groups-rings-fields

[worthy haven]

how do you feel about axiom of choice -> LEM

[kind temple]

not even surprised atp

AoC might as well imply existence of god. it’s just that broken.

[south storm]

Why do you think it‘s "broken"

[chilly ocean]

needs nerf

[oblique river]

Patch 1.2.1
-- AoC was too strong before, so it now only applies to countable sets, thus functionally merging it with the Axiom of Countable Choice.
-- In light of the above, AoCC has been removed from mathematics for now to be reworked

[chilly ocean]

Patch 1.2.2
-- AoC has been removed, being replaced by Zorn's Lemma
-- Zermelo Theorem is now obviously false

[kind temple]

am happy now. thank you math mods

[south storm]

Patch 1.2.3
--recognized previous mistakes, added AoGC

[stark obsidian]

Hi, can someone give some hints/tips to proof that $\mathbb{Q} \not\cong \mathbb{Q}^2 $ as $\mathbb{Z}$-Modules

[cloud walrus]

Polux12

[oblique river]

do you know about tensor products?

I guess even without that, you could try to prove a more general statement

suppose V and W are two Q-vector spaces which are isomorphic as Z-modules. Then they are also isomorphic as Q-vector spaces

that's my hint :^)

[stark obsidian]

i will work on it Thank u!

[chilly ocean]

cute

Let $G$ abelian group of order $n$, and $n = \prod p_i^{n_i}$ prime factorization. Prove $G = \sum S(p_i)$

[cloud walrus]

Charlie++

[chilly ocean]

Does anyone know what S means in this context?

[gritty sparrow]

S(p) is probably the p torsion elements of G, but that is just a guess

[chilly ocean]

what does p torsion element mean

[gritty sparrow]

The elements g st p^i(g)=0 for some i

[unreal portal]

I'm a bit rusty on my commutative algebra. I generated the alexander ideal for a knot, which is an ideal in the ring of laurant polynomials Z[t, t^{-1}]. How can I tell if the ideal is principally generated (in sage preferably, but an abstract idea could work as well)? I took a class a year ago and know that Groebner bases are a thing, but my understanding is that the algorithm only works for Z[x1, ..., xn] with non-negative exponents because of how it uses polynomial division.

[amber stag]

Z[t,t^{-1}] is the same thing as Z[x,y] mod out by xy=1. Looks like this can be done in macaulay2
https://faculty.math.illinois.edu/Macaulay2/doc/Macaulay2-1.17/share/doc/Macaulay2/Macaulay2Doc/html/_simple_sp__Groebner_spbasis_spcomputations_spover_spvarious_springs.html

Actually I think there's another way to do it without using multivariate groebner basis. Let I be the ideal and J=I intersect Z[t]. Say J, as an ideal of Z[t], is generated by a_1,a_2, ..., a_n, then these polynomials also generate I. Then it remains to check if the number of generators can be reduced with t inverted.
Might be wrong tho since I didn't write down all details.

[maiden ocean]

Hm

How would i actually demonstrate that the sheaf of regular functions on an affine variety is a sheaf?

Namely how do i show gluing?

[fossil shuttle]

what is your definition of a regular function

my definition is that being regular is a local property, so it trivially satisfies the gluing property

because it can be checked on an open cover

[maiden ocean]

Uh an element of the intersection of the local rings

I think my worry was/is like picking a representation

[fossil shuttle]

ok, I see.

so it's a section of the bundle of germs?

[maiden ocean]

if we have like f1, ..., fn in O(U1), ..., O(Un) then at each point p in U can we literally just pick a representative f_p/g_p for some f_i and then define f by being f_p/g_p in each O_X, p?

[fossil shuttle]

Yeah, I'm not sure what context you're working in but like, if this is in the sense of chapter 1 of Hartshorne, regular functions are functions, in the usual set-theoretic sense

[maiden ocean]

Im seeing them as elements of the quotient of k[x1, ..., xn] by an ideal I

Where X = V(I)

[fossil shuttle]

Ok

Ok, I see what you're saying.

yeah then this is a commutative algebra exercise. first of all what does it mean for f_i/g_i and f_j/g_j to agree on U_i \cap U_j?

let's say for convenience that all of the open sets U_i we are talking about are of the form D(f) for various f. then later on we can reduce the general case to the case where the opens belong to the standard basis

[maiden ocean]

Yeah I think thats where my confusion lies, im not 100% sure what it means for these functions to agree

It means that theres like an f/g with f_i/g_i = f/g = f_j/g_j on U so like. f_i(p) g_j(p) - f_j(p) g_i(p) lies in (f_i cap f_j)...?

Actually that does not sound right

[fossil shuttle]

Note that if f, g are functions in k[x1....xn]/I, then D(f)\cap D(g) = D(fg)

does that sound right?

[maiden ocean]

Yes

[fossil shuttle]

We define the ring of functions over D(f) to be the localization (k[x1...xn]/I)_f

does that make sense? If you're excising the closed set where f vanishes, then f becomes invertible on the complement

[maiden ocean]

Uh im not 100% on what you mean by invertible on the complement here. does (k[x1, ..., xn]/I)_f mean like the functions g/f^n for some n?

[fossil shuttle]

yeah, I define this to mean ((k[x1....xn]/I) [z])/(zf-1)

in this ring f has an inverse, and this is the free-est possible way of adjoining an inverse to f

[maiden ocean]

I see

[fossil shuttle]

i'm saying that g/f^n is undefined exactly where f is zero

if you remove the set V(f) where f is zero, the function becomes defined everywhere on D(f)

[maiden ocean]

so the invertible functions are precisely f and powers of f

Okay yeah this makes sense

[fossil shuttle]

So there's an obvious map $R\to R_f$, this is called the localization map, it sends $x$ to $x/1$, and if $f\mid g$, say $f\cdot s= g$, then it's not hard to show that $(R_f)_s = R_g$

well, up to natural isomorphism, yadda yadda

so the relevance to this conversation is that

[maiden ocean]

Uh is that supposed to be x mapsto x/1

[fossil shuttle]

yep

lol

[maiden ocean]

haha makes sense

[cloud walrus]

diligentClerk

[maiden ocean]

this also makes sense i think

[fossil shuttle]

if $t_i \in \mathcal{O}(U_i) = D(f_i) = (k[x_1,\dots, x_n]/I){f_i}$, and if $t_j \in \mathcal{O}(U_j) = D(f_j) = (k[x_1,\dots, x_n]/I){f_j}$, then we say that $t_i,t_j$ agree on the overlap $U_i\cap U_j=D(f_if_j)$ if the images of two elements are the same via the homomorphisms

[cloud walrus]

diligentClerk

[fossil shuttle]

$(k[x_1,\dots, x_n]/I){f_i} \to (k[x_1,\dots, x_n]/I){f_if_j}$

and

[cloud walrus]

diligentClerk

[fossil shuttle]

$(k[x_1,\dots, x_n]/I){f_j} \to (k[x_1,\dots, x_n]/I){f_if_j}$

[maiden ocean]

(uh is this also supposed to be O(D(f_i)))

[cloud walrus]

diligentClerk

[fossil shuttle]

yeah sorry

[maiden ocean]

np

[fossil shuttle]

i meant $U_i = D(f_i)$

[cloud walrus]

diligentClerk

[fossil shuttle]

so that's a definition you can work with.

it's now a commutative algebra exercise to show that

if $R$ is a commutative ring, and if $\left{U_i \right}$ is a family of opens which cover the space, say all principal so that $U_i = R_{f_i}$ for some $f_i\in R$, and $t_i \in R_{f_i}$ such that $t_i=t_j$ when they're both mapped into $R_{f_if_j}$, then there's a unique element $t$ of $R$ whose restriction/localization to each $f_i$ is indeed $t_i$

[cloud walrus]

diligentClerk

[maiden ocean]

is the space here supposed to be Spec(A)

[fossil shuttle]

yeah.

[maiden ocean]

I see

[fossil shuttle]

broad hints are that Spec(A) is compact, so you should be able to take a finite cover wlog, this is also a commutative algebra exercise

[maiden ocean]

I have seen that yeah

i did like all the exercises in the first 3(?) chapters of AM im not totally incompetent at comm alg i promise

[fossil shuttle]

cool. i think this exercise is in Eisenbud's commutative algebra somewhere on. then, like I said, the D(f) only form a basis, so you have to show that proving this is a "sheaf on a basis" is enough to prove that it satisfies the sheaf condition with regards to arbitrary open subsets

Also this same argument works for modules over a ring to show that an $A$-module is a sheaf over $Spec(A)$, which is pretty cool

[cloud walrus]

diligentClerk

[maiden ocean]

Wait I have seen this actually

it is exercise 24 in ch 3 of AM

Ok i remember this now

[fossil shuttle]

nice!

