groups-rings-fields

[cloud walrus]

Stephen

[past temple]

Let A be a commutative ring, and let G be a subgroup of Aut(L), where L is a field containing A

is it true that (Frac A)^G = Frac A^G?

where A^G, (Frac A)^G are the elements of A, Frac A, respectively, that are fixed by every element in G

[thorn delta]

don't really have time to help but when G is a galois group A^G is a field for any subset A of L iirc. But idk if that's true more generally, and idk if it'll help you here

[barren sierra]

what is a field extension

I think the isomorphism I want is $\phi\left(\frac{x}{y}\right) = xy^{-1}$

[cloud walrus]

Spamakin🎷

[barren sierra]

but I'm not sure what a field extension is

[rose axle]

ok so

if I have a subgroup H <= G

and H^{ab} = 0

then does the normal envelope bar{H} <= G have abelianization zero?

[barren sierra]

is F an extension of R is there is an injective homomorphism from R to F?

[chilly ocean]

I was trying to solve an exercise from a book but i got stuck in my calculation.
I'm trying to show that foundamental groups of a class of orientable 3-manifold is not left orderable where by this i mean that they admits a strict ordere invariant for left multiplication.
The groups are Seifert manifold built over a surface where the surface is $\mathbb{R}P^{2}.$

They admit the following presentation: $$\pi{1}(M)=\langle\gamma{1},...,\gamma{n},y,h;|$$
$$yhy^{-1}=h^{-1},\gamma{j}^{\alpha{j}}=h^{-\beta{j}},\gamma{j}h\gamma{j}^{-1}=h,y^{2}\gamma{1}\cdots\gamma{n}=1\rangle.$$

With some calculation I got to show that if $k>0,h>1$ the following holds: $h^{-k}<y^2<h^k$ and $h^{-k}<y^{-2}<h^{k}.$ In the book I was given the hint to show that $yhy^{-1}>1()$ and this in facts concludes as it is a contradiction with the assumption that $h>1$ given the first relation but I could not find a way to prove the key fact $()$. Any suggestion in what kind of relations I should look for?
Thanks in advance.

[cloud walrus]

Stephen

[thorn delta]

Field extension is just a field that contains another field (or ring in your case). So yea there would be an inclusion map of R into F

[deep nova]

For case number 4, imagine if G were not cyclic but H is cyclic, then I can conclude that this is not an isomorphism right?

i just want to make sure such a statement can be inversible if that makes sense

[thorn delta]

Sure, because phi^-1 is an iso in the other direction

[deep nova]

ah ok

[past temple]

is it true that Z[sqrt(-5)] = (2, 1+sqrt(-5))

?

[next obsidian]

What

Oh you’re asking if that’s the unit ideal

[prisma thunder]

Suppose that $L$ is an extension of $F$ and that $K_1, K_2$ are two intermediate subfields such that $L = F(K_1,K_2).$ Prove that $[L : F] ≤ [K_1 : F] · [K_2 : F]$.

[cloud walrus]

eM

[prisma thunder]

I'm trying to make sense of the notation for L since I haven't seen that before. Is that our field generated by elements in those two intermediate subfields? Is it the composite field of our two intermediate subfields?

[past temple]

if J is a maximal ideal of a ring R

and I is an ideal of R that is NOT contained in J

does that mean that I is also a maximal ideal of R??

[gritty sparrow]

No, here is an example, (x,y) is maximal in k[x,y] (here k is some field). The ideal (x-1) is not contained in this ideal, but is is also not maximal as it is properly contained in the ideal (x-1,y)

[deep nova]

if I have $\mathbb{Z}_4 \times \mathbb{Z}_2$, and I'm trying to find the direct external product, would the elements of this group just be the elements of the set of the cartesian product?

[cloud walrus]

Zoro

[deep nova]

i did a couple examples and they seem to give me the same pairs as i do the binary operation

[next obsidian]

It’s the smallest sub field containing K_1 and K_2

You can explicitly view it as taking like F(alpha_i) where that’s like rational functions in alpha_i over all alpha_i in K1 and K2

Which yeah is the composite of them

[modest zenith]

IS anyone around to help clear up something for me?

[hidden haven]

Ask 🔫 no promises

"Does (Q_+, ÷) form a group?"

ie does ÷ satisfy the group axioms

[modest zenith]

Ok. That clears it, thanks

[obsidian sleet]

just b

the a is not in math mode

so it means "there exists b in G such that..."

[modest zenith]

Kay

Damn, I might need glasses

[barren sierra]

so I'm considering the ideal generated by 85 in Z_204

the generators of this ideal are {17, 85, 119, 187} which I found by code

but how do I figure this out without using code and a brute force search

so 17 = 17 * 1, 85 = 17 * 5, 119 = 17 * 7, and 187 = 17 * 11

but I don't see a pattern

[rustic crown]

wait isn't every ideal of Z/nZ principal?

[barren sierra]

yes but they can be generated by different numbers

[rustic crown]

oh isee what you mean

[barren sierra]

so $\langle 17 \rangle = \langle 85 \rangle = \langle 119 \rangle = \langle 187 \rangle$

[cloud walrus]

Spamakin🎷

[rustic crown]

(a) = (b) if and only if a = bu where u is a unit?

[barren sierra]

?

[rustic crown]

a unit is an invertible element of a ring

so just find all the units in Z/204Z

and multiply them with your generator

[barren sierra]

Hm I see

[rustic crown]

units in Z/204Z correspond to the numbers which are coprime to 204

[barren sierra]

there are alot more than 4 numbrers in Z_204 that are coprime to 204

[proud bear]

When you mod by 204 you probably just get those generators still

[barren sierra]

ohhhh

I see

[rustic crown]

yea but probably the point is can we filter out which units to look for

[barren sierra]

ones such that u*17 <= 204?

[barren sierra]

still not sure how to show this

[rustic crown]

gcd(85, 204) = 17
so we're looking for distinct values of 17u where u is a unit mod 204. So need to understand the units which do no work, i.e. 17u = 17
this is same as 17(u-1) = 0(mod 204) which is same as u = 1 (mod 12)
so we're going to look at the map U(204) --> U(12)
the kernel is the units that do nothing.

[barren sierra]

oh wait

I see

[rustic crown]

U(12) has 4 elements which is why there were 4 generators

[barren sierra]

U(204) = the units in Z_204?

[rustic crown]

yea

[barren sierra]

aight

[rustic crown]

so if you wanna find the nice units take inverse image for each units mod 12

so find a unit in U(204) which is 1, 5, 7, 11 mod 12

no need to find all of them

but lucikly 1, 5, 7, 11 are units in U(204)

we're done lol

so the generators are 17, 17 * 5, 17 * 7, 17 * 11

[barren sierra]

ok cool

[tawdry crystal]

What is the name of the ring D referred to here?

[chilly ocean]

can you ive context

[tawdry crystal]

I have no idea what they’re talking about until I know what D is

I assume D is a standard notation like R or C or Z

[chilly ocean]

i've never seen it. maybe it was defined earlier in the text?

[tawdry crystal]

In fact it was I just missed it

[chilly ocean]

[modest zenith]

Anyone around to answer this one question that I've been sitting around thinking about?

