$ or $\mathbb R[X]/(X^2+1)$

[cloud walrus]

Trivial Lemma

[dull ginkgo]

mfw isomorphism

[delicate orchid]

mfw isomorphism

[arctic trail]

randomly thought about asking y'alls* favorite construction of the complex numbers

[chilly ocean]

Representation

[dull ginkgo]

$i^2 = -1 =$ \rotatebox{90}{$i$}

[cloud walrus]

The Library of Babble

[arctic trail]

that would be the second although the meme is quite funny

[coral spindle]

I have often thought that if $-x$ is really just shorthand for $0-x$, then $\frac{}x$ should be shorthand for $\frac1x$

[cloud walrus]

Boytjie

[dull ginkgo]

$\mathbb{R}[C_4]/(1 + \bar{2})$

[cloud walrus]

The Library of Babble

[arctic trail]

I like that one

probably my favorite

unironically

[dull ginkgo]

I just realized that works because we equate the scalar -1 with the square of i

You can do the same for Q_8

[arctic trail]

yeah

that's how I like constructing the quaternions

[chilly ocean]

If you only want to construct the field part, algebraic closure of Q adjoin continuum many variables is probably the simplest construction of C

[arctic trail]

that's cursed

C as a topological field then

[chilly ocean]

I'd just go with R[x]/(x^2+1) then

[dull ginkgo]

Normed ring completions: take ring of Cauchy sequences, quotient by null sequences :3

Algebraic closures: TAKE POLYNOMIAL RING OVER EVERY IRREDUCIBLE POLYNOMIAL, QUOTIENT BY MAXIMAL IDEAL CONTAINING FACTORIZATION RELATORS

[arctic trail]

real

no wait

complex

[grave sedge]

Ultraproduct of algebraic closures of Fp is also reasonably funny

[dull ginkgo]

There’s a lot of interesting BS about the ideals of Cauchy sequence rings in respect to norms apparently

[rain grove]

So in $\mathbb{H}_{\mathbb{C}}$, If I multiply for example: $(1+i_C)(1-i_H) = 1-i_H+i_C+i_Ci_H$, Can I simplify further?

[cloud walrus]

OHHELLNAH

[rain grove]

This is scalar multiplication

[dull ginkgo]

not really,

It would be (1 + i_C) - (1 + i_C)i_H in basis form

[rain grove]

Ok, then what would be a divisor of zero? I can't find any

[dull ginkgo]

as i mentioned

(i_C + i_H)(i_C - i_H)

= i_C^2 - i_H^2 = -1 - -1 = 0

[rain grove]

okk nice

[dull ginkgo]

Both become the -1 scalar

[rain grove]

i see

they are 1_c and 1_h but you can multiply the equation with 1_h and 1_c1_h = 1_h so they cancel out

[dull ginkgo]

1_c and 1_h we equate

By def

I’ll ask you this

How many dimensions is H_C over the reals

[rain grove]

8 dimensional

right?

[dull ginkgo]

:3 yep

[rain grove]

Its a tiny detail but I don't agree they are equal... from definition of the vector space I just see 1_c1_h = 1_h

[dull ginkgo]

It’s fucky

It really doesn’t matter in a weird way

1_h is “technically” a basis vector but we call it a scalar

Do we have the “scalar and the vector part”

[shell pilot]

Is there an example for which k > n?

[dull ginkgo]

what do you mean

[shell pilot]

Is there an example where it takes more than n products of a to get the full group?

[dull ginkgo]

That’s impossible

Order of an element must divide the order of the group

(Because the element generates a cyclic subgroup, which would divide the order of the whole group by Legrange’s theorem)

[shell pilot]

Yeah, but it says the group has order n and that n divides k. So the only way for n to divide k is if k >= n

[dull ginkgo]

what

It’s saying x^k = e iff ord(x) | k

[chilly ocean]

a^k = e doesn't mean the order of a is k

The order of a is the smallest k such that a^k = e

There could be a k such that a^k = e which is not the smallest

[shell pilot]

such as 2k?

or any multiple of k

[dull ginkgo]

Yes

[shell pilot]

Ok, thank you both!

[lean sail]

this should be a pretty easy one, just looking for some verification. i got an answer that is way different than the back of the book, but i think it may actually be correct. in the back of the book, it simply says "first note that if 1 is in H then by closure all five elements are in H. Then since gcd(p^q, q^p) = 1 and gcd(p + q, pq) = 1, theorem 0.2 shows that 1 belongs to H in all cases except case e"

[grave sedge]

I think the question means p^q and q^p as the actual numbers p^q and q^p

Else you have pq three times on the set and it's not really meaningful (especially because q^p would be pq as well so a and d would also be right with your interpretation)

[lean sail]

i think i understand, so in this situation, unless i explicitly see $+$, i can assume they are not referring to the binary operation?

[cloud walrus]

proofman

[lean sail]

so $a + b$ is a binary operation, but $a^2$ is actually just $a$ squared?

[cloud walrus]

proofman

[grave sedge]

Integers have more than one binary operation, each with its standard notation

Examples being addition, multiplication, subtraction, exponentiation, etcetera

Only one of those operations makes the integers a group, namely addition

If it's specified that H is a group, its elements are integers and the operation is addition (like in this case) it would be quite bad to denote it with anything other than +

Unless explicitly mentioned of course

Sadly the usual notation for abstract groups is multiplication so it can be confusing

[lean sail]

I see, so nothing really changes, addition is just addition, multiplication is multiplication

[grave sedge]

But again here there was also a "pq" which didn't seem to confuse you

[lean sail]

right

[grave sedge]

Yes it's usually best to assume that notations are reasonable (which is why it's very bad when they are not)

[lean sail]

You are alluding to there might be a situation where they are not so reasonable?

[lean sail]

does anyone know why in the solution, we would want to let 1 be an element of H?

i guess, my question is, why would we start with that, as the author does? i would prefer to start with saying, "consider gcd(p^q, q^p) = 1 since p and q are relatively prime... (fill in the algebra step)... it follows that 1 is an element of H "

thus, $p^q$ and $q^p$ cannot both be in group $H$, since $1$ is not in $H$

[glad osprey]

I like how they establish the implication 1 is in H => every element is in H first, because it breaks up the proof a bit, but it's just a matter of taste

[delicate bloom]

hmm fun problem, I guess the main trick to me is to know the euclidean algorithm

since a,b,c,d all have relatively prime paris it means you can make 1, so must be e

[dull ginkgo]

Hey Mero

You are rather involved with algebra right

Recently pondering this (final bit of my Jacobson density bs adventure)

Let R be a left Artinian ring,
If U is a simple left R-module, then U is a finite-rank left End_R(U)-module

If U is instead a finite sum/product of simple left R-modules, then by this logic would U still be a finite-rank End_R(U)-module

[hidden cairn]

i am wondering if there is an obvious way to see this. i proved two sets are the same by showing inclusion in both directions. but it feels like there should be a more basic way to show this, maybe using the propositions of conjugacy classes?