[maiden ocean]

I think I was confused because ive only worked in the case where the varieties were defined w/ solutions to the polynomial rather than directly with Spec but I see now

if you think of it as A_f then this is the gluing of the structure sheaf

as you might expect because this is literally a special case of the structure sheaf 🤔

[fossil shuttle]

yeah, if you're curious you might want to try showing how this relates to the local rings at the primes. If $p\subset$ A is prime, then $A_p$ is the stalk of the structure sheaf at the point $p$, i.e. the filtered colimit of the $\mathcal{O}(D(f))$ over all $D(f)$ containing $p$. that gets you more closely in touch with the previous definition you were trying to work with, with the local rings

[cloud walrus]

diligentClerk

[maiden ocean]

yea i proved that exercise

[fossil shuttle]

nOICE

[maiden ocean]

you take the limit of the A_f for f notin p basically I think

[fossil shuttle]

Yeah, because the principal opens form a basis it suffices to consider them when forming the filtered colimit

[maiden ocean]

okay i see

thank you!

[maiden ocean]

Ok quick question:

if A is a dedekind domain finitely generated over a field k and L is a field extension of k such that A otimes L is integral, is A otimes L also dedekind?

[vestal snow]

What is a transition function?

What does it mean for U_1 and U_2 to be related by an element T_12 of GL_2? Why must they be related by such an element?

[hidden haven]

Do you have any hypotheses on M other than it being a topological space? If f_i are the trivializing homeomorphisms from inverse image of U_i to U_i x F_i, then you can complete the triangle made by these 2 restricted to the intersection, by putting a map f_2 ∘ f_1^{-1} in between. I'm guessing that's what T_12 is, and this would satisfy the cocycle condition that's given, but without some associated algebraic structure you won't get linearity

[fossil shuttle]

This is a slightly dense way of wording it imo. If you want to see a good presentation of fiber bundles I recommend Steenrod's Topology of Fiber Bundles

I think it's fine here for M to be an arbitrary topological space, it's just that the fibers need to be vector spaces. The linearity can be expressed that way

[hidden haven]

The book's wording, or mine?

[fossil shuttle]

the book's

why would you say "an element of GL_n with entries consisting of continuous functions"

just say a continuous map into GL_n

[stark obsidian]

Hi, can someone help me to prove that (N:P) =Ann((N+P)/N)? $N,P\leq M$ where $M$ is an R-module

[cloud walrus]

Polux12

[gritty sparrow]

Well if x is in the LHS, then xP is a subset of N, or concretely for any p in P, xp is in N. So now consider an element n+p+N in N+P/N. multiplying by x we get nx+px+N. But both nx and px are in N so this is the same as the coset 0+N or the zero element of N+P/P, so x is in the annihilator. Can you try the other direction now?

[stark obsidian]

Let $x\in\operatorname{Ann}((N+P)/N)$ and $p\in P$. Then $p+N\in(N+P)/N$, so $xp+N=x(p+N)=0+N$ by our choice of $x$, which means that $xp\in N$. Since this is true for every $p\in P$, we get $xP\subseteq N$, so $x\in (N:P)$

[cloud walrus]

Polux12

[stark obsidian]

I'm sorry for the delay, but it was a bit difficult for me

u think its rigth?

[gritty sparrow]

Yep

[eager yew]

Hello, pls what defines a subring

[chilly ocean]

it's a subset of a ring that's also a ring itself with the operations restricted to that subset (and, of course, also contains the 0 and 1 of the ring)

[eager yew]

How can I prove for one

[chilly ocean]

by checking that definition

there's also probably some kind of funny "subring test" you could use

[ivory dust]

Would the solutions to cubic function fall under this channel? its a ring/field theory course

[eager yew]

Ok

[unique juniper]

so i understood that our d embeddings will extend to n/d things so we get a total of n embeddings as expected

but

not why they are multiplying by n/d, $T(\alpha) = \frac{n}{d}t(\alpha)$

[cloud walrus]

Yes ツ

[upper pivot]

oh hey number fields nice

[unique juniper]

lol

[upper pivot]

so the idea is that theres d conjugates of alpha, but n total embeddings

[chilly ocean]

nuts/deez

[upper pivot]

so you will repeat each conjugate n/d times

[unique juniper]

oh

but there would be n distinct embeddings if we unrestricted their domain?

[upper pivot]

right for the whole field theres n different ones, but if we restricted to the subfield Q[alpha] theres d

[unique juniper]

yep ok, that makes sense

[upper pivot]

think of an example, like take the embeddings of Q[(2)^(1/4)] and sqrt(2) in it

also we are doing a number fields reading group in voice chat semi-regularly, if you are interested

[unique juniper]

oh nice

how far are you guys in?

[upper pivot]

we finished reading ch2

will probably work on exercises for a bit

[unique juniper]

oh im not that behind

[upper pivot]

yeah you could join in, ill ping u next time we meet

[unique juniper]

sure, ty

[humble surge]

Is $\mathbb{Z}_n\times\mathbb{Z}_m\cong\mathbb{Z}_m\times\mathbb{Z}_n$ ?

[cloud walrus]

modus ponens

[humble surge]

In general, are all direct products isomorphic to a reordering of their factors?

[upper inlet]

yes

can you construct the isomorphism?

[humble surge]

Yea, I did the proof for a product of two. The obvious isomorphism that sends (a,b) to (b,a).

[upper inlet]

you can get the general direct product by proving the associative property

oh i assumed they meant finite

sorry should have been more specific with 'general'

[stark obsidian]

Let $R=\mathbb{Z}[x]$ and $M=(2,x)$ the ideal generated by 2 and $x$ seing as R-module

Why wouldn't it {2,x} be a base of $M$?

[cloud walrus]

Polux12

Polux12

[stark obsidian]

I really believed that it was

[vast quiver]

so what is a base/basis? what needs to be true for a subset to be a basis?

[stark obsidian]

generate the set

[vast quiver]

so if it isn’t a generating set, then it must be not generating some element

so you need to find an element that is not generated by {2, x}

do you know how to describe the set (2, x) in set builder notation?

like for example in the maybe more familiar situation of a vector space R^n, the span of some vectors is all linear combos

[stark obsidian]

this is what really confuses me

[gritty sparrow]

But M is generated by {2,x} over R, that is pretty much the definition

Ofcourse we don’t call a generating set for a module a basis, but the issue won’t be wether it generates M right?

[vast quiver]

yeah you can’t even generate M though

oh oops

my b i thought M was R. yeah M=(2,x), so by definition it is generated by {2,x}. you are right, i do be goofing.

but yeah, to show {2,x} is not a basis for M, you could also show it is not linearly independent

so if you wanna show {2, x} is not linearly independent, you need to find a Z[x] combination of 2, x that sums to 0

for example x^3*2 + 7*x would be a Z[x] combination of 2, x, but this one isn’t 0

[stark obsidian]

if i am understanding then 4+2x would work?

[vast quiver]

we're trying to do a combination with non-zero scalars that ends up getting zero

so when you say 4+2x, do you mean 2*2 + 2*x? that would be a Z[x] linear combination of {2,x}, but you don't get zero

this is really similar to linear algebra

for example, in R^1, is it possible to have 2 linearly independent vectors?

[stark obsidian]

nope

[vast quiver]

so tell me, are 2 and 4 as real numbers R-linearly independent?

no, right?

so if they aren't linearly independent, then that means that there is a way to combine them to 0

for example you can write 4*2 + (-2)*4=0, which is one way to prove that 2 and 4 as real number are not linearly independent over R

you can do a really similar thing in M=(2,x) as a Z[x]-module

you want to show that {2,x} are not linearly independent, so you want to write a Z[x]-linear combination to get them to add to 0

[stark obsidian]

I understand, thank you very much!

[vast quiver]

sick!

if you want a cute problem that does the same trick, you could try this bad boy out

assume R is an integral domain, and that every ideal J of R is a free R-module (i.e. J has a basis over R). Prove that R is a PID

[bronze jay]

this is from Dummit & Foote, why is dot not a binary operation here?

isn't $g \cdot a $ a binary operation?

[chilly ocean]

binary operation would be like G x G -> G or A x A -> A, but a group action is G x A -> A

i think

[bronze jay]

hmm, how is g dot a defined even?

[delicate bloom]

yeah I think so too, I believe they call it an 'external binary operation' if they're from different sets but I never see this terminology used so not sure

[chilly ocean]

it depends on the context, since it's just notation to represent a group action (defined in your picture)

g dot a is whatever (g, a) gets sent to by the group action map

[bronze jay]

how can we be sure that g dot a is in A? g and a are two distinct elements from two different sets?

[chilly ocean]

by definition

[delicate bloom]

maybe skip ahead to the part where they give examples

like, common example is take a set A = {a,b,c} and then g might act on the elements by switching them

[bronze jay]

so g is a permutation

[delicate bloom]

yeah in this example

and I guess by cayley's theorem you could see all of them that way

but might be like a rotation of a shape or something

[bronze jay]

is sigma_g a group action here?

[delicate bloom]

idk didn't read

[chilly ocean]

sigma_g is specifically the action of the individual element g of the group

the actual group action is the map G x A -> A

[bronze jay]

in the next page they write this

but isn't that basically the same as what they've wrote above?

sigma_g(a)=g dot a=phi(g)(a)

[chilly ocean]

these are all different ways of looking at the same thing, if that's what you're asking

[chilly radish]

These are equivalent ways to define a group action

[wooden ember]

here they say the roots are distinct, but i was stuck on the exercise for a while thinking that might not be necessarily true: how would one prove the roots to be distinct?

