groups-rings-fields

[chilly ocean]

?

it isnt?

[rustic crown]

what's the identity?

[chilly ocean]

ring without unity

😎

[rustic crown]

oh

[chilly ocean]

lol

[glossy grotto]

Look at what the boundary map is on $H_0(\mathbb{Q})\to H_0(\mathbb{R})$

[chilly ocean]

its joke

he means with unity

it isnt a ring ig

[cloud walrus]

blackiris

[glossy grotto]

But this is not abstract algebra

[chilly ocean]

which homology theory we using

because idk boundary map possed to be

[glossy grotto]

one from singular simplicial complexes I guess?

Is that what you're asking?

One induced by the injection i:Q to R

[chilly ocean]

so what is H_0(Q) supposed to be? , Q has subspace topology?

[glossy grotto]

Yes, Q has subspace topology

[chilly ocean]

oh thats why you are asking about the map

cool

im curious, why asking this?

[glossy grotto]

I'm answering an exercise about proving that $H_1(R,Q)$ is a free-abelian group. Everything will be nice if I can just convince myself that the induced map from $H_0(Q)$ to $H_0(R)$ is essentially what I gave above.

[cloud walrus]

blackiris

[glossy grotto]

I'll try asking in the other channel

[chilly ocean]

i think the map you gave is correct up to multiples between the sets as groups

because maps from Z to Z are uniquely determined by where 1 is sent to

[glossy grotto]

Yeah, up to multiples

[chilly ocean]

similarly Z^2 by where (1,0) and (0,1) sent to

and for Z^N the pattern continues

[glossy grotto]

I see, I'll think about this more. Thanks!

[chilly ocean]

np

where

did you find this exercise

[glossy grotto]

Yeah, I think I'm close to convincing myself about this

Hatcher's book

[chilly ocean]

🤮

[glossy grotto]

lmao, same

guy talks too much

[barren sierra]

I may be dumb but we know all polynomials over $\mathbb{R}$ are reducible if they have degree greater than 2. But how do I factor $x^4 + 1$

[cloud walrus]

Spamakin🎷

[chilly ocean]

assume its factorization is (ax^2 +bx +c) (dx^2 + ex +f). Solve for coefficients

obviously a lot of coefficients you can guess right away

[rustic crown]

factor over C, combine (x-alpha)(x-alpha bar)

[woven delta]

You factor it as (x^2+i)(x^2-i), then further factor it

[rustic crown]

you can also cheat if you want

write it like (x^2+1)^2 - 2x^2

use difference of squares

[woven delta]

You can also use roots of unity

To get the factorization

That's pretty good det

[chilly ocean]

yeah i remember this x^4+1 gave me good lesson

[lunar spruce]

what is this notation? Right side is a multiplicative group modulo q.

[dull root]

Semidirect product?

[lunar spruce]

Let me check.

[barren sierra]

is the quotient ring of a PID a PID itself?

[fresh vessel]

Anybody have any idea how to solve this? Since we have to show that the Galois group is Z/6Z and the extension is Galois, it must be the case that g is reducible. Rational root theorem tells us it has no rational roots so linear polynomial divides it. Then I tried forcing x^2 + ax + b to divide it by requiring that the remainder when you divide g by that is 0, but very quickly the computations for that get out of hand. I can check using desmos or something graphing calculator like that sqrt(2) is a root, so x^2 - 2 must divide it. So I guess I have two questions. First, how would I know to check if sqrt(2) is a root? Just...guess things that we hope will help? Finally, once I know this, I have that g(x) = (x^2 - 2)(x^3 - 3x + 1). So then Gal group = Z2 times Z3 = Z6?

[next obsidian]

Indeed

[barren sierra]

dope

[dull root]

Why is true that if I have an infinite torison-free group G and a element a s.t the index G: <a> is finite, then G is cyclic

[winter mesa]

Determine the size of the smallest subgroup of Z177 containing 57 and 516. anyone know how to do this? im lost

[past temple]

if i have an ideal I in C[x1,...,xn]

and i consider the radical ideal \sqrt(I)

how can i show that \sqrt(I) ^m is a subset of I

for some m

@hidden haven

[next obsidian]

This follows by the ring being Noetherian

If sqrt(I) is finitely generated by x1,…,xn

Then for each xi there is a number n so that xi^n is in I

So you can take a max and then get one N that works for all i

Consider an element in sqrt(I)

It looks like (a1x1 + … + anxn)

If you take a product of nN many of these (this is what a generic element in sqrt(I)^nN looks like)

Expand this into some humongous sum of products

each term will have at least one xi^N inside of it

Which is in I

So it becomes a sum of a bunch of shit in I, so it’s in I

[past temple]

hmm wait so doesnt the fact that A is Noetherian just mean that sqrt(I) is generated by some finite set of polynomials? how do we know that its generated by the xi

[next obsidian]

Sorry Lmfaooo

xi aren’t the like

Variables

I was proving t for a general Noetherian ring

xi there are polynomials

I just forgot the specific case was one where xi are also the indeterminates

[past temple]

ah okay

ok so ig ill just call them fi

instead

[next obsidian]

Sure

[past temple]

the a_i here are also polynomials right

[next obsidian]

Yeah

[past temple]

wait isnt an arbitrary element of sqrt(I)^nN going to look like

finite sums of this?

[next obsidian]

Sure but who cares

It’ll be finite sums of stuff in I

So still in I

[past temple]

ahhh ok

so is it going to look something like

$$ \prod_{j = 1}^{nN} \left(\sum_{i=1}^n a_{ij}f_i\right)$$

[cloud walrus]

pdk

[past temple]

@next obsidian

[next obsidian]

Uh

Yeah sure

I never bothered to write t down but yeah

[past temple]

im a little confused as to why its necessary to take the product of nN many

and not just N

[next obsidian]

Because

Think about the term that’s just like

Okay let’s say N = 2

You could have

f1f2

With some coefficients

You don’t have a high enough power of any fi

You have n things and need to guarantee that anyway to distribute k of them means one of them gets used N times

Distributing N of them won’t guarantee this

You need nN - 1 I think is the lowest you can go

Or maybe (n-1)N or something

n(N-1)?

But whatever who cares, nN works

[dull root]

If I have G, and an abelian normal subgroup N, is G/N always abelian?

[next obsidian]

No lmao

Take N = {e}

You can construct more interesting examples

[dull root]

yea lol

So I have the following situation. G has order 3x5x19 and so the sylow 5 and 19 groups are unique. Let N be the sylow 19 group, and I have in my notes that triviall G/N is abelian, but Im not seeing it now that im revewing it

[past temple]

ahhh okay i understand this reasoning

tyty

[subtle ivy]

so if N has order 19 then G/N has order 15. all groups of order 15 are cyclic and hence abelian

[barren swift]

how can I determine whether (R-{0}, *) and (R+, *) x Z2 are isomorphic or not?

my first instinct is to check whether theyre each cyclic

neither is cyclic, so isomorphism is still possible

oh, i think they might not be isomorphic because no element exists in the latter group with order 2

the identity is (1, 0) and in order for an element a to have order 2, a^2 must equal (1, 0) and also not be the identity. the only possibility is (-1, 0) which is not in the group

but the former group has -1 of order 2

...is this logic correct? im not 100% sure

[amber stag]

They are. Try to construct an isomorphism

Think about an operation that turns multiplication into addition

or the other way aroudn

[barren swift]

does there exist an element of order 2 in the latter group?