[quiet pelican]

In both cases if you get one inclusion, you get the other automatically by replacing S with g^-1 S g, then letting h = g^-1

But I don’t think there’s a nice way around showing at least one of the inclusions

[dull ginkgo]

Do you have access to group actions yet

[hidden cairn]

yes

[dull ginkgo]

okay cool so when G acts on its power set (collection of subsets) by conjugation, stabilizers are normalizers

[hidden cairn]

yep

[dull ginkgo]

I mean to use a shortcut is to basically already have shown some two way inclusion

[south patrol]

Real

[quiet pelican]

Yeah
What you’re suggesting is essentially to have already proven a more general form of this 🙃

(Or seems to be)

[dull ginkgo]

I thought it would be an easier way and i realized that lol

In particular, centralizer of a union is the intersection of centralizers, so you could do it quickly that way

(That’s intuitive, it’s a forall quantification by def)

[hidden cairn]

wait so how do i relate this fact to the problem

[dull ginkgo]

I realized it wasn’t really necessary

[hidden cairn]

ah ic

[dull ginkgo]

For the centralizer though, by definition the centralizer of a set S is the intersection of the centralizers of its elements (elements that fix all of them, i.e in each of the centralizers, as conjugators). This lets you prove the second one in a single line, since conjugation is a bijection

[delicate orchid]

what are you doing

it's immediate from the definition of normaliser/centraliser

[hidden cairn]

so is the single line argument is C(gsg^-1) = gC(s)g^-1, for s in S?

[dull ginkgo]

Brain damage

[delicate orchid]

pink panther, write out what an element of gN_G(S)g^{-1} looks like, it's very quick to see that it normalises gSg^{-1}

an identical argument will hold for the centralisers

[hidden cairn]

yes this is the way i did it to show inclusion of sets both ways

but i wanted to know if i'm missing the big picture reasoning by just checking this the usual way

[dull ginkgo]

Any other route would just be you having already shown a two way inclusion

[hidden cairn]

yeah looks like it

thank you both!

[dull ginkgo]

:3

Hey wew i am being an idiot right about this

So like

Let’s say we have two left R-modules A and B, and their direct sum. If A and B both have the structure of like, a K-module, then we can naturally equip the direct sum with it too

Am i a goddamn idiot or is this just the direct sum of End(A) End(B) (abelian groups) being a subgroup of End(A (+) B) and using the coproduct univ prop

I can’t figure out that if M is a semisimple left module of a left Artinian ring then it satisfies double centralizer

CAN I JUST FIND A FACT WITHOUT BEING JUMPSCARED BY REP THEORY

[abstract pulsar]

I need some help with division of polynomials under Z5, is this the right channel to ask?

[lone niche]

Yeah go for it

[abstract pulsar]

My bad, I didn't realize specific examples should go to #1021175428326633542. [Here](#1274917913207246881 message) is the thread if anyone's up for it 🙂

[tardy hedge]

Post it here its fine

[tardy hedge]

Poo

[next obsidian]

Peepee

[rough beacon]

If $R$ is a ring such that $(ab)^2=a^2b^2;\forall a, b\in R$, how can we prove that $R$ will be commutative? I saw a solution on Stack Exchange that had something to do with swapping say $x$ with $x+1$, and there is a good discussion about it there. Is there some other strategy to follow in order to prove this too?

[cloud walrus]

n11

[rough beacon]

It is proving $abab=aabb\implies ab=ba$ but that would require existence of multiplicative inverses. I can't grasp how we're getting a workaround to show this existence when they don't even necessarily exist...

[cloud walrus]

n11

[lone niche]

This is my personal opinion, but the strategy to solving some problems similar to this, is by throwing random b.s. at it to see if something works.

[tribal moss]

Introducing an addition seems to be the kind of thing you need to do to prove things for rings in particular. You'll need a way to bring distributivity into the picture.

[tribal moss]

Just about the simplest say to introduce an addition is to set a = 1+x in aabb=abab, then multiply out and remove like terms. One of the terms that disappear is xxbb=xbxb, and what is left is just xbb=bxb for any x and b. The same argument from the other side gives aax=axa, so now we know that in a product of three factors two of which are equal, order doesn't matter.

Now we're ready to try unfolding (1+x)²(1+y)² = ((1+x)(1+y))².
There are 16 terms on each side, which is a lot to keep track of. But we can group them into cases where most of them cancel pairwise when we match them according to which of 1 factors we pick in each term, when we label them like: (1°+x)(1¹+x)(1²+y)(1³+y) = (1°+x)(1²+y)(1¹+x)(1³+y)

• When we choose no 1, then we're looking at xxyy and xyxy, which we know cancel.
• When we choose one 1, then we're in one of the three-factor cases from above which we now know cancel.
• When we choose three or four 1s, then each side obviously commutes.
• When we choose two 1s, the terms still cancel except when the chosen "letter" factors come in a different order in (1°+x)(1¹+x)(1²+y)(1³+y) vs (1°+x)(1²+y)(1¹+x)(1³+y) -- which happens just for 1·x·y·1 vs 1·y·x·1.
  Thus, (1+x)²(1+y)² = ((1+x)(1+y))² reduces to just xy=yx, as desired.

[toxic zephyr]

this is awesome

Troposphere going hard

[rough beacon]

Thankyou troposphere and kiyoshitakeuchi for your inputs...
I have begun Ring Theory this semester, so I think I'll get to see more such trickeries that we have to do. Throwing random things at problems is what used to be mostly done in school... 😆

Troposphere, thanks for the elaborate working. I see the thought process, and I'll study it in detail. I have to solve this question in an assignment. I was able to derive xbb = bxb, but after that I was pretty confused what to do, with even the SE discussion at hand. You make it clearer

[arctic trail]

😭

math is a videogame

[tribal moss]

At the risk of being a bit dishonest -- because I actually found the argument by blindly trying a few simple things until one of them hit paydirt! -- here's a hindsight explanation:
We're looking at aabb = abab, and we'd like the initial a to go away. If we had multiplicative inverses, that would be easy, but we don't. So instead we make this a go away by making it have been 1 from the beginning. We can do that by saying (1+x) instead of a in the original identity, in which case one of the terms that come out when we distribute is 1xbb = 1bxb. Unfortunately, doing so gives us three additional terms on each side that we need to deal with, but since they are terms rather than factors, we can hope to get rid of them by additive cancellation, and that does work in a ring.

[rough beacon]

I might be misunderstanding, but does it perhaps mean that in such a ring, all elements are 1? Singleton ring...?

[tribal moss]

No, that's too literal an understanding of what I'm trying to say.

I'm saying something like: We're looking at aabb = abab, and we would like it to say abb=bab.
But if we say (1+a)(1+a)bb = (1+a)b(1+a)b instead, at least one of the things that come out on each side when we multiply out will be the 1abb and 1bab that we want, and then we'd have
1abb + other gunk = 1bab + yet other gunk
Sure, that doesn't immediately look like progress, but since we're in a ring and the other gunk is added rather than multiplied onto what we want, there's a hope that we can get it to cancel in a way we couldn't do with an unwanted factor.

[rough beacon]

Okay! Ah, I see now. That best explains the motive behind doing this +1 actually... and since we also already have abab = aabb

[tribal moss]

Exactly.

[rough beacon]

Again, thanks a lot for your explanation. I'm sure this is going to help me tons with further problems as well 🙏
I'll be sure to come here with further doubts I encounter

[arctic trail]

multiplicative inverses are relatively strong

simply cancelability is enough

(I hadn't read everything)

[tribal moss]

Sure, but we don't have that in a general ring either.

