groups-rings-fields

[past path]

I see.

[latent anvil]

hmm

If γ : G -> Inn(G) sends g to conjugation by g

then this implies γ(γ(g)(h)) = γ(g) ° γ(h) ° γ(g)^-1 < K

[past path]

We do have that (H',G) is a normal subgroup of G thanks to part (a).

[latent anvil]

No it doesn't ugh

This is very confusing

[past path]

Yeah, I'm normally super on top of homework in this class and in number theory, so it's a little unfamiliar to be so totally stopped by a problem.

But yeah, we do have that (H',G) is a normal subgroup of G.

An element of $(H',G)$ looks like $h_1h_2h_1^{-1}h_2^{-1}g^{-1}h_2h_1h_2^{-1}h_1^{-1}g^{-1}$.

Oops

Typo

[cloud walrus]

Amizar

[past path]

There we go.

I wonder if the fact that H' is in the center of G' and in the center of H can be useful.

I guess that begs the question, what elements of G are left that are allowed to not commute with H'?

[latent anvil]

Why is H' in the center of H?

[past path]

H' is a subgroup of G' so it commutes with H
H' is a subgroup of H so it commutes with G'
H' is a subgroup of both so it is central in both.

[latent anvil]

Oh right, we're assuming (G', H) = 1

Okay yes, H' is central in H

[past path]

So elements of G which do not commute with H' must be in $(G\backslash H)\cap(G\backslash G')$. Is this the same as $G\backslash (H\lor G')$?

[cloud walrus]

Amizar

[latent anvil]

no

That would be saying H v G' is H union G'

[past path]

Right, but at the same time

We do commute with everything in H

And we commute with everything in G'

So we should commute with everything generated by their union, right?

[latent anvil]

oh, I thought you were asking if G \ H cap G\G' is the same as G \ (H v G')

I see now

[past path]

Meaning, I think we can make the claim that anything that doesn't commute with some element of H' must be in the complement of H join G'

[latent anvil]

No I don't think I understand

[past path]

If we commute with everything in a set, we should commute with everything generated by that set, right?

Because we're really just taking products of elements in the set, which we can individually commute with.

[latent anvil]

I think I'm just having difficulty parsing this claim

[past path]

Perhaps the wording was a bit careless.

[latent anvil]

Elements of G which do not commute with H' must lie outside of both H and G', I agree

[past path]

If we commute with elements of H and we commute with elements of G'

Then we should commute with elements of H union G'

Meaning we should commute with elements of the group generated by H union G', which is H v G'.

[latent anvil]

x commutes with both H and G' iff x commutes with H v G', I agree with this

[past path]

Okay.

So if we commute with everything in H v G', then the only elements we don't commute with must lie in G\(HvG')

I grant that I'm not sure where to go next with this, but I think this is still a worthwhile claim.

We do also conveniently know that G/(HvG') is abelian, but I'm not sure what to do with that information.

[latent anvil]

Why is H v G' normal?

G' is normal but H isn't known to be, yet

[past path]

Hmm...I'm not sure, now that you mention it.

Is a subgroup that contains the commutator subgroup necessarily normal?

[latent anvil]

No

Take Sn

For n >= 3 this has trivial center

Err

Shoot sorry

Take An for n >= 5

[past path]

Any subgroup that contains the alternating group has to be normal in Sn.

Oh okay.

What's the commutator subgroup for An?

[latent anvil]

The commutator subgroup is trivial

[past path]

Oh, I see.

[latent anvil]

No I fucked up

Sorry

[An, An] = An

Okay one more time

[past path]

Wait yeah that makes sense.

I was thinking "Wait hold on that's not abelian"

[latent anvil]

Sorry haha

[past path]

Oh wait no that is true.

[latent anvil]

It is?

[past path]

Any subgroup containing the commutator subgroup is normal.

[latent anvil]

Oh yeah duh

The quotient by G' is abelian

Duh duh

[past path]

So that means (HvG') is normal in G.

[latent anvil]

Yes

[past path]

Can we do anything with the fact that HvG' is normal in G and that their quotient is abelian?

And the fact that H' commutes with everything in HvG'

[latent anvil]

I don't really see it

[past path]

Let's see what other information the chapter offers us.

[latent anvil]

Ah hm so (H', H') < (H, G') = 1

This in particular tells us H is a "solvable group"

[past path]

We also have (H,H') = 1 as well.

[latent anvil]

This means H' is abelian

It has trivial commutator subgroup

So hmm

[past path]

It's abelian because it's in the center of (H v G')

[latent anvil]

H is an extension of an abelian group by an abelian group but what does that tell us

What's (H v G', H v G')

We know H v G' = H G' = G' H

[past path]

Wait, how do we know that's true?

[latent anvil]

What part

[past path]

I know G' is normal, but do we know H is normal?

[latent anvil]

H and G' commute

[past path]

Or do we only need one to be normal

Oh

Right

[latent anvil]

elementwise

[past path]

Duh

[latent anvil]

Yeah so in general like

HK is a subgroup iff HK = KH

[past path]

So HG' = G'H

[latent anvil]

Normality of either ensures HK = KH

[past path]

Right.

[latent anvil]

So x = gh, y = ba, g, a in G' and h, b in H

xyx^-1y^-1 = ghbah^-1g^-1a^-1b^-1

Messy

g(hb)ah^-1(ag)^-1b^-1

messy

[past path]

If we took (HG',HG'), we would get $h_1g_1'h_2g_2'g_1'^{-1}h_1^{-1}g_2'^{-1}h_2^{-1}$

[cloud walrus]

Amizar

[past path]

I mean, the group of elements of that form.

And we know the h's commute with the g's

[latent anvil]

Not generated by elements of that form?

Oh right!

So it's not too bad

[past path]

So we would really have something of the form h'g'', I think.

[latent anvil]

Yes

Oh nice nice

[gh, ab] = [g, a] [h, b]

With g, a in G' and h, b in H

The bracket is just the commutator of elements, [x, y] = xyx^-1y^-1

Wait this is wild hm

It's a homomorphism

[past path]

Meaning (HG')' = H'G''

[latent anvil]

G'×H -> G' H

Weird

Anyways yeah

So I wanted to think about this because uhh

H' is abelian

(HG')'' = G'''

[past path]

Excuse me what lol

Oh I see.

[latent anvil]

?

Yeah

[past path]

Yeah that's kinda comical actually.

[latent anvil]

So this subgroup is a large part of G

I guess

[past path]

In fact, (HG')' is a subgroup of G'

[latent anvil]

Or like

It almost just is G

[past path]

Wait, that's obvious

[latent anvil]

Yeah

[past path]

Don't more apostrophes make the group smaller?

[latent anvil]

Yeah

[past path]

We have a theorem that says the following

I'm not sure if this applies to the problem we're trying to solve or if it just is what it is.

Also, a bit more theoreming:

We can't guarantee anything about G. That much is sure.

Because if we simply choose H to be in the center of G, both parts of the proposition follow immediately.

We just want H' to commute with everything in H v G'

But what's stopping an element of G that's not contained in that subgroup from refusing to commute with H'?

[past path]

An element of $(H,G')$ looks like\
$hg_1g_2g_1^{-1}g_2^{-1}h^{-1}g_2g_1g_2^{-1}g_1^{-1}=h[g_1,g_2]h^{-1}[g_2,g_1]$\
An element of $(G,H')$ looks like\
$gh_1h_2h_1^{-1}h_2^{-1}g^{-1}h_2h_1h_2^{-1}h_1^{-1}=g[h_1,h_2]g^{-1}[h_2,h_1]$\

[cloud walrus]

Amizar

[past path]

I found an answer on stackexchange. I'll see if I can understand it, take a break from it, and make sense of the concept on a fresh mind. The previous exercise gave some information that combines with the commutativity of H and G' that eventually prove what we wanted to show.

It's a really slick proof and I'm honestly not even mad at it.

