groups-rings-fields

[open sluice]

oh I forgot about that

[coral spindle]

Here's a nice example of a relatively natural operation that is commutative, but isn't associative. The operation is taking the midpoint of two vectors. If your space is at least two dimensional, then this will not be associative despite being commutative.

In fact now that I think of it, I don't think you even need the space to be two-dimensional

[tulip glacier]

The operation is the midpoint of two vectors, as in the position vector of the midpoint ?

[coral shale]

In some sense, naturalness to me has to do with functions. categories.

[open sluice]

it takes in two vectors and spits out another vector

[coral shale]

yes

[tribal moss]

You do need it to be at least one dimensional :-)

[tulip glacier]

I see. Thanks everybody!

[elder wave]

Categories were invented to give a proper meaning to naturality

Roughly speaking

[crystal turtle]

to talk about natural transformations, you needed functors

to talk about functors, you needed categories

[chilly ocean]

Let k be a perfect field of char = p. Is k( t^(1/p) ) a purely inseparable extension of k(t) (the field of rational functions) ?

[rocky cloak]

Yes

[chilly ocean]

Why is that? It's not obvious to me from the definition that requires every element having some pth-power in k(t). It's clear for the base field since k is perfect (forgot to mention), and it's clear for the pth-root of t, since the pth-power is t and so in k(t). I guess one can, from those two facts, show that any other element has some pth-power in k(t)?

[rocky cloak]

In characteristic p you have (f(x)/g(x))^p = f(x^p)/g(x^p)

Because x |-> x^p is a homomorphism

[chilly ocean]

I understand, thank you

[rocky cloak]

(the coefficients will also be raised to the pth power, but you get the idea)

[tardy hedge]

The connection between normal subgroups and homomorphisms

Every homomorphism out of G is given by a normal subgroup of G?

Is that true?

Any homomorphism out of G has ker(phi) as a normal subgroup of G

[delicate orchid]

the kernel of a homomorphism is always normal, yes

[tardy hedge]

So there is a connection between factor groups and homomorphic images?

[delicate orchid]

that's the first isomorphism theorem

[tardy hedge]

Right , i see that

My book did some things out of order that i may want clarification on

My break is over tho (im at work)

[small ingot]

The sum of the ideals (x^2-y^2)+(x^2-2y^2) is just (y^2)?

Because we have like a negative copy of the first which sort of gets rid of the x^2 part

But I feel uncomfortable with this answer because I don’t believe (y^2) is the smallest ideal containing both of these ideals

I actually do believe this answer sorry

[delicate orchid]

a good way to check is to just write down what it actually means to add up two ideals, taking all finite sums of elements in those ideals

[small ingot]

I thought about that but like r(x^2-y^2) + (r-1)(x^2-2y^2) gives you a term of x^2.

It’s possible I’m thinking about it wrong

[delicate orchid]

what's r here

[small ingot]

scalar in the underlying field

[delicate orchid]

how does that give you x^2?

ah I see

just an x^2 on it's own

[small ingot]

Ye

[delicate orchid]

yeah, personally I really cannot see why it should be y^2

[rocky cloak]

The sun should be (x^2, y^2)

[delicate orchid]

yeah that seems a lot more likely

[small ingot]

I sort of wanted to say that like ideals are abelian groups, by definition we have 2 sub abelian groups which are the ideals we want to find the sum of, for every element in one of them we can find one in the other that cancels the x squared term

[delicate orchid]

you can get an x^2 by just doing 2(x^2-y^2)-(x^2-2y^2)

[summer path]

I thought it was that but I felt like I was going to get gaslit

[rocky cloak]

Sure, but you can also find stuff that doesn't cancel x^2

[crystal turtle]

I didn't know the sun was an ideal.

[small ingot]

True that’s where my confusion was because I thought I was going to get something clean.

Thank you

[delicate orchid]

same I just didn't have the confidence to admit it

[summer path]

If you throw stuff at the sun, it gets absorbed

I'm easily gaslit

[crystal turtle]

The ring of the cosmos

[rocky cloak]

We're all gaslit by the sun

Btw, do we have to move our shitposting here now?

[summer path]

Consequences of no algechill i guess

[rocky cloak]

Alright, then I will move my shitposting here

[summer path]

It's just like aa before algechill

[small ingot]

Ok wait so what’s the smallest ideal containing (x^2-y^2) + (y), just (y,x^2)

Gotta be true

[summer path]

Yeah i think (I'm gonna get gaslit)

[rocky cloak]

A good test to check if (f, g) equals (h, k) is if h and k are in (f, g) and f and g are in (h, k)

[small ingot]

Right thats what I did just want external validation to avoid potential incompetency

[chilly ocean]

what a self deprecating way to say "sanity check"

😵‍💫

[rocky cloak]

Alright, then I will validate you:
Good job, you did it!

:3c is wrong though (even though they agree)

[small ingot]

I dont know about self deprecating felt more corporate type beat

[chilly ocean]

same thing

[delicate orchid]

never seen jagr like this.. very scary..

[rocky cloak]

I forgot to set my calendar. Currently October 31st for me

[summer path]

,ti jagr2808

[cloud walrus]

This user hasn't set their timezone! Ask them to set it using ,ti --set.

[summer path]

Oh

[crystal turtle]

eternally halloween

[summer path]

Halloween scary

[rocky cloak]

,ti --set halloweentown

[cloud walrus]

No matching timezones were found!

Need help setting your timezone? One of the following might help!

Try using our interactive timezone picker with ,ti --set
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[rustic crown]

alg-chill gone?

[summer path]

Lawnmower ran it over

[crystal turtle]

twasn't lawnmower

[summer path]

Oh

Oops

[delicate orchid]

I ardly knew 'er!

[summer path]

Wow ryx is fast

[crystal turtle]

Ryx is many things.

[sonic coral]

How can i show that if H <= G with G finite then n_p(H) <= n_p(G)

[lusty marlin]

What is n_p here?

[sonic coral]

number of sylow-p subgroups

[delicate orchid]

assume that n_p(H) > n_p(G) and arrive at a contradiction from the fact that there has to be a sylow subgroup of G containing multiple sylow subgroups of H

for a hint, assume T, T' are two sylow subgroups of H contained in a sylow subgroup S of G, then consider the order of <T, T'>, which is both a subgroup of S and H

[sonic coral]

thank you i’ll try this

[sonic coral]

so we know that the order of the subgroup generated by T and T’ divides the order is S and H

is there anything else we know

[delicate orchid]

it's a subgroup of S, so it has to be a p-group, but since T and T' are different sylow subgroups <T, T'> is strictly bigger than both of them

but <T, T'> is therefore a p-subgroup of H bigger than sylow subgroups of H.,.... uh oh!!!

[sonic coral]

are we using the assumption that n_p(H) > n_p(G) to know there are atleast two sylow p subgroups of H

[delicate orchid]

no, we're using that assumption to know that there are at least two sylow p-subgroups of H contained in a particular sylow p-subgroup of G

[white oxide]

is there any non-trial and error way to see that $\mathbb{Q}(\sqrt{2}, \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2} + \sqrt{3})$

[cloud walrus]

okeyokay

[white oxide]

because i'm trying to hack away at elements in Q(sqrt(2) + sqrt(3))

to get to sqrt(2) or sqrt(3)

i mean i got sqrt(6) in there

[daring nova]

https://en.wikipedia.org/wiki/Primitive_element_theorem#Example

just linalg

[delicate orchid]

(3sqrt(2)+3sqrt(3))/sqrt(6) = 1/6(3sqrt(2)+2sqrt(3)) then subtract the correct multiple of sqrt(2)+sqrt(3)

[white oxide]

oh ok

can i get a hint for this question pls

[south patrol]

think about the splitting field. what does f factorise as

[white oxide]

yeah i looked up a hint and they said write f as (x - c)^p and then assume that f is reducible

so that f has a factor (x - c)^r for 1 \leq r < p

[south patrol]

ye

[white oxide]

so now we want to show that c is in F

and that if f(x) = (x - c)^rg(x) that g(x) splits

[south patrol]

well you just need to show c is in F

[white oxide]

wait y

[south patrol]

you're showing that if f is reducible then it splits

[white oxide]

yea

so don't i need to also compose g(x) into linear factors in F

to get the entire thing to be a product of linear factors

oh

nvm

wait f(x) = (x - c)^;p

[white oxide]

i just had a startling realization

quotienting a group by its commutators is akin to the construction of the tensor product

🤯

[river shadow]

damn

[white oxide]

save some for the rest of us right?

