groups-rings-fields

[formal ermine]

yeah I was staring at those for a while

[elder wave]

stare at them for a little longer

[formal ermine]

r.(-a) + r.a = r.(a - a) = r.0 = r(0.0) = (r0).0 = 0.0 = 0 so r.(-a) is the inverse to r.a and because inverses are unique r.(-a) = -(r.a)

does that look gucci?

[rustic crown]

ooh lol you just meant torsion elements >.< i've been seeing a lot of derived functors lately, got confused >.<

[chilly ocean]

Oh lol

[harsh onyx]

Moving this from #discrete-math:
So I asked my prof this question and I'm not 100% sure why he replied yes, since he didn't justify it:
The question is: if $\alpha\in \overline{\mathbb{Q}}$, and $f$ is the minimal polynomial of $\alpha$ in $\mathbb{Q}[x]$ and $f\cdot h \in \mathbb{Z}[x]$ is irreducible and $f\cdot g \in \mathbb{Z}[x]$ is irreducible, then is $h \pm g$?
My reasoning goes: suppose $x \in \overline{\mathbb{Q}}$, then we can construct a minimal polynomial $f_x \in \mathbb{Q}[x]$ for it. We can always multiply $f_x$ by $\frac{gcd(numerators)}{lcm(denominators)}$ to get something in $\mathbb{Z}[x]$, and this must be irreducible.

However I'm not convinced. Thoughts?

This question stemmed from trying to extend Euclid's lemma to elements in $\mathbb{Z}[x]$.

[cloud walrus]

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

[south patrol]

If f is min poly and fh is irreducible, isn't h just 1 or -1? Since f is a factor

Or am I missing something

[next obsidian]

Minimal polynomial over Q, then h and h have to be polynomial in Z

They remove denominators such that the polynomial is irreducible

[south patrol]

Okay yes fair enough lol

[formal ermine]

is every polynomial primitive in Q[x]

I think yes?

because Q is a field

[next obsidian]

What’s a primitive polynomial

[formal ermine]

gcd of its coefficients is a unit

[next obsidian]

Then yes lol

[formal ermine]

yeah

$3x^7 + 5$ is irreducible by eisenstein's criterion, right? cuz we can just choose any prime number $\not \in \Set{ 3, 5 }$ and it will work, ye?

[cloud walrus]

yes yes yes no

[quiet pelican]

It has to be 5 as it must divide the last coef, but p^2 must not

[formal ermine]

ah ye

[formal ermine]

Let $I_1, \ldots, I_n \subseteq R$ be pairwise coprime ideals. Then there must exist $x_k \in I_k$ and $y_k \in I_j$ with $x_k + y_k = 1$. Thus $$1 = \prod_{k \neq j} (x_k + y_k) = \prod_{k \neq j} x_k + : (\text{terms with at least one factor } y_k)$$
Because the first summand lies in $\prod_{k \neq j} I_k$ and every other summand lies in $I_j$ (as $y_k \in I_j$), we get that $1 \in (\prod_{k \neq j} I_k) + I_j$. Thereby showing that $I_j$ and $\prod_{k \neq j} I_k$ are coprime.

is this proof correct?

[cloud walrus]

yes yes yes no

[delicate bloom]

just for funsies to show one trick that's unnecessary here, you can say for $f(x)=3x^7+5$ then $g(x)=x^7f(1/x)=5x^7+3$, which is eisenstein for p=3. f factors if and only if g factors. Point is you can apply eisenstein forwards or backwards on the coefficients basically.

[cloud walrus]

Merosity

[formal ermine]

will come back to this tomorrow

[delicate bloom]

another common trick is translating too, like if f(x) is irreducible so is f(x+1), specifically have showing cyclotomic polynomials are irreducible in mind as a good example here

just squeezing eisenstein for all its worth

[formal ermine]

what are cyclotomic polynomials

[pastel cliff]

superrrr late but - isn't this the whole field?

to repost the question, i was asking about showing the Q or Z/p are contained in every field

[delicate bloom]

minimal polynomials for roots of unity, like x^2+1 is the minimal polynomial for the 4th roots of unity

[formal ermine]

what's a minimal polynomial

[chilly ocean]

the subfield generated by 1 is the smallest subfield containing 1. R contains the subfield Q which contains 1, so no it doesn't have to be the whole field

[pastel cliff]

ooh that makes sense

ok brb then

[solar glacier]

question

in the ring of Gaussian integers, how do you find the GCD of an integer an a complex valued number

i.e., gcd of 85 and 1+13i

[formal ermine]

euclidean algorithm

[solar glacier]

are q and r pulled from C or Z

[formal ermine]

Z[i]

[solar glacier]

so of the form ni where n in Z

[formal ermine]

a+bi

[pastel cliff]

$a + bi, : a,b \in \Z$

[cloud walrus]

sebbb

[solar glacier]

so my r and q of of this form then ^

[formal ermine]

Z[i] are the gaussian integers

[pastel cliff]

it's integers adjoined i

[formal ermine]

just do the modulo variant

,, z_{j + 1} = z_{j - 1} \text{ mod } z_j

[cloud walrus]

yes yes yes no

[formal ermine]

with $z_0 = x$ and $z_1 = y$

[cloud walrus]

yes yes yes no

[formal ermine]

you can easily see that this is just the euclidean algorithm

[rustic crown]

(>.<)

[pastel cliff]

[formal ermine]

(a mod b means the minimal n under the norm s.t. a cong n mod b)

[rustic crown]

it's not "the" unless you're working with something like polynomials over fields >.<

[formal ermine]

the gaussian integers usually use a+bi -> a^2 + b^2, no?

[rustic crown]

yep

[formal ermine]

so wdym with this

[rustic crown]

like you give notation to something only if it exists and is unique

so there maybe like multiple n such that a = n (mod b) and they all have the same minimal norm

like say in Z[i] you'r dividing by 2 and the remainder is i, should you call it i or -i?

both have same norm 🙈

[formal ermine]

the gcd differs by a unit (-1 or 1 in Z[i]), so wouldn't it not matter?

[rustic crown]

the remainder doesn't though

wait are you saying the choice won't change the gcd (up to unit)? that's completely true

[formal ermine]

yeah only by a unit

[delicate bloom]

I just mean irreducible, like x^2+1 is the smallest polynomial with 4th roots of unity as roots, but of course x^4-1 also has them as roots but is higher degree

[formal ermine]

maybe I wasn't formal enough in my definition lol

[rustic crown]

right, but the point i was making was just the notation is weird :p

[formal ermine]

ye

[rustic crown]

ofc everybody reading it will understand what you mean >.<

[formal ermine]

ah

okk

[rustic crown]

but it's similar to writing sqrt(2) and say you're not working in something like reals where there is a way to distinguish the two roots and prefer one for the notation

[solar glacier]

so for GCD of 85 and 1+13i do I write 85=q(1+13i)+r then work backwards ?

[rustic crown]

do you also wanna solve 85 * x + (1+13i) * y = 1?

[formal ermine]

do 85/(1+13i) and round both components to the nearest integer

then plug that in as q and solve for r

[solar glacier]

so mult by its complex conj ?

