groups-rings-fields

[woven delta]

Okay suppose you have some x with x^n=1 and x^m = 1. Then x^{m-n}=1, and you can show that x^gcd(n,m) =1 (by doing the euclidean algorithm)

The order of the element o(x) should be the minimal number m which gives x^m=1, so gcd(o(x), k)=o(x) for any k with x^k=1

Sorry for my inconsistent notation lol

@frail summit does this make sense?

[frail summit]

It does, yes

Though I'm not quite sure how it shows that d|n

[woven delta]

d here means o(x)

[frail summit]

But d is not assumed to be the order. zeta is just some dth root of unity

[rocky cloak]

Say z^d = 1, but d doesn't divide n, then there is a k such that n-dk < d.

Then z^n-dk = z^n / (z^d)^k = 1

[frail summit]

Indeed, but again, d isn't assumed to be minimal?

[woven delta]

Oh I'm being silly

[rocky cloak]

Well, then it's just wrong right. Like (-1)^4 = 1

Even though 4 doesn't divide for example 6

[frail summit]

Yeah

Ok, good. Sorry about the confusion. Should've just looked for a counterexample

Thanks 🙂

[rocky cloak]

But the question says d is the order, which does mean d should be minimal

[frail summit]

Where does it say d is the order?

[woven delta]

Okay yeah what I was saying was correct because d has to be the order, otherwise you can take eg n = 8, d = 6, and take a 2nd root of unity. Then that would satisfy the requirements

Which I guess you were saying

Sorry it's 6 am so I'm not coherent

[rocky cloak]

Actually you're right, it doesn't say that. But it seems that's what they meant to say, like they start talking about the order, then they want to add in this variable d and they mess up the wording

[opaque finch]

I checked that it is indeed reducible into quadratics. But the factorisation is not good at all

[rocky cloak]

If you know exactly what the finite fields are and how they fit together, there is a pretty easy solution to this

More specifically ||adjoining a root of x^4 + x + 1 to F2 gives you F16||

||So the splitting field would have to contain both F16 and F64||

||F_2^n contains F_2^m iff m divides n||

[frail summit]

Hi again. Confused about another thing. If K is some radical extension over a field F of characteristic 0, and L is the Galois closure of K, then I see why the composite of the conjugates of the radical extensions here must be mapped to itself under any automorphism, but I don't see why that implies that it should be a Galois extension (this is part of lemma 38 chapter 14.7 in D&F)

[rustic crown]

separability should be easy to see, and what they say is one of the criterions to check normality

[frail summit]

Hm, I was trying to write my thoughts out to see if I had understood correctly, but I don't think I have...

Separability seems to come from char F != 0

[rustic crown]

you mean char = 0

[frail summit]

But normaility I'm not sure of. What are the polynomials that split?

[frail summit]

Oh yeah

[rustic crown]

okie so say a is an element in the composite of those extensions which i'll call K'.

for any root b of the minimal polynomial of a over F, there must be an automorphism in Gal(L/F) which sends a to b

but that automorphism sends K' onto itself

so b must also lie in K'

this shows that all roots of the minimal polynomial of a over F lie in K'

so the minimal polynomial splits over K'

which means K'/F is normal :3

[frail summit]

Hm, and all elements of K' are algebraic because K' is a finite extension right?

[rustic crown]

yep yep :3

[frail summit]

Ok. That makes sense

[rustic crown]

[frail summit]

Thank u

[chilly ocean]

Hi det

[rustic crown]

hewwo

[chilly ocean]

I proved 1=>2, now 2=>1, since im f = im f•f so for every x in N, f(f(k)) = f(x), so let x = x-f(k) + f(k), x - f(k) is in ker f.

Correct?

got it thank you

[rustic crown]

yepp

[whole hemlock]

Does there exist a maximal proper subgroup of (R,+)?

[dull marsh]

A maximal proper subgroup of any group would need to contain the subgroup generated by x for any x in the group, can you see why this yields a contradiction?

[coral spindle]

What? That's just not true

Did you make a typo there?

Maximal proper subgroups of groups do exist in general.

[dull marsh]

What's the issue?

[coral spindle]

Your claim that any maximal proper subgroup H of a group G must contain <x> for any x in G is false

[tough raven]

Consider the trivial subgroup of ℤ/pℤ...

[dull marsh]

Are we talking about maximal by inclusion or cardinality?

[coral spindle]

Maximal usually means by inclusion unless otherwise specified

But as it happens, you can choose either in this case and your claim will still be false.

[dull marsh]

Ah so proper subgroup that contains all other proper subgroups?

[coral spindle]

No!

[dull marsh]

I was thinking about proper subgroup that contains all subgroups

[coral spindle]

Maximal is not greatest!

Maximal means there is no other larger proper subgroup

Greatest means it contains all proper subgroup

[dull marsh]

Damn, I always confuse maximal and greatest

I see

[rocky cloak]

Nope, a maximal proper subgroup would give you a map onto an abelian simple group (i.e. Z/p)

But R is divisible, so this isn't possible

[coral spindle]

Nice proof. I'm trying to argue this directly using the elements but there is a funny barrier

[tardy hedge]

I feel like dummit and foote’s introduction to the tensor product is way too messy and confusing

They just go on and on about extension of scalars and all this stuff to try and give motivation but i think it just ends up making things too confusing and all over the place

[dull marsh]

Would it be correct to say that a proper subgroup cannot contain all of (0, 1], so you always just extend by an absent element in the proper subgroup?

No I don't think

[rocky cloak]

I guess the tricky part is how do you argue that you don't get everything when you add "one more element"

[dull marsh]

Yeah

[coral spindle]

You can argue that extending by one single element x does not include one of x/2, x/3, x/4 etc

But you have to consider all of those

If you don't the argument fails since any one of them could be included in some Abelian simple group

Neat

[rocky cloak]

I guess sort of unpacking the Z/p argument:

Let N be a maximal subgroup and x not in N.

Let n be the smallest number such that nx is in N.

Since Zx + N is everything, x/n = kx + y for y in N, so
x + nkx is in N.

Since nkx is in N, so is x contradiction.

Then there's also the case where Zx and N are disjoint. In which case you can just use x/2 for example x/2 to get a contradiction

[coral spindle]

Nice nice

[rocky cloak]

It's like a weird "anti-Noetherian" property, where every quotient module is infinitely generated

[whole hemlock]

Thanks for elaborating this.

[tardy hedge]

Hom being a left exact functor means it takes s.e.s to sequences that are exact on the left, but does it go the other way too?

Is it like an if and only if thing?

I know for R modules its an iff thing, but i guess im wondering more generally in terms of category theory language, is saying a functor is “left exact” is that an iff thing

s.e.s <-> s.e.s

[tawny dune]

Hard disagree, I think it’s the best intro to tensor products

[coral spindle]

Consider Hom(Z/2Z, -) on the non-s.e.s Z -> Z -> Z -> Z -> Z with any nonsense maps you like

[tardy hedge]

Im not totally sure what i should be thinking of

[coral spindle]

Well what's the image of this sequence

[tardy hedge]

This sounds silly but im kind of not totally sure about homomorphisms Z/2Z to Z. Is it just 0 map ??

Im not sure why i stalled on this

[coral spindle]

Let g be a generator of Z/2Z. What's it's order, and what could it be sent to by a homomorphism to Z

[tardy hedge]

The generator has order 2, umm and it needs to be sent to an element of Z that has order that is a divisor of 2? Damn am i even remembering this crap right. I think i havent digested this basic stuff as well as i should have

[coral spindle]

So you should know know all maps Z/2Z -> Z

[tardy hedge]

Yeah i see. It must send everything to 0 in Z because every element besides 0 does not have finite order in Z

[coral spindle]

So Hom(Z/2Z, Z) is what?

[tardy hedge]

Thanks for this

{e}

[coral spindle]

It's 0, ye

[coral spindle]

So what's the deal here.

[tardy hedge]

In general , characterizing all homomorphisms between two structures in surely not typically able to be done tho right ?

[coral spindle]

No

[tardy hedge]

Its because cyclic groups are just nice

[coral spindle]

That is hard in general

[tardy hedge]

Thats weird. The hom sequence is a s.e.s because its just a bunch of 0 maps …. Right?

But the original is not a ses

[coral spindle]

Yes

YEs

So clearly this is not an iff.

[tardy hedge]

Dang

You have enlightened me

Wait a moment though

In this case there is some iff thing

[coral spindle]

for all R-modules D

[tardy hedge]

I should have sent the pic with Hom(D, -) instead for consistency

[coral spindle]

N.b. it is not

For all R-modules D, seq 1 is exact iff seq 2 is

it is

seq 1 is exact for all R-modules D iff seq 2 is

[tardy hedge]

Ok um hmmm

So applying Hom(D,-) to the sequence u original said for some other module D may make the Hom sequence NOT exact?

[coral spindle]

Can you think of a module that might work?

[tardy hedge]

Kneejerk reaction is Hom(Z,-) but i havent even thought about it

[coral spindle]

It works

[tardy hedge]

Yeah if the maps are just id then the hom sequence also is just id maps, so not exact

Thats ok right?

I appreciate this rn cause in lecture we did not do any concrete examples

[south rain]

I should show that if $F$ is a free group of rank 2, then it contains a subgroup of infinite rank. The obvious choice is the subgroup $H = \langle {a^n, b^n}_{n\in \mathbb{N}} \rangle$. Now, we haven't covered the theorem of Nielsen-Schreier for general free groups, only for free abelian groups, so I'm unsure if I can use it, which means I probably should show that this subgroup is free explicitly. But I'm entirely new to proving that an explicit group is free, so I don't even know what I should show and how to start. So any hints would be appreciated.

[cloud walrus]

dellinger

[charred iris]

I'm a little confused on what H is... do you mean the subgroup generated by a, b, a^2, b^2, a^3, b^3, ... where a and b are generators of F?

because that's just F again

regardless of what H is, one way you can show a group is free is by showing it's the image of an injective homomorphism from a free group of (presumably you mean countably) infinite rank

[south rain]

Oh I'm sorry, thats a typo, what I mean was ${a^nb^n}_{n\in\mathbb{N}}$

[cloud walrus]

dellinger

[south rain]

That makes sense, but we have quite literally not have had any examples of infinite rank (non-abelian) free groups.

And I suppose I can't take an abelian free group as the domain, since my free group isn't abelian.

