groups-rings-fields

[coral spindle]

I don't know what you mean when you say revise the definition here

[muted comet]

uh yeah? i agreed i understood why she would say that.

[coral spindle]

Oh right, I think you may have phrased it as a question originally

This definition is the correct one. If your prof decided not to present it like this, and instead presented R[X] as formal sums in a symbol X, this is just as valid, although it hides these rigorous but confusing set-theoretic details.

They are the same definition in every way that matters.

[olive parrot]

is there a fast and generalized way to figure out if a conjugacy class contains an element and its inverse?

i know from here that this is not the case for abelian groups and i know from here that is it the case for symmetric groups

I'm working with the GAP CAS so would prefer an answer in that context

[rocky cloak]

In GAP you can produce a list of all the conjugacy classes of a group, and easily check if an element is contained in one.

So just looping over the elements in the conjugacy class and check if it contains it's inverse

[olive parrot]

That's O(n) some of my conjugacy classes have billions of elements

Not n^2 because iirc \in is O(1)

[coral spindle]

It sounds like you need to produce specific techniques for your group

Perhaps you are looking at a finite group of Lie type for example. There are, iirc, techniques specifically for these groups which exploit the geometry.

[olive parrot]

Oh wait I think I might have figured it out

[coral spindle]

Great

[rocky cloak]

I'm curious

[olive parrot]

One sec

I'm sure it's just Representative(c)^-1 in c because if it's true for one element it's probably true for all elements

[coral spindle]

That is indeed true.

[olive parrot]

Why?

[coral spindle]

I really suggest you try proving it yourself, it's very straightforward.

It might help you understand the situation a little better

[olive parrot]

I have little background in math what's a good resource to read to try and prove it myself

[coral spindle]

Do you know the definition of a conjugacy class? If so, you're equipped enough.

[olive parrot]

Oh

Ok I'll try to work it out

[surreal dagger]

Given a closed subset X of A^n wrt to the zariski topology
We can write it X=V(a) for some polynomial ideal
Now a closed set that is contained in X is of the form Y=V(b) for some polynomial ideal b.
We assume k to be algebraically closed.
We can say that rad(a) is contained in rad(b) via the Nullstellensatz.
Can we say that rad(a) is contained in b?

[rocky cloak]

If you assume b is already radical, otherwise just consider a=b as a counterexample

[surreal dagger]

Right, did I missunderstand this paragraph, given the context of my question, then? To me its unclear how we can assume that I(X) lies in a

[rocky cloak]

Well if Y = V(a) < X, then V(a) = V(rad(a)) and rad(a) > I(X), so it is true as they say that Y is the zerolocus of an ideal that contains I(X)

[surreal dagger]

Oh my bad, so this hinges on V(rad(a))=V(a). Thanks for clearing it up!

[candid dove]

Hey I was able to do (i) and (ii) thanks to you guys

But how do I conclude (iii) from there

[rocky cloak]

Well you have to determine which ideals are both of the form I_W and J_W'

Because those will be the two sided ideals

[candid dove]

Yeah I am not able to do that I'm afraid

I have absolutely no control over W and W'

[rocky cloak]

Well, say I_W is a twosided ideal.

If W = V, then I_W is (0). So let's say W is not V. Then there is some nonzero A in I_W. What could the image of A be, can you control it by multiplying by things on the left?

[candid dove]

Yeah so projections can be used to control that

But the problem is this image thing need to be surjective

[rocky cloak]

Why do you say that?

[candid dove]

Like you want me to get the identity right?

[rocky cloak]

Not necessarily, no

[candid dove]

Okay then idk

[candid dove]

I meant need not

Be surjective

[rocky cloak]

Like you have I_W = J_W' for some W'.

If you can show that W' = V, you'd be done, agree?

So that's all you need to do

[candid dove]

Ohkay

I mean if W'=V then J_W' has identity so....

But how exactly are we showing W'=V

[rocky cloak]

Well, what happened if W' isn't equal to V

[candid dove]

I dk tbh

[rocky cloak]

Then there is a vector in V not in W'

[candid dove]

Sure

[rocky cloak]

Alright, so now think about that a little

[candid dove]

Sure

All the stuff in the two sided ideal don't have v vector in their image

So uhh

[rocky cloak]

Like let's go over what we know so far:

• We have this nonzero A in I_W = J_W'
• So BA is in J_W' for all B
• Which means the image of BA is in W'
• Then we have a vector outside of W'
  How are we gonna put this together?

[candid dove]

So use B to move it to v

So like B is any transformation (with kerB= W) which takes Av to v should work?

No

That bracket thing is also not necessary

Any such B would work

Cause ideal right

[rocky cloak]

Yeah, you just need to move Av outside of W'

[candid dove]

Right

Rhanks

Thanks

Lmao I was so focused on other stupidly complicated manipulations that I forgot to see the simple things

Thanks

[dull ginkgo]

what paper/book is this because this is a result i kinda need lo

[candid dove]

I said thanks more than 3 times lol

[candid dove]

It's my assignment 😭

[dull ginkgo]

doom

me on my way to do more sporatic noncomm geo shit

I should actually get back to proving double centralizer for semisimple rings lo

[dull ginkgo]

noncomm alg?

[candid dove]

I mean the course name is just ALG-1

[dull ginkgo]

i n t r o a l g

[candid dove]

Yeah true

I ve had cats

[dull ginkgo]

welp I'm doomed

[candid dove]

I've had forms

I've had homological Algebra

I've lost all my will to live

[dull ginkgo]

:c

do analysis :p

[candid dove]

I wish

But i would even suck at that

I have 0 self esteem as of now

[dull ginkgo]

Ah

Welp we can both suck at algebra then

If someone has a cool/interesting Jacobson radical problem please ping me I'm bored and i like the jacobson radical

[candid dove]

Okay

I probably do

Wait

Look at 2(i)

It's enough to show rad(R)=0

rad here meaning Jacobson radical

K bar be algebraic closure

[candid dove]

How do dis

[dull ginkgo]

hm what's abs semisimple

[candid dove]

Like he's just naming that phenomena that

[dull ginkgo]

oh misread

I'll give that a shot when I stop procrastinating on my C++ project and do a bit of work on it

I haven't actually worked with the ring tensor product before, might need to try out some stuff with it

[candid dove]

Oh okay

Well it's really not a good excercise for specifically Jacobson radical tbh 💀

It just be using some corollary or sth

[dull ginkgo]

Might still give it a shot

Is the overline k the algebraic closure

[candid dove]

Yeah

[rocky cloak]

I'm guessing it's something you've done before, but just proving that all the equivalent definitions is kinda fun.

Like intersection of maximal left ideals, intersection of maximal right ideals, annihilators of simple modules, 1-xy has a left inverse, 1-xy invertible.

And in the artinian case, the sum of all nilideals.

[dull ginkgo]

I've done so many goddamn equivalences lmao

but I can try to refine it

[rocky cloak]

The conditions for a ring to be perfect also give some nice equivalences involving the radical
https://en.m.wikipedia.org/wiki/Perfect_ring

[opaque finch]

I have two questions

1. Let $L$ be a Galois extension of $\mathbb{Q}$ with the galois group isomorphic to dihedral group containing $8$ elements. Then we have to show $L$ is a radical extension of $\mathbb{Q}$.

My approach : Dihedral group is clearly solvable and it can be proved. But I cam clueless how I can prove that L is a radical extension.

2. Suppose that $L$ is a normal radical extension of $\mathbb{Q}$ and $[L : \mathbb{Q}]$ is not a 2-power. Prove that L must contain a primitive $p^{\text{th}}$ root of unity for some odd
   prime $p$.

Can anyone help me with these ?

[cloud walrus]

mycroftholmes1703

[rocky cloak]

So if E/F had degree 2 (and the characteristic of F is not 2), then E = F(sqrt(a)) for some a. This you can see from the quadratic formula.

So if you can find a sequence of subgroups in the dihedral group each with index 2 in the next you'd be done.

[coarse stag]

Im having a bit of a hard time wrapping my head around the concept of Wreath Products. I've heard it loosely described as a group G being acted upon by a permutation group H. The example I am aware of is if you have n cubes with a symmetry group G, then the ways to rotate these cubes and permute the order is described by the wreath product G with S_n. However, in this case S_n can also be a permutation group on the set S_n itself which has n! elements and thus has structure G^{n!} semidirect product S_n. But the former representation implies a structure of G^n semidirect product S_n. Im confused as to which it is.

[coral spindle]

It can be either. You need to choose an action when you pick the group

So if I'm writing G \wr H, I am also choosing an H-set X and in reality forming a semidirect product G^X \rtimes H

And yes, you are right, there is no surefire way to tell what H-set you're choosing when someone writes that.

However there are standard actions, and when these are written we just agree that that's what we're referring to. When we write S_n, we mean S_n acting on {1, ..., n}. When we say, e.g., C_n, we mean the cyclic action on {1, ..., n} also.

If there is no written caveat, you should assume that the action is this standard one.

[opaque finch]

Is the first comment for he second problem and the second one for the first problem?

[rocky cloak]

Just the first problem

[tough raven]

Let V be an infinite-dimensional vector space over a field k. What is the Jacobson radical of End_k(V)? (IDK the answer but I'm curious now.)