[formal ermine]

if the deg of r doesn't fulfill the property then round a different way (you have 4 choices at max)

[formal ermine]

yes

[rustic crown]

it will :p

because any alpha/beta is at most 1/sqrt(2) away from something in Z[i].

which gives you |alpha - beta * q| <= |beta|/sqrt(2)

so N(alpha - beta*q) <= N(beta)/2

[formal ermine]

ah right

[rustic crown]

and if you're rounding you're gonna get the closest thing in Z[i], so it's sure to satisfy it

if you only take floors/ceiling, then you may have to fiddle it a bit

[solar glacier]

so I get that 85/(1+13i) = (85-1105i)/170

so I write this as 1-7i?

[rustic crown]

,w 85/(1+13i)

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by det#7067

[rustic crown]

yee 1-7i works

[solar glacier]

and I take that and plug it in for q in 85=q(1+13i)+r and solve for r?

[rustic crown]

yee

[solar glacier]

i get -7+6i

[rustic crown]

that is the division algorithm, now you repeat this with (1+13i) and r

and keep doing it until the remainder at some point becomes 0

[pastel cliff]

is there a ring which is neither artinian nor noetherian

i cant seem to find one so im leaning towards no

[chilly ocean]

polynomial ring over a field with infinitely many variables

[pastel cliff]

i know that's not noetherian

but i thought it would be aritinian

[chilly ocean]

commutative + artinian -> noetherian

iirc

been a while since i did commutative algebra

[rustic crown]

hi terra :3

[chilly ocean]

hi

[pastel cliff]

oh shit yeah

[chilly ocean]

all rings are commutative anyways

[chilly ocean]

i don't know anything about non-commutative rings so i'm going to reduce this to just artinian -> noetherian

[pastel cliff]

im assuming commutative rings anyways

[rustic crown]

is the proof of that easy?

i never did much commie alg

[pastel cliff]

i mean it makes sense? i guess?

i dont get artinianness that much tbh

[chilly ocean]

let me crack open a&m and see

[pastel cliff]

like i get DCC yeah

but does it have a nicer analogue like finitely generated ideals for noetherian rings

[chilly ocean]

"A ring A is Artin if and only if it is Noetherian and of dimension zero.

the proof is not particularly long

6.11 says "let A be a ring in which hte zero ring is a product of maximal ideals. then A is noetherian iff artinian"

so maybe it takes a tiny bit of building up to

[pastel cliff]

most of my proof is "as shown in class, R[x, ...] is not noetherian" and then contrapositive of that

"as shown in class" is more powerful than "proof is left to reader"

[chilly ocean]

i use "as shown in class" on my homework all the time

i don't go to class

[pastel cliff]

tbf my prof did show it i just forgor

[pastel cliff]

going back to this - am i right in saying that 1 generates Q (and in general, something isomorphic to Q in every (probably not actually every) field

[chilly ocean]

a finite field isn't gonna have a copy of Q

just for cardinality reasons

[pastel cliff]

ah then every infnite field

[rustic crown]

nu

F_p(x)

[chilly ocean]

the point is to look at the characteristic of the field

[pastel cliff]

ngl i dont think prof defined characteristic of a field

like i know what it is

[chilly ocean]

maybe in more concrete terms, look at the map from Z into your field that adds 1 a bunch of times

[pastel cliff]

but dont think i can use it

[chilly ocean]

if you follow my hint you'll be using it anyways

bigger hint

[rustic crown]

[chilly ocean]

(because the characteristic of a field comes from the map i mentioned above)

oh wait you said you already know what it is

oop

[chilly radish]

It's a pretty difficult proof iirc

[chilly ocean]

i discovered that

classic deceptive a&m

[unique valve]

Isn’t this not a field

[rustic crown]

it is the fraction field of F_p[x]

[unique valve]

Oh that’s what that means

Ok I thought that was just polynomials in Fp

[rustic crown]

yee round parenthesis are usually used for fields

like you would see Q(sqrt2) instead of Q[sqrt2] even though they mean the same thing

[unique valve]

Ok

So is the fraction field just polynomial fractions?

[rustic crown]

yup

in general for an integral domain D, you can carry of the "fraction" construction to put it inside its fraction field

[rapid junco]

How could one disprove that Z[i] form a field?

I get that all elements besides the units arent invertible but how could one prove this

[chilly ocean]

disprove that it's not a field? so prove that it's a field?

[pastel cliff]

find a nonunit

[chilly ocean]

🤔

[rapid junco]

sorry

[pastel cliff]

oh wait

[south patrol]

this is the definition of unit

[rapid junco]

prove its not a field

[south patrol]

find an element without an inverse

[rapid junco]

yeah but im in a linear algebra class and we havent done anything whatsoever with rings and this was one of my questions

[south patrol]

a cheating way to do it is you know there must be a multiplicative inverse in like C

and so just pick a number whose multiplicative inverse in C no longer lies in Z[i]

[rapid junco]

ah

What I said is that when we have Im(z) = 0 then the gaussian integers are isomorphic to the integers

[pastel cliff]

also remember that the integers alone dont have mult. inverses

[rapid junco]

thus there are no inverses besides the units

would this suffice?

[pastel cliff]

pick a specific one

[south patrol]

again, this is just by the definition of unit

but yes you can do something like 2, if that's what you mean

[rapid junco]

I just said fix Re(z) != 1, -1 and fix Im(z) = 0

[pastel cliff]

huh

[rapid junco]

I didnt explicitely pick anything

[pastel cliff]

the easiest way to disprove a "for all" statement is to prove a "there exists" one

[rapid junco]

yeah

[pastel cliff]

if you prove there exists an element without an inverse you disprove that all elements have inverses

so just pick one, you dont need to be general

[rapid junco]

okay i get this but with what I said have I not just done that?

[pastel cliff]

ngl, i dont know what Re and lm stand for, maybe prof specific notation

[rapid junco]

Real part of z

[pastel cliff]

unless im having a goober moment

[rapid junco]

imaginary part of z

z = 5 + 4i

Re(z) = 5 Im(z) = 4

[pastel cliff]

oh you never said this

[rapid junco]

wdym?

oh sorry yeah I meant fix the imaginary part to be zero and the real part to be anything but 1, -1. Then you can never find an inverse

[pastel cliff]

oh nvm i thought you were using that z in your answer, it was just an example

im just saying that you're overcomplicating it

[rapid junco]

yeah okay

[pastel cliff]

like yeah that's probably fine

but just show that 2 doesnt have an inverse

[rapid junco]

just next time just pick something

[unique valve]

Yeah almost nothing has an inverse

[pastel cliff]

shorter usually = better

[unique valve]

Only 1,-1,i, and -i have inverses right?

[rapid junco]

yeah thats what i noticed

[pastel cliff]

i can see how this is relevant but what is the kernel of this homomorphism?

[chilly ocean]

[pastel cliff]

i'll rephrase

ok well the image should be Q

[rustic crown]

.<

[pastel cliff]

is that a good or bad face

[rustic crown]

neither :p

the image is generated by 1, so can't be all of Q

like if you keep adding 1 to itself, there are two possible cases. one where you loop back to 0, or other where you just keep going bigger and bigger

[pastel cliff]

yeah im assuming that's the finite and infinite case, respectively

[chilly ocean]

the kernel is (characteristic)Z

and the characteristic is gonna be zero or prime

in one case the quotient by the kernel is a field, and in the other a bit more argument is needed

that's more of a proof outline than a hint

sorry if it spoils a lot

[pastel cliff]

i mean i've been stuck on it since yesterday, sleeping on it didnt help

ohhhh wait so the Z/p case isn't necessarily only for finite fields

[rustic crown]

yee

[rustic crown]

that's the example i gave here

[pastel cliff]

it just has to be an infinite field where the subfield generated by 1 is also finite

(im saying that's the kind of field that contains Z/p)

[rustic crown]

yea

you mean the image of Z --> your field is finite/inifnite

instead of the actual field being finite or infinite

[unique valve]

I contain Z/p

Is there a nice way to represent the algebraic closure of a finite field

Like I know it’s just the union of them

But that is not nice

[rustic crown]

represent in what sense?