[arctic trail]

I like this discussion because I've studied a lot about a representation theory that comes from adding 1 to specific non-unital rings to create a group

So it gives some weird intuition on this

[dull ginkgo]

Could you possibly derive Ferrari’s formula for the quadric knowing the subnormal series of S_4

[dull ginkgo]

I’ll ask this again but with more context:

Let’s say we have a left-Artinian ring R

If we have a simple left R-module U, U also has the structure of a (faithful) left End(U)-module.

A theorem of Jacobson is that if D = End_R(U) (U is a left D-module) then any D-linear map is necessarily left multiplication by an element of R (Double Centralizer Theorem).

Does this generalize to semisimple modules (finite sum/product of simple modules) because the biproduct splits up endomorphisms?

[junior badge]

"Let R be a graded ring with the following property: if f and g are homogeneous elements and fg = 0, then either f or g is 0. Show that R is an integral domain."

So I've tried induction on deg fg, as well as induction on deg f. But in both cases, there's no obvious way to carry out the induction step. Is induction the right way to go? Cuz I don't see what would otherwise work...

[grave sedge]

Looking at the minimal (or maximal) degree homogeneous component of f and g should be enough

[junior badge]

Ah yesss, thanks.

[arctic trail]

commutative graded ring I assume?

else how do you show commutativity through that?

[lean sail]

Gallian Contemporary Abstract Algebra 6E Chapter 2 Problem 9

[tulip glacier]

Suppose we have an algebraic structure (eg. group, ring, module, anything).

Suppose, now, that we want to freely adjoin a new element to it. That is, it is the same algebraic structure, plus a new element which satisfies only those relations required by the algebraic structure itself.

Can this be defined functorially via an adjunction (up to isomorphism, of course) ? (eg. just like any free algebraic structure is the result of applying the left adjoint to the forgetful functor of the category of that algebraic structure).

[chilly ocean]

I believe so

Given a structure A, maybe you can consider the full subcategory of structures containing A (and maybe require morphisms to fix A, not sure)

And maybe you'll get the adjoint of the forgetful functor to do that

I'm not sure though, but I believe some idea of this kind might work

[knotty badger]

Coproduct with Z?

Or in general, take the coproduct with the free structure on the singleton

That sounds like what you want

[rocky cloak]

Yeah, that should do it.

Forgetful functor on the over category should probably also work

[knotty badger]

So it’s like

If the original structure is A

And the new one is A’

[tulip glacier]

Hmmm, but doesn't this only define the new algebraic structure inside this subcategory ? Like, we only know how it behaves relative to this subcategory, not relative to the whole category of the algebraic structure (which means we cannot apply Yoneda to uniquely identify the structure in this whole category), no ?

[knotty badger]

Then morphisms A’ -> B naturally correspond to morphisms A -> B together with a choice of element of B

[tulip glacier]

Oh, silly me. Yeah that works.

[knotty badger]

That’s the universal property

And if you wanna do initial/final stuff, take the category of elements

It’ll be the initial object in that

[tulip glacier]

What exactly do you mean by this ? I've never seen a forgetful functor on an overcategory. How does it work ?

[tulip glacier]

Mhm

[rocky cloak]

The over category has objects as morphisms A -> X, and morphisms commutative triangles.

Then you have a forgetful functor (A->X) |-> X -> U(X)
where U is the forgetful functor in your original category

And this should again have an adjoint being what you're after

This would then be a composition with the adjunction taking X to the inclusion A -> A+X

and the usual free functor

So exactly like Pseudo described

[chilly ocean]

Is the over category is equivalent to the category of objects containing a copy of A with morphisms fixing A?

[rocky cloak]

Almost, like you don't need the morphism to be a monomorphism.

But if you did, that would be equivalent yeah

[chilly ocean]

Oh of course

Will free objects in an over category (on a category of algebraic structures) necessarily be monomorphisms?

[rocky cloak]

Well, you have the adjunction
(A -> X) |-> X
X |-> A -> A+X

But I guess if X is the 0 ring, this might not be a monomorphism

The 0 ring is not free though

So I guess it depends what you mean exactly

[tulip glacier]

Uhhh, I don't understand from the notation what objects (A -> X) are taken to.

[rocky cloak]

So (A->X) is an object in the over category, which you map to X in the original category (so groups, rings, whatever)

So you have a forgetful functor from the over category to groups for example, and then from groups to Set. And both of these have adjoints. So composing them you get an adjoint to the composition

[tulip glacier]

Oh, I see. Thanks!

[knotty badger]

Hmm hang on

I’m a little confused by what you’re doing here

So you fix an object A in the original category

You can then consider the over category on A, whose objects are morphisms A -> X, and whose morphisms are commuting triangles

This has a forgetful functor which sends (A -> X) to just X

I’m trying to see what the free functor is here

[rocky cloak]

Yes, and the adjoint is
X |-> (A -> A+X)

[knotty badger]

Right, that makes sense

Ok, I think I get what you’re doing now

[rocky cloak]

Also I think I have my terminology mixed, and this is called the under category. But whatever

[rain grove]

Let the matrix $A \in M_2(\mathbb{C})$ not be a scalar multiple of the identity. show
that $C(A) = {\lambda A + \mu I | \lambda, \mu \in \mathbb{C}}$

[cloud walrus]

OHHELLNAH

[chilly ocean]

What is C(A)?

[rain grove]

hint is to use theorem about Jordan form of matrix that says that for nay A there exists a invertible P such that PAP^{-1} is either (matrices in pictures)

[rain grove]

C(A) is centralizator of A

[rocky cloak]

Notice that
C(PAP^-1) = P C(A) P^-1

[rain grove]

no how

[rocky cloak]

Well if
B commutes with PAP^-1 that means
B PAP^-1 = PAP^-1 B

conjugating by P^-1 gives
P^-1 B P A = A P^-1 B P

In other words P^-1 B P is in C(A), so B is in P C(A) P^-1

The other direction is similar

[rain grove]

Ok I have an idea from here ty

[junior badge]

Yeah, R is assumed commutative.

[dull ginkgo]

So let’s say we have a cyclic left R-module U, and there exists an ideal J such that Jx = U

If we have any other R-module K, it’s not hard to show that an element y in K is the image of x of some homomorphism if and only if J intersect Ann(x) is a subset of J intersect Ann(y)i

Anyway

So we have two simple left R-modules A and B

The annihilator of any element in each must be left-maximal, and from our prior lemma, does that mean that an element a in A gets mapped to an element b in B under some homomorphism IFF they have equal annihilators

I.e allowing you to characterize the endomorphism division ring off of the left-maximal ideals of R

[tribal moss]

That won't work for fields, since neither free fields nor coproducts exist in general in the category of fields -- but the extension we intuitively want, namely rational functions over whatever field, does.

[cobalt heath]

Can discrete subgroup of SL(2, C) generated by trace=2 elements have trace=-2 element?

This seems so simple but I dunno what to look for

[tribal moss]

Yes. because $$\begin{bmatrix}1&1\0&1\end{bmatrix} \begin{bmatrix}1&0\-1&1\end{bmatrix} \begin{bmatrix}1&1\0&1\end{bmatrix} = \begin{bmatrix}0&1\-1&0\end{bmatrix}$$ and the square of the RHS is $-I$.