[past path]

My professor's policy is basically "Use whatever resources you want, but if you look up a proof, don't submit it unless you can reproduce it on your own without referencing it."

So I took about half an hour to digest the ideas in the above proof and churned out this atrocity:

I have one more homework question.

It's honestly probably not worth the time it would take to complete the problem, given how busy my tomorrow is, but I'm too stubborn to not complete the assignment.

Is there anyone who might be able to help me with II.7.6?

[past path]

I have one direction down already.

If G is nilpotent, I've proven that every maximal proper subgroup of G is normal and has prime index.

I ignored the hint.

That might not be the best idea; I'm not sure what to do with the hint and the forward direction made enough sense without it.

Is anyone around who might be able to provide some insight towards why the backwards direction is true?

[past path]

The direction I'm trying to prove now is "If every maximal subgroup of G is normal in G, then G is nilpotent."

An equivalent condition to nilpotent would be that G is the direct product of its Sylow p-subgroups.

Though I'm not sure which might be more approachable at the moment.

[agile burrow]

I think the Sylow p-subgroups one seems more approachable

[past path]

Yeah, I was thinking the same, though I'm not sure how to show that it's the direct product of those groups.

We can start by assuming all maximal proper subgroups of G are normal in G.

[agile burrow]

actually, I have no idea lol

sorry

[past path]

I'm not sure if the hint is that helpful.

[agile burrow]

there's something about proper subgroups not being self-normalizing that I can recall

[past path]

Maybe it's supremely helpful, but I've not got anything from it so far.

[agile burrow]

which makes the hint seem very relevant

but I suppose if you haven't heard of what I'm struggling to recall then it isn't helpful at all

[past path]

Right, though the fact that the forward direction fell out without it makes me wonder.

[zenith seal]

I don't know why the order of p-group must be the power of p (p is a prime).(My textbook says it is easy to prove but I can't. )

[past path]

From the hint we can derive "If N_G(P) is not G for some Sylow p-subgroup P, then any maximal proper subgroup containing N_G(P) is its own normalizer and is therefore not normal in G."

Oh wait shoot.

So all I need to do is actually prove the hint. The "Help me figure out pity party sentence" literally solves the rest of the problem.

Oh wait

I see what I was missing though.

The part I'm missing is "If G is not the direct product of its Sylow p-subgroups, then there exists a Sylow p-subgroup of G which is not normal in G"

Is that true?

If it's not, the pity party sentence remains a pity party sentence.

[carmine fossil]

https://math.stackexchange.com/questions/250941/g-is-the-product-of-its-sylow-subgroups

[past path]

Oh wait

If they're all normal

Then by the definition of inner direct product, G has to be an inner direct product.

[agile burrow]

that sounds right

yeah, like recognition theorem thing

[agile burrow]

how do you define a p-group

[zenith seal]

G is a group, if every element's order is a power of p, then the group G is defined as p-group.

[agile burrow]

ah ok

then this holds iff G is finite

the proof is an application of Cauchy's theorem, if you know of that

[zenith seal]

oh the p should be a prime

[zenith seal]

Well, I am not familiar with that , thanks for your hint!

[agile burrow]

hmm, I'm not sure off the top of my head how I would prove it without that but I'll think about it a bit longer

[zenith seal]

I find the Cauchy's theorem, it is "if p(a prime) could divides the order of a group G, then there is an element's order is p." I don't think it could be used to solve my question, since my question is for every element instead of a specific one.

[agile burrow]

Right, if this is your first time seeing the theorem then it might not be obvious how to apply it immediately

[edgy vessel]

oh that's clever

[agile burrow]

The idea is to suppose that |G| is not a power of p

Then there exists another prime q which divides |G|. Can you see how the theorem applies now?

[zenith seal]

Oh ,I see. Thank you!

[past path]

Ok, so I'm most of the way to what I'm looking for but I now see that I have a small bug in my argument.

I need to prove the hint and I'm not sure how to go about doing so.

[carmine fossil]

Is this wrt proving the hint or using the hint?

[past path]

We have a theorem that says the normalizer of the normalizer of a Sylow p-subgroup P is the normalizer of P in G.

It's with respect to proving the hint.

[carmine fossil]

Well just consider
P_1 P_2... P_k where P_i are sylow groups

[past path]

Okay.

[carmine fossil]

This is a well defined group since all P_i are normal

[past path]

Not necessarily.

If all P_i are normal we have a different set of conditions.

We're examining the case where there is at least one Sylow p-subgroup which is not a normal subgroup of G.

[carmine fossil]

Try to prove the contrapositive

[past path]

So the goal here is to go from N_G(N_G(P)) = N_G(P) to N_G(H) = H for any H containing N_G(P).

[carmine fossil]

That's much easier

[past path]

That if there is a non-normal maximal proper subgroup then a Sylow group has to be not normal as well?

I mean, I have the whole argument as long as I can prove the hint.

The only issue I'm having is that I'm not sure how to prove the hint.

I have a non-normal maximal proper subgroup already defined assuming the hint.

I need to connect the last dots to make sure it actually works.

I already have that part.

The only thing I'm currently missing is the hint.

Sorry, I spent some time on Overleaf.

[next obsidian]

P is a Sylow p-subgroup of H if H > N_G(P)

Try using the part of the Sylow theorems which state that Sylow subgroups are conjugate, but working inside of H

I can say more if you want

[past path]

P < N_G(P), so it must be the only Sylow p-subgroup of N_G(P).
If x is not in N_G(P), then xPx^{-1} is not equal to P. By the second Sylow theorem, however, it is a Sylow p-subgroup of G and is therefore not in N_G(P).
Therefore, N_G(N_G(P)) = N_G(P).
If N_G(P) is a proper subgroup of H, then P is a Sylow p-subgroup of H, but not the only Sylow p-subgroup of H.
Therefore, P is not normal in H.

Ay I just turned blue

[edgy vessel]

stop holding your breath

[past path]

lol

Anyway, H is trivially contained in N_G(H).

How can we show the opposite containment?

If we can do that then the final dominoes fall into place.

[next obsidian]

Do you want a full rooof?

Or an idea

[past path]

I prefer an idea, but we'll see how close I am to understanding once an idea happens lol

[next obsidian]

Well I’m going to sleep soon

[past path]

Idk what time zone you're in but it's winding towards 3AM here and I have a 9AM exam.

[next obsidian]

My suggestion is as I said already

You have to compare conjugates of P conjugating by something in N_G(H)

And something in H

How you do that is group theory

[past path]

If you're going to bed you're welcome to prove that statement before you leave. I'm confident that I can at least understand the mechanics of it.

So far I have H is a normal subgroup of its normalizer (woo)

And I have N_G(N_G(P))=N_G(P)

Meaning anything outside of N_G(P) non-trivially conjugates P to something else in H, but I'm not sure how to show that anything outside H would conjugate P to something even further outside H.

Does this imply that if a group contains N_G(P) and all conjugates of P, then it's necessarily the whole group G?

[next obsidian]

Let x in N_G(H), and consider P^x (^ means conjugate). This necessarily lands in H because x in H’s normalizer and P < H. This means P^x is a Sylow-p of H, so by Sylow’s theorems it’s conjugate to P inside of H so there’s a y in H with P^x = P^y.
xPx^-1 = yPy^-1 => y^-1xPx^-1y = P, so y^-1x is in N_G(P) < H. As y^-1 is in H, this says that x is in H

[past path]

Okay, that makes sense. I can see how P is conjugated to something inside H.

There's a y in H which conjugates P the same way x does

Meaning you can normalize P by ... ohhhhh I see.

H contains the normalizer so you can treat xy^{-1} as an element of the normalizer

Meaning it's also in H

meaning x is in H

Thank you. You've helped me a lot.

