groups-rings-fields

[woven delta]

Do you get why it's true?

[shell brook]

I see where the kernel comes from

uh

I have read the proof a few times

[woven delta]

This is what I meant

Okay good

[shell brook]

I just don't get like

How we claim to know stuff about the contents of the kernel

I mean I guess you could write them down

[woven delta]

Well we do know some stuff

So we have our generators for E and also sqrt(5)

And an automorphism is determined by the generators

Right

[shell brook]

hm

yes

[woven delta]

So if we are fixing E

Then we are fixing all of the generators of E

[shell brook]

yes

oh

[woven delta]

So the only generator left

Is sqrt(5)

And we know that this group has order 2

So we know what the nontrivial element is

[shell brook]

sqrt5 -> -sqrt5 :d

[woven delta]

Yes

Okay yeah now do you see

We actually can look at the automorphisms very explicitly

Via the generators

[shell brook]

But isn't {id, sqrt5 -> -sqrt5} also the kernel of r_{E(sqrt5)}?

Like it's Gal(E(sqrt5)/E)

[woven delta]

No

[shell brook]

oh

[woven delta]

This is a little off

Try again

[shell brook]

I am reading here

[woven delta]

Yeah

That was for E

Not for F

I also made some mistakes

Earlier

[shell brook]

I am confused okay wait

[woven delta]

Okay read this again

But yeah also think about what the kernel of the restriction map means

[shell brook]

ohhh

fuck

we're not supposed to be talking about ker(r_{E(sqrt5)})

but ker(r_E)

thats why im mixed up

[woven delta]

Well yeah

The first one doesn't really make sense

[shell brook]

[woven delta]

You're restricting to yourself

Lol

[shell brook]

ker(r_{Q(sqrt5)}) = Gal(E(sqrt5)/Q(sqrt5))
ker(r_E) = Gal(E(sqrt5)/E))

is that right

[woven delta]

Yes

That's correct

[shell brook]

Okay and then
Gal(E(sqrt5)/Q(sqrt5)) = {id, sqrt5 -> -sqrt5}

and so why isn't this true:
Gal(E(sqrt5)/E) = {id, sqrt5 -> -sqrt5}

[woven delta]

This one isn't true

[shell brook]

[woven delta]

But anyway, you don't have to analyze that one

Just show the kernels intersect trivially

[shell brook]

Wait so
Gal(E(sqrt(5)/E) = {id, sqrt5 -> -sqrt5}
is true

[woven delta]

Yes

[shell brook]

okay that make sense

[woven delta]

Yeah it should actually

[shell brook]

how do i show sqrt5 -> -sqrt5 notin Gal(E(sqrt5)/Q(sqrt5))

like why is that not true

[woven delta]

I was overcomplicating the explanation a little

Because you are fixing Q(sqrt(5))

[shell brook]

Right any aut must have s(Q(sqrt5)) = Q(sqrt5)

[woven delta]

Not just that

[shell brook]

I thought this was up to permutation?

[woven delta]

That's too weak

[shell brook]

oh okay

nice

[woven delta]

It fixes pointwise

Fun fact

If a subgroup of a Galois group is normal it is fixed by every automorphism of the Galois group

[shell brook]

[woven delta]

But that's not pointwise

Sorry one sec

[shell brook]

I do not see why it has to fix pointwise

[woven delta]

Let me rephrase that

That's the definition

Gal(E/F) is defined to be the automorphisms of E that fix F pointwise

[shell brook]

oh I thought we just needed s(F) = F for any aut

I never realized that meant

pointwise

[woven delta]

That's good you're realizing it now then

Better late than never

[shell brook]

the proof for the 2.9 doesnt seem to say that is my confusion point maybe

like it says s(B_i) = B_j and so it preserves the set

[woven delta]

Yes

[shell brook]

im reading preserves as "permuting"

[woven delta]

Are those generators for F

Preserves means fixes

Oh one sec

Let me verify

[shell brook]

but how do you conclude fixes from the s(Bi) = Bj

[woven delta]

Oh lol

No those beta are for E

Not for F

Anyway here's a definition

Hey Dackid

So yeah

This is pretty important to know Jesse

[shell brook]

wow it is crazy I didnt know this lmao

[woven delta]

Yeah

[lethal cipher]

So question, let's say two groups have the save cardinality, and we are given that they have the same number of generators with corresponding order.
Is this enough to say the groups are isomorphic?

[woven delta]

I was very surprised

No Dackid

For example S_3 and Z/6

Can have the same number of generators

[lethal cipher]

Ah, Abelian is an important thing to bring up

[shell brook]

ah .

[woven delta]

of generators is bad, because it's not really well defined

Yeah jesse

I was wondering why you weren't understanding

[shell brook]

I just can't read

[woven delta]

And I guess we found the hole

This is good to figure out

[shell brook]

I just internalized s(F) = F and not the sentence before LMAO

[woven delta]

And now you can make progress

[shell brook]

yeah now i get it

tysm

[woven delta]

Wild

[old hollow]

What’s the deal with a “Galois extension” vs an ordinary field extension, why are the conditions normal and separable

[shell brook]

a galois extension E/F is one where |Aut(E/F)| = |E:F|

[lethal cipher]

Hmm okay. Well I need to show that the Galois group of the spitting field of x^4-2 is isomorphic to D_8.

The reason I mention the generator thing is because there are two elements that generate the galois group. One of order 4 and order 2. So it is easy to see how the isomorphism would map the generators to R and F in D_8

[old hollow]

Oh that’s the important part

[shell brook]

equivalently, if E is a splitting field for a separable polynomial over F, then E is Galois

[old hollow]

I assume it also has to do with the fundamental theorem of Galois theory

[woven delta]

A really nice definition is a Galois extension K over F is one whose automorphism group Aut(K/F) has fixed field K

Does D_8 have 8 elements or 16?

Cause there's 2 conventions

D_n vs D_2n

Sorry 4 or 8

Wait no

[lethal cipher]

Since the Galois group has order 8, I'm gonna say D_8 has 8 elements.

[woven delta]

Okay cool

[lethal cipher]

I was taught the other convention in my other class

[woven delta]

So symmetries of a square

[lethal cipher]

So I know what you're saying

Yes

[woven delta]

Oh and that's generated by rotations and a flip

Right

It's been a while

Okay so you have the fourth roots of unity

Right?

How are you looking at the Galois group

[lethal cipher]

Yea.
The first map fixes $i$ and sends $\sqrt[4]{2}$ to $i\sqrt[4]{2}$. The second map is the one that fixes $\sqrt[4]{2}$ and sends $i$ to $-i$.