[amber stag]

yes

(0,1)

[barren swift]

(0, 1)^2 = (0, 0) which isnt the identity?

[amber stag]

Oh, sorry, never mind. I misread the question

[barren swift]

ok

[proud bear]

(1,1) is an element of order 2

[barren swift]

oh, youre right

hm

[lunar lintel]

how do i prove ideal I of Z[x] that contains 1+x^3 and x^2 contains 1? i think that finding y in Z[x] st (1+x^3)y=1 or x^2y=1 or doing some algebra to show some combination of (1+x^3) and x^2 is 1 but idk how to do the algebra

[proud bear]

if it contains x^2, it contains -x(x^2)=-x^3. so it also contains 1+x^3-x^3=1

[lunar lintel]

lol true im idiot ty

[barren swift]

does anyone know an approach to my problem?

im trying to think of an isomorphism but blanking

[amber stag]

This is actually how we really multiply real numbers: numbers and signs are multiplied separately. ( ,1) models the negative sign.

[proud bear]

maybe it's easier if you think of Z2 as the group {1,-1} with multiplication as the operation

[barren swift]

so i could map -x to (x, 1) and positive x to (x, 0) ?

[proud bear]

yes

[barren swift]

what would get mapped to 0

[proud bear]

?

[barren swift]

i mean

what would get mapped to (0, 0)

[proud bear]

R+ doesn't have 0

[barren swift]

oh

._.

and there's my problem

lol

thanks

[dire bramble]

I really just have no idea how to approach this

[barren swift]

how could i determine if (Z, +) and (Z, @) where a @ b = a + b - 1 are isomorphic? my intuition says they aren't but i dont know how to prove it

as far as i can tell, every element has infinite order in both so that doesnt get me anywhere

[tardy yacht]

Isomoprhisms take Id to Id while maintaining addition and @, think about what happens to 1 for example

[barren swift]

0 to 0? do you mean the identity to the identity? so 0 to 1 in this case?

[tardy yacht]

Yeah my bad

[barren swift]

what do you mean by think about what happens to 1

[tardy yacht]

Actually give me a second I think I though wrong about this lol

[barren swift]

ok

[tardy yacht]

Think about what happens when you shift by 1

[barren swift]

not sure what you mean

[tardy yacht]

Phi(x)=x+1

[barren swift]

hm

interesting

that does seem to be a homomorphism

now i need to determine if its bijective

hm

[tardy yacht]

It's easier if you can just conclude there is an inverse

[barren swift]

well there is

its just x - 1

right?

[tardy yacht]

yep

[barren swift]

ah

so they are isomorphic

still, i wonder if i can determine bijectivity more... manually

oh

im thinking about this wrong

alright

thanks for the help :)

[tardy yacht]

Yw

[barren swift]

ive been trying for the longest time on this problem, and i cant seem to get anywhere.
let G = Z2 x Z2 x Z2... (one copy of Z2 for every positive integer)

is there an isomorphism between G and G x G?

truly do not know how to approach this

[dire bramble]

You can enumerate the copies of Z2 in GxG the same way you can enumerate them in G

the isomorphism essentially looks like the identity map

[barren swift]

as in, phi(x) = (x, x)?

but thats not onto

not sure what you mean

[barren swift]

anyone have an idea?

[thorn delta]

think about the bijection of the natural numbers N with N x N

an element of GxG is a pair of sequences ({xn}, {yn}). How can you take two sequences and combine them into one sequence (in a way which doesn't destroy "information")? That will be your isomorphism

[barren swift]

i cant even see a bijection here

[thorn delta]

(someone else is going to have to explain im too busy rn )

[barren swift]

if an isomorphism means two groups are exactly the same, what does there being a monomorphism mean? ...that more than one element in G is "the same" as H?

[next obsidian]

It means that one group embeds isomorphically as a subgroup

You get this from the first isomorphism theorem

[paper flint]

Pain

[paper flint]

Tagged the wrong message, but this is the idea behind showing a bijection between N and N×N

[next obsidian]

If it’s not clear you map 0 to the top left

1 to the thing that arrow points to

2 to the thing that points to

…

This is a countable enumeration of every element in N x N

[paper flint]

Do it all the way to \omega

[wooden ember]

Anyone know how one could show whether or not an infinite group can have a trivial automorphism group ?

Of course you can take it to be abelian but I asked my algebra teacher and he wasn’t sure

All I was able to conjure is that it can if it is simple but I don’t think there is any simple infinite abelian group

And if it’s finitely generated classification of finitely generated abelian groups gives us an answer

If someone has an idea pls lemme know

[next obsidian]

What about (Z/2Z)^N

I am not sure if for infinite products that the automorphism group of the product is the product of the automorphism groups

But the automorphism group of Z/2Z is trivial, so if it does then this gives you an example

https://stacks.math.columbia.edu/tag/032P

I don’t see why the ideal $\mathfrak{q} = yS$ in this proof

[cloud walrus]

Kanga Gang Boss Chmonkey

[next obsidian]

Indeed if you have f in S so that f^q in xR we can write
f^q = y^qr for some r in R

But unless r is a q-th power, I don’t see how we write f as some multiple of y????

Unless you can show that yS is radical or something, I’m not sure how this completes anything

Oh I think you can just show that it’s radical

Fuck my life

f^n = ys

Raise it by q

f^nq = xs^q, note that s^q is in R

So f = xr

Now you see f = y^qr in yS

[wooden ember]

Yeah I’m tying with $\prod_{i\in \mathbb{R}}\mathbb{Z}/2\mathbb{Z}$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

But struggling a bit

[next obsidian]

I mean I was just doing a countable product

[wooden ember]

Trying*

[next obsidian]

I know Aut(G x H) = Aut(G) x Aut(H)

[wooden ember]

Oh right countable isn’t finitely generated

[next obsidian]

So if you can extend this to the countable case you’re done

[wooden ember]

In my head it was

[next obsidian]

Lol

[wooden ember]

Yeah it’s what I’m trying

But I was trying uncountable countable is probs easier

[next obsidian]

Try a countable direct sum instead maybe

[wooden ember]

Does a countable product satisfy the same universal property?

[wooden ember]

Mm fair

[next obsidian]

I think you have a better chance to have the automorpjism group become the product

[sturdy marsh]

am I being dumb or are all of your examples Z/2 vector spaces

[wooden ember]

Yeah they are

[sturdy marsh]

so they have automorphisms

[wooden ember]

Because g^2 = e

They do?