[hidden haven]

Use the semisimplicity of the ring ℂG

V is a quotient of ℂG by some sumodule W. Any submodule of ℂG is a summand of it

That is Maschke's theorem

Then the complement of W in ℂG is isomorphic to V

The general principle is that semisimplicity of a module implies that whenever that module appears in the middle of a short exact sequence, the sequence is split. Thus every SES of modules over a semisimple ring is split

[lethal dune]

probably a basic question but I do not see why in the diagram
\begin{tikzcd}
A\times _C B \ar[r, "\varphi'"] \ar[d, "\psi'"] & B \ar[d, "\psi"] \
A \ar[r, "\varphi"] & C
\end{tikzcd}
if $\varphi$ is surjective then $\varphi'$ also surjective

[lethal dune]

say in R-mod

nvm I am dumb

[versed orbit]

So if i have a polynomial in let say Z_5 (or any other field for that matter) is there an algorithm to break it down to irreducible factors or am i just suppose to check for roots then break those away until all possible roots i have used them all. and then just double check that whatever i have left can be broken further in to factors of degree 2 or larger?

[coral shale]

theres no general algorithm afaik

that u can mindlessly apply

[tribal moss]

Over a finite field you can -- if everything else fails -- simply test divide with all polynomials of smaller degree. There are finitely many of them, so that qualifies as an algorithm.

[chilly radish]

Idea anyone

[versed orbit]

okay thank you.

[chilly ocean]

Given an algebraic structure S that has a 0 is there a canonical name for the { (a,b) in SxS | ab = 0 }?
( The set of zero divisors? )

And what about the set of (a,b) s.t. ab =/=0?

[ruby sundial]

no

also calling it set of zero divisors seems wrong because this means (a,0) is zero divisor for all a.

Also zero divisors are only a thing for rings,algebras because otherwise they are called inverses for a group

or annihilators I guess

[tough raven]

https://ncatlab.org/nlab/show/suplattice says yes!

This seems simple enough that I should have thought of this

But also unmotivated enough that I have no idea how I would have.

Oh and apparently this is a special case of the Yoneda embedding being a free cocompletion.

Ah.

[chilly ocean]

I'm not sure I follow.
But there are algebras that lack associativity and commutativity, but we can still talk about division in them, specifically 0 division as well - for example, the Octonions, Sedenions, etc... all have zero divisors, but are not associative via their multiplication. We can do the same for R^n with dot products whenever our vectors are non-zero, but orthogonal: I mean, that goes to R^1, but we could modify it to just come back to R^n. We can also do R^3, with cross products and still divide by 0, just keeping the vectors collinear.

[chilly ocean]

(a,0) = 0 would mean a0 = 0 for all a.
But that's just multiplication by 0 in general.

I'm not sure what I've asked is coming across.

They are saying that the set you wrote down, {(a, b) : ab = 0}, is not the set of zero-divisors.

[chilly ocean]

I'm aware that they've said this...

Then don't call it the set of zero-divisors.

[chilly ocean]

"In abstract algebra, an element a of a ring R is called a left zero divisor if there exists a nonzero x in R such that ax = 0,[1] or equivalently if the map from R to R that sends x to ax is not injective."

In other words a in R is a left zero divisor if there exists non-zero x in R s.t. (a,x) = 0.
If a is non-zero, then we call x a right zero divisor.

So the set I've described earlier is identifying in structures more general than a ring.
If I specify that the left element is non-zero, then I've identified a way to go about division of 0 from the right.
If I specify the element on the right is non-zero, then I've identified a way to go about division of 0 from the left.

I don't see why I shouldn't call this a zero divisor, even though I recognize that the user has suggested otherwise.

I don't know that the user has understood what I'm asking.
It seems like the user is just trying to say that there's a canonical, historic association of "zero division" with rings. But what I pointed out above is that many non-ring structures have binary operators ( products ) that constitute division of a zero-element ( some element that, when we collapse the binary operator to a unary one via currying, it becomes a constant map to that element ). In the case of Octonions and Sedenions, those elements also function as identities under the additive map. So there's a strong sense where I can said element a "zero". Since I can multiply non-zero terms to get to that element, there's a sense where I can say I have zero divisors.

If all the user is saying is that this is confusing because people tend to only think of "zero division" in the context of rings, okay, that's understandable. Though idk that it's agreeable.

If the user has identified a more serious mathematical error, I haven't seen it.

For these reasons, I don't see why I ought not refer the adjusted definition of the set as the set of (left, right, or both ) zero divisors.

I appreciate the effort, but I'm not going to read all of that.

[chilly ocean]

Okay. Let's end communication from here. Thank you.

I was only hoping I could clarify their message. I don't have any particular interest in the content itself.

Sounds good to me.

[midnight summit]

hello

can anyone help with minkowski space

the question asks to find 4 linearly independent null vectors in minkowski space

in relativity terms we are looking for lightlike vectors

how do you find such vectors

[delicate bloom]

I think something like gram-schmidt should work

[midnight summit]

I am not really familiar with that

but I'll look into it

thank you

[south patrol]

It's not really necessary - to me, the picture to keep in mind is the light cone

For the case of 2+1 dimensions, that is

You can (hopefully) see there are clearly 3 linearly independent vectors lying on the light cone - then you just need to generalise that to 3+1 dimensions

[midnight summit]

in 3D

there are ijk vectors for basis

do you mean similar basis vectors for light cone?

[south patrol]

Well, so let's take it even simpler and consider the case of 1+1 dimensions

What are two linearly independent null vectors? (Take easy ones)

(Here by n+1 I mean n space dimensions and then 1 time)

[midnight summit]

null vector in euclidean space?

isn't it just a zero vector by definition?

[south patrol]

No, I mean 1+1 dimensional Minkowski space

So vectors (x,t) with the product given by the metric (1,-1) instead of (1,-,1,-1,-1)

So here (x,t) is null iff x^2 - t^2 = 0 (if we use c = 1 for simplicity)

[midnight summit]

c^2t^2=x^2

yes

[south patrol]

Is that clear?

Cool

So can you find two linearly independent vectors lying on this light "cone" x^2 - t^2 = 0?

[midnight summit]

a(t,-x)+b(t,-x)=(m,n)

[south patrol]

Not sure what you mean by that

[midnight summit]

linear independence means that there should be a combination possible so we will get a new vector right?

[south patrol]

uh not really no

[cloud walrus]

potato

[pliant forge]

Why is there no function of non-isomorphic minimal left ideals

[midnight summit]

yes right

so t and x are components of s vector

[south patrol]

I was using x^2 -t^2 = 0 here as shorthand (in the usual way) for the set { (x,t) in R^2 : x^2 - t^2 = 0} - you can/should think of them as labels for the axes, rather than components of the vector

[midnight summit]

that made it even worse

[midnight summit]

wait

so on diagram

is it going to be a straigth line

t=x?

so linearly independent vectors will be et and ex unit vectors

[south patrol]

t = x or t= -x

But neither e_t nor e_x lies on the light cone, so they won't work

[midnight summit]

their linear combination does?

e_t+e_x

[south patrol]

Yup, that's one of them

Find another linear combination of them on the light cone which is linearly independent from that ^

[midnight summit]

well same with minus

e_t-e_x

it's gonna match the left side of the cone

[south patrol]

ye

exactly

Now you can generalise this to 3+1 dimensions similarly

[midnight summit]

e_z+e_t and e_y+e_t

THANK YOU SO MUCH!!!

[south patrol]

Well you need to get 4 which are all linearly independent but sure, that gives you three

(if we use notation of 4-tuples, these are (1,0,0,1),(0,1,0,1) and (0,0,1,1))

But yes actually I'm pretty sure, say, (-1,0,0,1) - which you already had - works, so maybe that's what you meant

Np

[elder wave]

collatz?

[viscid pewter]

how does this work?

i mean commutativity is obvious but what even is the inverse of something in F[x]/<f(x)>?

the identity is like <f(x)> so say we need an inverse for some g(x) + <f(x)>

what do you multiply by to make it the identity?