[cloud walrus]

dackid

[lethal cipher]

These two generate the Galois group

[woven delta]

Okay

So you can show that they satisfy the relations of D_8

Which is a pretty small list

Or really you can make an explicit isomorphism

If you want

[lethal cipher]

Yes. The first map is of order 4 and the second map is of order 2.
So send the first to R and send the second map to F

[woven delta]

If you show that two finite groups have the same set of generators, one satisfies the relations of the others, and they are the same cardinality then they are isomorphic

So this is sort of what you were saying before

[lethal cipher]

What do you mean by satisfies the relations?

[woven delta]

But you also need to know how the generators interact

Do you know about presentations of groups

[lethal cipher]

No I do not

[woven delta]

Oh lol okay nevermind then

So you need to know how to multiply the generators

That's what relations are

Some word made up of a product of the generators and the inverses of the generators = e

That's a relation in a group

A presentation of a group is a set of generators of the group and a set of relations between those generators so that the group is the maximal group that is generated by those generators and satisfies those relations

That's what I was talking about

Which is I think what you were getting at before

[lethal cipher]

Let's maybe step away from this word stuff. It's late, my brain isn't having it, and it doesnt need to for the problem.

[woven delta]

Yeah I get it

Just wanted to give exposition about what you asked initially

But okay

Yeah in that case there's just 8 elements so just do an isomorphism

Explicitly

You can describe the elements in terms of the generators pretty easily

And then just do the same for D8

And verify that the isomorphism works

I guess that would be my suggestion

If you want to just get this done

[lethal cipher]

I guess showing that it is a homomorphism is a bit trickier since you'd have to check the elements individually.

[woven delta]

Not really

Okay just do this for the generators actually

Okay so let's name the generators

[lethal cipher]

Sigma will be the one with order 4 and tau can be the order 2

[woven delta]

Sure

Okay now let's name the generators for D_8 to be alpha and beta

Alpha is rotation

[lethal cipher]

Oh, those are usually R and F

[woven delta]

Sure

Okay R and F

[lethal cipher]

R for rotation and F for flip. I know, it's super original

[woven delta]

Okay so now what are the elements of the Galois group in terms of sigma and Tau?

{e, sigma, sigma^2, sigma^3, Tau...}

4 more

[lethal cipher]

${ e, \sigma, \sigma^2, \sigma^3, \tau, \sigma \tau, \sigma^2\tau, \sigma^3\tau}$

[woven delta]

3 rather

[cloud walrus]

dackid

[woven delta]

Okay cool

And there's a relation between sigma and tau

Right

[lethal cipher]

Then if we let $\phi$ be our map, we map $\phi(\sigma)=R$ and $\phi(\tau)=F$

[cloud walrus]

dackid

[woven delta]

Yes

What is sigma tau

In terms of Tau sigma^i

[lethal cipher]

Sigma tau= tau sigma^3

[woven delta]

Okay cool

[lethal cipher]

Iirc, $R^k F=FR^{n-k}$

[cloud walrus]

dackid

[woven delta]

Cool

So yeah that actually should basically do it lol

Because that relation let's you transform any product

Into something of the form sigma^i tau

[lethal cipher]

Why is knowing this necessary?

[woven delta]

Okay so if you know this you can transform any product of two elements into one in the standard form Tau sigma^i

And then you show that phi(tau sigma ^i) = FR^i

Basically

Lol okay there's a picture in my head you can't see

[lethal cipher]

Okay. So we dont know it's a homomorphism yet. Then how do we determine $\phi(\sigma \tau)$?

[cloud walrus]

dackid

[lethal cipher]

Actually, we define the map to do so. Duh

[woven delta]

No so I'm saying we define phi(tau sigma^i) = F R^i

Yeah

So let's say we have a product ab

Then we can apply a bunch of moves using that relation to turn it into that form

And then it gets mapped

And then we unapply those moves

[lethal cipher]

Yea, I understand what you're saying

[woven delta]

So like for example phi(Tau sigma Tau) = phi(Tau Tau sigma^3) = phi(sigma^3)

And then that is R^3

And then R^3 = FF R^3 = F RF

Yeah

So that's what a relation does

[lethal cipher]

Ah I see what you're saying

[woven delta]

Yeah it's nice

[lethal cipher]

Im gonna call it for the night. Thank you for the help

[woven delta]

So I guess the map is phi(tau^i sigma^j) = F^i R^j

Yeah no problem

[chilly ocean]

Lick my ass

Is 0 a constant?

[lethal dune]

that used to be my pubg user name

[frank fiber]

hi, in the semidirect product there is a map $k:H \rightarrow G \rtimes_{\varphi} H$, is this map given by $k(h)=(1,h)$?

[cloud walrus]

Or x1

[frank fiber]

or $k$ is given by $k(h)=(1,f(h))$ with $f$ a derivation?

[cloud walrus]

Or x1

[frank fiber]

$G$ and $H$ are groups and $\varphi:H \rightarrow Aut(G)$ an action

[cloud walrus]

Or x1

[patent ocean]

Need help with 5bii)

For the first one I got 4 homomorphisms but im not too sure

i got this from mapping 1 to either : 0,5,10,15

[tribal moss]

Almost right. Except 15 is the same as 0 in Z_15.

[patent ocean]

Oh yes

what is the reasoning of mapping the generator

I used that method but idk why that works

[south patrol]

The image of a map is determined uniquely by the images of the generators, so it reduces the necessary work

[patent ocean]

I was thinking an identity map would be a homomorphism but apparently not

[tribal moss]

"Identity map" works only if the domain is a subset of the codomain.

[south patrol]

S4 has cyclic subgroup of order 9,because it has no elements of order 9 right

[tribal moss]

Z_3×Z_3 has no elements of order 9 either, but does have a subgroup of order 9 (namely the entire group).

[chilly ocean]

9 doesnt divide 24

[south patrol]

I mdan its order is not divisible by 9

[chilly ocean]

@delicate orchid ?

[tribal moss]

The problem here didn't ask for injetive homomorphisms, though, so we're not just looking for subgroups isomorphic to Z_9.

[south patrol]

Sure, that was sloppy of me, I mean no cyclic subgroup

[patent ocean]

so if it was Z15 to Z18 would that work?

[tribal moss]

No, because the elements of Z_15 are conjugacy classes modulo 15, and the elements of Z_18 are conjugacy classes modulo 18.

[patent ocean]

oh

[tribal moss]

However, even if the elements of one group are literally also elements of the other group, the group operations also need to agree before the identity map is a homomorphism.

[patent ocean]

oh ok

so you know how I said that a homomorphism for the 1st one is mapping 1 to either 0,5 or 10

what are the other elements of Z18 mapped too

[tribal moss]

Each of the elements of Z_18 can be writtten as 1+1+1+...+1 for an appropriate number of ones.