[cloud walrus]

Troposphere

[cobalt heath]

Ah, thanks, so it is possible

Guess what I am looking for is more specific then

How did you find this example?

[tribal moss]

My train of thought went more or less like:

In order for a matrix to have determinant 1 and trace 2, its eigenvalues will have to be 1 and 1, so in an appropriate basis it will be an upper or lower diagonal matrix with 1s on the diagonal. So a prototypical example will be the elementary matrix of a row operation that adds a multiple of one row to another. Hmm ... I already know this is enough to swap the two rows at the cost of negating one of them, and doing that twice will put each of them back where it was, but now with both negated ...

[cobalt heath]

Ahh, I did not think of them in terms of elementary matrices

Consider
X = {{1, 1}, {0, 1}}
Y = {{1, 0}, {a, 1}}
I am basically looking for if tr(X^e1 Y^e2 .. *^ek) = -2 is possible for e1 + .. + ek = 0. But it is not as nice.

[lean sail]

i am wondering if someone wouldn't mind taking a quick look at this problem. i feel like i am missing something here. or is this proof that easy? if i got it wrong, can i have a hint?

[cobalt heath]

I think this is supposed to be an easy exercise, yes.

[tribal moss]

Well, if you set e1=1 and e2=-1, then the trace of XY^-1 is 2-a, so if you have freedom to choose a, you can make that trace whatever you want ...

[cobalt heath]

Hmmm

Damn, I should have known this..

[tribal moss]

(And with a=1, basically my example from above can also be written X Y^-1 X Y^-1 X Y^-1).

[cobalt heath]

So the approach I thought of does not work, as a = 1 is likely allowed.

[tribal moss]

What are you trying to do?

[cobalt heath]

It's way more specific. I basically want to see that a polynomial coming out of a representation in SL(2,C) has nonvanishing coefficient.

[tribal moss]

It is indeed that easy.

[cobalt heath]

In one case, it reduces to checking if I + W is nonsingular.

Where W is a specific word

So I basically wanted to see if I can take more general word and get the same conclusion; guess I cannot, and this problem is way more specific and in-grained.

[charred iris]

as already said twice, yeah it is this easy but also I would personally actually say what c is rather than leave it to a reader with the "choose"

[lean sail]

Will update, thanks

[rocky cloak]

Is it obvious that this group is discrete though?

I guess it's a subgroup of GL(Z), so that would make it discreet

[lean sail]

is this maybe what you are referring to?

[tribal moss]

I think Edward's proposal was simply to say "let c = aba^-1" instead of "choose c such that".
But this way works too and is arguably neater.
(It's a little verbose though -- I definitely wouldn't have the line breaks before the "=y", and I'd consider writing on a single line

ab = x^-1xy = y = yxx^-1 = ca

[rain grove]

What is an example of a subspace of R^2 that is not a subalgebra of R^2 (with component multiplication)

Wait I think I cooked: $P_a = {(ax,x+y) \in \mathbb{R}^2 | x,y \in \mathbb{R}}$

[cloud walrus]

OHHELLNAH

[rain grove]

And this is also a subring that is not a subalgebra cool

[tribal moss]

Better. (But the initial "Let a, b, and c be elements of G" is now pointless and confusing, since you actually specify exactly what they are in the next sentence).

[chilly ocean]

When you say algebra here, you mean algebra over R?

[rain grove]

yes

[chilly ocean]

And by subspace you mean vector subspace?

[rain grove]

Yeah

Is my set not good?

[chilly ocean]

Why is it not a subring?

[rain grove]

It is a subring, that is not an algebra

[chilly ocean]

Doesn't Subring + vector subspace imply subalgebra?

[rain grove]

you are right, its not closed under multiplication so its not a subring

[chilly ocean]

Why is it not closed under multiplication?

[coral spindle]

For any nonzero a, this space is just R^2

[rain grove]

$(ax,x+y)(ak,k+l) = (axak,xk+xl+yk+yl)$, now this is not in the form $(aX,X+Y)$ because $X = xak$ and there is no $xak$ in that sum.

[cloud walrus]

OHHELLNAH

[coral spindle]

This is incorrect

Thrre is xak in any sum

Y = xak + (-xak + Y)

[coral spindle]

Again…

[rain grove]

Yeah I see that

ohh wow so then its not even a subspace

[coral spindle]

What no

It’s a subspace

[rain grove]

a proper subspace?

[coral spindle]

R^2 is a subspace of R^2

[coral spindle]

This was never required

[chilly ocean]

For zero a it is not an (unital) subalgebra though, since it equals 0xR and (1,1) which is the identity of the algebra R^2 is not in it

[coral spindle]

But indeed you will need a proper subspace if you want to find a subspace that isn’t a subalgebra

[rain grove]

Im still thinking why this is not right

[chilly ocean]

there may be an y such that xk+xl+yk+yl = xak+y

[coral spindle]

I explained why you can put it in that form here

[rain grove]

Yeah there might be, but its def. not for every y

does that matter?

[coral spindle]

That was never the claim

[rain grove]

If it has to be closed under multiplication then for any two elements (ax,x+y), the product should be in the same form no?

[coral spindle]

Yes, and you haven’t proven or disproven that yet with your argument

[chilly ocean]

That just means that for (ax, x+y)*(az, z+w) = (ak, k+l) for some k,l

[coral spindle]

Proof that 2 is not even: even numbers are of the form 2k for some k, but 2 = 1+1 is not of that form

[chilly ocean]

That's a good analogy

[coral spindle]

This is the same argument as the one you made. Do you see the flaw?

[lean sail]

proof for law of exponents for abelian groups, i think it's mostly right, having some trouble ironing out the last part where n < 0

[rain grove]

Okk I see

[chilly ocean]

It looks good to me, assuming you know that a^(-n) = (a^n)^-1

[rain grove]

Then this set is the whole algebra written in a bad way lol?

[coral spindle]

Like I said, when a is nonzero, this is just R^2

[lean sail]

that's a good question... how do i know that? i guess i just assumed it

[rain grove]

Right, what is an example of a subspace that is not subalgebra?

[chilly ocean]

Induction

[lean sail]

i dont think i proved that by induction

[chilly ocean]

How did you prove it?

[lean sail]

i dont think i did prove it

[lean sail]

are you pointing this out because i said that $b^{-n} a^{-n}$ is the inverse of $a^n b^n$?

[cloud walrus]

proofman

[chilly ocean]

Yes

[lean sail]

i read a paragraph in the chapter that said "familiar laws of exponents hold for groups"

also, something like, ``when $n$ is negative, we define $g^n = (g^{-1})^{\abs{n}}$''

[chilly ocean]

Yes, you can use induction to prove (g^n)^-1 = g^(-n) from that

[cloud walrus]

proofman

[lean sail]

yes i see what you're saying

[lean sail]

do you think this is a worthwhile exercise? i think maybe we should assume some things, or else this problem could really grow too big in size

[chilly ocean]

If it feels super obvious to you and you're sure you could do it if asked, then you can skip it imo

you could say "it's easy to show by induction that.."

