groups-rings-fields

[tawdry venture]

no it's not but I was curious so I'd thought to ask lol

additive identity

omg I cant spell

[kind temple]

hmm. what is your exact definition of an ideal?

[woeful sage]

what is your profs definition of a ring? does it include identity or not?

[tawdry venture]

It does include identity

[woeful sage]

then I think you can easily prove this

[tawdry venture]

ohh ok

[woeful sage]

I mean try to prove it

[kind temple]

right. that was slick. usually, i see it defined as a subgroup of the additive group of R which is absorbative

[tawdry venture]

Since it's closed under addition and it has the 0 element, it must have its additive inverse

[woeful sage]

right yea, that's what I learned as well

[kind temple]

not quite. you need to use the fact that your ring has an identity, like Neamesis suggested

[tawdry venture]

because the ring has additive inverses

the ideal absorbs the additive inverses which belong in the ideal

[woeful sage]

ring has additive inverse -a, but I don't see how that implies -a belongs to our Ideal I

put your proof into symbols, that would be more helpful (Only 14th century mathematicians did math just using words alone 😔)

identity is also in ideal? how do you justify that?

[tawdry venture]

nevermind. I thought of a counterexample

[woeful sage]

hint: ||try to use the fact that if 1 \in R then -1 \in R as well||

[tawdry venture]

oh so ideal can absorb -1 from R

to get its additive inverse

[woeful sage]

yup

so if a \in I, then (-1)(a) is also in the ideal

I forgot how you prove (-1)(a) = -a

[kind temple]

for completeness: ||0 = (1 + (-1))a = a + (-1)a so (-1)a = -a by uniqueness of the additive identity||

[somber sleet]

Has some of you ever seen the notation: P.R[X]? If some of you knows what it means, can you tell me? We have never done this in class and now I am confused

[wide surge]

It is the ideal generated by elements of the form $$rf(x)$$ where $$r \in I$$ and $$ f(x) \in R[x]$$

[cloud walrus]

Alex

[wide surge]

Another way to describe it would be to just consider the map $$ \phi : R[x] \to \frac{R}{P}[x]$$ where $$ \phi(\sum_{i=1}^{n} a_{i}x^{i}) = \sum_{i = 1 }^{n} (a_{i} + P)x^{i}$$

[cloud walrus]

Alex

[wide surge]

which you can check is a surjective homomorphism and its kernel is the polynomials whose coefficients are elements of P, in other words are elements of PR[x]

[somber sleet]

oh okay so it is just the multiplication of the ideal with the ring, I don't know why he changed notation, before we always used pR[X] without the dot and I was not sure of what was meant

thank youu

[wide surge]

I mean i gather that this is what they mean. With that interpretation the isomorphisms they list make sense

[rocky cloak]

Why do you say it doesn't work?

[somber sleet]

I am pretty sure you're right, since all the operations they do are coherent with the product

I have a banal question about this proof. Why does it hold that p is a maximal ideal? shouldn't it only be prime?

or in other words, why is A/p a field?

[kind temple]

after replacing A with A_P, P becomes the unique maximal ideal in A

[crystal vale]

How can I show that if M has a finitely generated submodule N such that the quotient M/N is finitely generated, then M is finitely generated?

I got it

Idea:

Take N is finitely generated by {n_1,...,n_a} and M/N is finitely generated by {m_1+N,..., m_b + N }.

So I have to show M is finitely generated by {n_1,..,n_a, m_1,.., m_b}.

Let m in M, if m is in N then we can write m as a linear combination of {n_1,...,n_a}.

If m is not in N, then we can write x + N as a linear combination of {m_1 +N,..., m_b + N }, and then x = r_1m_1 +..+r_bm_b + n for some r_i in R and n in N, clearly we can write n as linear combination of {n_1,.., n_a}.

Hence we can write m as a linear combination of {n_1,.., n_a, m_1,..., m_b}.

Is it correct?

[wide surge]

I belive this is correct. This is essentially the same construction you would use by considering the short exact sequence $$ 0 \to N \to M \to \frac{M}{N} \to 0$$

[cloud walrus]

Alex

[somber sleet]

isn't the maximal ideal p^e={a/s| a in p and s not in p}?

[crystal vale]

Yeah I thought of the exact sequence but the book doesn't introduce it yet

[wide surge]

ah ok. What you did is correct still

[crystal vale]

Thank you

[kind temple]

yes. i was just identifying A with A_P and P with P_P ( = p^e)

[somber sleet]

okay that's good then, thank you. I was just confused on how the identification worked

[crystal vale]

How can I show that any R-module M can be expressed as a quotient of a free R-module F.

I am thinking about let m be a maximal ideal in R then R/m is a field then somehow we can use that every vector space has a basis, any hint?

[coral spindle]

No, you're overcomplicating it a lot

Hint: think of a map from the free module with basis M

[crystal vale]

With basis M?

[coral spindle]

Yes, with basis M.

You can form a free module with a basis any set. So choose M.

[crystal vale]

Oh

[crystal vale]

You mean if I let any set $S$ then we can make a free module $F = \bigoplus_{s\in S} Rs$?

[cloud walrus]

Notknow🙇

[coral spindle]

I would just write $\bigoplus_{s \in S} R$.

[cloud walrus]

$\mathbf{Boytjie}$

[crystal vale]

Is it different?

[coral spindle]

What does Rs mean

[crystal vale]

Rs = {rs | r in R }

[coral spindle]

But s is just an element of some set

rs has no meaning

[crystal vale]

I see

Yeah I mean s copy of R

[kind temple]

Rs without the subscript is throwing it off lol

[crystal vale]

Actually I thought as cyclic module

[kind temple]

how are you defining (+)_{s in S} R?

[crystal vale]

(+)R?

Components addition?

[kind temple]

i mean $\bigoplus_{s \in S}R_s$

[cloud walrus]

c squared

[kind temple]

like, what do those symbols mean to you?

[crystal vale]

Set of tuples (r_1,r_2,...,r_s) where r_i = 0 for all but finitely many

S is not finite

[kind temple]

this works if S is finite.

if S is not finite, this does not work. instead, you want all functions f : S —> R such which are zero for all but finitely many S.
this works for any set S

addition and multiplication are done component-wise

[crystal vale]

Yes

[crystal vale]

I know I was not sure how to explain but now I know

[kind temple]

great. so if you have an $R$-module $M$, you are putting $F(M) = \bigoplus_{m \in M}R$

[cloud walrus]

c squared

[kind temple]

this is the free module with basis M, or the free module generated by M

[crystal vale]

Yes

I think we can identify for which m, r_m is non-zero and then we can map to their sum

[somber sleet]

is the radical of an ideal always the intersection of the maximal ideals containing it?

[crystal vale]

I think the intersection of all prime ideals containing it

[somber sleet]

that's for sure correct

I was wondering about the other statement as well

[rocky cloak]

Not necessarily no

[somber sleet]

mmh okay

does then rad(I)=QR_Q follow from the primary decomp of I

QR_Q is the maximal ideal of R_Q

[wide surge]

Sorry but isn't the definition of a base a set of elements S of a module such that every element of the module can be written as a finite sum of multiples of the set S. If i recall correctly $$\bigoplus_{s\in S}R $$ is the set of all sequences of R indexed by S who have almost all their terms equal to zero. The cartesian product of free modules is generally not free

[cloud walrus]

Alex

[rocky cloak]

rad(I) is the intersection of all primes containing I.

Since Q is a minimal prime over I and is maximal, it is the only prime over I.

[wide surge]

If you take an infinite family of free modules, of course

[somber sleet]

hahahahha wow, I get lost so easily, that's clear, thank you very much

[rocky cloak]

Yes, S is a basis if every element can be uniquely written as a finite linear combination of elements in S.

The direct product of free modules is in general not free. For commutative rings I think this happens iff the ring is artinian iirc

[wide surge]

So you should consider as the free module to be the direct sum of copies of R indexed by elements of M. It's description is precisely the one you gave @crystal vale. Sequences of R indexed by M whose terms are almost all zero, except for a finite set.

Then you consider the map $$ \phi: \bigoplus_{m \in M} R \to M $$, which you can easily see is surjective and by the first ismoorphism theorem you can get the desired result

[cloud walrus]

Alex

[crystal vale]

So my function works, thanks @wide surge @kind temple

[coral spindle]

I'm teaching an intro linear algebra thing tomorrow. I want to show my students some 'weird' examples of fields to really show them that the definition isn't just number-like things. Any suggestions?

So far I want to show them:

• The field Z/2Z, but presented as {True, False} with And and Xor operations.
• The field Q(sqrt(2)).
• The field of quotients of polynomials.

Not really satisfied with these, so I'd be interested in suggestions

[kind temple]

F((x)) the field of formal laurent series over F, maybe F((x^1/2)) as well.

the p-adic numbers (although this may seem too number-like and definitionally out of reach)

but exotic examples i feel like are hard to come by since fields are pretty tame in general. they are all extensions of Z/pZ or Q

[coral spindle]

Sure, I guess I just don't want them to get the idea that fields are somehow "just" numbers

I love the Laurent series idea!

I have no idea how I'd introduce the p-adics though... :(

[frigid epoch]

Honestly those three examples are excellent

[kind temple]

i think the vector spaces can get kinda funky. for example, the power set of a set is a vector space over Z/2Z

the integers with bitwise xor is a Z/2Z vector space. i believe it should work with R as well, modulo some base 2 representation stuff

[frigid epoch]

"Infinite dimensional" fields are a great class of examples

So... fields of rational functions, yeah

Meromorphic functions would be a good example too but a bit out of reach methinks

[coral spindle]

Yeah, unfortunately they haven't seen vector spaces yet :( I have a few fun ideas to get this "direction and magnitude" idea stamped out ASAP

[kind temple]

ah, that is unfortunate

[coral spindle]

I honestly don't know why the course is set up the way it is -- they see matrices before vector spaces -- but whatever, I have to deal with it

[kind temple]

woah that’s kinda wild. what do they do with them? row reduction?