[next obsidian]

What

Uhhh

Hmm

Wtf

Swap the basis vectors

Is what he’s saying

But isn’t Aut(G x H) the same as product of automorphism groups?

Lol

[wooden ember]

I’m not used to working with infinite bases that’ll work?

[next obsidian]

Maybe this is only true when they’re coprime

Ohhhhh right

That’s the issue I think

Lmfao

[sturdy marsh]

I dont think there's an infinite example

[wooden ember]

Oh yeah lol

[next obsidian]

Okay I have no fucking idea

[wooden ember]

You got me confused

[sturdy marsh]

it must be a Z/2 vector space

[next obsidian]

Well what about the general question

[sturdy marsh]

otherwise inversion will be a nontrivial auto

[next obsidian]

Is there an infinite group with no automorpjisms

[wooden ember]

That does it

[next obsidian]

What?

[wooden ember]

It can’t be nonabelian

[next obsidian]

Yeah

[sturdy marsh]

im saying there isnt

[wooden ember]

Because conjugation

[next obsidian]

But abelian doesn’t mean Z/2

Wait what

[wooden ember]

And the abelian ones are F_2 vector spaces

[next obsidian]

Wait what

[wooden ember]

Because g^2=e

[sturdy marsh]

if abelian, then inversion is an auto

[next obsidian]

The duck?

[wooden ember]

Since inversion is an automorphism

[sturdy marsh]

it needs to be trivial

[next obsidian]

What the hell are you saying

[sturdy marsh]

so Z/2 vect

[next obsidian]

Why is g^2 = e

[sturdy marsh]

we need g = g^-1

[wooden ember]

Because inversion is an automorphism that must be trivial

[next obsidian]

Ohhhh

Fugg

I see

[sturdy marsh]

hurb

[next obsidian]

Hurb indeed

Wtf

[sturdy marsh]

yeah there are no examples

[next obsidian]

Insanity

[wooden ember]

Thanks a lot @sturdy marsh

[next obsidian]

Doesn’t this like

Wait why does this need infinite lol

Like don’t you just need to not be Z/2Z

[sturdy marsh]

they asked for an infinite example

[next obsidian]

Yeah but like

Isn’t this way stronger

If I’m not massive hurbung rn

[sturdy marsh]

yeah G =1 or Z/2

if aut(G) is trivial

[next obsidian]

Okay

[wooden ember]

If it’s finitely generated you get nontrivial automorphisms

[next obsidian]

[wooden ember]

Except for Z/2Z

[next obsidian]

I mean I’m point out

This shit proves it

For literally any group

That isn’t trivial or Z/2Z

[sturdy marsh]

yes

[next obsidian]

Oh i see

You’re saying the finite case was handled

By classification

[wooden ember]

Yeah

[next obsidian]

But I guess my point was

[sturdy marsh]

what

[next obsidian]

This just handles everything

[sturdy marsh]

yes

[next obsidian]

Right

[wooden ember]

Oh fair enough

[next obsidian]

Okay

[wooden ember]

Of course yeah

[sturdy marsh]

I am confusion

[next obsidian]

It is simpler I think than doing classification as well

[wooden ember]

Lol

[next obsidian]

No worries Brofib

[sturdy marsh]

I am do grad appa

[wooden ember]

Yeah it definitely is

[next obsidian]

Yeah

This is why Brofib is a young god

[wooden ember]

It’s just the first time I had classification in mind I hadn’t realized g^2=e

[next obsidian]

Yeah

[dull prawn]

hm maybe a dumb question but I was discussing a problem from my school's most recent algebra qual and someone proposed an iso between Z[x]/(f(x)) and Z[a], where f(x) is an irreducible polynomial having a as a root. but this seems a bit fishy because I know the usual iso when adjoining roots depends on division with remainder. If f is not monic, perhaps it's possible to find another integer polynomial g with a as a root but which is not divisible by f in Z[x]. yet I can't find any examples.

is there maybe some quick Gauss's lemma argument that explains why I'm an idiot and there are no examples?

[waxen hedge]

Since f is irreducible, any polynomial ≠ 0 in Z[X] with $a$ as a root will be a multiple of f.
You can look at the kernel of the evaluation to $a$:

$\mathbb{Q}[X]\rightarrow\mathbb{C},\ U\mapsto U(a)$, the kernel is generated by a monic irreducible polynomial $P\in \mathbb{Q}[X]$, and here $a$ is a root of monic polynomial in $\mathbb{Z}$ so $P$ will be in $\mathbb{Z}[X]$
Now, any monic $Q\in\mathbb{Z}[X]$ such that $Q(a)=0$ will be a multiple of $P$ in $\mathbb{Q}[X]$, and by working a little more you can show that $Q$ will also be a multiple of $P$ in $\mathbb{Z}[X]$

[cloud walrus]

Adrien

[fresh field]

I've read that since the space $H([0,1]^3)$ can be viewed as a tensor product of $H([0, 1])$ like $H([0,1]^3)=H([0, 1])\otimes H([0, 1])\otimes H([0, 1])$, the Laplace operator is the kronecker product of lower dimensional Laplace operators. https://scicomp.stackexchange.com/questions/35165/kronecker-product-representation-of-the-finite-difference-laplacian

[cloud walrus]

Casper

[fresh field]

Does anyone have a source where I can read more about this?

[subtle ivy]

cycle length and cycle order are not always the same, no?

wait i think my brain is melting a little... wtf even is the length of a cycle? artin doesn't even define it

[subtle ivy]

oh yeah you're right lol

[summer notch]

What is the cardinality of the dual basis to the space R^N of infinite sequences of real numbers? I know it must be greater than uncountable, because the space of infinite sequences has uncountable basis, so the dual basis must be larger, but I was wondering if anyone knew if it had some well-known cardinality that has applications elsewhere.

[thorn delta]

by "cardinality of the dual basis," do you just mean dimension of the dual?

because "dual basis" sometimes refers to a particular basis, which is ofc not a basis in the infinite dimensional scenario

[summer notch]

yes, sorry

the "dual basis" probably refers to the set that takes the basis of R^N to (1,0,0....) which will have same uncountable cardinality and will not span (R^N)*

also probably i was imprecise when i said "greater than uncountable", when really i meant "greater than cardinality of the reals"

i'm guessing that the answer to my question is |2^R| or beth-two or whatever you want to call it

cardinality of power set of reals

but i'm not sure how to prove it (not really looking for formal proof, just intuition)

[languid meteor]

hey quick question: If N <= M <= G where M is maximal and f:G->G/N is the natural projection then is f(M) maximal in G/N?

and then more generally: if f:G->H is a surjective homomorphism s.t ker(f) <= M where M <= G is maximal then is f(M) maximal in H

cant find either result anywhere

[thorn delta]