[lean sail]

right, and actually what i've end up discovering is that it wasn't super obvious to me lol

[chilly ocean]

If it's not super obvious maybe you should try it

[lean sail]

so i think i will try it

[lean sail]

my problem is, there are a lot of situations like this, for me... i dont want to get too bogged down in trying to prove every single thing, i want to balance getting through the chapters and doing a fair amount of problems

and this can be a very hard process

[chilly ocean]

You can keep going the chapters and once in a while come back and prove the thing you skipped

Eventually some of them will probably become obvious to you

[lean sail]

ive thought about this

like, if i did all 90 problems or whatever in every chapter, it would take so long to get through the entire book

most of those are proofs, it would take so long to do

[chilly ocean]

You don't have to do the whole book linearly, and it's probably best not to

[rain grove]

Oh fk thats what im doing

I think it works for me idk why not

[chilly ocean]

Coming back to review things helps the brain get comfortable with them and learn better, I think

[lean sail]

it's a good way to get burnt out

[chilly ocean]

Have you tried doing it the other way? It could be that what you are doing works but the other option works even better. Or I could also be wrong

If something works for you don't change just because I say it

[rain grove]

I think ill stick to this method with abstract algebra just cause i find it very cool. But im def. not doing analysis this way

Anyways, If you find a subspace R^2 that is not a subaglebra, i'd love to know cause I can't find it.

[chilly ocean]

You already found one

[chilly ocean]

@rain grove

It is a nonunital subalgebra though

[rain grove]

ohh unit!! ty I missed that msg

[chilly ocean]

For an example of something that is not even a nonunital subalgebra

Consider {(x,2x) : x in R}

Then (1,2)*(1,2) = (1,4) but it is not of the form (x,2x) because there is no x that solves the system of equations x=1 and 2x=4

[cloud lynx]

if f is irreducible in Quot(R), is it also irreducible in R? and if not is there an example?

[chilly ocean]

By f do you mean a polynomial over R?

If yes then it's false: consider the polynomial 2x. It is irreducible over Q but not over Z

[cloud lynx]

yes I mean a polynomial sorry

can u explain why?

[delicate orchid]

2 is a unit in Q but not in Z

so 2x = 2*x is a factorisation into irreducibles in Z but not in Q

[cloud lynx]

ah right because Q is a field und Z has only -1 und 1 as units right?

[delicate orchid]

yus

[cloud lynx]

and how can i faktorisate in Z?

ty

[chilly ocean]

2x factorizes as 2*x in Z

Note that in general rings there aren't always factorizations into irreducibles

[cloud lynx]

[chilly ocean]

One thing that is true is that if the gcd of the coefficients of f is 1, then f is irreducible over Z iff it is irreducible over Q

And this is more generally true whenever R is a gcd domain

[cloud lynx]

gcd domain = factorial ring?

[chilly ocean]

What is factorial ring?

If factorial ring means the same as unique factorization domain

Then no, gcd domain is more general

But every unique factorization domain is a gcd domain

[cloud lynx]

when u can factorise every element into a product of prime elements where as the element is non zero and not a unit

[chilly ocean]

Yes, that is usually called unique factorization domain

[cloud lynx]

ah ok ty^^i will do some research about gcd domain

thanks for helping me

[chilly ocean]

The only example of a gcd domain that isn't an unique factorization that I know of is Z + xQ[x]

And, well I could build other ones similar to that one I guess

I wonder if every gcd domain is of this form: a subring of an UFD

[delicate orchid]

this is kind of nuts, I like it

[rotund aurora]

trivially yes: take the field of fractions

[rotund aurora]

the ring of all algebraic integers is also, but it's not UFD

[lethal cipher]

I have a potentially silly question. Suppose $R$ is a ring and $K$ is the corresponding quotient field for it. Can every non-zero element $f\in K$ be expressed as $f=\frac{a}{b}$ where $a,b \in R$ and they have no common factors?
This is true for a UFD, but can we still do this if R is not a UFD?

[cloud walrus]

dackid

[lethal cipher]

The key thing I am interested in is the existence of "coprime" a and b.

[arctic trail]

What's Quot?

a quick google search told me it's the field of fractions

first time I see someone using it over Frac

[next obsidian]

You can’t make sense of the statement for a general R, at least not in a satisfactory way.

[arctic trail]

Are you assuming R is an integral domain*?

[rotund aurora]

why not? "a and b have no common factors if and only if d divides a and d divides b implies d is a unit"

[arctic trail]

Cause if not I'm not sure the quotient field is well defined

[next obsidian]

I guess you can make sense of the gcd being a unit even when gcds don’t exist in general raaaaaaaaaaa.

I think the answer is no though

[rotund aurora]

I think you can still define it, but yeah it's better to assume it's an integral domain

[next obsidian]

It won’t be a field

You want to look at the total ring of quotients where you localize at the set of nonzero divisors

That’s the replacement for a non integral domain

[arctic trail]

How? If it's not commutative then unless you mean like the field of fractions of the ring?
Or maybe the localization by R\{0} if it's a domain, but it'd still be non-commutative

[arctic trail]

ah alr

[arctic trail]

In the case of unital commutative rings you can define a and b being coprime by Ra+Rb = R

[rotund aurora]

I'm not sure what this means, I was thinking something along the lines of what you said

[next obsidian]

Like it won’t be a quotient “field”

But you can make a quotient “ring”

A PID is a UFD

So this is going backwards

[arctic trail]

oh 🤦‍♂️

you're right lmao

well on gcd domains* then

[lethal cipher]

I didn't realize the subtlety here, so sure. Let's have it be an integral domain.

[arctic trail]

Yeah I think that an integral domain with that property is exactly a gcd domain?

really, nice

[lethal cipher]

Okay, let me give some context. I'm doing algebraic geometry. So in my case, R is a coordinate ring on some irreducible algebraic set V (R is an integral domain in that case)

[rotund aurora]

according to wikipedia, a GCD domain must be integrally closed

so non-integrally closed domains give examples of what you asked for. Coordinate rings are not automatically integrally closed

[lone niche]

I think you can uncross this lol

[lone niche]

Let Gcd(a,b)=d (which is unique up to units). Then, let pd=a, qd=b. It remains to check that if r|p and r|q, then r must be a unit. One can check rd is a Gcd of a,b. Since Gcd is unique up to units rd=ud for some unit u. By cancellation in domains r=u.

[tropic spade]

If I have sigma in S_n with cycle type consisting of distinct odd integers, how do I know it commutes exactly with elements of the set generated by cycles from its cycle decomp? I see how sigma commutes with all elements of this set, but not how it doesn't commute with anything else?

For context, I'm trying to prove 21 here:

But I am stuck on the backwards direction given in the hint.

[lone niche]

You can prove a domain R having those properties is exactly a GCD domain. I.e. those conditions are equivalent to the condition of a domain being a GCD domain.

[tropic spade]

Is it something like this:

Suppose $\sigma=\prod_i \sigma_i$ is the cycle decomp of $\sigma$ and let $|\sigma_i |=n_i$, $\sigma_i = (a_{i1}, ... , a_{in_i } )$. Let $\alpha$ commute with $\sigma$. It follows $\sigma = \prod_i (\alpha(a_{i1}), ... , \alpha(a_{in_i }))$. So, since the cycles of $\sigma$ partition the set ${1, ... , n}$, by comparing cycles on either side of this equation we get that the cycles of $\alpha$ are cycles of $\sigma$ also.