Since f is a homomorphism, you have f(1+1+1+...+1) = f(1)+f(1)+f(1)+...+f(1).

On the RHS of this, the additions happen in Z_15.

[patent ocean]

oh yes

so the mappings of the other elements are just dependent on whatever the generator is mapped to

[tribal moss]

Exactly.

[patent ocean]

makes sense

thanks

[tribal moss]

And you can use that in the other two subquestions too.

[patent ocean]

appreciate it

[tribal moss]

(For (iii) you need to be a bit lucky in your choice of generators for S_5).

[patent ocean]

what do you mean by 'lucky'

[tribal moss]

There are several different ways to select a few elements of S5 that generate all of S5. Some of these will make (iii) very easy; others will send you on a wild goose chase.

[patent ocean]

oh ok

[tribal moss]

Fortunately I think the most natural choice is one of the easy ones.

[south patrol]

Another way to think about iii) is to think about subgroups of Z_5

altho the above is quick^

[patent ocean]

Thanks will keep this in mind

[patent ocean]

Been tackling the final bit for quite a while and I don't think I've come up with the right generators

i used (5,4,3,2) cycle type

[south patrol]

there is a generating set with only two elements so try that

[tribal moss]

My suggestion would be to use a generating set composed entirely of transpositions.

There'll be more of them in your generating set, but that is compensated for by the simplicity of mapping them into Z5.

[prisma shuttle]

can someone explain what the quotient ring $\mathbb{Z}_4/(x^2-2)$ looks like

like intuitively

[cloud walrus]

JustKeepRunning

[barren sierra]

How do I do c and d?

c I feel is really simple

I want to say the only elements of finite order are +- 1

and since 1 + sqrt(2) =/= +- 1 it can't be of finite order

[stone fulcrum]

@prisma shuttle
You might mean Z4[x]/(x² - 2)
Then this is just Z4 except with a new element x such that x² = 2

[prisma shuttle]

by Z4 u mean $\mathbb{Z}/4\mathbb{Z}$ right?

[cloud walrus]

JustKeepRunning

[stone fulcrum]

Yeye. The ring, anyway

[prisma shuttle]

oh so u just like add a new element?

[stone fulcrum]

That's what it will end up looking like. You add the solution to x² - 2 = 0

[prisma shuttle]

ok thx u

[stone fulcrum]

@barren sierra
The unit is, of course, 1/(1 + √2)

[barren sierra]

Oh I meant the last part of c

I showed the unit lol

should have specified

like showing that u is of infinite order in R^x (meaning it is of infinite order in S^x)

[stone fulcrum]

Sorry, I said "unit" when I meant "multiplicative inverse"

But you got that too? Okay. Just show it has infinite order

Take the magnitude

Since the magnitude is above 1, the magnitude will grow without bound as you multiply. Can never become 1

[barren sierra]

oh that simple?

[stone fulcrum]

Takes advantage of R, so I figure that's what they're getting at

[barren sierra]

yea

and then d, I know of classifications of irreducibles in something like Z[i]

but not for general Z[sqrt(D)] like Z[sqrt(2)] here

I know 1 + sqrt(2) is not irreducible since it is a unit

but unsure about ii and the rest

nvm I think I got it

norms and prime elements

actually what defines a norm map lol

I've only just seen "yea this function is a norm map"

[shell brook]

In this context the norm of Z[sqrt(d)] is N(a + b*sqrt(d)) = |a^2 + d*b^2|

thats the actual definition

[barren sierra]

hm

ok

[shell brook]

maybe you can define it for other rings but idk

ive only seen it for Z[sqrt(d)]

i guess you could transfer it around to other integral domains and keep the cool + sick properties it has

anyways i have a galois theory question nwo

this is part 4 of my 4 part question where I have solved the other three parts.
I think this is relatively easy because all you need to do is determine the automorphisms that send sqrt5 + cbrt2 to itself. But since we have a decomposition of the Gal(K/Q) group into components, we really just need to find the automorphisms that send sqrt5 -> sqrt5, and the ones that send cbrt2 -> cbrt2.

Clearly for sqrt5 it is easy since there is only one non-trivial a.m., and it can't work. So identity is the only permissible am.

For cbrt2->cbrt2, we know that E \cong S_3, and we also know that E=Q(a, wa, w^2 a), where a = cbrt2, w = 1/2(-1 + sqrt(-3)). So I think all we really care about are the cycles in S_3 that fix a. So if we identify the elemtns of S_3 as {1, 2, 3} with 1 being a, then there are only two cycles that fix 1, those being identity, and (2 3). So there is only one non-trivial cycle that fixes cbrt2.

So now I think we can say that the automorphisms that fix sqrt5 + cbrt2 are identity, and the automorphism that is given by (2 3) in S_3. That is, there are only two such automorphisms. So Gal(K/L) has order 2, and so is isomorphic to Z_2.

i just need sanity checking

esp this part

So I think all we really care about are the cycles in S_3 that fix a.

[latent anvil]

@shell brook wgats your problem?

[shell brook]

its (iv)

in the image

[latent anvil]

kk

what have you done so far?

[shell brook]

im just sanity check my answer

i think i solved it

lol

[latent anvil]

oh okay

lmao

sorry

[shell brook]

Gal(K/L) cong Z_2

[latent anvil]

reading

That makes sense to me!

[shell brook]

lets goo

this is the first time ive solved a galois theory question without emma help

[latent anvil]

pog

[chilly ocean]

If you have H be quaternions and E the subset of imaginary quaternions, x such that x*=-x and a function a:E->E defined by fixing element g such that g is a unit quaternion and a(x)=gxg-1.
What is transpose of a?

because a can be represented as a matrix

im trying to show a*a(transpose) is identity

[latent anvil]

@chilly ocean so a basis for the vector space E is i, j, k right?

and you can write g = p i + q j + r k for some real p, q, r

then you can compute a(i), a(j), a(k) and they'll look like u i + v j + w k

these will give you the columns of the matrix representing a

however the better way to think about this is that a matrix A is orthogonal, i.e. A^T A = A A^T = I, if and only if A preserves norms, i.e. |A v| = |v| for all v

[chilly ocean]

o

[latent anvil]

(this is with respect to an orthonormal basis, which i,j,k is)

and we know |a b| = |a| |b| for the quaternion multiplication/norm right?

[chilly ocean]

yes

we know the norm is preserved

[latent anvil]

right

so

here's a way to think about this

[chilly ocean]

i feel like computing the matrix would be a waste of tkme

[latent anvil]

you can think of vectors in E as things in R^3, by looking at them as a i + b j + c k

it would

this means E has an inner product on it, just by taking the dot product of the coefficients

with me so far?

[chilly ocean]

yes

[latent anvil]

and this inner product induces a norm

[chilly ocean]

we know the inner product

[latent anvil]

and that's the usual norm on E

rigth?