[south patrol]

in what way

[white oxide]

what's an example of a homomorphism which doesn't preserve elements in the centralizer the groups

i.e. f: G --> H g in centralizer of G but f(g) not in centralizer of H

[white oxide]

like you quotient by the relations you want to obtain

cuz u wanna make the group abelian in some sense

and for tensor products you want the properties that satisfy bilinear maps

[crystal turtle]

"centralizer of G" you mean the center?

consider the inclusion {id, (1,2)} --> S_3

[white oxide]

oh right

[white oxide]

aren't they the same thing

[crystal turtle]

Yes

[night onyx]

first (general) thought is where G is abelian and H has trivial center (a non-abelian simple group would work)

and where f isn't trivial

even easier take H a group with trivial center, G a cyclic subgroup, and just embed it

[white oxide]

oh right lol forgot about trivial centers

[mighty spade]

Isn’t the centralizer of a subset of a group is the set of elements that commute with it?

Edit: each element of it

[white oxide]

yea but i was talking about the centralizer of G and not any particular subset

[mighty spade]

Ah, okay

Yeah

[white oxide]

oh y ea well

i meant centralizer of G is the same as center of G

[mighty spade]

that’s only if G is abelian

[white oxide]

am i mixing up definition

s

[south patrol]

sorry

[night onyx]

yeah the centralizer of G is the center of G

[mighty spade]

yeah

[south patrol]

this is sort of misleading

well commute with each element

rather than the whole subset as a thing

[mighty spade]

oh, right

[south patrol]

ngl i've never used centraliser for subsets rather than elements lol

😭

[faint fractal]

Suppose that A is a 4×4 matrix with rational entries and whose characteristic polynomial is x^2(x^2 + 1). Produce a finite collection S of explicit matrices and show that, for some B in GL4(Q), BAB^−1 belongs to S.

I know the idea is form someting like this

[alpine island]

Look at the definition of diagonalization

[faint fractal]

right, I want to show that A is diagonalizable right?

[alpine island]

Yeah

Then look at the BAB^{-1} term

in the context of diagonalization

[faint fractal]

what do you mean?

[alpine island]

Oh wait no, it's not that simple

I'm used to working over GL(n, C)

If this was over C, these matrices would be a conjugacy class, with the one element of S being a diagonal matrix with entries 0, 0, i, -i

[faint fractal]

I think we figured it out

[alpine island]

Yeah sorry I disappeared for a bit

[faint fractal]

I have another problem that is in Gln,C)

Let G be a finite subgroup of GLn(C), i.e. the group of invertible n × n matrices with entries in C. Let D be the subgroup of diagonal matrices in GLn(C). Show that if G is abelian then it is conjugate to a subgroup of D. (Hint: feel free to use that every simple C[G]-module is one-dimensional when G is abelian)

I think it has something to do with jordan decomposition of a matrix in Gl(n,C)

and then the hint means that the jordan blocks are 1d

[alpine island]

Yeah I'm not seeing the trick to that one, good luck though, someone else here should be able to help

[chilly radish]

correct, that would be a normaliser

[cobalt heath]

This is about semisimplicity, I guess

Just learned about the concept and this one seems very close

Maybe C[G] itself is semisimple? (Idk)

[rocky cloak]

Q(sq2, sq3) has degree 4, and Q(sq2 + sq3) is a sub extension, so will either have degree 2 or 4. If the degree is 4, there equal.

So you just have to show it's not degree 2. Squaring sq2+sq3 gives something involving sq6, so it can't be in the span of 1 and sq2+sq3. Hence it's not degree 2.

[rocky cloak]

A matrix is determined to to similarity by its invariant factors, so you just have to consider the different ways x^2 (x^2 + 1) can be divided into invariant factors.

Thinking about it through the structure theorem for PIDs can be useful

[rocky cloak]

Fun fact: if G is a finite group and K is a field whose characteristic doesn't divide |G|, then K[G] is semisimple. (This is Maschke's theorem)

[delicate bloom]

here's what I'd do in a kind of general way,
x = sqrt(a)+sqrt(b)
(x^2 - a - b)/2 = sqrt(ab)
now notice if you multiply sqrt(ab) by x, you get a new linear combination,
x(x^2-a-b)/2 = b sqrt(a) + a sqrt(b)
now since the determinant of the 2x2 matrix is a-b !=0 it implies we can solve it.

so Q(sqrt(a), sqrt(b)) is contained in Q(sqrt(a)+sqrt(b))

[delicate orchid]

6 hours late but idc I was sleeping. Take your group of matrices G, which can also be thought of a representation of G, and thus decomposes into a direct sum of 1 dimensional reps. What this decomposition actually looks like on the matrices is simultaneous diagonalisation of the entire group (which is possible because G is abelian, so the matrices commute with one another), hence the entire thing is conjugate to a subgroup of D, cause diagonalisation is just conjugating by a specific matrix

[cobalt heath]

But that does not use the property of C[G]

Also, isn't this problem basically showing that simultaneous diagonalization is possible?

[delicate orchid]

That’s literally what I just said…

[cobalt heath]

Sorry, I forgot to read the above part.

Where does the "it decomposes into a direct sum of 1 dimensional reps" come from?

[rocky cloak]

The hint said you could freely use that the irreducible representations were 1d

[cobalt heath]

Oh. C[G] is the representation 💀

[rocky cloak]

Well, C^n is a representation of C[G]

[cobalt heath]

Sorry for my ignorance, I am yet to properly learn representation theory.

[delicate orchid]

Then maybe don’t comment on a representation theory question

[cobalt heath]

Okay sorry

[rocky cloak]

Quickest way to get an answer to a question is always to post the wrong answer online

[coral shale]

algebra is useless

maybe its too early in the morning...

[dim widget]

You could use algebra to help understand this projection

Simone Weil called algebra one of the great evils of 20th century life

[forest eagle]

Does anyone know what the group given by permutation on two-set are called?
I.e. g in S_n acts on { {i, j} | i, j in {1,2,...,n} } by g{i,j} = {gi, gj}

[lime oriole]

The permutations of a set of cardinality 2 is S_2={e,(12)}

I think I’m misunderstanding the question though

[forest eagle]

I think in the case of n=2 its just the trivial group

[lime oriole]

Are you sure? Is the arrangement 12 the same as 21?

[forest eagle]

Yes because it's sets

{1,2} = {2,1}

[lime oriole]

No because we’re talking about permutations

[forest eagle]

it acts on the set {{1,2}}, right?

[lime oriole]

By your logic wouldn’t all symmetric groups be trivial

Because it’s unordered

[forest eagle]

The group should have cardinality n! and degree n choose 2

No no

S_n is permutations of 1,2,...,n

[lime oriole]

Right, and 2!=2

And (12)^2=e

[forest eagle]

and my group acts on all two sets {i, j}

so (1,2,3) acts on {1,2} and gives {2,3}

and (1,3)(2) acts on {1,2} and gives {3,2} = {2,3}

(if my cycle notation is correct (i hate cycles))

[lime oriole]

I am pretty sure you are misunderstanding symmetric groups

[forest eagle]

Won't it be a subgroup of S_{n choose 2}

[lime oriole]

Huh

What’s n

[forest eagle]

Im speaking general n

n = 2 should be the trivial group

Maybe im completly misunderstanding

[lime oriole]

Oh you mean Sn is a subgroup of S(nC2)?