[delicate bloom]

suppose $x_i =x_j$, then raise both sides to the ij power and you get $a^j=a^i$. Multiply by the inverse on one side to get $e=a^{j-i}$ but since $i \ne j$ and $a \ne e$, contradiction

[cloud walrus]

Merosity

[delicate bloom]

actually that doesn't work does it, I think you need to put some kind of relatively prime condition on i,j

[nova plank]

This seems really obvious but I have forgotten everything from group theory apparently

[delicate bloom]

there's no reason a^{j-i}=e can't hold true if j-i is a multiple of the order of a

[nova plank]

I guess it's not the case that they are always distinct, but that you can choose them to be distinct.

I think a nicer proof is just: ||a has no nth root or that would imply a is the identity||

[wooden ember]

because the way i did it was i just took an infinitely countable amount of "distinct" roots but i couldnt figure out if they had to be distinct, exactly for the reason mero exhibited

[chilly radish]

How do you prove uniqueness in the fundamental theorem of finitely generated abelian groups?

We only proved existence in my course

[hidden haven]

The group uniquely breaks into a torsion part and a free part. For the torsion part, use Chinese remainder theorem to uniquely split it into a direct sum of p-groups and prove uniqueness of the part associated to each p separately

[chilly ocean]

pee group

[hidden haven]

[bronze jay]

must a homomorphism between groups map the identity of one group to the identity of the other?

[hidden haven]

Yes

[bronze jay]

how do i prove that

[hidden haven]

Start with f(x) = f(x ⋅ 1)

See if you can continue

[bronze jay]

thanks

[hidden haven]

Yep

[bronze jay]

i guess i also have to do f(1 dot x)

but it's the same

[hidden haven]

Not necessarily, either one of the 2 in enough

is*

[bronze jay]

so if you prove one then you're done?

[hidden haven]

Yeah you got phi(1) = 1

[chilly radish]

Do you prove the uniqueness using Sylow?

[urban ice]

(either one of the two works bc you just multiply by the inverse on one side)

[hidden haven]

Uniqueness of each pee group decomposition?

[chilly radish]

yea

[hidden haven]

Shouldn't need sylow, you look at all elements with order p, separate it out, then order p², and so on

[chilly radish]

yea you're right I just thought about ti

it

don't need sylow

[bronze jay]

to prove that f(x^{-n})=f(x)^{-n}, given that we know that f(x^n)=f(x)^n for a homomorphism f

[bronze jay]

right?

for a homomorphism the order |x|=|f(x)|

[chilly ocean]

no

[hidden haven]

Nope, Z → Z/2Z

Order of 2 is not preserved

[chilly ocean]

you only showed one divides the other

[bronze jay]

oh

because f(x)^n=(f(x)^k)^m=e^m where n=km and f(x) can have order k?

[hidden haven]

Yep

Order is the smallest exponent which makes the element into e

So now you have to show that no smaller exponent d makes x^d = e

[bronze jay]

i guess like this?

[dusty river]

Yep

[bronze jay]

what about this?

[chilly ocean]

homie fix your writing

first direction I think you shouldd write the last thing before phi(xy) but yea

second direction seems okay

[bronze jay]

lol yea

thanks

but if phi is only a homomorphism, what constraints must be satisfied in order for "G abelian => H abelian" to be true?

[chilly ocean]

"surjective" is one sufficient condition

the image of an abelian group under a homomorphism is always an abelian group

[bronze jay]

oh so it's a subgroup?

[chilly ocean]

yes

[bronze jay]

and if that subgroup is the entire group then we're done

[thorny flame]

How do I find the order of element in infinite group?

[chilly ocean]

do you have an example in mind

[carmine fossil]

I don't think there is a systematic approach

[thorny flame]

Group

Element :

[chilly ocean]

just keep adding it to itself

[thorny flame]

function is addition

[carmine fossil]

It's 35

Assume 2k/5=14m

Which gives k=35m

Which gives k=35 is the least value for which k(2/5)+14Z=0

[thorny flame]

Really?

14 + 14m where m is interger is in G?

[carmine fossil]

14+14m=14k=0

[thorny flame]

I see thanks

How can I find all the homomorphisms of this?

[carmine fossil]

Choose phi(1)=k

And see if phi(0)=0 holds

[thorny flame]

What's k?

[carmine fossil]

An element of Z_21

[delicate bloom]

consider the order of the elements

[thorny flame]

one of the guys said to find via gcd(14,21)

In cosets what does it mean to have a representative?

[rustic crown]

phi is determined from where it sends 1 in Z_14. And 1 can be mapped to any element whose order divides 14.

[thorny flame]

that leave only 2 3 and 7

[rustic crown]

hm?

2 has order 21 on the right!!

21 doesn't divide 14

[bronze jay]

if we have a group G and elements e, g \in G where eg=g, can we necessarily conclude that ge=g and therefore e is the identity?

[carmine fossil]

Yes

Assuming you mean eg=g for all g

[bronze jay]

yes

how do you prove this?

[dusty river]

Is your question

if an element in a group is a left identity, is it also a right identity?
or is it
if you change the group axioms to say "left identity exists" instead of "2-sided identity" exists, do you still get the same notion of a group?

Answer to both is yes, but it will look very different in both cases

The first one is just

Suppose e is a left identity. Let 1 be the 2 sided identity of the group (which exists by the group axioms). Then 1 = e1 = e.

The second one is

Suppose 1 is a left identity. Then we need to prove g1 = g for all g. Start with g'g = 1, where g' is the inverse of g, and multiply both sides by g on the left. You get
(gg')g = g1
1g = g1

I am taking "inverse of g" to mean an element g' such that gg' = g'g = 1, the left sided identity here.

[maiden ocean]

so to clarify

This means that if the intersection of all of the points in U is a finitely generated k algebra then the intersection of all of the points in phi^-1(U) is too

and furthermore that the intersection of all of the points in phi^-1(U) is a finitely generated module over the intersection of all the points in U

I dont really understand this argument...

why does R have to contain O(U)?

Unless im misreading and its saying L contains O(U)

[ivory dust]

How do we show field ext E = Q(2^1/2, 2^1/3, 2^1/4...) is algebraic extension of Q?

take a generic element a in E and show it can be expressed as root of monic polynomials in Q[x]?

[gritty sparrow]

Do you know the lemma that if a and b are algebraic over a field F, then F[a,b] is algebraic over F?

[ivory dust]

Hmm noo

ik [F(a):F] = degree(p(x)) where p(x) is irreducible polyn with root a

p(x) monic and unique

i also know F<B<E, [E:B], [B:F] then E/F finite and
[E:F] = [E:B] * [B:F]

soo im guessing i can do infintie product?

[gritty sparrow]

Ah, don’t do that, reduce to the finite case

So if say a is the element, it is a finite combination of the 2^(1/n)’s

[ivory dust]

linear comb?

[gritty sparrow]

So we only have to show that Q[..2(1/n)..] is finite for the n’s appearing in the expression for a

[gritty sparrow]

Polynomial combination

[ivory dust]

a = sum of pi(2^1/i)?

Oh my bad i shoulda mentioned i need to prove its infinite extension, aswell as algebraic over Q altho i think ik how to prove infinite

[gritty sparrow]

Yes (you have multiple 2^(1/i) in your product right?, just making sure)

[ivory dust]

sum not product right

[gritty sparrow]

It is a sum of products

[ivory dust]

sum of polynomials

of 2^i

where i = 2, 3 ...

correct?

To show algebraic i just need to show all generic elements of it are algebraic over Q (i think?)

[gritty sparrow]

Yes, by products i meant expressions that look like rationalx2x2^(1/2)x..2(^(1/n)

[gritty sparrow]

Yes if by generic you mean all elements

In any case, did you understand the reduction to the finite case?

[ivory dust]

umm just to make sure itd be a = p1 + p2(sqrt2) + p3(sqrt3) + ...
where p3 is a polynomial with coefficients in Q(sqrt2), p4 is a polynomial with coefficients in Q(sqrt2, sqrt3) ... etc.

[gritty sparrow]

Yes that is true but it is a very weird way of putting it, to be quicker you would say, a=pn(2^(1/n) where pn has coefficients in 2^1,…,2^(1/n). Another way of saying it is a=p(2^1,…2^(1/n)) where p is a multivariable polynomial over Q

Does this make sense

Also, small typo you wrote sqrt(3) when i think you meant cube root of 2

[ivory dust]

where pn has coefficients in Q(2^1, 2^1/2, 2^1/3, ... 2^1/n)?

oh and yea sorry meant 2^1/3 not sqrt3

and is this logic correct:

[gritty sparrow]

Sorry upto 2^(1/n-1)) not n, that would male it trivial

[ivory dust]

oo yea

also using Eisenstein criterion x^n - 2 is irred so

Q/(x^n-2) is degree n

so [Q(2^1/n):Q] = n

and if we define finite version of E

say En = Q(2^1/2, 2^1/3, ... 2^1/n)

[En : Q] = product of [Ei+1 : Ei] from i = 0 to n-1

and [Ei+1 : Ei] = [Ei(2^(1/i+1) : Ei] = i+1?

or maybe not necessarily i+1

since im not sure x^(i+1) - 2 is irred over Q(2^1/2, 2^1/3, ... 2^1/i)

[small bison]

just being a finite extension should be good enough

[gritty sparrow]

We can directly show that En has dimension greater than n because x^n-2 has a root in E and this is an irreducible of degree n over Q.