[chilly ocean]

yes i think so

[latent anvil]

the norm of ai + bj + ck is a^2 + b^2 + c^2

i.e. (a,b,c) dot (a,b,c)

[chilly ocean]

it should be

[latent anvil]

so now let's think about R^3

or R^n

why does A satisfy A^T A = A A^T = I iff A preserves norms?

well <Av, w> = <v, A^T w>

right?

[chilly ocean]

yes

thats part of definition

[latent anvil]

yep

and you can write the inner product in terms of the norm

this is sometimes called the polarization identity

<v + w, v + w> = <v, v> + 2 <v, w> + <w, w>, so <v, w> = (|v+w|^2 - |v|^2 - |w|^2)/2

[chilly ocean]

yes

[latent anvil]

try to use these two facts to prove this statement

then you can prove the thing about E pretty easily

bc obviously multiplication with a unit quaternion is norm preserving

Does that all make sense?

[chilly ocean]

the two facts being this and this?

[chilly ocean]

here

[latent anvil]

Yep, those two

Oh and also <v, w> = 0 for all w iff v = 0

That might come up

Very useful fact

So the idea this is reasonable is because

• the inner product can be written in terms of the norm
• the transpose can be defined in terms of the inner product

Right?

So norm preserving should tell us things about the transpose

But you can also just compute the matrix

[chilly ocean]

uh

[latent anvil]

?

[chilly ocean]

ig doing converse is weird imo

but ill spend 2 min thinking about it

[latent anvil]

Right, the converse is the hard part

And it's the part you need for the quaternion problem

What was your proof for the forward direction?

[chilly ocean]

idk but intuition telling me something along lines of A^-1=A^T and <Av,w>=<v,A^-1w>. So im guessing we look at <Iv,w> or something and then expand out?

not actually sure at this stage

oh lol

nvm

Im trying to show <Av,Av>=<v,v>

so <Av,Av>=<v,A^-1Av>=<v,v>

other direction probably uses that polarization thingy

given <Av,Av>=<v,v>
<Av,w>=<v,A^Tw>= |Av+w|^2-|Av|^2-|w|^2 / 2=|v+A^-1w|^2-|v|^2-|A^-1w|^2 /2
this simplifies
|Av+w|^2/2=|v+A^-1v|^2/2

even further we get |Av+w|=|v+A^-1w|

So<Av+w,Av+w>=<v+A^-1w,v+A^-1w>

ah nvm

i probably fucked up

[latent anvil]

Sorry I forgot about this

[latent anvil]

Yeah this looks good!

[latent anvil]

Not sure what's going on here

[chilly ocean]

yeah i messed up

[latent anvil]

So just applying A to v isn't super helpful because you end up with the mixed term |Av+w|^2

[chilly ocean]

yeah

[latent anvil]

what if you apply A to both?

What does that tell you?

[chilly ocean]

|A(v+w)|^2=<Av+Aw,Av + Aw>

uh

[latent anvil]

yep

That's true

[chilly ocean]

im trying to see how A^-1 comes in

or A^T i should say

[latent anvil]

Just try expanding things

[chilly ocean]

hmm

[latent anvil]

We know the inner product is bilinear, right?

[chilly ocean]

okay

yeah

[latent anvil]

And symmetric

[latent anvil]

So you can expand this out

What do you get?

[chilly ocean]

one term we get becomes a <v,A^Tw>

[latent anvil]

I'm not sure what you mean

[chilly ocean]

ill expand it

[latent anvil]

That doesn't sound right to me

due

*sure

[chilly ocean]

<Av+Aw,Av+Aw>=<Av,Av+Aw>+<Aw,Av+Aw>=<Av,Av>+2<Av,Aw>+<Aw,Aw>=<v,v>+2<Av,Aw>+<w,w>

[latent anvil]

Yep!

And the left hand is <v+w, v+w>, right?

[chilly ocean]

yes

[latent anvil]

What do you get if you expand that?

[chilly ocean]

same thing but mixed term is 2<v,w>

so <Av,Aw>=<v,w>

[latent anvil]

Right!

[chilly ocean]

and then

okay

i sorta see whats next

<Av,Aw>=<v,A^TAw>=<v,w>

tasa

[latent anvil]

Yeah not quite

Right!

Now what do you want to know about A^TA?

[chilly ocean]

it has to equal identity i think

[latent anvil]

Right

So what does that mean in terms of vectors?

[chilly ocean]

AtAv=v

[latent anvil]

Right

[broken stirrup]

Hi everyone, can someone elaborate on this proof?

[latent anvil]

So try to deduce that from <v, A^T A w> = <v, w> for all v, w

Yeah?

[chilly ocean]

yeah i get it

<v,w>=<v,z> iff z= w?

nah

[latent anvil]

There are more assumptions here than what you've shown (eg is A has a zero radical rad(A)N = 0 is automatically true)

Is A A semisimple ring?

[chilly ocean]

oh yeah

<v,AtAw-w>=0

[latent anvil]

Yeah!!

[chilly ocean]

if we supposed that v is nonzero then rhs is

okay

[latent anvil]

What do you mean?

v could be nonzero but orthogonal to A^TA w-w

Right?

[chilly ocean]

if v nonzero then AtAv -w=0

[latent anvil]

Why?

[chilly ocean]

because <v,w>=0 iff v=0 or w=0

[latent anvil]

That's not true

Take two orthogonal vectors

[chilly ocean]

oh true

you said this

[latent anvil]

I said something else

I said that for any v, if <v, w> = 0 is true for all w then v = 0

Does that make sense?

[broken stirrup]

this is another formulation of the problem

[chilly ocean]

yeah

because v cant be orthogonal to all w unless it is zero vector

[latent anvil]

A/J(A) isn't semisimple for every ring though, right?

What about A = k[x]? If this were a semisimple ring then the ideal (x) would have to be a direct summand

[broken stirrup]

so you mean we must have J(A)=0 i.e. A is semisimple

[latent anvil]

That's not what semisimple means

[broken stirrup]

i thought for an algebra A they are equivalent

[latent anvil]

Maybe I'm using a different definition than you? To me semisimple means you're a direct sum of modules that have no proper submodules

[latent anvil]

Is A is a finite dimensional algebra over a field?

I guess I could see it in that case?

[chilly ocean]

ty shamrock

i am a bad student

i need to get better

[latent anvil]

What's your definition of rad(A) and semisimple? Maybe that's where we disagree?

No problem!

You're learning :)

[broken stirrup]

intersection of annihilators over set of representatives of irreducible modules

[latent anvil]

so if A = Z then the irreducible modules will be Z/p for p a prime, right? And the annihilators of generators of theses are the ideals (p) for p a prime, which have intersection 0

So rad(Z) = 0

And in particular rad(Z) Z = 0

So your thing would say Z is semisimple, which it isn't

Does my confusion make sense?