[sinful nebula]

how would I do this

[tough raven]

Depending on how D_2n has been defined, you could just take an arbitrary element x and check whether x has order 2, i.e, whether x^2 = identity but x ≠ identity.

[night bison]

Also recall $D_{2n}$ is usually defined by the presentation [D_{2n}\coloneqq{r,s\mid r^n,s^2,srs^{-1}=r^{-1}}]

[cloud walrus]

SelahW

[night bison]

This way it's not too difficult to find the elements of order 2

[chilly ocean]

Last inequality, is the generalization of Chinese remainder Theorem?

Because if A + B = R then it gives the Chinese remainder theorem

But I need R to be commutative

[tough raven]

It's not a problem if the special case (CRT) assumes more hypotheses than the general case.

Alternatively, you can apply this problem to the (R,R)-bimodule R under the additional hypothesis that A + B = R to get CRT for two two-sided ideals of a non-commutative ring.

BTW, are these from Blyth's textbook on module theory?

[chilly ocean]

(R, R) - bimodule R ?

[chilly ocean]

yes

[tough raven]

You can construct a ring S (technically the tensor product R (⨯) R^op) and put an S-module structure on R such that the S-submodules of R are precisely the two-sided ideals of R.

If you take this for granted, you can make CRT a special case of this.

[tough raven]

Nice (I like that book).

[chilly ocean]

okay i will think about it, thank you

any hint?

[tough raven]

In diagram chasing arguments like this, to define a function on one space (Z in this case), you usually pick an element and take preimages under various functions and then apply others.

[chilly ocean]

got it

actually there is a theorem in theory so we can directly apply it

[tough raven]

For example, in this case, to define h, take any z in Z and ||choose y in Y such that g(y) = z (why can you do this?) and define h(z) to be v(y). Show that h(z) does not depend on which y you chose and then that it is linear. The property v = h . g will be easy.||

[chilly ocean]

Can't we apply this?

[chilly ocean]

g is epimorphism therefore there exists g(y) = z

if g(y1) = g(y2) then y1-y2 in im f implies v(y1) = v(y2)

[flint cave]

Does the group <a,b,c|a^2,b^2,c^2,(abc)^2> have a name? (Please tag if you reply)

[nimble folio]

this question might not make sense

i can elaborate more if needed

but if $(f \circ g) \circ h = h$, does that mean $f \circ g$ is the identity function?

[cloud walrus]

clubsoda14

[alpine island]

Not without some more restrictions on which functions you care about

Consider if h is a constant function

[serene dune]

klein four

Z2 x Z2 with 3 equivalent generators

2d vector space over binary field

[alpine island]

Klien 4 is also the smallest noncyclic group

[tough raven]

I don't think that the given relations imply that abc = 1...

[serene dune]

did i fuck up

[alpine island]

yeah I think so

[tough raven]

For example, we can map the group to F_2^3 by mapping a, b, c to the standard basis.

[serene dune]

oh shi how did i see that was equal to 1

mb i was blind

[tough raven]

Can we show that ab has infinite order?

Oh if we add the relations acbc = 1, this is the semidirect product of <a, b | a^2 = b^2 = 1> with <c | c^2 = 1> by the action c.a := b, c.b := a.

In particular, any words in a, b which are distinct in <a, b | a^2 = b^2 = 1> are distinct in this group. By symmetry, the same holds for a, c and b, c.

[flint cave]

Yes, the group arised from three involutions in SLn(Z) with the same "fixed point" and "generic 1 forms" , by this I mean think of an involution R as Rx = -x + <q,x>p, here p is the fixed point and q is the one form such that <q,p> = 2

The product of two of these involutions is actually a rank one unipotent

Meaning a unipotent u such thst rk(u-1) = 1

Its a quotient of Z_2 \star Z_2 \star Z_2 of course... it also contains copies of Z^2 (corresponding to (say) ab, ca as generators)

[tough raven]

Does this come from root systems?

[flint cave]

Root groups yes

By some results of Venkatarama this bad boy cannot live inside any thin subgroup of SLn(Z)

[tough raven]

Seems like you're the one most likely to know what this group is called. 😂

[flint cave]

Lol I was hoping this group was studied classically in combinatorial group theory

Cus the presentation looks good

[tough raven]

This is true for involutions with the same one-form?

After all, otherwise you could get a nice Weyl group symmetry

[flint cave]

Yes! you are right

[tough raven]

It is cute.

[tough raven]

Nice. I never thought about composing reflections with the same form/hyperplane.

[chilly ocean]

the three lemma, took an hour to prove

chasing is cool

[olive parrot]

hi just wanted to confirm if this seems right

if a and a^-1 are in the class, and for all b, (a is conjugate with b) implies (a^-1 is conjugate with b^-1), then you can apply that for all b in the conjugacy class and say that because a^-1 is in the conjugacy class, b^-1 is in the conjugacy class.

ive been thinking about this for too long i just asked chatgpt right now 😢

[night bison]

We also have $(ab)^2=a^2b^2=e$ and similarly for $bc$, right?

[cloud walrus]

SelahW

[flint cave]

No? I believe ab is of infinite order

[night bison]

Hmm

Ok

Ah

$(ab)^2=abab\neq a^2b^2$ in general

[cloud walrus]

SelahW

[flint cave]

Freshmans dream haha

[night bison]

So do we contain a free algebra on a and b?

[flint cave]

It's just a group, no vector space structure

[night bison]

That's with plus! This is just because the ring not necessarily commutative

[flint cave]

There's no ring here?

[night bison]

Group*

Not necessarily abelian*

[flint cave]

Yeah

[night bison]

I read above, are we looking at the representation of this group?

[flint cave]

Faithful representations into SLn(Z)

[night bison]

Ahh

[flint cave]

I am interested in that

[night bison]

Big sl, not little? Ik people have different notations

[flint cave]

sl is the Lie algebra and SL is the Lie group

So yes, big SL

[night bison]

Yes yes

Okok I will read more closely above bc I think other people probably know more about this than I 😄

[thorn jay]

Freshmans dream group edition

[chilly ocean]

I want to prove this by myself

1 => 2, let's say contrary there is a non-empty subcollection such that there is no maximal element, now pick any submodule in it say N, then there exists submodule N_1 such that N \subset N_1, similarly for N_1, eventually we get ascending sequence N \subset N_1 \subset N_2 \subset....\subset N_k \subset....

But since M is a Noetherian Module therefore it will stop and give us contradiction.

2 => 1, since M follows maximal condition therefore for the collection of ascending sequence there must exist M_k such that M_n = M_k for all n≥k.

If there is any mistake please let me know

[chilly ocean]

any hint for 3 part?

[knotty pilot]

For 1 -> 3, if M isn't fg you consider what happens if you start from the zero module then toss in generators one by one
For 3 -> 1 start from an ascending chain, take the grand union and observe the union is a submodule

[opaque finch]

Ok so suppose $M$ is Noetherian and $N$ be a $R$-submodule of $M$. Choose $n_0 \in N$ and consider the module $<n_0> = N_0 \leq N$. If $N_0$ is all of $N$ then we are done if not then choose $n_1 \in N/N_0$, and keep repetaing the same argument. You will eventually have produced a chain modules contained, and your Noetherian condition tells it must stabilise.
Conversely, let every submodule is finitely generated. Then we have to show $N_0 \subset N_1 \subset N_2 \subset \dots$ must stabilise. Consider $N = \bigcup N_i$ and argue

[cloud walrus]

mycroftholmes1703

[chilly ocean]

yes, and if A\subset B, such that | A | = | B | with finite cardinality, A = B

[opaque finch]

Sorry I missed the part $N$ is a submodule of $M$ and hence must be finitely generated

[cloud walrus]

mycroftholmes1703

[chilly ocean]

so my first submodule is < n0>, then < n0 + n1 > ?

[opaque finch]

Almost

but not quite. The second submodule will be $<n_0,n_1>$

[cloud walrus]

mycroftholmes1703

[opaque finch]

the third will be $<n_0,n_1,n_2>$ and so on

[chilly ocean]

so how these structures are different?

[cloud walrus]

mycroftholmes1703

[chilly ocean]

oh got it

[opaque finch]

all these are different by the way you are choosing it

[chilly ocean]

no i was talking about < n0 + n1> different from <n0, n1> but i got it

[opaque finch]

yes

they are not the same

[chilly ocean]

got it, thank you

can you please verify my proof of 1 is equivalent 2?

[opaque finch]

lemme see

[chilly ocean]

yes so at some point we get N_k = N_n for all k=> n

[chilly ocean]

got it, thank you

[opaque finch]

Seems okay to me. If I am not missing out any finer details

[chilly ocean]

ah, thanks

[chilly ocean]

in converse part, say M is finitely generated by {n1,...,nk}, so at most k steps after we have N_s = N_k, for all s=>k

[dull ginkgo]

Root systems are funny because random ass algebra with them gives horrible identities like this

[chilly ocean]

if M is R-module and if N is submodule of M satisfy the chain condition and M/ N satisfy same chain condition then M also satisfy that chain condition.
any hint?

[dull ginkgo]

Images and preimages of submodules are submodules (this should be relatively obvious). Consider the quotient epimorphism x |-> x + N, what happens to the module chain conditions?