[south patrol]

Yup

[viscid pewter]

we want a h(x) + <f(x)> such that g(x)h(x) + <f(x)> = <f(x)>, ie. g(x)h(x) is in <f(x)>

[south patrol]

Yeah so the start of sheet 1 is weird cause they don't prepare you enough lol

The multiplicative identity is NOT <f(x)> tho

Thats the additive one

[viscid pewter]

wait

[south patrol]

It is the 0 of the field

[viscid pewter]

fuck

ok that makes sense

1 + <f(x)>?

[south patrol]

Yup

[viscid pewter]

ok, and it's irreducible so shares no roots so can euclid it or whatever

[south patrol]

F is a subfield of this in a natural way etc

[viscid pewter]

fuck

bezout's

[south patrol]

Yeah exactly ^

But ye dw like q1 and q2 are like anomalies lol

you'll do this again in rings and modules assuming ur taking it

[viscid pewter]

literally my first time ever interacting with an abstract field

[south patrol]

Yeah fair

[viscid pewter]

i think so?

[south patrol]

cool ye i mean i'd highly recommend it

[solar glacier]

hi can someone help me with a question on finitely generated abelian groups

[south patrol]

Sure, what's up

[solar glacier]

its solution verification really

ok group of order 270

first off it has 3 iso classes

i use the first and second versions of the fundementsal theorem to obrain them

from version I out of dummit and foote I get

[sage spruce]

Iso classes ?

[south patrol]

isomorphism classes

[solar glacier]

$\Bbb{Z}{270}, \Bbb{Z}{30} \times \Bbb{Z}_3 \times \Bbb{Z}3, \Bbb{Z}{90} \times \Bbb{Z}_3 $

[south patrol]

But yeah sure I get 3 classes as well

[solar glacier]

lemme write that without tex

z_270, z_30 x z_3 x z_3, z_90 x z_3

and using the second version i get

z_27 x z_2 x z_5 which is iso to z_270

then i get z_9 x z_3 x z_2 x z_5 which is iso to the last one i got from first version

and lastly from version to which is iso to the second group i got from version I of theorem

[south patrol]

is there a reason you're moving between the two versions?

or do you have to use each of them

[solar glacier]

z_3 x z_3 x z_3 x z_2 x z_5

i need to use both and determine which is iso to which

[south patrol]

ah

[solar glacier]

so my first ones are iso to eachother i claim

and my second from second version iso to third on frist list

and 3rd from second version iso to second on original list

[south patrol]

yh seems fine

[solar glacier]

cool cool

thx

[south patrol]

np

[solar glacier]

preparing for a midterm so

just trying random problems out the text

[south patrol]

Yeah fair enough - gl with prep!

[solar glacier]

thanks!!

[celest cairn]

I sent this picture a few minutes ago, but it dissapeared for some reason. What went wrong here? I’m using a formula to calculate the 4th cyclotomic polynomial.

[chilly ocean]

In Algebra we have various ways for talking about "how close" an algebraic structure is to having certain properties.

For example, we have centralizers, normalizers, commutators, anti-commutators, associators, etc.

Are there any canonical ways to go about describing such sets ( "measurements" ) for any arbitrary algebraic property or perhaps certain, notable such properties?

( for example, how easy it is to remove or add-in zero divisors, idempotents, involutionaries, alternatives, anti-alternatives, finite products, fibres, inverse elements, etc )

[chilly radish]

Your primitive roots are wrong. The primitive roots of unity are i and -i

You will get x^2+1, unsurprisingly

I think you should review the definition of roots of unity and in particular primitive roots of unity

[celest cairn]

Okie, thanks much 🙂

[dire grotto]

For Q1 do I need to find all A s.t [a b;c d]A = I

[bleak abyss]

rishi answer is, you want to find matrices such that az+b/cz+d = z

Now let's do some algebra

Notation test: $\mathfrak{k}$

Uh oh

[pastel cliff]

tex

[bleak abyss]

Still we shall proceed

Googled and that's the right one

[ruby sundial]

not like this

why did you ask though? curious?

its useful to have a notion of this for sure

[pastel cliff]

i had my own quick question tho - im being asked if there exists a homomorphism $\varphi: R \to S$ of rings so that $\varphi(1_R) \neq 1_S$ - the title of a SO post spoiled that it does exist but im having trouble coming up with such a homomorphism (i didnt read the post) - my main idea rn would be to find a homo from a ring with inverses to one without inverses but it hasnt gotten me anywhere yet

[ruby sundial]

what is 1r

[delicate bloom]

identity in R

[ruby sundial]

like identity

o

nah m8

this is weird

there do?

[pastel cliff]

i admittedly close the SO page before reading forward just bc i didnt want spoilers

[ruby sundial]

how about the inverse

[bleak abyss]

Hmm, trying to think of how to suggest an example without spoiling

[delicate bloom]

I'd suggest looking at modular arithmetic modulo a composite number

[ruby sundial]

maybe that counts

[bleak abyss]

Do you define rings as not having a 1?

Or must they have 1?

[pastel cliff]

yeah i was trying to find a map Z/5 -> Z/4 but nothing i tried work

im 99% sure they're defined to have 1 in my class

[ruby sundial]

a map cant exist between those

[pastel cliff]

i realized as much lol

[ruby sundial]

well

[bleak abyss]

No need even for modular arithmetic I think

What's the first example you think of when you think of a ring?

[pastel cliff]

Z/p

[bleak abyss]

Can't be the first, I mean the first first first

Okay not 0 since I know someone's gonna say that as a troll

[pastel cliff]

i was about to

[bleak abyss]

Hint: say Z/p but stop halfway through

[ruby sundial]

lol?

[pastel cliff]

Z lol

[ruby sundial]

Z/

[bleak abyss]

Yup there's a Z->Z example

[pastel cliff]

oh

[bleak abyss]

Yup

Yougot it

[pastel cliff]

map everything to 0

[bleak abyss]

Or okay I was thinking double but that works too

Lmfao

[pastel cliff]

lmao i appreciate the nudges anywayss

was def putting too much work into it

[bleak abyss]

👍

[delicate bloom]

I was gonna suggest f(x)=2x from Z/5 to Z/10

[bleak abyss]

Lmfao

Anyway let's do some of my kinda algebra

Setup time

We have a field $\mathfrak{k}$ with a discrete valuation. Ring of integers is $\mathfrak{o}$, its maximal ideal is $\mathfrak{p}$, generated by $\overline{\omega}$, and $q$ is the order of the finite field $\mathfrak{o}/\mathfrak{p}$

[ruby sundial]

does latex work?

[bleak abyss]

Seems the bot is down

[ruby sundial]

I forget what a Dvr isn’t

the D part

what is discrete

[bleak abyss]

If K is a field, a discrete valuation is a map v:K->Z cup {infinity}

[ruby sundial]

yeah ok

[bleak abyss]

Such that v(xy) = v(x) + v(y), v(x+y) \ge min(v(x),v(y)), and v(x) = infinity iff x = 0

Ring of integers is gonna be shit whose valuation is positive

[ruby sundial]

a valuation not being discrete has Z being an ordered abelian group or something

yeah i remember now ty

[bleak abyss]

Something like that yeah

[ruby sundial]

i had a question today

in AG we proved valuation criterion for seperatedness

i dont intuitively know what it means for a map to be proper or seperated

ive been told seperated is an analouge of hausdorff

[bleak abyss]

The relevant point-set topology exercise is this

[ruby sundial]

but seperated is on a map not a scheme

it comes from image of diagonal being closed ive heard

[bleak abyss]

A space X is Hausdorff iff the diagonal is closed in X x X

[ruby sundial]

yrah

however this is a diagonal map, not a morphism of schemes

[bleak abyss]

According to stacks, separated is a thing for maps of spaces

[ruby sundial]

so is it like f seperated means domain is ‘hausdorff’ when viewed inside codomain?

[bleak abyss]

Seems like you have the space X \times_Y X

[ruby sundial]

yeah the diagonal factors through that

but i mean more geometrically

i think i just need examples to see the visuals

[bleak abyss]

Probably. Idk much AG unfortunately :/

[ruby sundial]

what is your specialty?

