groups-rings-fields

[hollow mica]

I'm not sure how it'd help here

[coral spindle]

Oh mrean there is a problem though

This is false

[hollow mica]

really

[formal ermine]

boytjie, I gave up proving the well definedness of that morphism and looked it up online

and that thing is used in its proof

[coral spindle]

Like what about the cartesian inner product

That’s positive-definite

[hollow mica]

maybe I should've said k is algebraically closed

[coral spindle]

Yeah that is good to know.

I’ll give this another look later

[hollow mica]

just to be clear, k being algebraically closed prevents this from happening or?

[coral spindle]

There is typically no order on an algebraically closed field

So talking about things being positive is silly

[hollow mica]

ah

[coral spindle]

I'm really disappointed, illum. I think this proof is well within your ability and I think instead of giving up you should continue to try and prove this.

[formal ermine]

is it possible to prove it without the fact that $aH = bH$ iff $a\inv b \in H$?

[cloud walrus]

illuminator3 👻(#eric4honorable)

[hollow mica]

another way to phrase my question is

"every symmetric bilinear form has at least one degenerate form"

[coral spindle]

It is. However this fact that you cite is also very easy.

Prove it.

[formal ermine]

OH I JUST NOTICED LMAO

I was so confused by the proof given by the wiki

if $aH = bH$ then $a\inv bH = H$, let $x = a\inv b$, now for all $h \in H$, $xh = h'$ for some $h' \in H \implies x = h'h\inv$ and because the operation must be closed, x must be in $H$

is that correct?

[cloud walrus]

illuminator3 👻(#eric4honorable)

[coral spindle]

Sure that works.

Now prove the other direction

[formal ermine]

if $Ha = Hb$ then $H = Hb\inv a$, let $x = b\inv a$, now for all $h \in H$, $hx = h'$ for some $h' \in H \implies x = h\inv h'$ and because the operation must be closed, x must be in $H$

[cloud walrus]

illuminator3 👻(#eric4honorable)

[formal ermine]

for right cosets

[hollow mica]

This is the first part to a problem asking to prove that there exists a k-dimensional subspace that is killed by any symmetric bilinear form for any k <= n/2 (killed meaning B(v, w) = 0 for any v, w in the subspace)

[formal ermine]

same proof basically

[coral spindle]

When I said "the other direction," I didn't mean prove it for right cosets

You stated an iff, but you've only shown the "only if" part.

Show the "if" part.

[formal ermine]

ahhh

[hollow mica]

wait this is the question I'm trying to solve but no way this is right

Intuitively this makes no sense right? @coral spindle

that there always exists a n/2-dimensional subspace that is killed by a symmetric bilinear form

[coral spindle]

Can't really look at this rn sorry

[hollow mica]

ok all good

[formal ermine]

lemma: let $H$ be a subgroup, then for all $h \in H$ you have $hH = H$ \
proof: for any $h, h' \in H$ you have $h' = h(h\inv h') \in hH$, therefore $H \subseteq hH$. but also $hh' \in H$ for all $h' \in H$, so $hH \subseteq H$, therefore $hH = H$ for all $h \in H$. \
\
now if $a\inv b \in H \implies a\inv b H = H \implies aH = bH$

[cloud walrus]

illuminator3 👻(#eric4honorable)

[formal ermine]

is this setminus or right cosets?

[coral spindle]

Setminus.

[formal ermine]

right

thanks

[formal ermine]

is this correct btw?

[coral spindle]

Looks so.

[formal ermine]

epic

[shy turtle]

Hello! Are there books or lecture notes that review the theory of rings from the category theoretic side, other than Algebra: Chapter 0?

[chilly radish]

Right cosets Would be H\G

[formal ermine]

ohhh

thanks

[chilly ocean]

I am trying to classify some ideals in Z[x] as prime but not maximal, maximal, or not prime. My progress for 10Z[x] (since Z[x] is commutative ring), is it is maximal iff Z[x]/10Z[x] = Z/10Z is a field but 10 is not a prime hence it is not a field so 10Z[x] is not maximal. I also want to say it is not prime since 2x and 5 are both in Z[x], and 5(2x) = 10x is in 10Z[x] but neither 2x nor 5 is in 10Z[x]

does this make sense?

[chilly ocean]

I have a permutation problem that says sigma = (1 3 4)(2 3 5 7) I am so confused as to how this can exist

Because doesn't 3 map to 4 and 3 map to 5?

[coral spindle]

When we put two cycles next to each other, this is composition

So you compose the permutations as you would any other function.

Your course may write it as left-to-right composition rather than the more standard right-to-left, though.

[chilly ocean]

i think this is supposed to be one cycle

i take this cycle and multiply it with another one

[coral spindle]

.

Indeed I agree that the (left-to-right) composition of these two cycles appears to be a cycle too.

[celest cairn]

Have I found the minimal polynomial correctly?

[uncut girder]

You foiled incorrectly

#prealg-and-algebra

🗿

[celest cairn]

that’s akward…

[tribal moss]

"foiled". 🤮

[hollow mica]

did you figure anything out about this @coral spindle

I still can’t really believe it

that “half” of a vector space (n/2 dimension subspace) is always killed by a symmetric bilinear form

[maiden heath]

Why are group algebras so important in representation theory, what properties does the group algebra have that provides additional information on the representation of a group? Why can't we just construct representations of groups like we usually would all the time?

[tribal moss]

The usual dot product on R^n would seem to be a counterexample to that claim.

[hollow mica]

yeah which is so weird

is there even a 1 dimensional subspace inRn for which it is true

[tribal moss]

No.

[hollow mica]

wait isn’t this a proof that it has to exist though

<a, b> = 0 for some a, b in V (Gram schmidt). Now scale b so it equals a

[coral spindle]

We've already talked about how we require an algebraically closed field for the thing you're looking at

I don't know the answer and I'm doing sth else right now, so don't expect me to help I'm afraid

[coral spindle]

To be clear, this was information you provided of course

[hollow mica]

oh yeah i forgot abt this

[hollow mica]

and yeah this only works when a,b are in the same one-d subspace anyways

so doesn’t work in general rip

I guess if <a, b> = 0 and a, b are linearly dependent we win

nvm that doesn’t work either

damn

[tribal moss]

This sounds like it would only have a fighting chance of being true IF you replace the arbitrary field k with the complex numbers AND specify that V has dimension at least 2.

Or at least let k be some field in which all elements have square roots.

[hollow mica]

dim 2 thing I’m allowed to do I think

but the problem is asking over a general algebraically closed field

[tribal moss]

But then every element does have a square root.

Choose two linear independent vectors. If one of them is killed by your bilinear form, then you're done.

[hollow mica]

how do we know this

[tribal moss]

That's part of what "algebraically closed" means.

All polynomials with coefficients in k have roots in k. In particular, x²-a does for every a in k.

[hollow mica]

that’s the field though, right

[tribal moss]

Yes.

[hollow mica]

oh I thought you meant elements in the vector space had square roots

like for every v there exists w such that <v, w> = 1

how do square roots in the underlying field help us

[tribal moss]

Whenever we have a vector v with <v,v> != 0, we can choose c to be a square root of 1/<v,v>, and then <cv, cv>=1.

[hollow mica]

oh I see

[hollow mica]

here what do you mean by “one of them is killed”

certainly not <v, v> = 0 because we’re trying to find some v that satisfied it

oh do you mean b(v, -) is trivial

[tribal moss]

Yes, exactly <v,v>=0 -- I'm saying if we randomly found a solution, then we're already done, so the rest of the argument can get away with assuming we weren't that lucky.

In particular, if v and w are linearly independent and <v,v> and <w,w> are both nonzero, then we can do Gram-Schmidt on them and get v' and w' with
<v',v'>=1
<w',w'>=1
<v',w'>=0

(Or possibly we might end up with <w',w'>=0, in which case we can cash out our winnings early again).