[charred iris]

Did you not construct free groups as having elements (equivalence classes of) strings?

At the very least as an example

like take a set S, then F(S) is the group of strings with concatenation up to making inverses exist for letters in S?

[south rain]

Yea our construction was of this sort.

[charred iris]

If S is infinite you get an infinite rank free group

[south rain]

Yea that sounds right.

[charred iris]

in fact now I'm wondering what your definition of rank was, because I'm pretty sure I had it as being just the size of S

[south rain]

Thats exactly the case.

[charred iris]

So if you take S = N, there's a pretty obvious map you can extend to a homomorphism on F(N) a free group of countably infinite rank, and then all that's left is showing it is injective

[south rain]

Alright that makes sense but now I'm confused again because it now seems as if I have a tautology, because if is just take the set $S = { a^nb^n}_{n \in \mathbb{N}}$ for which it holds tha $S = \infty$, I already have the free group $H$ and a subgroup of $F$.

[cloud walrus]

dellinger

[south rain]

What I mean is I can just start with $S$, construct a free group from it and then have $H = \langle S \rangle$ which is a subgroup, but that doesn't seem right or am I not understanding something properly.

[cloud walrus]

dellinger

[charred iris]

How do you know H is free?

[south rain]

By construction no?

[charred iris]

As a subgroup generated by elements?

That's a very different thing to saying 'equivalence classes of words etc.'

The fact that the generated subgroup is a free group at all is the Nielsen Schreier theorem

[south rain]

Aw man

Alright

So I just construct a free group from $\mathbb{N}$ and find a monomorphism onto my subgroup?

[cloud walrus]

dellinger

[charred iris]

And even then, if I take S = a^2, a^3 in F({a}) (which is a free group of rank 1) it doesn't follow that the subgroup generated by S is rank 2 (it's just rank 1 F({a}) again) so you wouldn't know you have the correct rank subgroup

[charred iris]

That's what I did back when I did this as an exercise

[south rain]

Alright, thanks, I'll try that, seems to be the most straightforward way.

[tardy hedge]

For a splitting homomorphism, is there anything to be said about composing the maps in the other way?

If h a splitting hom and so phi o h = identity, what about h o phi?

[opaque finch]

I need help in a convention for affine maps. I know affine maps are maps such that $x \to ax + b$. Now I don't understand what one means when they say such a thing
"Let $f:\mathbb{F}_q \to \mathbb{F}_q$ be an affine map over $\mathbb{F}_2$, where $q$ is a power of 2". Does it mean that the coefficients $a,b$ come from $\mathbb{F}_2$ ?

[cloud walrus]

mycroftholmes1703

[tidal berry]

Is anyone studying ring theory - field theory on their own? I feel like making a study group

[opaque finch]

please do

[tidal berry]

@opaque finch you seem to know a lot already tho. I just know the basics of ring theory

Well I'm not sure if only the basics, I studied from Dummmit-Foote and Chapter 0

[opaque finch]

Chapter 0 ?

[tidal berry]

Algebra: Chapter 0

[opaque finch]

Chapter 0 in Dummit and Foote book ? 😅 sorry I don't get the reference

[tidal berry]

Noo

Algebra: Chapter 0 by Paolo Aluffi

[opaque finch]

that is a real book ? Unique name though

[tidal berry]

It's a huge book about abstract algebra from a category theory perspective

You should Google it, it's quite good

[opaque finch]

sounds fun

[tidal berry]

Like, I passed the subject about groups, rings and modules at my university

And I'd like to study a bit of field theory before taking that subject

[fossil shore]

Hey, why can we assume f to be monic here?

[opaque finch]

hmmnn, well the book I usually follow for my ongoing course is Field and Galois theory by Patrick Morandi

[fossil shore]

I assume it's an application of Gauss' lemma, but how can we make sure that f remains monic after applying the lemma?...

[opaque finch]

What is the definition of cyclotomic polynomial ?

[fossil shore]

Product of (x-zeta) as zeta spans over the primitive nth roots of unity

[opaque finch]

and the leading coefficient of such a polynomial is 1

[fossil shore]

Yes

[opaque finch]

so in $\mathbb{Z}[x]$ can you have two non-monic polynomials multiplying to give leading $1$ ?

[cloud walrus]

mycroftholmes1703

[fossil shore]

Oh, I didn't see the theorem assumes we want to show irreducibility over Z

Nevermind

Thanks

[dull ginkgo]

In polynomial rings of integral domains, the leading coefficient of a product is the product of the leading coefficients. Monic means having a leading coefficient of 1 (a unit), so really the only other option for leading coefficients here is -1 since that’s the only other unit in Z

[opaque finch]

True but even if it is -1 in both cases, one can just factor out the -1 and hence no harm in assuming that is monic

[dull ginkgo]

No longer monic

[glad osprey]

If F is a field, V is a vector space and T : V -> V is a linear transformation then V is a module over F[x] as described here: https://math.stackexchange.com/questions/2458918/difficulty-in-understanding-fx-modules

The scalar multiplication is p(T)v for p in F[x] and v in V. I don't understand how (pq)v = p(qv) holds though? Multiplication of polynomials is not equal to the composition of polynomials. There's probably something really simple I'm missing, can someone explain?

[opaque finch]

From where did you get such a relation?

[glad osprey]

You mean (pq)v = p(qv)? That's one of the axioms of modules, number 3 here: https://en.wikipedia.org/wiki/Module_(mathematics)#Formal_definition

[opaque finch]

Ahh I see now. So
(pq).v = (pq)(T)v = p(T)(q(T)v)
The relation just means that applying v first to q(T) and then the resultant to p(T).

[grave sedge]

What's (T²oT³)(v)?

More generally, think about the case when p and q are both powers of x

[glad osprey]

It's T⁵(v) Just before you wrote it I think I realized why it works - when you do p(T) you turn exponents into repeated composition, so multiplication becomes composition. Even multiplying p(T) and q(T) is probably not even well defined, since we're restricted to composition and addition

so it's sufficient to verify that T^a o T^b = T^(a + b) and since addition is pointwise everything works out. Thanks

[grizzled spindle]

are the quaternions isomorphic to R[x]/(p(x)) for some polynomial?

[glad osprey]

I don't think so, since the quaternions are not commutative

[restive birch]

"prove that there is no simple group of order 6545"
what i have:
take n5(G) \ne 1, and so we need n5 = 1 mod 5, n5 divides 6545--the only solution to this is n5 = 11. Then take a sylow 5-subgroup P of G, and take the normalizer of P having index 11, meaning that it admits a homomorphism phi: G -> S11 by the left action on cosets. the kernel of this action has to be {1} since otherwise it would be normal, so that G is isomorphic to a subgroup of S11, a contradiction since 6545 does not divide 11!

not sure how to show without just making a program that 11 is the only solution, it might not be but idk

okay it is the only solution, but i dont have a clean way to prove it, but the math isnt that bad it doesnt have that many factors

Want a hint:
"Let |G| = 231, and show that there is a normal sylow 7-subgroup of G and that Z(G) contains a sylow 11-subgroup of G"
ive shown that it contains a normal 7-subgroup and a normal 11-subgroup but idk how to show that the latter is in the center

[restive birch]

advanced help channels moment

[wraith nexus]

well if the 7 subgroup and 11 subgroup are normal

then the subgroup generated has order 77

and is the product of the two abelian groups, so is abelian

[restive birch]

you mean like P1P2

where P1, P2 are the 7 and 11 subgroups?

yea i agree

[wraith nexus]

I mean <P1 union P2> is equal to the product P1P2

you can then verify said group is normal in G

dunno where to go from there

[restive birch]

yea me neither

should i use the help channels for these? ive seen some people using them for advanced math

[wraith nexus]

the number of Sylow 3 subgroup is equal to 7 if G nonabelian

[wraith nexus]

maybe I assume solid amnt of people who help there would be able to help

[wraith nexus]

oh I see how to do it

you can quotient out by P1

and get a group of order 33 which is abelian

[restive birch]

okay

[wraith nexus]

picking g in G and h in P2, see

gh k = hg for some k in P1

then using normality of P2 we find k=1

fuck nvm

[restive birch]

well P2P1 can be identified as a subgroup of G/P1 right

does that help?

[wraith nexus]

can it?

G/P1 is order 33

and P1P2 is order 77

[restive birch]

youre right sorry i meant P2 (via P1P2 and lattice iso)

[wraith nexus]

yeah P2 we can

[wraith nexus]

oh wait no we are done here

k = h^-1 (g^-1 hg)

[restive birch]

wait sorry how do we conclude k = 1?

[wraith nexus]

the intersection of P1 and P2 is trivial

and P2 is normal

[restive birch]

i dont follow

can you write it out more explicitly?

[wraith nexus]

we have g^-1 hg in P2 as h in P2 and P2 normal

thus,k= h^-1(g^-1 hg) is in P2

note k is in P1 as well

so, the order of k divides 7 and 11

i.e. it has order 1, so k=1

[restive birch]

okay i get it

how do you figure out this method? seems pretty random to me

[wraith nexus]

taking the quotient?

I remember it being used occasionally when proving stuff related to Sylow thm and whatnot in my alg class and hw

[vast stump]

accurate description of mathematics problem solutions

[wraith nexus]

@restive birch actually there is a way to do it without taking the quotient

using semi direct products

[restive birch]

hasnt been introduced

its next chapter

[wraith nexus]

ah

okay

[nimble folio]

Let $t: \mathbb{R}^3 \times \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be defined by $t(v, w)=v \times w$. Let $\mathcal{E}3$ be the standard basis of $\mathbb{R}^3$ and $\mathcal{B}=\left{e_i \otimes e_j\right}{1 \leq i, j \leq 3}$. Let $T \in \operatorname{Hom}F\left(\mathbb{R}^3 \otimes{\mathbb{R}} \mathbb{R}^3, \mathbb{R}^3\right)$ be the linear map associated to $t$. Calculate $[T]_{\mathcal{B}}^{\mathcal{E}3}$.
\vspace{20pt}
The matrix I got was:
\begin{equation*}
[T]{\mathcal{B}}^{\mathcal{E}_3} = \left(\begin{matrix} 0 & 0 & 0 & 0 & 1 & 0 \ 0 & 0 & -1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 & 0 \end{matrix}\right)
\end{equation*}

[cloud walrus]

clubsoda14

[nimble folio]

Can anyone elude to why this is what [T]_B^E looks like?