[broken stirrup]

maybe we should assume artinian condition

[latent anvil]

On A?

[broken stirrup]

yes

[latent anvil]

Sure, this seems more believable

[broken stirrup]

okay let's just assume A is left artinian

[latent anvil]

So if M is completely reducible then M is a sum of simple modules, and J(A) * simple module = zero

Yeah?

So J(A) M = 0

[broken stirrup]

yes

[latent anvil]

For the other direction

hmm

Right your thing said to look at A/rad(A)

So why is this a semisimple ring?

[broken stirrup]

and appearantly A/J(A) is semisimple

[latent anvil]

Yep

So let's think about what simple modules over it look like

They're simple A-modules N such that rad(A) N = 0, right?

[broken stirrup]

right

[latent anvil]

So rad(A/rad(A)) = 0

At least I think so

rad(A) N = 0 is always true, right?

So you have all the same simple modules

[broken stirrup]

oh yes

[latent anvil]

And anything which annihilates all of those was already in rad(A)

Cool

So if A is an artinian (not necessarily commutative) ring with rad(A) = 0, is A semisimple?

I think this should be the case

[broken stirrup]

i think so

[latent anvil]

In the commutative case it's true by the Chinese remainder theorem + structure theorem for artinian rings

Sure

It seems like you're using some different definitions so I'll trust you

Well then M is a module over a semisimple ring

So it's a direct sum of simple modules

We get simple A/rad(A) modules M1,...,Mn and an A/rad(A) linear bijection f : M -> M1 (+)... (+) Mn

Yeah?

Well that bijection will still be A linear

You can think of all the Mi as A modules, right? Just by a.x = (a+ann(a))x

And they're still simple

So this shows M is semismple over A

[chilly ocean]

(For all a, b real numbers)

Ik this is very easily proved but I was wondering if there's more of an abstract algebra/ring theory approach to proving this?

[latent anvil]

So the issue is that inequalities aren't really algebraic

[broken stirrup]

thank you Hom(-,Shamrock)

[latent anvil]

The way you can link the algebra and the order on the real numbers is by x >= 0 iff x = y^2 for some y

So that suggests a way to prove this

And it's really the standard way

Does that make sense @chilly ocean?

A general ring won't have a notion of <=

[chilly ocean]

Hmm I think I understand you here but not what you said about the order of real numbers

[latent anvil]

So for a real number x we have x>= 0 iff x is a square

Does that make sense?

[chilly ocean]

Yeah

[latent anvil]

(and this tells you about when x >= y in general, it's when x-y is a square)

So that's sort of a way to talk about order algebraically

[hollow parrot]

You can just expand (a-b)^2 >=0

And manipulate

[latent anvil]

They were explicitly asking about whether there was an "abstract algebra/ring theory approach"

They know how to prove it

[chilly ocean]

Ik
I don't want that proof haha

[hollow parrot]

Anything else is unnecessary cmv

[latent anvil]

Nobody needs to change your view because you're not the one asking the question lol

[chilly ocean]

I see!!
Thank you, I get this now

[latent anvil]

My answer is basically no though, catfood

If you do what I'm suggesting you'll just get the standard proof

[chilly ocean]

I mean I was just wondering lol and couldn't find much on it online so thought I'd ask here. But I'm aware that it's very easily solved (I said that before I asked the question)

[hollow parrot]

I'm just messing

Interesting question

[chilly ocean]

You would immediately get x-y>=0 right? If I'm following correctly

Which I mean the case was that (x-y)² was >=0 anyway

[latent anvil]

I don't think you'd immediately get x-y>=0?

You're trying to show (x^2+y^2)/2 - xy >= 0, right?

So you'd want to write that as a square somehow

[chilly ocean]

Sorry I misread this as we're assuming that x>=y in general, not specifically when x-y is square

rip

[latent anvil]

No, sorry

I'm saying x - y >= 0 iff x - y = z^2 for some z

[chilly ocean]

That's almost the same as starting with (x-y)²>=0,no? I mean as the standard proof

[chilly ocean]

Lol wut

[latent anvil]

Yes, I'm saying you'll end up with the same thing as the standard proof

[chilly ocean]

I see!
Then I guess there's not really a much different approach to it lol

Thx!!

[chilly ocean]

nothing

[shell brook]

i am trying to construct a galois extension over Q of degree 11.

the idea i have is to take a 23rd primitive root of unity, call it z. Then |Gal(E(z)/Q)| = phi(23) = 22 by some theorem. Let G = E(z). Then I can get a subgroup H < G such that |G:H| = 2.

Then I can take the fixed field E^H, and by fund theorem of galosi theory, |E:E^H| = 2, and so |E^H:Q| = 11

But

I dont think E^H/Q is galois

b/c H is not like

a priori normal in G

oh hm maybe I just need to make an order argument?

[woven delta]

oh so I believe that galois group is abelian

lol

so every subgroup is normal

[shell brook]

is there a reason that this is true

[woven delta]

yes

[shell brook]

oh like specifically an extension of a primitive root is abelian

[woven delta]

yes

it's cyclic actually I believe

let me verify

[shell brook]

oh wait its literally isomorphic to U(n) and i use that in my proof lmfao

U(n) is the (Z/nZ)^x

okay yeah nvm pog

[woven delta]

yeah not cyclic

just abelian

okay yeah that's it

[shell brook]

okay hype

thank u

[latent anvil]

you're welcome :)

[barren sierra]

b is (x - 3, x + 3) right?

and then c the only examples I can come up with are things like Z/6Z = Z/2Z x Z/3Z

but I'm having trouble generalizing to product rings in general

[chilly ocean]

could someone point me in the direction for part b? I have been staring at it for 20 mins and I cannot figure out how to go about it directly to show (H cap K)h(H cap K)^-1 belongs in H

[next obsidian]

You’re multiplying three elements which belong to H

So it lies in H because H is closed under products

[chilly ocean]

is that legal? do I not need to use the fact K is a normal subgroup of G??

[next obsidian]

Do it on an element by element basis

Take g in K\cap H and h in H

You have to show hgh^-1 is in K\cap H, but this is clearly in H

And also in K because it’s normal

I’m not sure what you’re trying to do

[chilly ocean]

I may have been overthinking this then

oh wait

I should've been more specific when I said (H cap K)h(H cap K)^-1, I meant x in H cap K s.t xhx^-1 belongs in H

okay nvm I am silly

I figured it out thanks for the help

normal subgroups are new to me sorry lol

[next obsidian]

It’s no problem

I just stopped replying because it seemed like you were working through the stuff on your own and it’s best to just let you stew on the stuff for a bit by yourself IMO

[shell brook]

emma

reason I want a sanity check is that I sort of want |L:Q| = 4 but it comes out to 6 b/c I got Gal(K/L) iso Z_2

but 4 seems like the right answer??