[ivory dust]

[Q(2^(1/n)) : Q] = n so [E: Q] >= [Q(2^(1/n)) : Q] = n for all n?

[gritty sparrow]

Yes

[ivory dust]

so [E:Q] is infinite

okk

[gritty sparrow]

So this proves the other part of your question I suppose

[maiden ocean]

What exactly does "represented by quotients of homogenous polynomials of the same degree" mean

[dusty river]

should be = p({x_i})/q({x_i}) where p and q are homogenous with same degree

[maiden ocean]

Wait nvm

I get it

Yeah

[dusty river]

[maiden ocean]

I was awookend

when it said quotients i thought it meant like. equivalence classes of a quotient ring

I now understand

What

Why is this true

[dusty river]

That sentence doesn't make sense K doesn't have transcendence degree 1 over K

[maiden ocean]

Yeah i think they mean over some base field k

with k|K and k|L both finitely generated and of trans. deg 1

[dusty river]

Yeah then it seems legit, probably a similar argument as finitely generated algebraic extension → finite extension

[maiden ocean]

I dont even see how you can regard L as an extension of K here without finiteness

of the morphism

in the affine case thats what you do right... finite morphism implies injective map on the coordinate rings which gets an inclusion of function fields

[dusty river]

idk much AG I was just trying to reason from field theory lol

which doesn't seem to work

[maiden ocean]

a surjective morphism here means like. ok so Y^L = {R | k subset R subset L, Frac(R) = L, R a discrete valuation ring} and X^K is same thing but with K in place of L

and the morphism is a continuous map from Y^L -> Y^K with a map of sheaves i.e from O^K(U) = intersection of R in U onto O^L(preimage of U)

So... where does this imply a map K -> L

the curves have the cofinite topology... is there some reason getting non constant maps on all the intersection of all but finitely many discrete valuation rings should induce a map on the fields

Certified szamuely moment

[maiden ocean]

@prisma ibex sorry for pinging you can just ignore if you want but uhhh

can you help me understand this (that surjective morphisms Y^L -> X^K of proper normal curves induce a map K -> L)

[prisma ibex]

Okay so these sets you are defining are in terms of valuation rings yea?

[maiden ocean]

Yes

we have a morphism of sheaves mapping $\mathcal{O}^K(U) = \bigcap_{R \in U} R \to \mathcal{O}^L(\varphi^{-1}(U)) = \bigcap_{\varphi(R) \in U} R$

[cloud walrus]

Moth In Shambles

[maiden ocean]

for all cofinite U

[prisma ibex]

Okay so you’re defining the Zariski Riemann space

Packing rn so I can’t explain but if you search for Zariski Riemann space you’ll find the statement you need

I don’t like Zariski Riemann spaces so the way I would personally do this is go from Zariski Riemann spaces to algebraic curves and then from algebraic curves to function fields is more standard

[maiden ocean]

Ok

[maiden ocean]

@prisma ibex whenever u can if u can link me a reference on this thatd be helpful because when i google its all like k theory and birational geometry and stuff

[austere roost]

Can anyone give me a hint how to plot a co-amoeba? I was able to plot a amoeba using maple to apply the log(abs()) map grid-wise on the variety. But when I replace the log(abs()) map by arg() map, I dont get a meaningful plot.

[novel parrot]

would we need G to be commutative?

for example, a,b have order n but ab wouldnt unless its in the center?

[dusty river]

nope, its true for any group

[chilly ocean]

i don't think we need the group to be commutative, no

i was unsure before and that's why i deleted but now im sure

coldilocks

[dusty river]

[chilly ocean]

||coldicocks||

never not funny

[dusty river]

[cursive temple]

Is coldilocks bad ending moldi

[dusty river]

perhaps

[cursive temple]

If so then

[novel parrot]

if a and b had order n then (ab) will have order n ?

[dusty river]

nope not necessarily

You don't need that

[chilly ocean]

||conjugation fixes the set of elements of order n, so it ought to fix N i hope||

[chilly radish]

Isn't the problem closure though?

[dusty river]

You can also explicitly compute the order of any conjugate

[chilly radish]

I think I have a counterexample

[novel parrot]

oh yep, N is generated by order n elements

[chilly radish]

Wait

Generated vy

Nvm

[chilly ocean]

[chilly radish]

I thought it said the set of order n elements

[novel parrot]

same lol

[dusty river]

ok you can't explicitly compute order of conjugate, but you can explicitly write it down as product of order n things

[chilly radish]

I've done an exercise to disprove this so I was primed to read it like that

[dusty river]

[chilly radish]

Who cares about the order of the conjugate

[dusty river]

Won't you need to do that for some elements though

[chilly radish]

🤔

I don't think so

Conjugation fixes all the generators

So it should fix N

[chilly ocean]

[dusty river]

||if a has order n, then so does g'ag. Then any product of order n things, for example abc, maps to g'abcg = g'agg'bgg'cg||

[chilly radish]

Yea that's what I said

[dusty river]

Ok yeah that's a better way of phrasing it

Yeah so you have to show that order is preserved under conjugation

for the first step

[chilly ocean]

[chilly radish]

|| Just say conjugation is an automorphism and so preserves order ||

[dusty river]

right

[chilly radish]

Or show it directly it's not hard

[dusty river]

that's what I was doing

[chilly radish]

But then you need to use induction which is gross

[chilly ocean]

left-facing pfp moment

[dusty river]

bruh

[chilly ocean]

why would u ever explicitly perform induction

[dusty river]

That's extremely formal

[chilly radish]

Well yea

That's what my profs expect tbh

[chilly ocean]

my condolences

[chilly radish]

Dw next year i'm starting to take advanced(er) classes so all formalities will obviously be dropped

[chilly ocean]

sadly this is not always true

[chilly radish]

Ok yea anal is gonna be very formal cuz introductory measure theory

[chilly ocean]

analysis class where a visual proof of "topologist sine circle is path connected but not locally path connected" got 1/10

ok this isnt the place to seethe sry

[chilly radish]

At least i'll only have to do analysis bullshit and epsilonautics in 1/2 classes

[hidden haven]

epsilonautics

[bronze jay]

Am I just supposed to show that the group action definition holds here?

seems trivial

[chilly ocean]

ye

[bronze jay]

is this correct?

[chilly ocean]

ye

[hidden haven]

yedog

[bronze jay]

i shouldn't have written a \in kern i think

that's wrong lol

[chilly ocean]

ye

[hidden haven]

Btw you don't have to check that the identity is there in the subgroup, ther you get that with inverses and products automatically

[bronze jay]

ah

how in general does one find generators and the presentation for a group?

[gray comet]

Technically, you have to show it's nonempty

[hidden haven]

ah true

[gray comet]

And the easiest way is with the identity

[bronze jay]

but every statement about elements of the empty set is automatically true?

[carmine fossil]

You don't

[gray comet]

Which is why you have to show it's nonempty, lmao

[hidden haven]

I don't know if there's an easy way to find the best presentation, but the "canonical" presentation of any group has all the group elements as the generators and all the equations that are true in the group as its set of relations

Canonical in the sense of having a universal property

[gray comet]

Like, it's true that $\forall a, b \in \varnothing$, $ab \in \varnothing$ and $a^{-1} \in \varnothing$, but that doesn't imply $1 \in \varnothing$, catch my drift @bronze jay?

[cloud walrus]

Tormeson

[bronze jay]

oh i see

[bronze jay]

From dummit and foote they write that you should be able to derive all of the relations in the group from the group presentation, but if you were given the task of finding the presentation of a group, it seems that you have to just guess a presentation and then check if the relations can be derived from there

[chilly ocean]

Its hard in general yeah

[hidden haven]

Yeah pretty much. Most of the time you'd include whatever feels like a defining set of relations, and then prove that it is

[bronze jay]

but sometimes you don't know if the relations are redundant and that's fine?

[hidden haven]

Yeah redundant relations are ok

[hidden haven]

Otherwise the canonical presentation wouldn't work

Almost everything is redundant there lol

[bronze jay]

i see

[chilly radish]

Finding a 'good' presentation of a group is a really hard problem

[bronze jay]

the proof for (b) is basically the same as (a)?

[chilly ocean]

ye

[chilly radish]

In fact you can prove a from b

It can be shown that the kernel of the action is the intersection of all satbilisers

[bronze jay]

I see

how do I go about solving this problem?

what are pairs of opposite vertices?

and what action we talking about?