[hidden haven]

nope

Do you know about free groups and relations

😵‍💫

Wait that would be way too annoying to work with anyway

[hot lake]

looks like you need a group that's not commutative

[next obsidian]

Every group embeds into a symmetric group (don’t worry how to prove this, you’ll definitely see it in your class if you cover group actions)

[hot lake]

what's the simplest one you can think of

[next obsidian]

So you’ll definitely be able to find an example in a symmetric group

I’d maybe just start writing down some permutations and try to find an example

¯_(ツ)_/¯

[modest zenith]

That ain't the best answer but it might be the one I have to go with

[next obsidian]

I mean it’s good experience I think to just get your hands on some stuff

[modest zenith]

Since I'm stumped af on this one problem. Everything else just need a quick clearance on the question then everything snaps into place

[next obsidian]

And just try to notice what goes on

You might end up with stuff that after cubing the things are pretty close to the same so you think about how you might fudge it a little bit

I think there’s value in just writing some stuff down and trying to notice patterns and get closer to what you’re trying to find

[hidden haven]

oh brilliant

[modest zenith]

I'll try it I suppose

nothing better to do; Thanks fellas

[next obsidian]

The only other suggestion is to look at dihedral groups

[modest zenith]

I'll do that

[next obsidian]

Those are a lot simpler than symmetric groups but still are non-commutative

[modest zenith]

Professor likely like a clean proof than a messy one

[next obsidian]

Well in the end the proof is probably just a computation so

It won’t be too messy once you write it up

Just messy to find xD

[hidden haven]

well if you want a strong hint for the symmetric group approach then ||product of 2-cycles in a symmetric group is a 3-cycle (sometimes)||

[viscid pewter]

why is 0 != 1 a field axiom

[willow mason]

It’s not

[rustic crown]

because we don't like 0 ring to be a field

[viscid pewter]

why not

[chilly ocean]

BECAUSE WE SAY SO

[rustic crown]

a lot of things... say every map now becomes injective!

K --> L, 1 maps to 1, not to 0, so kernel has to be trivial

[hidden haven]

Imagine having to open every theorem about fields with a Let K be a non zero field

But yeah just removing the 0 ring from fields makes every field homomorphism injective, makes fields easier to study

Also study of 0 ring as a field would be very nothing to say study

No proper field extensions, I guess it's automatically alg closed? Lol

[viscid pewter]

i see

just convenience

[rustic crown]

group of units of a field is K\{0}

[robust pollen]

I'm asked to give an example of a non-projective $R$-module $P$ for some ring $R$, and a surjective $R$-module homomorphism $f \colon M \to N$ such that $\operatorname{Hom}_{R}(P, f)$ is not onto.
Am I correct in saying that $R = \mathbb{Z}$, $P = N = \mathbb{Z}_2$, $M = \mathbb{Z}_4$ and $f (x \mod 4) = x \mod 2$ would give such an example? It seems like here $Hom_R(P, f)$ would just be the zero map

[cloud walrus]

expectTheUnexpected

[hidden haven]

Where diagram

(I'm kidding don't actually spend time on it)

[cloud walrus]

expectTheUnexpected

[robust pollen]

lel

[hidden haven]

[robust pollen]

felt cute, might delet later

[hidden haven]

Ye that seems legit

[robust pollen]

Nice, ty moldi

[gray scroll]

what does this notation stand for? K[X]/f

[shrewd ravine]

how can I construct a monoid with a certain number of units?

[robust pollen]

polynomial ring modulo ideal generated by f?

[hidden haven]

Construct group of that cardinality

[gray scroll]

idts

[robust pollen]

context then?

[gray scroll]

quotient fields

[shrewd ravine]

I can't think of any examples

integers modulo 48 under addition comes to mind but idk

[hidden haven]

Yeah that's a monoid with 48 units

[shrewd ravine]

oh okay

thank you

[rustic crown]

can we construct a group of any cardinality?

[robust pollen]

so, what is f? I would have thought it's some irreducible poly

[rustic crown]

oh F(X)

lol

[gray scroll]

they havent mentioned what f is

[hidden haven]

Yeah and Z/nZ for finite lol

[robust pollen]

Can you post a sentence where that notation appears? @gray scroll

[hidden haven]

And model theory proof too 😌

[gray scroll]

yes, just a min

[robust pollen]

Yeah, f will be the minimal polynomial of alpha here

[shrewd ravine]

is it possible to find (up to isomorphism) all the abelian groups given a certain order?

[hidden haven]

Finite order?

[shrewd ravine]

yes

[gray scroll]

i still dont understand the notation

[hidden haven]

See the classification of finite abelian groups

Should be a wikipedia page or something

There's an easy classification

[robust pollen]

So f is some irreducible polynomial, and (f) is the ideal generated by f. f is irreducible, then (f) will be a maximal ideal, and so the quotient by (f) is a field.

Maybe you should look at chapter 2, where the notes apparently show this statement or something similar

[shrewd ravine]

is that talking about writing it as a sum of cyclic groups?

[hidden haven]

Yes

[robust pollen]

You know about quotients, right?

[gray scroll]

i tried looking, but cannot find this statement

[gray scroll]

i know quotient spaces

[chilly ocean]

I think the statement is: R/I is a field iff I is a maximal ideal in R

[robust pollen]

Then I'd suggest you look into ideals in rings and quotients of rings by ideals

[shrewd ravine]

how many abelian groups of order x are there?

is there a way to count it?

[void cosmos]

yea

check out fundamental theorem of finitely generated abelian groups

[hidden haven]

Yes Z/xZ is one, and for 2 integers p and q,
Z/pZ ⊕ Z/qZ is Z/pqZ iff p and q are co prime (Chinese remainder theorem)
Prime factorize x and count

[shrewd ravine]

I'm having trouble applying it to an example

if I wanted to find the all the abelian groups of order, say 180

[hidden haven]

Big number

[shrewd ravine]

180 is prime decomposed into 2x2x3x3x5

yeah for fun

[hidden haven]

Right

[shrewd ravine]

I know it has 48 generators

by the euler totient function

[hidden haven]

That's only if it's cyclic right?

[shrewd ravine]

right

[hidden haven]

So any group of that cardinality is a direct sum of groups of cardinalities 2², 3² and 5 by CRT

And using the classification theorem of course

You can just separate out the parts corresponding to different primes

So it reduces to counting number of abelian groups of order p^n for a prime p and natural n

That's just the number of ways of writing n as
a + 2b + 3c + 4d + ...

[gray scroll]

can someone give a plain and rigorous definition of k[X]/f where small k is a field and f belongs k[X]. A definition not using ideals but construction of what the elements look like

[hidden haven]

a tells you the number of Z/pZ, b the number of Z/p²Z and so on

[shrewd ravine]

I'm a bit confused with this Z/pZ notation

what does it mean?

[hidden haven]

Cyclic group of order p

[shrewd ravine]

ah ok

my lecturer uses notation C_p

[hidden haven]

Yeah both are standard, Z means the group of integers

[hidden haven]

This is Z/48Z

[shrewd ravine]

ah I see

okay time to try this for 180

[shrewd ravine]

trying to make sense of all the theory and I got C_2 x C_2 + C_3 x C_3 + C_5 for all the abelian groups of order 180 which doesn't seem right to me

[hidden haven]

What is C_n here?

[shrewd ravine]

a cyclic group of order n

[hidden haven]

But then it's not a number

[hot lake]

is x different from +

[shrewd ravine]

x is for direct product

[shrewd ravine]

oh wait right

[hidden haven]

Finite direct sums and direct products are the same for abelian groups

[shrewd ravine]

I see

I don't understand how to find more than one of these groups

isn't there only one prime factorisation of 180?