[lime oriole]

Well

[forest eagle]

nono it shouldn't be Sn

i think

[lime oriole]

Shouldn’t it have cardinality 2!

[forest eagle]

Hmm

idk n=2 is a weird case

[lime oriole]

Not really

It’s the permutations of two elements

[forest eagle]

but the set it acts on has 1 element

[lime oriole]

Which there are 2 of (12 and 21)

[forest eagle]

I want the representation of $S_n$ thats given as permutation of the indeterminants $x_{{i,j}}$ by the action $g*x_{{i,j}} = x_{{gi,gj}}$

[cloud walrus]

Hibana

[lime oriole]

Hmm

So your class defined groups in terms of actions?

[forest eagle]

no this is from self-study

[lime oriole]

I’m not as familiar with how Sn is defined that way

[lime oriole]

Or book

Same diff

[forest eagle]

yea but idk this isn't really from a book in that way

It's because this action is the same as permuting the edges of graph on n vertices

and i want to do some experimentation on this with sage

but idk how to encode this group or action or whatever it is

[lime oriole]

I think this is better thought of as the set of bijective functions from a set of cardinality n to itself

[forest eagle]

Well for example i want to compute the molien series for a given n

and sage has this function

https://doc.sagemath.org/html/en/reference/groups/sage/groups/perm_gps/permgroup.html#sage.groups.perm_gps.permgroup.PermutationGroup_generic.molien_series

[delicate orchid]

you can view this as S_n acting on vector space with basis x_{i,j} and then generate permutation matices

[forest eagle]

Yea!

That's what im already doing

[delicate orchid]

ok then what's the problem lol

[forest eagle]

and those matrices must combine to a group

what group is that?

[delicate orchid]

S_n?

like, by construction?

[forest eagle]

as a permutation group

[delicate orchid]

What is S_n if not a permutation group

[forest eagle]

Well any group is a permutation group

[lime oriole]

This source literally defines it in terms of bijective functions

[delicate orchid]

this representation is faithful so you can map the permutation matrices back to the permutations

[lime oriole]

No??

[delicate orchid]

yes, cayley's theorem

[lime oriole]

Or

No

[delicate orchid]

they're not permutation groups apriori though

[lime oriole]

I thought that said something else nvm

[forest eagle]

has a representation as a permutation group

besides the point

[delicate orchid]

yur the regular rep

[delicate orchid]

I still don't understand what this means

[forest eagle]

Idk either lol

[delicate orchid]

the map sending permutations to permutation matrices is a group isomorphism, just take the inverse map

[forest eagle]

You're probably right it's just S_n

But the moliens series i wish to compute is not that of S_n (as given as the example in the link)

[delicate orchid]

embed your group G into S_n for some n, then take the permutation matrices

the image of this composition is isomorphic to G by first iso because everything is injective

[forest eagle]

Ok looking back at one of the article i was reading i think it's what's confusing me

he calls it a subgroup

[delicate orchid]

well yeah, S_10 does not have order 120

[forest eagle]

He writes G = TransitiveGroup(10,12)

[delicate orchid]

the english in this extract does not fill me with joy

[forest eagle]

Yea okay so he considers S_5 within S_10

[delicate orchid]

yur

[forest eagle]

Hahah no me neither

It's a fr*nch article

Also i don't even think that TransitiveGroup(10,12) is what I want

When i compute the molien series of it with sage i get something much different from my own molien series

(but that's besides the point again)

[runic galleon]

can someone explain why the orange note is true?

[rocky cloak]

Moliens series depends on the representation, not just the group though. So if you let your group act on a different set you should expect a different series

[forest eagle]

Yea i think that's my main blunder here. I should've known that by now lol

[forest eagle]

So I guess what i want is a name (if it exists) to the specific representation i gave here

Or just some way to express it (either as a matrix group or a permutation group (of degree (n choose 2))) in SageMath without having to write down the cycles by hand

[lusty marlin]

#proofs-and-logic

[alpine island]

There is no orange

Like, which of those colors are you calling orange

Because I don't see anything orange there at all

[delicate orchid]

yeah nothing there is orange

[open sluice]

language diff?

or maybe she’s like me and has a color-shifted display

[summer path]

color-shifted?

[runic galleon]

there is lmao

i noted with red then i added yellow

yellow+red=orange

the orange note is the last one

[alpine island]

That's red

[coral shale]

that might be a dark orange on some screens i see now, but fairly red

onto aka, surjective, (being more sensible with naming)

==
"function into A" means A is the codomain
"function onto A" means A is also the image

[coral spindle]

Another problem that could've been avoided if only people weren't so scared of using the word "surjective"

[coral shale]

Also, this isnt #groups-rings-fields but more like #proofs-and-logic

"one-to-one" vs "one-to-one correspondence" moment and more bullshit

[coral spindle]

But Shuri you don't understand... a word with more than three syllables is just too much for undergrads!!!

[coral shale]

then we have people who apparently use range to mean codomain rather than image!

absolutely bananas

[coral spindle]

Criminal

10 000 years jail

[mighty spade]

“correspondence”

[coral spindle]

[coral shale]

Now wait until you hear about the french who agree with us that the naturals are the positive integers.

[coral spindle]

Tbh I do like that we can call them positif and not nonnegative. That word is annoying.

[coral shale]

But yeah, then we have increasing meaning non-strictly

[coral spindle]

Aha, but we can just say monotone

[coral shale]

wait.

i thought monotone means non strictly

it just means your sequence is either all increasing or all decreasing

anyways, that's for the analysts to worry

[warm urchin]

call them weakly positive

[chilly ocean]

Hiya, if i take an invertible matrix A how can i see that it's in the orbit of diag(det(A),1,1,…,1) the action being left-multiplication by SL(n)??

[delicate orchid]

SL(n) is the kernel of the determinant map, hence GL(n) is partitioned into cosets of SL(n), one for each value of the determinant. These cosets are clearly fixed by left-multiplication by SL(n), and SL(n) acts transitively within each one, hence these are orbits of the action of SL(n) on GL(n). From here it's you can see that because each matrix of determinant x is in the coset diag(x, 1, 1, ..., 1)SL(n) that your result holds.

[lime oriole]

It includes zero

[hollow mica]

Where in the field of fractions construction is the integral domain condition used?

i.e. where does it go wrong for a plain ring with no other guaranteed structure

[chilly radish]

If R\{0} has zero divisors in it the construction will give you back the 0 ring

Or sorru

Not necessarily the 0 ring

Let me think about the specifics for a minute

[delicate orchid]

it's just not local right, or rather not a field in this case

[chilly radish]

The point is that inverting zero divisors is not possible

Without fucking stuff up

In particular you'll not get a field

Oh yea sorry, it'll definitely be the 0 ring in this case

[hollow mica]

Let R be a ring, and let F = R x (R \ {0}).

• Define the equivalence relation ~ on F by declaring (a, b) ~ (c, d) iff ad = bc.
• Define addition on F by declaring (a, b) + (c, d) = (ad + bc, bd)
• Define multiplication on F by declaring (a, b) * (c, d) = (ac, bd)
• Declare the additive identity to be (0, 1)
• Declare the multiplicative identity to be (1, 1)
• Define the additive inverse of (a, b) to be (-a, b)
• Define the multiplicative inverse of (a, b) where (a, b) not ~ to (0, 1) to be (b, a)

[chilly radish]

So, what happens is, if you have zero divisors, R\{0} is not multiplicativelu closed

So the construction doesn't even go through

[hollow mica]

oh like

when you multiply fractions

you get zero in the denominator sometimes

[chilly radish]

Yea

[hollow mica]

yea that'd be pretty bad

[chilly radish]

And that's not possible

So you can't even go through with the construction

[delicate orchid]

localise at R

[chilly radish]

Also, if you try and do a similar construction eith a smaller set than R\{0}, and your ring isn't an integral domain, then you need to modify the relation

So that it'll actually be an equivalence relation

You can generally take a "total ring of quotients", which is a similar construction localising (inverting) all nonzerodivisors

And this always works, as long as you modify the equivalence relation

The modification is such: (a,b)~(c,d) if there is n in the set we're localising (so in this case a nonzerodivisor) such that adn=bcn

When your ring is an integral domain this reduces to what we had before since we can just cancel n on both sides

In general, because we can't cancel showing transitivity is not possible unless we add this modification

You can think of this is the fact that you can expand the numerator and denominator by multiplying them by the same element

Or cancel a common element

[hollow mica]

hm ok I don't know about localization besides what you wrote here and some Wikipedia

But it's saying you can localize wrt any multiplicative set

meaning 0 could be inside that set

so I guess it's fundamentally different than the field of fractions construction

[coral spindle]

Yes, zero could be in that set, and you'd get the zero ring.