[hollow mica]

gram-schmidt just says “give me some linearly independent vectors and i will give them back as an orthonormal set”

so if we give gram-schmidt v, w it will return v’, w’

I don’t see how you got to the last line, <v’, w’> = 0; nvm this is just a consequence of gram schmidt

[tribal moss]

Yes, but as usually presented, Gram-Schmidt assumes that we have an inner product instead of just a symmetric bilinear form. So we need to dissect its inner workings a bit to make sure that "every scalar has a square root" can make up for us not knowing it is positive definite.

[hollow mica]

I see

ah so basically

if we could use gram schmidt from the beginning we would be done

but we can’t since we are working with a symmetric bilinear form, so it may not be positive definite

but existence of square roots implies that it is positive definite?

[tribal moss]

No, not quite.

But we can still use the idea behind Gram-Schmidt to see that there must be some scalar a such that <v,w+av>=0.
Namely by bilinearity <v,w+av> = a<v,v>+<v,w>, so we can set a = -<v,w>/<v,v>.

Because v and w were linearly independent, v and w+av will also be linearly independent.

Now normalize each of v and w+av to have square 1, and we get the v' and w' from my description above.

[hollow mica]

oh I see

and 1/<v, v> is guaranteed to exist because of your earlier argument

[tribal moss]

Yes.

[hollow mica]

and if <v, v> = 0 we cash out

wait I’m not too familiar with gram schmidt but isn’t it much different

oh actually i guess a = -proj_v(w)

[tribal moss]

The basic idea of Gram-Schmidt can be summed up as:

Each time we get a new input vector, add or subtract a combination of the basis vectors we've already found to make sure the result is orthogonal to all of them. Normalize the basis vectors as they're created.

The rest is just notation and bookkeeping.

[hollow mica]

I see

I’m now seeing if your argument can be extended either inductively or directly to make an n/2 dimensional subspace that is killed

and somehow it shouldn’t work for n/2 + 1

[tribal moss]

We haven't quite gotten to a vector that is killed in the n=2 case yet (but perhaps you've already seen how that goes).

[hollow mica]

I have a general idea

say V has dimension 4

pick v, w same as above

now pick w, k in a similar fashion, where k is linearly independent of v, w

so <v, w> = 0 and <w, k> = 0

oh here i thought by symmetry we could say <v, k> = 0 but this is not true

[tribal moss]

Just to be on the same page, in the n=2 case, are you sure you see how having v and w with <v,v>=<w,w>=1 and <v,w>=0 gives us an 1-dimensional subspace that is killed by <-,->?

[hollow mica]

oh i’m confused

didn’t we just show that

we found <v’, w’> = 0

now we can scale w’ or v’ by anything

and by symmetry this space is dimension 1

right?

[tribal moss]

We're looking for a single (nonzero) vector u such that <u,u>=0.

[hollow mica]

yeah so we can scale either v’ or w’ appropriately

wait fuck

they’re not in the same 1d subspace necessarily

i keep making this mistake lol

uhh

so we have some <v’, w’> = 0

and the v’, w’ are orthonormal

[tribal moss]

They're definitely not in the same subspace -- on the contrary they were constructed to be linearly independent.

[hollow mica]

yeah

[tribal moss]

But try to compute <v'+iw', v'+iw'>.

[hollow mica]

wait doesn’t <v’ + w’, v’ + w’> work

yeah

[tribal moss]

<v'+w', v'+w'> would be 2, not 0.

[hollow mica]

huh

oh

wait is that i an element of the field

[hollow mica]

here

[tribal moss]

By "i" I mean one of the square roots of -1 in the field. Remember that everything has square roots!

[hollow mica]

Oh right!

[tribal moss]

This is the step that doesn't work for the dot product on R^2.

[hollow mica]

<v’ + iw’, v’ + iw’> = <v’, v’> + 2i<v’, w’> - <w’, w’> = 1 + 0 - 1 = 0

[tribal moss]

👍

[hollow mica]

wait this is pure magic

how’d you come up with this lol

[tribal moss]

Ain't my first rodeo. :-)

[hollow mica]

lol ok

ok so now if V has dimension 4

do the same thing with v1, v2 and v3, v4

[tribal moss]

Yeah.

[hollow mica]

and you get two linearly independent vectors such that <u1, u1> = <u2, u2> = 0

[tribal moss]

Exactly.

[hollow mica]

wow this problem is trippy

would you have thought the result was intuitive

i thought it was blasphemy when i first read it

though i don’t know too much yet about this stuff

[tribal moss]

There's some threat left from the "cash out early" cases along the way though, and it's not immediately clear to me how much footwork it will take to deal with that. But I think that as an overarching intuition this is sound.

[hollow mica]

wait cash out early just means some <vi, vi> = 0

and everything fine in that case right

we still get a linearly independent vector that is killed by the bilinear form

[tribal moss]

Yeah, but now we want more than one of them, and we want them all to be independent and orthogonal, so it's too early to quit and "cash out" the first time we see one.

You stated this was singled out as a first step in getting to higher n, so it's possible you have a hint available for using this as a building block in an induction, rather than directly generalizing the proof of this base case.

(As said, I'm not immediately sure how each strategy would play out).

[hollow mica]

No I mean, given a basis for V, say {v_1, v_2, …, v_{2n}}, perform the same process for {v_1, v_2}, {v_3, v_4}, …, {v_{2n-1}, v_{2n}}. Now we just need a single vector killed from each two element set, so if one <v_i, v_i> = 0, that can serve as the vector killed for whatever two element set it’s in

[hollow mica]

am I crazy for thinking this works

[tribal moss]

Hmm, that sounds intriguing.
You'll get n linearly independent u's such that <u_j,u_j>=0 all right, But I'm not sure you will also automatically get, for example <u1+u2,u1+u2>=0 -- and without that, you won't have an entire n-dimensional subspace that's killed.

[hollow mica]

oh shit i just assumed that those u_j could serve as building blocks for that subspace, but I guess that’s not true

[tribal moss]

For example, in C^4 we can consider the bilinear form given by <(a,b,c,d),(a',b',c',d')> = aa'+bb'-cc'-dd'.
Then for u1=(1,0,1,0) and u2=(1,0,-1,0) we have <u1,u1>=0 and <u2,u2>=0 but <u1+u2,u1+u2>=4.

Induction seems like a more promising way to get to higher n will all the letters dotted and crossed.

[hollow mica]

I am trying induction yeah

[rotund aurora]

How seriously should the adjective "universal" be taken? As in the context "universal property of quotients". I'm still not sure what it is supposed to convey.

Is this just category theory?

[south patrol]

Yeah - in brief, universal properties boil down to limits/colimits which you can study in cat theory, which are unique up to isomorphism

[tribal moss]

It's not about "universes", more about "tool that can do all the things in question". A quotient together with its projection map is "universal" in the sense that it is, so to speak, the only homomorphism from G that sends all such-and-such elements to zero you'll only need. All the others can be made by tacking on additional maps after the quotient,

[rotund aurora]

thanks

[hollow mica]

I was thinking about the induction and I got a basic layout:

If every vector space of dimension 2k had a k dimensional isotropic subspace, we must show that every 2k + 2 dimension vector space has a k + 1 dimensional subspace.

Let V be a 2k + 2 dimension vector space, and let a_1, ..., a_{2k + 2} be a basis. Note that the subspace S = <a_1, ..., a_{2k}> has a k dimensional isotropic subspace. Now we just need a vector v in V\S = <a_{2k + 1}, a_{2k + 2}> such that Sv = {0}, or equivalently, a_1 s = a_2 s = ... = a_{2k} s = 0

[tribal moss]

Hmm, I'm not sure that way around will be easy. The solution I can see starts by finding a single null vector, and then uses that to define a 2k-dimensional subspace where we can look for the rest of them using the induction hypothesis.

[hollow mica]

null vector?

[tribal moss]

vector with <v,v>=0.

(That name is used in relativity; I don't know how common it is otherwise).

On the other hand, I'm not positive there will be any problems with your tactic, so take this with a grain of salt.