The places an am could send sqrt5 + cbrt2 to:
sqrt5 + cbrt2
sqrt5 - cbrt2
-sqrt5 + cbrt2
-sqrt5 - cbrt2

I'm making some dumb mistake

[woven delta]

Yeah your initial argument was good, and I'll look at what you're doing now to figure out why it should be 6

What is the minimal polynomial of sqrt(5)+cbrt(2)?

[strange parcel]

can someone recommend introductory materials for abstract algebra
i only know engineering maths

[woven delta]

I can show you how to find the minimal polynomial

[shell brook]

it should be (x^2 - 5)(x^3 - 2)?

idk im just guessing lol

[woven delta]

No

It's going to be irreducible

[shell brook]

oh yeah fair enough

[woven delta]

Okay here's an idea

First square sqrt(5)+cbrt(2)

[shell brook]

oh ive done this stuff

[woven delta]

Okay good

[shell brook]

we had one of these on the term test

lol

yeah okay I can work it out

[woven delta]

So yeah if you find the minimal polynomial

[next obsidian]

Try #book-recommendations

Or uh…

Can someone type the channel

[shell brook]

,w minimal polynomial sqrt(5)+cbrt(2)

[next obsidian]

I can’t see it cuz perma-study lol

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by jesse#3659

[woven delta]

Then the degree of the extension will be the degree of the minimal polynomial

Yeah

So it's 6

[shell brook]

hm

[woven delta]

Cool

[shell brook]

so what did I miss

[woven delta]

No

That's good

[shell brook]

no here

whatd I miss lol

[woven delta]

Ah okay

So look for the roots

[strange parcel]

ah i see the channel @next obsidian ty

[woven delta]

Of the minimal polynomial

[shell brook]

OH

lmfao

there are complex roots?

ih ope

no nvm I dont even know if any of those are right

[woven delta]

Yeah

You don't really know

You know one root

[shell brook]

damn i got mad lucky it sounds like

no id idnt even get lucky i just wrote gibberish

[woven delta]

I think your argument was good

[shell brook]

yeah

[woven delta]

Above

Yeah

[shell brook]

I think I can just call that stuff scratch work

cz thats what it was lol

it made me realize the solution so it was good

[woven delta]

Remember degree of extension by alpha is degree of minimal polynomial of alpha

It's nice

Good to keep in mind

[shell brook]

oki

thank u

[woven delta]

Lol I never think to ask Wolfram alpha

I should keep that in mind

Before doing tedious computations

It actually is interesting that you can draw the conclusion that the minimal polynomial has degree 6 from the fact that Gal(K/L) has order 2, so L is degree 6 over F

And then if you find a degree 6 polynomial that sqrt(5)+cbrt(2) satisfies you have a proof that that's irreducible

@shell brook

So you can use Galois theory to show a polynomial is irreducible

[shell brook]

oh that is interesting actually

[woven delta]

Yeah

Kinda weird

[shell brook]

very good

im warming up to Galois shit

[woven delta]

But pretty cool

[somber thorn]

my aops group theory book just arrived, from the table of contents does it seem to be sufficient for a first course in abstract algebra, or should i work through a more conventional textbook like artin or something to make sure i have all details correct?

i'm wondering if this and mendelson's topology over the summer would give me enough background to maybe start working through hatcher's algebraic topology

[thorn delta]

looks fine

[lethal arch]

Hi! I got a quick question so I was wondering when someone says,"so(3,1)-valued 1-form", so(3,1) being the algebra. What does so(3,1)-valued mean? I know that real-valued means that the components are real, what could it mean to take values on an algebra? Further, what is the implication of that?

[heavy dagger]

Is there any way to understand the cardinality of the set of isomorphism classes of groups?

[heavy dagger]

I get that it will be at least countably inf but not able to decide if it's uncountably inf

[upper pivot]

this is not going to be a set

[thorn delta]

How do we know “set of isomorphism classes of groups” is a set?

[upper pivot]

you have atleast one group for each cardinality

[heavy dagger]

Oh? Why?

[upper pivot]

because of this

[heavy dagger]

Okay I'm not sure why that means it can't be a set

because I haven't learnt set theory axioms maybe

[upper pivot]

right its because like, the collection of all cardinals is not a set

so like, the collections of groups is going to be bigger than any given cardinality right

so it cannot be a set

[heavy dagger]

Ohhh

[upper pivot]

(ill ask a set theorist to verify this, im kinda going off intuition lol)

[heavy dagger]

How do we know that a set of a certain cardinality can be endowed with a group structure?

[upper pivot]

well ill ask you to work that out, good check of group intuition

[heavy dagger]

Wait got it

[upper pivot]

nice

[heavy dagger]

The set of bijections on the set would form a group, that's why right?

[upper pivot]

yeah sure i think that works

i was thinking free group on a set

[heavy dagger]

Oh right

thanks mate

[upper pivot]

mhm

[south patrol]

Apparently that you can turn any set into a group is equivalent to AC

Interesting

[upper pivot]

oh wow

no it makes sense actually

[south patrol]

Oh ye

If cardinality k identify with direct sum of k copies of Z

and just use the group structure on that apparently

[upper pivot]

right, which one of these needs choice again? i know to do things with cardinality you usually need choice

[south patrol]

I'm not knowledgeable about this kinda thing honestly

Maybe it's just to show that the direct sum of k copies of Z has same cardinality as the set

[sturdy marsh]

empty set

the empty set cant have a group structure

[south patrol]

well OK yes

Every nonempty set

[upper pivot]

lol

[south patrol]

Actually assuming the axiom of choice you can prove that groups can be empty

by applying Zorn's lemma to whether i asked

lol

[sturdy marsh]

brofibration disproves AC

[broken stirrup]

guys how do i show that if a module is semisimple over A\J(A) it's also semisimple over A where J(A) denotes Jacobson radical of an algebra A

[barren sierra]

How do I do c?

(from there I'll attempt d)
My first thought is just it's determined by where 1 and x go? 1 is determined so then I'm not sure what the choices for x are

[tribal moss]

It's a somewhat unusual terminology, but it ought to mean a linear transformation that takes in a single vector (from some unspecified vector space) and produces an element of the Lie algebra so(3,1).

[tribal moss]

That definitely needs choice. For example if A is a Dedekind-finite infinite set, then the set of functions A->Z with finite support is not Dedekind-finite, and therefore is not in bijection with A.