Or if this looks familiar to something cross product related

I can show my work if needed but if anyone knows it off the top of their head that'd be great

Also I have a feeling this is completely wrong

Im confused

[fresh gate]

is there any value to having vector fields of p-adics?

[wild jasper]

Let $\mathbb{Z}_2 \times \mathbb{Z}_2$ as subgroups of $S_4$ , given by $\langle (1 2) \rangle\times \langle ( 3 4)\rangle$ then this is not transtive since these are disjoint cycles, is that correct?

[cloud walrus]

mh_le

[coral spindle]

Indeed it does not act transitively on {1,2,3,4}

[wild jasper]

but $\langle (1 2) \rangle\times \langle ( 2 4)\rangle$ does?

[cloud walrus]

mh_le

[coral spindle]

Ignoring the misuse of \times…. every element of that group fixes 3. What does this tell you.

[wild jasper]

I don't know

[coral spindle]

It does not act transitively on {1,2,3,4}.

[wild jasper]

I see

ahha

how can it be transitive?

[coral spindle]

It isn’t, as I said

[wild jasper]

I meant, how can I even represent $\mathbb{Z}_2 \times \mathbb{Z}_2$ as a transitive subgroup of $S_4$? If it is even possible

[cloud walrus]

mh_le

[coral spindle]

I think you need to think about the orbit-stabiliser theorem

[wild jasper]

ok

I am having an anxiety attack and my brain is fried

could you tell me, please?

[coral spindle]

You're having an anxiety attack? Please go to someone in person right now so they can look after you

[wild jasper]

there isn't any

[coral spindle]

Do you have any people you know that live nearby?

[wild jasper]

no

[coral spindle]

Any neighbours? Do you live in a flat? Please visit them right now

[wild jasper]

I don't know any of them

[coral spindle]

OK what are your symptoms? High heart rate, difficulty breathing, does you head hurt, is your vision going? You may need medical attention -- I sure as hell did when I had a panic attack

[wild jasper]

heart racing, panicing and lots of thoughts going 1000 km/h

[coral spindle]

Call your parents, call a friend, call anyone who knows you personally and can help calm you down right now

[wild jasper]

yeah

[glad osprey]

I want to show that the polynomial ring $F[x_1, x_2, ...]$ in infinitely many variables over a field F is neither Noetherian or Artinian. Showing that it's not Noetherian is easy since $(x_1) \subset (x_1, x_2) \subset \dots$ is an infinite ascending chain. How do I show that it's not Artinian?

[cloud walrus]

sheddow

[rocky cloak]

The idea is very similar.

So notice what you did is you made the ideal bigger and bigger by adding one variable at a time.

So maybe you can start with an ideal an make it smaller and smaller by removing one variable at a time

Do you see how you could do that?

There's also the heavy hitting theorem that all artinian rings are Noetherian.

Or you can use the same argument you would use to argue that F[x] isn't artinian

[chilly ocean]

Can someone give me resources for Ring theory for self study?

[chilly sparrow]

At what level? I'm enjoying Aluffi's Algebra Notes From the Underground. It's an intro to algebra which does rings first

[chilly ocean]

level 1

doing it after group theory

:>

[glad osprey]

Ah I see, just take the infinite descending chain (x1, x2, x3, ...) > (x2, x3, ...) > ... Thanks

[wild jasper]

We used dummit and foote

[tardy hedge]

How do we know that the 6 groups presented here, any one of them arent isomorphic to the other?

This book just presented the structure theorem for fin gen abelian groups and now just used it here

[wild jasper]

they look like to have different cardinalities

[wraith cargo]

they all have the same cardinality

[wild jasper]

you're right

[wraith cargo]

you essentially see they're different by looking at the order of each elements

1. has no element of order 4

but 2. does

[wild jasper]

but they all have elements of orders there arn't in the others

[terse crystal]

Just calculate elementary divisors

For example the first group

Or just the smith form, the first group has smith form diag{2,12,360}

They are distinct iff their smith form (determinant divisors) / elementary divisors/ invariant factors are different

[wild jasper]

how does the RHS account for all of the n'th roots of 1?

[rocky cloak]

If x is a nth root of unity, then it's order must divide n.

The roots of phi_d are exactly the elements of order d

[wild jasper]

ah

I see

[dawn quest]

This isnt really algebra but all the proof of the first sylow theorem I see make use of Kummer's theorem

Not sure how to prove it using Legendre's

[dull ginkgo]

Jacobson provides an alternate proof that i know by memory

[dull ginkgo]

Would you like me to walk you through it?

[rocky cloak]

The only proof I've seen uses properties of group actions by p-groups.

How would you use kummers theorem? Not really seeing where binomial coefficients come in...

[south rain]

I need to show that $\mathbb{Q}/\mathbb{Z} = \prod_{p} Z(p^\infty)$ where $Z(p^\infty)$ is the Prüfer group defined by $\mathbb{Z}(p^{\infty}) := {\left[\frac{a}{p_i}\right] | a \in \mathbb{Z}, i \geq 0}$. So far I've shown that the $Z(p^\infty)$ are subgroups, that if $\langle c_i \rangle := \langle \left[1/p^i\right] $ I have that $\bigcup_{i\geq 0} \langle c_i \rangle = Z(p^\infty)$ and that every finitely generated subgroup is cyclic and the Prüfer group itself is not. The exercise is structured in a way that proving the above is the last one, suggesting to me that I should use the other obtained results. Sadly I don't even know how to begin really.

[cloud walrus]

dellinger

[dull ginkgo]

that is the most intuitive route

[wraith cargo]

what is?

with or without group actions

[dull ginkgo]

with

[dawn quest]

The proof on wikipedia uses group actions, but it ends by showing that the v_p of (p^k * m) choose p^k is r, where v_p(m) = r. For this they use Kummer's theorem

[rocky cloak]

Well it doesn't use anything about group actions of p-groups.

The lemma right below could be relevant for example.

But I guess cool proof, even though it's kinda unnecessarily complicated

[dawn quest]

That there are easier proofs I didn't know

Wikipedia is messed up for that

[rocky cloak]

Anyway, for your actual question:

Write
n = n0 + n1p + n2p^2 + ...
then
n =
n0 + n1 + n2 + ... +
n1(p-1) + n2p(p-1) + n3p^2(p-1) + ... +
n2(p-1) + n3p(p-1) + ... +
...

Write out the same thing for m and n-m, and substitute into the equation n - m - (n-m) = 0

Moving the
(n0 + n1 + n2 + ...) -
(m0 + m1 + m2 + ...) -
((n-m)0 + (n-m)1 + ...)

Over to the right hand side and dividing by p-1 gives the desired formula on the the right and the result of Legendre's formula on the left

[dull ginkgo]

Here's a long winded explanation that explains it for me.

If we have a group action of G on some set X, then naturally we can create an action on the power set of X, P(X), by saying xS is just the orbit of subset S under x, i.e {xs, s in S}.

Well obviously, G acts on itself by left multiplication transitively and faithfully (i.e left multiplication is a bijection for each g, since g^-1 cancells it), so we can naturally consider how G acts on its own subsets. If S is a subset, by bijectivity, |xS| = |S|, so multiplication preserves the order of subsets.

From there, one might ponder about orbits and stabilizers. So if we have a subset S, it's natural to ask about Stab(S), in other words, g : gS = S. In that case Stab(S) acts on S since it's closed under left multiplication by Stab(S). We can then partition S into orbits by Stab(S)... all of the form Stab(S)s for s in S. All of these are cosets of Stab(S)s, so S is a disjoint union of right Stab(S) cosets, so |Stab(S)| divides |S|.

[rocky cloak]

I guess you can simplify this with some induction:

The relevant formula is
n = Sp(n) + (p-1) vp(n!)

Clearly true for n=1. Now
vp((n+1)!) = vp(n!) + vp(n)

So you just need to show that
Sp(n+1) - Sp(n) = 1 - (p-1) vp(n+1)

[dull ginkgo]

How is this useful?

Well, lets say |G| = p^k * m, where p doesn't divide m, and we consider K_p to be the collection of all subsets of size p^k. This is closed under the G-action since that preserves cardinality. If we take any set S in K_p, so |S| = p^k, we have |Stab(S)| divides |S| = p^k, so the stabilizers ALL have cardinalities being powers of p.

So lets partition K_p into subcollections, P_1 ... P_k, consisting of subsets S that all have the same power of p (or say they're equivalent if they have the same stabilizers of the same order).

From this point, you can derive the sylow statements by analizing the sizes of these subcollections and their constituents' G-orbits, and how they partition up these subcollections

Namely, if |Stab(S)| = p^n, then |GS| = mp^(k - n) by orbit stabilizer, so mp^(k - n) divides |P_n|

Incredibly, these partitions are related to Krummer's theorem

if you feel like it you can connect them, as the number of subsets of order n is |G| choose n :3

Actually i think it would be funny if you proved Krummer's via this bat shit insane algebraic route applied to a cyclic group of order mp^k

[rocky cloak]

As for the proof I'm familiar with:

Lemma [Cauchys theorem]
If G has order a multiple of p, then G has an element of order p.

Proof: Let X be the subset of G^p where the elements multiply to the identity. Note |X| = |G|^p-1 is a multiple of p.

Cp acts on X by just shifting entries. Note that a fixed point of this action is a tuple where all the elements are the same. So an element with g^p = 1.

Since Cp is a p-group the number of fixed points is congruent to |X| mod p, so is a multiple of p. Since the tuple where every element is the identity is a fixed point the number of fixed points is not 0. So G has at least p-1 elements of order p.

Lemma: If H < G is a p-group and N(H) is the normalizer of H, then [N(H):H] is congruent to [G:H] mod p
proof: consider the set G/H. H acts on it by left multiplication. A fixed point of the action is a coset gH with HgH = gH.

This can be rewritten as g^-1HgH = H which means g is in the normalizer of H. So the fixed points are exactly N(H)/H. Since H is a p-group |N(H)/H| = |G/H| mod p.

Theorem: if H < G is a p-group such that |G|/|H| is a multiple of p, then there is a p-group H' < G with H a proper normal subgroup
Proof: [G:H] is a multiple of p means [N(H):H] is a multiple of p. So N(H)/H is a group whose order is a multiple of p. By Cauchys theorem it has an element of order p giving rise to a cyclic subgroup of order p.

A subgroup of N(H)/H corresponds to a subgroup of N(H) containing H as a normal subgroup. Calling this subgroup H' we have our results.