[bleak abyss]

Let's say automorphic forms

Right now I'm trying to learn about Bruhat-Tits tree

[ruby sundial]

woah

ive heard of automorphic forms but idk what they really are

ik there is something with them and laglands or something

langlands*

[bleak abyss]

Yeah

Anyway I'll continue what I was doing since I have an advisor meeting tomorrow lol

Time crunch 😛

[ruby sundial]

good luck

[bleak abyss]

But I'll try to go through these notes in a way that's interesting for someone who wants to follow

[ruby sundial]

oh you are gonna stream of consciousness

[bleak abyss]

Yup

[cloud walrus]

Sloth King Daminark

Happosai 八宝斎

[verbal harness]

eg, f(x) ≠ x

f(x) = x+1 is a valid example

[pastel cliff]

tex

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

[pastel cliff]

sorry to interrupt dami, but this isn't a ring homomorphism is it?

[bleak abyss]

Studying what does it break?

[pastel cliff]

$\phi(a)\phi(b) = (2a)(2b) = 4ab \neq \phi(ab)$ no?

[cloud walrus]

stμ₂dying

[bleak abyss]

Ah good point

Yeah 0 map then

[delicate bloom]

oh for that matter mine is wrong too, I should have said phi(x)=6x from Z/5 to Z/10

phi(x)phi(y)=6x6y=6xy=phi(xy

[pastel cliff]

im pretty sure this map also shows that ideals aren't necessarily preserved by a homomorphism

[delicate bloom]

this is why mapping id to id is morally good for ring homos yup

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

[ruby sundial]

how similar is to 2x?

because 5=5,10=2*5,6= 3*2

[delicate bloom]

well 2x isn't gonna work cause it breaks that property

2*2=4 while 6^2=6 (in Z/10Z)

[ruby sundial]

yea

but how similar are they to eachother

[delicate bloom]

the key is 6=0 mod 2 and 6=1 mod 5, so you're sorta sending it to 0 or 1 in different primes

if you want to try to work out a more general version for yourself to better understand it

[ruby sundial]

lol i dont know what goal is

[cloud walrus]

Sloth King Daminark

[bleak abyss]

Now

[cloud walrus]

Sloth King Daminark

[bleak abyss]

Okay sorry for the public disappointment but I think I've hit a point where it makes a lot more sense to blackboard it than to tex it

But I'll post highlights in here

Just not super detailed examples

[wicked zephyr]

i

HI

why is the group $F^{\times}\cap K^{\times n}/F^{\times n}$ finite?

[cloud walrus]

Potitov06

[chilly ocean]

Hello

Are people who are algebra experts also better Rubik's cubes solver?

[wicked zephyr]

no

[bleak abyss]

@wicked zephyr probably use the fact that K = F[\sqrt[n]{a}] for some a in F^{times}/F^{times n}

[wicked zephyr]

I think it is because $F^{\times}/F^{\times n}$ is a $\mathbb{Z}/n\mathbb{Z}$-module and the groups in question are finitely generated (in fact, by the classes of $a_1,\ldots,a_n$, where $K=F(\sqrt[n]{a_1},\ldots,\sqrt[n]{a_n}$).

[cloud walrus]

Potitov06

[cloud walrus]

Hausdorff

Hausdorff

[median pawn]

So any homomorphism must map these elements to a fourth root of unity and a sixth root of unity, respectively

[south patrol]

Lcm moment right

[bleak abyss]

Subgroup generated by the two is contained in 12th roots of unity

[median pawn]

Why is that? I know it seems obvious

[cloud walrus]

Hausdorff

[bleak abyss]

Well I'm saying all you need is in C^{times}

[median pawn]

but I don't see why any homomorphism maps elements of SL_2(Z) to a 12th root of unity

[bleak abyss]

So here's the idea

[median pawn]

The multiplication on SL_2(Z) isn't commutative

[bleak abyss]

Let's say phi:SL_2(Z) -> C^{times} is a homomorphism

phi(a) is a 4th root of unity, call it z

phi(b) is a 6th root of unity, call it w

[median pawn]

Agreed so far

[bleak abyss]

Give me anything in SL_2(Z), write it as a word in a and b

phi(a^{n_1}b^{n_2}... a^{n_k}b^{n_{k+1}})

Using that we're a homomorphism

That just gives you a word in z and w

This is the calculational way of saying this

The conceptual way is

If H is a subgroup of C^{times}

[median pawn]

Ahhhhhhh I see it now

[bleak abyss]

And phi:G->C^{times} is a homomorphism

If S is a generating set for G, then phi(S) is a generating set for phi(G)

So if that lives in H, the image of G lives in it as well

Out with it potato

[prime dock]

because it has a first cusp form in eisenstein series

[bleak abyss]

Huh

[south patrol]

Lol what

[bleak abyss]

It showed you as typing a few times

[chilly ocean]

There are many sorts of algebras where you have a zero and you have ways to take products of non-zero elements to get that 0. I'm interested in those structures.

[south patrol]

Oh sorry

[prime dock]

for what

[south patrol]

Yeah I typed up an answer but you gave more detail

[chilly ocean]

I don't understand what you're asking.

[prime dock]

.... why

[chilly ocean]

Why am I interested in such algebras that have non-zero elements that multiply to 0?

[median pawn]

Maybe elaborate some more? I'm familiar with modular forms to some extent

[median pawn]

I think the real reason this works is because multiplication in C^times is commutative

So a word in z and w is just of the form z^m w^n

Right?

or do you not need commutativity?

I presume such a result will not continue to hold if the codomain is not an abelian group, for instance

[bleak abyss]

Depends on what you mean by "such a result"

In general, if phi:G->H is a group homomorphism, then phi(G) is the subgroup of H generated by phi(S) where S is a generating set for G

[median pawn]

Did you use this, though? or do you just let a word be of the form z^{n_1} w^{n_2} z^{n_3} ....

[bleak abyss]

The key in this case is that the subgroup of C^{times} generated by a 4th and 6th root of unity is contained in the 12th roots of unity

[bleak abyss]

This is the fact which gets fucky depending on the codomain

[median pawn]

Ahh, I see, and that doesn't assume abelian-ness on H

[bleak abyss]

If you give me two finite subgroups maybe the subgroup they generate is infinite

(Excellent example: take the two generators of SL_2(Z), each generates a finite subgroup :P)

[median pawn]

Indeed! Thanks 😄

[median pawn]

I've a question

Why is the size of the subgroup generated by the two elements of orders 4 (say w) and 6 (say z) each, equal to 12?

Perhaps this is too basic

but here's what I'm thinking: the elements of the codomain group are commutative, so any element of the subgroup generated by w and z looks like w^{p} z^{q}

so elements in the image look precisely like ab, where a \in <w> and b \in <z>

[cloud walrus]

Hausdorff

Hausdorff

Hausdorff

Hausdorff

[median pawn]

and the said formula follows

@bleak abyss what do you think? i'm quite confused why the 12 appears

[bleak abyss]

What's H cap K in this case?

[median pawn]

on second thoughts, it's hard to say, isn't it? i think it depends on which roots of unity we pick

or we need to show that |H \cap K| = 2 in any case, H = <w> and K = <z>

[bleak abyss]

H is the subgroup generated by a 4th root of unity

But it's also the set of x such that x^n = 1 for some n dividing 4

Similar for K and 6th roots

[median pawn]

ok trying to see why this is true

[bleak abyss]

They're complex numbers

Just do the computation straight

[median pawn]

yeah i can do that

in fact i did but it gives no insight into why the equality is true

[bleak abyss]

mu_4 is the group {1, zeta, zeta^2, zeta^3} where zeta = e^{2pi i /4}

If you take any of those four elements and raise them to the 4th power you get 1, so they're all solutions to x^4 = 1. But polys of degree n have at most n solutions

So this is everything

But guess what? e^{2pi i/4} = i

So really it's just {+/- 1, +/- i}

Now you do something similar for mu_6

And it turns out the intersection is {1,-1}

This is the complex numbers way

You can also just say oh if zeta_n is a primitive nth root of unity and zeta_m is a primitive mth root of unity, what's the order of (zeta_n zeta_m)?