In fact you can show that the natural map from R into the localisation of at some multiplicative set is injective iff the set contains no zero divisors — if I remember correctly at least.

[delicate orchid]

this sounds correct, because if a is a zero divisor you have the unit messing things up in the equivalence relation

[coral spindle]

It's really not fundamentally different from the field of fractions construction though. With a very small adjustment the construction looks exactly the same, and in fact as mentioned if you localise at a particular prime of an integral domain (i.e. 0) then you just get the field of fractions

And in fact localisations of integral domains are just particular subrings of their fields of fractions

[delicate orchid]

wait I wanna try and do this properly

[coral spindle]

The map is r → r/1 and r/1 ~ 0/1 iff there is some element s such that s(r - 0) = 0 right

So there you go

[delicate orchid]

some unit s, yeah

[coral spindle]

If there's a zero divisor s in the set, the map isn't injective.

[delicate orchid]

I think it's a unit

[coral spindle]

[delicate orchid]

googling...

[coral spindle]

Yeah it's any element of the multiplicative set

so it must become a unit

[delicate orchid]

ok good

then it works :wholesomechungus:

[coral spindle]

I misunderstood bc I thought you meant some unit in the original ring

[coral spindle]

don't make me bring out the 🇺 🇵 🇴 🇳 🇹 🇭 🇪 🇼 ℹ️ ✝️ ♑ 3️⃣ 🇸 💲 🇮 🇳 🗜️

[delicate orchid]

stolen

[coral spindle]

Indeed I am a thief

[hollow mica]

Ah so then it doesn't embed?

[coral spindle]

Yup

Not in the natural way at least

[hollow mica]

Oh it could still embed in a fucky way lol that's scary

[coral spindle]

Well idk maybe

[rocky cloak]

I mean you could do something stupid like take your ring to be R^N then localize with respect to elements where only the first coordinate is 0. Then the localization is again R^N, but the localization map just shifts everything down and kills the first coordinate

[delicate orchid]

that's awful

[crystal turtle]

jagr honourable

[rocky cloak]

Woopwoop

[coral spindle]

You can never leave this server just like us

[crystal turtle]

ryx was wondering when jagr would ascend

[rocky cloak]

Not sure what that entails, but I appreciate it

Still not G+ though

Maybe next year

[coral spindle]

I imagine the backlog is immense

[crystal turtle]

idk there are fresh accounts with G+

[elder wave]

your application probably didn't go through

[hollow mica]

I rejected it

[rocky cloak]

Should I like toggle the button or...?
Doesn't really matter anyway I guess

[elder wave]

fixed it

[chilly radish]

It's not too bad actually

We have a few small logistics problems to iron out

[rocky cloak]

Alright, now I can enact my secret plan of spamming the graduate lounge with links to my sick pyramid schemes

[chilly radish]

It was really bad when we opened it up to the public

[elder wave]

indeed

[coral spindle]

Sick like "sick bro!" or sick like "twisted" I NEED to know bc I need money thanks

[delicate orchid]

sick cohomology

[coral spindle]

Aho-Corasick

[elder wave]

the average application was/is "math" or "book" in all three questions

[coral spindle]

Bet someone spammed it

[torn warren]

I have a question here, the $\sigma(\alpha_2)$ is undefined? Maybe $\alpha_2$ is not in $L_1$

[cloud walrus]

WT

[rocky cloak]

Should be sigma(alpha_2^n_2)

[blissful zinc]

Hi, dumb question (don't hit me pls):
Let A a ring, B a subring of A (assuming all conditions are verified).
Can we see B as a ring too? 🤔

[torn warren]

thank you!

[dusky patio]

@blissful zinc B must be a ring in that case. Subring conditions are just those ring properties that one needs to verify because they are not already implied by A (?)

[torn warren]

subring is a ring

[blissful zinc]

okay ty for not hitting me because of that question

[delicate orchid]

subobjects are objects yeah

[torn warren]

subway is way?

[white oxide]

lol i remember struggling with this proof

took me like 5 reads to understand it

[torn warren]

I compare different editions and this typo seems is always there

[white oxide]

interesting

how are you finding fraleigh for galois theory

[rocky cloak]

I mean usually you'd expect
sigma(a)^n = sigma(a^n)
So from that perspective it makes sense I guess

[torn warren]

does that mean the author wants to extend sigma from G(L1/F) to G(F-bar/F) ? But in this step, he only mentions sigma is in G(L1/F), then sigma(alpha2) is undefined..

[torn warren]

there is a chapter?

[rocky cloak]

Nah, I just meant from like a notation perspective

[grizzled crane]

What are the importance of Klien-4 group? Other than elements being self invertable?

[open sluice]

it's the only other group with 4 elements

[torn warren]

but can we interpret sigma as some automorphism tau in G(Fbar/F) and sigma is hence the restriction of tau in G(L1/F) ? but even by this way, the sigma(alpha2) is still undefined in G(L1/F) , but the value do equals sigma(a)^n = sigma(a^n)

[rocky cloak]

Sigma can be extended to Fbar, but the extension isn't unique, so sigma(alpha_2) still wouldn't really be well defined

[rocky cloak]

Smallest non-cyclic group,
normal subgroup of S4 (thus only non-trivial normal subgroup of Sn that isn't An),
galois group of x^4 + 1
Not sure why it gets a special name though. Think that's just historical reasons

[delicate orchid]

it is just historical iirc

[grizzled crane]

how do I prove, it's the smallest?

Any group G whose cardinality is less than 4 is always cyclic ?

[delicate orchid]

it's small enough you could literally just exhaust every option

or use the fact that 2,3 are prime

[grizzled crane]

say for |G|= 3

[delicate orchid]

do you have lagrange's theorem

[grizzled crane]

G={e,a,b}

[grizzled crane]

not yet

[delicate orchid]

ok fair enough

[grizzled crane]

clearly, a and b must be inverse pair

else it will contradict |G|=3

right?

[delicate orchid]

yeah so you do have lagrange's theorem

[delicate orchid]

and this completely determines the structure of the entire group!

[grizzled crane]

i haven't studied it yet

i just know the statement

[delicate orchid]

you can use the same train of logic to show that any group of order p is cyclic

[grizzled crane]

The order of a subgroup must divide the order of the group

[delicate orchid]

yeah exactly

[grizzled crane]

because only subgroups are {0} and p ?

[delicate orchid]

yup, or since each element is order 1 or p, each element is a generator!

[grizzled crane]

yes

oh so you mean those elements generates G

[delicate orchid]

everything non-trivial generates G yeah

[grizzled crane]

is there a cyclic group of 3 elements other than Z3?