[chilly radish]

If you look at pairs of vertices, you can look at symmetries of an n-gon as permuting the order of the pairs

That's the action

It's easier imo when you think of the other definition of the action

Which is a homomorphism into the symmetry group of the set being acted om

[bronze jay]

but what are opposite vertices?

[chilly radish]

In this case it'd be a homomorphism $\rho:D_{2n} \rightarrow S_{\frac{n}{2}}$

[cloud walrus]

ShiN

[chilly radish]

Where you identify a pair of opposite vertices with some number

[bronze jay]

don't you just mean S_n ?

[chilly radish]

No, since you have an element for each pair of vertices

Not for each vertex

[bronze jay]

oh

[chilly radish]

This is why $n$ must be even

[cloud walrus]

ShiN

[bronze jay]

but the group action is not uniquely determined from the group right?

you have to define the group action after you've picked your group and the set it acts upon

[chilly radish]

Right, but it's implies here what the action is

There's usually a 'natural' action for groups like Dn or GLn(F)

Or most simply Sn

Like if I told you GLn(F) is acting on F^n and didn't specify further then you'd assume it's acting by martiz multiplication

Same here, Dn acts on the vertices of the polygon by permuting them

So in particular it can act on pairs by permuting then

[bronze jay]

I am not such how i find the kernel of this action

it seems that the kernel is only the identity?

[chilly radish]

I'm actually not sure if they meant ordered or unordered pairs

It would make sense this action is faithful, but I don't wanna say for sure since I haven't checmed

If it is unordered pairs though it wouldn't be faithful

Since reflections about any axis would keep it invariant

[bronze jay]

what about the kernel?

[chilly radish]

That would be the kernel in this case

All the reflections about some axis

If the pairs are unordered

I think maybe also 180 degree rotations but i'm not sure

Again, don't wanna state anything as fact since I haven't checked

There may be more elements

[bronze jay]

I don't quite get it, the kernel is the set of elements , g, where ga=a for all a in the set, isn't that just the identity in this case?

[chilly ocean]

no, I think r^2 also

by pairs they mean unordered

pretty sure

[bronze jay]

but why would r^2 be a part of the kernel?

[chilly radish]

If the pairs are unordered, then reflections about any axis would not do anything since they just switch the order of opposite vertices

[chilly radish]

I think r^{n/2} generally not r^2

[chilly ocean]

yes obviously

oh I thought its for square

[chilly radish]

Ah lol

[bronze jay]

r^{n/2}, isn't that just s?

[wooden ember]

how would i go about doing the last part for S an infinite set? When finite you can just do an induction on the first case done, but im not sure about dealing with an infinite set. I suppose a more elegant way to prove it than i did for the first case

[hidden haven]

The map G → G/N, x → x bar, is a homomorphism, so a finite product of elements in G maps to the same thing but with bars on top

[lapis temple]

I want to learn representations of finite groups (actually in particular symmetric and alternating groups) over finite fields. For now, I am particularly interested in the semisimple case. Can anybody recommend me a suitable text or any other kind of source (notes etc.) to start with?

[fossil shuttle]

Serre has a book called "Linear Representations of Finite Groups."

I like it.

There is a book "Representation Theory: A First Course" by Fulton and Harris. I'm less familiar

[prisma ibex]

This is pretty much only over C

[prisma ibex]

If you’re looking at representations of finite groups over characteristic p fields then you will need a book on modular representation theory.

Asking for the semisimple case alone is kinda an awkward ask, since most sources on modular representation theory are written with an eye towards the non semisimple case, while most sources on more basic representation theory of finite groups don’t treat even the semisimple modular setting at all

A good middle ground might be to read through something like Fulton Harris that works over characteristic 0 and as an exercise reprove everything you can in the book over \bar{F_p} or whatever

[maiden ocean]

possibly a silly question but why exactly is the tensor product of a separable extension with k(P) etale over k(P)?

[sturdy marsh]

L/k is separable, so it is iso to k[t]/f for some separable polynomial f. Then the tensor product is k(p)[t]/f.

[maiden ocean]

Oh that makes sense

It was a silly question

[fossil shuttle]

Whoops, I didn't see that they asked for finite fields

[cloud walrus]

Coldilocks

[maiden ocean]

I havent checked the details but iirc this is erratum lol

Yeah it is

[dusty river]

oh epic lol

[maiden ocean]

the actual map sends a to 1_B otimes g(a) = f(a) otimes 1_C

[dusty river]

I see, thanks

[maiden ocean]

Np

https://mathoverflow.net/questions/42241/errata-for-atiyah-macdonald

@dusty river heres a thread

[dusty river]

ty

[untold cloud]

Hi everyone, i am not so sure what is the meaning of the highlighted part. I think 0 is always the unique trivial zero divisor in any nontrivial ring, not just commutative ring. it seems like this sentence is redundant？

[scarlet estuary]

well, they havent defined zero divisors for noncommutative rings yet

(for noncommutative rings, theres also the ba = 0 case)

but you are right that, once they are defined, 0 is the unique trivial zero divisor in any ring.

(caveat: except the trivial ring)

[untold cloud]

ok, thank you!

[quaint tree]

@upper pivot and I were working on this, from Guillot's Local Class Field Theory, and we think that this is wrong, just given a dimension argument considering it as real vector spaces. But $\mathbb{H}$ is 4-dimensional over $\mathbb{R}$, so shouldn't $\mathbb{H}\otimes_{\mathbb{R}}\mathbb{H}$ be 16-dimensional instead of 4-dimensional over $\mathbb{R}$, which contradicts the given proof?

[cloud walrus]

ℤ(𝜁23) is a UFD

[lapis temple]

hi... please help me grasp the tensor product of modules in the most intuitive way

[bronze jay]

"Since r and s generate D_8, every element of D_8 commutes with r^2" - this doesn't seem like a trivial assertion, is there a proof for this?

[carmine fossil]

you know r r^2=r^2 r

and sr^2=r^(-2)s=r^2s

[bronze jay]

yes but just because s and r generate D_8, how can you be sure that r^2 commutes with the rest of the elements?

[carmine fossil]

Recall the definition of "generate"

[lapis temple]

hi.. can anybody tell me when the representation of a group, which is induced from an irreducible representation of its subgroup, is still an irreducible one for the whole group itself

[chilly radish]

In the explicit construction of the free group as the equivalence classes of words, how does one prove the uniqueness of the reduced representative? My prof chose to give an example instead of a full proof in class and i'm not sure how one would go about doing it rigorously

[dusty river]

Look at the universal property of a tensor product. That is the way to think about tensor products

[cosmic oriole]

How is it "induced" exactly?

[lapis temple]

it is defined so that one can make a representation of a subgroup into a representation of the whole group

@cosmic oriole please lemme know if you need the detailed construction.

[cosmic oriole]

Yes. That would be helpful. I haven't done representation theory in a while.

[lapis temple]

Suppose that 𝐻 is a subgroup of the finite group 𝐺 and 𝜎:𝐻→GL(𝑊) is a representation of 𝐻. Then the induced representation from 𝐻 up to 𝐺 denoted 𝜋 is a group homomorphism 𝜋:𝐺→GL(𝑉), where
𝑉={𝑓:𝐺→𝑊∣𝑓(ℎ𝑔)=𝜎(ℎ)𝑓(𝑔), for all ℎ∈𝐻, 𝑔∈𝐺}.
The representation 𝜋(𝑔) is then defined on 𝑓∈𝑉 by
𝜋(𝑔)𝑓=𝑓(𝑥𝑔), for all 𝑥,𝑔∈𝐺

[dusty river]

How did you type this

[urban ice]

ɡ∈G

close

but a combination of good unicode and discord formatting

it's easy with latex though

[chilly ocean]

if t is trans, lets say over Q, how can I show Q(t^2) is strict subset of Q(t)? Or I guess how can I show t isnt in Q(t^2)

[dusty river]

if p(t^2)/q(t^2) = t, multiply both sides by q(t^2) and you get a contradiction by degrees

[chilly ocean]

wait I dont get it

can every element in Q(t^2) be written as p(t^2)/q(t^2)?

for some rel prime p,q

[dusty river]

t is transcental → t^2 is too

and if t is transcental then Q(t) is isomorphic to Q(x) where x is a formal symbol

[chilly ocean]

hmm okay ye I think

[dusty river]

Yes because it is the smallest field containing Q and t^2

All those elements must be contained in every such field, and the collection of those elements is closed under the operations

[chilly ocean]

wait but rather than degree contradiction, by mutipliying and subtracting wouldnt it imply t is algebraic over Q?

[dusty river]

Right that too

By degree I meant you get an equation of formal polynomials in Q(x)

but one has odd degree and the other has even degree

[chilly ocean]

okay ye

thx

catking

[dusty river]

[bronze jay]

How does lagrange's theorem simplify things here? Is it because we can check if our centralizer has an order that divides the order of the group?