Hint: ||Take the preimages of the chains, which is a chain||

[chilly ocean]

yes images and preimages of submodules are submodules, do i need correspondence theorem?

[dull ginkgo]

actually the correspondence theorem is that applied to the x aforementioned map since the image of a submodule B is B/N, and the preimage of B/N is thus B by surjectivity

[chilly ocean]

say M_0\subset M_1\subset M_2\subset....\subset M_m\subset...

then taking the f(M_0)\subset f(M_1)\subset...\subset f(M_m)\subset....., which are chain in M/N

[dull ginkgo]

yep, so you would apply your given chain condition

then what can you do?

[chilly ocean]

taking preimage?

[dull ginkgo]

yep!

Actually i realized there's a small hickup give me a moment

[chilly ocean]

assume f(M_k)\subset f(M_n) for all k=>n, then taking preimage gives us M_1/N \subset M_2/N\subset..., M_k/N \subset M_n/N for all k=>n

[dull ginkgo]

should work

[chilly ocean]

but where i used N satisfy chain condition?

[dull ginkgo]

because we apply it to the f(M_k) chain

[chilly ocean]

but f(M_k) in M/N

[dull ginkgo]

Yeah you said M/N satisfies the chain condition

[chilly ocean]

yes

[dull ginkgo]

so you apply it to that chain

[chilly ocean]

yes

and N satisfy chain condition where i used >?

[dull ginkgo]

Yep, that's the last part to check, but in that case it's considering what happens relative to N...

[dull ginkgo]

I think the correspondence theorem might be a bit more obvious here for ya

So we have a chain M_n in M. Taking the quotient gives a chain K_n = M_n/N, which must stabilize in some way. The only case that it doesn't is when ||It's entirely within N, in which you can ALSO apply the chain condition||

[chilly ocean]

got it, thank you

[dull ginkgo]

sorry for the roundabout way

[chilly ocean]

its fine

[candid dove]

Ok so stupid question but why does the minimal polynomial of a transformation not change when u take it over an algebraically closed field

Like i know an explanation using matrices but like I want a more abstract formulation or sth

[somber sleet]

Can you give me an example of why the following ring is not reduced: $\frac{\mathbb{C}[x]}{(x^2)} \times \mathbb{C}$

[cloud walrus]

damn_guuurl

[somber sleet]

Would be taking (x,0) suffice, since $x\cdot x = x^2 \equiv 0$

[cloud walrus]

damn_guuurl

[hollow topaz]

hey fellas, is there an open course with video lectures that follows tom judson's abstract algebra: theory and applications?

[glad osprey]

yep that works

[dull ginkgo]

what do you mean by "not change"

[candid dove]

Like

It's the same poly but over the closed field

[dull ginkgo]

Like some sort of invariance or preserved under the End(k^N) -> End(\overline{k}^N) embedding

[chilly ocean]

Can someone help me to understand the converse part?

[dull ginkgo]

which one aaaa

[prisma ibex]

ahh yes what a nice theta function... 56 is the dimension of the miniscule representation of E_7... 126 is the number of roots... yes of course

[dull ginkgo]

doom

[wild jasper]

Does this argument work? Suppose that $K$ is finite extension of a field $F$. and suppose that $K/F$ is Galois, then if $E$ is a subfield of $K$ and $E=\sigma(E)$ where $\sigma \in Gal(K/F)$. Then since $K$ is Galois we have that $K=F(\alpha)$ if $p$ is the
minimal polynmial of $\alpha$ over $E$. then $\simga(p(x))=p'(x)$ the minimal polynomial of $\alpha$ over $E'$, hence $[K:E]=[K:E']$ and it folows that $[E:F]=[E':F]$

[cloud walrus]

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

[prisma ibex]

actually I'm pretty sure the example you posted with E8 is of basically the exact same flavor

[dull ginkgo]

I like the idea that evaluating it probably gives extremely small numbers

[prisma ibex]

theta functions of lattices my beloved

[dull ginkgo]

shattices

[prisma ibex]

nooo lattices are so good

[wild jasper]

note that both $p$ and $p' $ divides the minimial polynomial $f$ of $\alpha$ over $F$ so $\sigma$ will map any root of p to a root of $f$

[cloud walrus]

mh_le

[opaque finch]

Ok so I have the sequence like this
${e} \triangleleft \mathbb{Z}/2\mathbb{Z} \triangleleft \mathbb{Z}/4\mathbb{Z} \triangleleft D_8$.

[cloud walrus]

mycroftholmes1703

[opaque finch]

But how are you saying we are done from here @rocky cloak ?

[rocky cloak]

So by Galois correspondence these correspond to field extensions, and each of those are given by appending a square root

[opaque finch]

Ahhh yes. So they are basically quadratic extensions of 2.
The field extension corresponding to ${e} \triangleleft \mathbb{Z}/2\mathbb{Z}$ say $F$ is a degree 2 extension and hence $F(\sqrt{\alpha}) = \mathbb{Q}$ and consequently we have for all the extension $\mathbb{Q} \subset F(\sqrt{\alpha}) \subset F(\sqrt{\alpha},\sqrt{\beta}) \subset F(\sqrt{\alpha},\sqrt{\beta}, \sqrt{\gamma})$

[cloud walrus]

mycroftholmes1703

[rocky cloak]

As for your second question, consider first the case where L is the splitting field of x^p - a for some a

[opaque finch]

Ok.

[chilly ocean]

My doubt is how C is non-empty?

[somber sleet]

can someone tell me, why $P= mP$? I don't see why this would hold

[cloud walrus]

damn_guuurl

[rocky cloak]

So there seems to be some typos, but if P = M/im(phi).

Then N/mN -> M/mM being surjective means that mM + im(phi) = M. Which means mM/im(phi) = M/im(phi)

[glad osprey]

This example also works with x² and y² instead of x³ and y³, right?

[toxic zephyr]

this was easy to prove as a subgroup. i found that

$a^\ell b^k a^{-\ell}=\m{1&2^\ell k\0&1}$

which is in $H$ for all $k,\ell\in\bZ$. so isn't this subset isomorphic to all rationals with a power of 2 in the denominator? can i argue that set isn't finitely generated and then i'm done?

[cloud walrus]

eigentaylor(got that eigenvalor)

[toxic zephyr]

also what is the name/symbol for that subset of the rationals?

also is it wrong to say $\gen{a}\gen{b}\gen{a}^{-1}\subset H$? lol

[cloud walrus]

eigentaylor(got that eigenvalor)

[toxic zephyr]

to denote the set of all a^l b^k a^-l

[spice whale]

interesting problem i thought of during a group theory supervision:
is there a group with exactly 5 elements of order 2?

[grave sedge]

||D_5||

[tough raven]

Huh, that's a good question. I guess it's in some way because the dimension (say, the dimension of k[A] over k, where A is the linear map) doesn't change. But that's not a good answer.

[tough raven]

Start at M and keep taking proper submodules while remaining non-zero. Eventually you are forced to stop, and the submodule you have must be a simple submodule - call it M_1. If M_1 is strictly smaller than M, start at M again and keep taking proper submodules while strictly containing M_1. Again eventually you are forced to stop at some M_2, and M_2/M_1 must be simple because there are no submodules between M_1 and M_2. Keep going in this way to get M_3, M_4, etc. If you never get M_n = M, the M_n would form an infinite ascending chain, which is a contradiction. So M = M_n for some n, and M_0 := 0, M_1, ..., M_n = M is a decomposition series.

[tough raven]

BTW, when you get to the proof that series of submodules have a common refinement, I would recommend the proof in Ravi Vakil's algebraic geometry notes The Rising Sea: Foundations of Algebraic Geometry over Blyth's. They come with a picture that makes the argument much easier to follow.

[rocky cloak]

I guess one way to think about it is how if T is a transformation on V, then V becomes a k[x] module retaj to
k[x]/(f1) x k[x]/(f2) ...
where fi are the invariant factors of T. So f1 is the minimal polynomial and f[i+1] divides fi.

Then it just comes down to k[x]/(f) tensored with a field extension K being equal to K[x]/(f)

[tough raven]

That seems correct.

[tough raven]

I've heard it called the dyadic rationals. Notationally, you can use ℤ[1/2] and nobody should object (although I don't know if they would know it).

[tough raven]

I think there would be too much risk of confusion with {a^l b^k a^m | l, k, m ∈ ℤ}.

[candid dove]

Okay this sounds way more promising to show

Thanks

[toxic zephyr]

would this work as a proof sketch? just need a sanity check

suppose G is a finite abelian group that is not cyclic. then G is isomorphic to Zn1xZn2x...xZnk such that at least one pair gcd(ni,nj)>1 (WLOG, suppose i=1,j=2). suppose n1=ap and n2=bp for some prime p>1. then H=<(a,0,0,...,0),(0,b,0,...,0)> is isomorphic to ZpxZp. by the map (ak,bl,0,...,0) maps to (k,l) (pretty sure my proof this is an isomorphism is correct)

[toxic zephyr]

ah yeah that's familiar. yeah makes sense

[toxic zephyr]

totally fair, yeah that also makes sense

[toxic zephyr]

the step i'm the least sure of is that there is at least one pair that is not coprime. but if they were all coprime, then (1,1,...,1) would be a generator, wouldn't it?