It's lcm(m,n)

Since we're in an abelian group

[median pawn]

ahh thank you! that's cool

what's up

btw here's a solution i saw

i wonder how they can specifically say i and (1 + root(-3))/2 are the (fourth and sixth) roots of unity involved

[tough raven]

S -> some 4th root of unity
All 4th roots of unity are in subgroup generated by i
So image of S is in subgroup generated by i (though not necessarily equal to i)
etc.

If $M'$ is a \emph{free} sub-module of a (right) module $M$ and $N$ is a (left) module over a ring $R$, is the map $$(\text{inclusion}) \otimes_R 1_N \colon M' \otimes_R N \to M \otimes_R N$$ injective?

[cloud walrus]

Raghuram

[median pawn]

subgroup generated by phi(i), i believe

but yeah the idea is clear now, thanks!

we're sort of upper estimating the image

if i am to say it in analytic terms lol

[tough raven]

?
i is in C^×
i is not in SL_2(Z) as I understand it

[median pawn]

oh right my bad

sorry

[tough raven]

Yes, (partially) in symbols
phi(SL_2(Z)) = <phi(S),phi(ST)> subset <i, other thing>
because phi(S) in <i>, phi(ST) in <other thing>, etc.
where <> = subgroup generated by

[tough raven]

If $U, V$ are modules over a ring $R$, when is the canonical map $U^{*} \otimes V \to \Hom_R(U, V) \colon f \otimes_R v \mapsto (u \mapsto f(u)v)$ injective (or more generally, what is its kernel)? When is it surjective (or more generally, what is its range)?

For example, when $R$ is a field, it is always injective and its range is the finite-rank linear maps from $U$ to $V$, so it is surjective iff $U$ or $V$ is finite-dimensional.

[cloud walrus]

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

[rustic minnow]

Im confused. Is there a typo in the last sentence? Shouldn't it be "since both f(x) and g(x) are of lower degree than p(x)?

[tribal moss]

Yes.

[rustic minnow]

Thanks, Im also wondering about the field extension $\mathbb{Q}_p/\mathbb{Q}$ I have seen around. Is the p here supposed to be prime? I guess we could factor the rationals as primes by factoring the numerator and putting negative exponents for the primes in the denominator.

[cloud walrus]

Fredrikpiano

[rustic minnow]

Since we cannot span over Q_p with a finite set of elements from Q I then take it that Q_p/Q is infinite dimensional.

[lethal dune]

any hint on how to show $A \to B$ is injective iff $\operatorname{Hom}(B, Q) \to \operatorname{Hom}(A, Q)$ is surjective for all injective module $Q$

[lethal dune]

one direction is obvious, any hint for the converse?

[agile burrow]

Try looking at an injective resolution of A

[lethal dune]

I tried

I don't see what to do after I get the resolution and a map from B →I^•

nvm got it

but do share your solution

[agile burrow]

If Q is the first nonzero term in the resolution, you have an injective map f: A -> Q and a map g: B -> Q such that A -> B -> Q is f. Looking at kernels, we have that ker(A -> B) is in ker f = 0

[formal ermine]

what's that symbol between zn and z2

is it a variant of the cartesian product?

[chilly ocean]

Semidirect product.

https://en.wikipedia.org/wiki/Semidirect_product

[lethal dune]

Can anyone give me an intuition behind the mapping cone of complexes?

I do not see why it's related to topological mapping cone

if at all

[next obsidian]

@latent anvil do u have thoughts

Maybe something about something being a homotopy equiv or something if it’s contractible or something

Also it’s used to make some LES right?

[lethal dune]

it says if the mapping cone is exact then the maps are quasi-iso

[agile burrow]

Ryu sama, you're working through Aluffi, right?

[lethal dune]

yes

[agile burrow]

Check out the exercises at the end of section 4, there's decent motivation there

[lethal dune]

ah it'll be long till I reach there

[agile burrow]

I mean, you can just read them

[lethal dune]

that's what I'm going to do now

the definition feels super artificial as of now

[agile burrow]

Yeah, once you see how it fits into certain exact sequences and specific factorization properties, I think it starts to make a bit more sense

Oops, you might want to read the definition of chain homotopy too if you aren't familiar with that

[lethal dune]

maybe doing AT alongside will give more insights

[agile burrow]

I'm in class rn but I can try to talk a bit more about this later

[latent anvil]

What's the question?

Ah

The mapping cone of both complexes and topological spaces is sort of a pseudo-cokernel

So in the case of topological spaces, the mapping of f : X -> Y is what you get if you make the image of f in Y contractible

@lethal dune does that make sense?

[lethal dune]

actually no

[latent anvil]

Sorry I got distracted by something irl

[latent anvil]

So let's look at this picture of the mapping cone from Wikipedia

If you look at just the f(X) part, that can be contracted to to the vertex of the cone

So sort of "up to homotopy" we've made f(X) trivial

(pls ping me in case I get distracted again)

[chilly ocean]

very nice interpretation

[latent anvil]

Yeah also the model category/infinity category perspective allows you to say exactly what sense in which this is true

The ordinary cokernel of f : X -> Y is the pushout of 0 <- X -> Y

The mapping cone is the homotopy pushout

(in the case of topological spaces, take everything to be pointed and interpret 0 as the singleton space)

This is also called the "homotopy cofiber"

[lethal dune]

so how does it relate to the cone construction of complexes?

tbf I haven't read homotopy of complexes yet

[latent anvil]

Ah yeah so I think you really can't understand the mapping cone without thinking about quasi isomorphisms

Have you seen the long exact sequence in homology of a short exact of complexes?

[lethal dune]

yes

[chilly ocean]

https://gyazo.com/7a85b53c7936102c096a73febbc9df72

how do i go about this?

[coral spindle]

Please don't spam your question in multiple channels. One is enough.

[lethal dune]

@latent anvil pinged because you asked to

[latent anvil]

Oops, sorry haha

So I think one motivation for the mapping cone of complexes is that it "explains" the LES in homology

[lethal dune]

okay so like Li -> Mi --0--> L(i+1) -> M(i+1).. now apply homology functor?

say something similar to this, $M^{\circ} \to MC(\alpha)^{\circ} \to L[1]^{\circ}$

[lethal dune]

How to draw M^•

[chilly ocean]

To help you out a bit, in any group, if x,y is in the group, then (xy) is in the group.
x^-1 and y^-1 and the inverse of xy is in the group.

Given the inverse of x and y, how do you compute the inverse of xy?

[latent anvil]

Omg I'm really sorry ryu, I've spent all day dealing with medical & insurance stuff and I keep getting distracted!!

I'll type something up

so say we have a chain map $f : C_\bullet \to D_\bullet$

[cloud walrus]

shamMy rock

[latent anvil]

the mapping cone is explicitly defined by $M_n = C_{n-1} \oplus D_n$, with a certain differential

[cloud walrus]

shamMy rock

[latent anvil]

we then have a short exact sequence $0 \to D \to M \to C[-1] \to 0$

[cloud walrus]

shamMy rock

[latent anvil]

given by inclusion and projection maps objectwise

now it's not hard to see that given a short exact sequence $0 \to X \to Y \to Z \to 0$ of complexes, we get for every $i$ an exact sequence $H_i(X) \to H_i(Y) \to H_i(Z)$ (ie there is no snake lemma-ing required)

[cloud walrus]

shamMy rock

[latent anvil]

in particular the mapping cone gives us an exact sequence $H_i(D) \to H_i(M) \to H_{i-1}(C)$

[cloud walrus]

shamMy rock

[latent anvil]

here's the thing I'm building up to

say $f$ is a monomorphism, so we have a short exact sequence $0 \to C \to D \to \mathrm{coker}\ f \to 0$

[cloud walrus]

shamMy rock

[latent anvil]

I claim that in this case, the homology of $M$ and $\mathrm{coker} f$ is the same

[cloud walrus]

shamMy rock

[latent anvil]

*are

actually, I have chosen not to explain this lol

Look at 1.58 of Weibel

It's an application of the 5 lemma and it's not super hard, but i don't want to talk about the mapping cylinder

but the point is, if f is mono then we have a map M -> coker f that induces an isomorphism in homology

and then we can fit together the exact sequences H_i(C) -> H_i(D) -> H_i(coker f) and H_i(D) -> H_i(M) -> H_{i-1}(C)

so the connecting homomorphism is really arising from an identification of the mapping cone with the cokernel! (in homology/"up to quasi isomorphism")

@lethal dune hopefully this helps explain why you might care about the mapping cone

[lethal dune]

I see

thanks for the time

[latent anvil]

tbh when I first learned this stuff the mapping cone made no sense to me

but I've been working a lot with triangulated categories recently and it's all clicked

unfortunately that means I dont have a great low tech explanation haha

[lethal dune]

you recommend weibel over aluffi for homological alb?