[coral spindle]

I think they just learn how to multiply them first

Yeah. I'm trying to think of good ways to motivate the way they're multiplied...

[mint seal]

counterpoint: fields are just numberlike things

[coral spindle]

Grumble grumble

[mint seal]

but a field of rational functions over a finite field is pretty interesting

[coral spindle]

I want them to realise that linear algebra isn't just the real numbers

This is mostly the point

[mint seal]

it has infinitely many elements and finite characteristic

do you mean you want to show there’s more to linear algebra than real vector spaces, or that there’s more to linear algebra than real numbers? Because R^n has plenty of geometry and stuff that isn’t covered by just thinking about real numbers

[kind temple]

i think that the examples you have, with potentially a few more, would get the idea across. i remember seeing fields like this for the first time and certainly not thinking of them as number-like things

[mint seal]

the surreal numbers is another really cool weird field

but it’s obviously numberlike

[south patrol]

Lol me marking linear algebra last year

Kept having to cross stuff out cause people falsely assume characteristic 0

[coral spindle]

Then I will really drum in the idea that things might not be char 0

This is a great catch I think

What proportion would you say got that wrong....?

[south patrol]

Oh I mean it wasn't even like meant to be about characteristic lol

Just a couple of people made that assumption without realising which made stuff easier

It was funny though cause I took a linear algebra practice exam and it turns out it was false if a matrix had an inseparable minimal polynomial

So like at one point I got stuck cause I needed that to not be the case lol

[coral spindle]

That's kinda cool

[frigid epoch]

Nonseparable irreducible polynomials are such bs

[coral spindle]

Well the minimal polynomial of a matrix may not be irreducible, but yeah

they are definitely pretty weird

[kind temple]

i want there to be a real world analogy of this

it’s so real

[frigid epoch]

Does anyone have a wealth of examples of nonseparable polynomials?

It's the kind of thing you encounter but aren't given enough intuition/examples for

[coral spindle]

Come to think of it, the only examples I know are the minimal polynomials for x in the extension F_p(x)/F_p(x^2)

or over F_p(x^n) in general

[frigid epoch]

We need an inseparability hero

[kind temple]

my prediction is jagr or chmonkey to the rescue

[rocky cloak]

More or less any example reduces to this I think.

But you can do
Fp(x) / Fp(f(x^p))

for whatever f you like

[coral spindle]

I guess I should have expected this

[rocky cloak]

It's at least easy to show that any irreducible non-seperable polynomial is off the form f(x^p) where p is the characteristic of the field.

[frigid epoch]

Can you give a good ol' example

[coral spindle]

He just gave an example 😅

The minimal polynomial of x over F_p(x^2) is X^2 - x^2. For p=2 this is inseparable

Similarly for x over F_p(x^p)

[rocky cloak]

Of a polynomial?

Okay, say f(t) = t^2 and p=3.

Then F3(x) / F3(x^6) is an non-seperable extension, x has minimal polynomial
t^6 - x^6

[frigid epoch]

The / sign confuses me hella hard here

[coral spindle]

it's just denoting an extension of fields

We're not doing a quotient

[frigid epoch]

Aight

[coral spindle]

This is totally similar to the notation Gal(E/F).

[frigid epoch]

Fp(x^6) seems highkey isomorphic to Fp(x)

[coral spindle]

It is

[frigid epoch]

Oh so it's more about the embedding

[coral spindle]

That's right

[frigid epoch]

It's like adjoining roots of x

[coral spindle]

Precisely!

[frigid epoch]

Yasss!

[rocky cloak]

So I guess the examples can be slightly more interesting if you sort of mix algebraic and transcendental stuff.

Like say F9 = F3(i) where i^2 = -1

Then F9(x) / F3(x^3) is non-seperable and x+i has minimal polynomial
t^6 + x^3 t^3 + x^6 + 1

[frigid epoch]

oh dear

are the elements of this F9 basically $\varepsilon 1 + \eta i$ where $\varepsilon, \eta \in {-1,0,1}$?

[cloud walrus]

PKThoron

[frigid epoch]

actually yeah, it's a 2D vector space where the basis vectors are just 1 and i, like in C over R

cool cool

[crystal vale]

but then my function is not a module homomorphism

what should that function looks?

[rocky cloak]

Alright so here is a characteristion of all of them:

Take some algebraic extension K/F of fields with characteristic p.

Take an element x in K that does not have a p^n-th root in K.

If f is the minimal polynomial of x, then f(t^p^n) is irreducible and non-seperable, and all such polynomials appear in this way.

[crystal vale]

I think we can identify non-zero elements, say for example $(r^1_{m_1}, r^2_{m_2},...,r^n_{m_n},0,0,......)$ then we map this element to $r^{1}m_1 +...+r^{n}m_n$

Is it correct?

[cloud walrus]

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

[crystal vale]

Yeah it looks so bad

[coral spindle]

Yes.

[crystal vale]

How can I properly define this function?

Wait I had a very active tag but now I haven't, why?

[naive oracle]

this is a dumb question , but associativity is always defined for just one operation each time, right?

i,e:
the rule
(A#BxC = A#(BxC) is not associativity (for binary operations # and x)

[somber sleet]

are powers of primes, prime?

[naive oracle]

no

[somber sleet]

do you have a counter example in mind, because I don't see one

[naive oracle]

look

[rocky cloak]

This wouldn't usually be called associativity, I don't think it has its own name

[rocky cloak]

Any prime

[naive oracle]

2^3 is divisible by 2^2 , it is instantly not prime by definition

[rocky cloak]

Do you have an example in mind? I would be interested to see that

[somber sleet]

oh wait I was thinking about prime ideals

[rocky cloak]

Same thing

[naive oracle]

so I am dumb, xD, sorry

[somber sleet]

which works the same

[crystal vale]

no take 2Z in Z

[rocky cloak]

I guess (0)^n = (0) would be a possible example

[naive oracle]

thanks, I was thinking that but I wasnt sure if that had a name

[somber sleet]

wait this confuses me more

[rocky cloak]

I was just saying basically every example is a counterexample.

But yeah there are a few examples that actually work, (0) being one of them

[crystal vale]

when ring is integral domain

[somber sleet]

If I consider a prime ideal P and its power P^n, is it enough to say that p^n = p*p^n-1 is in P^n, however p, p^n-1 both are not in P^n?

[somber sleet]

okay that makes sense, thanks

[crystal vale]

yeah

[rocky cloak]

For Noetherian integral domain (0) is the only example.

But in for example
k[x^2^-n : n in N] then rad(x) is a prime that squares to itself.

[crystal vale]

oh

[rocky cloak]

And then I guess boolean rings will give a bunch of examples (but they're not integral domains)

[somber sleet]

does someone have a proof of why for a commutative ring R and a prime ideal p Quot(R)=Quot(R_p)?

I kind of don't see it, I thought about dividing elements of R_p, but then I get (a/b)/(c/d) with a,c in P and b,d in R\P, doesn't this lead to a°d/b°c which then is in P since it is an ideal?

[rocky cloak]

So you need R to be an integral domain for the field of fractions to really make sense here, but it's just

(a/b) / (c/d) = (ad / bc)

You don't need a or c to be in P, that's not relevant

[somber sleet]

it is an integral domain, forgot to mention it

[rocky cloak]

Right so then every element of Quot(R_p) is in Quot(R) and conversely

a / b = (a/1) / (b/1)

[tough raven]

(Maybe to keep things unambiguous) it's usually not called associativity AFAIK, but IMO it's sensible to think of it as some kind of associativity.

[tough raven]

Is this by Nakayama's lemma applied to pR_p?

[rocky cloak]

Yeah. You don't need to localize either, depending on your formulation of Nakayama.

If I is finitely generated and I^2 = I, then I = (e) for an idempotent e.

[tough raven]

Ah, I see.

Idempotent prime

[tough raven]

Valuation domains my beloved

[somber sleet]

okay if you do that identification it makes sense

thanks a lot

[arctic trail]

funny how one of the questions

I posed the other day

the counterexamples people gave were like subgroups of affine groups over finite vector spaces

I just proved that essentially those were the minimal counterexamples (as in, other counterexamples degenerate into one of those in some meaningful way to me)

[coral spindle]

Nice!

[arctic trail]

yeah that was cool

So like the minimal counterexamples come from a representation of a group H into V (fin. dim. vector space over F_p) which is both irreducible and faithful

and just Aff(H)

[somber sleet]

does someone know how the dimension of R and it's localization S^-1R are connected? (S mult closed and R comm ring)

I was looking through gathmann but I don't find anything

[rocky cloak]

Krull dimension? It will either decrease or stay the same, since you have a correspondence between prime ideals of S^-1R and prime ideals of R not intersecting S

[somber sleet]

okay that makes sense, that's why it can0t be bigger then

thank you

(thank you for being so patient all the time)

[worthy solar]

Maybe a dumb question but when does a ring have that notion of a "division algorithm". I suppose presumably we want elements to have inverses so a field but my confusion would be on the ordering of these elements or what is the defining characteristics for order between two elements.