So, lets call the space of real valued sequences $\bR^\infty$. Assuming the dimension of this space is $\mathfrak c$, we can write $\bR^\infty \simeq \bR^{\oplus \mathfrak c}$. Then you can write the dual as
$$ (\bR^\infty)^* \simeq (\bR^{\oplus \mathfrak c})^* = Hom(\bR^{\oplus \mathfrak c}, \bR) \simeq \prod_{\mathfrak c} Hom(\bR, \bR) \simeq \bR^{\Pi\mathfrak c}.$$
From here, $$ \prod_{x \in \bR} \bR_x = \prod_{x \in \bZ} \left(\prod_{y \in \bZ} \bR_y\right) = \prod_{x \in \bZ} \bR^\infty.$$
I think in general, $|dim(\prod )| = \prod |dim|$, so it would follow that
$$|dim((\bR^\infty)^*)| = |dim(\prod_{x \in \bZ} \bR_x^\infty)| = \prod_{x \in \bZ} |dim(\bR^\infty)| = {\mathfrak c}^{\aleph_0}$$

idk hopefully there is some truth to what i wrote

[cloud walrus]

kxrider

[thorn delta]

not sure if there are any applications of this

[summer notch]

i will try to digest this over my lunch

thank you for your help!

[thorn delta]

npnp

[summer notch]

just boosting this since i kind of covered it up

[languid meteor]

actually this is immediate if correspondence theorem respects maximality

which it does it respects inclusions ok cool

if anyone knows anything about the general situation let me know!

[thorn delta]

Lets say $\varphi : G \to G'$ is surjective and $M \leq G$ is maximal. Suppose $\varphi (M) \leq N \leq G'$. Then $M \leq \varphi^{-1}(\varphi(M)) \leq \varphi^{-1}(N)$, so by maximality of $M$, $M = \varphi^{-1}(N)$. Since $\varphi$ is surjective, $\varphi(M) = N$.

[cloud walrus]

kxrider

[median pawn]

Could you give an example of a ring homomorphism, which doesn't send identity to identity?

[thorn delta]

f: Z -->Z, f(n) = 0

[median pawn]

Ah yes! That just struck me lol

[thorn delta]

there are def less dumb ones tho

[median pawn]

Why do some authors have this assumption? It is only restricting the choices of homomorphisms for us. Why would we do that?

[thorn delta]

in order to preserve the multiplicative monoid structure of rings with identity

groups do preserve the identity automatically but monoids don't

[median pawn]

hmm okay fair

but wait!

[cloud walrus]

Hausdorff

[median pawn]

What's wrong?

it clearly maps the identity to the identity

[thorn delta]

are you trying to apply cancellation here or something?

[median pawn]

no that's just how identity is defined right

[delicate bloom]

x=xy doesn't mean y=1

[median pawn]

no but

if x= xy for all x

then y is the identity?

[cloud walrus]

Hausdorff

[median pawn]

look at (b)^^

[thorn delta]

ah okay, ye

[cloud walrus]

Hausdorff

[next obsidian]

This just by convention isn’t a ring map because 1 doesn’t go to 1 lol

[median pawn]

^ the author decides not to put that condition

[next obsidian]

I guess the thing here is that

When you distinguish 1 as important

{0,3} isn’t a subring of Z/6Z

Because the 1 in the subring has to be the same as the 1 in the super ring

[median pawn]

no

that happens in integral domains

but not necessarily

[next obsidian]

No I mean

When you care about 1

You define this

To be the case

[median pawn]

ahh okay, cool

[next obsidian]

I don’t consider {0,3} a subring of Z/6Z

Because 1 matters to me

[median pawn]

lol maybe the author doesn't care about 1

😦

[next obsidian]

I think so

I mean if ring maps don’t have to take 1 to 1

Then you certainly have much less of an emphasis on 1

In commutative algebra you basically need this assumption

So it’s just implicit for me

[median pawn]

oohh cool

[next obsidian]

I have no idea how rng theory works

Or what breaks when you don’t have a 1 lmfao

Anyone that can keep straight in their head what exactly needs a 1 and what doesn’t is gigachad

Their brain is much bigger than mine lol

[cyan marten]

It's minor but the image of the identity of a rng hom is an idempotent, so when you're working with integral domains (or rings with no nontrivial idempotent), the theories coincide. (Well, there is the zero morphism..)

[thorn delta]

there must be something wrong with my solution there, because c^aleph0 = c and this would imply R^\infty is isomorphic to its dual hmm

yea the problem is definitely on the second centered equation there

because that product in the second (incorrect) equality only gives Z x Z many copies of R, not prod_1^infty |Z| many copies

i tried

[summer notch]

you could ask the same question, why assume that a homomorphism preserves multiplication or addition? it restricts the amount of functions for us. The reason is simply that we see interesting properties that pop up when we add this restriction that aren't there for general functions

[summer notch]

careful, it is true that x=xy for all y means y is the identity, but the ring homomorphism might not be surjective, so not every element s in S may be expressed as phi(r)

phi(1_R)phi(r)=phi(r), but there might be some s not in the image of phi such that phi(1_R)phi(s) =/= phi(s)

therefore without explicitly making it an axiom, its not true in general that a homomorphism will send the identity to the identity

[delicate bloom]

keep reading, they settled this already and came up with an example even

[summer notch]

i saw that they had a counterexample to the statement, but I didn't see whether or not they realized why their proof didn't work

there was already a counterexample (the trivial one mapping everything to 0) before they gave their proof, so i didn't know if they were still confused

[median quartz]

I have a quick question about a symbol, not exactly sure where to ask it, it's about quandles

The top part says definition, the top blue underlined word is automorphism group, the bottom left one is homogeneous, and the right word is transitive

Can someone tell me what the half circle arrow means?

[thorn delta]

Aut(Q,s) "acts on" Q

[median quartz]

ahhh thats right, thank you so much! for the life of me I could not find it lol

[deep nova]

Got my algebra exam in an hour or two

And everything I read on ring theory is just leaving me

[next obsidian]

Good luck

[dull prawn]

Yes, I'm aware of how this works when f is monic. My point is that there are irreducibles in Z[x] that aren't monic

actually hold on a sec maybe i misread

@waxen hedge okay no I don't buy your argument. you're assuming a satisfies a monic integer polynomial. I'm not making any assumptions on a other than it being a root of some irreducible integer polynomial, not necessarily monic

[fossil shuttle]

Gallian's abstract algebra book doesn't assume rings have unity

Ideals are my favorite kind of ring

[next obsidian]

Yes and technically D&F doesn’t either but then they start assuming it because the theory is much improved with a 1

Keeping straight what does and doesn’t need a 1 is something I gave up on long, long ago

[fossil shuttle]

yeah i wasn't advocating for it

[next obsidian]

Imagine having a non quasi compact Spec

[prisma thunder]

We have recently talked about Kummer extensions

Let $F$ be a field containing a primitive $n$-th root of unity and let $a \in F$ be nonzero. Show that any irreducible factors of $x^n-a \in F[x]$ are given as $x^m-b$ , $m , | , n$ and $b^{n} = a^m$.