[hollow mica]

well, I can't seem to get anywhere with it

is this along the lines of your solution? :

Let v be such that <v, v> = 0 and v \in {a_{2k + 1}, a_{2k + 2}}, and consider the 2k - 1 subspaces S_j = <v, a_1, ..., a_{j - 1}, a_{j + 1}, ..., a_{2k - 1}> for each 1 <= j <= 2k - 1 (span of v and every a_1, ..., a_{2k - 1} except a_j). Note that each S_j has dimension 2k

[tribal moss]

Hint: I'm considering ||the subspace { w | <v,w>=0 }|| and then ||extend v to a basis for that||.

[hollow mica]

Hmm

what can I say about K = { w | <v, w>=0}, where <v, v> = 0

for starters, v in K

[tribal moss]

Right.

[hollow mica]

it's easy to check that K is a subspace (by bilinearity)

so K has a basis v, w_1, ..., w_j

[tribal moss]

You can also bound the dimension of K.

[hollow mica]

Note to myself: K is not necessarily isotropic

K = ker (<v, ->)

so dim K = 1 or 0 ?

I'm confused now

[tribal moss]

No that's the codimension.

[hollow mica]

oh, I see why

[tribal moss]

For a linear transformation we have dim(domain) = dim(kernel) + dim(range).
And the range has dimension 0 or 1 here.

[hollow mica]

so dim(kernel) = 2k + 2 or 2k + 1 ?

[hollow mica]

under this hypothesis (V is of dimension 2k + 2)

[tribal moss]

Yes.

[hollow mica]

ok so K has basis v, w_2, ..., w_{2k}, and possibly also a w_{2k + 1} as well

[tribal moss]

Right.

[hollow mica]

wait K isn't isotropic so what's the whole point of this

[tribal moss]

We're looking for a 2k-dimensional subspace to apply the induction hypothesis to.

[hollow mica]

oh I see

[hollow mica]

I am off by one here I think

[tribal moss]

Yeah. Didn't notice w_1 was missing. :-)

[hollow mica]

it should be v, w_1, ..., w_{2k} and possibly also a v_{2k + 1}

so now apply the induction hypothesis to <v, w_1, ..., w_{2k - 1}>

[tribal moss]

Hmmm....

[hollow mica]

or <w_1, ..., w_{2k}

idk we brought v this far along I don't want to throw it away

[tribal moss]

We're not throwing it away. We're setting it aside.

The induction hypothesis will only gives us k isotropic vectors, and we need to end with k+1.

[hollow mica]

ok so we can have v rest for the moment, and let's now apply the induction hypothesis to W = <w_1, ..., w_{2k}>

it gives us a k dimension subspace with basis p_1, ..., p_k

now we substitute v back into the game, and consider <v, p_i>?

I don't see why <v, p_i> has to be 0

oh wait

[tribal moss]

Because p_i lies in K.

[hollow mica]

yeah

so now

<v, w_1, ..., w_{2k}> works!

[tribal moss]

Yes!

Um, no.

[hollow mica]

wait I'm tripping again

why is this thing of dimension 2k + 1

[tribal moss]

<v, p_1, ..., p_k>

[hollow mica]

oh yeah

why'd I just change those

lol

man

this was a cool problem

besides for the part where you had to spoonfeed me through the solution

I will get good at these!

[tribal moss]

Well, more "slightly dense hinting" than "spoonfeed".

You did find many of the reasoning steps yourself.

[hollow mica]

pat on my back

ok

are you up for one more rodeo

it's a problem my professor proved in class but I don't understand the proof

so less of a rodeo more of a "why is statement x true"

[tribal moss]

I might not have time to get all the way through. It's getting late.

[hollow mica]

No worries, I will just post the problem and proof

Let V is a finite-dimensional vector space over an algebraically closed field k with char k = 0, and suppose T: V -> V is a linear operator such that T^k = I for some k > 0. Prove that T is diagonalizable.

We can't use fancy splitting field stuff because we haven't learned it, so the proof relies on Jordan Normal Form.

If T^k = I_n, then T is invertible, with no eigenvalue equal to 0. Consider the Jordan Normal Form of T. Let A be a Jordan block of size m. Note that T^k = I_n implies A^k = I_m. Moreover, consider the power of Jordan blocks in general. No positive power (> 2) of A can equal I_m, because the jth row jth column element of A^k is of the form kλ^{n - 1} for the corresponding eigenvalue λ. So we must have k = 1, which means m = 1, and thus the JNF of T is diagonal.

Why does T^k = I_n imply A^k = I_m ?

I think I understand the "no non-trivial power of a Jordan block is the identity" part, but that part ^ is tripping me up

[tribal moss]

Because when you multiply block diagonal matrices (with matching block sizes) you can do it block for block.

In particular a power of a block diagonal matrix is simply made of the same power of the blocks.

[hollow mica]

wait, does T^k = I_n the JNF of T also raised to the kth power is I_n

[tribal moss]

Yes.

Recall that the JNF is just the operator expressed in a particularly convenient basis. And the identity map is represented by I_n in every basis.

[hollow mica]

I see

If V is a n dimension vector space over the complex numbers C, and T: V -> V is a linear operator such that T^k = I_n for some k > 0 (same as above), prove that there exists a Hermitian inner product on V that is unitary with respect to T.

The converse seems easy to prove

because hermitian + unitary means the eigenvalues are 0 and 1

[hollow mica]

wait do you know why we need 0 to not be an eigenvalue @tribal moss

[agile burrow]

If 0 is an eigenvalue then T has non-trivial kernel, hence it isn't invertible

[hollow mica]

yeah but why do we need T to be invertible anyways

at least I don't see it being used in the proof

nvm I figured it out

[hollow mica]

anyone have hints for this problem

[upper pivot]

my hint would be ||say you already have an inner product on V, can you think of a combination of these inner products and T that would be unitary for T?||

[hollow mica]

By this result, T is diagonalizable, so let v_1, ..., v_n be an eigenbasis where||v_i|| = 1 (can normalize because C is alg. closed), then if we show there exists a hermitian inner product so that these v_i are mutually orthogonal, by the spectral theorem, T is unitary wrt this inner product

[hollow mica]

so ...

[hollow mica]

is that it @upper pivot

sorry for the double ping

wait actually

I need to find a hermitian inner product so that the v_i are mutually orthogonal

[upper pivot]

yeah i am not sure how what you stated there solves it.

[hollow mica]

this is what's left to do right

then the proof is complete

[upper pivot]

hmm yeah right that would be enough i think

[hollow mica]

modified so it's right this time

[upper pivot]

(its not how i'd have done the problem but i think this might work)

[hollow mica]

yeah but finding this hermitian inner product is kinda hard lol

what was your approach if you don't mind me asking

[tribal moss]

Just take the standard hermitian product in C^n, according to your eigenbasis.

[hollow mica]

whats the standard one in C^n

like

[tribal moss]

$\overline{x_1}y_1 + \overline{x_2}y_2 + \cdots + \overline{x_n}y_n$

[cloud walrus]

Troposphere

[tribal moss]

(Or possibly its complex conjugate, depending on your conventions for hermitian products).

[chilly ocean]

physicist spotted

[hollow mica]

our definition requires that h(v, w) is conjugate to h(w, v)

[hollow mica]

did you not mean the complex conjugate here?

[tribal moss]

That's what I wrote, isn't it?

[hollow mica]

idk you said "or possibly"

ohhh

do you mean the complex conjugate of the entire expression

[tribal moss]

Either $\overline{x_1}y_1+\cdots$ or $x_1\overline{y_1}+\cdots$, depending on which side your convention wants ordinary and conjugate linearity on.

[cloud walrus]

Troposphere

[hollow mica]

because people choose the "conjugate linear" argument to be different

right

so isn't the standard hermitian product the same regardless of choice of basis

like we didn't have to choose a basis before defining it

[next obsidian]

How do u have conjugation?