[opaque finch]

and what is the other case ?

[chilly ocean]

Okay thank you

[rocky cloak]

Well, you can see if you can find an intermediate extension L/K of degree p

[chilly ocean]

I will look into it

808 pages

If R is the principal ideal domain then R is finitely generated, correct?

[restive birch]

checking my calculation, the thompson subgroups of QD16 and M16 (modular group, uv = vu^5) both have 8 elements right

one of them is iso to Z2 x Z4 and the other to D8

[rocky cloak]

It's unclear what you mean. R is always finitely generated as an R-module (generated by 1). This is true for any ring.

If you're asking if it's finitely generated as a ring, then this is not necessarily true

[chilly ocean]

As finitely generated as a ring, why?

[rocky cloak]

Well for example Q is not finitely generated as a ring

[golden turtle]

I'm struggling to think of an example of a commutative ring R, and an injective R-module M such that M is not divisible over R.

I know that R cannot be a PID because then an R-module is injective <=> it is divisible, so I'm trying to think of a simple commutative ring, that is not a PID, that is also not a Zmodule,

but I can't think of any simple examples

[chilly ocean]

Principal ideal domain implies every ideal generated by a single element?

[rocky cloak]

That is the definition yeah

[chilly ocean]

Isn't Q the ideal of itself?

[rocky cloak]

Q ( = (1)) and (0) are the only ideals in Q, that's correct

[chilly ocean]

Yes

[chilly ocean]

So can't I say Q is finitely generated by 1 ?

[rocky cloak]

The subring generated by 1 is Z

[chilly ocean]

I got it

[rocky cloak]

If you're not talking about being generated as a ring, but something else that's fine

[chilly ocean]

But generated as a ring we need x such that {rx | r in R } = R, right?

No

That's not a definition of generated subring

[opaque finch]

As @rocky cloak points out, there is indeed a typo. But if you are interested in the solution then
Let's see
For a local ring, the maximal ideal is it's Jacobson radical. Let $\phi : N \to M$ be given morphism then you can see that, since tensoring is right exact you got $(\text{coker} f) \otimes_{R} R/m \cong \text{coker}(f \otimes R/m) = coker(N/nM \to M/mM) = 0$, since the map is surjective. So you have easily $\text{coker} f = m (\text{coker}f)$. And now it is the perfect time to apply Nakayama's lemma as then you have $\text{coker} f = 0$ and you have $f$ is indeed surjective.

[cloud walrus]

mycroftholmes1703

[rocky cloak]

If R is an integral domain, then any injective module is divisible.

So it would have to be a non-domain.

For example R = k[x]/(x^2) then M=R is injective, but you can't divide by x.

R = Z/4 would also work

[next obsidian]

What if you took a semisimple ring

Are all modules divisible over a semisimple ring?

[white remnant]

@grizzled plover

can you help me with my anal ring?

[chilly ocean]

a => b, yes A is Noetherian because jagr point out A is finitely generated because A -module, A is a ring so r = r•1.

But the second sum of two principal ideals is the principal, I am not sure but isn't the sum of principal ideals is ideal and A is principal ideal domain therefore it is principal ideal?

[next obsidian]

Yeah

[spice whale]

oh yeah
i am stupid

[rocky cloak]

In general if r is not a zero divisor, then (r) = R so if M is injective and m is any element of M, then we can define a homomorphism (r) -> M sending r to m.

Lifting the inclusion (r) -> R by the injective property shows that we can "divide" by r in M

[grim hawk]

I am currently working on extending Maschke’s Theorem to continuous compact groups. Is it as simple as stating that the group elements are bounded and the space between elements is infinitesimal so we have an integral instead of a sum and a range rather than the order of the group? Or am I missing some important intricacies

[rocky cloak]

Maybe it depends on how exactly divisible is defined.

Like you can't divide by (1,0) in kxk

[candid dove]

What equations are they talking about

Why is k suddenly algebraically closed

I am supposed to use this apparently

[rocky cloak]

I guess there is some work in showing that you can do integration on your group at all. Like the existence of Haar measure or whatever

[grim hawk]

That is actually what the question is titled “Haar Measure” however it has never been introduced to us. Is there a short definition you could provide?

[rocky cloak]

Well, the Haar measure is just the measure (way to integrate) that is invariant under left multiplication by elements of the group.

[toxic zephyr]

In(S4) iso S4/C(S4)=S4/<e>=S4
im just trying to understand how S4 can be iso to In(S4). is this saying that S4 acting on itself by conjugation is the same as acting on itself by multiplication? i.e. that conjugating by some element is the same as left multiplying by a different element? i'm struggling to understand what this means

[grim hawk]

Hmm…. Okay. I’ll have to do more reading. This course requires a lot more formalism than I am familiar with so it’s making things difficult. I appreciate it!

[toxic zephyr]

or is this more trivial than i'm making it out to be. just that conjugating by each element gives a unique inner automorphism which preserves the group structure phi_(g1g2)=phi_g1 o phi_g2

[next obsidian]

Conjugation by x and conjugation by y are different when x and y are different

Gives u two ways for the group to act upon itself faithfully

Left mult and by conjugation

[toxic zephyr]

and normally, phi_(xc)=phi(x) if c is in the center, so a nontrivial center means that that these conjugate maps are no longer unique (because there are "fewer" of them than elements in the group)?

[next obsidian]

It just says that different elements give the same conjugation action

When they differ by something in the center

[toxic zephyr]

yeah that's basically what i'm thinking

that's why it's iso to the quotient by the center

okay thank you chmonkey ❤️

[rocky cloak]

For a compact group it's not so bad.

Consider an open neighborhood of the identity U. Then for an element g in your group gU = {gu | u in U}.

The set of all such gU covers your group, and since the group is compact you only need finitely many. Call the minimal number you need [G:U].

For any subset S of G. Let [S:U] be the minimal number you need to cover S. Then [S:U] <= [G:U].

Now define the measure of S to be the limit of [S:U]/[G:U] for U smaller and smaller neighborhoods of the identity.

[wild jasper]

suppose that K/F and L/F are finite extensions where each basis is linearly independent over the other, then K intersected with L = F?

[next obsidian]

How can you even make sense of the bases being linearly independent?

You need to embed K and L into a super field and that isn’t canonical

I can’t say right off the top of my head, but I’m 99% sure you can embed them in different ways which have different intersections

Actually yeah, take Q and look at quadratic extensions

You can do say Q(i) and Q(i) again

Or like Q(i) and Q(sqrt(2))

[next obsidian]

This is kinda silly but if you work with quartic extensions you can get intersections of degree 2 or degree 1 or degree 4

Q(e^pi/4) and Q(i, sqrt(2)) for example

Or Q(e^pi/4) and like, Q(sqrt(2),sqrt(3))

[wild jasper]

I see

[restive birch]

?

[dull ginkgo]

This is a pain to show additivity, it's a bit easier with the Reisz Markov way but eh

[night bison]

yeah, choosing a basis requires choice

[next obsidian]

This isn’t the important part

[night bison]

well, it requires a larger field

as well

[next obsidian]

For one they’ve started with a choice of basis

[next obsidian]

Vector space, but yes that was my point

[night bison]

to compare the two bases

[rocky cloak]

You need a common superset for the intersection to make sense anyway. So I would take that as given

[spice whale]

?

abelian groups???

[next obsidian]

Idk. To make sense of it yes, but it isn’t uncommon for people to not realize stuff like this when first learning stuff

[night bison]

question was about $S_4$, right?

[cloud walrus]

SelahW

[next obsidian]

I was saying what inn(G) = G means

[spice whale]

oh

i can't read

sorry

[next obsidian]

Just kidding ily aly 🫂

[next obsidian]

Good

Wow those sent out of order… awkward 😂😂😂😂

[proud canopy]

is there an elementary proof that even and odd permutations are well defined that doesnt involve constructing a homomorphism that takes functions and spits out functions of functions of whatever

like a more elegant one

[next obsidian]

How are you defining even and odd?

[rocky cloak]

I'm not sure what kind of proof you're alluding to, but the one I'm familiar with goes like:

• every permutation is a product of transpositions.
• if p can be written as a product of an even number of transpositions and as a product of an odd number, then e = p * p^-1 can be written as a product of an odd number.
• so it's enough to show that the identity can not be written as the product of an odd number of transpositions. This you can do by induction and using the relations between transpositions

[next obsidian]

You can have S_n act on the polynomial f= Prod (xi-xj) for i < j and then it’s clear that the action produces f or -f and use that to define even and odd

This is maybe what you’re saying with functions of functions idk, but when you phrase it this way it’s very direct

and it’s clear the notion is well-defined here

[proud canopy]

okay thanks thats a billion times simpler

[proud canopy]

It was kind of like that but the proof used a specific example of a homomorphism that uses directed graphs in order to get one that goes from Sn to Z2 where every 2 cycle goes to the non identity

[next obsidian]

Lmao

[rocky cloak]

Similar to what chmonkey said you could also define parity by the number of inversions (pairs i<j such that p(j) < p(i)).