[delicate orchid]

it's a very trivial result that all cyclic groups of the same order are isomorphic

[torn warren]

when you say not-well defined, can I interpret is as, given two conjugate mappings, $\tau_1$ and $\tau_2$ in Fbar, where $\tau_1(\alpha_2)\neq \tau_2(\alpha_2)$, and the restriction $\tau_1=\tau_2=\sigma$ in $L_1$, but we have $\tau_1(\alpha_2^{n_2})= \tau_2(\alpha_2^{n_2})=\sigma(\alpha_2^{n_2})$

[grizzled crane]

I don't think there is any multiplicative group of order 3, because totient functions are always even

[delicate orchid]

? confusing statement

there are plenty of groups of order 3, they're just all isomorphic

[grizzled crane]

I mean, U(n)

[delicate orchid]

take for example the rotations of a triangle

[dim widget]

I think they mean Z/n^x

[cloud walrus]

WT

[delicate orchid]

my point is who cares

[grizzled crane]

this cannot have an group having even order

[delicate orchid]

if it's order 3 it has to be cyclic that's what we've just shown

[delicate orchid]

you mean odd?

[grizzled crane]

yes

[grizzled crane]

^

[dim widget]

Ah I see

[delicate orchid]

|Z/2Z*| = 1?

other than that though I buy it

[dim widget]

@grizzled crane . Do you care about groups or groups up to isomorphism

?

[grizzled crane]

upto isomorphism

just looking for some solid examples

[dim widget]

I would say if not the latter wew can teach you why it should be the latter

[delicate orchid]

all categories are skeletal

[dim widget]

There is only one group of order 3 up to isomorphism

It is Z/3

It is never the multiplicative group of Z/n but it is a subgroup of Z/7^*

Generated by 2

[torn warren]

hello?

[delicate orchid]

HAI!! :3

[rocky cloak]

Yes, indeed

[torn warren]

big big thank you!

[torn warren]

you changed name?

[delicate orchid]

I have been called this since 2013 on every platform you can think of

[torn warren]

but I remember your mid name seems changed?

[crystal turtle]

Wew lids tbh

[delicate orchid]

oh yeah that was for halloween

[delicate orchid]

MODS

[torn warren]

every platform? Does it include this one?

[delicate orchid]

especially that one

[summer path]

what's a parental kidnapping

a parent napping with their kid?

[torn warren]

send me your address then I will be rich

[delicate orchid]

:uponthewitnessing:

[torn warren]

oh maybe the reversed? kid nap parent?

[summer path]

[fair lantern]

Hello! Could someone guide me through the following exercise? Let $M = \mathbb{R}^2$ be considered as an $\mathbb{R}[x]$-module through the linear transformation $T(x, y) = (0, y)$. Find all the submodules of $\mathbb{R}^2$.

[cloud walrus]

Kenny

[delicate orchid]

ok well first do you understand how R[x] is acting on R^2?

[tardy hedge]

How can i interpret the fact that if a group is simple, the only homomorphisms out of the group are either one to one or trivial

[delicate orchid]

all non-trival representations of a simple group are faithful

[void cosmos]

consider the kernel

[delicate orchid]

or even better, if there's a map from a simple group to anothr group, that map has to be injective

I really don't understand what you're asking tbh

[tardy hedge]

I guess, a way to interpret it

What i said was correct right?

[void cosmos]

consider the kernel

[tardy hedge]

In a simple way

[delicate orchid]

This is a meaningless question

Are you asking for a proof?

[karmic moat]

you are a meaningful statement

[delicate orchid]

Like, you could interpret it as viewing simple groups as not being able to be split into any smaller parts?

[tardy hedge]

Yes something like that^ is what i was trying to think of

[delicate orchid]

I’ll let tropo post cause he’s clearly got something in mind

[tardy hedge]

Yeah i was thinking of them as this

[void cosmos]

XD?

consider the kernel?

[tardy hedge]

Yeah the kernel either all of G or its just e

[void cosmos]

what about a magnet

[void cosmos]

which gives u what ur saying

thats it

[void cosmos]

what reference is this

[tribal moss]

There definitely are rings whose multiplicative group has order 3: the field with 4 elements is an example. (This can e.g. be constructed as the quotient ring Z[X]/<2, x²+x+1>).

[karmic moat]

this is how i feel when jagr starts typing

[tardy hedge]

Ok fair, i guess im just trying to have an intuitive sense of things as well

[delicate orchid]

He just meant Z/nZ* tropo

Even there I pointed out n = 2 but it didn’t seem to bother him LOL

[tribal moss]

Yeah, but it might be interesting to him all the same.

[delicate orchid]

true

[tribal moss]

Sorry I didn't get up the the later discussion. What was I supposed to be typing? :-)

[delicate orchid]

Some mind bending insight into simple groups

[void cosmos]

||not cool||

[tribal moss]

I'll let you know if I ever understand the Monster.

[tardy hedge]

@delicate orchid i was thinking of something like this fyi

So if i were to ask a question like this again how should i phrase it

I guess to people who have studied it for a while this type of thinking is trivial

But not for someone whos learning for first time

[delicate orchid]

Ok then yes there is an analogue to prime numbers using decomposition series

[tardy hedge]

Thats cool

[tribal moss]

The slight trouble with that explanation is that even if we know that G/N = H for some normal subgroup N, and we know all about N and H themselves, that is not in itself enough to know which group G is. That situation is different from building the integers from primes.
It is still better than nothing, though.

[tardy hedge]

Cause u dont know still the specific elements and operation that make up G ?

So u dont know G in that sense?

Oh its more than that?

[tribal moss]

For example suppose I tell you I'm thinking of a group that is "built from" the cyclic groups of order 7 and 2 in the sense that my group has normal subgroup isomorphic to C_7 and its quotient is C_2.
This does tell you something about my group, but it's not enough for you to know if I'm thinking of the cyclic group of order 14, or the dihedral group of order 14.

Figuring out exactly what we can say about a group when we know a normal subgroup and its quotient is the group extension problem, which as far as I know is not yet satisfyingly solved in general.

[sonic coral]

So i have shown that if G is a finite group of order n and there is some subgroup of G which has index k in G, then if n does not divide k! then G is not simple

how can i adapt the proof so that if n does not divide k!/2 then G is not simple

for the proof i have done, I acted on the left cosets with a permutation representation with a map from G to S_k and considered the kernel of the map

but i guess I can do the same thing but into A_k?

[chilly radish]

Ermm askchually it's COMPOSITION series not DECOMPOSITION series

[chilly radish]

I'm pretty sure the extension problem in general is undecidable

Or something to that effect

[tribal moss]

Hmm, if it is, then at least the Wikipedia article doesn't seem to deign to mention it.

The group-related problem that first comes to mind as undecidable is the word problem for groups, but that's not obviously the same as the extension problem.

[chilly radish]

I'm trying to find something on it rn because I remember hearing this from a credible source, but I can't seem to find anything

At least, the theory of noncentral extensions is really hard to develop in general because we can't apply regular homological algebra

[tribal moss]

"Really hard" does match what I believe I've heard. :-)

[chilly radish]

Wdym?

But even for central extensions the problem is not solved in any satisfying manner

[icy bear]

no odd order group is abelian????!?!!!??

[chilly radish]

What?

All cyclic groups of prime order

[icy bear]

wait no sorry

[tribal moss]

I mean I have no objection to calling it "really hard" rather than calling it "undecidable".

[chilly radish]

All odd order groups are solvable

[chilly radish]

Oh I misread does as doesn't

[chilly radish]

It's interesting tho, I can't find anything about the decidability of this problem

I should maybe do a more thorough literature review

If only I had mathscinet outside of campus

[tribal moss]

Whether a given finite group is an extension of this group by that group, should be trivially decidable, at least. The infinite case I have no idea about, and I could buy that could be undecidable under some appropriate input encoding.

[icy bear]

this may be a weird ass question but is it true that every cyclic group has at max one element other than the identity that is its own inverse?

[tribal moss]

Yes.

[icy bear]

oh shit

I was expecting a no

actually I think I know how to prove it

OH I KNOW

it's true in addition modn

because everytime n is even, n/2 + n/2 = 0modn, when it's odd it's just the identity

and every ciclic group is addition modn in desguise because it's addition modn with the ""exponents""

[tribal moss]

Yes.