[cloud walrus]

eM

[prisma thunder]

Guess I can't put my code here lol I'll have to write it down here then

[waxen hedge]

Ok then
We can characterize the irreducibles of Z[X] as the union of irreducibles of Z (i.e. the primes numbers) and as polynomials of Z[X], irreducibles over Q[X], and such that coefficients of such a polynomial has no common divisor in Z; that's a standard result about irreducibles of A[X] with A an UFD

Now let f, g in Z[X], irreducibles in Z[X] and both having a as an root, this implies that f and g are not prime numbers.
They are both irreducible over Q[X] and having a as a root, so there are both the minimal polynomial of a over Q[X], up to a unit of Q[X]
So we have f=rg with r a rational, r≠0
Since coefficients of f (or g) don't have any common divisor, r has to be 1 or -1

[waxen hedge]

This is related https://en.m.wikipedia.org/wiki/Gauss's_lemma_(polynomials)

[dull prawn]

Oh okay thanks. Now I'm convinced.

[waxen hedge]

Great !

[prisma thunder]

Suppose $x^n-a$ is reducible over $F[x]$ and let $g(x) \in F[x]$ denote its irreducible factor; then we may write $g$ as
$$g(x) = (x-\sqrt[n]{a})(x-\omega^{k_2}\sqrt[n]{a})\cdots (x-\omega^{k_m}\sqrt[n]{a}) =x^m - \cdots + (-1)^m \omega^l \sqrt[n]{a}^m \in F[x]$$
where $l = k_2+\cdots + k_m$. Since $F$ contains a primitive $n$-th root of unity, observe that $\sqrt[n]{a}^m \in F$, i.e., $\sqrt[n]{a}^m = b$ for some $b \in F$, for which we can rewrite as $a= b^{n/m}$. I understand where the $b$ comes from now. Now, I want to make sure $m , | , n$ and that the middle coefficients are all zero--as an example, from the theory of elementary symmetric polynomials, the first term after the leading coefficient is
$$(-1)^m\sqrt[n]{\theta}\sum_{i=2}^m \omega^{k_i},$$
but I'm not exactly sure what to deduce here. Hmm.

[cloud walrus]

eM

[prisma thunder]

Should be a, not theta. Since g is irreducible over F, my guess would be the nth root of a cannot be in the base field $F$, so that the only way for this term to be in F is if its zero, i.e., the sum of those powers of primitive nth roots of unity are zero.

oooohhhhhh

I think I got it now

[chilly ocean]

Ahat do they mean by group of p-th roots of unity here

do they mean Q(Z_p)?

or do they mean adjoing each root of unity

[small bison]

Sounds like it means that

Which is the same as adjoining any generator

[barren swift]

is there a simple algorithmic way of generating all elements of Sn in their r-cycle forms?

for example for S4

i could think of each permutation of 1234 and then figure out what it looks like as an r-cycle composition

but is there a better way?

[median pawn]

yep, got it!

thanks

[chilly ocean]

is there a field extension E/Q where gal(E/Q) is a lie group?

wait lol

context is missing ig

[next obsidian]

Finite groups are all Lie groups

So you can find a lot, but this isn’t really well-posed

[chilly ocean]

yeah i messed up my bad

[chilly ocean]

also is that true?

i thought you need to give the group the topology to be a manifold

[next obsidian]

Yes just give it the discrete topology

And it’s a compact 0-dim manifold

[chilly ocean]

oh lol

[next obsidian]

But this is why I said it isn’t really well-posed

Usually these groups don’t really have a topology

[chilly ocean]

rip

[next obsidian]

If they’re finite it’s the discrete and if it isn’t then it’s some fucked up inverse limit but

I don’t think you’re really getting anything good here

[chilly ocean]

yeah i didnt pose question correctly

also is there a way to think of galois groups in a context of regular covers?

the phrasing might be weird

but i know for topologies E->X is regular if Aut(pi) acts on p-1(x) freely

so it preserves fibers and it only stays fixed if you are considering the identity

wait lol

we cant think of a projection map from E->F since field homs are injective

lol nvm its going to seem like rambling

but im wondering if there is a covering space perspective we could take

[next obsidian]

There is lol

This is like the idea of a Galois connection and Grothendieck’s Galois theory

[chilly ocean]

oh man

ok

so its just something i need to look into deeper

will do it later unimprtant rn

[hidden haven]

Krull topology on the finite Galois groups should be discrete

So putting the discrete topology on it wouldn't be an arbitrary choice

[chilly ocean]

wtf krull topology

profinite groups o

[hidden haven]

Ye my understanding is that Aut denotes the aut group and Gal denotes the topological group when you put the Krull topology on it. At least this seemed to be the implicit convention from where I studied

[next obsidian]

@hidden haven I said that here yeah

So like in the finite case I mean the discrete topology is just always the canonical topology if you had to put one on

[hidden haven]

oh right

[next obsidian]

But I just mean in general like…

Idk the question just doesn’t seem like the right question

I guess

I want to say the Galois group naturally exists as a group w/o a topology but this isn’t really true

It just doesn’t seem like the right sort of thing to be asking I guess

[fallow plume]

How do you extend Z_3 into Z_9?
My understanding is because Z_3 is a normal group (and the quotient group is isomorphic to it too) there should be some way of extending Z_3 by Z_3 into a group isomorphic to Z_9

But obviously it can't be the semidirect extension since the only possible homomorphism into the automorphism group is the trivial one

and everyone knows the direct product isn't isomorphic

[lethal dune]

what's the problem with Z3 x Z3?

[fallow plume]

It isn't isomorphic to Z_9.

[lethal dune]

then some kind of field extension ig

[fallow plume]

My question is how you extend Z_3 by itself into an isomorphic group to Z_9

oof okay so I still got a lot of reading to do

[lethal dune]

Z3[x]/<x^2+1>

tho this is not isomorphic either

[fallow plume]

I've only read up to chapter 6 in Dummit & Foote, so apologies if the question is dumb lol

[lethal dune]

there are only 1 groups iso to Z9, so

[fallow plume]

Those edits lmao

[lethal dune]

yeah I realized that it's now what I was trying to say

imma just delete that lol

[fallow plume]

Is the Extension Problem still more or less an open question?

[coarse stag]

are integral domains and fields identical on finite sets?

[hot lake]

yes

[coarse stag]

but not nessesarily for infinite because Z is an integral domain but not a field right?

[thorn delta]

true

[thorn delta]

this basically follows from the fact that an injection from a finite set into itself is a bijection

so it doesn't have any nice generalization to infinite domains

[coarse stag]

where does the injection come from?