That needs a choice of basis

I’m pretty sure

🤔

[hollow mica]

a + bi -> a - bi

[next obsidian]

But

[hollow mica]

i just flip flopped the + to a -

[next obsidian]

…

If V is an abstract vector space

[tribal moss]

He's already chosen an eigenbasis for T.

[next obsidian]

Right, but

[tribal moss]

Ah, I missed that post.

[tribal moss]

Not anymore than the dot product is the same for a real vector space regardless of the choice of basis ...

Which is to say, not at all.

[next obsidian]

I’m sure they all induce the same topology or whatever

[hollow mica]

oh

I see

wait for concreteness

if I have a basis of magnitude one vectors v_1, ..., v_n

and I want a hermitian inner product H such that H(v_i, v_j) = 0 for all i, j

we just talked on how the standard hermitian product wrt the basis works

[hollow mica]

how are we sure that there exists a standard hermitian product that satisfies this

like before you said "just take it according to the basis"

[hollow mica]

but concretely how do you do that

[tribal moss]

I'm not sure what else I can say here.

The basis lets you express each element of V as something from C^n.

[hollow mica]

yeah

[tribal moss]

So take the standard hermitian product of those two elements of C^n.

[hollow mica]

I guess my misunderstanding is where the choice of basis is used

[tribal moss]

You need a basis before you can get from an abstract element of V to a tuple of coordinates.

[hollow mica]

okay so we have to choose a basis

but how can we just say H(v_i, v_j) = 0 for all i, j without worrying about anything

[tribal moss]

Um, because that's what the standard hermitian product does?

[hollow mica]

oh because we chose the v_i to be the basis

I seeeeeeeeeeeee

Thank you so much for tolerating my bs lol

[agile burrow]

I'll try to give you a place to start: you know that a generator maps to an endomorphism A such that A^n = Id. What are the eigenvalues of A? Is A diagonalizable? What are the dimensions of the eigenspaces of A?

[rotund aurora]

Let F be a field, and let kappa be some cardinal number. What is the dimension of the vector space F^kappa over F?

[wary yoke]

I have no idea whether this problem is solvable or not because it came up unexpectedly somewhere else and I'm not an algebra person

Suppose I have a Ring R of integers mod 10 (under usual operations) and 16 unknowns x_1 to x_16 which satisfy a system of 16 equations (non homogeneous)

when does a solution exist and what is it (if it exists?)

[cloud walrus]

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

[rotund aurora]

no there is not lol

the question is when kappa is infinite

[coral shale]

Could you use a tower law argument to assert equality?

[gusty thistle]

I did it by using the fact that alpha_2 and alpha_2^3 are complex but the intersection is a subset of the real numbers

[coral shale]

sure that works

I was thinking along the lines of: the degree of extensions are 4, 4, 2 respectively, so the degree of intersection is either 2 or 4. Then you show it can't be 4 by showing the intersection is a proper subset of the two. Think this argument works as well

[gusty thistle]

Yeah I think so too

[cursive edge]

are there any prerequisites for intro to abstract algebra other than proof writing

[long nebula]

I'm confused, couldn't G also be cyclic if no section exists?

[long nebula]

not really, but some basic understanding of number theory might help

[coral shale]

Probably not (just check it doesn't expect group theory as a prereq or anything)

[coral shale]

f: G/N = C_4 -> C_8; f(x) = x, isn't f a section?

[long nebula]

G/N is C_2

[coral shale]

oop

but yeah, that?

f: G/N = C_2 -> C_8; f(x) = x

ok lemme edit

[long nebula]

wait is that a well-defined homomorphism tho?

[coral shale]

no thats just poorly written by me lmao

[long nebula]

Lol

[coral shale]

f: G/N = C_2 -> C_8; f(e) = e, f(a) = b^4

where a, b are respective generators

sry if i misunderstood

[long nebula]

but then f(a) = b^4, pi(b^4) = e, no?

because b^4 is contained in N

[coral shale]

right, ofc

hmmm

[long nebula]

I think this problem is just wrong

I'll email my prof about it lol

[next obsidian]

@long nebula wait why do you think so?

Oh lmao, the map backwards isn’t actually a section

Kekw

[long nebula]

Lol

[chilly ocean]

I think they mean for N to be a normal subgroup

[long nebula]

it says N is a normal subgroup of order 4 tho?

but either way G -> G/N has no section if G is cyclic

I think

[chilly ocean]

it doesn't say it's normal

but yeah I think it has to do with being a semidirect product of Z_2

and Q_8 and Z_8 are the only groups of order 8 which aren't

Because existence of such section is equivalent to there existing a short exact sequence $$0\to\mathbb{Z}_4\to G\to\mathbb{Z}_2\to 0$$

[cloud walrus]

Blitz

[chilly ocean]

that is, G is a semidirect product of Z_4 and Z_2

maybe they forgot to write that Z_8 is also excluded, or the word "non-abelian" was forgotten

I think the latter

wait. I mean the existence of such section implies that it's the semidirect product

[formal ermine]

what does it mean for a diagram to commute?

[chilly ocean]

https://kconrad.math.uconn.edu/blurbs/grouptheory/splittinggp.pdf
see here, theorem 3.3

that is, the sequence splits

[flint cave]

I was literally about to type that

Such great notes

[formal ermine]

ah wait I think I get it

if we have this diagram:

then it commutes iff h = f \circ g?

[chilly ocean]

wrong way around

g circ f

[formal ermine]

ah yeah

but that's the definition of a commutative diagram like that?

[chilly ocean]

yes

[formal ermine]

oke thanks

[chilly ocean]

commutative just means that the order you follow the arrows in from one point to another doesn't matter

[formal ermine]

ahhhhhhhhhhhh

that makes sense

thanks!

what does a dashed line mean?

[chilly ocean]

I think it means unique

But there's no standard convention about dashed lines that I know of

[formal ermine]

oh yeah uniqueness would make the most sense here

thanks

if A is isomorphic to B then A and B are the same structures but "viewed through different lenses"?

[coral shale]

sometimes dashed means 'implied existence of'?

[chilly ocean]

It means you're just relabeling the elements, and all the structure stays the same

[formal ermine]

thanks

[chilly ocean]

At least all the common notions of isomorphism.
In category theory I wouldn't give that interpretation

Isomorphisms kinda lose that part of their meaning

[wooden ember]

Yeah I feel like this may be more frequent than unique

[formal ermine]

this means phi bar is an iso from g/ker phi to image of phi?

[wooden ember]

Though it’s usually for there exists unique

[formal ermine]

specifically the tilde

[wooden ember]

Yep

[chilly ocean]

I'd write $\simeq$ or $\cong$

[wooden ember]

No but they’re showing that phi bar is doing the mapping

[cloud walrus]

Blitz

[wooden ember]

So you put the tilde above the arrow

My profs do that too

[chilly ocean]

I'd still write it that way

[wooden ember]

Fair enough

[chilly ocean]

Instead of an arrow

[formal ermine]

does $\bar{g} \coloneqq gH$ have a specific name

[cloud walrus]

illuminator3 👻(#eric4honorable)

[formal ermine]

the g bar

[wooden ember]

Left closet

[chilly ocean]

coset

[wooden ember]

Coset *

Sniped

[formal ermine]

I mean in the context of a quotient group

still just left coset?

[wooden ember]

Just an element

[chilly ocean]

Yes

[formal ermine]

oke

@chilly ocean why the

[chilly ocean]

Some people are like that. Just don't pay them attention

[celest cairn]

How do I find a cyclotomic polynomial using this formula?
Also what does the n’ mean?

[long nebula]

it's of index 2, so it has to be

[long nebula]

yea this is what we went over in class, and I just classified all the possible actions of C2 on C4 and C2 on C2xC2, so G has to be either D8, C4xC2, or C2xC2xC2

currently waiting for a response from the prof as to clarification on the question

ty y'all for the sanity check :)

[chilly ocean]

ah yeah. My bad

[long nebula]

dw I was also confused about that for a while until I realized xD

[long nebula]

the n' probably means the euler totient function of n, (\phi(n))

[cloud walrus]

Ei ao (e/i) e

[long nebula]

and then the zeta are the primitive nth roots of unity numbered from 1 to phi(n)

[celest cairn]

So how do I find (\phi(4))?