[icy bear]

I noticed a pattern suddenly and my heart told me this and now I proved it I love math so much

[torn warren]

Does the auther skip some steps here? if $x^n-1=(x-a_1)...(x-a_k)(x-b_1)...(x-b_t)$, and all these $a_i$ are primitive roots, and all $b_j$ are non-primitive roots, since all automorphisms permute primitive roots with primitive roots, hence, all automorphisms permute non-primitive roots with non-primitive roots, this means the product $g(x)=(x-b_1)...(x-b_t)$ is invariant under all automorphisms, so $g(x)$ is in $Q[x]$

[cloud walrus]

WT

[chilly radish]

I was thinking more given two groups how many extensions are there up to isomorphism

[chilly radish]

The author says this above, not sure why you're asking if they're skipping steps

[torn warren]

the author says the $(x-a_1)...(x-a_k)$ is in $Q[x]$, but why he can immediately claim it is a divisor? we need to show the rest $(x-b_1)...(x-b_t)$ is also in $Q[x]$, right?

[cloud walrus]

WT

[chilly radish]

He specifically says the cyclotomic polynomial is on Q[x]

The point is that the galois group not only sends roots of unity to roots of unity

It sends primitives to primitives

[tardy hedge]

Ikr

Math is awesome

[torn warren]

but he claims the cyclotomic poly is a divisor of x^n-1, that's what I am asking for

if $x^n-1=\Phi(x) g(x)$, and he shows $\Phi(x)$ is in Q[x], but we still need to show g(x) is in Q[x] then it is a divisor

[cloud walrus]

WT

[chilly radish]

Well, if the galois group permutes all the roots of unity, and it permutes all the primitive roots of unity, where do the nonprimitive roots go?

[torn warren]

nonprimitive permute with nonprimitive

[chilly radish]

Yea

So the other part will also be a polynomial in Q

[torn warren]

yes, so this is what I am asking for, he skips this...

[chilly radish]

Sure

I wouldn't call it skipping

It's just the same argument again

[torn warren]

is there an elementary way to prove this? I mean if some poly f, g, h satisfying f=gh, and if we know f and g is in Q[x], then h is in Q[x]

[chilly radish]

Generally if g,f are in F[x] and f divides g in some extension, then just do division with remainder in the original field, the remainder has to be 0 in the extension so the other divisor is also in F[x]

Another way: gcd is invariant under extensions, so if f|g in some extension then gcd(f,g)=f, but this is also true in the original field, so f|g in the original field

[torn warren]

Let E be the extension field, and F be the oringinal field, and f, g are in F, so in E we have f=gh, and also we do the division in E, and get f=gq+r, so we get gh=gq+r, gives g(h-q)=r holds in E, but the degree r< degree g, so h-q=0, hence r=0, but this is still on E, we still need to show h is in F ...

[chilly radish]

Uh no, do the division with remainder in F

So q is in F

Then use the fact that division with remainder is unique

So f=gq+r with q,r in F, but also f=gh, so over E we have both forms and by uniqueness h=q\in F and r=0

[torn warren]

ah, right , i see, that's smart

[chilly radish]

(You argued this well enough)

It's just one of those standard tricks

[next frost]

hello

Consider a set $X$ equipped with a unary operation $\tau$. We define the equivalence class $[x]:={y\in X \left| \langle y\rangle=\langle x \rangle \right.}$.

[cloud walrus]

huchaney

[next frost]

where $\langle x\rangle$ is the orbit of $x$ under $\tau$
I wish to prove that for orbits with $\langle x \rangle = [x]$, $\tau$ is bijective on $\langle x \rangle$

[cloud walrus]

huchaney

[next frost]

surjectivity is pretty simple

injectivity seems very simple but I can't write a proof

help

it makes so much intuitive sense

I don't know if this is true or not

but it seems like it should

[void cosmos]

<x> = {t^n(x) | n is a positive integer}?

where t^n is compositoin?

composition*

@next frost

[next frost]

what is t?

tau?

my prof won't let us use integers

[void cosmos]

yea

[next frost]

lmfao

[void cosmos]

what

wdym

[next frost]

literally

[void cosmos]

whats <x> then

[next frost]

he says notation tau^n is "informal"

[void cosmos]

ig he is right

[next frost]

orbit of x

formally defined

[void cosmos]

yea whats that

[next frost]

as the smallest invariant subset of a $(X,\tau)$

[cloud walrus]

huchaney

[void cosmos]

all elements y such that t(y) = x or what

[next frost]

containing x

[crystal turtle]

what

[void cosmos]

ohh

yeah this class seems cool

not for me tho

so its the smallest subset ,<x> such that t(<x>) = <x> ?

whats <x> tho

what are elements of <x>

[next frost]

"tau^n" is formally defined by the unique homomorphism from an initial object of the category of nullary unary structures to the relevant set you are considering

[chilly ocean]

[next frost]

well <x> isn't an element of X so t(<x>) isn't defined

[void cosmos]

im sorry

[next frost]

it'd be $\tau_* \langle x \rangle \subseteq \langle x \rangle$

[void cosmos]

i meant whats <x>

[next frost]

woop

[cloud walrus]

huchaney

[next frost]

and $x\in \langle x \rangle$

[cloud walrus]

huchaney

[next frost]

the smallest subset satisfying this property

[void cosmos]

<x> is the orbit of x under t , so its the set of all y , such that t(y) = x?

[void cosmos]

whats t_*

[next frost]

direct image function

[void cosmos]

yeah this seems out of my scope

sorry

its a jagr is typing kind of stuff we need

[topaz solar]

This @void cosmos?

[void cosmos]

yeah

deal with that

[chilly ocean]

this prof sounds like a funny guy

[void cosmos]

once i saw this i knew this isnt for me

lmfaaao

[next frost]

funny indeed

I am about to pull all my hair out

[void cosmos]

ig too much doxy but seems a nice guy man

[next frost]

I have a midterm tmrw

[void cosmos]

looks like someone is fucked

[topaz solar]

noncommutative

Ok yeah checks out

Ok what exactly do you want chaney?

[next frost]

lmao an answer to my original question

[topaz solar]

Restate it pls

[next frost]

hold on

[void cosmos]

he wants to prove that if <x> =[x] then tau is bijecitive on <x>

where <x> is the orbit of tao

and [x] is = {y in X | <y> = <x>}

[topaz solar]

<x> the set of things which can be reached by t?

Do we consider anything in <x> as like, y such that ty = x

Consider X = N, t(x) = x+1

Each natural has a different orbit if we don’t including things which map to it, so [x] = <x> is essential

Clearly there exists y in <x> such that ty = x, otherwise y in <x> but not in [x]

[void cosmos]

sorry for interrupting just curious , what is this course's name?

[topaz solar]

Garbo complaint btw, just define $\tau^\omega[A] = \bigcup_{n<\omega} \tau^n[A]$ where $\tau^0=$id tbh

[cloud walrus]

Dragonslayer Sharp

[topaz solar]

And each t^n is easily definable in itself

By the obvious method ofc

[next frost]

it's just introduction to abstract algebra

[void cosmos]

sad

[next frost]

but we spent half the semester doing category theory

[void cosmos]

would be cool if it was ilke

advanced algebra or something

or idk maybe u guys already know basic algebra

seems cool tho gl hf

[topaz solar]

Well, you know it’s surjective ye?

The silly tau thing

For injective, literally just suppose y, z both map to some w, then how does that doom you since functions (tau) are single valued

[torn warren]

thank you!

[chilly radish]

of course!

[hollow imp]

you have wodzicki as a professor?

[cobalt heath]

Yea this is next level 💀

[cobalt heath]

But can they use the notion "Injectivity"?