Then what you need to prove is that this behaves the way you'd expect with composition

[golden turtle]

What would be the non-injective module, I tried Z2 as a Z4 module and thought it would work but it seems like Z2 is divisible as a Z4 module

[rocky cloak]

What's your definition of divisible that makes Z2 divisible?

In any case Z/2 is not injective as a Z/4-module

[golden turtle]

Oh what I thought I had shown that it was injective

Do you know what I did wrong?

[rocky cloak]

g(1) = 1 implies g(2) = 1+1

[golden turtle]

oh equals 0

[golden turtle]

I wasn't sure where it was in the book, so I went with this one, maybe they are not the same or maybe I mis-used it

[next obsidian]

This is the definition

[golden turtle]

but then I thought the non-zero divisors were 1 and 3, and we can solve
0=1*a
1=1*b
0=3*c
1=3*d
for a,b,c,d in Z2

[next obsidian]

But I’m not sure about the non-zero divisor part I guess I’ve only really seen it for integral domains 🫥

[rocky cloak]

I see, in that case any injective module is divisible, but the converse is not true as the example Z2 shows

You can also have counterexamples for domains, but they will have to be more complicated

[golden turtle]

any injective module is divisible?

I don't think so, its asking me to make a counterexample

[rocky cloak]

Proof here

[golden turtle]

oh in an integral domain

yes

I misunderstood what you mean

[rocky cloak]

Not integral domain no

Or what do you mean?

[golden turtle]

"If R is an integral domain, then any injective module is divisible."

[rocky cloak]

This must be using a different definition of divisible then

[rocky cloak]

Sure, but you can replace "integral domain" by commutative ring

Don't need commutative either

[rocky cloak]

This definition is only stated for integers, but I suppose you could imagine extending it to
A = rA
for all nonzero r in R

[golden turtle]

or maybe all non zero divisors r in R?

[rocky cloak]

Sure, but then your exercise would be false

[golden turtle]

so an R module A is divisible if rA=A for all nonzero r in R
and then we have to use a non-integral domain to find an example of an injective but not divisible module

[proud canopy]

yeah shit was ass

i ended up with 4 diff homomorphisms that build on eachother

[rocky cloak]

Yes, though it's a little silly because if R has zero divisors then it's impossible for A = rA for all nonzero r, unless A=0

[golden turtle]

so the example has to be 0 ?

[rocky cloak]

Well you wanted an example where it wasn't divisible, so you'd just take any non-zero injective

Could also be that the exercise meant to ask about a divisible module that wasn't injective...
Maybe worth checking erata on that

[golden turtle]

oh we let R have zero divisors and then any non zero module over R is not divisible, that's what you meant by that?

so we just find a ring R with zero divisors and any injective module over it

then it can't be divisible, because of the zero divisors

[rocky cloak]

Yeah, I mean, you've already done that

[golden turtle]

but I thought Z2 was not injective as a Z4 module

[rocky cloak]

Z2 isn't, but Z4 is

In general Z/n is an injective Z/n-module

(except when n=0)

[golden turtle]

I see

thank you

[golden turtle]

I am looking at exercise 10 like it says, and struggling to see the difference between both problems part (a)

now M is a left R module instead of being a ring like S, is that the only difference?

[opaque finch]

Maybe I will need slme more hints.

In this case we have $L = \mathbb{Q}(a^{1/p},\zeta_p)$. The degree of the extension is definitely $p(p-1)$. But what more?

[cloud walrus]

mycroftholmes1703

[next obsidian]

They're really different, you should take a closer look

The former says that if you take linear maps out of a ring R you can turn this naturally into an R-module

or well okay

they're not too different

if you let R = Z and S = R (from the first image) then the second image's a is the same thing as the first one

The thing that made the top image work is that an R-module M is also a (Z,R)-bimodule

[restive birch]

?

like J(QD16)

[amber cradle]

How do I calculate $Z[\sqrt{-5}] / (1 + \sqrt{-5})$

[next obsidian]

Can you express 1 + sqrt(5) in terms of sqrt(-5) first?

Say you can do that, and say that 1 + sqrt(5) = p(sqrt(-5)) where p(x) is a polynomial

Then Z[sqrt(-5)] = Z[x]/(x^2 + 5) where you send x to sqrt(-5)

So Z[sqrt(-5)]/(1 + sqrt(5)) = Z[x]/(x^2 + 5,p(x)) by the third iso

Now you can maybe try to quotient by p(x) first, and then see what x^2 + 5 becomes under that quotient

[amber cradle]

Oops

That's a typo

[next obsidian]

Bruh

[cloud walrus]

aNDY

[amber cradle]

My bad

[next obsidian]

This is just Z

Lol

(1 + sqrt(-5)) = (sqrt(-5))

Wait

[amber cradle]

Im pretty confident it should $C_6$

[next obsidian]

No wtf am I smoking

[cloud walrus]

aNDY

[next obsidian]

Okay well

My previous comments basically stand

This is

Z[x]/(x^2 + 5, 1 + x)

If you quotient by the latter first you get Z where x = -1

So you have Z/((-1)^2 + 5) = Z/6Z

[amber cradle]

Lemme think about this for a second

Need to verify

Im so poor with commutative algebra

[next obsidian]

My point is that R/(f,g) = (R/(f))/(g-bar)

By the third iso

[amber cradle]

Sorry, what's g - bar

[next obsidian]

So you can use that Z[sqrt(-5)] = Z[x]/(x^2 + 5)

[next obsidian]

The image of g in R/f

I’m basically saying you can do a quotient by two things by quotienting by them 1 at a time

[next obsidian]

So when you observe this and then compute Z[sqrt(-5)]/(1 + sqrt(-5)) you’re choosing to quotient by f and then g where f = x^2 + 5 and g = 1 + x

But you can do it in the other order

[amber cradle]

I see

[next obsidian]

Note that g-bar in this case is 1 + sqrt(-5) because once you quotient by f you’re saying that x is sqrt(-5)

[amber cradle]

What's the algorithm to compute Z[x]/(x+1)

[next obsidian]

Just send x to -1

[amber cradle]

Why does that work

[next obsidian]

This is a surjection to Z with kernel x + 1

[amber cradle]

Ohhhh

That makes sense

[next obsidian]

But like when you quotient you’re just saying

x + 1 = 0

So x = -1

That’s the intuitive thing, the first iso is the way to justify it

This is the same reason why Z[x]/(x^2 + 5) is Z[sqrt(-5)]

[amber cradle]

Yeah im familiar with the abstraction, just haven't done enough examples yet

[next obsidian]

Send x to sqrt(-5) and the kernel is (x^2 + 5)

But intuitively yeah, you’re saying x = sqrt(-5) cuz algebra

Yeah I’m just saying what you should be thinking to recognize how to compute it

Then you justify it after

You gotta know your idea before you can go and prove it yknow?

[amber cradle]

Facts

So Z[x]/(x+1) should just be Z, right

[next obsidian]

Yeah, but where x = -1

This is important to compute Z[x]/(x^2 + 5, x + 1)

[amber cradle]

What does x = -1 mean

The set is Z(?)

[next obsidian]

You set x = -1 in x^2 + 5 lol

Yeah so okay

Z[x]/(x^2 + 5, x + 1) = (Z[x]/(x+1))/(x^2 + 5-bar)

By the third iso

But the top guy there is iso to Z where you send x to -1 right?

[amber cradle]

Yeah

[next obsidian]

So now what I’m saying is that if R is iso to S by f, then R/I is iso to S/f(I)

You can apply your isomorphism to the ideal too and take the quotient on the other side

Apply that to (Z[x]/(x+1))/(x^2 + 5-bar) under the iso Z[x]/(x+1) -> Z where x goes to -1

And then x^2 + 5-bar goes to (-1)^2 + 5

So (Z[x]/(x+1))/(x^2 + 5-bar) ≈ Z/((-1)^2 + 5) = Z/(6)

So because the iso Z[x]/(x+1) to Z was obtained by setting x = -1, you just write x = -1 in x^2 + 5 to find out what to quotient by in Z

[golden turtle]

I am stuck trying to do (b)

Specifically I am stuck here

I want to show F' is an R-module homomorphism first of all

So I want to show for all b1, b2 in B, for all r1, r2 in R, that the equality holds

but I don't know how to simplify the LHS anymore because R might not be commutative, so I feel like I can't get the two sides to be equal

[amber cradle]

I get it now

So our map $\mathbb{Z}[x] \to \mathbb{Z}[x]/(x+1)$ is given by $x \mapsto -1$, since this makes $(x+1)$ the kernel of the map

And by 3rd isomorphism theorem

[cloud walrus]

aNDY

[amber cradle]

We have that $\mathbb{Z}[x]/(x^2+5, x+1) \cong (\mathbb{Z}[x]/(x+1))/((-1)^2 + 5) = \mathbb{Z}/(6) = \mathbb{Z}/6\mathbb{Z}$

[cloud walrus]

aNDY

[rich hull]

i have a question from aluffi

this looks to be almost trivial?

im not sure if i am missing anything because once you identify the cayley graph with the group itself then you get that this is a free action immediately, no?

i am slightly confused on what the "action" defined in the parentheses is trying to convey

orrr maybe i am just bad at reading and the actions are on sets of vertices like it says.......