Soft question related to F[x] and how the division algorithm exists and we have the order where we talk about the deg of the polynomials

Like for regular integers you are just looking at the value of r and making sure r < |b|
If you have something like a=bq+r, where b != 0

And for the polynomials it is about degree

[rocky cloak]

You may want to look into "Euclidean domains"

[worthy solar]

Okay thanks 🙏

[velvet hull]

as the other guy said, a Euclidean domain is exactly a ring where you have some kind of division algorithm

[worthy solar]

My only slight confusion is what type of order it would be.. looking at the degree of polynimals isn't a total order because different polynimals can have the same degree

Hmm they are not even Fields

[velvet hull]

the Euclidean function doesn't have to have an explicit formula or be dependent on any specific value

it's just a vague measure of "size"

[worthy solar]

Oh I realized we get over that issue with r=0

[velvet hull]

something being a field is actually a really, really strong condition for a ring to have. if it interests you here's the tower of inclusions for common classes of rings

[worthy solar]

Yeah I noticed that

Hmm

[velvet hull]

but turns out, a Euclidean domain is actually "really close" to a field

[chilly ocean]

Dang, “tower” is right lmao

[worthy solar]

Ohh looks like we are going to cover this topic in the next few chapters

[velvet hull]

yeah they are important notions in ring theory, you'll see them a lot

[thorn jay]

Whole ass descending filtration

[worthy solar]

You algebra folks got everything already covered and named

[thorn jay]

Not if your idea is stupid enough

And then it turns out that it already has a name

AND APPEARS IN 3 FUCKING PAPERS IN TOTAL

[worthy solar]

sounds like some personal experience

[thorn jay]

Very

U-exact sequences are goofy as hell

[elfin wraith]

Towers of rings is a classic in algebra, you even have little sub towers, you can look at universal catenary rings for another chain

[thorn jay]

With tower of rings you mean
R_1 -> R_2 -> ... -> R_n
Where every map is an embedding?

As in, like a tower of field extensions

[elfin wraith]

No I was meaning the chain of inclusions as Hchan said

[thorn jay]

Mm

[elfin wraith]

I’m aware tower of rings isn’t really the appropriate term, I was just using the words they used,

Cohen Maculey rings are cool

[barren sierra]

I should learn what those are at some point

(like past the definition, more like why I should care about them)

[elfin wraith]

This is my chance to shill for Miller and Strumfels again

You like combinatorics and comalg, they’ve got just the book for you

[barren sierra]

that has been on my "I should get to this at some point" list for a while

perhaps that'll be the next adventure after I finally properly learn commalg this summer

[elfin wraith]

For what it’s worth I’ve only read part 1, so potentially the book takes a nose dive in quality, but at least those 6 chapters are fantastic

[barren sierra]

I doubt it'll take a nosedive tbh

I've read other stuff by Sturmfels and his writing is

[elfin wraith]

Yeah it’s a super well written book, I’d really like to finish it off if I ever get the chance

[barren sierra]

this semester I'm working through Manivel's Symmetric Functions, Schubert Polynomials, and Degeneracy Loci

summer will be commutative algebra time

and then next fall idk

but Miller, Sturmfels is up there on the list

Sturmfels' Algorithms in Invariant Theory also looks really nice

and not too long

but part III in Miller, Sturmfels really looks up my alley

[elfin wraith]

Hilbert schemes of points do seem quite cool, applied for an REU project on those last year

So much maths to learn, so little time, just unfair really

[tough raven]

Is the following proof of lying over for integral extensions correct? If A → B is injective and integral, then for any prime p of A, A_p → B_p is injective (exactness of localisation) and integral; in particular, the A_p-module B_p is non-zero, so B_p/pB_p is non-zero by the local-global principle. And B_p/pB_p = B (⨯)_A A_p/pA_p is such that Spec B_p/pB_p = preimage(p). So we are done.

[trail cave]

Hi, I have a homework question: "Let R be a ring with identity. Show that R is a division ring if and only if the only left ideals of R are {0} and R". I showed the first direction, but how do I show that R is a division ring from the assumption on left ideals?

[velvet hull]

for some a in R, what can <a> be? what does that tell you about a?

[crystal vale]

Take non-zero element r, now generate the left ideal by r, it can be (0) or R but it cannot be 0 so it will be R

[crystal vale]

I think the ring has to be domain first

[velvet hull]

what do you mean?

[crystal vale]

I mean what if there exists some non-zero elements x and y such that xy = 0

[velvet hull]

having 2 left ideals only should be sufficient to prove it, even w/o assuming domain

[crystal vale]

How?

What if (r) = 0, then ?

[velvet hull]

well, <r> contains r...

[crystal vale]

Yeah my bad

Got it, thanks

[trail cave]

okay so if a is a nonzero element of R then Ra = <a> = R (since Ra = 0 if and only if a = 0). So R is a principle ideal. Does this tell me that it's a division ring somehow?

[velvet hull]

an equivalent characterization of a priciple ideal is this: <a> = {r in R: a | r}{r in R: r = xa}. can you prove that?

[trail cave]

I'm not sure how

[velvet hull]

you can prove it by showing inclusion in both ways

[trail cave]

I know I mean I don't know how to go from x in Ra implies x = ra to a | r

sorry this seems obvious I've been writing the solutions to this HW for the past 6 hours and generators are my weakest topic

[velvet hull]

okay I just realised since we're not assuming commutativity this isn't entirely true, fixed

[velvet hull]

but anyways, what part are you stuck on

(a | r is not true, since we can't assume commutativity, by the way)

[trail cave]

Okay so <a> = {ra | r in R}. Let x be in <a>. Then x = ra. By cancellation xa^-1 = r. But that requires R to already be a division ring

[velvet hull]

by definition, <a> is the smallest left ideal containing a. Now the set S = {ra | r in R} is a left ideal (check that), which indeed contains a, so <a> is in S.
Now conversely, since a is in S, every element of the form ra must be in <a>, which is exactly the form of every element in S, so S is in <a>. what does that mean

[trail cave]

Okay I think my confusion is that the way my prof. introduced it I was confusing the definition of generator. I also found a theorem that says a ring with 1 is a division ring if and only if it has no nontrivial left ideals

[velvet hull]

yeah, that is exactly the theorem you are proving right now

[trail cave]

I'm silly

But I don't think the proof in the book uses this equality between S and Ra

[velvet hull]

it's just what I immediately thought of right now, im sure there are more proofs

[trail cave]

I appreciate it regardless I understand now

[crystal vale]

can i say module homomorphism of Z- modules are exactly group homomorphism?

[mighty kiln]

Yes

[lusty marlin]

Abelian groups can be naturally identified with ℤ-modules

[crystal vale]

yes

[trail cave]

Part iii here is the last hw problem I need. I don't really see how to get Z and 3Z here; from the 2nd iso theorem so far what I think I could get is something like R/I = Q/I

I thought about trying to show Z is a subring of R and then showing 3Z is an ideal of Z, where 3Z and Z = 3Z, but I don't see why he would make me do (i) and (ii) here if they didn't pertain to the solution for (iii)

[crystal vale]

How do you get R/I = Q/I from 2nd iso theorem?

[trail cave]

that was basically nonsense my thought process was that for R a subring of Q, I an ideal of R, R/(I and R) = Q/I. But I is a subset of R so (I and R) = I

[crystal vale]

What exactly does the 2nd iso theorem say?

[trail cave]

Let R be a ring, I an ideal of R. Support R' is a subring of R such that R'+I = R. Then (I and R) is an ideal of R' and R'/(I and R') = R/I

[crystal vale]

Why just don't we try to construct a mapping?

[trail cave]

a mapping for the second isomorphism theorem?

[crystal vale]

No simple homomorphism where kernel is I and image is Z/3Z

I am thinking of constructing a homomorphism between R to Z/3Z with kernel I

What if I send a/2^n to a2^n

Just you have to check it is well defined

And I think it is

[trail cave]

I will try like this, thank you

[crystal vale]

For any $R$-module $M$ and an exact sequence
[
0 \rightarrow N' \xrightarrow{f} N \xrightarrow{g} N'' \rightarrow 0
]
the induced sequence
[
0 \rightarrow \text{Hom}_R(M, N') \xrightarrow{f^} \text{Hom}_R(M, N) \xrightarrow{g^} \text{Hom}_R(M, N'')
]
is exact.

[cloud walrus]

Notknow🙇

[crystal vale]

I got it but don't get the significance of it

[chilly ocean]

Hey

Hamiltonicity of cayley graphs

Important how?

[finite turtle]

is there such thing as a complete (totally ordered) group like a complete field, ie satisfies least upper bound property?

[chilly ocean]

Yeah but idk about satisfying lub

[storm kiln]

Z maybe?

[chilly ocean]

Is there an example of a crossed B-Module $p:E\to{B}$ where $B$ is Abelian but $E$ isn’t? I’ve been thinking about it for a while, so far I’ve only been able to deduce that $E$ must be Metabelian, and p can’t be constant, so the kernel of $p$ must be some normal subgroup with $[E, E]\subseteq{ker(p)}\subsetneq{E}$

[cloud walrus]

NAT Enthusiast

[rocky cloak]

I guess you can do something like E=Q8, B = Q8/Z(Q8) = C2xC2

Where B acts on E by conjugation.

[somber sleet]

Why is the kernel of this map I_1 \cap ... \cap I_n? Is there some general statement?

[kind temple]

any element in the kernel of this map must be projected into each I_k

and if pi_k(r) = I_k in the quotient, then r in I_k

since this happens for all k, then r is in the intersection of all I_k

[rocky cloak]

I guess the general statement would be that the kernel of a homomorphism
R -> Prod R_i
is the intersection of the kernels of the corresponding
R -> R_i
maps

[crystal vale]

How can I show if $P_\alpha \text{ is projective for each}\alpha$ then $P = \bigoplus_{\alpha } P_\alpha$ is projective?

[cloud walrus]

Notknow🙇

[rocky cloak]

Write down your definition of projective and check that P satisfies it.