[chilly ocean]

ye

like for S3 it has order 1,2,3 or 6 and you can probably eliminate some options

[bronze jay]

this correct then?

i see

[bronze jay]

what about C_D_8(r)? I don't see to be able to figure out what it is, I've written that C_D_8(r) = {1,r,r^3} but that doesn't have an order which divides |D_8|

[chilly ocean]

i mean its a subgroup right so also product of those should be in there no?

@bronze jay

[bronze jay]

oh ofc

[runic hemlock]

In Vakil's The Rising Sea, the author stats that an $A$-module $M$ ($A$ is a domain) is torsion iff $M\otimes_A K(A)=0$. I understand one of the implications, but why does $M\otimes_A K(A)=0$ imply $M$ is torsion?

[cloud walrus]

Phorphyrion

[sturdy marsh]

Try proving this: If $S \subset A$ is a multiplicatively closed subset, and if $M$ is an $A$-module, then $S^{-1}M \simeq S^{-1}A \otimes_A M$.

[cloud walrus]

Brofibration

[sturdy marsh]

once you have that, then $M\otimes_A K(A) \simeq S^{-1}M$ where $S = A-0$

[cloud walrus]

Brofibration

[runic hemlock]

Thank you 🙂

[ivory dust]

Does a polynomial being reducible in a field imply it has a root in that field?

i swear i may have asked this long time ago

or is it just that the polynomial can be expressed as two lesser degree polyn multiplying one another

[upper pivot]

The latter

[ivory dust]

ohh ok tyty if d(f(x)) = p prime, then f(x) = g(x)h(x) what can i say abt deg g,h

they are both greater/equal to 1 less than p?

[upper pivot]

Think about like, (x^2-2)(x^2-3) in the rationals, reducible but no roots

[ivory dust]

Oh yea

[upper pivot]

They are both less than p is all you can conclude there

[ivory dust]

can they be 0 degree?

[upper pivot]

And that their degrees add up to p

[ivory dust]

err i guess they can but then im not expressing as two polynomials x eachtoher

but a constant

[upper pivot]

Well if you say something is reducible you usually don’t allow the factors to be constant

Yeah

[ivory dust]

so id say both greater than equal to 1 so its a useful statement (trying to arrive at contradiction)

okk tysm

[shell brook]

can i get a hint for this one

[chilly ocean]

Lol kes askign for a hint from df xdd

[shell brook]

it should be "disjoint p-cycles" not commutative, btw

[ivory dust]

if K = Zp(t) how do I show f(x) = x^p - t is irreducible in K[x]

the hint is if E/K is splitting field of f(x) then for some a in E f(x) = (x-a)^p

splitting field assumes it splits over E/K? but not any proper subfield of E?

[gritty sparrow]

Well, don’t the size of the cycles all have to divide p? Intuitively that makes sense to me because when you take the power of a permutation we can just look at what happens in each cycle separately

[ivory dust]

I assumed for contradiction f(x) = g(x)h(x) both w degree less than p, greater/equal to 1

not sure how to utilize hint

i know K = {f(t)/g(t), g nonzero, f,g e Zp[x]}

[gritty sparrow]

Hint: If it is reducible in K[x] then K is an extension of Zp and we can construct an extension of K,E where the polynomial splits. Now find the degree of E over Zp and see what possible degrees K can have

[ivory dust]

wait but if its reducible in K[x]

err

in any extensoin* it will be rdubile

reducible*

[gritty sparrow]

Splitting is different from irreducibility. I’m constructing E where it completely splits into linear factors

So E is the splitting field of the polynomial over K

[gritty sparrow]

Sorry I should rephrase, this is wrong. Basically any factor of your polynomial in K will be (x-a)^s for some s<p. And this is impossible as a^s is in K and there will be a number k such that sk=1 mod p, so a^(sk)=a so a in in K

Really sorry for the error

@ivory dust does this make sense?

[ivory dust]

Oh by fermat?

Hm

Sorry i was afk

just read it

[gritty sparrow]

Yes by fermat’s

So now all you have to do is show that the polynomial has no root in K

[ivory dust]

hmm but doesnt that just show a is in K

so its a root

err

i was gnna argue f(x) = x^p - t = (x-a)^p for some a in K

(not sure why this holds but just using hint)

[ivory dust]

are you first assuming a is not in K?

i said if d(g(x)) = n then g(x) = anx^n + ... + a0, 1<=n<p also gcd(n,p) = 1

[gritty sparrow]

Yes i was assuming that

[green locust]

Quick question - this is from Altman and Kleiman commutative algebra - what does the exponent e and c mean in Exercise 1.13

[ivory dust]

https://puu.sh/HWWvr/7eb9d4f376.png

@gritty sparrow i saw this sln on stack exchange

they assumed a smaller degree minimial polyn exists

u(x) w degree r <= p

do u understand how the expanded (x-u)^r

im so confused abt that

[gritty sparrow]

They expanded using the binomial theorem, and then they collected all of the terms of x with degree more than 2 into x^2P(x)

[ivory dust]

shouldnt it be -u^r?

or doesnt matter right

[gritty sparrow]

It doesn’t matter. It depends on wether r is even or odd i suppose

[wooden ember]

commutative p-cycles are necessarily disjoint as you pointed out. For a single p-cycle the order is clearly p. A product of disjoint p-cycles will also clearly have order p. For the converse you consider the disjoint cycle decomposition of your permutation assume it's not a product of p-cycles. You'll see it's order is the lcm of the length of each cycle, which definitely can't be p since it's prime (this also hints at what type of counter example to take for the second part of the exercise)

[azure plinth]

Alright I need help understanding a particular solution to a problem. (From Dummit and Foote, Ch 14 : Galois Theory)

This is the problem and bit a) is the one in question

And here's the solution I am talking about

My question is where do we get the $[\bar L: K]\mid [L_1:K]\ldots[L_n:K]$ part from? Can someone explain that properly? Sorry if this is a dumb question

[cloud walrus]

AidenM27

[azure plinth]

<@&286206848099549185>

[azure plinth]

Anyone?

[fossil shuttle]

i think it's notation they introduce in that section to save space - something like "extension" and "contraction". i.e. when a is an ideal in R, then a^e is the ideal of R' generated by phi(a), and if b is an ideal of R', then b^c is the preimage of b in R via the map phi

[fossil shuttle]

^

[green locust]

@fossil shuttle that makes sense, thanks. I wonder why they don’t define it specifically first

[fossil shuttle]

page 4 in this copy

[green locust]

Hah wow I can’t believe I missed that, thanks for clearing it up!

[mild laurel]

@azure plinth I believe it's the fact that the Galois group of Lbar over K is a subgroup of the product of the Galois groups of L_i over K

https://math.stackexchange.com/questions/507671/the-galois-group-of-a-composite-of-galois-extensions

[azure plinth]

Oh I should have paid more attention to this SE answer then, that's pretty neat! Thanks @mild laurel. It is kind of vaguely intuitive but the exact reason why eluded me

[sullen island]

hi, im working through the lie algebras book by humphreys and im not really understanding how exactly is lagrange interpolation used here - lagrange interpolation is based on the a_i, but here, he interpolates based on a_i - a_j, could someone walk me through what's going on here?

this is a page talking about lagrange interpolation from hoffman-kunze's linear algebra book, for ease of reference if needed

[carmine fossil]

Define b_{ij}=a_i-a_j and do interpolation on those b_{ij}s

[sullen island]

oh, which brings me to another question - how do you guarantee the a_i's (or well, b_{ij}'s) are distinct? since the eigenvalues may not be distinct and certainly their differences may not be distinct

my first thought is simply pick scalar multiples of the eigenvectors such that the eigenvalues are all distinct, or is that too naive?

[carmine fossil]

For the function to be well defined you don't need distinctness

You need if b_{i_1}{j_1}=b_{i_2}{j_2} then both get mapped to the same thing

[sullen island]

but for lagrange interpolation, we require them to be distinct, right?

[carmine fossil]

Yes

But you can just ignore the duplicates

[sullen island]

hmm i see

thanks 🙂

[wooden ember]

I’m in dire need of a quick yes or no as I’m getting frustrated: if h has finite index in g, does it have finite index in a subgroup of g that contains h?

If say yes but at this point I don’t want to prove it I just need to know

Nvm I found what I needed

[rustic crown]

you can show that H < K < G then G/H is bijective to G/K x K/H (where these are just sets of left cosets)

[clever mountain]

I'm working on this assignment:

I have managed to show it is a well-defined right-action and in (c), I am studying a homomorphism
phi: N_G (H) -> S_{G/H}

I have managed to show it has H as kernel, but when I tried to verify it is a homomorphism, I found to my surprise that it is an antihomomorphism, namely f(ab) = f(b)f(a). Right-actions typically induce antihomomorphism, so maybe it should not be all too surprising. But I can't for the life of me think how else we can find a homomorphism with kernel H.

My homomorphism is such that phi takes g in N_G(H) to a permutation pi_g.
pi_g is such that it takes every left coset xH to the left coset xgH, which mirrors a right-action.
Using this definition of pi_g, I managed to show ker phi = H. But phi is not a homomorphism, which puzzles me.
Any hints on what approach to take?