[latent anvil]

I think aluffi is a better introduction, but weibel covers far far more material

[lethal dune]

I'll look into triangulated cats soon

🗿

[latent anvil]

Eh, don't sweat it for right now

[lethal dune]

well we only had 4 classes on HA

[latent anvil]

if you have heard people say the words "derived category" that's all I'm thinking of really

[lethal dune]

and it'll be done on 5th

so like

[latent anvil]

but I would still suggest not rushing to learn about it

[lethal dune]

[latent anvil]

wait wdym?

they're talking about triangulated categories 5 lectures in?

[lethal dune]

exactly what I wrote

[latent anvil]

that seems like questionable pedagogy....

[lethal dune]

well no he didn't talk abt it

we won't be covering a lot but derived functors ab cats homotpies cones etc

our instructor is rushing through the course so yeah

[latent anvil]

I think a good introduction should do abstract abelian categories and derived functors, but it's okay not to dig too deep into the more homotopically flavored stuff

like the derived category, mapping cones, etc

oh sorry I misread what you said

[latent anvil]

yeah this is all standard for a first course

derived functors != the derived category

[lethal dune]

will they be useful in topology?

[latent anvil]

what specifically?

[lethal dune]

derived cats

[latent anvil]

hmm, not really. I would say they're more of an algebraic geometry thing

[lethal dune]

or trig cats

[latent anvil]

triangulated cats definitely yes though

the stable homotopy category is a triangulated category

[lethal dune]

okay I see

next is rep theory

god

[latent anvil]

nice!!

[lethal dune]

It's a lot

[latent anvil]

it sounds like the algebra class I took

[lethal dune]

we covered 2sem worth or material till now

Lol

[latent anvil]

we did groups/rings/modules in the first quarter, rep theory/galois theory in the 2nd, then homological and more advanced commutative algebra/baby AG in the third

you can do it!!

[lethal dune]

Lol

we did like grp/ring/modules over pid/proj-injective stuff/quadratic forms/miltilinear alg/homological and next rep

[latent anvil]

yee sounds very similar

[lethal dune]

comm algebra as well

you had that in your first year?

[latent anvil]

second year

wait

first year of what?

[lethal dune]

masters/phd?

[latent anvil]

ahh

[lethal dune]

definitely not ug so

[latent anvil]

not so fast lol

I took it in my 2nd year of my bachelors but it was a 1st year phd qualifying exam course

[agile burrow]

WOAH your algebra courses are crazy

[lethal dune]

Lol

you took that in your Bsc

[agile burrow]

I second aluffi for exposition, I think it gives a nice overview of how the pieces all fit together on the way to defining derived functors and all that

[latent anvil]

(I am currently in my first year of my PhD)

[lethal dune]

I see

[tough raven]

Is there a way of measuring the failure of a linearly independent subset of a module to be part of a basis?

For example, consider {(2,0,0), (0,1,0)} as a subset of Z^3. No linearly independent superset can span (1,0,0) because then (2,0,0) would be a linear combination with even coefficients which it isn't. On the other hand, {(2,0,0), (0,1,0), (0,0,1)} is a basis for 2Z^3, so in some sense we could measure the failure by 2. (In this case, 2 is the volume of the parallelopiped spanned by those two vectors; that might be generalisable to a suitable measure of failure-to-extend-to-basis.)

[coral spindle]

Sure, I would say you could do so as follows

[latent anvil]

I would say that the measure of any set's failure to span is the quotient by the span

[coral spindle]

Lol well yes

but there's a dual to that

[latent anvil]

you can then take numerical invariants like the cardinality or length, if you want

[coral spindle]

The syzygies

If you have an $R$-module $M$, and you have any set of elements $E \subset R$ (usually we choose these to be generators, but whatever) we then have a natural map from the free module on $E$ which I will notate $R^{(E)}$ into $M$

[cloud walrus]

Boytjie

[chilly ocean]

https://gyazo.com/d55cc47e55908110569f9b3b96f25d44

In a) the ans is there is identity in Z but not in E, I'm not sure how it proves there is no isomorphism, can someone elaborate on it?

[coral spindle]

The kernel of this natural map $R^{(E)} \to M$ is then called the syzygy module

[latent anvil]

Hey thrinity, please wait until the current discussion has finished to post your question

[coral spindle]

whoops

[chilly ocean]

sorry I didn't know about this rule

[cloud walrus]

Boytjie

[coral spindle]

It's not a rule, it's just etiquette

[chilly ocean]

got it

[coral spindle]

And well, this syzygy module is gonna be 0 exactly when E is a basis* (*for its image).

[latent anvil]

won't the syzygy always vanish, since they're asking about linearly independent sets E?

[coral spindle]

Oh, my bad

I misunderstood the question 🤡

[latent anvil]

no worries haha, I think this is also good

[coral spindle]

But yes I think this is relevant at least lol

[latent anvil]

but it's sort of dual, syzygy's measure the failure to be linearly independent

quotients measure failure to span

[coral spindle]

So I think

The quotient being free will do it

[latent anvil]

do what?

[coral spindle]

Like the quotient is free iff you can extend the basis

[latent anvil]

Yep

you can think of this as the fact that an SES ending in a projective always splits

[coral spindle]

Right right ofc

Well that answers the question in totality I think

[latent anvil]

my answer at least is "take the quotient by the span"

[coral spindle]

Just look at the freeness/projectiveness of the quotient

[latent anvil]

that's only a binary yes/no though, yeah?

[coral spindle]

And I suppose if you want to measure how far away that is, you can look at the projective dimension, but this is overcomplicating it a bit

[latent anvil]

whereas they were interested in measuring the failure to span

[coral spindle]

yeah

[latent anvil]

oh that's an interesting idea

I think I like that

hmm

thinking

right so umm

@tough raven, are you interested in linearly independent subsets of free modules or modules in general?

for a free module the projective dim of the quotient is at most 2

if I have my indexing right lol

[coral spindle]

Good point, if we have the module is free then this simplifies things a lot haha

[latent anvil]

if E is free and we have a linearly independent subset S, and the quotient if C, we get an SES 0 -> R^(S) -> E -> C -> 0

and this is a free resolution of C

[coral spindle]

I'd love to assume the ring is Noetherian too, I'm getting the heebie jeebies thinking about the nasty cases

[latent anvil]

oh yeah i mean every ring I deal with is uhh

"quasi-noetherian"

this is a technical term ive invented

[coral spindle]

LOL

Let me guess

Noetherian unless we say otherwise?

[latent anvil]

other way around :P

[coral spindle]

Hahahaha

[latent anvil]

non noetherian until literally the first opportunity where it would simplify things to assume noetherian

[coral spindle]

I'm gonna steal that

[latent anvil]

im giving back to the mathematical community

I should pr this defintion into the stacks project

[latent anvil]

If φ : R -> S is an isomorphism and 1 is an identity for R, then φ(1) is an identity for S. Try to prove this!