[long nebula]

you would calculate all of the primitive 4th roots of unity

(there's only 2)

and then multiply [(x-\zeta_1)(x-\zeta_2)]

[cloud walrus]

Ei ao (e/i) e

[celest cairn]

Thanks much! 🙂

[long nebula]

np!

[chilly radish]

N is index 2 hence normal

[next obsidian]

This was already said

Nerd

[long nebula]

Lol my professor wrote back to me and said that he forgot that abelian groups existed since classification for finite abelian groups is easy

[next obsidian]

Hurb

[next obsidian]

This is such a meme

Literally only 2 options then

[long nebula]

hahahahaha

now time to figure out how to show G must be Q8

[next obsidian]

Just classify groups of order 8 4head

[long nebula]

Lol "G must be Q8 by checking the table in Dummit-Foote"

[untold basin]

I know that the polynomial has a degree of 2 but do I have to test some polynomials to find it ? My prof told me that I need to see some results first

[long nebula]

isn't it just (x - \zeta_3\sqrt[3]2)

[cloud walrus]

Ei ao (e/i) e

[untold basin]

Wait I messed up

[cloud walrus]

Christophe*

[untold basin]

In that case, yes XD

[long nebula]

Lol

I don't know any field theory, but try computing zeta_3 cbrt(2) squared

and then seeing if you can get some sort of linear dependence relation with zeta_3 cbrt(2)

[untold basin]

I tried this but I feel like I just test things stupidly

[long nebula]

what did you get?

[chilly radish]

It should divide the minimal Polynomial of it in Q, which is x^3-2. Look at the factorisation of x^3-2 over Q(cbrt(2)) into irreducibles and conclude what it is

[delicate bloom]

further hint if you absolutely can't figure it out: ||x^3-y^3=(x-y)(x^2+xy+y^2)||

[tribal moss]

Well, for one thing, if you take the representation of a generator and look at the Jordan normal form over C, you'll find that you can't get T^n to be I without the JNF actually being diagonal.

So it is diagonalizable over C but not necessarily over R.

Since the characteristic polynomial is real, the non-real eigenvalues must come in conjugate pairs.

And remember that two real matrices that are similar over C are also similar over R.

See e.g. https://math.stackexchange.com/questions/57242/similar-matrices-and-field-extensions

If you follow the train of thought I'm sort of hinting at, you will get that the representation is reducible unless it has one a few very special forms that are each easily seen to be irreducible.

A key fact is that $\begin{pmatrix} e^{i\theta} \ & e^{-i\theta} \end{pmatrix}$ is similar (over $\mathbb C$) to $\begin{pmatrix}\cos \theta & \sin\theta \ -\sin\theta & \cos\theta\end{pmatrix}$.

[hollow mica]

I really like this self-contained run-through of (basic) galois theory https://www.maths.ed.ac.uk/~tl/gt/gt.pdf

[cloud walrus]

Troposphere

[tribal moss]

Eyup.

[sweet sundial]

hello

[tribal moss]

Hello and welcome.

[sweet sundial]

hi :))

okay here goes

this is the hw problem im stumped on

and this is what i have written down so far

i would just like some guidance not a solution because there are two more of these

[tribal moss]

0 is in Q, so what is f(0)?

[next obsidian]

What’s your ultimate claim?

Is it well-defined or no?

[sweet sundial]

i said no

that would be undefined

[next obsidian]

Erase everything and write down a single counterexample then

[sweet sundial]

so proof by counterexample?

kk

thank you

should prolly not use pen then

[uneven jackal]

use whatever but do a draft first

[sweet sundial]

that's a good idea

[next obsidian]

You could also learn LaTeX and then do scratch work and type up the final draft

[long nebula]

yea I usually type everything in latex

and then I can just delete the wrong ideas I have, copy and paste the stuff I need copy and pasted, move text around, etc

[solar inlet]

so when I'm finding simple reps for Wedderburn, and then I end up with two isomorphic reps, what do I do with them? Count them as just a single rep? My prof's proof makes it seem like I should treat them as the direct product of everything in the isomorphism class, which can't be true right?

[wicked zephyr]

shouldn't it be p instead of p^2 at the beggining

?

[red imp]

if $M$ is an $R$-module then we can interpret $R\otimes_R M$ as being $M$ by the isomorphism $r\otimes m\mapsto rm$. Is there a similar reduction that we can make for $(R\oplus R)\otimes_R M$?

[next obsidian]

That’s just M^2 now

Because tensor product distributes over sums

So you get (R (x) M) (+) (R (x) M)

[hollow mica]

I am reading something about why the isomorphism between a vector space and its dual is not canonically isomorphic

what does the last sentence mean?

"duality involves the linear transformations between the space"

ok... but what difference does it make if there's more than one isomorphism

[next obsidian]

Really this is saying you want the functor * to be naturally isomorphic to the identity functor

So you can find isomorphisms from V and V*

Given maps V -> W you get maps V* -> W*

[hollow mica]

are you saying

[next obsidian]

There’s no way to make the isomorphism between V and V*, and W and W* such that the square you get commutes for all possible maps, all possible V and W

You want this to play nice with maps too

[hollow mica]

hmm

why doesn't this commute?

can't we just pick a basis for V and W to get the maps V -> V* and W -> W*

and then S* can be appropriately chosen probably? or are you saying there won't exist an S*, in which case I don't see why

[next obsidian]

There does exist S*

The issue is that you S* is fixed

S* is going to be precomposition with S

It sends f:W-> R to V -> W -> R

The issue is the isomorphisms from V to V* and W to W*

You have only one of these for any particular V, and you want this square to commute for those choices of isomorphisms for ALL S

This isn’t possible

[hollow mica]

You have only one of these for any particular V

wait what

I thought the point was that there's an infinite number of isomorphisms from V to V*

[next obsidian]

Yeah and you have to fix one

The issue is that the one you fixed is supposed to make this square commute

But the fixed choice has to make every such possible square commute

If this doesn’t make sense just ignore it or look up what functors and natural transformations are

[red imp]

Ahhh thank you

[hollow mica]

why is (L*f_j)e_i = f*_j(Le_i)

this the second equality on the last line

it is a proof for why the matrix of a linear transformation is the transpose of the matrix of the transpose map W* -> V*

[red imp]

why there so much lin alg in the algebra channel

[hollow mica]

nvm i got it

[hollow mica]

it is true that a vector space is isomorphic to its dual

but what is so special about the isomorphism between inner product spaces and duals

[chilly ocean]

For finite-dimensional vector spaces, yes. For infinite-dimensional vector spaces, no.

[hollow mica]

yea sure

[red imp]

because you dont get that sort of niceness in all Banach spaces

[chilly ocean]

I don't have a satisfying general answer, but I can name some places where it comes up: differential geometry (in raising and lowering tensor indices), functional analysis (the Riesz representation theorem; this generalizes the isomorphism in question)

[hollow mica]

I see

[thorn delta]

the analytic consequence of this is that hilbert spaces are 'isomorphic' to their duals

[next obsidian]

It also gives you a specific isomorphism in terms of your inner product

[hollow mica]

ok I now finally understand what you meant, but I don't understand why the isomorphisms (V <-> V*) and (W <-> W*) don't work (make the diagram commute) for any linear transformation S, it's not like we constructed the isomorphisms with respect to S, all we did was make them relative to some choice of basis for each of V and W

[next obsidian]

Which doesn’t depend on your basis (I think)

[chilly ocean]

It doesn't depend on the basis.

[hollow mica]

huh don't we have to pick a basis in order to get the isomorphism between a space and it's dual

[chilly ocean]

The map v -> <v, -> is perfectly well-defined and independent of any choices.