[cloud walrus]

DootDooter

[tropic spade]

The latter product being from conjugating sigma by alpha.

I'm a bit worried I'm messing this up though.

"Comparing cycles on either side" specifically.

Ah yeah this feels wrong

The cycles here don't tell me anything immediately about the cycles of alpha.

[lone niche]

I think I have an idea for a proof, but I haven't expanded it out completely. So please take it with a grain of salt. We more or less follow a similar approach to what you did. From the formula you wrote, $\alpha$ must move the ${a_{i1},\ldots, a_{in_i}}$ within a cycle in the cyclic decomposition of $\sigma$, cyclicly in order for $\alpha\sigma\alpha^{-1}=\sigma$. And this is the same (I think) as $\alpha$ having a cycle the power of the cycle $(a_{i1},\ldots, a_{in_i})$.

[cloud walrus]

kiyoshi

[tropic spade]

I'll muck about and give that a shot. Thanks for the hint!

[cloud lynx]

Quotient field for example the quotient field of Z is Q by adding the inverse

[arctic trail]

yeah that's why I crossed it

I realized it was a different name for field of fractions

[cloud lynx]

Sorry😭😭

[arctic trail]

np!

[abstract pulsar]

TA is trying to prove this, albeit with sqrt(2) and sqrt(5). As I understand it, to prove that Q(sqrt(2), sqrt(5)) is a subset Q(sqrt(2) + sqrt(5)), we need to do operations using only elements we know are in Q(sqrt(2) + sqrt(5)) until we derive these two roots as standalone values. is that correct?

[lone niche]

yes

[abstract pulsar]

Great! Sorry in advance for the ranty question, but I really need a straight answer to not flunk this subject

Here's what the TA does

Because the Q extension is a field, the inverse of the sum (rationalized) will also belong to the field

Then, he writes sqrt(5) as this monstrosity

Three separate students have pushed him to explain how he arrives at this representation of sqrt(5) and why he introduces the halves, and he avoids the question every time

could he have just put 3 and 1 instead of 3/2 and 1/2?

I just don't see the thought process here

[vapid vale]

then it would be 2sqrt(5) (still fine, but the point was to show that sqrt(5) is expressible in this way, which is done so directly)

[rocky cloak]

The 3 is to cancel the 3 under the denominator.

The 1/2 is because the average of sqrt5-sqrt2 and sqrt5+sqrt2 is sqrt5

[abstract pulsar]

oh my god I get it

[rocky cloak]

I don't think the thought process is particularly deep. They've just manipulated the equation until they got sqrt5

[abstract pulsar]

you end up with 2 * sqrt(5) so you multiply the whole expression by 1/2 so it cancels out when you do the addition

[vapid vale]

yeah, i mean theres a number of ways to "justify" the arithmetic but the best justification is "its right there"

[lean sail]

i was looking back at this, i don't see where i did that in the proof

[abstract pulsar]

The problem is that this is literally the whole solution, I just posted it piecemeal for clarity

he really just pulls the representation out of thin air

but I see it now

[lone niche]

There's another way to solve this kind of problem which I personally like more

Here

[vapid vale]

i dont really think a TA for an abstract algebra class needs to justify why one came up with some sort of arithmetic manipulation

[rocky cloak]

With some theory you can do it with relatively little calculation:

Like Q(sqrt2, sqrt5) is a degree 4 Galois extension and the automorphisms are given by flipping signs of the square roots. None of the automorphisms fix sqrt2+sqrt5, so Q(sqrt2+sqrt5) must be everything.

Best part is that it works for basically any pair of square roots

[lone niche]

smart! I'll add it to my toolbox

[lean sail]

trying to show that if $a \in G$ and $G$ is Abelian, then $a^{-n} = (a^n)^{-1}$. getting stuck on the inductive step, any hints?

[cloud walrus]

proofman

[lone niche]

lol I was worried for a second... G doesn't need to be abelian.

anyways I think you can show that both satisfy being the inverse of a^n directly

[lean sail]

ah so no induction

[south patrol]

Well it depends how rigorous you want to be, as technically that still requires induction

[lean sail]

@lone niche @south patrol close? or way off? nothing's coming to me 😦

[delicate orchid]

Line 3 to line 4 makes me go 🤨

[glad osprey]

I think you need to think about how a^{-n} is defined, compared to a^n

[hidden wind]

i am very mechanical about these kinds of proofs and don’t bother with induction as long as it’s clear where the induction would take place

[lean sail]

Yes I was considering that also, the book gives a definition like when $n$ is negative, $g^n = (g^{-1})^{\abs{n}}$

[cloud walrus]

proofman

[tropic spade]

It took me a while but $\alpha(a_{i1})=a_{ij}=\sigma_{i}^{j-1} (a_{i1}) $ can be used to show the cycle of $\alpha$ corresponding to the shared orbit ${a_{i1},...,a_{in_{i}}}$ is just $\sigma_{i}^{j-1} $ like you said. This shows the group generated by the cycles in the decomp of $\sigma$ is exactly the elements of $S_n $ that commute with $\sigma$. Each $\sigma_i$ is an even permutation (as an odd cycle), so the elements that commute with $\sigma$ are all in $A_n$, hence none are odd. The rest of the work ends up being by past exercises.

[cloud walrus]

DootDooter

[tropic spade]

We know that first equality has to work for at least one j by the conjugation identity way from the start and the definition of the cycles from the decomp of sigma.

Iterating it to work out the cycle in the decomp of alpha for that orbit is basically what does it.

[glad osprey]

So for positive n, $g^{-n} = (g^{-1})^n$. The proof I'm thinking of is really primitive, but you should just try to write out the definition of $g^n$ for positive n and see what you get.

[cloud walrus]

sheddow

[lean sail]

Did you mean to say ``write out the definition of $g^n$ for positive $n$’’?

[cloud walrus]

proofman

[glad osprey]

I think so, did I not write that or did I mess up?

[lean sail]

Definition of $g^n$ is $g g \cdots g$? Feels like I’m missing something

[cloud walrus]

proofman

[glad osprey]

yes, and do the same for g^{-n}

[lean sail]

$g^{-n} = g^{-1} g^{-1} \cdots g^{-1}$

[cloud walrus]

proofman

[glad osprey]

Yep, now multiply them 🙂

[lean sail]

$= g^{-1 \cdot n}$

[cloud walrus]

proofman

[glad osprey]

$g^n g^{-n}$ is not $g^{-1\cdot n}$. Remember that you want to prove that $g^n$ and $g^{-n}$ are inverses of each other

[cloud walrus]

sheddow

[lean sail]

Oh, my bad I thought that you meant multiply the terms… I was just thinking, but we just factored it

You get $e$ since all the $1$s and $-1$s exponents combine to $0$s

[cloud walrus]

proofman

[glad osprey]

Yep, and if $g^{-n} g^n = e$ what does this mean?

[cloud walrus]

sheddow

[glad osprey]

actually, I guess the easiest way is just to use the exponent law $g^{a+b} = g^a g^b$ for this problem, but that's assuming you have proved that already

[cloud walrus]

sheddow

[lean sail]

It means that $g^n$ and $g^{-n}$ are inverses

[cloud walrus]

proofman

[lean sail]

Cool, I am still highly interested in what the inductive proof looks like, I think I was close

[lean sail]

.