Sylow theorem: apply previous theorem until |G/H| is not a multiple of p.

[dawn quest]

Thank you @dull ginkgo @rocky cloak , I will carefully read over your messages now

[tardy hedge]

If a module has nonzero annihilator, then Tor(M) = M right

Sanity check kind of thing

[rocky cloak]

Yes, if an element is annihilated by something it's torsion

[tardy hedge]

Taking M an A-module, whats the interpretation of considering it as an A/Ann(M)-module

[dull ginkgo]

quotienting out the redundancy

[tardy hedge]

It has no annihilator now?

[dull ginkgo]

of elements that don't have any effect on the action

so now it's faithful :3

[tardy hedge]

Cool

[dull ginkgo]

I interpret the annihilator as kind of the redundancy of the "ring action on M", (kernel of the map from R to End(M)) and so quotienting it out, it's surjective and faithful

[rocky cloak]

One interpretation is that an A module is a homomorphism
A -> End(M)

Then you're just factoring this map by modding out the kernel

[dull ginkgo]

Also do note, annihilators of subsets of M and shit aren't typically two-way ideals, but they are left/right ideals.

[dull ginkgo]

I say this literally because the whole idea of a module being a ring action on an abelian group (map from R to End(M) or R^op to End(M)) is similar to group actions being G to Symm(X)

same gist, quotient out stabilizer, faithful

[dull ginkgo]

Yo if we have a left R-module M, would the annihilator map from submodules of M to left-ideals of R, and the annihilated module map from subideals of R to submodules of M be a contravariant galois correspondence?

actually i think that's due to the heterogeneous relation where rRm iff rm = 0

[pastel condor]

according to this text, wouldn't x(3) = 1 and x(4) = 3 mean that 1 and 3 swap places? but rather we see that 2 and 5 swapped places...

[rocky cloak]

The annihilator of a submodule would give you a two-sided ideal containing the annihilator of M.

And this should be a Galois connection between two sided ideals and submodules of M yeah

[dull ginkgo]

Oh wait yeah I was thinking of general subsets of both

[rocky cloak]

Probably still a Galois connection

[dull ginkgo]

Yeah still a Galois connection due to the relation

[pastel condor]

wait maybe im confused what does x(3) mean? Isn't it where 3 gets sent to by the permutation x?

[rocky cloak]

x(3) = 5 and x(4) = 2
then they swap yielding
x(3) = 2 and x(4) = 5

[fading summit]

super discombobulated on how to define the bijective homomorphisms to prove these two problems

idk how to define the rule that maps the pairs in G to even permutations in S4

or the rule that maps the pairs in G to the rotations and reflections in D6

[rocky cloak]

Notice that if sigma1 is a 3-cycle and sigma2 is a 2-cycle, then (sigma1, sigma2) has order 6.

Is there such an element in G? If so that's a good candidate for where to map the rotation in D6

[rocky cloak]

There's something weird here though. Because G doesn't appear to be a subgroup.

Like if s1 is the 3-cycle (1, 2, 3) and s2 is the 2-cycle (1, 2).

Then (s1, s2)^2 = ((1, 3, 2), ()) which is not in G.

[pastel condor]

wait what? does x(i) denote the ith position in x? but isn't this not well-defined since (123) = (231) = (312)

[rocky cloak]

x is a function maping {1, 2, 3, 4, 5} to itself.

Then you specify the function by saying where everything maps.

In this case x = 31524 means
x(1) = 3, x(2) = 1, x(3) = 5, x(4) = 2, x(5) = 4

This your text calls "complete notation"

For example the identity would be specified as 12345

[pastel condor]

oh

thank you for the clarification

ive never seen that before

i guess the other notation is usually denoted by brackets

like (123)

[restive birch]

problem was "let G be a group whose order is the product of three distinct primes--show that G has a normal sylow p-subgroup for some prime p dividing its order"

i went for contradiction and showed it would require too many elements--does this work?

[wintry sluice]

for group actions is there any kind of restriction on the set and the group?

like does it matter what kind of set and what kind of group? how do you know what the binary operator does to the elements?

how would a permutation group act on a set of matricies?

[chilly ocean]

Can I say if V is finite dimensional vector space then V is a F-module of finite height?

[void cosmos]

what does finite height mean

[void cosmos]

haven't read the details but probably

overcounting arguments are common with these problems

if not then you must have n_p n_q n_r not != 1

each dividing their orders so u would have (q-1)*p + ..

yeah probably what u did is okay

[restive birch]

yea so n_r = pq

[void cosmos]

yes

[restive birch]

so there are pqr - pq elements of order r

[void cosmos]

of order r?

[restive birch]

yes?

[void cosmos]

numbr of subgroups of order r is pq

[restive birch]

right each one has r - 1 elements of order r

since r prime

[void cosmos]

then the number of order r elements would be pq(r-1) right?

because each has r-1 elements

[restive birch]

yes...? thats pqr - pq

[void cosmos]

yeah my bad

i thought this would be the expression of number of elements not order r

sorry just came back from the gym ;D

continue

[restive birch]

n_q >= r > p so there are more than p(q - 1) = pq - p elements of order q
and n_p >= 3, so there are at least 3(p - 1) > p elements of order p

terms cancel out and its >pqr

[void cosmos]

yeah

good 😄

works

[restive birch]

what?

[void cosmos]

btw you know exactly what n_q is

or no u don't

yeah mb whatever

forget me

[restive birch]

its r or pr

[void cosmos]

ur proof is correct

[void cosmos]

yeah thats what i meant

[serene dune]

n_q \neq 1 for the hypothesis of contradiction right

[restive birch]

yes

[serene dune]

oh yeah solid proof

btw i was thinking can we generalise it for p1, p2,.....pk

[void cosmos]

i mean probably not cuz literally every finite group has order of product of primes

Z_n is not simple for n not prime so u can probably find counterexamples quick

[serene dune]

15

we have to prove that it is not a simple group no ?

[void cosmos]

im off to sleep so idk

also 15 =3*5 it is infact cyclic

[serene dune]

...

[rocky cloak]

It is at least true that no group of square free order is simple, whether or not this proof generalizes

[serene dune]

what is square free

[void cosmos]

its isomorphic to Z_15

[rocky cloak]

Not divisible by a square

So 15 is square free, while 45 is not for example

[serene dune]

15 was a mistake, i realised that we have to quite prove that it has a non trivial normal subgroup

ok so the power of the distinct primes are 1

[spark veldt]

Hi im not sure how to proceed after this..

Like i know phi(g) can belong in overbarN, but so can all g that is a membrr of ker phi. Question is, why don’t we just use ker psi = ker phi and instead we have to choose ker phi = N?

[tardy hedge]

7.2.8, i dont really understand how we can say Z(G) is divisible by p. I understand why each [G : C(x)] is divisible by p

[rocky cloak]

The class equation says that
|G| = |Z(G)| + sum [G:C(x)]
where x ranges over representatives of conjugacy classes not in the center.

In other words
|Z(G)| = |G| - sum [G:C(x)] is a multiple of p

[tardy hedge]

Ty

[south rain]

Let $G$ be a finite abelian group and let $t$ be a divisor of $|G| = r$. We prove that $G$ has a subgroup of order $t$. I used the structure theorem of finite abelian groups but with invariants, where $n_i | n_{i+1}$ and we then have that
$G \cong \mathbb{Z}{n_1} \times ... \times \mathbb{Z}{n_r}$. My proof is the following:

Let $G$ be of order $r = |G|$ and let $t$ divide $r$. By the theorem of finite abelian groups we have that $$ G \cong \mathbb{Z}{n_1} \times ... \times \mathbb{Z}{n_r} $$ We can now take the subgroup $$\mathbb{Z}{n_1} \times ... \times \mathbb{Z}{n_t} $$ which works since $t$ divides $r$ and thus we have a subgroup of rank $t$.

This seems a bit short, but is this correct?

[cloud walrus]

dellinger

[tardy hedge]

Whats the situation when a ring hom doesnt map prime to prime? Is it just when the map isnt surjective?

[wraith cargo]

Everything maps to 0 and the codomain isn't an integral domain

That's sort of a trivial example tho lol

[rocky cloak]

Well surjective maps don't have to map primes to primes either.

Like Z -> Z/4, then (0) and (3) are both prime ideals in Z, but don't map to prime ideals in Z/4.

If the map isn't surjective then the image of an ideal doesn't even have to be an ideal

[fossil shore]

Hey, if we have some irreducible cubic x^3 + ax^2 + ... and then substitute x = y - a/3, must the resulting polynomial of y then also be irreducible?

[rocky cloak]

Yes. You could always substitute back by y = x + a/3

[tardy hedge]

Ohh god ugh

I always mess these facts up

[tardy hedge]

Surjective maps map prime ideals bigger than ker(f) to prime ideals right

[rocky cloak]

Yes, there is a nice correspondence theorem there

[tardy hedge]

I think thats why i forget / get confused on that

Cause im remembering that one

[fossil shore]

Oh, my bad

I mixed up two different g's

Still stuck though. Could someone help me figure out what the galois group of x^4 +2x +2 is? I believe it reduces to figuring out whether x^3 - 8x -4 is irreducible over Q which I don't know how to approach

[rocky cloak]

You can use the rational root theorem if you want to figure out if x^3 - 8x - 4 is irreducible

[fossil shore]

What's that?

[rocky cloak]

https://en.m.wikipedia.org/wiki/Rational_root_theorem

[fossil shore]

Awesome, thank you!

[wild jasper]

Just got the top grade in my Galois theory exam 🙂🙂🙂🙂

[vapid dew]

cold, how many people are in your class?

[wild jasper]

39 I think

[vapid dew]

damn I thought postgrad classes were small

gj though im still getting cooked by intro group theory

[wild jasper]

Well if you have a qualifying bsc in math, you are legally entitled to admission to the master degree program…

[wild jasper]

And it’s a common course to start with, so there is still a bit of culling to do

[tidal torrent]

just to clarify but

x^4+1 is irreducible over Q and reducible over R correct

[coral spindle]

If a polynomial is reducible over Q then it must be reducible over R, since in particular a factorisation in Q would also be a factorisation in R.

So no.

[tidal torrent]

oh ok

[coral spindle]

OK you edited your message.