[hollow mica]

oh there I was talking about in general, not inner product spaces

[next obsidian]

https://mathoverflow.net/questions/345136/is-a-vector-space-naturally-isomorphic-to-its-dual

[chilly ocean]

I meant to respond to the other person.

[hollow mica]

.

[hollow mica]

oh @next obsidian this was supposed to be in response to your earlier message (replied above)

[hollow mica]

I don't see how the fixed choice doesn't make each possible square commute (each choice of transformation)

like the only information we used to define the isomorphism between V and V* was a basis, we didn't use anything inherent to the linear transformation S

[bleak abyss]

Let's do some math

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

Stabilizer of the identity is the image of the orthogonal group

[agile burrow]

I like this action

[bleak abyss]

Yup, so in fact this action is transitive. You see why?

[agile burrow]

Yeah, there's like a standard form for positive quadratic forms

Or it's just the sum of squares x_i^2

But I forgot how that works :(

[bleak abyss]

Hmm I guess I was just thinking, symmetric matrices are orthogonally diagonalizable

[agile burrow]

I guess it becomes the same thing

You diagonalize and then just scale to normalize eigenvalues

[bleak abyss]

So you can turn S into a diagonal matrix and then yeah hit it with more diagonal matrices to get to identity

Yup

[agile burrow]

And the action is proper

So if you have discrete torsion-free subgroup of GL(n, R)

[bleak abyss]

So we identify P(C) with PGL(2,R)/PO(2). Another way to see it is

SL(2,R) embeds into GL(2,R)

[agile burrow]

You can quotient out space with action to get a K(G, 1) if you're interested in that

[bleak abyss]

Now if we write S = XDX^T, then D = X^TSX where X is orthogonal

But then diagonal matrices commute

So I can replace X with {{1,0},{0,-1}}

What's the point? The point is that the action of SL(2,R) is still transitive

And now this is just SL(2,R)/SO(2)

[agile burrow]

I don't understand replacing X

Why/how can you do that?

[cloud walrus]

Sloth King Daminark

[bleak abyss]

And diagonal matrices commute

[cloud walrus]

Sloth King Daminark

[agile burrow]

Ah ok

[red imp]

this is all in the category of abelian groups btw

or at least I want to know something about the structure of A if possible

[tough raven]

A = Z_np × A' should give the same result as Z_np for any torsion-free (or even "p-torsion free") A'.

[red imp]

ah yeah true

the full question is that I need to find the structure of $A$ if there exists a short exact sequence $$0\to \bZ_p\to A\to \bZ_p\to 0$$ and $A\not\cong \bZ_p\oplus\bZ_p$

[red imp]

i.e. the class of the extension in Ext(Zp, Zp) is not 0

so I applied the long exact sequence for Ext and all that I could get out of it was that Hom(Z_p , A)=Z_p

[tribal moss]

Same reasoning that Mrean used the day before in a different context. Let me find it ...

[tribal moss]

Here.

Not sure offhand.

[formal ermine]

having a brain fart right now. how does $\pmod{p}$ relate to $\bZ_p^{\cross}$? I can see how it relates to $\bZ_p$ but $\bZ_p^{\cross}$ only has $p - 1$ order?

[cloud walrus]

illuminator3 👻(#eric4honorable)

[formal ermine]

so like in Z_3 \ {0} 5 "is" 1 but 5 = 2 (mod 3)

[delicate bloom]

It's multiplication mod p, you remove 0 because it isn't invertible

[delicate bloom]

This is wrong, 5=2mod3 in Z_3{0} still

[formal ermine]

oh modulo is always wrt some group?

[coral spindle]

No

Really this is ring theoretic.

There is no notion of 'multiplicative part' of a group in general

[coral spindle]

Oh just realised

if you really wanted a purely group-theoretic description of Z_p^\times without reference to any ring theory...

You could describe it as Aut(Z_p), the group of automorphisms of Z_p.

[agile burrow]

Here's a hint/direction:you know by the fundamental theorem of finite abelian groups that A is either Z/pZ + Z/pZ or Z/p^2Z. Isomorphism classes of extensions are in bijection with Ext^1(Z/pZ, Z/pZ). If you compute this group, you can ignore one element corresponding to the trivial extension and all you're left with are the non-trivial extensions which you're looking to compute. This will at least give you the number, and you can think about how to compute them explicitly or I can give another hint at that point.

[formal ermine]

wait but we have {1, 2} as Z_3 \ { 0 } and 5 = 2 + 2 + 1 = 2 + 1 = 1, no?

[coral spindle]

2 + 2 + 1 = 2 + 1

[formal ermine]

like

if we add 1 to 2 we go back to the beginning because it's cyclic, no?

ohhhh

wait

we have multiplication here not addition

[coral spindle]

The group operation for Z_3^\times is multiplication, yes.

[formal ermine]

then how do you get the result 5 = 2 mod 3

[coral spindle]

5 = 2 + 3 = 2 mod 3.

I do not know where this confusion is arising

[formal ermine]

I mean in Z_3\{0} with multiplication

[coral spindle]

5 is still 2 mod 3

[formal ermine]

uhh the context is fermat's little theorem

like it's x^(p - 1) \conguent 1 mod p but we set G = Z_p \ { 0 }

[coral spindle]

You ought to forget about groups for a moment.

Like I said earlier, this is really about rings, but you won't know what that is.

We are simply working in modulo 3 arithmetic, so 5 = 2 mod 3.

That is all.

[formal ermine]

uhh okay

[formal ermine]

orbit are all possible outcome values and stabilizer are all values that act like a neutral element?

[chilly ocean]

It's imprecise, but that's one way to look at them.

[next obsidian]

How much do you want?

A hint or like the solution

[formal ermine]

thanks. just got introduced to them and need like a rough overview of what they are when they appear, so yeah

[next obsidian]

You can show any finite group center which admits a faithful representation has cyclic center

Faithful irreducible

[chilly ocean]

You will see examples and form intuition as you keep studying.

[formal ermine]

yip

[next obsidian]

Ah fuck

That doesn’t do anything here!

[formal ermine]

$G/G_x \cong Gx$?

[next obsidian]

🤔

[cloud walrus]

illuminator3 👻(#eric4honorable)

[next obsidian]

Sorry!

I guess try to classify all the irreducible representations?

Do you have enough character theory to know when you’ve found all of them?

The stuff relating degrees of reps to the size of the group

Oof

[chilly ocean]

I wouldn't use the \cong symbol, because there's no apparent structure that's being preserved by the obvious bijection. I would also let the other two proceed with their conversation, and tell you to go read up on the orbit-stabilizer theorem some.

[next obsidian]

Idk lol maybe @delicate orchid knows

He knows a lot of rep theory I think

I’m not gonna lie, my rep theory knowledge is insanely weak

By that what I mean is, I have no idea

[quiet pelican]

phi(z) is a g isomorphism of the space if z is in the centre

[celest cairn]

Why do I keep getting -1 when finding the 3rd cyclotomic polynomial using this formula?

[quiet pelican]

what does schur's lemma say about g automorphisms of an irreducible rep?

And in particular of the identity

so they commute with everything

so if its restriction is faithful on H...

Yes, as phi(z) in G is equal to the value of its restriction on H

So that means that phi(z) commutes with everything in phi(G), if the restriction is faithful on H, but z doesn't commute with everything in G

Z(G) is trivial, but we've shown that Z(phi(G)) is not

if it is faithful on H

I'm doing the contrapositive, if a representation of G restricts to a faithful, irreducible rep of H, then the rep of G is not faithful, so that a faithful rep of G restricting to a faithful rep of H, can't restrict to an irreducible rep

[celest cairn]

Why do I keep getting -1 when finding the 3rd cyclotomic polynomial using this formula?

[chilly radish]

why are you substituting x-1 and x+1 for the primitive roots of unity

x is a variable here

[celest cairn]

What would we substitute?

[next obsidian]

You don’t…

[chilly radish]

the values of the roots of unity?