[unique valve]

Like the elements of finite fields can be represented by polynomials in Fp mod irreducible polynomials

[rustic crown]

oh represent the elements?

[unique valve]

Yeah

It’s another example of infinite characteristic p

But I have like no idea what it is

Even though I know what it is

You know what I mean?

[rustic crown]

[rustic crown]

you formalize this by saying it's a filtered colimit, if that's what you're asking

[unique valve]

I just don’t have much understanding of the structure of it

[rustic crown]

but in any case working with elements of any algebraic field extension can be done by working with the corresponding minimal polynomials or something

[unique valve]

Yeah

Like I assume it’s not isomorphic to the fraction field

[rustic crown]

fraction field of?

[unique valve]

F_p(x)

[rustic crown]

yea, because this guy has a transcendental element

while algebraic closures are algebraic by definition

[unique valve]

You mean transcendental with respect to Fp right?

[rustic crown]

yep

[unique valve]

I see

Everything is transcendental

Other than Fp

[rustic crown]

yee

[unique valve]

Actually Fp/Fp

[tender bough]

I'm manually computing the roots of the Lie algebra sl(3) and this question came to me:
In a semi-simple Lie algebra, do two linearly independent simultaneous eigenvector of the adjoint actions of the Cartan subalgebra necessarily generate different roots?
A similar but not identical statement is false: "two linearly independent eigenvectors have different eigenvalues". This can be false for example, when the eigenvalue has geometric multiplicity 2

[pastel cliff]

question about that discussion from earlier about Z/p and Q in a field - what about negative values of Z

actually wait

char cant be negative

[south patrol]

Indeed

[muted epoch]

Can I get some help with this problem?

[pastel cliff]

ring without identity

[unique valve]

rng

[chilly ocean]

An ideal generated by a is intersection of all ideals containing a

[slender hamlet]

That's true for non-commutative too. Artinian rings are noetherian
(More specifically left-Artinian rings are left-Noetherian, right-Artinian rings are right Noetherian)

[chilly ocean]

Therefore, given that a is in this ideal (intersection of ideals is an ideal), derive what needs necessarily to be a subset of it

Then show that what's left is indeed an ideal

This will show it

[slender hamlet]

So the proof starts showing that for an artinian ring R, J(R) is nilpotent

Then R/J(R) is a semisimple ring, and is both Noetherian and Artinian by Artin-Wedderburn

And over a semisimple ring, being a Noetherian module and Artinian module are equivalent to being finitely generated, so since each J(R)^i/J(R)^{i+1} is an Artinian R-module and thus an Artinian R/J(R)-module, it is also a Noetherian R/J(R) module

Now if 0->M->N->P->0 is an exact sequence with M, P Noetherian then so is N

Inductively show that for each j in N, n≥0, that J(R)^n/J(R)^{n+j} is a Noetherian R-module

(Here J(R)^0 is taken as R)

Now take r such that J(R)^r=0 and then R=J(R)^0/J(R)^r is Noetherian as well

So if we started with left Artinian ring, R is Noetherian as a left R module, so left Noetherian, similarly for the right Artinian case

[pastel cliff]

im being asked to determine if there exists a ring $R$ and a multiplicatively closed subset $S$ of $R$ so that $S^{-1}R$ is noetherian but $R$ is not noetherian.

[cloud walrus]

sebbb

[pastel cliff]

oh wait

uhhh

no nvm, but i do think it can exist

[agile burrow]

Sure, take a non-noetherian domain and its field of fractions

[pastel cliff]

Q[x, ...] is my goto non-noetherian ring

[agile burrow]

Mine too 🙂

[pastel cliff]

oh i remember proving a long time ago that polynomials over a domain are also a domain

i think

[agile burrow]

Yeah, that's right

[pastel cliff]

and Frac(R) is noetherian bc it's a field right...?

[agile burrow]

Yeah ! 😀

[chilly ocean]

Frac(R)? i hardly know her!

[pastel cliff]

nice meme

i'll be taking that thank you very much

[restive birch]

i was just thinking, is it true that if a group has at least one element of order 2, then the order of the group is even? because if |g|=2, then {1, g} should be a subgroup of G, and then lagrange's theorem

[pastel cliff]

well you have to assume finite group first

[restive birch]

right

assuming a finite group, then

[pastel cliff]

then probably yeah

[restive birch]

does this also hold for any |g|=n?

[pastel cliff]

the converse is also true, which might be a nice exercise, though it's a direct consequence of a thm you might not know

[south patrol]

Yes this is correct

[restive birch]

whoops

[south patrol]

This is lagrange applied to <g>

[pastel cliff]

i.e. a group of even order must have an element of order 2

might also be a nice exercise to show that an abelian group has even order

[restive birch]

my one problem with DF is that its like "alright, heres what blah blah blah means" and then throws you directly into 54 very hard exercises

[pastel cliff]

lol i tried and failed to self study out of dnf this past summer

and i didnt like the linear algebra in artin

[restive birch]

ive spent a month reading ~40 pages

[pastel cliff]

i spent the better part of two months reading 100 pages of munkres

this is the painfully universal experience of trying to learn math

especially alone

stick with it

the ability to stick with things like that is a skill

[restive birch]

oh i am entirely resigned to losing interest in the next month

[pastel cliff]

still, even if you dont feel like you make much progress, you're still at least developing the skill of sticking with hard things

[restive birch]

yea, i just tend to have short-lasting hobbies

[pastel cliff]

depends on what you want out of it that's just my two cents

dnf is probably not the most engaging thing for a "short" burst of studying though

@next obsidian can i get that hint from yesterday im embarrassed to say i didn't get it

it was showing that for $R$ a noetherian ring and assuming that $\phi: R \to R$ is a surjective homomorphism, this is iso

[cloud walrus]

sebbb

[pastel cliff]

there's probably some short pf by contradiction

[next obsidian]

Let f be a surjective morphism with non-trivial kernel. Find a surjective morphism g whose kernel strictly contains the kernel of f

Then conclude

[pastel cliff]

[slender hamlet]

Yeah it takes like hours to understand a single page when reading new stuff

[restive birch]

totally and entirely stuck on this problem. I just have no idea how to start - normalizers confuse me

[chilly ocean]

how do you normally show that two sets are equal

[restive birch]

G (upside down U) H = G -> G = H?

[chilly ocean]

i'm looking for a simpler answer

forget groups for a second

how do you show two sets are equal?

[restive birch]

by showing they have the same elements?

[chilly ocean]

yes

that's where you can start

show that N_H(A) and N_G(A) cap H have the same elements

so an element of the first is an element of the second, and an element of the second is an element of the first

each is a subset of the other

etc.