[robust pollen]

roughly, k[x]/f is like k where you add another element alpha that satisfies f(alpha) = 0. For example, if k = Q = rational numbers and f(x) = x^2 - 2, then k[x]/(f) is a two-dimensional vector space over Q with basis 1 and alpha, where alpha also satisfies the equation alpha^2 - 2 = 0. In other words, you have added a root of the polynomial f (which doesn't have roots in Q) to Q

[hot lake]

well Z/180Z is an abelian group of order 180

for example

[shrewd ravine]

there are more?

[hot lake]

yes

there should be 4 different abelian groups of order 180

[shrewd ravine]

why 4?

[hidden haven]

See this lol

[hot lake]

did you do the classification of finite abelian groups already or not ?

[shrewd ravine]

yes but it wasn't explained well

so I'm trying to learn

in more detail

[hidden haven]

Fair

[hot lake]

can you prove that C180 is different from C2 + C2 + C3 + C3 + C5 ?

[shrewd ravine]

I'm not sure about the repeated ones

because C2 x C2 is not cyclic

[hot lake]

yeah C2 + C2 is different from C4

so there are 2 sorts of abelian groups of size 4

despite 4 having a unique prime factorisation 4 = 2 * 2

[shrewd ravine]

I don't think I can prove this

[hot lake]

how did you prove that C2 + C2 is not cyclic

[hidden haven]

That's exactly the Chinese remainder theorem

[shrewd ravine]

isn't it isomorphic to the Klein 4 group

which is not cyclic

[hot lake]

how do you know that the Klein 4 group is not cyclic

besides "someone told me so"

[shrewd ravine]

you can check the subgroups made by each generator

none of them are all the elements

[hot lake]

well you can kinda do that too for C2 + C2 + C3 + C3 + C5

any element in there will have their order be a divisor of 30

[shrewd ravine]

there are 4 abelian groups with order 180 because 2^2 x 3^2 x 5 = 180, and the powers are 2x2

so correspond to partitions of 2 ?

[hot lake]

yeah ?

[gray scroll]

thank you so much

also what is the difference between k[alpha] and k(alpha)

[chilly ocean]

k[alpha] is the ring of polynomials in alpha, k(alpha) is the ring of elements p(alpha)/q(alpha) where p and q=/=0 are polynomials. k(alpha) happens to be a field.

(You can also say that k[alpha] is the smallest ring containing both k and alpha and k(alpha) smallest field containing both k and alpha)

[hidden haven]

And it's the second definition with subring and subfield in place of ring and field when alpha is some element in a larger ring/field

[shrewd ravine]

C2 x C2 x C3 x C3 x C5, C4 x C9 x C5, C2 x C2 x C9 x C5, C4 x C3 x C3 x C5

are these all of them?

[hot lake]

yeah

which one among those is C180 ?

[shrewd ravine]

C4 x C9 x C5

[hot lake]

yep

[shrewd ravine]

because coprime

I think I somewhat get it now

thank you

for all the help

[hot lake]

uh you're welcome

[shrewd ravine]

why does every element of a group lie in a subgroup of order of one of the prime factors of the order of the original group?

[hot lake]

um no ?

[shrewd ravine]

of a non-abelian group

[hot lake]

a generator of C4 doesn't lie in a subgroup of order 2

[shrewd ravine]

hm

like if we have G

a non-abelian group of order pq

p,q are both prime

doesn't every element of G have to lie in a subgroup of order p or q?

[hot lake]

then yeah

because if you have an element or order pq then the group is cyclic but it can't be cyclic if it's not abelian

so every element has order 1,p, or q

in this very specific case

[shrewd ravine]

yeah that makes sense

because every element has to have order dividing pq

[vocal wolf]

thats equivalent to asking whether a subgroup of a normal subgroup is normal

no

[willow mason]

no

[chilly ocean]

suppose I have rings

A and B

If i have a ring morphism A to B

B is then a module over A

[willow mason]

yea

[chilly ocean]

and then if i have 3 rings

A to B to C

C has two module structures,

[willow mason]

yes

[chilly ocean]

and these should be compatible in some sense

[willow mason]

yep

[chilly ocean]

what is the sense

[hidden haven]

(A,B)-bimodule

But also stronger lol

So maybe you can say more like one is the restriction of the other

[chilly ocean]

what i am looking to say is

[rustic crown]

you need some sort of commutativity if both are left modules

[chilly ocean]

the support of C as a B module is contained inthe support of C as an A module

[hidden haven]

I assumed we were in commutative rings

[chilly ocean]

we are but thank you det thats good to know

[rustic crown]

usually for an (A, B) bimodule M, you want A to act on left and B to act on right

if you're working with non-commutative stuff

if you have a ring map A --> B, then we have an exact functor B-Mod --> A-Mod which just "restricts" the scalars. So a B-Mod M is same as a ring hom B --> End(M), all you do is precompose by the ring map A --> B, to get the A-Mod structure... this means the underlying abelian group doesn't change, and so exactness.

[chilly ocean]

Sorry im a still a little bit confused

maybe ill explain the situation as best as i understand maybe im going wrong some where

Let M be an A module, then Supp_A(M) is all the prime p ideals of A such that the localization M_p is nonzero

Simillarly Supp_B(M) are all the prime ideals q of B such that the localization of M_q is non zero

there must be some statement im missing about

f:A to B a ring morphism then every prime ideal from one comes from an extension or restriction from the other?

[rustic crown]

idk much algebraic geo, so can't say for sure... I'll assume the two module structure are compatible... as in
A --> B --> End(M) gives the A-Mod structure on M.
So if q is a prime such that M_q is non-zero, this should mean that there is an element m in M such that b * m = 0 => b in q, (might have to check my reasoning again)
but if this is the case, then look the the inverse image of the prime q, say it's p. Then a * m = 0 means that f(a) * m = 0 which means that f(a) in q, so a in p. which would mean that M_p is non-zero.
So it seems like this a map from Supp_B(M) --> Supp_A(M) which is just taking the inverse image of a prime, so same as restriction of Spec B --> Spec A

[deep nova]

Just a real quick question regarding direct external products of two groups G and H;.... if the two groups are defined under binary operation of addition, their product G x H is just the cartesian product right?

[rustic crown]

yep, it's cartesian product whether you use additive or multiplicative notation

[deep nova]

ok thanks

[chilly ocean]

okay thanks

ill try think about this some more

as in

"b * m = 0 => b in q, (might have to check my reasoning again) "

but it seem slike this works

[wooden ember]

i dont really get why the centralizer of pi is a direct product of wreath products

basically C_ci here would "be" a single cycle and S_k_i just permutes the disjoint cycles of length c_1?

ah right yeah that's it, an element of the centralizer acts on the cycle decomposition of pi by permuting disjoint cycles of the same length

makes sense

[robust pollen]

Let $R, S$ be any rings, let $P$ be an $(R,S)$-bimodule, and $Q$ a left $B$-module.
If $P$ and $Q$ are both generators as left modules (over their respective rings), is $P \otimes_S Q$ also a generator in $A\text{-mod}$?
My idea to show this was the following: let $p \colon Q^n \to B$ be an epimorphism, and let for $M \in A\text{-mod}$ also $e \colon P^m \to M$ be epi.
Then
[
(P \otimes_S Q)^{n m} \cong P^m \otimes_S Q^n \xrightarrow{P^m \otimes p} P^m \otimes_B B \cong P^m \xrightarrow{e} M
]
is an epi, and so $P \otimes_S Q$ is a generator.
Is that correct?