[thorn delta]

so, pick an element x in your integral domain A. we want to find an inverse for it, so define f : A --> A by
f(y) = xy. f is injective

[coarse stag]

I dont see how that falls apart for infinite sets

[delicate bloom]

map integers into even integers

[thorn delta]

well from here, since A is finite, f is also surjective

so in particular there is z such that f(z) = 1, and z is the inverse of x

yea and mero has a good counterexample for why injectivity does not imply surjectivity for maps from an infinite set into itself

[coarse stag]

i see

ty

[thorn delta]

npnp

[simple mulch]

what would be the group generated by <e> ? (e the identity), I am afraid to say is G = {e}

[thorn delta]

<e> = {e}

[simple mulch]

yeah

lol

[prisma thunder]

Quick question

When does the semi direct permute over Cartesian products of groups? (e.g. (G x H) < K is in the same isomorphism class as (G < H) x K).

I want to say something about (Z2 x Z2 x Z3) < Z3 and (Z2 x Z2) < (Z3 x Z3)

[tame grove]

I just have a notation question

$M_n(R)$ is an n by n matrix over the real numbers?

[cloud walrus]

CaesiumIsFake

[chilly ocean]

the set of n x n matrices with entries in R. whatever R is.

[tame grove]

Thanks. my prof means real numbers by that. He does that a lot with Z and R.

and $GL_n(R)$ is the general linear matrix on reals then? As in not needed to be n by n? and $SL_n(R)$ is the special linear group where the determinant is 1 right?

[cloud walrus]

CaesiumIsFake

[chilly ocean]

GL_n(R) is the set of invertible n x n matrices with entries in R

[tame grove]

Ah that makes sense

[chilly ocean]

SL_n(R) those elements of GL_n(R) with det 1, yes.

[tame grove]

Alright, thanks again. This would be 30 minutes of googling, thanks this helps me a lot.

[chilly ocean]

30 minutes is a bit of a stretch lol

[tame grove]

true, but it feels like 30 minutes when you cant find the one definition you need for notation.

[simple mulch]

Hey guys, I want to show T is linear but I don't understand how

I would need to have $T(v_j + v_k) = T(v_j) + T(v_k)$ but I don't understand how to represent the image of $T(v_j + v_k)$

[cloud walrus]

mns

[simple mulch]

I thought about $T(v_j+v_k) = \sum^m_{i=1}a_{ij}(w_i + w_c)$ but I am 100% sure this is wrong

[cloud walrus]

mns

[chilly ocean]

T is assumed to be linear

[simple mulch]

Well ok but if it is assumed, shouldn't I be able to confirm it?

[chilly ocean]

Well you dont have formula for T, you know that on basis elements the image is lin combination of w_i's so by linearity, T(v1+v2)= sum (ai1+ai2)wi

[simple mulch]

hum I see

[hidden haven]

But if you haven't done this before, then you should check that this map is linear

The theorem is that given a basis B of V, any set map from B to a vector space W extends to a unique linear map from V to W

It's called the universal property of a free vector space, it's a big reason we care about bases

[simple mulch]

gotcha

thanks

[prisma thunder]

Been working on this the entire day

There has to be a better way than determining all groups of order 36 (15) and shifting through which one these groups can’t be (anything Z_4 or Z_9 are auto-eliminated)

Hhhmmmmmmm wait a second

All of those groups look like they have (at least?) three elements of order 3 and that no normal subgroups of order 9 are in any of them

[prisma thunder]

Gonna need to look into this more, I think I found a more efficient working path

[next obsidian]

Anyone have any idea why X can’t have a submodule isomorphic to k?

X looks like a quotient of a free module, so if I write it as like F/K, then having a submodule isomorphic to k means that F has a submodule T such that T/K ≈ A/m

So there’s a submodule such that (K:T) = m, but idk how that contradicts anything

For showing M doesn’t have a submodule isomorphic to k this is easy, M is a submodule of L’ and L’ doesn’t have one since it has depth 2, and depth 0 <==> submodule isomorphic to k

[maiden ocean]

cant you just say that X cong L'/M so if k is a submodule of X then M can be embedded in k? maybe this is allowed but i would assume not

@next obsidian

[untold cloud]

Hi, guys, does the ideal for algebra only makes sense if the algebra is associative algebra?

[sly storm]

Hello. Is there an easy and intuitive approach showing that a finite p-group having a unique subgroup of order p is either cyclic or generalized quaternion?

[fickle brook]

define generalized quaternion?

@sly storm

nevermind, looked it up.

[lethal dune]

is there an easy way to know which of the quadratic integer rings are UFD/PID/ED other than memorizing?

[upper pivot]

ideal in a general sense means "subobject closed under some action from the entire algebra", and this still makes sense in non-associative algebras

i dont have much experience on non-associative algebras except for lie algebras

and there as you can imagine its a subspace closed under [a,-] for all a in the algebra

[untold cloud]

thanks, but doesn't the definition of ideal I for an algebra A is: ax in I for all a in A and x in I? But for the ideal generated by b, Ab, i need a(xb)＝(ax)b to let Ab to be an ideal?

[upper pivot]

well you want it to be closed under multiplication by A

so if these are not equal

you would want both a(xb) and (ax)b in your ideal

[untold cloud]

oh, i see, thanks

[untold cloud]

Hi, guys, how to define an ideal generated by infinitely many elements?

[next obsidian]

No, you get a submodule of M’ so that (M:M’) = m

[maiden ocean]

What is M'?

the screencap u sent just has 0 -> M -> L' -> X -> 0

[next obsidian]

So like

X = L’/M

Then if that has a submodule iso to k

It looks like M’/M

But actually maybe this is fine

Since maybe I can do something like

Make a composition series involving M < M’

Then I’ll have to check but

M’/M has m as an associated prime

So maybe this implies L’ has m as an associated prime

I don’t think this is necessarily true?

I forget tho

Anyway I have to go back to sleep why am I awake rn

[hidden haven]

[thing] generated by a set of elements inside a structure is always(?) defined as smallest [thing] inside the structure that contains that set

This may not always exist, but in the case of ideals it does, intersection of any collection of ideals is an ideal

[untold cloud]

i see, thanks

[lethal dune]

[lunar lintel]

To prove Fermats little theorem (n^p=n mod p), my idea is to take group Z_p*. Z_p has order p-1 hence n^p-1 = 0modp. Then n^{p-1} n=n mod p. Does this seem correct

[covert vector]

Yea that works for all n that are in Z_p*

Then you can do 0 as a separate case

[chilly ocean]

Other than checking their gcd being 1?

cause that's the one that falls out of the definition fastest @ember field

I was gonna type another trick but then double checking it realized there's a huge mistake in the report I sent to my old research prof related to that I did not realize until now

[simple mulch]

Hey
for each $r \in R$, consider $A_r = {(x,y) \in R \times R : y = 3x + r}$

[cloud walrus]

mns

[simple mulch]

I need to show that ${A_r : r \in R}$ is a family of lateral classes of $R \times R$ relatively to $A_0$

what does the "relatively to A_0" means?