[white oxide]

damn bro he’s a goner

[topaz solar]

Well he mentioned it earlier and wants to show it’s bijective so

[cobalt heath]

Yea he mentioned the prof was extra picky, so I thought maybe prof does not like taking an element

[fair bay]

struggling on this one

not sure how to bring the n across k_1 in n_1k_1 nh

since we only have normality of K in H

[mighty spade]

yeah, but N is normal in G

so kN=Nk for all k in g, a fortiori k in K

[fair bay]

but wouldn't that change n to some other n'? don't we want nh NK = NK nh for any n, h

and we would want nh nk = n'k' nh

[mighty spade]

It would in general, yes.

[mighty spade]

Here’s a hint:

||Show NK=KN||

[teal vessel]

proof by multiplication table

every distinct product of elements of order 2 produce an element of order 3, and since the two elements of order 3 are inverses of each other, both must be included. Further, the directional products of any 3-element with a 2-element produce the other 2-elements.

I'm sure there's some symmetry that's more precise here that lets me prove this without being exhaustive about it, but when the group is small enough, sometimes brute force is worth it.

[delicate orchid]

S_3 is rank 2 as a relfection group :pack:

[teal vessel]

sounds interesting, that'll be cool to get to.

[chilly ocean]

Can we really use nakayama here? It's to show that a semisimple ring is semiprimitive

[delicate orchid]

Nakyama's states that J(R)M = M => M = 0, so yeah

[chilly ocean]

but M is finitely generated no?

[delicate orchid]

R semisimple => R is artinian, so R has a finite length over itself, hence every maximal ideal is f.g, and again because R is artinian their intersections are f.g.

thus J(R) is f.g. by induction

[chilly ocean]

ohh thanks!

[delicate orchid]

no worries

[tardy hedge]

Helloo im the nostalgia critic

[teal vessel]

question to talk through things:
the group generated by a subset of elements from some group G must be a subgroup, yes?

1. it has to be a group, and uses the same operation
2. the generators are part of G, and therefore all products of them have to be part of G for G to be a group
3. inverses of the generators and their products must also be in G same as 2. (along with identity)
   I'm not missing anything special am I?

[south patrol]

I'd do lagrange

And that sort of trickery

[teal vessel]

I'm more asking basically that there's no way to generate a group that is larger than the group from which the generators are taken

[south patrol]

Once you have two elements of order 2, you have a subgroup of at least 4 elements, so it must be whole thing

[south patrol]

That they form a group is essentially by definition

[teal vessel]

the group formation is not in question, it's the size and membership, but just typing it up got me there I think

[south patrol]

Oh so what is the first line doing then sure

[south patrol]

No

Again this often holds by definition

How are you defining the subgroup generated by a subset of a group

[delicate orchid]

if you could, then this immediately contradicts G being closed

[teal vessel]

yeah, thought so. I don't trust my own head for some reason

[delicate orchid]

but potato is right in that you can see it by the definition as well

[south patrol]

I would define it as the intersection over all subgroups of G containing that subset

[delicate orchid]

I'm presuming you're using the "intersection of all subgroups that contain this set"

yeah ok

[south patrol]

In which case it is by definition a subset of the group

[delicate orchid]

intersection of all groups that contain the subset

[crystal turtle]

Intersecting a class of sets

[delicate orchid]

skill issue if u can't do it

[delicate orchid]

holy smokes that's a lot of inequalities

[chilly ocean]

I'm thinking you can do something similar to this hint

to get a linearly independent wedge product

but I'm not sure where to go from there, and then I don't know if that's actually related because I don't really get out coefficents

[crystal turtle]

Holy moly is that a hint or the entire proof

[chilly ocean]

basically the entire proof

because the only thing that's missing is like show that the two wedge products aren't 0

at the end

is anyone able to help?

[dim widget]

this is not really group theory @chilly ocean

you would have more luck in #diff-geo-diff-top or #advanced-algebra

[chilly ocean]

that's fair, this is just from my algebra class

okie

[dim widget]

it's not really a #diff-geo-diff-top question but probably is most likely to be answered there

[delicate orchid]

thie channel is fine for it

It's just that I ain't reading all that! (not like I'd be capable of solving it anyway)

[chilly ocean]

the hint isn't relevant

It's just this problem if you have any idea how I should start

[delicate orchid]

so simple means that a wedge a = 0 right? Inferring from context

[chilly ocean]

yea

[dim widget]

okay I will bite. So if $\alpha$ is not simple then $\alpha = \beta \wedge \gamma$, write $\beta, \gamma$ as $\sum_{i} b_i e_i, \sum_j c_j e_j$ and write $\alpha = \sum_{i < j} a_{i, j} e_i \wedge e_j$

[cloud walrus]

n-Coskeletal E-Girl

[dim widget]

this is always true though

[delicate orchid]

right but you've said that a = b wedge g, which is a way more natural thing to call "non-simple"

good

[chilly ocean]

so a_ij = bi \wedge cj?

[dim widget]

no

[delicate orchid]

you can't wedge scalars

[chilly ocean]

oh right

so just the product

[dim widget]

the wedge product yes

[delicate orchid]

i have a feeling this is just going to boil down to some annoying swapping of e_i and e_j

[dim widget]

so now we forget about $\alpha, \beta, \gamma$ and instead think of $\alpha, \beta$ in terms of a pair of functions $b, c: {1, \dots, n} =: [n] \to K$ and $a: [n] \wedge [n] = {(i, j)|i, j \in [n], i < j} \to K$

[chilly ocean]

woah wtf lol

[delicate orchid]

yeah what the fuck

[dim widget]

well the identity is purely combinatorial so i was going to prove it purely combinatorially

[chilly ocean]

hmm

I don't really get what's going on in that formulation

[dim widget]

i'm just thinking of the $a_{i, j}$ as a function on pairs $1 \leq i < j \leq n$

[chilly ocean]

oh ok

[cloud walrus]

n-Coskeletal E-Girl

[chilly ocean]

uhh

[cloud walrus]

n-Coskeletal E-Girl

[delicate orchid]

what's K? the base field?

[dim widget]

wait im actually really confused here

[chilly ocean]

oh shit

sorry

[dim widget]

what in the world is your definition of $\wedge^2 V$ if $v \wedge v$ is not always 0

[chilly ocean]

that hint was for a previous problem

[cloud walrus]

n-Coskeletal E-Girl

[delicate orchid]

they're doing proof by contradiction in that hint?

[delicate orchid]

wait

what

-x = x doesn't imply x = 0 in a field moment

[cloud walrus]

n-Coskeletal E-Girl

[chilly ocean]

like in our class?

[dim widget]

Ah I seeeeeee

okay yes this makes sense

[chilly ocean]

This was the first problem we did, I've already done this one

[dim widget]

yes okay

[chilly ocean]

I figured the hint from part a would be helpful since it has a similar formulation, but it's not really interesting since we're not necessarily working in Lambda^2 here right?

[dim widget]

sorry wew and I were being dumb

[delicate orchid]

I still don't get how we can have alpha \wedge alpha be non-zero

[chilly ocean]

if it's not simple

[delicate orchid]

is there something here that is different to the normal exterior algebra I'm missing?

what

hold on

[dim widget]

so the thing is that \wedge^2 \wedge^2 is not \wedge^4

[tribal moss]

Suppose V is 4-dimensional, and consider alpha = (e1 wedge e2) + (e3 wedge e4)

[delicate orchid]

v_1 \wedge v_2 + v_3 \wedge v_4

yeah ok

duh

[delicate orchid]

I'm stuff

[chilly ocean]

so true

[dim widget]

yeah the only relation is that you have wedges of odd degree things with themselves have to vanish

even degree is fine

[delicate orchid]

these were fun to do at least!