Remember that a map from a direct sum is the same as specifying a map from each summand

[crystal vale]

I want to use the result M is projective if and only if it is a direct summand of free module F

[crystal vale]

Okay

[rocky cloak]

This would also work

[crystal vale]

I am using this definition

[rocky cloak]

Just remember that a direct sum of free modules is free

[crystal vale]

So you mean I can take $F_\alpha = P_\alpha \bigoplus Q_\alpha$ for each $\alpha$ then $\bigoplus_{\alpha} F_\alpha = P \bigoplus_{\alpha} Q_\alpha$, is it correct?

[cloud walrus]

Notknow🙇

[crystal vale]

Thank you @rocky cloak

[serene dune]

Aye aye, richard walked me through this

[crystal vale]

If P is a direct summand of a free module of finite rank, then how can I show P is finitely generated projective? Yes it is projective by definition but finitely generated, hint?

[proper jolt]

im learning about localization right now

is this basically a general version of a fraction field but not in the integral domain?

did they choose S to be multiplicative so that the equivalence class (x, s) has addition and multiplication operations that resembles fractions and the localization is still a ring?

[kind temple]

yes

[crystal vale]

Anyone?

[coral spindle]

Hint: something is finitely generated if and only if it is a quotient of a free module of finite rank

[crystal vale]

I got the one direction

Quotient of a free module of finite rank is finitely generated

The quotient of finitely generated is finitely generated

[coral spindle]

Well a hint for showing the other way is that it's a very similar argument to the proof that every module is a quotient of a free module

[crystal vale]

Oh yes

[boreal inlet]

You can kind of say that

[chilly ocean]

Thanks!

[crystal vale]

Is it correct that if M is finitely generated and M/N is finitely generated then N is finitely generated?

M is finitely generated and M/ker \phi is isomorphic to free module now I have to show ker \phi is finitely generated

I got it, thanks MSE

[pliant forge]

just for sanity check, for a finite dimensional algebra A, the duality functor Hom_k(-,k) : mod A \to mod A^{op} sends right/left A-modules to right/left A^{op}-modules, not left/right A^{op}-modules

I think I get associativity issues if i assume right modules get sent to left A^op -modules

[rocky cloak]

A left A-module is the same as a right A^op module and vica versa.

So you can think of the duality either as sending left A-modules to left A^op-modules or you can think of it as sending left A-modules to right A-modules.

[pliant forge]

yea

thanks, that makes sense

[rocky cloak]

Typically one picks a convention so that mod R always means for example left modules.

And then if you want to talk about right modules you just say mod R^op

[pliant forge]

right right, the R probably also varies in location sometimes to distinguish the two

[rocky cloak]

Yeah, some people do that
R-mod vs mod-R

[frigid epoch]

Is the multiplicative set S in localizations precisely the same as a multiplicative submonoid?

[tardy hedge]

Whats the point here? Is the fact that det(A) is 1 means A-1 has integer entries so we can make sense of A-1 as a linear combo of group elements?

[crystal vale]

I did this exercise yesterday in module theory

[tardy hedge]

Haha bruh

[rocky cloak]

So a matrix with integer entries exactly describes a homomorphism G -> G.

A homomorphism is an isomorphism if it has an inverse, so we need A^-1 to be an integer matrix.

[tardy hedge]

Exactly describes a homomorphism G->G when G is free right

Or just finitely generated

[rocky cloak]

Yeah, you have to choose a basis

You need G to be free to have a basis

[tardy hedge]

Ok

And an isomorphism sends a basis to a basis

[crystal vale]

Let ( n = d_1 d_2 ) be a positive integer. Now let ( M = \mathbb{Z}/n\mathbb{Z} ), and consider the exact sequence
[
0 \to d_1 M \to M \to d_2 M \to 0.
]
This sequence is split if and only if ( d_1 ) and ( d_2 ) are relatively prime.

[cloud walrus]

Notknow🙇

[crystal vale]

Now let ( d = \gcd(d_1, d_2) ). Then
[
d_1 \mathbb{Z}/n\mathbb{Z} \oplus d_2 \mathbb{Z}/n\mathbb{Z} = d \mathbb{Z}/n\mathbb{Z}.
]

Right?

[cloud walrus]

Notknow🙇

[rocky cloak]

Consider d1=d2 = 2 for example

[crystal vale]

Oh so remove direct summand

No

Then I think it works

[rocky cloak]

Yeah, in general
d1Z + d2Z = gcd(d1, d2)Z

so d1Z/nZ + d2Z/nZ = gcd(d1, d2, n)Z/nZ

[crystal vale]

I see

But n = d_1d_2 so gcd(d_1,d_2, n ) = gcd(d_1, d_2), I am not sure

[frigid epoch]

Bros just multiplied by (nZ)^-1

[coral spindle]

What do you mean by this?

[crystal vale]

For the first direction let exact sequence split then Z/nZ = d1 Z/nZ \bigoplus d2 Z/nZ hence dZ/nZ = Z/nZ, and since d divides n so d = 1, is it correct?

And for second direction, since gcd(d_1, d_2 ) = 1, so d1 Z/nZ \cap d2 Z/nZ = {0}, hence we get split exact sequence

[frigid epoch]

Just poked fun at what's going on in here

[rocky cloak]

So you can identify the field generated by the transcende basis with the field of rational functions in that many variables.

After that you just need to show that the algebraic closure is unique. The usual argument involves extending the homomorphisms with Zorn's lemma or transfinite induction

[crystal vale]

For the first direction, let the exact sequence split. Then we have:

[
\mathbb{Z}/n\mathbb{Z} \cong d_1 \mathbb{Z}/n\mathbb{Z} \oplus d_2 \mathbb{Z}/n\mathbb{Z}.
]

Hence, we get:

[
d \mathbb{Z}/n\mathbb{Z} = \mathbb{Z}/n\mathbb{Z}.
]

Since (d) divides (n), it follows that (d = 1).

[cloud walrus]

Notknow🙇

[crystal vale]

Is it correct? @rocky cloak sorry to ping

[crystal vale]

No it is wrong

[rocky cloak]

Say E/F is an algebraic extension and K is algebraically closed.

If you have a map from F -> K you can extend it to a map from E like this:

for x in E let f be is minimal polynomial. Using the map F -> K we can turn f into a polynomial with coefficients in K. Since K is algebraically closed it has a root. Define a map by sending x to one of these roots. This gives a well defined map F[x] -> K.

Repeat inductively for E/F[x].

[tardy hedge]

If G is free group of rank n, and you have a set {ai} of n Z-linearly independent elements, it is not necessarily a basis?

[rocky cloak]

That's correct. For example
{2} is not a basis for Z

[tardy hedge]

Hmmm ok im trying to connect that with the module theory stuff i learned last term

Z has the invariant rank property, which just means that given a free Z-module, any basis is the same size. But that does NOT mean that a set of lin ind of size of the rank will be a basis

[rocky cloak]

Yeah that's right.

Not every set of linearly independent elements can be extendeded to a basis

[thorn jay]

Culture shock

[arctic trail]

man

I'm stuck trying to find a solvable group H with a faithful irreducible representation over some F_p

[rocky cloak]

Like is there something more you need about H.

Like you can just use a cyclic group or even the trivial group

[arctic trail]

can't be abelian

or rather the affine group Aff(H) can't be supersolvable

[rocky cloak]

I'm not sure what Aff(H) is, but if you want non-abelian and easy thing to do would be take V a vector space and take H = GL(V)

[arctic trail]

that won't be solvable

[rocky cloak]

Certainly sometimes it will.

Take V = (F2)^2 for example

Then H = S3

[arctic trail]

PSL is simple innit for most n,p

[rocky cloak]

Like you want an example for all n and p?

[arctic trail]

not really

what's the representation

of S3 in V?

which is faithful

[rocky cloak]

S3 = GL(V)

So that's the representation

Just permuting the 3 non-zero vectors in V

[arctic trail]

3?

[rocky cloak]

Yes

[arctic trail]

ah right mb

I'm tired

I'm confused

cause an element of V is just something like (0,1)

[rocky cloak]

V is the 2-dimensional F2 vector space.

G is the group of invertible 2x2 matrices with coefficients in F2. It acts on V in the obvious way making an irreducible faithful representation

This will permute the nonzero vectors
(1, 0), (0, 1) and (1, 1)

[arctic trail]

ah that's what you meant

[rocky cloak]

And in fact all permutations arise making G isomorphic to S3

You can also have S3 act on 2-d spaces over any field in much the same way to get an irreducible faithful representation

(but then GL(V) would be bigger)

[tardy hedge]

i wont get involved lol

[rocky cloak]

F would be that and then E it's algebraic closure

Make an isomorphism Fp(X) -> Fp(Y)

Field with p elements, the prime field of your fields

Well I assumed you wanted both fields to have characteristic p

The isomorphism Fp(X) -> Fp(Y)?

Well elements of Fp(X) looks like rational functions in X, so you just replace the variables using f

To get a rational function in Y

Well you can add, multiply and divide stuff

Say X just consisted of one element x.

Then you would also contain x^2 and x+1 and (x+1)/x^2 etc

[wooden fulcrum]

Hello this is kind throwing me for a loop

[rocky cloak]

prime field of K = smallest field contained in K (will always be Fp or Q depending on characteristic)

Frac(R) = field of fractions, completely different thing

[wooden fulcrum]

[rocky cloak]

Yes, Fp(X) will be equal to Frac(F[X]), I guess that's one way to see that it equals the field of rational functions

So Fp(X) is defined to be the smallest field contained Fp and X, and when X is an algebraically independent set this will be the set of rational functions.

Fp[X] is defined to be the smallest ring containing Fp and X, and when X is algebraically independent it will be the ring of polynomials.

Frac(Fp[X]) will be the smallest field contained Fp[X], hence will be Fp(X)

I think I would probably take a few steps back and review some material before attempting this problem

You wake a rational function

p(x1, x2, .. ) / q(x1, x2, ...)

and map it to
p(f(x1), f(x2), .. ) / q(f(x1), f(x2), ...)