[chilly radish]

By defining the function to $S_{\sfrac G H}^{op}$ instead, the antihomomorphism becomes a homomorphism, from there you just compose it with the natural isomorphism from a group to its opposite

[cloud walrus]

ShiN

[chilly radish]

You could also define your function to just take g to pi_g^-1 and this would also fix the problem similarly

[ancient night]

How canonical is it to differentiate ${ (x_1,x_2) \mapsto (x_1,x_2), (x_1,x_2) \mapsto (x_2,x_1) }$ and ${ (1,2), (2,1) }$? When it happens, how one emphasizes either case?

[cloud walrus]

JohnDark

[ancient night]

As I think about it, the first case must be a set of functions of one element but it isn't a big deal

[dire trellis]

Hello! How can I know if $x_1=\sqrt{1-4i}$ and $x_2=\sqrt{1+4i}$ can be obtained from each other algebraically in a field?

[cloud walrus]

segrendel

[dire trellis]

at the end of the day, what I want to know is if $\mathbb{Q}(x1)=\mathbb{Q}(x1,x2)$

[cloud walrus]

segrendel

[stone leaf]

A question, how to proof this, I am stuck...

[low surge]

i think we are happy to say x^pk is identity, so the order definitely divides pk

the work is in showing it cant be less

so let's suppose the order isnt pk

try and get a contradiction on the minimality of the order of a (can the order be a multiple of p without being pk itself?)

@stone leaf

[stone leaf]

I see... Thank you @low surge !

[low surge]

did u get it?

[stone leaf]

I can show p | order(x) from order(x^k) = p.

This part causes me headache most.

[low surge]

i dont think i tried that approach

that seems more difficult

also x^kp=e could potetnially mean x^k could be identity right so im not sure if thats right

maybe im missing something easier

[hidden haven]

Did you assume that N≠kp means N is some other multiple of p?

[low surge]

that was a further assumption that i showed wasnt possible but then later on dealt ith the other possibility

thats why i said the bit with N<k i think

[hidden haven]

ah I see

[cursive temple]

this is given as an exercise in jacobson, and i was wondering what is meant by "An abstract characterization", and what uses such characterizations might have

[clever mountain]

@chilly radish : Brilliant, that was the clincher, big thanks!

[bronze jay]

how is an infinite cyclic group "cyclic"?

[fossil shuttle]

i guess an abstract characterization is one in the language of pure group theory. S_5 is a "concrete" group in the sense of being an automorphism group or subgroup, (the group of automorphisms of {1,2,3,4,5} in the category of sets). Other "concrete" groups might be the general linear group, the special linear group, other matrix groups. Cayley's theorem shows that every group can be regarded as concrete in this sense, but this turned out to be slightly less important than was anticipated at the time it was proved.

There are some theorems of group theory which work better if you take cyclic to mean 'generated by a single element', which Z is.

For example, the fundamental structure theorem for finitely generated Abelian groups states that every finitely generated Abelian groups is the finite direct sum of cyclic groups

[bronze jay]

hmm, so it's a some form of convention?

[fossil shuttle]

the convention is that cyclic means generated by one element, yes.

[bronze jay]

is x^inf=1 then, if |x|=inf

[fossil shuttle]

lol. that's a good question. I don't know how to answer that. I haven't seen that convention in the literature. it's certainly not true in any literal sense. but we use the phrase "characteristic zero" to talk about a field where 1 has infinite order in the additive group

the same proof works for various theorems about cyclic groups, maybe? in spite of the fact that all but one of them are finite and the other is infinite

for example proving that cyclic groups are abelian doesn't require you to split into the finite and infinite cases

[cursive temple]

why are we interested in abstract characterizations then

these characterizations dont seem too useful in my opinion

[fossil shuttle]

idk i don't really have good examples of applications. i think if you have a philosophy where you don't want to necessarily think of groups as being automorphism groups, this contributes a little bit to another way of developing group theory

[cursive temple]

sure, not everything has to be done for some grand reason

its just that it seems like a quite a bit of work

[fossil shuttle]

Oof. yeah. lmao

[clever mountain]

In part (d), the wording is slightly confusing. "The centralizer of (the image of) G in S_{G/H}".
If ϕ: N_G(H) -> S _ {G/H} is the homomorphism in (c) with kernel H, the first isomorphism theorem tells us

N_G(H)/H ≅ im ϕ

I suspect im ϕ is what I should uncover to be "the centralizer of (the image of) G in S_{G/H}".
But I'm struggling to parse what this centralizer is.
ϕ sends N_G(H) to S_{G/H}, not the entire G.
So im ϕ is definitely not the image of G, rather the image of N_G(H).

And how I understand "the centralizer of (the image of) G in S_{G/H}" is that
it is the set of permutations in S_{G/H} that commute with elements in im ϕ.
But I haven't been able to progress from here.

Any thoughts or hints on what I am missing here?

[wraith obsidian]

Because the right level of abstraction makes things easier because it lifts the weight of unnecessary technical detail.
Imagine having to define the free group a as a permutation group just because your definition of group required that
Or actually caring about how you would construct such a group (e.g. irreducible words vs equivalence classes of words in L_grp vs the cayley representation over itself)

[bronze jay]

why by the minimality of d?

what does that mean exactly

H is a cyclic group <x>

but why is r=0 just because d is the least element?

[hidden haven]

Because r satisfies the other condition of being in P (which is x^r ∈ K), but is less than d

So it shouldn't be in Z+, because then d couldn't be minimal

[cursive temple]

Okay, im completely lost on part (iii). The numbers s_1 and s_2 are defined here

[bronze jay]

ah

[cursive temple]

The three classes of involutions can just be seen by computing the conjugacy classes of D_8 directly, and the fact that x = u_2 is conjugate to xu_2 is also clear due to x \in C_2

I just have no clue on how to relate these to s_2

[bronze jay]

how does this follow from proposition 5(3)?

[dusty river]

is H = <x>?

[bronze jay]

yes

[dusty river]

n/a divides n

|x^d| = n/d = a

[bronze jay]

oh i see

and since |x|=|G| for any x \in G and G is a subgroup of H then G must have order a

[wooden ember]

im stuck on the implication iii) => iv). Im not exactly sure what they suggest you induct on in the hint? What i was trying to do was to take the minimal nontrivial subgroup, quotient by it, then take the minimal nontrivial subgroup of that and show that the subgroup of G that it's in bijective correspondence with (by 4th isomorphism theorem) is abelian so that we expand the chain and the same argument could be repeated to go from the ith group in the chain to the (i+1)th group in the chain

cant get it to work though

[wooden ember]

aight looked it up found what to induct on, the proof is simpler than i thought

[thorn delta]

real quick, how do we get y = 0 on the underlined part?

wait nvm im dumb lol

[maiden ocean]

AM moment

[uncut girder]

Is this canonical isomorphism?

What does that mean

[mild laurel]

what does what mean

[uncut girder]

Canonical isomorphism

Any Why is it canonically isomorphic

[hidden haven]

Natural isomorphism?

[uncut girder]

What does that mean

Does it mean there is only 1 isomorphism from the group on the left to the group on the right?

[fossil shuttle]

if A is isomorphic to B, an isomorphism $\sigma :A \cong B$ is said to be canonical if when you tell two different mathematicians to write down an isomorphism $A\cong B$, they both write down $\sigma$.

[cloud walrus]

diligentClerk

[hidden haven]

You can apply this isomorphism on any n, in a nice "canonical" way that doesn't depend on n. Suppose you gave a map from Z/nZ to Z/mZ. Then this induces a map from Z/nZ* to Z/mZ*, call it f, and a map on the other side as well, call it g. Naturality means that you can either apply f first then this isomorphism for m, or isomorphism for n then g. Both should give you the same result

So it's like the isomorphism doesn't really depend on n, it's essentially the same for all integers

[uncut girder]

I feel like that's not really what canonical isomorphism means

[hidden haven]

I gave the definition of a natural one

[uncut girder]

People use the word canonical isomorphism when there are no parameters in the groups involved

[hidden haven]

I'm not sure if canonical is the same thing though I've seen these used interchangeably

What do you mean by parameters?

[uncut girder]

What you called n or m, and what's N in the screenshot

[hidden haven]

Natural does depend on those

In the above thing you don't have just 1 isomorphism

It's a family of isomorphisms, one for each N

That behave nicely

[uncut girder]

That's not really the point

[hidden haven]

Idk then lol

Where is this used?