("cap" means the upsidedown u)

[restive birch]

N_H(A) ≤ N_G(A), right?

wait, this problem is way easier than it seemed

[chilly ocean]

i wouldn't write the symbol ≤ until you actually know that N_H(A) is a subgroup of something (first part of the problem), but yes, that is true

again all you really need to do to prove it is show that the two sets have the same elements

[restive birch]

well, in previous problems, it been shown that N_G(A) is always a subgroup of G

and the definition of N_H(A) necessitates only elements in H

[chilly ocean]

"(note that A need not be a subset of H)"

what do you want to use N_H(A) ≤ N_G(A) for anyways?

[restive birch]

ah, true

[lapis trail]

Group actions; I get them but I don't get them. Is it just the permutations of elements of a set? Can someone give me an example of a group acting on a a set that is not a group. I know the definition but I don;t really see why it's important

[restive birch]

alright, i worked out the problem the long way and it makes sense. just out of curiosity, how could we prove that N_H(A) ≤ N_G(A)? obviously its a subset, but i cant think of how to show that its a group

[chilly ocean]

Is it just the permutations of elements of a set
you could look at it that way. a group action of a group G on a set X is a homomorphism G -> {bijections X -> X}. in practice we usually care about more than just bijections from X to X (we usually care about X with extra structure, and groups whose elements act by bijections on X preserving that structure).

Can someone give me an example of a group acting on a a set that is not a group
symmetry groups acting on {1, ..., n} are an immediate one. you've almost certainly seen other examples no?

but I don;t really see why it's important
group actions pop up literally everywhere and they help us understand a lot about groups

[chilly ocean]

the problem concludes that N_H(A) is a subgroup of H, so in particular it's a group

[restive birch]

right, i didnt see that part

so i'll work on that

[chilly ocean]

all you need to do to prove it is use the fact that N_H(A) = N_G(A) cap H (which you proved)

something something the intersection of two subgroups of G is a subgroup

[dusky birch]

Hi

what does the mapping phi(q) = q(i) mean?

I don't know what q(i) is.

[chilly ocean]

evaluate the polynomial q(x) at x = i

substitute i for x

[restive birch]

ah, i see how to do this

[next obsidian]

TTeppa

[restive birch]

thanks for the help

[chilly ocean]

Chmonkey.

[next obsidian]

Ur back

[chilly ocean]

indeed

joker mode arc is over

or something like that

u know how it be

[next obsidian]

Yeah

Can u help me with algebra

[dusky birch]

kt

kt

ty

sorry wrong layout

[chilly ocean]

probably not at your level

[next obsidian]

It’s not that hard

It’s finite group theory

[chilly ocean]

oh god

[next obsidian]

How do u show a group of odd order is solvable?

[chilly ocean]

seems easy lmao

have you tried just showing it???

[next obsidian]

No, maybe I should try that

I can do it when the order is prime

Do u think that’s a good first step?

[chilly ocean]

prolly

[next obsidian]

Ok

Thx

[chilly ocean]

👍

if u get stuck ping helpers 15 mins

[next obsidian]

Ok sounds good

[rotund aurora]

Thanks

[thorn delta]

think harder I guess? Idk

I don't want to be rude

[formal ermine]

start by showing that a finite group has a finite order

[delicate orchid]

Bro straight up asking for a proof of the odd order theorem on the heavens above

[rustic crown]

Try showing that every non-abelian finite simple group has even order

[lethal dune]

[maiden heath]

Hi, this is the proof i'm writing for the proposition stated in the screenshot. I'm trying to do the last part but i'm a bit stuck

[cloud walrus]

ArtyLeAardvark

[maiden heath]

I'm fairly certain that the only elements of $\mathbb{C}\mathcal{S}n$ that is invariant under conjugation of $\mathcal{S}{n-1}$ is $X_n + t$ where $t$ is some element in $\mathbb{C}Z_{n-1}$, but i'm struggling to show it

[cloud walrus]

ArtyLeAardvark

[median pawn]

https://en.wikipedia.org/wiki/Monothetic_group

Why are monothetic groups necessarily abelian?

It's probably a density argument but I need help formalizing it. Do we use nets?

Certainly the dense cyclic subgroup is abelian

[chilly ocean]

xy = yx for all x, y from the cyclic group

now assuming they mean Hausdorff topological groups

this means that g(x, y) = xy and f(y, x) = yx agree on a dense set

and as such are equal

this is commutativity

[chilly ocean]

you can use nets, yes, but it's a standard result for Hausdorff spaces

it's usually phrased as "a continuous function defined on a dense subset has at most one continuous extension to the whole space"

[median pawn]

ahh i see

so if the space G is hausdorff, and two continuous functions f and g agree on a dense subset of G, then f = g?

[chilly ocean]

yes

but note that here G isn't the group we are concerned about

but a product of it with itself

[median pawn]

yes G x G

but the codomain for the maps is still G

it seems we only need the codomain to be hausdorff: https://math.stackexchange.com/questions/1084018/if-two-continuous-maps-into-a-hausdorff-space-agree-on-a-dense-subset-they-are

[chilly ocean]

yeah

now if H is this dense subgroup, then note that H x H is dense in G x G

because cl(H x H) = cl(H) x cl(H) = G x G

closure of product is product of closures

[toxic zephyr]

can someone give me an example of a ring which does not satisfy the ascending chain condition?

[deep halo]

Say, I have a group, represented as a list of permutations. How can I find a minimal generating set for this group? Are there algorithms for that?

(Minimal, as in cannot be made smaller, but not necessarily the smallest)

[proud bear]

k[x_1,x_2,x_3,...]. Then take the ideals (x_1), (x_1, x_2), (x_1, x_2, x_3), ...

[toxic zephyr]

what is k?

[proud bear]

A field

[toxic zephyr]

ahh so all polynomials with coefficients in a field with an arbitrary number of variables?

[proud bear]

Yeah

[toxic zephyr]

that makes sense that it would not be finitely generated lol

[proud bear]

Indeed

[formal ermine]

doesn't have to be a field

can be a ring

[toxic zephyr]

probably should have realized that any example would be something like that haha. im trying to review noetherian rings.

[rustic crown]

(non-zero ring)

[formal ermine]

how would I show $$R \text{ is a field} \implies \text{every finitely generated } R\text{-module is free}$$

[rustic crown]

show vector spaces have basis

zorn it up

max linear indep set

[formal ermine]

it's sad that I barely know any linear algebra

[rustic crown]

oh you have finitely generated

in that case do "min spanning set"

so no need of zorn

[formal ermine]

especially because my course just started with field extensions and are assuming prior field theory knowledge

[formal ermine]

what's a spanning set

[rustic crown]

a generating set, but the word span is used in the context of vector spaces

[elder wave]

module theory without lin alg

[rustic crown]

basically start with a spanning set, keep on deleting elements until the set doesn't span it anymore

[formal ermine]

span it?

[rustic crown]

other elements of your vector space can't be written as linear combinations of these anymore

[formal ermine]

ah it doesn't generate it anymore

got it

[rustic crown]

you have to find the minimal spanning set

and show that it forms a basis

[formal ermine]

how do I show that something is linearly independent

[rustic crown]

assume a linear combi is zero

such that not all coefficinets are zero

get a contradiction

[formal ermine]

could I also just show that $f_{m_1, \ldots, m_n} : R^n \to M, (r_1, \ldots, r_n) \mapsto \sum_{i=1}^n r_im_i$ is an isomorphism?