[cloud walrus]

expectTheUnexpected

[elfin patrol]

I’m trying to find an isomorphism between (Z/6, +) and (Z^inv/7, *) but I am kind of at a loss

[waxen hedge]

Try to find a multiplicative generator of (Z/7)*

And map it to 1 in Z/6

[elfin patrol]

So I’ve gotta map 1 to 3 and 5 to 5 right? Seems like n mod 6 -> 3^n mod 7

[rustic crown]

looks right

[elfin patrol]

It works for the generators but it seems like it doesn’t map each element in the first group to the second

[rustic crown]

nah, it works
{0, 1, 2, 3, 4, 5} are maped to {1, 3, 2, 6, 4, 5}

in general if G is a group and g is an element of order n, you get a group homomorphism Z --> G which sends 1 to g, the kernel of this map is nZ and the image is <g>
so you get an isomorphism Z/nZ --> <g>

in our case, you found an element in (Z/7Z)* with order 6. the element is 3, so it also generates the group. which gives us the required isomorphism

let me know if that clears it up...

[elfin patrol]

That makes sense, but is there a way to explicitly find the isomorphism each time? This example I just guessed and got n mod 6 -> 3^n mod 7 but for other problems seems like it could be troublesome

[rustic crown]

as in finding an isomorphism between Z/(p-1)Z and (Z/pZ)*?

that's like a computationally hard problem, finding a generator modulo a prime isn't easy

[viscid pewter]

wait, really?

i thought any non-identity element (Z/pZ)* was a generator

and then you just map that to 1 in Z/(p-1)Z

wait no the first part def isn't true

ok

[rustic crown]

non identity of Z/pZ is a generator

[viscid pewter]

yeah

[last cargo]

Guys, I've some exercises about equivalence class and quotient set, but I didn't get the point yet. Can someone explain me the item A?

The exercise is asking for describing the equivalence class and quotient set in relation to ~ induced by the following functions.

Can someone explain it to me, pls?

[unreal portal]

I think the idea is that the equivalence classes are such that a ~ b if f(a) = f(b)? I'm not sure though, I've not seen problems like that before

[deep nova]

how do you take the order of an element that is a pair? Like so: |(a,b)|? This is in the context of direct products of two groups G and H

where say (a,b) is in G x H

$\mathbb{Z}_2 \times \mathbb{Z}_2$

[cloud walrus]

Zoro

[deep nova]

for this my book says there is 4 elements

but order two

im not sure why, but what i guessed was they divided the order of elements of the group (4) by 2 to get order 2 for a pair elements

[void cosmos]

compute (a,b)^n for n= 1,2,3,...

untill u get the identity

or if u cant then its infinity

just like you would do in any group

@deep nova

[deep nova]

ah ok

[wooden ember]

i know this is supposedly connected to the concept of colimits in category theory but i fail to see the point of it or even think of an example

i feel like the directed family {S_n} under inclusion might work but i have so much trouble even visualizing the operations that i dont see what group it might be isomorphic to

[next obsidian]

Direct limits are so useful

they’re exact in modules for example which is nice

And they commute with a lot of things like say tensor product

[wooden ember]

im sure they are but even though i managed to do the exercises without an issue it's really hard to get a feel for them

[next obsidian]

They’re not something you deal with explicitly

IMO

It’s less a feel for and just something you do to reduce to say, finite generation

[wooden ember]

the universal property is the main important part i suppose?

[next obsidian]

Yes, they are, but even moreso I use it mostly to reduce to FG

Cuz anything is the direct limit of its finitely generated submodules

[wooden ember]

do you have an example

[next obsidian]

This is literally saying

A module is the union of finitely generated submodules

So now if you want to show a module is flat let’s say

[wooden ember]

modulo the equivalence relation

[next obsidian]

No

Like

The fact it’s the union of its finitely generated submodules

Is how you’d show it satisfies the universal property for the direct limit

[wooden ember]

but how do you show it's the union of its finitely generated submodules

that sounds mad

[next obsidian]

Umm…

Let x in M

Then x in <x>

That’s finitely generated

[wooden ember]

yeah okay lmao

[next obsidian]

Hahah

[wooden ember]

i just overcomplicated things

[next obsidian]

Anyway so

Direct limits are exact and commute with tensor products

So to show M is flat

You only need to show it preserves injectivity for finitely generated modules

[wooden ember]

im not sure what exactness means or what a flat module is

[next obsidian]

Okay

All it means here is

Suppose you have a system of maps

M_i -> N_i

Which are injective / surjective

Then let M and N be the direct limits of both sides

By the universal property you get a map

M -> N

If all the original maps of the M_i and N_i were injective / surjective

Then the map M -> N is injective / surjective

So sometimes you want to know a map is injective say

And you can write M and N as the direct limit of some stuff

Then you can reduce to showing each of the M_i -> N_i is injective

And this often lets you reduce to assuming all the modules are say finitely generated

And then your proof becomes easier / actually fucking possible

[wooden ember]

hard to see when youd need this but alright sure

[next obsidian]

It will pop up

[wooden ember]

i guess ill know it when i see it

[next obsidian]

Another example is that any ring is the union of its finitely generated Z-subalgebras

This literally means

Take a finite set of element r1,…,rn

Take all polynomials in them with coefficients in Z

The upshot here is those sub rings are Noetherian

so again you can reduce to the Noetherian setting sometimes

And trust me

The Noetherian setting is 1000000000% better

[wooden ember]

hmm alright ill just trust you

[next obsidian]

It’s something you just have to see in practice

It’s usually just a sort of technical tool

[wooden ember]

yeah i get that

[next obsidian]

That you use to take a hard problem and make it easier

[wooden ember]

can i not use it to construct interesting new groups?

(groups/rings/modules whatever)

[next obsidian]

I mean

So yes

You can construct S_infinity this way

I believe

As a direct limit of all S_n or something

But it’s usually hard to get your hands on an explicit description of a direct limit without more info because the explicit construction kind of sucks

And verifying you have a direct system is kinda cringe

[wooden ember]

fair enough

[next obsidian]

However direct limits are used to construct the stalks of sheaves

Which are very very useful

But a bit further down the line

[wooden ember]

and i already have a construction of S_infinity anyways

[next obsidian]

Or maybe just never down the line

[wooden ember]

or at least i saw one i dont remember how it goes at all

[next obsidian]

Think of a direct limit as like

A generalized union

Lol

[wooden ember]

this particular example captures the idea of a ring of all ways that functions that can be continuous at p?

that's kind of cool

[wooden ember]

i thought that was the way we were meant ot think about coproducts

[next obsidian]

Well so

You’re right!

Those are like

[wooden ember]

category theory really makes me wanna oiwejfoiaf

[next obsidian]

A discreet union

As in like they don’t overlap at all

But a direct limit identifies some elements via that equivalence relation right?

[wooden ember]

fair enough

[next obsidian]

Think about the “any module is the direct limit of fg submodules”

If we did the coproduct it’s like we’re ignoring the fact some of them intersect

[wooden ember]

you get rid of all repetitions yeah

[next obsidian]

This is the stalk of a sheaf

[wooden ember]

oh lmao

[next obsidian]

But yes this is why these are important

If you just buy into the sheaf = some sets of functions we care about

The stalk = direct limit

Captures the behavior of those functions locally at that point

Like as a limit of all the nbds of p just like shrinking and shrinking and shrinking

So it says that f a function on U and g a function of V are equal in this stalk iff there’s a nbd of p, call it W, with W < U\cap V

Such that f|W = g|W

[wooden ember]

what's an nbd?