[south patrol]

Sure yeah

Now I need to google Dedekind-finiteness aha

[tribal moss]

"Dedekind finite" set is a set that is not in bijection with any proper subset of itself.

[south patrol]

Ah, sure

And so does choice imply no infinite Dedekind-finite sets exist i suppose?

[tribal moss]

Yes.

Countable choice is enough for that.

[south patrol]

Sure

I remember even having seen infinite sets defined as being ones in bijection with some proper subsets of themselves actually ig

sure, thanks

[barren sierra]

Ok so what I've gotten is there is a bijection between homomorphisms Q[x] -> C with kernel subset of (h) and homomorphisms Q[x] / (h) -> C

but again like what do I do with that

I still think it has to do with what x can map to but idk what x can map to

[tribal moss]

Whatever x maps to, the image of (x²+x+1)(x-1)² must be 0 in C, which means that phi(x) must be a root of the polynomial.

[barren sierra]

ahhh

ok

so there are 2 choices?

since (x - 1) and (x - 1)^2 have the same roots

hm no that just sounds wrong

[tribal moss]

The ² in (x-1)² becomes relevant in part d. It's not an error if it seems to do nothing in part c.

[barren sierra]

yea it's just trying to justify it does nothing for c

[barren sierra]

because this fact makes me feel like it should be different

[tribal moss]

(h) should be a subset of the kernel, not the other way around.

That means the kernel is the preimage of an ideal of Q[X]/(h).

[barren sierra]

Hm yea

for part d do we have that x must map to 1?

[heavy dagger]

I thought I understood but it seems I didn't, so why does the class of all cardinals have to have cardinality greater than all its elements?

[next obsidian]

It isn’t a set, it doesn’t have a cardinalitu

If it was a set you could create the set of all sets

And then :(

[tribal moss]

All the homomorphisms you already had will still work when the codomain is extended from C to C[eps]/(eps²). The question is whether you can find more places to send x to when the epsilon is available.

[barren sierra]

Hmmm

[barren sierra]

send x to eps + 1

can't really do anything with the roots of x^2 + x + 1

[tribal moss]

Indeed. And what about 1-eps?

[barren sierra]

oh yea

💀 I am so fucked for this midterm

if the actual one reflects the practice one I am praying for a curve

[barren sierra]

a) isn't I'm 99% sure
c) I think is true
b) I've got no idea

I want to say yes since x^2 - 2 =/= x^2 - 8 up to unit in Q

[coral shale]

If you think they are isomorphic, perhaps write down the isomorphism explicitly and check

[upper cape]

Anyone have tips for computing automorphism groups given a group presentation?

[acoustic fossil]

I WTS that all elements of G are of the form b^{2k}a^{r} for some integers k and r. Could someone help me, I don't know why I'm struggling so much with this. Given that a and b generate G it is clear that any element of G must be of form b^{i}a^{j} where i and j are integers and must fulfill the properties. Any suggestions?

[rocky gorge]

Given that a and b generate G it is clear that any element of G must be of form b^{i}a^{j} where i and j are integers and must fulfill the properties
that would only be true if G were abelian

unless you've shown that already

[acoustic fossil]

So the next part is actually to conclude that it is abelian

Given that it's the next part of the same problem I was under the impression that you could show any element is of the form b^{2k}a^r prior to proving it is abelian. However, I've been unable to make progress

[uncut girder]

Questuon: let G<H<K be a chain of subgroups.
If G is normal in K does that mean G is normal in H?

[rocky gorge]

yes: if G is stable under conjugation by elements of K then it's also stable under conjugation by elements of H, since every h in H is also in K

[rocky gorge]

how did you get that every element must be of the form b^i a^j?

[acoustic fossil]

I guess I was getting ahead of myself then as that would require it to be abelian. I'm very confused and feel like I should start at square 1

[rocky gorge]

hint: if you could prove that b is somehow related to an even power of itself, then ab^2 = b^2a would tell you that a and b commute

[uncut girder]

Thanks!

[acoustic fossil]

to do so should I only deal with a^{4}=b^{3}?

[rocky gorge]

that's how i did it

[barren sierra]

Note that 900 = 3^2 * (1 + i)^4 * (2 - i)^2 * (2 + i)^2

Ok so I need chains of ideals. So I guess one chain would be (300) is a subset of (3). So Z[i] / (3) x Z[i] / (300) is one decomposition in invarient factor form? This then yields the invarient factor form Z[i]/(3) x Z[i]/(3) x Z[i]/((1 + i)^4) x Z[i] /((2 - i)^2) x Z[i]/((2 + i)^2). Is this correct so far?

I've never actually had to find invariant factor form or elementary divisor decomp for modules. Just went through the proof in class.

So I think this is right cause it seems to fit all the conditions but I'm not sure.

(well right so far, I know this is just one such decomposition)

[acoustic fossil]

So naively if I solve for b by multiplying both sides on the left by b^{-2}. I get b^{-2}a^{4}=b. Seems straightforward to get b as an even power of its inverse but I feel like I'm missing something

[rocky gorge]

elements of G are products of the form a^i b^j a^k ... where i,j,k integers, right?

[acoustic fossil]

yes

[rocky gorge]

think about how b^-2 a^4 = b and the fact that a commutes with even powers of b can be used to simplify those products

[acoustic fossil]

Ok, slowly getting there. but still working If b^j is an even power we can commute it with a neighboring power of a^i. Then each time we have an a^k a^l we can keep simplifying and commuting so long as b^{even power}

[rocky gorge]

yep

[acoustic fossil]

Ok wow this is making a lot more sense. I've convinced myself that it holds that ab^{2k}=b^{2k}a using a couple of simple cases. However, I still feel that I need to prove it rigorously. It intuitively makes sense because every time I multiply on the right on both sides by some b^{even power}. I can keep factoring b^{2} and commuting ab^{2}. so that eventually ab^{2k}=b^2{k}a. So for example, trying ab^{2}b^2k=b^{2}ab^{2k} => ab^{2+2k}=b^{4}ab^{k}. I was thinking I could case it so that when factoring b^2 out of b^{even power} will always work or be equal to identity so then I can keep commuting. Edit: Ok nevermind this part wasn't too bad. Thank you so much for your help thus far

[rocky gorge]

glad you got it figured out

[lethal cipher]

Okay, so I am a little stuck. So I need to show that the subgroups in this lattice are the only subgroups of the Galois group. I understand exactly why they are subgroups, but I am not too sure how I eliminate other possibilities for a subgroup

[oblique leaf]

I need to compute the Galois group of x^n-d over Q with d a squarefree integer

I have thought about it for three days but have no idea where to start

can anyone give me a hint?

[barren sierra]

Can someone just confirm if this is right or not?

[lethal arch]

Let me see if I get you write. An (algebra)–valued quantity is one that takes in a vector as an input to return an element of the algebra which generate a transformation along that vector?