[cloud walrus]

yes yes yes no

[rustic crown]

yee, same thing

[formal ermine]

because then $\Set{m_1, \ldots, m_n}$ would be a basis

[cloud walrus]

yes yes yes no

[rustic crown]

surjetivity is equivalent to showing the set spans

and injectivity is same as showing linear independence

[formal ermine]

ah

lol

<sarcasm>
in the lecture

we had the example "Example 3.5.7: (a) If R is a field, then every R-Module is free, every vector space has a basis"

do you reckon I can just reference that and call it a day

</sarcasm>

ok back to the actual thing

[rustic crown]

but more importantly, why are you studying modules before studying vector spaces >.<

ik it doesn't matter, but still >.<

[formal ermine]

my course assumes linear algebra knowledge

but I didn't know that it does

[rustic crown]

ah

module theory is just spicy linear algebra

[formal ermine]

how do I know that such set exists

[toxic zephyr]

modules are so cool

[rustic crown]

(btw you don't need commutativity here, just assuming it's a division ring is enough)

[rustic crown]

cuz your vector space is finitely generated. which precisely means there is such a finite set

that's why the "deleting" algo will eventually end

[formal ermine]

I actually have to show iff (already did the other direction)

[rustic crown]

whut

[formal ermine]

right

so we have a finite generating set

[rustic crown]

if you're only looking at commutative rings then it explains the statement :p

[formal ermine]

they want me to show $R \text{ is a field} \iff \text{every finitely generated } R\text{-module is free}$ as a homework, I already managed to do $\if$

[formal ermine]

yes also that

[rustic crown]

is texit sad?

[formal ermine]

so we just have to show that it's linearly independent and we're done yes?

I don't understand the deleting elements to make it minimal part

[rustic crown]

you have a finite generating set. pick the subset of smallest size which still generates the module

(you can do that because natural numbers are well ordered)

[formal ermine]

why are we doing that

[rustic crown]

cause if you have redundancies, then it no way would be linearly independent

like consider R^2 with {(1,0), (0, 1), (1, 1)}

that's a finite generating set

[formal ermine]

yes

[rustic crown]

but not a basis

because you don't need all 3 elements

any two are good enough

[formal ermine]

(1,0) + (0,1) - (1,1) = 0

right

[rustic crown]

i said it like that earlier because that's how i think about it... "remove stuff until ..."

[formal ermine]

yeah

does the set being minimal already imply linear independence?

[rustic crown]

yee!

that's where you would need field stuff

[formal ermine]

I can intuitively see why but I can't put it into words

[rustic crown]

because if you consider the module Z, then {2, 3} is a minimal generating set as neither {2} nor {3} manages to generate it

but {2, 3} is still not a basis

[formal ermine]

does Z even have a basis

[rustic crown]

try showing linear dependence would imply there is a smoller generating set

[rustic crown]

yea, {1}

R^n is a free R-module in general

with the standard basis {(1,0,...) etc}

[formal ermine]

if k_1n_1 + k_2n_2 + ... + k_mn_m = 0 then k_mn_m = -k_1n_1 - k_2n_2 - ... and therefore the set would not be minimal as one of the elements is a linear combination of all/some of the other ones?

(k_m \neq 0)

[rustic crown]

not exactly

keep that above example in mind

M = Z, m = 2, n = 3

{m,n} is a minimal generating set

but it is dependent

3m-2n=0

in this case we can't remove elements to get a smaller generating set

but in the case of fields this can be done

[formal ermine]

why

[rustic crown]

think a little >.< you're almost there >.<

[rustic crown]

what does this and a field tell you

[formal ermine]

we can divide by k_m

[rustic crown]

[formal ermine]

isn't that what I said though

[rustic crown]

and we couldn't do that here in Z

[rustic crown]

ah i didn't read divide then >.<

[formal ermine]

oh I forgot to say that, sorry

well

thanks! I understand it now

what does $\mbb{L} \oplus \mbb{K}$ in the context of field theory mean

[solar glacier]

Question

How do you determine

[chilly ocean]

wolframalpha

[solar glacier]

$\Bbb{Z}[i] / (1+i)$

[chilly ocean]

it's like Z[i] but i equals -1 (i am joking)

[rustic crown]

because i think i is not a free variable

it's probably sqrt(-1)

.<

Z[i]/(1+i) should be F2 right

[solar glacier]

The field of two elements ?

[rustic crown]

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

[rotund aurora]

completelly ignore me

definitely this is useless

im gonna delete so the bs I said doesnt confuse anyone

[solar glacier]

Could you elaborate a bit on this plz lol

[rotund aurora]

correspondence theorem essentially

[rustic crown]

okie so, mainly i'm using the third iso theorem

yea same thing

[solar glacier]

Aha I see third iso theorem then is used here

[rustic crown]

if R is a ring and a, b are elements, then
R/(a, b) = (R/a)/(b)

basically one can quotient one after the other

[rotund aurora]

its so nice

[rustic crown]

so here if you look at Z[x]/(x^2+1, x+1)

and if you first quotient by x^2+1, you get
(Z[x]/(x^2+1))/(x+1) but under the isomorphism Z[x]/(x^2+1) the ideal (x+1) corresponds to (i+1)

so this is Z[i]/(1+i)

now try going the other way

[solar glacier]

This is from third iso theorem?

[rustic crown]

yep

[chilly ocean]

how do you remember what any but the first isomorphism theorems are

i never could

[rustic crown]

the more general version which you have seen says that if I is an ideal contained in J, then
R/J = (R/I)/(J/I)

third is fractions 🙈

[solar glacier]

Yes this is the version I’ve seen

[rustic crown]

yea, just take J = (a,b) and I=(a)

[solar glacier]

Aha!

[rustic crown]

and i'm just writing the ideal (a, b)/(a) in the ring R/(a) as the thing generated by b, so (b)

i should be more careful and write it ([b]) or something

but you get the point i hope >.<

[solar glacier]

Yeah (b) is clear enough

Thanks 🙏

[rustic crown]

[formal ermine]

is there an easier way of showing that the image and preimage are submodules other than just checking that all of the properties hold?

[chilly ocean]

that's already so easy though

[formal ermine]

it's too much effort for me

like

writing that much

it's at least 5 properties each

[chilly ocean]

if you have a module map M -> N you could say the pre-image of a submodule N' of N is the kernel of the composition M -> N -> N/N'

if you already know that kernels of module maps are submodules

[formal ermine]

nah haven't shown that yet lmao

[chilly ocean]

stop being lazy and just check the properties if you really have to

i dunno of a quick way to check that the image of M -> N will be a submodule of N like this. just do it lol

[formal ermine]

):

oke

[chilly ocean]

(and that quotients of modules by submodules are modules)

[rustic crown]

you already know most from when you probably studied abelian groups. you just erally have to check that they're stable under the action

but yea, stop being lazy >.<

[formal ermine]

as in rn in M for r in R and n in M?

[chilly ocean]

do people actually go through the lists of properties like these and check them outside of homework assignments

[rustic crown]

(i do >.<)

[chilly ocean]

()

[rustic crown]

(i checked in all gory details that the sheaf hom and sheaf tensor product are adjoints)

(but ofc took the naturality on faith uwu)

[rustic crown]

yee

[formal ermine]

okk

[next obsidian]

I think this is formal

Like I think you can derive this from the usual statements over opens combined with sheafification

Maybe

[rotund aurora]

So I get that in the case for algebras, introducing a group for grading is useful so that the degree function has some structure. But does that add anything in the context of vector spaces (without a vector product) ?