[next obsidian]

FWIW I’m reading it the same way

The thing in the parentheses just says what acting on a directed graph means

It says that when you move vertices which has an edge the place you moved it to also has to have an edge

But like you said, the Cayley graph just… clearly works

[golden turtle]

I'm very confused by part (a) of this

why does it just say [M is a Z-module..]

that makes part (a) completely pointless, as cor37 says "Every Z module is a submodule of an injective Z module"

so.. if M is a Z-module then its a submodule of an injective Z module, by cor37 and part (a) is pointless...

[next obsidian]

It’s just to setup the rest

It is truly pointless

[chilly ocean]

Proof: Take sequence M = M_0 \subset N \subset {0}.

By Theorem 5.9, for every Tower of submodule there exists refinement of tower such that it is Jordan Holder tower because M is R-module of finite height.

Correct?

[alpine island]

I've got a and b done, so I can assume those results. I would appreciate pointers for c, I'm not really sure how to approach it.

[rocky cloak]

You seem to have a mistake in the bottom right.

r1 f'(b1) = f'(b1)(r r1)
not r1 f(r b1)

[somber sleet]

where does here $A$ come from in the last part of the proof?

[cloud walrus]

damn_guuurl

[rocky cloak]

You have to remember that every R-module is in particular a Z-module. But yes the exercise is very easy, but it's also crucial to do (a) before (b), so that's the point

And I guess, if they didn't give you the solution in the hint it might be slightly harder

[rocky cloak]

A=R

[grizzled spindle]

Wrote down this proof in class but im unsure for the last line. Why do we have equality? I dont think p(x) | p(p(x)) if p(x) has a nonzero constant term

[rocky cloak]

You can write it out if it's hard to see, like say
p(x) = a0 + a1x + a2x^2 + ...
and write J = (p(x))

Then p(alpha) =
a0 + a1(x + J) + a2(x + J)^2 + .. =
a0 + a1x + a2x^2 + ... + J

[grizzled spindle]

oh, is it specific to alpha

like this does not hold for general q(x) + (p(x))?

[rocky cloak]

I'm not sure what you mean

Are you asking if p(q(x)) = p(x) for all q?

[grizzled spindle]

yeah that was basically it, seems like an obvious no

[rocky cloak]

That's a no yeah

[grizzled spindle]

yes for q(x) in R[x]/(p(x))

just to make sure

[rocky cloak]

In general p(q(x) + J) = p(q(x)) + J

[grizzled spindle]

for any ideal?

[rocky cloak]

Yes

[grizzled spindle]

oh ok yeah i had a misunderstanding about how the polynomials interacted i see

thx a lot

[rocky cloak]

It's just because a polynomial is given by addition and multiplication, which all work nicely with the quotient ring

[serene dune]

due the property in nilpotent maps that makes is an improper subset of Mn(R)\GLn(R) does there aries any algebraic property in the nilpotent set ?

i feel there is some thing cyclic about it, but i mightbe stupid

[rocky cloak]

Don't think so.

They're not closed under addition, not closed under multiplication, not closed under commutators.

Can't really think of anything else one would ask for

[serene dune]

aren't nilpotent maps closed?

i just argued to prove that

[rocky cloak]

What do you mean by them being closed?

[serene dune]

closed in the set of MnR

oh mb

i see your point

algebraic

[rocky cloak]

Like topologically?

[serene dune]

yeah I was doing that

[rocky cloak]

Seems plausible, but idk

[serene dune]

right two of em can produce good old GL

[rocky cloak]

What do you mean by feeling cyclic?

[serene dune]

now i think it's not really cyclic, its annihilation

but the fact that for a finite n and it's multiples the map would result in a 0 map

but 0times anything is zero

[rocky cloak]

Yeah, I guess it's closed under exponentiation

So that's something

[rocky cloak]

There's kind of a mistake in this exercise.

M shouldn't be an R-module, but just an abelian group. Or at least, every time they talk about a homomorphism to M it should be a homomorphism of abelian groups.

For example f defined by
f(a) = f'(a)(1)
is not R-linear.

[grizzled spindle]

I think I had a fundamental misunderstanding of quotient rings. I just got done doing a lot of groups and the way quotient group are defined depend on the fact that group elements permute the group

But the way the operations are defined on quotient rings vs quotient groups seem pretty different in nature as a result

[mighty kiln]

Are they

[grizzled spindle]

they seem to be at least

like aN * bN is a set equivalence to (ab)N from the usual group action

but if I take (a+I) + (b+I), if we just use set equivalence with the ring (not quotient) operation, we get (a+b) + 2I

which is not the same as (a+b) + I

[mighty kiln]

I+I is just I

Since 0 ∈ I

The same way N*N = N

The addition structure is just quotient of abelian groups

[grizzled spindle]

oh right woops for that

yeah I + I is not 2I

need to be careful about this stuff

[mighty kiln]

However (aI)(bI) ⊂ abI instead of =
(Assuming cring)

[grizzled spindle]

but I^2 is not I is it

right

okay so i think my point still makes sense

one of the operation doesnt rly follow from the set equivalence from the usual ring operation on it

[mighty kiln]

Yea

You can still do [a][b] = [ab] though

[grizzled spindle]

yeah its still well defined ofc

[rocky cloak]

So L is a radical extension, so you can find a tower
Q < Q(a1) < Q(a1, a2) < ... < L
where each time we adjoin some pth root.

If these ps are all 2, then L would have degree a power of 2. So one of these must be an odd prime.

So you have some K < L such that K(a) < L, a^p is in K, but a is not.

And you know that L is normal

[chilly ocean]

if M = M_0\subset M_1\subset M_2\subset....\subset M_n = {0}, h(M) = n+1?, where h is height of tower.

[rocky cloak]

You would usually define h(M) = n in this case, so that h(0) = 0

[somber sleet]

yeah okay, but why does this hold then?

[chilly ocean]

thank you

any hint for 5.3?
i proved 5.4 but in 5.5 it seems easy since M/ker f is isomoporphic to im f, so i can use 5.4, correct? it just outline

by 5.5, we can say that if M_1 and M_2 is isomorphic then h(M_1) = h(M_2), but can we proof h(M/ker f) = h(im f) without circularity?

[white oxide]

is there any jargon for homomorphisms which have f(a) < f(b) if and only if a < b

this is not increasing right since increasing is one direction (a < b => f(a) < f(b))

never mind it's an iff

[rocky cloak]

This is an order embedding

[white oxide]

nice thanks

[chilly ocean]

By 5.7, can I say that if G is a finitely generated abelian group and f:G -> G surjective homomorphism then f is also injective.

And for 5.8, Since f is injective therefore M is isomorphic to im f so im f is Noetherian and A cap im f is submodule of im f therefore it is Noetherian.
Correct?

[rocky cloak]

A finitely generated group doesn't necessarily have finite height, but if you know that fg abelian groups are Noetherian, then you can use the same argument

[chilly ocean]

Yes but I just need to use Noetherian property

I mean if G is a finitely abelian group then I just need Noetherian property

To show f surjective implies injective

[rocky cloak]

Yeah, in general for any Noetherian module M, if f:M -> M is surjective then it's also injective.

Dually if M is artinian and f:M -> M is injective, then it's also surjective.

[chilly ocean]

Yes so my argument is correct ?

[rocky cloak]

I don't know what your argument is.

You mean for 5.8?

[chilly ocean]

5.7

Also 5.8

[chilly ocean]

Sorry but I don't get if you know that the fg abelian group are Noetherian, fg abelian group?

[rocky cloak]

finitely generated abelian groups are Noetherian, yes

I'm not sure if that's your question

[chilly ocean]

My question was I gave an argument for 5.8 first part, I wanted to verify it

[chilly ocean]

Oh got it yes I know they are Noetherian

[rocky cloak]

im f \cap A is a submodule of im f yes, so it's Noetherian

Hence finitely generated

[chilly ocean]

Because Noetherian is equivalent to finitely generated R-module

[rocky cloak]

No, that's not correct

Noetherian is stronger

But for abelian groups it's the same

[chilly ocean]

Isn't?

[chilly ocean]

I can deduce that P is Noetherian because P/ker g is isomorphic to N so it is Noetherian and ker g = im f, so ker g is Noetherian.

So P is Noetherian but I have no idea about how to show existence of y_i ?

[next obsidian]

Wtf is height

Does that mean length?

[rocky cloak]

It means length yeah

[next obsidian]

Their hint is silly

They’re combining the proof of surjective => injective for Noetherian and the converse for Artinian

But it’s much better I think to demonstrate that length is additive on SESes

Grrrrr

[chilly ocean]

Yes

[chilly ocean]

@rocky cloak sorry to ping but theorem 5.1 implies it is equivalent

[next obsidian]

No

[coral spindle]

No, jagr was correct

[next obsidian]

Theorem 5.1 c) says every submodule is finitely generated

Take a ring such as k[x1,x2,…], as a module over itself its generated by 1

But the ideal (submodule) (x1,.x2,…) is not finitely generated

It’s a theorem that fg modules over a Noetherian ring (like Z, which is why it works for abelian groups) are Noetherian

[chilly ocean]

You mean A is finitely generated doesn't imply its submodules are finitely generated

[next obsidian]

Yes

As my example showed

[chilly ocean]

Yes

[next obsidian]

If this is true (which says all fg modules are Noetherian) then your ring R is actually Noetherian

Apply it to the submodules I < R and you get it

[chilly ocean]

Got it

[chilly ocean]

I haven't read the Noetherian ring but my guess is when you treat R as R-module and if it is Noetherian then R is the Noetherian ring, right?