[chilly radish]

When defining Brauer groups over rings, we define the brauer classes as similarity classes Where A\sim A' if A(x)M_n(R) \cong A'(x)M_m(R) for some m,n, where R is the base ring. Over a field, the brauer classes correspond to morita equivalence, is this still the case for azumaya algebras

[delicate orchid]

yeah pretty sure the 16 dim irreducible is faithful, its character has a trivial kernel ||iff the rep has a trivial kernel iff the rep is injective||

no clue if you're allowed to use characters

shit sorry you only wanted a hint

[next obsidian]

What 16 dimensional representation

What the fuck

[delicate orchid]

wait is it S_5

oh no it's S_6

if you want a smaller one I will have to google the character table

[agile burrow]

yung diagram

[delicate orchid]

huk formulah

ok yeah all four 5 dim reps are also faithful

whoops

[next obsidian]

Okay but

What 16 dim rep for S_5??

[delicate orchid]

it's S_6, chmonkey

[delicate orchid]

one of these is the standard rep if you want a concrete example

which is probably the one the question you wanted to find, in hindsight

[next obsidian]

Okay so

What is it for S_6?

Like what is the representation??? Why is it so big

[agile burrow]

hook length!

hook length formula gives dimensions of representations of symmetric group, they just get big p quickly

[next obsidian]

Nooooooooooo

[delicate orchid]

idk hahahahah!!!

lets look at the funny tetris pieces

[next obsidian]

I purposefully didn’t pay attention to that part of my combo class

[agile burrow]

silly question from a silly guy, I'm reading that if I have a projective resolution P of M and a projective resolution P' of N, then I can compute Tor(M, N) as the homology of P \otimes P'. Is this the homology of the total complex of P \otimes P'?

[next obsidian]

I would assume so, but this is maybe some hyper cohomology stuff

Idk the specifics

🙂

[agile burrow]

I can take it on faith, I never learned the double complex stuff

[chilly radish]

couldn't you do some silly spectarl sequence argument to show this

from like the homology of the double complex

maybe you need hyperhomology

sorry I forgot chmonkey only enjoys spectrral sequences passively

[agile burrow]

i'm hyperhomology

[oak hill]

Suppose I have some quadratic polynomial f(x) over $\mathbb{Z}_p$ which is irreducible. How do I find an extension field which contains all the roots of f(x)?

[cloud walrus]

F♯A♯ℵ0

[oak hill]

Because $\frac{\mathbb{Z}_p}{<f(x)>}$ doesn't necessarily contain all the roots right?

[cloud walrus]

F♯A♯ℵ0

[south patrol]

(\langle \rangle btw)

And you mean Zp[x] i guess

[south patrol]

But yes, the idea is to keep going until we get all the roots.

Suppose $f$ is of degree n over a field $F$ and take an irreducible factor $g$ of it; by the construction you gave, we have a field extension $K/F$ such that $g$ has a root $\theta$ in $K$. But this means that in $K[x]$,we can write $f(x) = (x-\theta)h(x)$ for some $h \in K[x]$ and by induction on degree we can find some extension $L/K$ in which $h$ splits

[cloud walrus]

potato

[south patrol]

You can then take the intersection of all subfields of L in which f splits to get a splitting field.

[oak hill]

Ok thanks 🙂

So if i took $f(x)=x^2+1\in\mathbb{Z}_3[x]$ then I could get the root $x+\langle x^2+1 \rangle \in E=\frac{\mathbb{Z}_3[x]}{\langle x^2+1\rangle}$. Then $f(x)=(x+\langle x^2+1 \rangle)h(x)$, so i get the second root $x+\langle h(x) \rangle \in \frac{E[x]}{\langle h(x) \rangle}$?

[cloud walrus]

F♯A♯ℵ0

[south patrol]

Well this is a bit different, since as soon as you have one root you have the other, since it's only quadratic

[oak hill]

Is that the case for any quadratic?

[south patrol]

Sure - work out why

[oak hill]

second root is just $x - \langle f(x) \rangle$?

[cloud walrus]

F♯A♯ℵ0

[oak hill]

which is in the first extension?

[south patrol]

Well my point was that h is degree 1

[oak hill]

ah right

well thank you for the help!

[sweet sundial]

hello

i need some help and will be posting my problem with what i have come up already

wait lemme crop my proposed solution

[next obsidian]

You should try #proofs-and-logic

[sweet sundial]

alright there we go

[next obsidian]

For naive set theory

[sweet sundial]

kk

[past path]

If $H$ and $K$ are subgroups of a group $G$,\
let $(H,K)$ be the subgroup of $G$\
generated by the elements ${hkh^{-1}k^{-1}\mid h\in H,k\in K}$.\
Also note that $G'$ denotes the commutator subgroup of $G$.\
Show that if $(H,G')=\langle e\rangle$, then $(H',G)=\langle e\rangle$.

[cloud walrus]

Amizar

[past path]

This has got all of us stumped so far.

It's a 3 part question and the other 2 parts were pretty solvable.

I've made progress in a few different directions, but I'm not sure how to tackle the problem altogether.

Information that I've got so far:

[latent anvil]

I feel like this is what the hall Witt identity is for, but I night be misremembering

[next obsidian]

I had a similar feeling

But I don’t remember the exact statement, and it seemed a little different

[past path]

I don't think that's the move in this case, because we haven't covered anything like that in the course.

Though it may be a reasonable route to solving the problem.

I did some research and it seemed appropriate but I'm not sure where to go with it.

The key point of $(H,G')=\langle e\rangle$ is that every element of $H$ commutes with every element of $G'$. The target statement is equivalent to $H'\leqslant C(G)$.

[cloud walrus]

Amizar

[past path]

$(G,G)=G'$\
$(G,H)=(H,G)$\
$(H,(H,H))\leqslant(G,(H,H))\leqslant(G,(G,H))\leqslant(G,(G,G))$\
$(H,(H,H))\leqslant(H,(G,H))\leqslant(G,(G,H))\leqslant(G,(G,G))$\
$(H,(H,H))\leqslant(H,(G,H))\leqslant(H,(G,G))\leqslant(G,(G,G))$

[cloud walrus]

Amizar

[latent anvil]

C(G) is the center?

[past path]

Yes.

[latent anvil]

So your conclusion is equivalent to H' < H cap C(G)

I'm just thinking

[past path]

I mean, I guess that's also true. That should be equivalent to H'<C(G)

[latent anvil]

Right sorry I remember why I was thinking this

A normal subgroup contains the commutator subgroup iff it has abelian quotient

H cap C(G) is clearly normal, its central in H

And H/(H cap C(G)) = H C(G)/C(G)

This is what I was thinking

hmm

[past path]

I guess that makes sense. In fact, H cap C(G) is central in G.

[latent anvil]

So why is this abelian?

It isn't obvious to me

[past path]

I mean

[latent anvil]

It's a subgroup of Inn(G)

Those automorphisms given by conjugation by H

[past path]

The normality is what checks out. A central subgroup is always normal, but the converse isn't necessarily true.

[latent anvil]

Wdym?

I meant C(G) cap H is normal in H because it's central in H

[past path]

Meaning, H\cap C(G) is central, not only in H, but in G as well.

So it's automatically central in H.

[latent anvil]

Sure

I am trying to figure out why H C(G)/C(G) is abelian

[past path]

We also know that H' < C(H), but that isn't necessarily contained in the center of G.

It's in the center of H and the center of G'

Because it's in H, it's in the centralizer of G'.
Because it's in G', it's in the centralizer of H.
Because it's in both, it's in the center of both.

I'm not sure how to go from there to center of G, though.

Wait, is the center of G' a subgroup of the center of G or am I crazy?

If G is abelian, then C(G) = G, and C(G') = G' = \langle e\rangle.

Just for an example, not a proof.

[latent anvil]

I don't see why this would be true

[past path]

Well, (G,C(G)) = <e> for any group G.

[next obsidian]

What’s the center of Q8?

[latent anvil]

It's ±1

[past path]

{1,-1}, right?

[next obsidian]

Just {+-1} yeah?