[arctic trail]

I like using the tower lemma

it's the fastest

Clearly Q(sqrt(2),sqrt(3)) contains Q(sqrt(2)+sqrt(3))

So we must have Q <= Q(sqrt(2)+sqrt(3)) <= Q(sqrt(2),sqrt(3)), and since (sqrt(2)+sqrt(3))^2 = 5 +2sqrt(6) then:
Q <= Q(sqrt(6)) <= Q(sqrt(2)+sqrt(3)) <= Q(sqrt(2),sqrt(3))
By the tower lemma one of these has extensions has degree 1 (as Q(sqrt(2),sqrt(3)) = 4 again using the tower lemma and inequalities), and since sqrt(2)+sqrt(3) is not in Q(sqrt(6)) then Q(sqrt(2),sqrt(3)) = Q(sqrt(2),sqrt(3))

[arctic trail]

and the most clean imo

[lone niche]

Use (ab)^-1=b^-1a^-1

[lone niche]

Jagr posted a slick two-line proof up there

[lone niche]

Here

[rain grove]

exercise: characteristic of nonzero pre-ring is either 0 or a prime number.

Not sure If im translating correctly but
pre-ring: a ring K where for each nonzero a,b in K there exists x in K so that axb != 0

I showed that if K is ring without zero divisor, with characteristic n>0 then n is a prime number

[rain grove]

Ok i got it, if n = rs then (r.1)x(s.1) = (n.1)(x) = 0 for all x so then n must be prime

[lean sail]

shoes and socks principle for groups?

[knotty badger]

Yep

[chilly ocean]

What is shoes and socks principle?

[rocky cloak]

To get dressed you first put on socks then shoes. To get undressed you first take off shoes then socks.

Illustrating that the inverse of ab is b^-1 a^-1

FILO = first in last out

[lean sail]

common name given to that particular property

[chilly ocean]

Ah

[rain grove]

Cool name, I didn't know it

[arctic trail]

yeah it works

[rain grove]

Let $a$ be such an invertible element of the algebra $A$ over the field $F$ that $a^2 \in Z(A)$. Show that $A$ is the direct sum of the subspaces $C(a)={x\in A|ax=xa}$ and $D(a)={x \ and A|ax=-xa}$, but only if the characteristic $F$ is different from $2$

[cloud walrus]

OHHELLNAH

[rain grove]

I just showed $ax+xa \in C(a)$ for all x and $ax-xa \in D(a)$ for all x. And then since subspaces are closed for scalar multiplication I can do $x = \frac{1}{2a}(ax+xa)+\frac{1}{2a}(ax-xa)$ for any $y$.

So i wrote any y as sum of two elements from these subspaces

oh nvm I thought I had some issue but I think im good

I thought ax-xa is also element of c(a) but a-b != b-a ofc

[cloud walrus]

OHHELLNAH

[arctic trail]

So C(a) is the centralizer?

[rain grove]

Yeah

[arctic trail]

yeah this ends the proof for characteristic different from 2

as C(a) intersection D(a) is trivial

ax + xa = ax - xa => 2xa = 0 => x = 0

although

@rain grove you should show that 1/2a C(a) = C(a) and respectively for D(a), which shouldn't be hard

the a isn't an element of the field

[rain grove]

ohh yes i see ty

[rain grove]

finite rings without zero divisors are fields

Is that true?

[chilly ocean]

Do you mean either finite commutative rings at the beginning or division rings at the end?

[rain grove]

I have that version in my book

[chilly ocean]

There's a nice trick here to try (these steps are for commutative rings, but you can do something similar for noncommutative ones):

1. Suppose a is not a zero divisor. Show that the function f_a(x) = a*x is injective.
2. If f:X->X is injective and X is finite, must f be bijective?
3. If f_a(x) is surjective (or bijective) then a is an unit.

[rain grove]

yeah for non commutative you just do 2 maps ax and xa and then left inverse is same to right inverse

[arctic trail]

Yes

Wedderburn's little theorem

there are no non-commutative finite rings without zero divisors. And all of the commutative ones are fields

[chilly ocean]

Oh oops you're right I forgot about that part

[knotty badger]

Oh I didn’t know about the non-commutative part

[arctic trail]

Well it justs means that in finite rings domains = fields (where domain just means no zero divisors)

The easier proof is the class equation

[knotty badger]

The class equation?

[rain grove]

nice ty… but looks a bit out of my league for now

[arctic trail]

The proof isn't super complicated

[arctic trail]

yeah, relating the order of a group with the indices of the conjugacy classes

It's a version of the orbit stabilizer theorem

can be written like this, where the CG(x_i) are the centralizers which aren't the centers

[knotty badger]

Oh you mean for the self-conjugation action

Huh I didn’t realise that special case had a name

[arctic trail]

Anyway, the proof of Wedderburn's little theorem is through induction, on the dimension of the vector space of the division ring over its center

[dull ginkgo]

Then there’s the proof of Wedderburn’s from Jacobson using the fact that any (even nonunital) where there is an n for each x such that x^n = x then the whole damn thing is commutative

That fact is a pain in the balls to prove

[arctic trail]

Since R is a division ring then its center Z is a field and thus R is a vector space over a field and denoting q = |Z| then |R| = q^n for some n >= 1. Clearly if n = 1 then R is a field.
Regardless R* will be a group of cardinality q^n-1, and its center is Z* of cardinality q-1.
The class equation applied to the group R* gives q^n-1 = q-1 + Sum_i |R*|/|C_R*(x_i)| where the xi are taken appropriately as to not repeat.
C_R*(x_i) union {0} is a subring of R, and since it contains the center it's a vector space over Z and so it has cardinality q^(d_i) -1. (where the exponent has to be strictly larger than 1 and divides n since by induction it'll be a field and therefore R is a vector space over that field).
All and all we have an equality of the form q^n -1 = q-1 + Sum(appropriate d) (q^n -1)/(q^d-1).

Anyway the nth cyclotomic polynomial in q divides the LHS and each (q^n-1)/(q^d-1).
And so Phi_n(d) divides q-1
Which implies |Phi_n(d)| < q-1 which we can prove to be false

[dull ginkgo]

I was bored yesterday and doing a bit of pondering

So let’s say we have a finitely generated module U over ring R.

[arctic trail]

unital ring?

[rain grove]

thats wedderburns little theorem?

[arctic trail]

a proof essentially

[dull ginkgo]

Ring with multiplicative identity

[arctic trail]

I don't like this type of proof though, I think of it as not to informative

It doesn't give any underlying reason for it not working

[dull ginkgo]

If we have a finitely generated module, we can consider the minimum size for a generating set, then if any generating set of that size has a linear dependence, the scalars cannot be units (otherwise one of the terms would be in the span of the others). Note this is not necessarily commutative

I wonder if we could characterize an ideal, ore left multiplicative set or smth unique based off the scalars of a minimal generating set

[rain grove]

|R| = q^d for some n >= 1... you mean d = n?

[knotty badger]

Do you have a proof you prefer

[arctic trail]

yes

[arctic trail]

Brauer group approach maybe

[knotty badger]

What’s that?