[coral spindle]

Yes

[chilly ocean]

Btw, thanks @rocky cloak and @next obsidian for pointing out my mistake

[chilly ocean]

To prove this, do I need any advanced tools?

[coral spindle]

You don't, no. It's fairly straightforward to show that f.g. free modules are Noetherian and that quotients of Noetherian modules are Noetherian.

[next obsidian]

No

Show that if you have a Noetherian submodule N < M, and M/N is also Noetherian, then M is Noetherian

Doing this lets you show that R^n is Noetherian by induction

And then a fg module is a quotient of R^n

[chilly ocean]

I proved that

[next obsidian]

Well then induction gives you R^n is Noetherian, and now M = R^n/N for some N so you win

[chilly ocean]

I got how induction works here, assuming for R^n-1 = R^n/R implies R^n is Noetherian, why M = R^n/N?

[next obsidian]

If you are generated by x1,…,xn define the map R^n sending ei to xi

This is surjective (because the xi generate M) so by first iso you win

[chilly ocean]

Here I am not sure about f.g. free modules, isn't diff from f.g. ?

[chilly ocean]

Thank you for making the winner❤️

[coral spindle]

I don't know what this means exactly. f.g. means finitely generated. A free module is a particular kind of module that I don't want to get into if you don't know. A finitely generated free module is something with both of these properties. In particular, they look like R^n for n being a natural, like Chmonkey said.

[void cosmos]

can i consider the stabilizer of the action of G on the set of sylow p subgroups as a subgroup of index n_p where n_p is the number of sylow p subgroups

G acts on sylow p ssubgroups with conjugation

[wild jasper]

Our professor mentioned that if (m,n)=1 where m=[K1 : F] and n=[K1 : F], then K1 intersection K2 = F. Is this because then [K1K2 : F] = mn (because lcm(m,n)=mn) and since we have by the translation theorem that [K1K2 : F] = mn/([K1 intersect K2 : F]) and hence the denominator is 1, we have that the intersection of K1 and K2 equal F. If this argument holds, this seems to imply that if [K1K2 : F] = mn (for any reason), we can then conclude that the intersection of the extensions equal F?

[toxic zephyr]

so for a p group, we know there's always an element of order p. but can we say that if |G|=p^k, then G has an element of order p^j for 0<=j<=k? and by extension, would that show G always has a subgroup of order p^j for 0<=j<=k?

[tawny dune]

I have no idea how to do this problem in Lang and don’t really understand the hint

any hints?

[toxic zephyr]

could we say something like, let g1 have order p. then G/<g1> has order p^(k-1) (say k>1), so it has an element of order p g2<g1>. then g2 has order p^2 in G. etc.?

[tawny dune]

I guess those polynomials listed are the same as functions

[sonic coral]

every p-group is not cyclic. consider klein 4

[toxic zephyr]

i know it's not, but i'm just asking if it has an element of order p^k. in a proof i'm doing, i need to show a p group |G|=p^n has a normal subgroup of order p^k for all k between 0 and n. i've proved it providing i can say G actually has a subgroup of order p^k. are you saying that i can't guarantee this by using cyclic subgroups? because the klein 4 group has cyclic subgroups of order 2^0 and 2^1, which is consistent with what i'm trying to say.

[sonic coral]

if youre trying to show that a p group has a subgroup of the size of every possible divisor of its order, then you should use induction

and the fact that p groups have a nontrivial center

[toxic zephyr]

thats kind of what i'm trying to do here. is this idea wrong? @sonic coral

[rotund aurora]

why do you think you can quotient by g1?

[toxic zephyr]

OOF

yeah that's a damn big problem

[sonic coral]

that’s what my suggestion will help with

[toxic zephyr]

so i remember we know the center is nontrivial

and so it must have order a power of p

are you suggesting to quotient by the center?

[sonic coral]

that doesn’t guarantee that your quotient group is p^(k-1)

[toxic zephyr]

yeah

[rotund aurora]

If G is nilpotent and H subseteq G then H subset N_G(H) (strict inclusion)

p-groups are nilpotent because every nontrivial p-group has nontrivial center

[toxic zephyr]

tbh we have not learned nilpotent yet

[sonic coral]

you don’t need it but it’s a way to do it

the argument i was thinking doesn’t require it

[rotund aurora]

How did you prove that a normal subgroup of order p^k exists given that one of order p^k exists?

[toxic zephyr]

but just to confirm, i was completely wrong: we can't guarantee an element of order p^j for all valid j?

[toxic zephyr]

group action by conjugation on the set of all subgroups of order p^k

[rotund aurora]

Jaxon already told you not all p-groups are cyclic

[toxic zephyr]

its not obvious to me that cyclic is implied by "there exists an element of order m for all m that divide the order of the group"

cyclic would imply the other, but i don't see the converse

[rotund aurora]

Your groups are finite,right?

[toxic zephyr]

yeah

[rotund aurora]

If |G|=n and g has order n then G=<g>

[sonic coral]

maybe you mean to restrict to proper divisors

[toxic zephyr]

hsdgsdf right i was thinking j<n

yeah that

[rotund aurora]

but just consider Z/pZ x Z/pZ x... x Z/pZ

all elements have order p or 1

[toxic zephyr]

okay that makes sense to me

[sonic coral]

i was hoping klein 4 would get that across but i should have been more explicit

[toxic zephyr]

yeah a direct sum of exactly two wouldn't be a counterexample for what i was thinking, we need at least 3

okay so im convinced we can't use cyclic subgroups

[sonic coral]

sure, i was working with your idea of any divisor at first

[toxic zephyr]

ye

im just not sure how to use the center here to guarantee existence of subgroups

with order p^k

k=0 and k=n are obvious. just <e> or G

[rotund aurora]

Look the correspondence theorem

and try to show that normalizers grow

[toxic zephyr]

we definitely didn't cover that yet

[sonic coral]

the center is an abelian subgroup that is also a p group, how can you show the existence of a subgroup of order p that will also be normal in G

[rotund aurora]

You can come up with it

also, in case it isn't obvious, note that a quotient of a p-group is also a p-group

[toxic zephyr]

yeah this is obvious

[rotund aurora]

But the idea is that knowing information about subgroups of G/N you can conclude things about subgroups of G containing N

[toxic zephyr]

so the center is abelian p group, which thus has an element g1 of order p. this element will be in the center, so <g1> is normal with order p?

[sonic coral]

how do you know there’s an element of order p

and why is the cyclic subgroup generated by said element normal in the whole group

[toxic zephyr]

cauchy's theorem

[sonic coral]

that’s the right argument, yeah

[toxic zephyr]

p divides the order means there's an element of order p. the center is nontrivial. if g1 commutes with everything then of course conjugation will cancel

so is the idea to then do G/<g1>? or is there something simpler im not seeing

[sonic coral]

yes

and then induction hypothesis and lattice iso theorem

[toxic zephyr]

what is the lattice iso theorem

[sonic coral]

i think it’s the forth one?

it’s essentially what croqueta was suggesting

[toxic zephyr]

i dont think we ever talked about that either. but is the idea to show that G/<g1> also has an element in the center of order p which will then correspond to a subgroup of order p^2 in G

like im sure the correspondence theorem provides a very clean proof, but im trying to keep it simple.

[sonic coral]

something like what you’re saying I think

you would need to argue that the subgroup is indeed a subgroup of G

which the lattice theorem gives, but it’s not clear to me otherwise

[toxic zephyr]

okay so apparently i'm a total idiot. the first sylow theorem tells us that any group with maximal prime divisor p^n has subgroups of order p^k for each 1<=k<=n. so i was proving this for nothing (and i was done like an hour ago LOL).

still i came up with an argument, and i was wondering if you can tell me if this is BS or not?
let H1=<g1> and G1=G/H1. then |G1|=p^(n-1). if n>1, then we can repeat the argument and say we have a g2H1 in the center of G1 with order p. let H2=<g2H1>. then (G/H1)/H2 is iso to G/K for some normal subgroup K by the third iso theorem, and |G|/|K|=|G|/(|H1| |H2|) implies |K|=|H1| |H2|=p^2. so actually i guess this argument sort of directly implies the normal subgroups because i think i can just repeat this process?

[opaque finch]

Thanks. I found a seperate solution, it's actually almost similar. I used the fact that finite extension of $\mathbb{Q}$ is seperable and normal in question and hence as such is Galois. That makes the question easier I guess although I think they are the same

[cloud walrus]

mycroftholmes1703

[opaque finch]

I am stuck with something else though

Find the splitting field of $x^4 + x + 1$ over $\mathbb{F}_{64}$

[cloud walrus]

mycroftholmes1703

[opaque finch]

Assume has no other $0$ other than the trivial one. Consider the polynomial $h(X) = 1 - f(X)^{q - 1}$. Then you have the degree of $h(X)$ to be $d(q - 1)$. Now consider the polynomial $q(X) = \prod_{i = 1}^{n} (1 - (X_i)^{q - 1})$ then the degree of $q(X)$ is $n(q - 1)$. Since the only root of $f(X)$ is the trivial root $0$, then $1 - f(X)^{q - 1}$ has $n(q - 1)$ roots. But $n > d$ which is a contradiction.