[restive birch]

i would just like to say, as someone who doesnt know most of these words, this sentence sounds ridiculous

math coins weird terms

"sheaf"

[chilly ocean]

the sheaf terminology comes from agriculture

[restive birch]

huh

[formal ermine]

just think about how weird group theory terminology (which you now know) must've sounded to you before you started learning it

[restive birch]

at least "centralizer" is better than "sheaf"

[rustic crown]

lol but i did that when i barely understood how to work with a sheaf, so felt like checking a huge mountain of details :p

[formal ermine]

how do I show that for any element in the preimage of a modul homo the stable property or w/e holds

like it'd be easy to show for the image

cuz r.f(x) = f(r.x)

but we have f^-1(x)

[rustic crown]

what is f^-1

like the map may not be an isomorphism

do you mean the whole coset mapping to x?

because that may not even contain 0 (unless x = 0)

[formal ermine]

Let M and N be two R-Modules and f : M -> N a module homomorphism of R-Modules. Let M' subseteq M and N' subseteq N be submodules. Let m in f^-1(N') (preimage of f under N') i.e. m = f^-1(x) for some x in N'

wait

that's wrong isn't it

the m = f^-1(x)

f(m) = x for x in N'

that's the correct one lol?

[rustic crown]

yee

all the m such that f(m) lies in N'

[formal ermine]

ah darn it

ok thanks

[rustic crown]

its the same

f(m) = x for some x in N' is equivalent to saying f(m) in N'

[formal ermine]

yeah but I said m = f^-1(x) everywhere lol

[rustic crown]

but you corrected :3

[formal ermine]

yeah but I already wrote it down in my notes lmao

[rustic crown]

oh oopsie

[formal ermine]

ok now I'm stuck at showing that it's closed

we have a,b in preimage of N' then f(a) = x, f(b) = y for x,y in N'

nvm

got it

I'm stupid

[rustic crown]

there's another simple way

since you've already proven that kernel is a submodule

i shouldn't call it simple, since it's exactly the same... just that you don't have to write these twice

[formal ermine]

no we haven't

[rustic crown]

ah okie, i thought that this question said image and kernel >.<

[rustic crown]

and terra has said what i was gonna say anyway :p

[uncut girder]

UWU

[rustic crown]

uwu

[chilly ocean]

sniped

[toxic zephyr]

can someone provide an example of a module that isn't free (doesn't have a basis)?
also my prof mentioned that there are noncommutative modules with bases that have a different number of elements. could someone provide an example of that too?
thanks 🙂

[next obsidian]

Z/nZ over Z

For the noncomm thing, I have no idea it is some cursed shit I pretend doesn’t exist

[agile burrow]

The Wikipedia page on the "invariant basis number" gives an example of such a ring, namely the ring of column finite matrices

[solar glacier]

Question

[chilly ocean]

the answer is an unfortunate negative

[solar glacier]

Is Z[i]/(i) just the trivial ring w 0

[chilly ocean]

what do you think?

[solar glacier]

Since if an ideal contains a unit it’s the whole@ring

And (i)(-i)=1

So i is a unit

So the ideal generated by i is all of the Gaussian integers

[chilly ocean]

nice

this is good

[solar glacier]

Ok cool

Thanks 🙏

[ripe basalt]

i was given the definition for the krull topology on Gal(K/F) by the open subsets U either are empty or their a union of sets of the form \sigma N where N is a subgroup of Gal(K/F) corresponding to an intermediate field and \sigma is an element of Gal(K/F). How can you show for example this is closed under unions?

[next obsidian]

…

You defined it as a union of sets of a certain form

A union of unions of sets of a certain form is in particular a union of sets of a certain form

[ripe basalt]

i'd need to know the union is a subgroup of the galois group

why are you saying "..."

or is this just an attempt at being rude

(i generally assume people are being nice here so sorry if i messed up that assumption)

anyways i guess you're right

i dont need to show it's a galois group

and it's simple as you pointed out

so thanks for that i guess, just no need to be rude man

do you klnow how to show these sets are also closed?

the \sigma N i mean

[rotund aurora]

Something like this

[toxic zephyr]

this is awesome! thanks 🙂

would V=F[x] work for this example?

[rotund aurora]

Any vector space with countably infinite basis would work

so yes

[slender hamlet]

Did the same a few days ago

[slender hamlet]

We can do this without Hausdorffness. The function f: G×G->G sending (a,b)->aba^-1b^-1 is continuous.
Let e be the identity of G. The preimage of e under f ,say K, is a closed set containing Z×Z, where Z denotes a dense cyclic subgroup of G. As Z dense in G, Z×Z is dense in G×G (easy to check) so K=G×G, thus for all a,b in G, f((a,b))=1.

Thus the commutator subgroup [G,G] is trivial, so G is abelian

[wooden ember]

Does someone have an example of a fg k-algebra which is a domain, and for which the integral closure is not isomorphic to a polynomial ring?

Unless I’ve been doing it wrong, all the examples I’ve had to work with I ended up with a polynomial ring

(My examples were k[x,y,z]/(y^3 + y^2x^2 + yx^2 + x^3z) and k[x,y]/(y^2 + x^3) if someone wants to tell me I did it wrong)

[slender hamlet]

k[x,y]/(xy-1)

[wooden ember]

Right, that’s integrally closed correct?

[chilly ocean]

You just used Hausdorffness though

{e} won't be closed in general

Unless G is Hausdorff

[gritty sparrow]

Yeah localisations of normal domains are normal.

[wooden ember]

Where is a localisation involved here 🤔

[gritty sparrow]

k[x,y]/(xy-1) is isomorphic to S^-1k[x] where S={1,x,x^2,…}

[wooden ember]

Ah right yeah

[slender hamlet]

Oh

Well I guess this works when G is T_1

[chilly ocean]

It's all equivalent for topological groups

I didn't say that G is Hausdorff because idk.

I said it to differ from the case where no separation axioms are assumed

[slender hamlet]

Oh

Ok I see the equivalence

[gritty sparrow]

Obvious point but I’ll say it anyway, hausdorffness is necessary because otherwise we could take any non-abelian group with the indiscrete topology, and not that 1 is dense in it.

[rotund aurora]

How useful is the result of Shafarevich that says that any finite solvable group is the galois group of a finite field extension of Q?

You may replace "solvable" by abelian if you want

So I think this is a beautiful result and is interesting on its own, exemplifying the complexity of the Gal(\ol Q,Q) (I think), but I was wondering about applications

[formal ermine]

why is $f(x)$ irreducible $\iff$ $f(x + 1)$ irreducible

[cloud walrus]

pure for president

[formal ermine]

nvm

lol that seems way easier now that I've written it out

yeah got it

y = x + 1

[rustic crown]

.<

[south patrol]

ye i guess the map R[x] -> R[x] evaluating polynomials at x+1 is a ring auto lol

to make it slightly more fancy

[chilly ocean]

Is there a good description of the automorphisms of R[x] in terms of the automorphisms of R?

[next obsidian]

No

[gritty sparrow]

When R is a field, then it should just be given by an automorphism of the field and a choice of linear polynomial to send x to. In general it should be pretty bad. For example if R=Z[t_1, t_2] then we can define an automorphism of R[x] by sending t_1 to x, x to t_1 and t_2 to istelf. The restriction of this map to R maps has image outside of R so it looks like it would be hard to describe in terms of aut(R).