More intuitively, two rings being isomorphic should mean all "ring theoretic properties" of them are the same

and having an identity is a "ring theoretic property"

You'll get a feel for this as you do more algebra

[chilly ocean]

https://gyazo.com/d10699346a4b1a46a997da7523eed0e5

Why does the solution assume degree of f(x) is 1? Why can't the degree of f(x) > 1?

[elder wave]

Because it divides polynomials of degree 1

[chilly ocean]

x^2+x / x no?

[latent anvil]

What do you mean thrinity?

Which one is f?

[chilly ocean]

f(x) = x^2 + x

[latent anvil]

I think you're mixing up "divides" and "is divisible by"

x divides x^2 + x, but x is not divisible by x^2 + x

And our condition on f would require the second thing to be true

Does that make sense?

[chilly ocean]

makes sense now, thanks

[chilly ocean]

If f(x) = cg(x), f(x) is reducible right?

[delicate bloom]

no, if you're saying c is just some scalar constant

[chilly ocean]

so when f(x) = g(x)h(x), the lowest possible deg of g(x) and h(x) is 1 right

[delicate bloom]

why? g(x)=1 and h(x)=1 is possible as far as I'm aware

I didn't read what came earlier if I'm missing context here

[chilly ocean]

like if degf(x) is 3, the possible deg of g(x) and h(x) can only be 1+2 since 3 + 0 is not possible

[delicate bloom]

why can't h(x)=1

what set is h(x) taken from? Q[x]?

[chilly ocean]

how does where it's taken from matter?

[delicate bloom]

you're telling me deg(h(x)) can't be 0

it depends on what h(x) can be

if it's an element of Q[x], it certainly can be

[chilly ocean]

so if it is Q[x], it can be 1+2 and 3+0 ?

The degree of g(x) and h(x) needs to be in Q[x] right?

[bleak abyss]

So analysis on the Bruhat-Tits tree time

[bleak abyss]

Yeah lol

Got distracted again

[agile burrow]

it's ok, I have to do probability hw rn

but if you get around to it, I'll come back and read it later

I wanna learn buildings and stuff eventually, idk a good entry point

[bleak abyss]

Troposphere spit it out

[tribal moss]

(Wasn't saying anything, just trying to type somewhere else when it turned out Discord had the keyboard focus instead)

[bleak abyss]

Okay so

[cloud walrus]

Sloth King Daminark

[tribal niche]

How can I classify all the homomorphisms from the dihedral group of order 2n to the cyclic group of order n?

I know any such homomorphism is uniquely defined by where it sends r and s (the generators), but I'm not sure how to show the resulting function is a homomorphism, e.g. how would I simplify r^a s^b r^c s^d?

[agile burrow]

If you have a presentation of a group G= <S | R>, then a map S -> H defines a homomorphism if the images of the generators satisfy the relations R

This should help you limit the possibilities of where you can map r and s in C_n and make it easier to classify the homomorphisms

[sonic parcel]

this is <x(x-1),y(y-1)> yh

?

[tough raven]

By taking a preimage of the basis, right?

[tough raven]

I originally came up with this question for free modules, but then thought it would be cool if there was something in general (so for a non-free module, it would always be “non-zero” because nothing can be extended to a basis).

[tough raven]

Identities are not necessarily preserved by ring homomorphisms unless it's required in the definition. (The image of the identity will be an identity for the image, but not necessarily for the entire codomain.)
As a counterexample for monoids rather than rings, consider N with the operation max and the endomorphism x->x+1.

[tough raven]

I think so.
||Consider f(x,y) as a polynomial in x with coefficients in C[y] and do long-division by x(x-1), getting f = x(x-1) q1 + g; by the Remainder Theorem, the remainder g(x,y) = g_0(y) + x g_1(y) vanishes for y = 0,1 so using the Remainder Theorem again, g_0, g_1 are divisible by y(y-1), say g_i = h_i y(y-1), so f = (q1)x(x-1) + (h_0(y) + h_1(y) x)y(y-1).||

[tough raven]

TBH I don't really know what that is.

[tough raven]

Can it measure something like this example?
Say, if I have a maximal linearly independent subset (which should be equivalent to all elements of the quotient having torsion), then there's no extending it, so the answer should be something like the size of the quotient. Does the projective dimension behave like that?

[chilly ocean]

hello

what is the operation of the group SO(3) called in math ?

the group SO(3) consists of all "rotations" of a sphere in R^3, so the operation of SO(3) can be called "composition" ?

and more over

can R x SO(3) represent the group of all "translations"/"transfers" of a point in R^3 ?

[wooden ember]

The operation is either matrix multiplication if you’re viewing this as matrices or composition if you’re viewing it as linear transformations (doesn’t really matter)

[wooden ember]

As for this, I’m not sure it’s the right group to consider

Affine transformations do not have a product that follows a direct product rule

[chilly ocean]

by R x SO(3) I am trying to represent: 1st move along an axis and 2nd do spherical rotation

[wooden ember]

The orthogonal direct affine transformations of R^3 would be represented by R^3 : SO(3) where : is a semidirect product if I’m not mistaken

[wooden ember]

Which axis though? This won’t have a nice geometric interpretation since multiplication won’t correspond to an element that is moving along that fixed axis and then rotating

[chilly ocean]

Maybe I should ask where or what material I can study to understand well about:

https://en.wikipedia.org/wiki/Isometry_group

https://en.wikipedia.org/wiki/Euclidean_group

[wooden ember]

Im not familiar with texts on geometry so we should wait around for someone else to give you good resources

[chilly ocean]

specifically, is SE(3) isomorphic to some "combination" of other isometry groups?

[tribal niche]

do you have a link to a proof of this fact?

[next obsidian]

It follows by what the definition of a presentation is

[coral spindle]

I can't vouch for this entirely by myself, but I've heard that "Introduction to Geometry" by Coxeter is a good text.

[wooden ember]

Eyy Boytje is yellow (well deserved 👏 )

[delicate orchid]

It’s called the substitution test iirc

[rotund aurora]

Geometry by Audin, Continuous symmetry by Barker, and maybe Elementary Geometry by Agricola

I think these may be good resources, it is what I'm currently reading on these topics

[coral spindle]

I've been told to try and prove that the reflection group G(2,2,3) is isomophic to the symmetric group on 4 elements. I can't think of what four things the action permutes – any hints on what to look for?

In case anyone doesn't know, the group G(2,2,3) is a subgroup of GL_3(C) generated by (1) the diagonal matrices with entries +/- 1 with determinant 1, and (2) the permutation matrices.

[pastel cliff]

Suppose that $R$ is a ring with identity and that $\phi : R \to S$ is a surjective ring homomorphism. Show that $S$ is also a ring with identity.

this is p simple but my idea rn is that $\phi(r)\phi(1) = \phi(r)$ implies that $s_1s_2 = s_1$ and this is only the case when $s_2 = 1$, but is this true?

[cloud walrus]

stμ₂dying

[coral spindle]

Have you tried proving it?

Ok well I'll take that as a no, so why not try proving it

[pastel cliff]

i mean i dont see how s1s2 = s1 where s2 \neq 1 breaks something but i also dont see when it would be the cat

and i feel like it should break something

[coral spindle]

well spoiler, that's false.

Try showing it with the specific setup you have.

[pastel cliff]

i guess 0 makes that obvious

s1s2 = s1 when s1 = 0 and s2 is anything

[coral spindle]

Now may be a good time to remind yourself of the definition of the unit.

By which I mean the multiplicative identity, of course.

[coral shale]

i aint done rep theory, but curious if S4 also being the group of cube rotations has any relevance at all. random thought

[coral spindle]

True, not entirely sure at this point

[coral shale]

Rip, well all i know is best way probably to find 'helpful' actions 😓, oh u wrote that already lol

[pastel cliff]

an element s \in S such that s*x = x for all x, nothing too complicated

[coral spindle]

Not quite.

Unless you're working with commutative rings only, this definition is not right.