[cloud walrus]

mycroftholmes1703

[white oxide]

what element would satisfy f^{-1}(y) = a kerf?

because if a in X and x in ker f, f(ax) = f(a)f(x) = f(a), but there can be multiple points in the preimage

[chilly ocean]

Yes so what's the problem? There can be multiple elements in the preimage

You can show that if f(y) = b then y in some coset of ker f

But I think for fixed b, f^-1(b) is a coset of ker f

[golden turtle]

this exercise felt really weird to me because it felt like we were assuming things about M that shouldnt be assumed, but I think I did it accoridng to the problem statement

I was hoping someone could check my work

[opaque finch]

can you please tell which text you are reading ?

[white oxide]

Rotman introduction to algebraic topology

[opaque finch]

Thank you

[opaque finch]

@rocky cloak any hints for this ?

[soft tiger]

Does the first #7 seem weird here? I don’t see how K is any different from G. If H is a subgroup of a group G, it automatically inherits the identity element e of G, So you have a tautology in the description for the set K, and every x in the group G would satisfy the conditions described for membership in K. Just take a=e. Or am I thinking about this wrong?

[chilly ocean]

there is "iff" statement, so may gag^-1 in H but a not in H, for some g in G

also you have to show if a in H then gag^-1 in H

S_3, H = {1, 12 }, a = 12, g = 13, gag^-1 not in H

so K is proper set of G

[lone niche]

First of all check if the polynomial is reducible or not.

[chilly ocean]

i want to prove this one by myself, any hint how can i proceed? I have no idea how to go 1 =>2, yes there exist \pie:P->N injective morphism such that g(\pie) = id on P

[void cosmos]

what do you know about f

and btw it is easier to do 1-->3

@chilly ocean

[chilly ocean]

yes wait let me do this

[void cosmos]

yeah do 1-->3

and i think you can do 3-->1 and 3-->2 aswell

good luck ;D

[chilly ocean]

1=>3, my guess is N is direct sum of ker g and im \pie.
first they have intersection {0}, and since g(\pie) = id thus g^2 = g on im g
so we can write n = (n-g(n)/2 + (n + g(n) )/2.
and (n-g(n) )/2 in ker g
but how can i show n+g(n) in im \pie

[void cosmos]

okay

now n-g(n) is in easily in ker(g)

[chilly ocean]

yes

[void cosmos]

without even the /2 right?

[chilly ocean]

yes

[void cosmos]

so now

ur having trouble with shiowing that the right hand is in the image

[chilly ocean]

yes

[void cosmos]

well what is the easiet thing that can be in the image

like in general

f(something) rihgt?

anythuing of the of the form f(something)

[chilly ocean]

f?

[void cosmos]

pie*

or whatever the maps name

[chilly ocean]

yeah

yes

so i have to find k such that\pie(k) = n + g(n)

[void cosmos]

well, use the splitting u have

also try to modify the way u wrote n abit

u can also do this by defining a map directly

constructing an excpliti isomoprhism

this is easier

but i can't tell u without ruinig the hwole problem for u ig

ig the hint is to draw a diagram

[chilly ocean]

no i am wrong g^2 = g on im g is incorrect

[tough raven]

It's almost correct - \pie \circ g = id on im(\pie).

[chilly ocean]

yes now how it helps me ?

[tough raven]

Suppose m = n + p with n in N and p in im(pi). What functions can you apply to m?

[chilly ocean]

g?

g(m) = g(n) + s, where \pie(s) = p

[tough raven]

But what is g(n)?

[chilly ocean]

sorry i don't get the question

[tough raven]

You can simplify g(n).

[chilly ocean]

how?

is my guess correct? N is direct sum of ker g and im \pie ?

[tough raven]

N = ker(g); I think you meant that M = N (+) im(\pie)?

[chilly ocean]

how N = ker g?

[tough raven]

Oops

M = ker(g) and N = M (+) im(\pie).

[tough raven]

So yes, that's correct.

[chilly ocean]

sorry, you mean M is not that M which is given in question

[tough raven]

Yeah, I got the letters wrong.

[tough raven]

This one is correct.

[tough raven]

This should also be with the first term in M, i.e., let's say n = m + p with m in M and p in im(\pie).

[chilly ocean]

then g(n) = s, but g is surjective

[tough raven]

where s is defined by \pie(s) = p?

[chilly ocean]

sorry i defined it here

yes

do i look at solution? i don't want

[tough raven]

I'll say this much.

If N = M (+) P, this means that every n in N is equal to m + p for unique (m, p) in M ⨯ P. Obviously you can get n from m and p as m + p, but for the direct sum property to hold, it must be possible to get back m and p from n. So you should look for an expression in (n+p) which gives you either n or p (once you get one, you can get the other by subtracting from n+p).

[chilly ocean]

thank you

[nimble folio]

can someone help me fix this definition

i really fudged something up

[cloud walrus]

clubsoda14

[void cosmos]

what

whats this v_I

is this the dual basis

[nimble folio]

v^\vee is the dual basis yes

[void cosmos]

what is w_J

shouldnt w_j be v_j

[nimble folio]

T(v_j)

why should it be w_j and not v_j

thats what i cant figure out too

[void cosmos]

T(v_j) is just an element in V

so you can write it out as a linear combination of your basis elements

[nimble folio]

oh right since T:V -> V

[void cosmos]

yes

infact; it should be a_i,jv_j

[void cosmos]

the coefficints u get from this to T(v_j) is exactly the entries in ur marix

matrix

[hidden cairn]

to show all 3-cycles are conjugate in A5, is it enough to just note a 3-cycle is a sylow 3-subgroup of A5 and all three cycles are therefore conjugate to each other?

most proofs i've seen explicitly show any three-cycle is conjugate to (123) in A5 so all 3-cycles are conjugates

[chilly ocean]

i know hint is enough but i don't get it, i am stuck at showing im f^n cap ker f^n = {0}, i concluded that if x in their intersection then f(x) must be in their intersection

[void cosmos]

do it by writing a similar decompositoin to the splitting lemma's proof

particulary use compositions of functions

u know from the chains that ker(f^p) is ker(f^n+p)

and similiary with the img

[chilly ocean]

yes

[chilly ocean]

sorry splitting lemma?

[void cosmos]

/.

[chilly ocean]

oh its called splitting lemma

[void cosmos]

yes

existence of an inverse of this map is called a "splitting" for this map

[chilly ocean]

yes

i don't get your point how can i use splitting lemma here

[void cosmos]

no

i mean't when u tried to do the proof

and tried to write n as something here + something there

not apply the splitting lemam

lemma*

sorry

[chilly ocean]

Yes so I am trying to show im f^n cap ker f^n = {0}

[void cosmos]

i think you can keep going down

like take x in im f^n

1 sec

im on bed and im stupid but im trying to show like

that the ker chain stops pre-maturaly

probably at n-q or something

[chilly ocean]

Yes

[void cosmos]

do u think this works

[chilly ocean]

Yes because the given hint has the same idea

[void cosmos]

well ig let us claim that ker(f^(n-q)) = ker(f^(n-q+k)) for all k

(under the assumption that the intersectino is non empty)

or no that won't do anything

[chilly ocean]

I think we need the same for im f

[void cosmos]

x = f^n(y) for some y and f^n(x) = 0

[chilly ocean]

Yes

[rocky cloak]

Looks great.

But Q is an injective Z-module, not an R-module.

And this was probably a typo in the previous problem as well.

[void cosmos]

f^2n(y) = 0 --> y is in ker(f^2n) which is ker(f^n) by noetherian

or whatever

the chain

but x = f^n(y)

so x =0

yeah this should work

right?

so ig we don't need the img to stablize

atleast for that

unless im wron

g

[rocky cloak]

I guess you have to also show that a 3-cycle is conjugate to its inverse.

But the advantage of the explicit proof is that it extends to An for n>5

[void cosmos]

@chilly ocean what do u thikn

[chilly ocean]

I think this part is correct

[void cosmos]

cool

me go sleep

good luck !

[chilly ocean]

@void cosmos sorry to ping, but i am again stuck M = im f^n (+) ker f^n

[void cosmos]

me on bed dead

sorry

[chilly ocean]

ah no problem i will wait

[void cosmos]

surely someone else cna help

[chilly ocean]

it seems tricky to find combination

[dense root]

how do i find the units of this Ring if anyone could give me any insights

[frail summit]

This is probably a stupid question, but I can't see why this is true, could someone help? (mu_n is the group of nth roots of unity over Q)

[woven delta]

So one way to see it is that an nth root of unity is of the form $e^{2\pi i k/n}$ where k is in ${0,..., n-1}$

[cloud walrus]

EmmaGhost

[frail summit]

Well, assume we are not splitting over C but over some abstract splitting field K

[woven delta]

We can always just work inside C here

You would just embed the splitting field into C

[frail summit]

Hm, that is true...

I don't like it though lol