[next obsidian]

Open neighborhood

[wooden ember]

ah right

[next obsidian]

So f and g are equal in the stalk if they’re “eventually the same”

Where eventually means we just zoom in closer around p

That’s what W is doing

[wooden ember]

so like every taylor polynomial gets identified with a corresponding analytic function?

[next obsidian]

I don’t know

Lol

My point is like

If you’re looking at like derivatives

At a point say 0

All you care about is behavior arbitrarily close to 0

So in that sense x^2 and some function that’s like

x^2 from (-0.000000001,0.0000001)

But then like idfk, e^x outside that

If all you care about is the derivative at 0

Then x^2 and this other function might as well be the same

[wooden ember]

right

and yet x^2 and the constant 0 function have the same derivative at 0 but arent identified

[next obsidian]

Well I never said this is like bijective

[wooden ember]

fair

feels a bit random though, like when do you ever come across a function that's defined differently in all small enough nbds of a point than outside of those

it's not really as much of a local property since there's some finite radius for which the function is what you want within an nbd of that radius

so idk it doesnt really feel as much like a limiting process?

im not sure what im saying tbf

[next obsidian]

For complex analytic functions germs determine the entire function

This is the identity principle

Note that the germ at p is strictly more information than the value at p

x^2 and constant 0 are not identified as germs at 0

Even though their value at 0 is the same

[wooden ember]

right

[next obsidian]

You want to have basically as local information at a point p which still captures behavior on open sets around p

[wooden ember]

but isnt it more work to determine a germ at a point than just the function straight away?

given the direct limit construction?

[next obsidian]

Because germs are more local and when we only care about local info that’s what we want

[wooden ember]

cause sure determining f on a small open set will give us all of f but it feels like germs are more complex than that

[next obsidian]

They are which is why we want them

[wooden ember]

fair enough

[lethal cipher]

Hmmm, so having a bit of trouble with this. I need to show that the intersection of two normal subgroups is normal.

[chilly ocean]

what is troubling you

[lethal cipher]

Well, we know that for any x in G, there exists, h,h' s.t. xh=h'x. However, what if this h' is not in HnK

Sorry, H and K are normal subgroups

We need to show HnK is also normal

[viscid pewter]

let b be in both H&K

[chilly ocean]

the $\forall g\in G, g(H\cap K)g^{-1} \subset H\cap K$ definition of normality might be easier to apply

[cloud walrus]

TTerra

[lethal cipher]

Speaking of, what would be a situation where those set is not equal?

[viscid pewter]

which sets?

[chilly ocean]

if that inclusion holds for all g then it's an equality for all g

[lethal cipher]

xHx^-1 and H

[chilly ocean]

when H is not normal, i suppose

[lethal cipher]

Lol, alright. My book says H of G is normal iff xHx^-1 is a subset of H for all x in G

Anyways, back to the problem at hand

[viscid pewter]

there's a: G -> A such that ker(a) = H
and b: G -> B such that ker(b) = K

let c: G -> AxB such that c(g) = (a(g), b(g))?

i think ker(c) will be H&K

[lethal cipher]

Kernels are discussed in the next chapter

[viscid pewter]

what the fuck

why would you do normal subgroups before kernels of homomorphisms

[chilly ocean]

let $x\in G$ be arbitrary. to show that $x(H\cap K)x^{-1} \subset H\cap K$, let $h \in H \cap K$. then $h \in H$ and $h \in K$. what does normality tell you?

[cloud walrus]

TTerra

[chilly ocean]

that should start it off

[viscid pewter]

tterra, does my proof work?

i don't think i've seen it before and it's v. elegant

[lethal cipher]

It tells us there exists an h' in H and an k' in K so that xh=h'x and xh=k'x.

[chilly ocean]

we're thinking of different definitions of normality

this will get confusing. let me think about yours

[lethal cipher]

Well, if you want, xHx^-1 is a subset of H and xKx^-1 is a subset of K

[chilly ocean]

yeah it's a nice one. "element-free" proofs are good

[chilly ocean]

so xhx^-1 is in H and in K

[lethal cipher]

Yes indeed

[chilly ocean]

so we're done

kaisheng's proof is the best here though

[lethal cipher]

What you are talking about here is a theorem relating the definition with this in my book.

Of course, they are equivalent notions, but that's why I didn't think of that right away

[chilly ocean]

actually now that i look at it

you can just $$x(H\cap K) = xH \cap xK = Hx \cap Kx = (H \cap K)x$$ or $$x(H\cap K)x^{-1} = xHx^{-1} \cap xKx^{-1} \subset H \cap K$$

[cloud walrus]

TTerra

[chilly ocean]

then you dont have to think about elements in H or K lol

(these set manipulations work because left/right multiplication and conjugation by x are bijections)

[lethal cipher]

Tbh, I really liked the element approach. Normally I prefer to stay away from them, but this version was very simple

Moreover, it's a very natural way to do this with an arbitrary number of sets

[chilly ocean]

[next obsidian]

yeh

well

is H a subgroup?

if so, then yes

[oak grove]

so this is pretty easy with restrictions on the groups

is this true without any restrictions at all?

im guessing yes just maybe grasping

i guess its just free

theres not a whole lot to prove its just machinery

nvm

[lethal cipher]

I assume ≈ means isomorphic?

If so, you have a pretty good guess at an isomorphism. $\phi((g_1,h_1))=(g_2,h_2)$.

[cloud walrus]

dackid

[oak grove]

i think its just definition putting

like its the same sort of thing as the direct product of groups is a group

it looked more interesting than it was 😄

[lethal cipher]

Yea, this shouldn't be too difficult.

Direct Products are pretty nice to work with

[oak grove]

they are honestly

i wish wed done them before FIT since its gives you limitless examples to practice with

[lethal cipher]

FIT?

[oak grove]

first isomorphism theorem

[visual wolf]

first isom theorem

[lethal cipher]

Oh, I dont think I have heard of that

I now see why

[oak grove]

its very pretty 😄

[lethal cipher]

We just started talking about kernels

[oak grove]

my teacher in class said

she didnt know why they chose the word kernel

and she couldnt think of any thing it relates to in other fields

but doing products before fit seems right

[lethal cipher]

Ah, kernel means seed

[oak grove]

products, fit, tfa

i guess you could say tafg

[lethal cipher]

Gotta admit, I am surprised I got an answer with my search xp

[oak grove]

oh its just funny since like

i mean its kernel in lin alg right

[lethal cipher]

Yep

[oak grove]

i thought she was joking but then no one in class noticed or laughed or anything

and she didnt seem like she was joking

it seems like theres so many reasons thats the name

but someone asked and shes like "i have no idea" and theyre like "oh weird"

[lethal cipher]

Well, linear algebra is just a subset of abstract algebra

[oak grove]

i was thinking of that too, yea

like if linear came first or

algebra strengthened linear

or theyre just the same

im guessing some forms of linear came first

[lethal cipher]

Well, it is definitely a mix in between.

To this day, more things are being developed to LA and Abstract Algebra

[oak grove]

it would be excited to go back and do linear algebra now

with more language

[lethal cipher]

And since LA has groups we know so much about, we can learn a lot from them

[oak grove]

la gives groups?

like invertible matrices of some size or something?

i havent really done lin alg other than a really intro course

[lethal cipher]

The group of invertible matrices of size n under matrix multiplication

And the set of matrices of size n under matrix addition/or vector multiplication

Iirc, the combination of matrix addition and vector multiplication gives a ring.