[cloud walrus]

mns

[simple mulch]

oh wait

nvm

I already proved A_0 is a subgroup of R x R

like, ${A_r : r \in R}$ this is a family of cosets if their union is R x R

[cloud walrus]

mns

[ionic ingot]

considering the finite field F₂⁵, and multiplication modulo the irreducible polynomial x⁴ + x³ + 1, what would be the multiplicative inverse of x³ + x + 1?

i know the formula to use is a(x) * b(x) + m(x) * c(x) = 1

with a(x) being x³ + x + 1 and m(x) being x⁴ + x³ + 1

and i'm trying to use the answer here https://math.stackexchange.com/questions/124300/finding-inverse-of-polynomial-in-a-field to figure it out but its kinda difficult

is there an easier way?

what should be my first step?

[delicate bloom]

use the extended euclidean algorithm

search that term up to try to find easier to understand examples, I'd walk you through it but I'm about to head out now

[ionic ingot]

i did, but i can't really understand it

can someone else help?

[obtuse kernel]

How would I find the rational form a matrix given the minimal polynomial?

I know I have to identify the invariant factors but I'm not certain of how they should be chosen

Sorry, is this not the appropriate channel?

[willow mason]

it is the okay channel

https://math.stackexchange.com/questions/1554530/rational-canonical-form-from-given-minimal-and-characteristic-polynomial

maybe this helps

[glossy grotto]

Quick question: If $G/N \oplus N \cong N$, must $G\cong N? $

[cloud walrus]

blackiris

[glossy grotto]

or G=N

[next obsidian]

No for equality

For isomorphism I’m unsure

Take a countable product of Z to be G

Let N be the same thing, except you omit the first coordinate

So it’s like Z (+) Z (+) … for G

And 0 (+) Z (+) Z (+) … for N

Then G/N = Z

And when you direct sum with N you still have a countable product of Z

Then you have G/N (+) N ≈ N because both sides are a countable product of Z

But clearly N ≠ G because it’s a proper subgroup

If I had to hazard a guess

It’s no for isomorphism as wel

But idk

Also: if G is finite

Then it is true that N = G in this situation just because of size stuff lol

[wise igloo]

sorry to interrupt but let's say we are talking about (finite) cyclic groups, so $C_n={i, \sigma, \sigma^2,..., \sigma^{n-1}}$. This makes me think that $i=\sigma^0$, is this true?

[cloud walrus]

gmod

[next obsidian]

Yeah I mean this is true for any element ever

[wise igloo]

alright just wanted to make sure

ty

[next obsidian]

Either by convention

Or to make it true that g^n•g^m = g^n+m

So whether it’s a result or a definition is kinda up to you

Also in general the identity is denoted e

Or 1

I would avoid using i

[glossy grotto]

I have $G/(Z^2)\oplus Z^2 \cong Z^2$ in particular. $Z^2 = Z\oplus Z$

[cloud walrus]

blackiris

[next obsidian]

Is G abelian?

[glossy grotto]

Sorry for interrupting!

Yes G is abelian

[next obsidian]

Then yes Z = G

The reason is like…

Invariant basis number

The point is, okay do you know what a module is?

[glossy grotto]

Z^2 =G

?

[next obsidian]

Yes

[glossy grotto]

Yeah, I know what a module is

[next obsidian]

Okay so you know how for vector spaces

They’re classified by cardinality of a basis

So like

V ≈ W

Iff their dimension is the same

[glossy grotto]

yeah

I was about to say this

[next obsidian]

The same is true for free modules over a commutative ring

An abelian group is a Z-module

So here you can see

G/Z^2 (+) Z^2 ≈ Z^2

[glossy grotto]

so we have '2-dimensional' free-modules here

[next obsidian]

Yup

[wise igloo]

oh okay thanks for the tip

[next obsidian]

Free of rank 2

So there’s one thing here that’s tricky

[glossy grotto]

thanks gang boss

[next obsidian]

You need to know G/Z^2 is free

To apply this

[wise igloo]

gang boss

[next obsidian]

Ohhhh hmm

Okay so

[glossy grotto]

I think I can go on from here.

[next obsidian]

If G is finitely generated

You get it for sure

[glossy grotto]

I just want a quick reality check

[next obsidian]

But I’m a little tiny bit worried

In the non finitely generated

Point was if G is finitely generated, if it isn’t free then it ha torsion

This is a contradiction

So it’s free

Then by that dimension shit it has to be 0

But…

[glossy grotto]

I think I'm working with G finitely generated

but I'll have to check

[next obsidian]

Actually I think you’re fine anyway… probably

I bet

You can show that like

The left hand side is

Free of rank 2 direct sum something

And it’s iso to free of rank 2

I bet you can somehow show that like

This means the “something” is 0

Ah yes!

Okay so G/Z^2 has to be finitely generated

Cuz if not then G/Z^2 (+) Z^2 isnt

But it’s iso to Z^2 which is

So now the argument goes through

[glossy grotto]

ah yeah,

[next obsidian]

[glossy grotto]

G/Z^2 finitely generated, Z^2 finitely generated, then G finitely generated right?

[next obsidian]

Yeah

[glossy grotto]

Okay baws

thanks a lot

[next obsidian]

Swag

😎

[glossy grotto]

gang gang

[obtuse kernel]

is it not required to have every factor in the min poly be in each of the invariant factors?

Ok, maybe as a concrete example, if I had the min poly $(x-a)^3(x-b)^3(x-c)$ what would the invariant factors be

[cloud walrus]

TylerUno

[obtuse kernel]

for a 7x7 matrix

wait is it just one companion matrix?

because the deg of the min poly is the same as the space?

[rustic crown]

yep, that's right.

[prisma thunder]

Some proofs I’ve seen for showing a symmetric group cannot have a subgroup of a given order usually makes use of the index of the subgroup being the smallest prime divisor and getting wonky stuff with normality.

I decided to look at S_7 and tried to see what goes wrong if I don’t have the above tool and supposed a group of order, say 45, exists. My gut tells me such a subgroup shouldn’t exist (GAP confirmed that but I don’t want to rely on it). 🤔 How would I argue this?

Hmmm. So it seems S_7 has no normal 3- or 5-Sylow subgroups and groups of order 45 are abelian.

[proud bear]

i think you can just say that groups of order 45 have an element of order 15, but S_7 has no element of order 15

[prisma thunder]

Yes

[next obsidian]

Anyone have any clue why C^(p)[1/f] = B_f??

[paper flint]

What is that typesetting

[sly storm]

Hello. Is there an easy and intuitive approach showing that a finite p-group having a unique subgroup of order p is either cyclic or generalized quaternion?

[wooden ember]

Have you seen that a finite p group with every abelian subgroup being cyclic is either cyclic itself or generalized quaternion?

[sly storm]

No.