[dim widget]

Anyway okay

[chilly ocean]

problem 1 made a lot more sense to me lol

[dim widget]

let's get back to the original problem

[chilly ocean]

okie

[dim widget]

So we have $\alpha = \sum_{i < j} a_{i, j} e_i \wedge e_j$

[cloud walrus]

n-Coskeletal E-Girl

[chilly ocean]

right

[dim widget]

So then $\alpha$ defines a nxn matric where $V$ has dimension $n$

[cloud walrus]

n-Coskeletal E-Girl

[dim widget]

specifically, it is the skew symmetric matrix with coefficients $a_{i, j}$ above the diagonal

[chilly ocean]

I agree, although I have a feeling this might turn into what question 3 is trying to show

kinda

[cloud walrus]

n-Coskeletal E-Girl

[dim widget]

what is question 3?

[chilly ocean]

actually rereading I don't think so, but I'll send it

[chilly ocean]

wait, why is this true?

[delicate orchid]

yeah this looks like the same thing

[dim widget]

this is the other direction

this is saying if you are reducible then you have rank < 2

or no maybe it is the same

ugh okay

[delicate orchid]

this question is an if and only if?

yeah ok

[dim widget]

ah i didn't see the question

[delicate orchid]

wedge product is alternating so the matrix has to be skew symmetric right?

[dim widget]

well also we cannot really define the a_{j, i} in a sensible way besides that

[chilly ocean]

ah fair

[dim widget]

Okay well I will explain how problem 3 implies problem 2

then do problem 3 with linear algebra

[chilly ocean]

lmfao

ok

[dim widget]

So $A = (a_{i, j})$ defines a matrix $A: V \to V$, and the equation $\alpha \wedge \alpha = 0$, worked out in coordinates, says that the 3x3 minors of this matrix all vanish, or that $A$ defines the $0$ map $\Lambda^3V \to \Lambda^3 V$

where the map is induced by taking $A(e_i \wedge e_j) = (Ae_i) \wedge (Ae_j)$

[cloud walrus]

n-Coskeletal E-Girl

[dim widget]

this is equivalent to this matrix having rank <= 2

by usual linear algbera

So in this case we can choose a basis in which the image of $A$ is contained within the span of the first two basis vectors $e_1, e_2$

[cloud walrus]

n-Coskeletal E-Girl

[chilly ocean]

what 2x2 minors does alpha wedge alpha mean? Like why specifically are we working in Lambda2?

[dim widget]

oh

[cloud walrus]

n-Coskeletal E-Girl

[dim widget]

so the mxm minors of a matrix are the determinants of the $mxm$ matrix obtained by deleting some k rows and k columns

[cloud walrus]

n-Coskeletal E-Girl

[dim widget]

where k = n-m

[chilly ocean]

right right

uhhh

I guess specifically where does 3 come from in the formulation of this question?

[dim widget]

so okay

let's think about this when V is dimension 4 maybe

[chilly ocean]

ok

sure

[dim widget]

okay so this comes from some linear algebra, but I didn't know this was unfamiliar to you

[chilly ocean]

ah

[dim widget]

it is definitely not always part of a linear algebra class

[chilly ocean]

I've taken an elementary linear algebra course, this course I'm taking now is abstract algebra so we don't really touch that much on like proof based LA

so like we only did computational eigenvalues, determinants, etc stuff

not actually theory

[dim widget]

I see

do you know that the rank of a skew symmetric matrix is even?

sorry typo

[chilly ocean]

ok haha, I was questioning it for a second. I think I can prove that

[dim widget]

okay

great

so there are some vectors $e_1, \dots, e_r, f_1, \dots, f_r$ such that the $e_i, f_i$ span the image of our matrix $A$

[cloud walrus]

n-Coskeletal E-Girl

[chilly ocean]

ok

I agree

[dim widget]

wait is k characteristic 0?

This effects the ammount of work we have to do

[chilly ocean]

all I have is the dimension of k, it's not necessarily finite so idk

[dim widget]

okay then

hmm this is hard to do without algebra

what I want to say is that if we base change to the algebraic closure then A^2 is symmetric

[delicate orchid]

k is characteristic 2 :thecosmoshumswithatunemostsweet:

I managed to prove the hint

[dim widget]

So it can be diagonalized

[south patrol]

base change

found the alg geometer

[dim widget]

then if A^2 is diagonal [possibly non scalar] we see that A is block diagonal with 2x2 blocks

so we can find vectors $e_1, \dots, e_r, f_1, \dots, f_r, g_1, \dots, g_k$ where $Ae_i = \lambda f_i$ and $A g_i = 0$

[cloud walrus]

n-Coskeletal E-Girl

[dim widget]

but there must be a better argument for this using just combinatorics

Another way to do it is to use the theory of the pfaffian but i doubt this makes sense if minors are too confusing

[chilly ocean]

sorry, I have to go to class real quick. Thanks for the help! I'll look back and see if I can understand what you're suggesting better afterwards

[delicate orchid]

I should really be doing research but this has properly nerdsniped me

if you do decide to write up your answer tteg can you put it in spoilers it for me

[summer path]

nceg?

[delicate orchid]

?

ok I think I've got it

let $\alpha = \sum_{i < j} c_{ij}e_i \wedge e_j$, then $\alpha \wedge \alpha = \sum_{i < j}\sum_{k < \ell} c_{ij}c_{k\ell}(e_i \wedge e_j \wedge e_k \wedge e_{\ell})$, we want to write this sum in the basis $e_i \wedge e_j \wedge e_k \wedge e_{\ell}$ with $i < j < k < \ell$, so if we have $k < j < \ell$ we need to swap $k$ and $j$ around, and if we have $j > \ell$ we need to permute $j \mapsto k \mapsto \ell \mapsto j$. Note the first permutation is odd and the second is even, so
$ \sum_{i < j}\sum_{k < \ell} c_{ij}c_{k\ell}(e_i \wedge e_j \wedge e_k \wedge e_{\ell}) = \sum_{i < j < k < \ell} (c_{ij}c_{k\ell}-c_{ik}c_{j\ell}+c_{i\ell}c_{jk})(e_i \wedge e_j \wedge e_k \wedge e_{\ell})$, if $\alpha$ is simple then this sum must be 0, but since the $e_i \wedge ... \wedge e_{\ell}$ are linearly independant, this sum is zero if and only if all of the $c_{ij}c_{k\ell}-c_{ik}c_{j\ell}+c_{i\ell}c_{jk}$ are zero, QED.

[cloud walrus]

Wew Lads Tbh

[dim widget]

But this just shows that that relation holds! You need to show why that relation implies that the matrix has vanishing 3x3 minors now

[delicate orchid]

oh I did that

I'm not writing it up, this is just for question 2

[dim widget]

But even for 2, did you show that this means that the tensor is not simple?

[delicate orchid]

I'm piggy backing off of the previous question. I'm taking that as granted

[tardy hedge]

So is the symmetric group Sn the maximum amount of symmetries possible on an entity defined by n “axes” or degrees of freedom?

Or smth like that?

[delicate orchid]

bit of a weird way of thinking about it

it's just the group of n points with no conditions about them, so yeah I guess it's the "most symmetry possible"

[tardy hedge]

Ok thx

[remote trout]

Are there any good youtube channels for studying group theory from the basics? Need it for a undergrad course

[lusty marlin]

Why not use a textbook?

[remote trout]

I will do that, but I just want to watch some introductory videos before that to get some motivation and basics in lol the textbook is really tedious

[lusty marlin]

I see

[remote trout]

https://youtube.com/playlist?list=PLztBpqftvzxVvdVmBMSM4PVeOsE5w1NnN&si=a2E_s1EVfKChD7Hd Any idea if this seems good?

[lusty marlin]

I haven't watched it, so I don't know

[white oxide]

what does lang mean by this?

namely can somebody explain more how we can always find a polynomial with nonzero constant term if alpha is algebraic over F

[round hull]

because α is nonzero

so $a_1\alpha + \cdots + a_n\alpha^n = 0$ implies $a_1 + \cdots + a_n\alpha^{n-1} = 0$

[cloud walrus]

quasi_semi_group

[tardy hedge]

Mathemaniac has a series similar to 3b1b

[round hull]

if a1 is zero you keep factoring α out