[warped viper]

Hi ! I'm interested in deeper topics in abstract algebra 🙂 If you are too I'd love to hear about what made you love this field so much

Join me here if you're interested, to avoid spamming #math-discussion message ❤️

[rotund aurora]

I love algebra so much

I have no idea why

[restive birch]

ive done all of this one but the last part- i cant think of an example

[warm wyvern]

I think it's likely you'll find an example with any two elements of a group that don't commute so think of a simple non-commutative group

even bigger hint: ||think of S_3||

[rotund aurora]

S3 is the smallest non-commutative group, so simplest examples will be there

[chilly ocean]

take any non identity x and look at y = x^{-1}

the problem only asks for x and y such that the order of xy does not equal ..., not does not divide ...

so you don't have to work hard

[formal ermine]

how do I find a homomorphism between $\bF_4$ and $\Set{\begin{pmatrix}a & b\b & b + a\end{pmatrix} | a,b \in \bZ/2\bZ}$

[cloud walrus]

i believe in mathemagic

[rustic crown]

use that F_4 is a 2-dimensional F_2 algebra

say F_2[x]/(x^2+x+1)

so every element in F4 looks like a+bx

and {1,x} is a basis for F_4 over F_2

you can write the multiplication by a+bx (which is a linear transformation on the vector space F_4) as a matrix

that should hopefully be the above matrix

[formal ermine]

what's an algebra

[rustic crown]

vector space where you can also multiply

or rings where the notion of "scalar multiplication" makes sense

[formal ermine]

this is for a homework and we haven't properly started field theory yet lol

[rustic crown]

basically when rings and modules hug each other

[rustic crown]

oh okie

[formal ermine]

the actual homework is to show that the set is a field under matrix addition/multiplication

wew gave me the hint of showing that a isomorphism between that and F_4 exists

that's where I got stuck

[rustic crown]

(btw you could also show the isomorphism by defining the map other way around, only problem is that you would need to first show that the given set of matrices is closed under + and * etc, and then you can define the map F_2[x] --> algebra of matrices, by sending x to the matrix corresponding to a=0,b=1 and then verifying that this matrix indeed satisfies x^2+x+1=0)

[formal ermine]

why x^2 + x + 1 = 0?

[rustic crown]

so that you get the quotient map F_2[x]/(x^2+x+1) --> the matrix algebra

this would be surjective and both sides will have the same size, so also injective

[rustic crown]

but i feel this is the morally correct way of doing it

[formal ermine]

was able to do it with more of wew's advice

just gotta show that x^2 + x + 1 over F4 is irreducible

not sure how to do that

[south patrol]

Just need to show it has no roots tbf

[formal ermine]

how?

[south patrol]

Like why is that enough?

[formal ermine]

no

[south patrol]

Any like proper factor of a quadratic is linear

Oh

[formal ermine]

I know why it's enough

just not how to show it has no roots lol

[south patrol]

Well f4 has only 4 elements

[formal ermine]

ye

ig I'll just check 'em all lol

it's not irreducible

x is a root

[rustic crown]

over F2

over F4 it factors

[formal ermine]

yes

I read the exercise wrong lol

"check whether x^2 + x + 1 in K[x] is reducible or not"

x^2 + x + 1 = (x + a)(x + a + 1)

[rustic crown]

🙈

[solar glacier]

question, why is it when working with Gaussian integers, when we divide we round

[rustic crown]

we round to get the nearest gaussian integer

think about it geometrically

when you divide two gaussian integers, the result lies in Q(i)

and to bound the remainder, getting the closest point in Z[i] would be better

if you just take the floor, the difference could be at most the diagonal of a unit square

which is sqrt(2)

but when you round, you can make sure to be within sqrt(2)/2 of some point in Z[i]

[solar glacier]

by the floor you mean round down?

[rustic crown]

when you translate this to norms, you get
N(remainder) <= 2N(divisor)
in the first case

and
N(remainder) <= N(divisor)/2 < N(divisor)
in the second case

[rustic crown]

yep

normally in Z, we just take the floors to keep the remainder positive

but you can reduce the size of the remainder if you're okay with getting a negative remainder

you can force the remainder to be in the range (-divisor/2, divisor/2]

which is "better" than just [0, divisor)

[solar glacier]

i see makes sense, cause when i rounded up I got a negative integer as the real part

[rustic crown]

[warm wyvern]

seriously doe

what do we gain from identifying monomorphisms/epimorphisms in Ring, Grp, Ab, etc?

[rustic crown]

nothing

Ab is nice, but ring and grp are mostly useless :p

[chilly ocean]

nothing

[warm wyvern]

[rustic crown]

the proof that epi iff surjective in Grp was weird

[warm wyvern]

aluffi says it's cumbersome

[rustic crown]

it is

you have to specifically construct the two weird homomorphisms

and they weren't easy in anyway

some stupid maps into weird symmetric groups

if G' --> G is epi, with image H, then the inclusion H --> G would be epi. another simple reduction is that G is the smallest normal subgroup containing H, else H --> G --> G/N is zero when G --> G/N is either the projection or 0

so if H was not G, then it's index is at least 3

now you you construct those weird maps

[warm wyvern]

[rustic crown]

iirc it was something to do with the symmetric group on the right cosets H\G, and one of the map would physically swap two distinct non-trivial cosets Ha and Hb

ig its nice to know that epi in Grp are surjective, but dunno if that's any useful

[warm wyvern]

is identifying mono/epis useful in general?

or are they just convenient terms and a nice generalizations?

[rustic crown]

i think they're nice

i can only give you one example though

do you know what an abelian category is?

(its basically a category where you have 0 objects, kernels, cokernels, monos are kernels, epis are cokernels, (finite) product and coproduct agree, etc, all nice things happen just like in Ab)

[elder wave]

[rustic crown]

but the thing is even though you would construct such a category by piecing together sets, the notions of epimorphisms and surjective may not be same

the only example i know and care about is that of sheaves on spaces

for example, if you let F be the sheaf of holomorphic functions and G be the sheaf of non-vanishing holomorphic functions both on the complex numbers

then you have the exponential map exp: F --> G

this map isn't surjective on sections as for F(C*) --> G(C*), you have the non-vanishing function z on the right, and you're asking if there is a global holomorphic Log defined on C*

but it is locally surjective

and epimorphisms in the category of sheaves exactly do that

[agile burrow]

is locally surjective just surjective on stalks?

[rustic crown]

yee

[agile burrow]

i should learn a little bit about sheaves

[rustic crown]

yeaaa

i think the above example also gives a reason why one should care about sheaf cohomology

[agile burrow]

running into situations where i want to compute homology with local coefficients

oh yeah, clerk talked to me about this not too long ago

i think it was like, if you have local surjectivity then the next question is when you have global surjectivity - that is, when can you glue your local sections together into something defined on the whole space

[rustic crown]

yee

and the first sheaf cohomology group gives you the obstruction to this

[warm wyvern]

det, I'm doing intro algebra 😭

wtf are sheaves 😭

[rustic crown]

.<

[agile burrow]

fwiw, there's an exercise in chapter 9 of aluffi that is exactly about this lol

but you've probably got a little while before that 🙂

[chilly ocean]

agriculture

do u know how 2 farm

[rustic crown]

informally, given a topological space X, a sheaf is like a data of all the nice functions on the all the open sets of X

[warm wyvern]

wait, isn't that the homological algebra chapter?