[uncut girder]

You dont have to specify a family of isomrophisms parameterized by n or some other index in order to call a specific isomorphism canonical

The map from a group G to its quotient G/H is called the canonical projection

In this case it's not an isomorphism but it's still canonical

[mild laurel]

my guess would be that canonical means that the isomorphism doesn't depend on the choice of primitive nth root of unity

[hidden haven]

Yeah there is a parametrization there

[uncut girder]

No

[hidden haven]

It's a natural transformation instead

The parametrization is the groups G and H themselves lol

[uncut girder]

[hidden haven]

Atiyah macdonald uses canonical isomorphism for natural

So like AxB ≈ BxA as sets

And this is canonical, meaning the isomorphism is the same for any choice of A and B

[uncut girder]

Oh right theres the canonical isomorphism from A(x)B = B(x)A for tensor products

[hidden haven]

So parametrization here is by the pair A and B

Yep

And this naturality is useful when you work with maps

So for example (A x B) x C

For tensor products let's say

And suppose you have a short exact sequence of modules of this kind

And maps also being of the kind (f x g) x h

Then naturality lets you apply associativity on the whole sequence

And it remains exact

If it were just an isomorphism it wouldn't, so the whole point is that natural maps work nicely with homomorphisms etc

[uncut girder]

[violet bane]

can I ask here where I can learn abstract algebra for free?

[hidden haven]

Books. Check pinned messages in #book-recommendations. You can probably also find lecture series on yt but it's best if you work through some book side by side

[violet bane]

do you have any recommendations for yt series?

[hidden haven]

Nope, haven't tried any

[violet bane]

okay, thank you for your help!

[hidden haven]

[thorn delta]

ok so on this proof, im a bit confused how they're justifying the claim that x \pm y an xy are integral over A with iii) of 5.1

What i would say is that A[x,y][xy] = A[x,y][x\pm y] = A[x,y] are finitely generated so xy and x \pm y are integral over A, but that doesn't seem to have much to do with iii, does it?

5.2 is just the statement that A[x1, x2, ..., xn] is finitely generated for x1,...., xn integral

[hidden haven]

Here you proved that xy is contained in a subring that is a finitely generated A-module, not that A[xy] itself is finitely generated as an A-module

So ii doesnt directly apply while iii does

Like you proved that xy ∈ A[x,y], and A[x,y] is finitely generated. But this doesn't directly tell you that A[xy] is too

[thorn delta]

Ah yea yea okay that makes sense

thanks

[hidden haven]

[cursive temple]

im trying to prove an extension of sylow 3 in jacobson, but i dont understand how S_0 is defined or why there exists an H_A with |H_A| = p^k

how does a subset of A affect the size of its stabilizer

[cursive temple]

i think its a mistake but i cant find errata anywhere

jacobson does say that the proof can be found in an edition of archiv der mathematik from 1967 which i cant find on sc*hub

so if anyone has a springer subscription or any clarification to the question that would be very cool

[cursive temple]

I think i figured it out; S_0 is defined to be the subset of S such that for A \in S_0 we have |H_A| = p^k

[wooden ember]

ive got just about everything done for this one except for one detail which is just preventing me from completion. To use the inductive hypothesis i construct two composition series of $H$ namely $1=N_{0}\cap M_{s-1}\leq N_{1}\cap M_{s-1}\leq ...\leq N_{r-1}\cap M_{s-1}=H$ and similarly but iterating over the M chain for the second one. Ive worked out everything except for the guarantee that subsequent elements of these series are not equal to eachother. Ive been stuck for hours and just cant figure this out, any hint would be appreciated

[cloud walrus]

Little Narwhal

[wooden ember]

i ended up doing it differently my original approach was flawed i suppose

[unreal portal]

Let's say I have a quotient polynomial ring $R=\mathbb{Q}[x,y]/I$ for some ideal $I$, and I want to find a Groebner basis for a different ideal $J$ in $R$. Is there a `right' way to do this? As in, does it make a difference if I try to compute the Groebner basis directly in $R$ vs. considering $J$ as an ideal in $\mathbb{Q}[x,y]$, finding the Groebner basis there, and then taking the quotient of the resulting basis by $I$? I'm relatively familiar with Groebner bases in general, just not when it comes to quotient rings

[cloud walrus]

cgodfrey

[uncut girder]

nonachimedian Local field gives a local ring consisting of those elements of norm leq 1

General Hensel's lemma:

[delicate bloom]

yes, but this is not what I would call the general hensel's lemma, I'd just call this hensel's lemma

there's more than one way to generalize hensel's lemma but the way I was referring to it was the way when this condition fails and f'(alpha) = 0 in the residue field

you can still lift, it just requires you to work more first

the convenient way I gave of $|f(\alpha)|<|f'(\alpha)|^2$ can be unpacked if you say $|f'(\alpha)|=p^{-n}$ then it means $f'(\alpha)=0 \mod p^n$ and $f'(\alpha) \ne 0 \mod p^{n+1}$. Then we can take the inequality $|f(\alpha)|<p^{-2n}$ and turn it into $|f(\alpha)|\le p^{-2n-1}$ since this nicely turns into $f(\alpha)=0\mod p^{2n+1}$

[cloud walrus]

Merosity

[delicate bloom]

So I haven't proven anything, just unpacked it into a nicer form for you to use,
$$f(\alpha)=0\mod p^{2n+1}$$
$$f'(\alpha) \ne 0 \mod p^{n+1}$$

[cloud walrus]

Merosity

[delicate bloom]

when n=0, that's the regular version, this is the generalized version

I think the converting back and forth is kind of confusing but really $|x-y|\le p^{-n}$ is the same as $x-y=0 \mod p^n$

[cloud walrus]

Merosity

[wooden ember]

Feeling a little unsure about a step in one of my proofs: if I have a map form <S> to <T> that ive shown to be a homomorphism, and ive shown that this map restricted to S is a bijection to T, is this map an isomorphism?

ive just sort of assumed this is true and i think ive proved it before but ive got about 210 pages of exercises form the past 3 weeks and i cant really parse them all

[rustic crown]

what's <S>?

free group on S?

[wooden ember]

Because im trying to show $S_p = <\sigma,\tau>$ for any transposition $\sigma$ and p-cycle $\tau$ and to do this I showed that if i wrote $\tau = (a_1 a_2 ... a_p)$ then we had $(a_i a_{i+1})\in <\sigma,\tau>$. Then via a the map from $<\sigma,\tau> \to <(1 2)(1 2 ... p)$ defined by simply taking each $a_i$ in a cycle to $i$ (and i show it to be an isomorphism by what i just asked) we have $\left{(i i+1), 1\leq i\leq p-1\right}\subset <(1 2), (1 2 ... p)>$ which we showed to generate $S_p$ in another exercise. So i can take the isomorphism the other way around for the result.

[cloud walrus]

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

[daring ibex]

this is true, consider f^-1(f(ab)) if u want to show that a bijective homo is an iso

[wooden ember]

all words formed from elements in S

oof latex doesnt like cycles

well im not retyping that

[wooden ember]

no i know a bijective homo is an iso

but im only showing the bijection between the generators

[daring ibex]

i see

yeah that should work too

[wooden ember]

what's the other definition for an isomorphism

[daring ibex]

Homo that has an inverse

[wooden ember]

oh

i mean funciton is bijection iff it has two sided inverse so i guess it's not too different

[daring ibex]

ya

u just need to show that the inverse isn't too gross and maintains structure to prove their equivalence

[rustic crown]

so just a monoid and not group?

[wooden ember]

it is a group?

oh right i mean you do take inverses as well

didnt remember the definition by heart

closure of S

[rustic crown]

oh okie, i thought you're abstractly taking words to form a group

like the free group on one letter is isomorphic to Z

[wooden ember]

i havent gotten i free groups yet could you elaborate?

oh nvm i see

because the letter doesnt have a particular order is that what you mean

[rustic crown]

they way you have written it is kinda wrong. we also want the relations to be respected by that bijection from S to T

in our case S = {sigma, tau} and T = {(12), (12..p)}

[wooden ember]

how would you show that since the presentation doesnt come with any relations (explicitly)

[hot lake]

what's the original question ?

[wooden ember]

gimme a sec gotta scroll back lmao

and i tried to show that as i stated earlier

[wooden ember]

though this isnt very readable ig

[hot lake]

it feels like you're way overthinking stuff

[wooden ember]

sounds about right

[hot lake]

what you would typically say to make things a bit cleaner

is something like "without loss of generality tau = (123...p)"

[wooden ember]

yeah i sort of did it weirdly in two steps

[hot lake]

that is, there is an inner automorphism of Sp (by conjugation) that sends tau to (123....p) and sigma to some random transposition

[wooden ember]

because i showed wlog sigma = (a1 a2) and tau = (a1 a2 ... ap)

[hot lake]

so the exercise becomes showing that for any transposition sigma, (12) is in <sigma, (123...p)>

and once you have done that it's over

[wooden ember]

ah okay that's a bit different

[hot lake]

and it looks like you have shown somehow that you can get (12) in <sigma, (123...p)>

when you said

[wooden ember]

yeah

wait no

[rustic crown]

consider this example G, S = {g} and T = {e} Then we have a homomophism from <S> --> <T> which sends everything to e, but its not injective if g is not e. Even though in this case the restriction to S --> T is a bijection.

[wooden ember]

fair enough

[hot lake]

just change the names of everyone so that ai = 1 ; ai+1=2 ; ...

then tau = (123...p), sigma is some random transposition and (ai ai+1) = (12)

[wooden ember]

yeah but this started from the assumption that sigma is the transposition of two subsequent elements of the cycle tau