[oak grove]

oh hrm

we are getting to rings later

and skipping galois

[lethal cipher]

Don't worry too much about it now. At its basic, rings are like groups with two operations isntead of 1.

I say 'like' because the multiplication operation removes some assumptions

[oak grove]

oh

we were talking about rings

[lethal cipher]

Sorry, the two operations are called addition and multiplication

[oak grove]

in the context of some game though

at math club

[lethal cipher]

Ah, does not surprise me tbh

[oak grove]

this game

https://awm-math.org/publications/playing-cards/evenquads/

the rules are really convoluted but the natural question was like

well when you play how many cards have to be on the table before a set exists, and then why the hell it seems like no matter which you recognize and which you dont, all cards can be setted at the end

[lethal cipher]

Hmm, interesting

[oak grove]

idk she cited an older game, i guess there are a lot of games like this

idk ill stop babbling

[lethal cipher]

You're fine Jan, I'm just distracted rn

[barren sierra]

I am having major trouble showing that this is surjective

I have shown that this is a homomorphism and that it is injective

but surjectivity got me down bad

original question

[clever dust]

Isn't an injective ring homomorphism an isomorphism onto its range?

so that $\phi(Frac(R))$ is the desired subfield of $F$

[cloud walrus]

IlIIllIIIlllIIIIllll

[past temple]

@barren sierra

define K in this way

it shouldn't be hard to prove that this map is an isomorphism

given any element in K

sorry wait K is generated by the elements in the set in my picture

[barren sierra]

I mean K is how I defined F' essentially

[barren sierra]

but don't I have to prove that my defined F' is the whole range?

which is the same problem just different words

[clever dust]

@barren sierra $F$ is a field extension of $R$, so we may as well assume that $R \subset F$.

[cloud walrus]

IlIIllIIIlllIIIIllll

[barren sierra]

yea

[clever dust]

your function $\phi$ is an injective ring homomorphism from $Frac(R)$ to $F$. So (I think so, but I have to verify) $\phi(Frac(R))$ is a subring of $F$ and $\phi$ is a ring isomorphism from $Frac(R)$ to $\phi(Frac(R)) \subset F$. Since $Frac(R)$ is a field, it follows that $\phi(Frac(R))$ is also a field, so you are done.

[barren sierra]

Hm

[cloud walrus]

IlIIllIIIlllIIIIllll

[next obsidian]

This is exactly the proof

Frac(R) embeds into F’ by universal property of Frac(R) but then this is a subfield of F containing R so F’ by minimalist is contained in it

[next obsidian]

@barren sierra have you managed to resolve your confusion with the proof?

[barren sierra]

yea

thanks

[next obsidian]

[wind sphinx]

any recommended books for some self-learning?

[next obsidian]

Hartshorne Algebraic Geometry

Just kiddingggff

I like Aluffi

[wind sphinx]

https://bookstore.ams.org/gsm-104/ this badboy?

[next obsidian]

Yuh

Although it’s a little nonstandard due to category theory being a bit more central

The standard one is probably dummit and Foote

[wind sphinx]

dropped Reals since it wasn't going well going to spend a little time with this book before Algebra next semester

[sturdy marsh]

aluffi + d&f

[next obsidian]

Or Artin

There’s a lot of books

[wind sphinx]

I figured the wizards in here were a good place to ask for reasonable undergrad level books

dummit and Foote cost $130 yeah this is probably the book most schools use if that's the price

[sturdy marsh]

l*bgen

[wind sphinx]

oh trust I'm not paying for a book for selfstudy

I just want to graduate so I want to spend my 30 hours of free time now that I'm sans-Real Analysis doing something to set me up for success next semester

[sturdy marsh]

rain destroyed my copy of aluffi

[wind sphinx]

Real talk bruh, fuck rain

[sturdy marsh]

the day I bought it

[wind sphinx]

Miss me with that shit

Rain is just the lame version of snow

[next obsidian]

Fucking rip dude

[gray scroll]

if i show that a finite extension of a perfect field is a separable extension, is it enough to say that it is perfect?

[next obsidian]

Umm

Are you trying to show that a finite extension is separable??

Or are you trying to show the field is perfect?

Because it sounds like you’re already assuming the field is perfect

[gray scroll]

i need to show that an algebraic extension of perfect field is perfect

[next obsidian]

Ah well

An algebraic extension of an algebraic extension is algebraic over the bottom thing

So like if K/L is algebraic, L perfect

You have some algebraic F/K then F/L is algebraic

So separable

Then that shows F/K is separable

Because well, if it weren’t then F/L has no chance of being separable

So I guess your question is

“If every finite extension of K is separable does that mean K is perfect”

And yes this is enough

[gray scroll]

yes

oh

thanks a lot

[regal carbon]

I am currently practising proofs and logic which are prerequisites of abstract algebra. I am referring to a popular book "mathematical proof: a transition to advanced mathematics"
Per a single topic, there are 15 practice problems and 15 examples I think are saturated and time-consuming. Of course, I have got less time in my hands.

Since I'm preparing for exam.

The quantified statements consists of more than 15 examples and problems.

How much are they required as far as my preperation is concerned?

[thorn flint]

Can someone help me with the process for using the euclidean algorithm to determine gcd(3+i,1-i) in Z[i]?

3+1 = (1-i)(-1) + 2
1-i = 2(??) + ?

I am stuck at the second part

[chilly ocean]

2= (1-i)(1+i)

[thorn flint]

yea but shouldn't I be dividing 1-i by 2?
$$\frac{1-i}{2} $$

[cloud walrus]

Mac

[chilly ocean]

well in the euclidean algorithm you want the remainder to be smaller than (1-i), you didnt factor correctly at first step

[thorn flint]

Oops

$$\frac{3+i}{1-i} = 1+2i $$

[cloud walrus]

Mac

[thorn flint]

But how should I write that in the euclidean algorithm?
3+i = (1-i)(1+2i) - 1?

[willow mason]

3+ i = (1-i)(1+2i) - i

[cloud walrus]

██▓▒░⡷⠂Braham Gödel⠐⢾░▒▓██

██▓▒░⡷⠂Braham Gödel⠐⢾░▒▓██

[last cargo]

Is it right?

I mean, I know the class will be a pair like (x, 1) for a given y, each y will have a class

But I'm not sure how to write it formally.

[hidden haven]

I am supposed to find all the intermediate extensions of the splitting field of x^4+3 over Q that are Galois over Q. I have an answer but I am pretty sure it is wrong and I can't find the mistake. So here is the sequence of claims I have in my answer:

• Let 4th root of -3 be c. Then the roots of the above polynomial are some power of i times c.
• The splitting field is then Q(c,i).
• This is a degree 8 extension of Q because i is not in Q(c).
• This is generated by the 2 automorphisms:
  r which maps c → c, i → -i
  s which maps c → ic, i → i
• These automorphisms satisfy r^2, s^4, rsrs = id, and by a cardinality argument the galois group is D_4. The normal subgroups of that are <s^2>, <r, s^2>, <rs^2>.
• ic is fixed by rs^2, thus Q(ic) is fixed by <rs^2>, and since this is degree 2 under the splitting field, this must be the fixed field.