It gives a homomorphism Fp(X) -> L
then you have to extend that to a map K -> L

[rocky cloak]

.

[rocky cloak]

Fp

Well, are you aware what the definition of transcendence basis is?

[rocky cloak]

Well you want this to have an inverse / be bijective

[rocky cloak]

Well, think about what happens to
1/(x1 - x2) for example

[velvet hull]

this actually sounds suspiciously like a universal property, but after looking it up apparently taking the algebraic closure just doesn't play nice that way

[rocky cloak]

There's no uniqueness no

[velvet hull]

there's still a guaranteed lift though, surely that's gotta mean for something

[rocky cloak]

Yeah it's nice, just not a universal property

Well it's algebraically closed, and it's an algebraic extension

[woven plover]

is there any special notation for the kernel of a group action without invoking the permutation representation?

[coral spindle]

Yeah, if X is your G-set you can write G_X, though this notation is sometimes used just for stabilisers...

I would maybe just say "let K be the kernel of the action of G on X" lmao

Or idk, define your own notation

[woven plover]

ig ill use this, thanks! just wanna prevent the inevitable hand cramps as much as possible by reducing the amount of stuff i need to write :p

[tardy hedge]

Yo how is 1 a root of x^4+4

[rocky cloak]

It's not

[tardy hedge]

What am i confused by

[rocky cloak]

The roots are 1+i, 1-i, -1+i and-1-i

[tardy hedge]

Oh thats what it meant? Lol

[rocky cloak]

Aka ±1±i

[tardy hedge]

Also, the splitting field of a degree n f(x) in F[x] would be n! Only when each intermediate field extension has p(x) irreducible in (x-a)p(x) ?

I probably didnt say that accurately enough

[coral steeple]

I'm looking for the original text by Noether where this proof is given. The trouble is I can't read German

[barren sierra]

Original article: https://link.springer.com/article/10.1007/BF01456821
Claimed translation: https://arxiv.org/pdf/1503.07849

I cannot read German either so I cannot verify that the translation is any good

perhaps better ones out there exist

[coral steeple]

Thanks!

[proper jolt]

does anyone know why showing the R' (11.7) satisfy the UMP in (11.3) implies that R_f is isomorphic to R[X]/<1-fX> ?

[crystal vale]

for any $R$ module $M$, i have to show, $Hom_R( R, M ) \cong M$

so i am sending $\phi \rightarrow \phi(1)$, it works, right?

Because $\phi(1) = 0$ implies $\phi = 0$ and also it is onto.

[cloud walrus]

Notknow🙇

[crystal vale]

If P is a projective module then I have to show there exists a free module F such that P\oplus F is free.

Any hint?

[proper jolt]

i think you can use exact sequences for this

[crystal vale]

No actually we have to take a big free module

[crystal vale]

How?

Maybe there is another way

[proper jolt]

like i think you can show that 0 -> F -> R^K -> P -> 0

splits

if P is projective module

[proper jolt]

maybe

F=ker(R^k -> P)

[crystal vale]

But how can I say R^k -> P is surjective?

[proper jolt]

i think you can just take the map e_i to an element in the generator of P

[crystal vale]

But P is not finitely generated

[proper jolt]

it doesn't have to be

when i say R^K

i just mean that R is the direct sum of rings R

[crystal vale]

Then how R^k -> P is surjective

Okay then it just gives me P is a direct summand of free module

[proper jolt]

yeh then you have to show this splits

0 -> F -> R^K -> P -> 0

R^K is free btw

[crystal vale]

Yes

But then what do you choose free module F such that P\oplus F is free?

[proper jolt]

F=ker(R^k -> P)

[crystal vale]

But ker not necessarily to be free

[proper jolt]

oh shoot yeh that's true i missed the part that F must be free rip

but maybe you can

do F to be R^N -> ker(R^k->P)

to get the exact sequence

R^N -> R^K -> P -> 0

then show that this splits

[crystal vale]

R^N -> R^k, I don't get it

[proper jolt]

like you use the surjective map R^N -> ker(R^k -> P)

to get the map R^N -> ker(R^k -> P) -> R^K

then get the exact sequence R^N -> R^K -> P -> 0

[crystal vale]

If P is a stably free module then if I add any free module to P then it gives me a free module, right?

In my textbook, it says that if P \oplus R^m = R^n for some m,n in N

[coral spindle]

The direct sum of free modules is indeed free.

[somber sleet]

Good morning guys, I wanted to show that if $0 \to N \to M \to P \to 0$ is an exact sequence, with $\iota: N\to M$ the inclusion and $\pi: M \to P $ the projection, then M Noetherian iff N and P Noetherian. Now I have =>, but what about <= ? Is it okay to define $N_k=M_k \cap N$ and $P_k = \pi(M_k + N)$ if $M_o \subseteq M_1 \subseteq M_2 \subseteq \ldots$ is an ascending chain of M?

[cloud walrus]

damn_guuurl

[rocky cloak]

Yes, that would be the right idea

[crystal vale]

No but my stable free module is not free

[somber sleet]

that's nice, thank you! 😄

[glad osprey]

Is it correct that R(pi) = R, but Q(pi) is isomorphic to Q(x) which cannot be embedded in C?

[frigid epoch]

Ofc Q(x) can be embedded in C, just send x to pi

Or any other transcendental

[glad osprey]

Oh, right 👍 I was thinking the field of rational functions are bigger than C, but if the base field is Q then I realize you can embed in C

[frigid epoch]

🙂

[rocky cloak]

In fact you can embed the field of rational functions in uncountably many variables

[frigid epoch]

Oh yeah, map one to pi, one to e... wait

[rocky cloak]

Give me a sec, just need to prove that pi and e are algebraically independent real quick 😉

[crystal vale]

My textbook used the definition that, a finitely generated projective R-module P is said to be stably free if P \oplus R^m = R^n for some m, n \in N.

Is it necessary that P to be finitely generated?

[rocky cloak]

Not really necessary I guess, but an infinitely generated module is stably free if and only if it is free

[crystal vale]

Can I get the definition of a stably free module?

I am very confused right now

i have to show stably free P is free iff P has unimodular extension property

And I get the equivalent definition of unimodular extension property that every surjective R^k -> P can be split, but it is always true since P is projective module

[rocky cloak]

P is stably free if P(+)R^n is free for some positive integer n

[rocky cloak]

What's the definition of P having the uep?

I would think the statement was R has uep iff every stably free module is free.

[crystal vale]

Every unimodular row can be completed into non - singular matrix

[rocky cloak]

Okay, but this statement doesn't mention P

[crystal vale]

Isn't the same what i said?

[rocky cloak]

Well maybe you had a typo, but you wrote that P should have uep

[crystal vale]

I have to show this one

[rocky cloak]

Alright, this makes sense to me

[crystal vale]

Oh my bad then

[crystal vale]

So if I let P to be finitely generated then my textbook definition is equivalent?

[crystal vale]

Is it true?

[rocky cloak]

No, this is just P being projective like you said.

So maybe you have copied it down wrong somehow...

[crystal vale]

Oh

Is it some terminology for a module M such that if add any free module F to M then M\oplus F is free ?

[rocky cloak]

That should be equivalent to being projective

[crystal vale]

One direction is trivial

[rocky cloak]

Like say P is projective, and P(+)Q is free, then F = (P(+)Q)^(N) is free and P(+)F = F

[crystal vale]

Yes

But I am saying if I add any free module F to M then F\oplus M is free

[rocky cloak]

Any free module, like including 0?

[crystal vale]

Oh I got your point

[crystal vale]

if $P \oplus R^m \cong R^n$ then $P\text{ is homomorphic image of } R^n$, hence $P$ is finitely generated

[cloud walrus]

Notknow🙇

[frigid epoch]

Wait, under which map

[crystal vale]

i am not sure but P \oplus R^m is isomorphic R^n, so i think there exist a mapping such that P will be homomorphic image of R^n

[rocky cloak]

R^n -> P(+)R^m -> P

[frigid epoch]

Oh cause P is projective

Duh

That's why, right?

[rocky cloak]

I mean, P is projective but it's sort of irrelevant.
X is always the homomorphic image of X(+)Y

Just by the canonical projection map X(+)Y -> X

[crystal vale]

i understand the proof of the later statement

[frigid epoch]

Mam I'm so rusty

You can map out of a PRODUCT, not a coproduct

[crystal vale]

so that's why we are in interested in finitely generated stably free module

[frigid epoch]

Does R-mod have a biproduct or what

[rocky cloak]

It does yeah

[frigid epoch]

Ah that makes sense

I somehow expected the average ring to make this potentially pathological

[crystal vale]

i am working with commutative ring, so is it also true that infinitely generated stably free module over non-commutative ring is free ?

[rocky cloak]

Yes, commutativity doesn't play into it

[crystal vale]

okay

[crystal vale]

any hint?

i think it helps

i got it this

but can anyone help me to understand this in matrix terminology?

so it means we extended our function f to f' such that if we make a matrix of f' then there is a submatrix which is associated to f

so author did ii => iii and iii => ii

but why?

[rocky cloak]

A row is the same as a map
R^n -> R
and a unimodular row is a surjective map R^n -> R.

So then finding more rows would mean extending the map to
R^n -> R (+) R^n-1

[rocky cloak]

Presumably they find it easier to prove
i <=> ii and ii <=> iii
as opposed to
i => ii => iii => i

[crystal vale]

So how ii) helps here, ii gives us P \oplus R \cong R^m for some m in N.

[rocky cloak]

Yes, that's right. So you can frame things in terms of maps R^m -> R, which correspond to rows

[crystal vale]

But it says every unimodular row, so isn't every surjective R^n -> R for every n, or just fix m ?