Also is it weird that I kinda like the class number proof

[arctic trail]

Essentially, you can grab the central simple algebras over a field, 'quotiented by Morita equivalence' and give that a group structure.
Two rings are morita equivalent if their category of (left) modules are equivalent

[knotty badger]

The central simple algebras?

Also huh was not expecting to see cat theory here

[arctic trail]

Yeah the finite dimensional algebras over the given field K, whose center is exactly K, and are simple.

Which is exactly the type of algebras that finite division rings are for their center

[knotty badger]

What does simple mean here…?

Also is an algebra a vector space with a bilinear product, or is there more than that?

[arctic trail]

f: A->B homomorphism of K-algebras, ker f = 0 or A.

[knotty badger]

Ahh cool cool, so no nontrivial quotients

[arctic trail]

algebra meaning a unital associative algebra

[arctic trail]

yeah

[knotty badger]

I see I see

Ok, got it

[arctic trail]

I'll be honest that I don't really know how to prove it, I don't know much about Brauer groups other than their definition

but I think the proof is from cohomology

[knotty badger]

Interesting interesting

That sounds pretty cool

As someone who hasn’t seen much algebra, it’s interesting how much homology seems to appear

[dull ginkgo]

Random algebraic extension structure try not to be a group challenge (impossible)

[delicate orchid]

Does the corrispondence between the second galois cohomology group and the Brauer group hold more generally than I thought it did or something?

[arctic trail]

probably

[delicate orchid]

I really like this anyway

[rotund aurora]

homology is algebra

[knotty badger]

Right, I’m more interested as to why a lot of algebra seems to be homology

[mighty kiln]

Is it mostly from derived functors

[shell pilot]

If we take an example such as 7/5 + 8/5 = 15/5, can we say in our proof that 15/5 = 3/1 which is in lowest terms so technically this group is closed under addition?

[chilly ocean]

"Lowest terms" is not a property of a rational number but of a representation of a rational number

15/5 and 3/1 represent the same rational number

[chilly ocean]

The rational number 15/5 is represented by 3/1 in lowest terms, and it has odd denominator in that representation, which is the requirement of the question, so it is in that set

[shell pilot]

Ok, thank you!

[formal kite]

Is anybody familiar with the theory of Group Contractions of Lie Groups?

[hidden wind]

oh i've seen this semidirect product showing up every now and then but i never bothered to look it up, finally did and i'm relieved it's a lot simpler than i feared hahha

[knotty badger]

Yeah semidirect product confused me for a long time

I think viewing it as a “group acting on another group” helped a little

As well as understanding the universal property

Also examples ofc

[hidden wind]

the example of the euclidean group $E(2) \cong T(2) \rtimes O(2)$ is enlightening

[cloud walrus]

rødbet

[knotty badger]

Yes yes very much so

Poincaré group too

[hidden wind]

i am not at all familiar with the geometry of spacetime

[knotty badger]

Right yeah

I’m a physicist so I’m a little more accustomed to it

[chilly ocean]

Also D_n being semidirect product of C_n and C_2

[formal kite]

Are you familiar with contractions of Lie Algebras? I have a doubt regarding them.

[knotty badger]

Not particularly

[formal kite]

would you know anyone here that might be familiar with it? I have some (probably) very basic conceptual-doubts regarding them

[knotty badger]

Don’t think so, since I’m not really an algebraist…

[formal kite]

I mean tbf it arose from physics 😅

Like to find a relation b/w special relativity and gallilean relativity (Poincare and Gallilean Groups respectively)

[knotty badger]

Oh interesting

Yeah having looked up group contractions they seem pretty cool

[formal kite]

I have been trying to look for a way to put all lie algebras in a category, with contractions as the arrow b/w them. Also to formalise some intuitive facts (which I think might be true but not sure) in the language of category theory.

[knotty badger]

I guess you’d have to define what a contraction is, and composition of contractions, and an identity contraction

[formal kite]

this is the standard definition of a contraction

the identity contraction is also easy as you can just do Cr(X) = X forall r.

the composition might be a bit more involved though

[formal kite]

also this leads to some weirdness bcs according to physicists this isn't a contraction (you can't have contraction b/w isomorphic algebras). But ig it satisfies this definition so it shouldn't matter??

[rain grove]

I will try to read this, is it okay if I ask you here or maybe somwhere else? I got question for almost every line 😅

[arctic trail]

sure

it's aight

[rain grove]

Is this proof for all rings or just R?

I mean for all finite rings

[arctic trail]

it's for finite rings

R is a finite ring

[rain grove]

Ok R is not a finite ring what am I talking about

wait how

[arctic trail]

I'm not talking about the real numbers

the letter R is commonly used to denote a ring

in this case finite as per the hypothesis (finite division ring = finite ring without zero divisors)

[rain grove]

so R is a vector space over the field Z(R) then

Why is |R| = q^n?

[south patrol]

All finite fields have cardinality a power of a prime

[next obsidian]

WHAT

[rain grove]

So q is a prime? But R is not a finite field (yet)

[south patrol]

A power of a prime

R is a finite dimensional vector field over a finite field Z(R)

Z(R) has cardinality a prime power

R has cardinality Z(R)^k for some k

so it's also got prime power cardinality

[rain grove]

Why?

[south patrol]

.

[arctic trail]

q is not necessarily prime

but a power of a prime

there is a cute proof

[rain grove]

Ok I see yes, I need to prove this now

[arctic trail]

suppose q is not a power of a prime, so q = pp'r where p,p' are distinct prime numbers and r>=1.
What happens is that by Cauchy's theorem there exists x!=0 such that p*x = 0 and y != 0 such that p'*y = 0.
Since R is a division ring this implies both p and p' are 0. Which is a contradiction as 1 can be written as a linear combination of p and p' by Bezout's theorem and 1 != 0 for any non-trivial ring.

As to why we can't just say q is prime, because not all fields have prime cardinality!
Take the set {0,1,a,a+1} with addition take similarly to mod 2, and multiplication given from a^2 = a+1. This forms a field of cardinality 4

[south patrol]

The way I would argue is this: if R is any ring there is a map Z -> R whose kernel is necessarily of the form nZ. So we have an embedding Z/nZ -> R. If R doesnt have any zero divisors then the kernel is either 0 or pZ since Z/nZ has zero divisors otherwise. In particular if R os a field k then this js the statement that fields have characterustic 0 or p, and it turns k into a vector space over Q or Z/pZ. If k is finite then we can't have the first so k is a finite dim vector space over Z/pZ and has cardinality p^n for some n

[rain grove]

How did you use Lagrange theorem here to get px = 0 and p'y = 0?

[arctic trail]

shit I meant Cauchy's

[south patrol]

This is Cauchy's theorem

[rain grove]

hmm okk Ill need to read cauchy theorem first

[arctic trail]

nvm

[arctic trail]

read this then

@rain grove

it avoids Cauchy's theorem

[dull ginkgo]

Me when simple module over itself

[rain grove]

why is kernel for some Z -> R necessarily in the form nZ

[arctic trail]

the kernel is a subgroup

[void cosmos]

cuz the kernel is a subgrouip

[arctic trail]

Z are the integers

The subgroups of the integers form an ideal