This last result is wrong because Q(ic)/Q is not normal (it doesn't contain c which is a conjugate of ic). I am pretty sure the mistake is somewhere near the end but I am just not seeing it

Pls just tell me I am doing the last step wrong and ic is somehow not fixed by rs^2 I don't wanna redo the whole thing

[rustic crown]

btw one thing before going forward.... who names s the rotation and r the swap??

okie, so i think the normal subgroups you claim are normal are false

[hidden haven]

is the convention the opposite one

[rustic crown]

<s^2> is the center
<s>, <s^2, r> and <s^2, rs> are subgroups of index 2

[rustic crown]

yea lol, r should be rotation and s should be swap 😛

[hidden haven]

damn I see

[rustic crown]

isn't rs^2 just another swap?

like any r*(s^i) is a swap

and swaps don't commute with rotations

[hidden haven]

oh shit I see

ok so its not too far up, neat

[rustic crown]

[hidden haven]

thank 😌

Ok what am I doing wrong now lol
Conjugating s by r we have
rsr = s^-1
So <s> can't be normal
But it has index 2 so it should be?

What is going on have I forgotten group theory

[rustic crown]

normal doesn't mean it commutes lol it just means that r<s>r = <s>

[hidden haven]

Fuck

I was thinking <s> = {e, s}

[rustic crown]

need to extend ag deadline lol

[hidden haven]

Pls mail him

[rustic crown]

yea lemme watch lecture 10 before mailing

[hidden haven]

Mapping cones 😌

[rustic crown]

another three hours 😶

[molten silo]

Any one know what the set X is?

[chilly ocean]

triples (a,b,c) where a,b,c are 0 or 1

[molten silo]

i see

[chilly ocean]

Can someone help me understand this

Suppose A and B are rings and f is a ring morphism

Then B is a module over A

Suppose now M is another ring

And there is a ring morphism from B to M

M gets the structure of an A module

From the composition of maps

[chilly ocean]

And M gets the structure of a B module from this map

What I want to know is

Is the M as a B module

Just the same as

$M \otimes_A B$

[cloud walrus]

lime_soup

[hot lake]

I don't think so

[chilly ocean]

Damn

What I am really trying to understand

Is in this situation I want to know that the support of M as an A module is comtained in the support of M as a B module

But I’m not sure how to see this

[hidden haven]

Can't you say that M ⊗ᴀ B ≅ M because
M x B → M defined by
(m,b) ↦ bm
Is A-linear so induces a map from the tensor to M
And M → M ⊗ᴀ B given by
m ↦ m ⊗ 1
Is a 2 sided inverse to this

[rustic crown]

these aren't inverses

consider A = R, B = M = C

M tensor B will be dim 4 real vector space

[hidden haven]

This doesn't seem semantically correct because the support of M as an A module is a subset of spec A while the other is a subset of spec B

[hidden haven]

Right, I see

[rustic crown]

i don't wanna hear Spec

[hidden haven]

Is that where this goes wrong

Or like is one of the maps not defined or something

[rustic crown]

yea just that

m --> m tensor 1 --> m is correct

but reverse isn't

[hidden haven]

ah

[rustic crown]

m tensor b --> bm --> bm tensor 1

[hidden haven]

Since you are only able to move around scalars of A

[bitter mauve]

how did u get these symbols

[rustic crown]

you can only juggle around stuff from A as it's tensor prouct over A

yea right

[chilly ocean]

okay yes but i think there is a way

just take the zerosets or something

[hidden haven]

google gboard latex dictionary (if you are using gboard on Android)

[bitter mauve]

thx

[chilly ocean]

sorry det this is the same question you answered the other day but im still confused

(just saying this in case you thought ignored what you said the other day)

[rustic crown]

lol i didn't mind that... i didn't prolly understand it well enough

moldi's emoji makes me more sad lol

[hidden haven]

Ignore det

[chilly ocean]

Okay just to double check

A to B to M maps of rings

B is an A module

M is a B module

M is an A module

the following is not true? : M as a B-module is M ⊗ᴀ B (where M is taken as an A module in this tensor product)

[hidden haven]

Nope

[chilly ocean]

chungus rip

ahh

but the following is true

if B is a subring of A

then the support of M as an A module is contained in the support of M as a B module

and this does make sense because now the prime ideals are living in the right place

[chilly ocean]

the ideals in a sub ring

are just the ideals that are contained in the subring

[rustic crown]

sorry kinda tired, have to watch a lecture...

[chilly ocean]

does this mean im wrong

[hot lake]

yes

[chilly ocean]

okay

was "then the support of M as an A module is contained in the support of M as a B module" correct for B a subring of A

[hot lake]

that doesn't make sense if you don't relate Spec B with Spec A

but if B is a subring of A then you have a ring map from B to A so this induces a map from Spec A to Spec B

which should be intersecting an ideal of A with B ?

[chilly ocean]

okay

[chilly ocean]

ill go try think about this map

[simple mulch]

hey guys, I should show the identity element of the group of permutations S_n is (1 1, 2 2, 3 3, ... ..., n n)

the identity * sigma (where sigma is an arbitrary permutation) yields the identity

However I am having trouble with sigma * identity because we have 1 -> 1 and 1 -> sigma(1) (where sigma(1) is the first entry for the arbitrary permutation sigma)

[chilly ocean]

you mean yields sigma?

[simple mulch]

oh wait, I am doing something wrong

what I mean in here is identity = inverse of sigma

oh lol

nvm

I get it

[obsidian sleet]

hello friends

thanks to you all i got a very high score on my algebra midterm

well sort of i know i didnt ask too many questions here but

the help is appreciated and im sure when i studied my homeworks again you all came back in that manner

[chilly ocean]

[south patrol]

Gg

[sturdy marsh]

you are now the cube @obsidian sleet

[obsidian sleet]

i am now the cube.

🧊

[sturdy marsh]

🧊

[obsidian sleet]

it has all been coming to this.

[untold cloud]

Hi guys, is there any clue to show that at least one element of {-1,2,-2} is a square in Z/pZ, for all prime p?

[wooden ember]

The multiplicative group of Z/pZ is isomorphic to Z/(p-1)Z. Considering some isomorphism between the two, how does being a square in the multiplicative group of Z/pZ translate to a property in Z/(p-1)Z? @untold cloud

You can either go down that path or note that x^(2m+1) * x^(2n+1) = x^2(m+n+1): both are equivalent but mine frames it more in terms of morphisms so I find it kind of neat

[untold cloud]

Thank you, but i do not have clue how to go down the isomorphism path. f is the isomorphism, f(a^2) = f(a)^2. And what should i do next?

[sharp sonnet]

Z/(p-1)Z is an additive group (and the isomorphism is from Z/pZ^* to Z/(p-1)Z)

[untold cloud]

Oh, Thank you. I solved it by considering f:Z/pZ --> Z/pZ, x-->x^2, and then the image of Z/pZ{0} --->Z/pZ{0} is a multiplicative subgroup of index 2, then if two of {-1,2,-2} is not a square, then the product is a square

[next obsidian]

Mf said he needs to get good but he already got good

[prisma thunder]

Suppose $F \subset L \subset L(\alpha)$ and that $[F(\alpha) : F]$ and $[L : F]$ are relatively prime. Prove that the minimal polynomial of $\alpha$ over $L$ is in $F[x]$.

[cloud walrus]

eM

[prisma thunder]

Should that be F(alpha) in that towers of extension?

I'm wondering if that is a typo.

In addition, I'm not quite sure how to approach this problem.