[rocky cloak]

I'm not sure what you're asking exactly. But if you have
P (+) R = R^m and you want to prove that P is free you want to look at R^m -> R

If your proving the other implication then you would just look at arbitrary n yes

[crystal vale]

say i want to show iii => ii, then by iii, is it means every surjective R^m ->R can be extended into R^m -> R ( + ) R^(m-1)?

oh i think it means, if P ( + ) R = R^m, then every surjective R^m -> R can be extended into bijective R^m -> R ( + ) R^ (m-1)

now i have to show P is free

correct?

then i think, proposition 3.4 will be help

[crystal vale]

right?

[hidden wind]

does someone have a hint for showing that the polynomial ring in n variables over an algebraically closed field is a jacobson ring? i’ve already been told to use the nullstellensatz but i don’t see it i feel i’m missing something really obvious :S

[wooden fulcrum]

Hello

If we construct the rees algebra then by definition of multiplication therein elements can never loop back into the lower layers right

So with graded structures the whole point is that we can separate the algebra out into subgroups a way where we always go up with multiplication?

[crystal vale]

I don't get how C is free A - module?

[tough raven]

Show that if I is an ideal and f is not in the radical of I, then there exists a maximal ideal containing I but not f. If you vary f, this shows that I is an intersection of maximal ideals - i.e., that your ring is Jacobson.

[tough raven]

In any ℕ-graded algebra A, A_m A_n ⊆ A_{m+n}. so yes.

[tough raven]

Well, it's hard to say what exactly the point is. AFAIK there are a lot of applications.

[tough raven]

C is defined as a free module.

[surreal dagger]

Also maybe consider V(J) for some proper Ideal J and leverage the fact that you have a correspondence between its points and maximal ideals

[wooden fulcrum]

Thank you

[wooden fulcrum]

I think it should be $A^{\oplus (M \times N)}$

Then it's the (M x N)-indexed sum of As (each of which is free over the 1-element set) so as a whole it's free over (M x N)

[molten bone]

Is it true that if "every subgroup of a cyclic group is cyclic" then is "every subgroup H is cyclic if and only if the group G is cyclic" also True?

[tough raven]

G is a subgroup of G, so yes.

If we change it to proper subgroup, I'm not sure.

[wooden fulcrum]

Wtf xD

[molten bone]

The question is asking me if we have no proper, nontrivial subgroups of G. Then show that G is cyclic.

I'm thinking take {e} the trivial subgroup and G itself, the non-proper one, and then show that these are both cyclic

Then use the theorem to show that G must be cyclic

[tough raven]

ℚ/ℤ is a non-cyclic group all of whose proper subgroups are finite cyclic.

[tough raven]

Wouldn't you have already shown that G is cyclic by that point?

[cloud walrus]

Fort Boyard

[molten bone]

Yea, you're right

Did not see that circular reasoning

Back to the drawing board i guess

[tough raven]

Try looking at <x> for some x in G.

[noble adder]

I have a problem asking me whether $2\mathbb Z$ is ring isomorphic to $3\mathbb Z$ and $4\mathbb Z$. The $4\mathbb Z$ case seems to be an obvious isomorphism, but I'm not sure about the $3\mathbb Z$ case. Would it be right to say that $\forall a \in 2\mathbb Z, a = 2b$ for some $b \in \mathbb Z$, then map that to $3\mathbb Z$ by multiplying by 3?

oops

[cloud walrus]

Silver

[noble adder]

actually ig that's not a homomorphism

hmm

that's what i was thinking but i dont think it satisfies multiplication

[molten bone]

No i see why

Yea it doesn't

[south patrol]

Yeah there are only two group isos and you can just check they aren't multiplicative

[noble adder]

ohh i didnt think about taking the group isos 😭

i saw the 3/2 one what's the other one

[south patrol]

-3/2

[noble adder]

ohhh

ok that makes sense tysm

[noble adder]

would i need to show that these are the only additive isomorphisms?

wait is 2\Z also not isomorphic to 4\Z???

nvm i figured it out 🙏

[tardy hedge]

Whats 2\Z?

Z/2Z?

[noble adder]

the ring of even integers

my book defines rings without unity

🙏

[vast stump]

this is normally written as simply 2Z

[noble adder]

i've been typing it in latex lol

so the backslash is there out of habit

[tough raven]

They are not. (2Z)(2Z) has index 2 in 2Z, while (4Z)(4Z) has index 4 in 4Z.

[cloud walrus]

meinebow

[surreal swan]

$F = { { g^i | 1 \leq i \leq p-1 } | g \in G \setminus 1 }$

^ how would you write that, it kinda feels ugly to me

[cloud walrus]

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

[surreal swan]

left and right for the curlys is the first thing that comes to mind

But the spacing still seems a bit off

[plain bloom]

I’m guessing this is true but just wanted to check, are our standard addition and multiplication on the natural numbers uniquely the only operations that satisfy the ordered field axioms on that set (minus the rule of having inverses for either)?

[tardy hedge]

Why do we consider the field fixed by ALL automorphisms in Aut(K)? If you take just one element in Aut(K), the set fixed by that morphism is still a subfield of K right? Whats the difference of considering the field fixed by all morphisms simultaneously vs just one particular one?

Im early in learning galois theory so i imagine these questions will become more clear later but

[frigid epoch]

Pick a countable set A and two bijections $A \to \bQ, A \to \bQ(\sqrt{2})$

[cloud walrus]

PKThoron

[frigid epoch]

That's too brainfry for me rn, sorry

[plain bloom]

Wait when I said natural numbers I didn’t mean any countable set, I’m assuming it’s natural numbers with our fixed standard order. and in that case I’m pretty sure standard addition and multiplication are the only two that satisfy the ordered field axioms

[frigid epoch]

Ah

Well you're not gonna get a field, only a... semiring?

[plain bloom]

Yeah as I said dropping the requirement for inverses of either addition or multiplication

[frigid epoch]

A commutative one

Hm

[plain bloom]

Not sure if that has a name

[tardy hedge]

“Even if they were we would have to check the maps are automorphisms”, the maps would certainly be homomorphisms because we just need to map generators, but is the point of that comment is that for some reason the map may not be bijective?

[frigid epoch]

You do want a multiplicative identity, right?

Otherwise you could grab 2N, which is order isomorphic to N

[plain bloom]

Yeah, multiplicative identity equal to second least element (idt it would even make sense to be 0 cuz that would probs contradict w distributivity)

[frigid epoch]

I'm not sure what the author means here either

Either way, the automorphisms of Q(cube root of 2 = y) can't tell Q(y) and Q apart

Not sure why that's important tho

[tough raven]

The field fixed by an automorphism g is related to <g> (the cyclic subgroup generated by g) in the same way the field fixed by a group G of automorphisms is related to G.

(Assuming <g> and G are finite for simplicity.)

[cloud walrus]

meinebow

[barren sierra]

I don't understand this notation. What is $x_g$? Do you mean $x \cdot g$?

[cloud walrus]

Spamakin🎷

[barren sierra]

like what is g(x_g)?

[surreal swan]

I think your R definition is misspelled, probably meant x_g instead of x{g} ?

I think g(x_g) is just shoving an invented variable x_g into the polynomial

If K is not generated by 1 (for example complex numbers), take any element that isn't in <1> (for example i)
Then the element is in R
Take I = {g1(x)=x}
Then i * g1 won't be in J, so J is not an ideal?
I might not understand what generation in this context means

[dense moss]

u tell us

[dense moss]

it isnt in J, but in R, u gotta prove it @grand ridge

[proper tide]

Can I say (P(A),symmetric difference ),whereA={1,2},a Klein's four group?

[coral spindle]

That is correct

[rocky cloak]

I mean for every set of automorphisms in Aut(K) you can consider the field fixed by those automorphisms.

This will be the same as the field fixed by the subgroup they generate. One of the central ideas in Galois theory is that when K is a Galois extension then you have a correspondence between subgroups and intermediate field extensions.

Yes, J is a proper ideal

It doesn't contain 1

I guess the easy argument would be, consider the algebraic closure of K.

Consider a map R -> algclosure(K) given by mapping x_g to a root of g.

R is clearly surjective and the kernel contains J. Hence J is contained in a maximal ideal

Well (x_g)^2 maps to the square of whatever x_g maps to for example

The map is surjective

Every element is the root of some irreducible polynomial.

I guess you need to account for polynomials having multiple roots. Which you can do, but the image is nonzero anyway, so for proving that J is proper you don't actually need it

Anyway, I guess maybe why you want to prove this is to prove algebraic closures exist (?). So maybe you don't want to assume that in the proof.

In which case you can just be a lot more explicit:

If R is a ring and f(x) is a monic polynomial, then R[x]/(f(x)) is nonzero (this should be straight forward). Say 1 is in J. Then R/J = 0. 1 is the linear combination of finitely many polynomials f1, f2, ..., fn. So R/(f1, ..., fn) = 0. So (K[x1]/(f1))[x2]/(f2) ... = 0

Contradiction

[noble adder]

y'all, why wouldn't $\mathbb R / \langle x^2 \rangle$ be the same as $\mathbb R / \langle x^2 + 1 \rangle$

[cloud walrus]

Silver

[noble adder]

i know the latter is isomorphic to the complex numbers, but can't you use the same idea for the former

where you have all polynomials of degree less than 2

[wraith cargo]

the former isn't a field

the latter is

R[x]/(x^2) isn't a field because x is a zero divisor

💀

[noble adder]

ohhh that makes sense

but what about R/x^2 + C where c is positive nonzero

i mean based off what my prof told me that would just be R[sqrt(-c)]??