[tidal torrent]

i think i mixed up the two idk

how is it that x^4+1 over Q has a dimension of 4

but x^4+1 over R has a dimension of 2

dont know if its just me

[coral spindle]

That would be the case iff x^4 + 1 is irreducible over Q.

[coral spindle]

This would certainly imply that x^4+1 is reducible over R

[rocky cloak]

x^4 + 1 = (x^2 + sqrt(2)x + 1)(x^2 - sqrt(2)x + 1)

[tidal torrent]

does there exist a splitting field over R? not sure how to find it

i found the splitting field over Q

[rocky cloak]

Splitting fields always exist. Over R it's the complex numbers

[tidal torrent]

Oh ok

[tardy hedge]

I dont know why this is confusing me

Maybe something about divisors?

[rocky cloak]

Are you still confused about it?

[tardy hedge]

Yeah, im not sure exactly why though

Each [G : C(x)] divides p^n, so it is a multiple of p. You are summing up them over a representstive for each nontrivial conjugacy class, so now u have a sum of things that are a multiple of p

[rocky cloak]

Yes, the sum of multiplies of p is again a multiple of p

[tardy hedge]

Ok and then the last thing i was thinking was p^n minus all that surely is not going to be negative

So you are left with a positive number i guess so |Z(G)| > 1?

[rocky cloak]

Well, you know that |Z(G)| isn't negative

It must be at least 1 since the identity is in there

[tardy hedge]

Yeah, right

[rocky cloak]

So it's at least 1 and it's a multiple of p, then it must be at least p

[tardy hedge]

Yes thanks i understand it a bit better now but now im trying to understand what breaks down when u dont have the order of group be like p^n

[rocky cloak]

Well, then it's not for sure that [G:C(x)] is a multiple of p

[tardy hedge]

Hmm ok

Im not really sure about it

The benefit that we knew the index was a multiple of p was because it guaranteed |Z(G)| was also a multiple of p, so it then has to be >1

Otherwise we could have a situation where |G| - sum of index = 1 i guess ?

[rocky cloak]

Yeah, take for example G = <s, r> to be the symmetry group of a triangle, with r a rotation and s a reflection.

Then |G| = 6,
|Z(G)| = 1
[G:C(r)] = 2
[G:C(s)] = 3

[tardy hedge]

Thank you. I will work this out

Appreciate the help

If a and b are in the same conjugacy class, is it true that C(a)=C(b)?

Maybe not right ?

But [G : C(a)] and [G : C(b)] have same number of elements

[next obsidian]

Those are numbers

[tardy hedge]

Same size*

Yes

[next obsidian]

They aren’t sets

[coral spindle]

Try thinking about this for e.g. the symmetric group

Like choose a = (123)(45) in S_5

Pick a centralising element, and look at some interesting conjugate b

try seeing what you get

[coral spindle]

N.b. this also means |C(a)| = |C(b)|.

[tardy hedge]

Oh yeah

[tardy hedge]

Ifk

Idk

A centralising element

[coral spindle]

An element that centralises a is an element of C(a).

So pick an element of C(a), is what I am saying. You will find out, if you choose it well, that it won't commute with a good choice of conjugate. Spoiler ig.

[vital birch]

One thing that tripped me up while working problems is that |C(a)| is not a number of conjugacy classes even though the index gives you the size of a particular conjugacy class. I don't think it is a directly computable value because the class equation admits multiple possibilities.

[tardy hedge]

Womt commute with a good choice of conjugate?

Im having trouble understanding that

[tardy hedge]

Ok so (45) is in C(a)

Oh ok so now we’ll pick a conjugate of a , so we are picking something else in the same conj class

[vital birch]

C(a) and cl(a) are associated by the orbit-stabilizer theorem. C(a) stabilizes a (same as centralizes a) whereas cl(a) is the orbit under conjugation so they are different things but very connected. I've confused them many times

[spark veldt]

hiya

i'm reading this proof on how every infinite group has nontrivial subgroup, and

i can't really comprehend the meaning of the last line.

how does x not in <x^2>, x^2 not e, mean that <x^2> is a proper subgroup of G?

[coral spindle]

Well what is a proper subgroup of G? It is a subgroup that is not {e} and is not G.

So why is <x^2> proper?

It contains x^2, so it cannot be {e}.

It does not contain x, so it cannot be G.

So it is proper.

[spark veldt]

ohh

got it thanks!!

eh wait

does <x^2> not give rise to the possibility that it might contain exactly the same elements as <x>

[coral spindle]

I think you need to read the proof once again, because that case has been dealt with.

The case you are thinking of is when x is in <x^2>.

[tardy hedge]

What is cl(a) ???

Like the conjugacy class of a?

[vital birch]

Yes, that's the notation some books use for conjugacy class

[south patrol]

What does the l even mean

[coral spindle]

Cl(ass) :3

[tardy hedge]

What about when G is abelian so Z(G) = G?

It looks like both cases are choosing an x not in Z(G)

[rocky cloak]

I mean, it's covered by the case "thus we may assume Z(G) = G"

So that would be case 1

[vagrant zinc]

Could someone give me a hint for this exercise? (4)

[coral spindle]

Hint: a homomorphism out of Z is determined entirely by where it sends 1.

[untold cloud]

terminal elements in the category Rings

[untold cloud]

that is true, G abelian iff G is equal to its center. If G finite abeliam, then G is a product of cyclic groups of prime power, so there is an element of order p

[coral spindle]

I think you mean initial element!

[tardy hedge]

yea the proof there tho is not assuming structure thm for finite abelian group

they prove that later

[untold cloud]

terminal includes initial right?

and also finial?

[cobalt vessel]

wait can you elaborate on this point, I still dont understand why is this the case

is this just due to the fact that <1> generates Z?

[coral spindle]

Yes

[cobalt vessel]

or is this bc 1 is the unity

i have my algebra midterm tmrw and im soooo cooked

[coral spindle]

No, the terminal object in the category of rings is the zero ring, which is certainly not Z

[cobalt vessel]

am i allowed to send questions from the practice midterm here so you know the content thats being covered

[coral spindle]

No, because other rings do not have this property

[tardy hedge]

is the initial object in rings Z because every ring is a Z-algeba or smth

[coral spindle]

I don't see why not

[coral spindle]

Sure, you could see the fact that rings are Z-algebras as a consequence of this. You'd need first to argue that the image is central ig, but it's not too hard to do so

[cobalt vessel]

ive been working on 5d for a minute

and im quite lost 😩

[untold cloud]

zero is not a ring. It is a rng

[tardy hedge]

lul

[coral spindle]

No, it is a ring. It has a unit, it just happens to also be the additive identity.

Some people define rings to be those in which 1 is necessarily not equal to 0, and this is fine, but in that case there is no terminal object of Ring.

[coral spindle]

Well this isn't a totally easy question. Talk me through what you get for a, b, c and tell me what you notice. E.g. if we set m = kn for some k what would happen

[untold cloud]

oh i see

yeah initial

[untold cloud]

i think case 1 includes the case when G is abelian

maybe refer to theorem 7.2.8

[fossil shore]

Could someone tell me where I could find a reference for this?

Here D is the discriminant of a quadratic polynomial over some field F

Oh. Wait... is this just the normal quadratic formula for a quadratic polynomial...

[void cosmos]

is the idea that

assuming you have a subnormal series

if its not composition , then H_i+1/H_i is not simple --> normal subgroup corresponds to a normal subgroup containg H_i, call it H' , insert it in the subnormal series?

1. gives us that this process will terminate the refinments

and 2) gives us that you can't insert infinitely many maximal normal subgroups? (ie that this "algorithm") terminates?

is this the idea?

or basically 2 gives us that like there aren't infinitely many choices of normal subgroups to choose from

basically 1) is 2) but for G in a way

would llike ur opinions

[grim hawk]

Anyone have any resources or advice for understanding character tables and how to construct them for arbitrary symmetry groups?

[wild jasper]

@rocky cloak thanks for your help and conversation this weekend, my exam yesterday went rather well 🙂

[void cosmos]

1. says that right end of the series is finite and 2) says the left end of the series is finite

[nimble folio]

Suppose $T \in \text{Hom}_F(V,W)$ and $S \in \text{Hom}_F(W,V)$. If I want to show that $S$ and $T$ are inverses, does it suffice to show that their composition is the identity on their respective basis elements?

[cloud walrus]

clubsoda14

[void cosmos]

to show that S and T are inverses you show that their compositions (left and right) are both the identity

[nimble folio]

Yes

[void cosmos]

to show that a map is the identity its enough to show that it acts by doing nothing on the basis elements

[nimble folio]

Okay thank you 👍

[void cosmos]

yw

[hot pebble]

Just a sanity check: Is it true that if G is finite, the Sylow p-subgroup (under the "maximal p-subgroup" definition) must be of order p^n where n is max s.t. p^n divide |G|, and the proof is by first sylow theorem?

[mighty kiln]

Yes

[hot pebble]

then if P is a Sylow p-subgroup with |P|=p^n, is it also true that the maximal m s.t. p^m divides |G| must be n?

[mighty kiln]

Yes

[hot pebble]

ok thank you

[inner harbor]

guys help plw

let G be a group and H a normal subgroup (i believe its called in eng), of index n, then for all g in G, g^n is in H

its works for cyclic groups but i think if i try with some weird one its wont

and need help cause im terrible at group theory

[serene dune]

take the quotient

or see the cosets of H which forms a subgroup of G

[south patrol]

It's worth pointing out that you need something like Lagrange's theorem to do this, because this includes the case where H = {1}

(And taking the quotient reduces you to that case)

[inner harbor]

so basically its true cause g^n = 1 in G/H

i dont rly get the def of index, i have to work on it tks guys

[serene dune]

lagrange applied on the quotient

[serene dune]

oh where is the problem?

try to see the set and the cardinality

[inner harbor]

never seen the def lmao, i guess its like the order of the subgroup

but no, its related to equivalence class and kinda missed the exercices to work on elliptic curves 💀

ty, ill be back with other questions guys

am back

they ask me to find a criteria over G so that every two elements that are conjugate have the same order

dont rly know, cyclic doesnt even seems to work with Z/3Z for eg

G is abelian btw

btw without this condition is it harder hmmm

so many questions

[serene dune]

[somber sleet]

Short question, if I have a composite series $0 = M_0 \subsetneq \ldots \subsetneq M_n =M$, why does it have to hold that $\frac{M_{i+1}}{M_i}$ is simple?