[small bison]

at a glance yeah seems like it should be $F \subseteq F(\alpha) \subseteq L$

[cloud walrus]

young_smasher

[small bison]

actualyl that doesn't make sense

[next obsidian]

I think it’s subset L(alpha)

If you want to make sense of that chain

[small bison]

maybe it should be $[L(\alpha) : F]$ instead of $F(\alpha)$? idk

[cloud walrus]

young_smasher

[next obsidian]

I think it’s F(alpha)

Give me a second

[small bison]

hmm maybe there's no type

typo*

i think consider the diagram

% https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZARgBoAmAXVJADcBDAGwFcYkQAxEAX1PU1z5CKAAyli1Ok1bsOACgA6CpmgAW9AJQ8+IDNjwEi5cZIYs2iEABlt-fUKJkRp6ReuLljNZp6SYUAHN4IlAAMwAnCABbJDEQHAgkcl4wyJjEOISkYhSQCOjsmizEAGZc-PTjeMTS7kpuIA
\begin{tikzcd}
& L(\alpha) & \
F(\alpha) \arrow[ru] & & L \arrow[lu] \
& F \arrow[ru] \arrow[lu] &
\end{tikzcd}

[cloud walrus]

young_smasher

[next obsidian]

So like

Here this shows that [L(alpha):F] >= mn by them being relatively prime

But I’m pretty sure explicitly you can show that it’s <= mn

By just like combining some polynomials

The upshot is that [L(alpha):L] = m

Which is the same as [F(alpha):F]

And this says that alpha’s min poly over F has the same degree as alpha’s min poly over L

[prisma thunder]

Oh so it's a diagram of extensions where one isn't contained in other.

[next obsidian]

But note that the min poly over F is also a poly over L

So that the min poly are the same

[prisma thunder]

Okay that makes more sense then.

[next obsidian]

The only thing I haven’t shown on here is that [L(alpha):F] <= mn

Like

Okay yes

This is easy

Because note that [L(alpha):L] <= m

Reason is

It’s the degree of alpha’s min poly over L

[small bison]

i think once you have the two towers of extensions it's easy to prove that [L(alpha) : L] = [F(alpha) : F]

[next obsidian]

But the degree of alpha’s min poly over F is degree m

And a poly over F

Is in particular a poly over L

So the min poly of alpha over L has degree <= m

So now we have [L(alpha):L] = [L(alpha):L][L:F] <= mn

But we also know it’s a multiple of m and n and since those are coprime it’s >= mn

[small bison]

this is a lil long winded

i don't think you need any inequalities

[next obsidian]

Idk ¯_(ツ)_/¯

The point is you want to show [F(alpha):F] = [L(alpha):L]

And I think you have to decompose the thing and use the fact it’s a multiple of coprime things to show it’s >= their product

[small bison]

hmm i think you can just do it directly

like from the diagram i drew it's like [F(alpha) : F][L(alpha) : F(alpha)] = [L(alpha : L][L : F] and i want to say that coprime just gives you things for free

[next obsidian]

Umm

[prisma thunder]

I feel like I did a group theory problem that resembles this but I can't remember it--using coprimality to deduce relationships/properties between groups.

[next obsidian]

I don’t think you drew that right

Or wrote that right

[next obsidian]

Like the LHS here

[next obsidian]

Maybe like order of GH is product of orders when coprime

[small bison]

i would just write out hte prime factorization lmao

and i think coprime should do the rest

[next obsidian]

But you need a way to relate [L(alpha):L] and [F(alpha):F]

And that is only ever going to be an inequality

[small bison]

my equation (which i fixed) should relate them

it's cause orders of extensions are multiplicative

[next obsidian]

No

Because you’re asserting something like

ab =cd with b,d coprime implies ab = bd

This isn’t true unless you know some more stuff

you need a relation between a and c

And in this specific case all you have is an equality between them

I mean if you can prove me wrong pls go ahead but I’m very doubtful

[small bison]

hmm maybe you're right then

i haven't put that much thought into

it

[next obsidian]

Like by coprime you know the value is a multiple of the products of the coprime things

But to get that it’s equal requires another argument

And I think this basically just boils down to an inequality since the only thing you can really get from this is that [L(alpha):L] <= [F(alpha):F]

[thorn delta]

ye

[acoustic turtle]

K thanks

[acoustic turtle]

How would I work this out

[kind temple]

try representing them as matrices

[acoustic turtle]

When i do that i get the following

@kind temple

[kind temple]

i don’t think you’re doing that the right way

[acoustic turtle]

Where are am i going wrong?

[kind temple]

do u mind explaining this?

i cannot follow what you’re doing

[acoustic turtle]

@kind temple

[long obsidian]

In the context of a group of order $p^2q$, p and q are prime, what does the notation $n_q=1$ usually denote? I believe it just means the exponent of q

[cloud walrus]

fajitas

[rustic crown]

if q doesn't divide p^2 - 1, then you can say that the number of sylow q subgroups is 1. n_q should be that.

[long obsidian]

Thank you!

[elfin patrol]

Trying to find an isomorphism from (F, +) to G where G={[1,x; 0,1] x in F} and I think it should just be mapping x to matrix [1,x;0,1]. Only problem I’m having is proving that phi(x+y)=phi(x)+phi(y)

It seems to fall apart as left hand is [1, x+y; 0, 1] and RHS is same but with 2 in place of 1s

I’m thinking I am using the wrong operation on the right side or something

[robust pollen]

@elfin patrol What is [1, x; 0, 1] * [1, y; 0, 1]?

[rustic crown]

operation on G is matrix multiplication and not addition

so you actually want to prove phi(x+y) = phi(x) * phi(y)

[lilac trench]

anyone here have professor nate

[long obsidian]

Can someone quickly summarize how to use the theory of finite abelian groups to classify groups?

[chilly ocean]

classification of groups

[long obsidian]

How does ya boy classify these finite abelian groups?

Is it literally just the different factorizations?

Like if I have a group of order $12=2^2 3$ then is $Z_2 + Z_2 + Z_3, Z_4 + Z_3, Z_2 + Z_6$?

[cloud walrus]

fajitas

[long obsidian]

Ahhh okay 😁

[upper pivot]

is the orbit that of conjugation?

well |orbit(x)| is a constant in a conjugacy class right

yes

so do you see how to do the sum now?

[next obsidian]

[upper pivot]

Yep

hmm use orbit-stabalizer i think

well depending on what you are summing over, use orbit-stabilizer to make 1/|C| nicer and then sum that

anyhow i gotta run for now

[long obsidian]

I see that a nilpotent group G can be defined by having it's upper central series terminate at G. Yet equivalently a nilpotent group may be defined by having it's lower central series terminate at the identity. How are these definitions equivalent?

[wild sapphire]

I'm trying to prove that if a polynomial over Q with nonzero constant and nonzero leading coefficient is irreducible, then so is the polynomial formed by reversing all the coefficients (i.e the constant value is now the leading coeff, the leading coeff becomes the constant, everything backwards).

I don't feel like it's intuitive at all

I was thinking of Eisenstiens Criterion, but that criterion is just an implication and not a "if and only if", so with the given setup I don't see how it could be used.

[next obsidian]

This is just an idea

I don’t know if you can turn this into a proof but

[wild sapphire]

I'm sure there's some fact stemming from irreducibility I need to work with

I'm listening, anything to get the brain going

[next obsidian]

If the poly is degree n the reversed poly is obtained by plugging in 1/x then multiplying by x^n

Maybe this is useful somehow?

Or an observation I suppose rather than an idea

[wild sapphire]

Hmm that does seem useful

it's really just 1/x there? I guess I'd have to work with it a bit