[wraith cargo]

okay firstly you should be careful because you're quotienting R[x] not R
Now R[x] is a PID so any quotient by an irreducible polynomial gives a field since the ideal generated by it is nonzero prime and hence maximal (because PID)
x^2 is reducible so you don't get anything good out of it lol

[noble adder]

oops idk why i said R
ok that makes a lot of sense tysm

[wraith cargo]

what doesn't make sense 😭

He's just using R as an example ring
Ig it's an abuse of notation
But what you have is like R (your ring)/(ideal generated by g_i's)
Now you can write this out as like (K[x_{g1}]/g1)[x_{g2}]/(g2)....
You keep going like this Now each of these quotients on their own are non zero
But you postulate they should be 0
Ergo contradiction

[crystal vale]

But it says that the elements of C are formal linear combination of elements of M × N, so what are the basis elements?

[heady acorn]

let k be a finite field. given any element a in the algebraic closure of k. how do you prove it's minimal polynomial has distinct roots?

The notes i took go something like this:

thus p_b(x) divides p_b'(x), implying that p_b' is zero.
now, consider p_b(x) = a_0+a_1x+...+a_rx^r; p_b'(x) = a_1+2a_2x+...+rx^r-1. there exists a_i (i>=1) such that a_i=/=0```

does this suffice? or is there some other subtle point

i should specify that p_a denotes the minimal polynomial of a over k

[crystal vale]

but the elements of C are the all finite linear combination of a_i(m_i, n_i)

i think they treat every (x,y) as linearly independent element

[tough raven]

The set M ⨯ N.

[tough raven]

Exactly.

[tough raven]

There is indeed a subtle point. Here's a hint: your proof actually works for k a field of characteristic 0, but not for k a finite field.

[crystal vale]

now i got it, thanks

[vivid kestrel]

in a presentation of a group, are the generators always assumed to not be the neutral element and not equal to each other?

[frigid epoch]

Yes

If you have n generators, they're assumed to not be in each other's span either

Imagine a map from $\bZ^{\ast n}$ to your group that just sends the relations in your presentation to $e$

[cloud walrus]

PKThoron

[frigid epoch]

(The free product of Z with itself)

Of course, it can happen that there will be relations like $a = e$ or $b = c^{-2}$ but then that's not a very economical relation, and people wouldn't really use those

[cloud walrus]

PKThoron

[velvet hull]

I've thought about that before, is that an implicit assumption or part of the definition? because otherwise there's no reason why a group presentation would not hold for all subgroups of that group

you can still characterize the group through its universal property or whatever

[frigid epoch]

If they all could happen to be the neutral element, then you couldn't tell whether it's describing the group you want or just the trivial group

[velvet hull]

like, you could still characterise the group being presented as the "largest" group that admits said relations is what I'm trying to get at here

it's not a problem of whether the presentation uniquely determines a group

it's just a technicality of, does the definition admit smaller groups

[frigid epoch]

Yeah, and in that case all the generators are a priori independent of another

If you want the "most general" group, they end up having to be independent

It's like asking whether the x in Z[x] has to be independent of Z!

[proper jolt]

hey this is a problem from hungerford anyone know what (R/M^n) is suppose to be?

[rocky cloak]

M^n is the n-fold product of M with itself (the ideal generated by elements that are product of n things from M).

R/M^n is the quotient ring R modulo M^n

[proper jolt]

oh ok thanks

[crystal vale]

i have to show $(A\otimes B) \times (A\otimes B) \rightarrow A\otimes B$ such that $(a\otimes b, c\otimes d)\mapsto ac\otimes bd$ is well defined

can i say, for fixed $a\otimes b\in A\otimes B$, let the mapping $f: A\times B \rightarrow A\otimes B$ such that $(c,d)\mapsto ac\otimes bd$, then it induce the R-linear $f_{ab}: A\otimes B \rightarrow A\otimes B$ such that $f'(c\otimes d) = ac\otimes bd$.

then let $g: (A\otimes B)\times (A\otimes B) \rightarrow A\otimes B$ such that $(a\otimes b, c\otimes d) \mapsto f_{ab}(c\otimes d)$
i think this is incomplete, help me

[cloud walrus]

Notknow🙇

[vivid kestrel]

thanks!

another question about semidirect products, there is if i understand correclty the case of a normal subgroup N being acted on by another subgroup U by conjugation such that G = NU, and then we say that G is the inner semidirect product of N and U. So is this just a special case of the outer direct product where our map $\varphi : U \rightarrow{} \mathbf{Aut}(N)$ is just conjugation?

[cloud walrus]

eggman

[coral spindle]

You need also that N n U = 1

[vivid kestrel]

ah, yeah, mb

[crystal vale]

if A is R-algebra , and N, P be A-module, can i say N,P are R-module too?

[coral spindle]

The outer semidirect product is isomorphic to the inner one in a way that respects N and U. It is a reconstruction of G, if you like, from the relevant information. Much like the inner and outer direct product, in fact.

And yes, the action is conjugation

[rocky cloak]

I don't think it's fair to say one is a special case of the other.

The outer semidirect product is exactly constructing a group with N and U as subgroups such that it is the inner semidirect product of those subgroups.

And conversely if G is the inner semidirect product it will be isomorphic to the outer semidirect product

[rocky cloak]

Yup

[crystal vale]

they defined R-algebra as, A is a ring, which is also an R-module satisfying the condition a(xy) = (ax)y = x(ay), a in R, x,y in A.

so if N is A-module, how can i make it into R-module? i am thinking that fixed a in A then rn = (ra)n

[rocky cloak]

Try replacing a with 1

[crystal vale]

But why? Does it change something?

[rocky cloak]

What's 1*n

[crystal vale]

I see

Yeah got it

Can I say N is (A, R) bi-module?

Here A and R both are commutative

To be N -(A, B) bi module, we need N to be A module and B module and also (an)b = a(nb), but here both ring Commutative and A is B-algebra

[crystal vale]

any comment?

i want to prove that, let $M$ be an $R$-module, $A$ an $R$-algebra and $N,P$ be $A$- modules. Prove that
$(M\otimes_{R} N)\otimes_A P \cong M\otimes_R(N\otimes_A P)$

[cloud walrus]

Notknow🙇

[crystal vale]

so to make $(M\otimes_R N)$ $A$-module, first i will fix $a\in A$ then the map $M\times N\rightarrow M\otimes_R N$ such that $(m,n)\mapsto m\otimes an$, now this is $R$-bilinear so it induce $\phi_a:M\otimes_R N \rightarrow M\otimes N$ such that $m\otimes n \mapsto m\otimes an$. Now in MSE, they said that $\phi_a \in End(M\otimes_R N)$ hence it induce the module structure

[cloud walrus]

Notknow🙇

[crystal vale]

https://math.stackexchange.com/questions/135978/how-to-show-that-m-b-b-otimes-a-m-is-a-b-module

[vivid kestrel]

in semidirect products, for the map $\varphi : U \xrightarrow{} \mathbf{Aut}N$, does composition with an automorphism on either side yield an isomorphic semidirect product?

[cloud walrus]

eggman

[crystal vale]

This definition and $A$ said to be $R$ - algebra, when $A$ is a ring and $A$ is a $R$ - module with $a(xy) = (ax)y = x(ay), a\in R, x,y \in A$.

Let $A$ follows above definition then we have to find the ring homomorphism $f:R \rightarrow A$ such that $ra = f(r)a$ so let $f(r) = r(1)$ then it is additive by module property of $A$ and it is also preserve multiplicativity by the property
$a(xy) = (ax)y = x(ay), a\in R, x,y \in A$.
And $f(r)a = r(1)a = 1(ra) = ra$ so it follows the Atiyah definition also.

[cloud walrus]

Notknow🙇

[crystal vale]

So I can say both definitions are equivalent, right?

[next obsidian]

Yes

In the commutative unital case, an R-algebra is just a ring map R -> A

[crystal vale]

Thank you

[tardy hedge]

ayy atiyah macdonald type shiii

[crystal vale]

Yes

[crystal vale]

I want to show that for the commutative ring $R$, $R[X] \otimes R[Y] \cong R[X,Y]$, so first I construct a bi-linear mapping such that $f: R[X] \times R[Y] \rightarrow R[X,Y]$ which maps $(p,q)\mapsto pq$. So it induce the unique R-linear $f': R[X] \otimes R[Y] \rightarrow R[X,Y]$ such that $f'(p\otimes q) = pq$.

But now I want an inverse of $f'$, any hint?

[cloud walrus]

Notknow🙇

[next obsidian]

This isn’t the right way to go about it IMO

You want to show that for any R-Algebra A that A (x) R[x] is isomorphic to A[x] as an A-algebra (which implies it holds as an R-algebra too, which means as a ring)

To do this you want to use universal properties, you want to show that

1: Maps R[x] -> B for an R-algebra B is a choice of an element in B (R[x] is the free R-algebra on one element)

2: Given an R-algebra B, and an A-algebra C, that R-algebra maps B -> C are the same as an A-algebra map A (x)_R B -> C [this is the extension of scalars adjunction]

When you combine this, you see that A (x)_R R[x] has the property that a map from this into an A-algebra C is just a choice of an element in C, so it’s the free A-algebra on one element, aka A[x]

[sturdy spear]

tensors

[crystal vale]

Yes

[sturdy spear]

interesting

[vivid kestrel]

anyone 😢

[rocky cloak]

So composing with an automorphism of U gives you something isomorphic, but on the other side it's not so clear what you mean exactly.

Something that is true is that you get something isomorphic if you conjugate by an automorphism of N

[vivid kestrel]

oh sorry, i meant an automorphism Aut N -> Aut N

[rocky cloak]

Yeah, so conjugation by automorphism of N would yield an (inner) automorphism of Aut(N). But an arbitrary automorphism will not do

If you want an example, consider N = C12 and H = C2.