[barren sierra]

for b is it the ideal generated by x + 3 and x - 3?

and then c I'm stuck

I've come up with this working for examples

so like if R = Z/6Z

and so R1 = Z/2Z and R2 = Z/3Z

but I can't generalize this to general product rings

[thorn delta]

ok as a hint, consider this: let R be a module over itself, and let r,s be in R and x, y nonzero such that rx = 0 and sy = 0. For T(R) to be a submodule, we would expect that there is some nonzero element of R which annihilates r + s. An obvious choice for this element is xy because (r + s)xy = rxy + sxy = 0
However, the problem is that we may have xy = 0. In a product ring, i think you should be able to find elements which force you into this situation

[barren sierra]

An obvious choice for this element is xy because (r + s)xy = rxy + sxy = 0
ok so this was my first thought

however

x and y may not commute

so that's one other issue

also ya xy = 0 I did run into earlier also and is what made me thing this was the wrong idea

so I'll try again

oh wait hm

[thorn delta]

these are commutative rings

[barren sierra]

I think I got it

ok ok

how do you know that?

cause I did not get that from the description

[thorn delta]

beginning of the problem

[barren sierra]

is it the same R?

[thorn delta]

no, but T(M) is defined for modules over a commutative ring R

[barren sierra]

Hm fair

[thorn delta]

even so, i think you just need identity for the example i have in mind to work

[barren sierra]

ok yea then I got it

yea

[thorn delta]

aight epic

[barren sierra]

oh @thorn delta

sorry but also

for part b of that question

is the ideal (x + 3, x -3) correct?

[thorn delta]

yes

[barren sierra]

dope

[thorn delta]

actually uh

[thorn delta]

it might not be an ideal. For example what would eliminate x + 3 + x - 3 = 2x?

i.e. we had the right idea initially, but it may require a slightly more careful description

i.e. we know it certainly contains $(x - 3) \cup (x + 3)$

[cloud walrus]

kxrider

[thorn delta]

and i think this should be equality, but you should check that

[barren sierra]

would x - 3 itself not eliminate it?

oh wait no nvm

hm

ya I'll check

[potent lantern]

Given this definition of an affine morphism

how should I approach this problem?

[kind temple]

if phi is injective then phi(A) = phi(B) implies A = B. if lin(phi)(AB) = phi(A)phi(B) = 0 show that AB = 0.
if lin(phi) is injective, then take phi(A) = phi(B). now 0 = phi(A)phi(A) = phi(A)phi(B) = lin(phi)(AB) so...

im assuming there is a law like AB = 0 iff A = B? if thats true, then the above approach should work

[potent lantern]

yes, AB = 0 <=> A=B
but isn't the condition for lin(phi) injective lin(phi)(AB) = lin(phi)(CD) => phi(A)phi(B) = phi(C)phi(D)?

[kind temple]

linear maps are injective iff their kernel is trivial

so what you have said is an equivalent condition

[potent lantern]

oh you're right i've had that in my course but i forgot

then you're right

actually they are bijective

i think

[kind temple]

not sure why, but AB is acting like the vector B - A. is there a reason you aren't using this notation?

[kind temple]

they are bijective onto their image, which is true of any injective function, but injective affine maps are not in general surjective

[kind temple]

yea all im seeing is subtraction lol

[potent lantern]

it's just a notation

[kind temple]

but i guess subtraction isn't necessarily defined in A so it makes sense to use this notation

its just the way i was interpreting AB when i was doing the problem

[fickle brook]

can i ask something unrelatecd

unrelated*

i seem to be lacking any sort of intuition for projective and injective modules and its driving me mad

the lifting properly definition just feels like word salad to me

it doesnt make any sort of intuitive sense

[tribal moss]

An X-valued <thing> generally describes a <thing> whose values are X or in X instead of real numbers (or instead of whatever kind of values a <thing> takes in general). A "one-form" is usually a linear transformation from a vector space to scalars.

[devout crow]

this definition of finitely presented doesn't make a lot of sense to me. from what I understand from other sources, a module is finitely presented if it is the linear image of a free module and the kernel of the map is also finitely generated

so shouldn't the sequence go $$R^{\oplus m} \to K \to R^{\oplus n} \to M \to 0$$

[cloud walrus]

xdres

[hidden haven]

Yes but then it is not exact

Because K injects into R^+n

The map from R^m to R^n in that is exactly the composite that you have drawn

[devout crow]

oh yeah

but then, is M iso to R^n/R^m ?

[hidden haven]

Nope

R^n/image of R^m = R^n/K

[devout crow]

oh sure because there's no 0 --> on the left

cool ty

[bright marsh]

Is it true that if G is a group, and K a normal subgroup of G, then EVERY homomorphism from G to G/K has kernel K? I'm able to find sources to show this is true for some specific homomorphism, but not EVERY homomorphism

[upper pivot]

its not

[proud bear]

No. There's always the trivial homomorphism mapping everything in G to the identity of G/K

[upper pivot]

yeah, and even weaker you can ask if the kernel contains K, and the answer is no

since

you can take ZxZ, and the subgroup Zx{0} right

the quotient is Z

[bright marsh]

oh the trivial mapping G to the identity of G/K means that the kernel of the homomorphism is G/K, not K

[next obsidian]

no

the kernel is G

[bright marsh]

oh wait

yes you're right

i forgot it's the domain, not the range

[upper pivot]

and the projection onto the first coordinate in a map ZxZ - > ZxZ/(Zx{0}) but its kernel is {0}xZ

[bright marsh]

gotcha

[chilly ocean]

my intution telling me this is only true for K=G

because given any K you have G->G/K for the projection as a homomorphism and then the trivial one

but if ur asking for nontrivial then ig it depends and it might be case by case

probably has something to do with chains of normal subgroups of K

but idk what specifically

[pastel cliff]

is a bilinear map just something combining elements from two vectors spaces and outputting em into a third

[lethal dune]

yes

it also has to be linear in both components

[pastel cliff]

is bilinear enough to justify the fact that the lie bracket of any v \in F and 0 must be 0

[delicate orchid]

oh that's what you wanted to ask me about

[pastel cliff]

kinda

im learning me some very basic lie stuff

[delicate orchid]

let [x,y] be a lie bracket, then [v, 0] = [v, 0*0] = 0[v, 0] = 0 should work

[pastel cliff]

i thought that'd be cheesy lol

gotcha tho

[delicate orchid]

yur

[pastel cliff]

also ik it's not true that [x,y] = [y,x] but is it true that [x,[y,z]] = [x,[z,y]]?

[delicate orchid]

[x, [y, z]] = [x, -[z, y]] = -[x, [z, y]]

apparently