[cloud walrus]

damn_guuurl

[somber sleet]

Does it directly follow because in case it was not, then there would be some submodules between M_{i+1} and M_i ?

[rocky cloak]

I mean, that's just the definition of composition series.

But yeah if Mi+1/Mi wasn't simple, then you could find some submodule between Mi and Mi+1

[somber sleet]

We defined it differently, hence why I wasn't sure, thank you❤️‍🔥

[tacit ginkgo]

I have that a 5-cycle (a,b,c,d,e) can be written as (a,b)(b,c)(c,d)(d,e)(e,a) but also as (a,e)(a,d)(a,c)(a,b)

I am working out the sign of it, the definition i have is
Definition 20. The sign of a permutation σ ∈ Sn is the value sgn(σ) :=(−1)^m where σ can be written as a product of m transpositions.

the cycle can be written as a product of 5 transpositions but also 4 tranpositions? How do i work out the sign

I know i use 4 but why do i use 4

[rocky cloak]

The sign of a permutation is well defined, so it should be impossible to write something both as the product of 5 and as a product of 4 transpositions.

So there must be a mistake in your computations.

If you try to verify where for example a is mapped by your two products you'll see that only one of them is correct.

[tacit ginkgo]

ah shit mb thanks

[tacit ginkgo]

Could the correct one be written in other ways with more transpositions?

what rules would they follow ?

[rocky cloak]

Yes for example you can always add something like (a, b)(a, b) to any product to make it longer by two.

Other than that there aren't really that many rules other than that the number of transpositions would always have to be even.

It should also be clear that you can't do it with fewer than 4 transpositions

[tacit ginkgo]

ok thank you

[tardy hedge]

so a multiple of x minus multiple of x cannot be 1? oh shit, it can only be one if x and x are coprime right? lol!

Yes I know I am still thinking about this proof lmao

I swear its the basic number theory that trips me up. Im so bad at it

[rocky cloak]

A multiple of x minus another multiple of x is still a multiple of x
ax - bx = (a-b)x

1 is not a multiple of x for x>1

[tardy hedge]

and for our purposes here it cant be true that a=b because Z(G) always at least contains one element

[rocky cloak]

Yes thats true

The center of a group cannot have 0 elements

[tardy hedge]

Thanks im finally slowly understanding it better

Btw, all this stuff is important for galois theory right?

Im taking galois next term but I have not learned this group theory stuff, like sylow theorems

im trying to get to that stuff so im familiar with it

[rocky cloak]

I guess it depends what you mean.

The main idea of Galois theory is to translate problems about field extensions to problems about groups.

So if you want to compute things about specific field extensions you usually need to know quite a bit about finite groups.

But just the general machinery of Galois theory doesn't really need much beyond the isomorphism theorems.

For the general machinery you want to know more about rings and fields

[vital birch]

I am having difficulty trying to "see" an example of what this problem is talking about.
Dummit Foote Sec4.3 #19
Let K be a conjugacy class in Sn and assume that K is in An. Deduce that K consists of two conjugacy classes in An if and only if the cycle type of an element of K consists of distinct odd integers.

These are the conjugacy classes of A7 per GAP
I see the 7-cycle has two conjugacy classes but why not the 3-cycles or 5-cycles, which are also distinct odd cycle lengths?

[coral spindle]

Suppose that we have some permutation k in A_n and suppose that this cycle type has two conjugacy classes in A_n

The other conjugacy class can be obtained by conjugating by an involution in S_n. You can see this in (1,2,3,4,5,6,7) and (1,2,3,4,5,7,6)

However, this may not be the only thing that allows us to get here. There may be, as it were, 'enough space' in A_n to reach this.

For example, if the cycle type (1,2,3) split into two conjugacy classes in A_n, then the other class would be represented by (1, 3, 2) by conjugating by (2, 3) for example. But notice that conjugating by (2, 3)(4,5) is possible in A_n.

As for why (1,2,3) fits in this particular regime that D&F are saying, it's because this is of cycle type 3, 1, 1, 1, 1.

[vital birch]

I see now how (1,2,3) is in one conjugacy class. There's enough room in An to tack on the (4,5) for an even permutation and this does not change the conjugation by (2,3).

But you pointing out all the ones... those are not distinct

I see the cycle type is not simply 3. Its 3,1,1,1,1

Thank you very much. That was driving me mad since last night

[mystic ether]

doing this problem from Lang's algebra. I'm confused about part (c): if the Galois group is S_3, how could the dimension be 2? Isn't the degree of a Galois extension equal to the size of the Galois group? So the dimension over k should have to be 6, no?

is the vector space generated by the roots somehow different than the splitting field?

[polar stone]

Hey, someone here have studied from Alufi's Algebre Cahpter 0?

[coral spindle]

Yes, in particular you will not have products of roots for example.

E.g. take the splitting field of x^2+1 over Q, which is Q(i). Note that this is a Q-vector space. What is the space spanned by (i.e., generated by) the roots of x^2+1? It is {k i | k in R}, and it is 1d.

[mystic ether]

ah true

thanks

[flat tree]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[tardy hedge]

For an R-module M, and an ideal I of R, there is not some canonical way to define an R/I-module structure on M, is there?

You could say the action of R/I on M to be from how R acts on M, but then well-definedness can be an issue right?

Is an R/I-module defined in that way well-defined iff I is a subset of the annihilator of M?

[coral spindle]

Let's take an example. Say R = Z and M = Z. Let's say I = 2Z. Is there any R/I-module structure on M?

[coral spindle]

If you want to define it as (r+I).m = r.m, then yes

[tardy hedge]

yea thats what i was thinking

[tardy hedge]

weird

[coral spindle]

Weird?

[tardy hedge]

Yes

[coral spindle]

Okay

[tardy hedge]

Ok so ill just think out loud then

I realized R/I is a field and so whatever R/I-module would be like a vector space

then i kind of paused and wasnt sure where to go

[coral spindle]

Well what do R/I-modules look like

R/I is Z/2Z

[tardy hedge]

i know it either kills everything or does nothing

is what im thinking

[coral spindle]

?

I don't know what that means

[tardy hedge]

Hahahaha ok sorry wait

[coral spindle]

What do vector spaces over Z/2Z look like

[tardy hedge]

Ok am I wrong? Its either 0m = 0 and 1m = 0 for all m, or 0m = 0 and 1m = m, right?

[tardy hedge]

i havent thought about vector spaces over finite fields its new to me

[coral spindle]

They behave exactly the same as vector spaces over infinite fields in terms of structure.

[coral spindle]

So what is m+m

[tardy hedge]

0

[coral spindle]

Wait what, 1m = 0 for all m? The identity sends things that are nonzero to 0? Whatever, let's move on for now.

[coral spindle]

So what is the order of any element of a Z/2Z-module?

[tardy hedge]

we normally cant talk about order of an element in a module, can we? I think of order as like a group thing

but its 2

[coral spindle]

Well all modules are, in particular, Abelian groups under addition

Are they not

[coral spindle]

But moving on, you can answer this now.

[tardy hedge]

wow the fact that I forgot this. I get too caught up on thinking about the ring multiplication that I forgot M itself is just an abelian group

[tardy hedge]

i mean like

yes?

is the m+m=0 thing a problem?

[coral spindle]

Can you name some element of M for me, like pick one other than zero

[tardy hedge]

3

[coral spindle]

Great

So let's suppose M is a Z/2Z-module under some choice of structure

The Abelian group structure must still be the same

So 3 + 3 = 0

I think the problem here is clear.

[tardy hedge]

Yeah what the heck.

Ik i just havent thought about it like this before

ik it seems obvious to you

[coral spindle]

We supposed the Abelian group underlying M had a Z/2Z-module structure, and we got a contradiction.

[tardy hedge]

Yeah what

this is interesting to me somehow

I think its because I somehow completely forgot the abelian group was there when thinking about modules

or more like i wasnt considering it in context properly

wait so the only Z/2Z-modules you could have are those where every element of the abelian group has order 2?

[coral spindle]

The Z/2Z-modules are precisely the Abelian groups in which every element is 2-torsion, yes

And, while we're at it, Z-modules are just Abelian groups.

There is no additional information that the Z-module structure endows the group with

[tardy hedge]

Yeah Ive known that but maybe dont consider it deeply enough

[tardy hedge]

ive gotten thrown off by this before, but why exactly ? WHy cant there be a different action of Z on M ?

oh because we specify that 1m = m?

[coral spindle]

So 1m = m

So what's 2m

[tardy hedge]

yeah its induced by 1 cause 1 generates Z

2m is just m+m in the group

[coral spindle]

2m = (1+1)m = 1m + 1m = m + m, yes

Etc.

[tardy hedge]

I see

[tardy hedge]

so, ring homomorphisms from Z/2Z -> End(M) do not exist if the elements of M don't all have order 2?

[coral spindle]

This is what we just showed, yes

[tardy hedge]

like i havent internalized how R-modules are just ring homomorphisms or like a "representation" of the ring

Is all the facts about the R-module somehow seen / can be described from that ring hom?

[coral spindle]

Yes, the scalar multiplication is described entirely by the ring hom R -> End(M)

In the same way that a group action of a group G on a set X is described entirely by the group hom G -> Sym(X)

[tardy hedge]

Thanks, i think ive thought more deeply about modules more now than the past whole semester

I still have a lot of thinking to do but im tired now

[tardy hedge]

Are there applications to this in further algebra or other contexts?

Or is this just a neat description kinda like cayleys theorem from group theory

[coral spindle]

It's not the most helpful description but it's there.

[tardy hedge]

I see

How is a free module illustrated in the ring hom?

Ok so first off the map is injective in this case

[coral spindle]

You won't get anything more illuminating than the general case I'm afraid

[tardy hedge]

So the fact that the module has a basis you cant say anything about the ring hom ?

[coral spindle]

You won't get any special formula

[tardy hedge]

Can we even interpret the module having a basis as something about the ring hom ?

Like i dont even know how to think about that

[coral spindle]

Unless you look at End_R(M) and not simply End(M) -- i.e. End_Z(M) -- you won't get a nice structure in general.

Like what is End_Z(R)

R being the real numbers

It's an absolute nightmare

[tardy hedge]

the problem here is only present because somehow the ring action says something about orders of elements in the group?

but this is only the case for Z/mZ modules isnt it?