[agile burrow]

yeah

[warm wyvern]

oh

[rustic crown]

i haven't read chapter 9 so far >.<

[warm wyvern]

ye

[rustic crown]

i gave up after he started talking about triangulated categories :p

[warm wyvern]

nope

[chilly ocean]

fix that

[agile burrow]

it's ok, I did everything except the last two sections of chapter 9 because I don't like his indexing for double complexes

or like, I get annoyed that the squares anticommute because it's too much stuff for me to keep track of

i just do my silly little total complex

[warm wyvern]

algebraic geometry here I come

or AT, sheaves are used there too, I think

[chilly ocean]

so you're an algebra person?

[warm wyvern]

I don't think so tbh

[chilly ocean]

what kind of person are you

[warm wyvern]

I like applications so I'll prolly lean to analysis

[chilly ocean]

implying algebra has no applications...

[warm wyvern]

but I haven't done enough math to be sure

[coral shale]

applied algebra -> algebraic number theory

[warm wyvern]

implying analysis has more

[chilly ocean]

🤨

[warm wyvern]

they're also more interesting

[chilly ocean]

||cope||

[warm wyvern]

(optimization and ML my beloved)

[chilly ocean]

you should learn design theory

or something useless like that

[rustic crown]

number theory is applied everything >.<

[chilly ocean]

it's actually not useless

[warm wyvern]

like...

graphic design?

[chilly ocean]

combinatorial design

sounds applied right

[warm wyvern]

oh, I actually hate combinatorics LOL

[chilly ocean]

you said you like applications

combinatorics and graph theory seem like they have the most applications, kinda

[warm wyvern]

bruh

[warm wyvern]

I would've imagined statistics would hold that title

altho this is pure math dominant server so you guys might not even consider that a part of math

[rustic crown]

🙈

[chilly ocean]

I don't, I hate statistics

mathematical statistics is part of math to me, though

and is kinda fun

[warm wyvern]

is this commutative algebra?

[rustic crown]

yea :p

[warm wyvern]

can't wait to learn about that lmao

[rustic crown]

imagine learning that in a lecture >.<

and the prof saying "assoicated primes of M" instead of "Ass(M)" everytime :3

[warm wyvern]

homomorphisms and Ass

[chilly ocean]

the prof better be saying ass(M)

[chilly ocean]

is there any sort of intuition that makes the semi-direct product construction feel less artificial?

like, understanding that it generalizes the direct product (they are equal when the map is trivial), what makes it the correct generalization

[rustic crown]

yee there are many intuitions

i remember discovering the construction on my own while trying to find out what Aut(D_n) was :p but don't ask me why i cared about Aut(D_n)

maybe a good way to think about that is by looking at affine linear transformations

you know that GL(n, k) is a nice group of linear automorphisms of the vector space k^n

but you would could also define other automorphisms which may not preserve 0, but are still linear-ish

basically a linear transformation followed by a translation

these would look like f(x) = Ax + b where A is a matrix in GL(n, k) and x, b live in k^n

a good exercise is to figure out a way to compose two such maps

[chilly ocean]

f(g(x)) = f(Ax + b) = B(Ax + b) + d = BAx + Bb + d ?

[rustic crown]

yep!

[chilly ocean]

is this what you mean by compose or am i misunderstanding

ok

[rustic crown]

and one nice way to write that is by looking at n+1 x n+1 matrices

[A b]
[0 1]

the product of matrices here, corresponds exactly to the group operation above

[slender hamlet]

Yes

[chilly ocean]

i have seen this in GL2 referred to as the the affine group

[rustic crown]

anyway, that was just a cute little observation

[slender hamlet]

I also got this same intuitive notion while understanding semidirect products

[rustic crown]

semi-direct products just do the exact same thing but more abstractly

so here you had two groups, (GL(n, k), *) and (k^n, +)

and one group naturally acts on the other!

[chilly ocean]

i think i see, on the second coordinate it is just multiplication in the subgroup K which in this case is GLn? which is here the BAx term

[slender hamlet]

So a nice example of semidirect prod is "translation maps" between groups, which are bijective maps f:G->G such that ab=cd implies f(a)f(b)=f(c)f(d)

[chilly ocean]

GLn acting on K^n by matrix multiplication?

[slender hamlet]

Each such map can be written as f(g)=a \phi(g), a in G, and \phi in Aut(G)

[slender hamlet]

Then from the definition it is clear that the set of such translation maps forms a group, say Tr(G)

[slender hamlet]

Given a, \phi for f and g, we try finding it for fg

And note that Tr(G) is a semidirect product of G and Aut(G)

[rustic crown]

so say more generally you had groups H and N, such that H acts on N, i.e. there is a group homomorphism H --> Aut(N).
then you can define a nice group operation on N x H, by taking
(n, h) * (n', h') = (n * (h.n'), hh')

[next obsidian]

Chmonkey

[rustic crown]

[chilly ocean]

I have a problem: Prove U(9) is isomorphic to Z(6). I get why it is, U(9) has 6 elements and is cyclic, but it just saying that it has 6 elements and is cyclic enough to rigorously prove it?

find a generator of U(9)

[south patrol]

Yes, all cyclic groups oc the same order are fairly trivially isomorphicn

So that should be fine but yes to show its cyclic you probably found a generator and giving that allows you to write an explicit iso

[chilly ocean]

this was my first thought, check that they have the same cardinality but then realized theres nothing obvious preventing U(9) from not being cyclic

[south patrol]

WellI mean they said like they know it is cyclic

[chilly ocean]

ah i did not read the whole sentence haha

okay yes all cyclic groups of the same order are isomorphic so thats a rigorous proof

more explicit is sending a generator to a generator

[wooden ember]

im not too sure how to show that extension of scalars is well defined for a tensor product

like how do i know that I have a well defined structure by just defining it on simple tensors

[wooden ember]

nvm i got it

[rustic crown]

maybe looking at bimodules would make the situation more clear for you. as they would be a little more rigid

[wooden ember]

i think i get it now

[rustic crown]

say you have an (A, B)-bimodule M and a (B, C)-bimodule N, then M ⊗_B N is naturally an (A, C)-bimodule

[wooden ember]

i just need to get used to thinking about tensor products in terms of their universal property

[wooden ember]

do you think this can help for this next one im on right now? That the R-tensor product of two R-algebras is naturally an R-algebra

[rustic crown]

nah, that's a little different

[wooden ember]

im honestly not sure how to even define the product

[rustic crown]

guess on elementary tensors and use universal property

[wooden ember]

if i try to make it seem natural it doesnt distribute

and if i try to make it distribute it doesnt seem natural

[rustic crown]

ah okie, take a look at an example then

k[x] ⊗_k k[y] = k[x,y]

an element on the left side is an k-linear combination of x^i⊗y^j

which under the isomorphism is mapped to x^iy^j

[wooden ember]

right but then the product would mandate a tensor b * a' tensor b' = aa' tensor bb'

is it really that naive?

[rustic crown]

so with this example what do you think should be the definition of (a⊗b) * (c⊗d) in A⊗_RB

[wooden ember]

cause this intuitively doesnt feel like it would play nice with bilinearity

[rustic crown]

yea lol

[wooden ember]

thing is even when im looking at well definedness on two simple tensors

it seems weird

like

[rustic crown]

yea, to get well definedness you always have to go back the the universal property

in this case you're constructing a bilinear map (A⊗B) x (A⊗B) --> (A⊗B)

so naturally one should consider (A⊗B) ⊗(A⊗B)

[wooden ember]

hmm okay ill give that a shot