Then Aut(N) = C2xC2, so from any nontrivial map H -> Aut(N) you can achieve every other by composing with automorphism of Aut(N), but
D4xC3, C4xD3 and D12 are all different

[vivid kestrel]

hmm, thanks!

[rocky cloak]

Also worth noting that you can have "exceptional" isomorphisms between semidirect products. Meaning you're not concerned with the specific embedding of N for example.

If you take something like N=S3 and H=C2, then any semidirect product is isomorphic to NxH, even though there are lots of nontrivial maps H -> Aut(N)

[tardy hedge]

for k a field, ideals of k[x1, ... xn] have a basis because you can view the ideal as a k-vector space?

whereas if R is just noetherian then its not clear that ideals of R[x1, ... xn] are f.g, thats where hilberts basis theorem comes in?

something seems off on what i said

im learning about monomial ideals

[crystal vale]

isn't R Noetherian implies R[x_1,..., x_n ] Noetherian ?

[rocky cloak]

Both of the things you said are correct

But you have the word "whereas" between two statements that have nothing to do with eachother.

[rocky cloak]

Yes that is the Hilbert basis theorem

[south patrol]

This is sort of irrelevant from the second poiny

Since 1) that is a k-basis, not a k[x1,...,xn] generating set and 2) the basis may be infinite

[rocky cloak]

You want to think of g1 as a polynomial in x1. Whether you write that as g1 or g1(x1) is up to you.

Well it's one of the isomorphism theorems.

(A/B)/C = A/(B+C)

I think it probably depends on the starting field

It's theorem C on Wikipedia
https://en.m.wikipedia.org/wiki/Isomorphism_theorems

N = B, K = B+C

[vivid kestrel]

is there a "simple" way besides just experience to easily see why the A4 would have no subgroup of order 6?

[rocky cloak]

Well, just looking at the group is fairly "simple" I guess.

There's only 12 elements after all.

[vivid kestrel]

fair point

[rocky cloak]

There are tricks to see it quickly though. Any subgroup of index 2 is normal, so a subgroup of order 6 is the same as a surjective map A4 -> C2.

So if you know that A4 is generated by 3-cycles or you know it's commutator subgroup that would do it

[vivid kestrel]

Hmm the commutator subgroup is the V4 iirc? So the argument here would just be that a subgroup of order 6 would need to contain the V4 which isn't possible?

[candid patrol]

Many ways

Hom(A4,C*) = Hom(A4,U3)

But U2 notsubset U3

[candid patrol]

Proove it 😜

Hom(A4,C*) ~ U3, so f : A3 —> U2 is trivial because of the order

[vivid kestrel]

sorry, i no neither U3, U3 nor C*

[rocky cloak]

Yup

Every element is a root of some polynomial

Of R/K

Because of how J is defined

[rocky cloak]

Maximal ideals don't contain 1

[empty perch]

Hey folks, just a quick reminder that the Abstract Algebra Lecture series is meeting regularly again ^_^
We will meet to discuss Group Presentations in about 10 minutes over in the ⁠⁠#1055201711679082516. More information can be found in our thread: ⁠⁠#1317307081535000606

[unkempt stream]

Not sure how to think about this to be honest

i haven't done abstract alg in a while and I can't parse how to do this even with the primary assumption since homomorphisms don't "distribute over direct sums"

Unless informally consider the intersection of the direct summands with the kernel but it doesn't really work out nicely like that

[rocky cloak]

I mean following the hint seems like a good start.

If M is p-primary you can tell everything you need about M from the dimensions of p^(i-1)M/p^iM.

[unkempt stream]

Unless I fully decompose the module into p-components, use the fact that the p-component of a subgroup should be a subgroup of the larger group's p-component and then "rebuild it" after quotienting by the broken down subgroup

[unkempt stream]

actually I'm pretty sure this works

since we'd be quotienting by a direct sum of the summand's subgroups

[unkempt stream]

I don't remember this nor get why this would be helpful

[vivid kestrel]

so for a field K, the ideals in K[x]/(g) for some polynomials g are just the principal ideals generated by the divisors of g right

[coral spindle]

Yes. In general the ideals of R/I correspond to ideals of R that contain R. So in particular, since every ideal of K[x] is of the form (f), and (f) contains (g) iff f | g, you have the answer

[vivid kestrel]

ah, enligthening, thanks

[neon river]

hi can i get a hint for b)

[wicked patio]

how did you do a?

[acoustic igloo]

hi

um i'll post my progress on this problem

i was working on this problem, i skipped part (a)

[neon river]

Suppose h in H and k in K. Then consider (hk)^n = e where n is minimal.
That means h^n = e and k^n = e. Then n must be a multiple of r and s and in that case n is minimal if n = lcm(r,s)

[wicked patio]

Why does it mean h^n and k^n = e?

[neon river]

because G is abelian

[wicked patio]

Just because two things multiply to the identity doesn't make them both the identity

Abelian or otherwise

[acoustic igloo]

yeah i think part (a) is handled by case 1

so you will have to use ab^gcd(r,s) rather than ab as a generator

but i can't articulate why this works because my brain turned to mush

[wicked patio]

Seems to be

Prove all the field axioms one by one

They should all be immediate unless I'm missing something

[acoustic igloo]

oh ab should still work

you say if (ab)^k = e then a^(rk) b^(rk) = e and so b^(rk) = e

but b^(rk) is not e, and this is the part i'm stuck on

why is b^(rk) not e

it should follow from the fact that r and s are coprime

[neon river]

But then if h^n and k^n were inverses of each other then h^n = k^(-n) which means k^(-n) is an element whose order divides r and s which implies the order of k^(-n) is 1 which implies k^(-n) = e

[wicked patio]

ok sure

n must be a multiple of the order of h and the order of k

then, you need to choose h and k to be generators and it should work

[neon river]

ok nic

[wicked patio]

so what happens if we do this in the case where r and s aren't coprime

[acoustic igloo]

i don't get it

[neon river]

i was kinda thinking that it's basically the same approach

[acoustic igloo]

but hk doesn't always work

in that case

[wicked patio]

it doesn't?

[acoustic igloo]

no

[wicked patio]

that's news to me, do you have a concrete counterexample?

[acoustic igloo]

yes

[neon river]

🍿

[acoustic igloo]

for example G=<Z30,+>
a=5, b=3, r=6, s=10
a+b = 8 has order 15 but lcm(6,10)=30

[wicked patio]

oh huh

[acoustic igloo]

that's why you need case 2a

sorry

i mean

that's why you need case 1

[wicked patio]

hmm

i'm guessing fundamental theorem of finite abelian groups is off limit to use?

[acoustic igloo]

idk, it came before that in the book

[acoustic igloo]

but can this be answered?

because i think that will unlock the rest of the proof for me

[wicked patio]

sorry idek what k is

[wicked patio]

you can kind of nuke the problem with it if it is allowed

[acoustic igloo]

k is any integer with 0 < k < lcm(r,s)

oh

[wicked patio]

visually i think of it like split G into it's Tor_p groups and show each one has the necessary prime power of the lcm

[wicked patio]

well if k=s there's a bit of a problem i think

unless s = lcm(r,s)

[neon river]

is the case work really necessary 😭

[acoustic igloo]

wait i will check this

ok i think i got confused because we started talking about part (a)

[dense moss]

not really

[neon river]

but maybe this only doesnt work because your task actually assumes a product?

[acoustic igloo]

no no

[dense moss]

since G is abelian

the product hk commutes, h & k are the generators

so (hk)^n = h^n k^n

so h^n k^n = e iff h^n = e & k^n = e

[acoustic igloo]

for part (a) my solution should use hk^gcd(r,s) as the generator

but i'm more interested in part (b)

[dense moss]

u said ur askin for help in (a)

[wicked patio]

in other works hk?

[acoustic igloo]

no i'm asking for help in (b)

[dense moss]

oke

[wicked patio]

HK is a quotient of (Z/rZ) x (Z/sZ) but i'm not sure if that helps

[acoustic igloo]

i only need to show that b^rk is not e in case 2a

[dense moss]

yeah i dont think thats a clever approach

[acoustic igloo]

no wait

ah i gotta eat dinner

i think it works

possibly there's another approach that doesn't need casework but you need a way to describe the generator, and (hk) doesn't work for all groups

[dense moss]

think of the prime factorization

$r = \prod_{p | r} p^{e_p}$ $s = \prod_{p | s} p^{f_p}$

[cloud walrus]

Goëtia

[dense moss]

thus $m = lcm(r,s) = \prod_{p | m} p^{\max(e_p,f_p)}$

[cloud walrus]

Goëtia

[dense moss]

which boils down to two cases both leading to a quite nice result

if $e_p > f_p$ then since H is cyclic u get the existence of unique subgroup of order $p^{e_p}$

[cloud walrus]

Goëtia

[dense moss]

like wise if $f_p > e_p$

[cloud walrus]

Goëtia

[neon river]

oh lord whats happening

[wicked patio]

oh

just find an element in H with order a maximal power of p

same with K

whatever the larger power is, that power has to exist in HK

same for every p and then combine them all together using part a

[dense moss]

correct, thats the idea of what i just said, basically it stems from prime factorization

[neon river]

u genius

[glad lance]

geniuses

[dense moss]

no u

[tardy hedge]

Who r u ppl

New ppl

[dense moss]

actually there is another way to do part (b), but that would be overkill

[dense moss]

im older than you here though OwO

[glad lance]

How old is you

50?

[dense moss]

18

[glad lance]

younger

respect your elders

[dense moss]

I meant that im older than him since i was in this server before him kekw

[dense moss]

which elders? XD

[glad lance]

me

[dense moss]

bet

[wicked patio]

[neon river]

i am older than Goetia yet i respect them equally

[dense moss]

OwO

age is just a number though Uwu

[glad lance]

..

I mean

Yeah age is just a number