[coral spindle]

Other rings will enforce other conditions

But yes, Z/mZ-modules are precisely those Abelian groups whose elements' orders divide m

[tardy hedge]

the Z-ness says something about order cause its like counting

i sound like a baby but ye

[tardy hedge]

will enforce other conditions on how the elements of the abelian group relate to each other, but not purely in a group theoretic way is it?

[coral spindle]

Probably not in general

[polar stone]

Hey, I'm working on Alufi's Algebra Chapter 0 and I have this problem that I don't know how to attack:
Prove that Q is not the direct product of two nontrivial groups.
Some help?

[coral spindle]

Well, there is an important property of the product of two groups.

G and H embed in G x H in such a way that

1. The two embedded subgroups have trivial intersection
2. The elements of the two subgroups commute
3. Together, they generate the entire group G x H

So if you can show that this never happens in Q, you're done.

You only need to prove that one of these fails for any pair of nontrivial subgroups, and you're done.

[next obsidian]

Look at where 1 maps to under an isomorphism and then write this as (g,h). If h is nonzero, then every element of G x H has the second component nonzero, because the element corresponding to p/q (for p ≠ 0) call it (g’,h’) satisfies q(g’,h’) = p(g,h) forcing h’ ≠ 0. This is obviously false. So h = 0, and similarly g = 0. This is a contradiction

I guess I am assuming that G and H are both torsion-free, but that’s obviously true cuz they embed in Q

[vital birch]

Is that at all related to this exercise? So if it were a product, it would have to be an infinite product?

[next obsidian]

I don’t see how that’s really related

I did kinda use divisibility of Q

But I do t see how that exercise is related

[vital birch]

ok thanks

[polar stone]

Thanks. I'll try to do the proof

[toxic zephyr]

so {e} is a fg abelian group. what's the structure theorem decomp for it? just Z_1?

also, I forget. is the multiplicative group of (units in) Zp always cyclic for p prime?

[serene dune]

yeah, from the finite field argument

[mighty kiln]

Yes

[hollow topaz]

is it sufficient to show that its associative for three operands i.e. x(yz) = (xy)z? also i dont know how to approach the second question

[coral spindle]

You have said "commutative" but written associativity!

But yes, the question asks you to prove associativity, so you should prove associativity.

Hint for second part: an identity e must satisfy eb = b for all a in S.

[hollow topaz]

oops sorry hehe

[hollow topaz]

is it every set of cardinality 1?

[coral spindle]

Yes

[tardy hedge]

Thanks for your help yesterday Boytjie it was great

[coral spindle]

No worries

[hollow topaz]

thanks from me as well

[tardy hedge]

Yeah ill always be grateful for how much knowledge ive acquired from help from this server

I hope one day I can repay the favour lol

[barren sierra]

I may be stupid why are there only 3 equations

I'm looking at a 4th one in my own work: B^T * C - D^T * A = -I

just from expanding out that condition

[rocky cloak]

This is just the transpose of the second equation (multiplied by -1)

[barren sierra]

Ah

ty

[tardy hedge]

They pulled out linear algebra out of nowhere and expect me to follow 😭

Module homomorphisms out of finitely generated modules can be represented by a matrix

?

In the same way just write where it sends the generators in the columns of the matrix?

So i am ok with believing that the double index sum there can be written in matrix form

[wraith cargo]

this is not what they're saying

they're saying that to each endomorphism we can associate a matrix in some ring like the matrix ring with elements in A adjoin the endomorphism

[vestal bobcat]

Can anyone give example of an module over an integral domain which is projective but not free

[tardy hedge]

yea i wasnt thinking they were saying that, i was just asking in general is this true?

[rocky cloak]

Over R = Z[sqrt(-5)], the ideal (2, 1-sqrt(-5)) is projective. In fact every ideal is projective.

[wraith cargo]

if the module is free then yeah
Otherwise not really

[tardy hedge]

finitely generated but not free?

[wraith cargo]

the issue arises from the fact that the decomposition of an element into linear combinations of generators doesn't have to be given uniquely

[rocky cloak]

I mean you can write out a matrix by having the columns represent where the generators are mapped, and matrix multiplication will correspond to composition.

But the choice of matrix is not necessarily unique, in particular the 0-map can be represented by something that isn't the 0 matrix.

[vestal bobcat]

It isn't obvious to me why?Can you explain

[wraith cargo]

an easier example might be Z^N as a Z-module where Z^N is the countably infinite direct product of Z's

[vestal bobcat]

But isn't over a PID every projective module is free

[rocky cloak]

Z^N is not projective

[wraith cargo]

hm okay that is true lol

I messed up

thought that was a good example lol

[rocky cloak]

So the easiest way to see it is probably to realize it as a direct summand of a free module. Calling the ideal J You have the obvious surjection
Z^2 -> J
sending (1,0) to 2 and (0,1) to 1-sqrt(-5).

You get a splitting by mapping 2 to (4, -(1+sqrt(-5)) and (1-sqrt(-5)) to (2(1-sqrt(-5)), -3)

This is an example of more general theorem, namely that a commutative Noetherian ring is hereditary (every ideal projective) iff it's a Dedekind domain

[vestal bobcat]

Thank you for clearing that up

[vestal bobcat]

I have just taken a first course in Commutative and have read until Ch 5 Atiyah MacDonald....and even that selectively....so am not aware of many theorems

[flat wharf]

What is the power group $E_2^{S_m}$ ?

[cloud walrus]

jonathan_fisherman

[toxic zephyr]

I'm not used to defining rings as not needing identity, but if you do, then an ideal is a subring (just possibly without identity)?

[rocky cloak]

That's right, but not all subrings are ideals

[serene dune]

do u mean unity ?

[toxic zephyr]

okay cool. i think I may still prefer ideals are a subrng lol

[serene dune]

the only distinction with a unity is
now the whole ring can be generated by an ideal, ig

[opaque finch]

Let $\mathbb{Z}(p^{\infty})$ be a the group ${z \in \mathbb{C} | z^{p^k} = 1, \text{for some} k \geq 0}$. Show that the only proper subgroups of the group is $T_n = {z \in \mathbb{Z}(p^{\infty}) | z^{p^n} = 1}$ and also $\mathbb{Z}(p^{\infty})/T_n$ is isomorphic to $\mathbb{Z}(p^{\infty})$ for each integer $n \geq 0$.

Can someone help me solve this problem ?

[cloud walrus]

mycroftholmes1703

[opaque finch]

Please add the set brackets, so for the inconvinience

[toxic zephyr]

can someone clarify the structure theorem decomp of finite abelian groups? there's the decomp in terms of prime powers (is that the primary decomp?), and then there's the other one where q1|q2|...|qk (what is that called? invariant factors?). the latter decomp is slightly more helpful because it excludes certain possibilities.

ex. if |G|=6 then we cannot have Z2xZ3 (since 2 does not divide 3, thus it must be cyclic and Z6). similarly if |G|=10, then we cant have Z2xZ5 so it must be Z10? but if |G|=12, then we cant be so sure. it could be Z12, but also Z2xZ6 (but not Z3xZ4 which would be Z12)? am i understanding this right?

if i was using the prime one, then is it possible for it to still be obvious some of these options are not possible? or do i just have to be more careful and the prime decomp is nice in its own way because it separates by the prime divisors of |G|?

[wraith cargo]

Both decomposition are equivalent

[toxic zephyr]

uhhhh... but Z6 itself isn't a primary decomp of itself tho

like i get theyre the same group

but the decomps are different

[rocky cloak]

For matrices it's called rational canonical form, but I'm not sure if it has a name for integers.

Both decompositions are uniquely determined though.

For a group of order 12 the options for primary decomposition are
Z3xZ4 or Z3xZ2xZ2

[serene dune]

check chinese remainder

[toxic zephyr]

right right. i forgot Z2xZ2xZ3. but that one is isomorphic to Z2xZ6, which would be the invariant factor decomp? and Z3xZ4 is iso Z12

[rocky cloak]

Yes that's correct

[toxic zephyr]

okay okay i think im getting it. so primary to invariant for |G| has only two options

• Z2xZ2xZ3 iso Z2xZ6
• Z4xZ3 iso Z12
  every option of the primary decomp technically exists/works (since it itself is an example of a group with that primary decomp). but for a generic finite abelian group, the invariant decomp is slightly more helpful because it excludes more possibilities?

[wraith cargo]

The invariant decomp doesn't exclude any possibilities I'm confused what you mean by that?

[toxic zephyr]

if G abelian has, say, 15 elements, then we have two possibilities

• Z15
• Z3xZ5
  but if we restrict ourselves to the invariant factor decomp, the last is not a valid structure theorem decomp (3 does not divide 5). so we find that it must be Z15. then we can deduce that any group of 15 elements is cyclic (this is true because Z3xZ5 is cyclic with generator (1,1))
  thats my thinking anyway. not sure if im 100% right

[serene dune]

every finite abelian group is a direct product of its Sylow p-groups

[rocky cloak]

But if you use the primary composition then only the second is valid

[toxic zephyr]

yeah exactly

[rocky cloak]

In both cases the choice is unique

[toxic zephyr]

was just about to say haha. glad i was right in thinking that

[wraith cargo]

Is this a good way to think about it tho
We're looking at these up to isomorphism anyways so why the need to separate these out if they're isomorphic?

[serene dune]

wouldnt gcd(m,n) be a better thing to say

[rocky cloak]

I mean it's just about choosing a canonical representative for each isomorphism class.

Both methods work, I don't see a general reason to prefer one over the other

[toxic zephyr]

this problem im doing is about the "possibilities" of structure theorem decomps for certain groups.

i suppose maybe the primary is better for that... not sure.

[rocky cloak]

They're the same. They exclude exactly as many possibilities

In both cases you have exactly one representative for each isomorphism class of groups

The only thing that would matter is if it's easier for you to write a number as a product of primes or as a product of numbers that divide eachother

[opaque finch]

any hints for this 😅

[tough raven]

You can prove it in steps like this: (i) {1} = T_0 ⊆ T_1 ⊆ ... ⊆ ℤ(p^∞) with the union of all T_n's being ℤ(p^∞). In particular, any x in ℤ(p^∞) lies in T_n for some n. (ii) For any x, if we pick the smallest n = n(x) such that x in T_n then the subgroup generated by x is T_n. (iii) Given any set of generators S for a subgroup H of ℤ(p^∞), if the respective n(x)'s are bounded above (i.e., {n(x) : x in S} is bounded above), then H = T_n0, where n0 is the maximum n(x) for x in S. If the n(x)'s are not bounded above then H ⊇ T_n for all n and so H = ℤ(p^∞).