Ah no, the commutator is order 2 I think

[latent anvil]

I think so too

yes

Q8/{1,-1} is abelian and Q8 isn't

[past path]

I think {±1} is also the commutator, right?

Yeah.

I feel like that would be a pretty relevant statement not to be known, but it seems like it should hold, for some reason.

[latent anvil]

So I was saying H C(G)/C(G) is isomorphic to the subgroup of Aut(G) given by conjugation by elements of H

and I was trying to figure out whether this was obviously abelian, at all

We know that any H-conjugation kills off G'

[next obsidian]

Okay

[latent anvil]

so for any automorphism γ in this subgroup, γ(ab) = γ(ba) for all a, b

[next obsidian]

S_3

Its commutator is A_3

Which is abelian

But it has trivial center

[latent anvil]

Why does this imply they mutually commute

[past path]

Hmm...

[latent anvil]

I think that works chm

[past path]

Yeah, I think you're right.

Dang.

[next obsidian]

I just went through small non-abelian groups on group props

Lol

[latent anvil]

So I want to say if h, k in H then for any x we have hkxk^-1h^1 = khxh^-1k^-1

[past path]

Every once in awhile those kinds of revelations actually lead to something productive, but more often than not they end with a simple counterexample.

[latent anvil]

if we rearrange this says [h, k] x = x [h, k]

ugh

[past path]

Wait, why is that the case, sham?

[latent anvil]

What part?

[next obsidian]

Wants to say

He didn’t claim it’s true

[latent anvil]

I mean I didn't get anywhere

[past path]

Oh, I see.

I didn't see want to say.

The thing that we're aiming for is equivalent to H' < C(G)

[latent anvil]

So this is equivalent to the assumption that H < C(G')

It says any automorphism γ of G given by conjugation against an element of H satisfied this relation

And the desired conclusion is that the group of these is abelian

[past path]

We haven't used that sort of notation in class. Doesn't mean it's off the table by any means, but we haven't gone in that direction so I imagine there's a route to solving the problem that doesn't require that kind of machinery.

[latent anvil]

What notation

[past path]

The things with the automorphisms gamma etc.

And a lot of stuff with [h,k]

[latent anvil]

What specifically

Have you seen that γ(x) = gxg^-1 is an automorphism?

[past path]

Particularly applying the commutator (function?) to elements rather than groups.

I mean, it's not hard to convince myself.

But it's not something we've done in this class.

Oh

wait yeah

conjugations and translations both produce automorphisms.

[latent anvil]

I'm just saying I'm not using any machinery here

I'm reframing things in terms of group K

[past path]

Ok, sounds good.

[latent anvil]

K is a subgroup of the automorphism group of G, consisting of all automorphisms given by conjugation by an element of H

I noticed that the assumption says that γ(ab) = γ(ba) for any γ in K, a, b in G

[next obsidian]

Translation does not, it’s not a homomorphism (if by that you mean x -> gx)

[latent anvil]

This says γ kills off the generators of (G, G)

and the conclusion says that K itself is abelian

[past path]

I thought it was an automorphism if you translate by an element in the group itself.

[next obsidian]

It isn’t a homomorphism

Just write it down and you end up with an extra copy of g

[past path]

Oh, wait. Yeah I see what you mean.

[latent anvil]

It is a bijection, and this is important

[next obsidian]

It gives a set-automorphism

[past path]

I guess that makes sense.

Ahhhh

[next obsidian]

So it defines a group action

[past path]

That might have been what I was thinking.

It's an action, which I might have mixed with automorphism because it's bijective.

[latent anvil]

so for any γ in K, im γ is an abelian subgroup of G

Ie

[past path]

Here's the problem statement.

[latent anvil]

ker γ contains G'

[past path]

part (a) was covered in class and is free game.

parts (c) and (d) I proved on my own.

[latent anvil]

What's the join here?

Smallest subgroup containing them?

[past path]

The subgroup generated by HK, the set of products hk where h\in H, k\in K.

[latent anvil]

Cool

[past path]

Note that the problem statement defines (H,K), meaning this is the first time the book has introduced the function at all.

[latent anvil]

Okay

[past path]

I'm pretty handy with picking up the intuition for an algebraic structure, but my gut tells me that there's not a whole lot of digging necessary within the definition of the function itself.

Going back to something from earlier, in case it can be of any use:

$(G,G)=G'$\
$(G,H)=(H,G)$\
$(H,(H,H))\leqslant(G,(H,H))\leqslant(G,(G,H))\leqslant(G,(G,G))$\
$(H,(H,H))\leqslant(H,(G,H))\leqslant(G,(G,H))\leqslant(G,(G,G))$\
$(H,(H,H))\leqslant(H,(G,H))\leqslant(H,(G,G))\leqslant(G,(G,G))$

[cloud walrus]

Amizar

[past path]

The second line gives us a notion of commutativity (shocking, I know) in the commutator operation.

If we had something like associativity, which I'm not sure how or in what capacity to establish it, then it would make the rest of the problem trivial.

[latent anvil]

The correct version of associativity is more like the Jacobi identity

[past path]

What is the Jacobi identity?

[latent anvil]

The axioms for a lie bracket

[past path]

We haven't been introduced to lie groups yet.

[latent anvil]

I was just sharing

[next obsidian]

You can just look up the Jacobi identity

[past path]

Fair enough. I've pulled it up in another tab; I was just providing context for where we are in the calss.

[latent anvil]

https://mathoverflow.net/q/81224/136523

See the question/comments/answers blog posts here

[past path]

The thing that stands out is the two-variable analogue, which I agree with the poster is nearly trivial: (a,b)(b,a)=1

[latent anvil]

Sorry to clarify, the lie bracket of a lie algebra is nonassociative

The correct version of associativity in that case is the Jacobi identity

[past path]

Ahhhh, that makes sense. Sounds sad.

[latent anvil]

That's why I was bringing this up

[past path]

Ahh, I see.

That makes sense.

But it does remain commutative, right?

[latent anvil]

It's actually anticommutative

[v, w] = - [w, v]

[past path]

Ah, that makes sense.

But for groups that's the same thing.

If we apply it to groups instead of elements, because groups contain their own inverses.

[latent anvil]

Ah so

The sort of fundamental examples of lie algebras are where you start with an associative algebra A (eg matrices over a field k) and look at the bracket [a, b] = ab - ba

In fact all lie algebras arise this way (although A may need to be infinite dimensional)

[past path]

I see.

[latent anvil]

So this is why there's an analogy/relationship with the commutator in group theory

[past path]

It seems like the following should be true: if H < G, then (H,K) < (G,K).

[latent anvil]

I very much don't see why this is true

[past path]

Right?

[latent anvil]

Yep

[latent anvil]

Sorry this was about the overall problem

[past path]

So the three compound inequalities are all sound, right?

[latent anvil]

Looks like it

[past path]

I eventually algebrad my way into $\forall g\in G, h_1,h_2\in H$,\
$gh_1g^{-1}h_2gh_1^{-1}g^{-1} = h_1h_2h_1^{-1}$

[latent anvil]

Okay I am now going to bring out machinery, sorry

[cloud walrus]

Amizar

[latent anvil]

Let P = H C(G)

hmm

[past path]

I can show you the trail, but it involved several substitutions.

[latent anvil]

maybe not

[past path]

I'd like to know if that implies anything about ghg^{-1} or g^{-1}h_2g

[latent anvil]

This says that my subgroup K is normal in Inn(G)

Interesting

[past path]

Or, in English, If you conjugate an element of H by another element of H, the result is not changed by first conjugating the second element of H by any element of G.

[latent anvil]

Oh no not exactly

It's more subtle

[past path]

Right.

I was thinking that might be the case.

If g commutes with h_1, then it's immediately true.

[latent anvil]

If you start with an H-inner automorphism

Then you conjugate by any inner automorphism

The result agrees with the automorphism you started with on H

[past path]

The same is true for individual elements?

[latent anvil]

What I said is literally this identity