[toxic zephyr]

one more trivial question: the trivial group has no elements of order 2 right? i'm given this class of groups (multiplicative group Zp) and one has only 1 element (Z2), and the problem says "find an element of order two".
my answer is that it's impossible for Z2 but p-1+pZ works for p>2. just want a sanity check

[rocky cloak]

You are still sane, not to worry

[toxic zephyr]

thank you haha 😅 ❤️
i really appreciate your continuing help, jagr

[tawdry venture]

For subrings: Does the identity element have to be in the subset for it to be unity?

[coral spindle]

That's what it means to be a unital subring

Whether or not a subring has to contain the identity unfortunately depends on author. You should check whatever course notes or textbook you are using for the relevant definition.

[dull ginkgo]

A lot of ring results and statements implicitly assume unity, and thus the unit is preserved.

Just depends

[cloud walrus]

gabi the ancient

[icy bear]

oh no this will stay here noo

[tardy hedge]

[toxic zephyr]

suppose we're taking Z_n under multiplication (which is not a group). is it wrong to still write <a+nZ> to represent {(a+nZ)^r : r in Z+}={a^r+nZ : r in Z+} as the (possibly not cyclic) set of all powers of the "coset(?)"?

[vestal bobcat]

Kind of like this.....let H be a proper subgroup of G(I am calling that group G)....
Now assume p^t to be the maximum order among all elements of H.....this exists as if we get an element of order p^m for all arbitrary large m.....then Tn will be in H for all n....and U Tn=G(Think about this)

Then if p^t is the maximum order show H=T_t.....use the fact x^(p^t) -1 has exactly p^t roots

[cloud walrus]

nastasya

nastasya

[serene dune]

is this fine ?

as a = gcd(m,n) would again result in 0 in Z_n

[languid trellis]

I'm working on this exercise, my best idea so far is to notice that we can consider E = F(u) as an overring of K, so there exists (by the universal property) a homomorphism K[x] -> F(u) that is the identity on K, and maps x -> u. Now, I just need to show that the kernel of this map is nontrivial. If it were trivial, then I have a suspicion that we need F = K, but I don't know how to show this. Any help would be appreciated ( :

Oh, if the kernel is 0, then we have K(x) \cong F(u) \cong F(x), which should imply that K = F?

[rocky cloak]

Not really related to the approach you're going for, but Do you know what the elements of F(u) looks like?

What happens if K contains an element of F(u) not in F?

[languid trellis]

Elements of F(u) are f-linear combinations of powers of u

[rocky cloak]

Not quite

That be F[u]

[languid trellis]

right, its the field of fractions of f-linear combinations of powers of u

[rocky cloak]

Indeed, so elements looks like f(u)/g(u) for polynomials f and g

So say k=f(u)/g(u) is an element of K. How can you construct a polynomial that has u as a root?

[languid trellis]

K is a field, so f(u)/g(u) has inverse, so we have f(u)/g(u) * g(u)/f(u) - 1 = 0, which is some polynomial which has u as a root?

Actually i dont htink that works

becuase that is identically 0

f(x) - xg(x) should have root u

Actually I'm not sure again

[rocky cloak]

Close, so plugging in u here would give you
f(u)/g(u) = u
for it to be 0. Which is not quite right.

But you're onto something. f is a polynomial and g is a polynomial, so if we took some linear combination of them with coefficients in K we would get a polynomial with coefficients in K

Now how could we do that to make u a root?

[languid trellis]

f(u)g(x)/g(u) - f(x) is a linear combination of f and g with coefficients in K such that u is a root

f(u)g(u)/g(u) - f(u) = f(u) - f(u) = 0

which shows that u is algebraic over K

and we know that there is some element of F(u) in K not in F as if not, then K would be identically F

[languid trellis]

thank you jagr

[icy bear]

what does being transitive mean here?

[coral spindle]

That the action is transitive. The definition of a transitive action is one with a single orbit.

[icy bear]

oh ok, I didn't read this definition

I was very confused

[tough raven]

Let M be an module over a ring with injective hull I(M). Any embedding of M in an injective module N extends to an embedding of I(M). Is the image of I(M) uniquely determined?

Oh, is M = ℤ, I(M) = ℚ, N = ℚ (+) ℚ/ℤ (ℤ-modules) with M embedding in the ℚ factor a counterexample with two extensions to I(M) being r ↦ (r, 0) and r ↦ (r, r mod ℤ)?

In fact, are all extensions to I(M) of the form r ↦ (r, nr mod ℤ) for some integer n?

[vagrant zinc]

What book are you using, because I need to look at the possible homomorphisms of Z-->Z_n

[serene dune]

oh brandon gave me this problem in #「the-obsidian-twink-jesus」

[void cosmos]

yeah

[serene dune]

so much modules in this channel 😿

[somber sleet]

I have a question regarding left exactness of the covariant functor hom. I'm looking at this given solution which is only confusing me. Where does it follow from that theta should factor through the kernel of psi?

[void cosmos]

theta is in the kernel of "fork sign dont remember the name" and hence in the image of phi

[somber sleet]

psi

[void cosmos]

i am not seeing the entire context tho but i think ur mis"remembering" how the Hom functor works on morphisms

they work by composition

and hence it being in the image meaning there is an element such that phi(element) is this theta

[void cosmos]

yeah

just unravel the definition of how Hom functor acts on your morphisms

and what it means to be in the image

so you have 0 --> M_1 --> M_2 --> M_3 --> 0

ur taking Hom(N,-) right?

[somber sleet]

yes

I just didn't really like the fact that I have to apply it to an arbitrary x

I actually remember the definition

[void cosmos]

im not following you honestly

[void cosmos]

did u get it

[serene dune]

i think its gonna be n
u only need to see where the generator=1 \in Z is being mapped to and there are n options

[somber sleet]

yeah

[regal zodiac]

If V1 and V2 are 2 affine varieties, we know the coordinate ring of V1×V2 is the tensor product of C[V1] and C[V2]
But is there a Formula for the isomorphism between these 2 rings ?

[rocky cloak]

For f(x)g in CV1C[V2] you make it into a regular function on V1xV2 by
f(x)g of (v1, v2) = f(v1)*g(v2)

[regal zodiac]

Yeah i got that , but i struggle to define the inverse of this map?

[tardy hedge]

By the time we see that ker(i) is a subset of ker(phi), cant we just stop there and state that N/ker(i) is the largest quotient?

Not sure why we need to say all that N/ker(phi) is a quotient of N/ker(i) etc etc

[rocky cloak]

Well, defining maps to tensor products tends to be kind of horrible.

I guess you would maybe pick a basis for each of the coordinate rings or something if you want an explicit inverse.

[rocky cloak]

What you may be able to do explicitly is that
C[x1, ..., xm]/I (x) C[y1, ..., yn]/J = C[x1, ..., xm, y1, ..., yn]/(I+J)

And then you can think of V1xV2 as a subset of C^m+n

[toxic zephyr]

suppose $|G|=p_1^{r_1}\ldots p_s^{r_s}$ (let's say $G$ is abelian for simplicity) and i want to show that $g\in G$ is a generator (so that $G$ is cyclic). what are the minimal powers i want to take of $g$ (verifying it is not the identity) such that i confirm $g$ has order $|G|$ (without calculating out every power of $g$)? for example if $|G|=12$, i could check all possibilities $g^2,g^3,g^4,g^6$, but can i pick an ``intelligent" subset to reduce the number of required checks?

[cloud walrus]

eigentaylor(got that eigenvalor)

[coral spindle]

You need to check all divisors since you can easily find examples of groups of the same order with the order of some element g being precisely that divisor

Actually, now that I am thinking about it...

You only need to check maximal (wrt divisibility) divisors, right?

So for |G| = 12 you need to check only g^6 and g^4. If it is the case that g^2 = 1 or g^3 = 1 then at least one of those vanish.

Yeah, you only need to check maximal divisors.

So for an arbitrary prime factorisation these look like $p_1^{r_1-1}p_2^{r_2}\cdots$, then $p_1^{r_1}p_2^{r_2-1}\cdots$ and so on.

[cloud walrus]

$\mathbf{Boytjie}$

[toxic zephyr]

yeah i think this is right. i decided to have some fun and code a program that finds non-primitive roots and raises them to the power of every nontrivial divisor of p-1 so i can see which combination of powers i need to check alongside the list of maximal divisors like you said, and it seems to work. (p-1)/2 is definitely a good choice because it's either 1 or -1, so then it's mostly checking which nonprimitive roots are -1 and which other powers are 1. for the smaller primes this is pretty trivial, but this is the first interesting case. looking at the ones under 15 which are -1, we see we need to check both 6 and 10 to make sure it is not a primitive root.
(i know this is not at all necessary, but it was fun to program)

[toxic zephyr]

gonna try to come up with a nice little proof that this is sufficient. it's probably not that hard

[coral spindle]

Indeed it's not

Hint: every proper divisor of |G| divides a maximal proper divisor of |G|

[toxic zephyr]

oh yeah nah that was easy

[coral spindle]

Haha

nice

[sly crescent]

Does every maximal subgroup of a p-group have index p?

[south patrol]

a p-group of order p^n has a subgroup of order p^k for all k with 0 <= k <= n

so ye

[rocky cloak]

I think Tarskis monster should give examples where every maximal subgroup has infinite index

[sly crescent]

And?

[rocky cloak]

For finite p-groups it's true though. If you mod out the Fratini subgroup (intersection of all maximal subgroups) you get an elementary abelian p-group

So in fact every maximal subgroup is normal

[rocky cloak]

Something a bit stronger you can say is that if H < G is a subgroup of order p^k and p^k+1 divides the order of G, then there is a subgroup H < H' < G such that H' has order p^k+1 and and H is normal in H'.

So in particular in a p-group, any group of order p^k is contained in a group of order p^n-1

(for k < n)

[opaque finch]

If it is finite then yes.

[rocky cloak]

Are we talking about different Tarskis monsters?

Like the only subgroups of Tariskis monster should be the cyclic subgroups (of order p), and they are all maximal.

[next obsidian]

I wish I had a pet monster

I’d name him jeffrey

[rocky cloak]

And PSL(2, 13) isn't even a p-group....

[opaque finch]

Apologies. I am not talking about the Tariski monster group, I am talking about just the Minster group M