[wooden ember]

I really don’t know an easy and intuitive approach to this

The only partial approach I know, which is certainly not easy and intuitive is to show that for an odd prime p a p group P that is not cyclic must have a normal subgroup isomorphic to Z_p x Z_p (by induction, not very easy to do)

But I’m not sure how to go about showing that if P is a 2-group with a single subgroup of order 2 then it is a generalized quaternion group

Oh actually I guess for the quaternion case you can show it must have a normal subgroup isomorphic to Z/4Z and one isomorphic to Z/2^nZ and then construct the semi direct product

[next obsidian]

It’s called a book originally published in 1969

[wooden ember]

Well it’s gross

Read another book

[next obsidian]

No

This has the info I need

It’s like the only book which does which isn’t EGA

Which is French

So even more gross

[wooden ember]

Then type up the book again properly tabs then read it

[next obsidian]

-_-

The book has been typed up again. It’s called “Commutative Ring Theory” instead of “Commutative Algebra”

It’s just that the info I need didn’t make it into that newer version

[wooden ember]

Oof

[next obsidian]

Because I already read all of the latter, so the info I need is exactly Commutative Algebra \ Commutative Ring Theory

[wooden ember]

Lmao

[waxen hedge]

That's weird

[next obsidian]

.

It would be impossible for it to not be the case

[lean moat]

Could someone explain me the intuition behind why Z/nZ represents the integers mod n?

[broken stirrup]

they are isomorphic. Also you can think Z/nZ have the elements equivalence classes of 0 and 1. You can interpret them as 0,1 which are elements of Z_2

[lean moat]

So they aren’t equal, only isomorphic?

Z/nZ and the set of integers mod n I mean

[weak oriole]

I like to think of it as, declaring all multiples of n as 0

This is same as caring only about remainders modulo n

[spiral wolf]

Let $\alpha = 2^{1/3}$. Show that the set of elements ${1,\alpha,\alpha^2}$ is linearly independent over $\mathbb{Q}$. In other words, prove that if $a,b,c \in \mathbb{Q}$, then $$a+b\alpha+c\alpha^2=0 \rightarrow a=b=c=0$$

[cloud walrus]

Espio

[spiral wolf]

This problem is related to field extensions. I'm not exactly sure how to show this, even though it's intuitively obvious.

[weak oriole]

its like $\alpha$ being a root of $cx^2+bx+a$

[cloud walrus]

Noob666

[weak oriole]

Now can you think of an approach?

[spiral wolf]

I would have thought we cared about alpha being a root of $x^3-2=0$

[cloud walrus]

Espio

[weak oriole]

did that help?

[weak oriole]

Yes, exactly

[spiral wolf]

Hm. So if I show that polynomial is irreducible over Q, I'm done?

[weak oriole]

$\alpha$ can't be a root of a quadratic rational polynonial

Yep

[cloud walrus]

Noob666

[spiral wolf]

Gotcha

Thanks

[frank fiber]

What is your definition of Z/nZ?

[lean moat]

{[a]| [a]={b|b is congruent to a mod n}}

[frank fiber]

you can prove that Z/nZ has n elements

[n]=[0]

[n+a]=[a]

[hidden haven]

[1]

[waxen hedge]

Yes !

You can find a proof here
https://stacks.math.columbia.edu/tag/00CT

[obtuse kernel]

I didn't find much success in the other channel, so I'm reposting my question here:
If $A$ is a fixed $n\times n$ matrix, and the function $f_A:V\rightarrow F$ is defined $f_A(X) = trace(A^tX)$. How do I show that $A\rightarrow f_A$ is an isomorphism from $V$ to $V^*$

[cloud walrus]

TylerUno

[obtuse kernel]

The linearity of the map is straightforward but I'm not sure how to approach showing the bijection

I'm thinking I have to come up with $n^2$ matrices $X_i$ such that $f_{X_i}$ spans $V^*$

[cloud walrus]

TylerUno

[wooden ember]

by V in this case do you mean square nxn matrices?

id intuitively say try evaluating linear forms on the elementary matrices that form a basis for V

that might get you somewhere not sure

[obtuse kernel]

yeah

That's also what I was thinking but it seems tedious

[wooden ember]

shouldnt be too tedious

just evaluate the dual basis for V^* (easy to evaluate) and then extend it to any linear form

[obtuse kernel]

maybe I'm misunderstanding. could you elaborate?

the dual is simply the basis s.t. $f_{i,j}(X_{i,j}) = \delta$ right?

[cloud walrus]

TylerUno

[obtuse kernel]

excuse the abuse of notation

what do you mean by extend it to any linear form

[hidden haven]

Take some f in the dual, look at its actions on each E_ij where E_ij is the matrix that is 0 everywhere other than the i,j entry where it is 1

I think those are what narwhal is calling elementary matrices

This will allow you to define a map backwards

and then check that the 2 maps are inverses of each other

[unique berry]

naw man I'm at my limit

Anyways my prof said that I shouldn't do ANT next semester because he could tell I was a bit lost during the second half of the semester and I also had a fairly low grade in the class (probably bottom 5)

[hidden haven]

Is this from some sort of quiz?

Given that you asked a similar question before and also deleted it later

[delicate bloom]

very sus

[final oasis]

if d divide the order of G does their necessarily exist a subgroup of order d

[viscid pewter]

no

try and find a counterexample

wait counterexample is trivial, any non-cyclic group

[south patrol]

Yeah, the converse of Lagrange's theorem is invalid, but there are some nice partial converses

with Cauchy's theorem being a nice starting point

[final oasis]

if G is cyclic then it's true?

[next obsidian]

It characterizes them

[chilly ocean]

The Sylow theorems deal with many of the cases where it is true, but it's not in general

[viscid pewter]

yes

many cases where it's true otherwise also

[south patrol]

if G is cyclic of order n then given d|n we can consider g^(n/d) where g is a generator

[final oasis]

now for G finite d divides its order
if the set of x such that x^d = 1 has size at most d, does the elements form a group

of order at must d

or does G need to be abelian for this to work

[viscid pewter]

if it's abelian, it definitely works, if not it can fail

in the group of all invertible 2x2 matrices there are matrices with order 2 that multiply to give an matrix with infinite order

[final oasis]

but G is finite

[viscid pewter]

there's probs still something where it fails

hmm

yeah just consider Sn

wait no

the sub won't have size at most d

[proud bear]

there may not be an element of order d. for example, S_4 doesn't have an element of order 15

[viscid pewter]

yeah that was resolved

[proud bear]

oh

[viscid pewter]

i feel like D_2n will disprove this

take all the elements with order n

or something

wait yeah let's just take S_3 = D_6, there are three 3-cycles but they don't form a group

[final oasis]

how many 3-cycles

[viscid pewter]

wait fuck

2

i'm a fool

damn

ok fine, D_8: r, r^3, rs, r^3s are all that have order 4, but they're not a group

[final oasis]

What about those with order dividing 4

[viscid pewter]

heh?

oh fuck

[final oasis]

D_8, 4 or 8 edges?

[viscid pewter]

8 elements

[final oasis]

So the Dihedral group with 8 vertices

[viscid pewter]

the group of the 8 symmetries of a square

[final oasis]

4 vertices

[obsidian sleet]

Sylow strong.

[final oasis]

So it's true then

[final oasis]

I am not familiar with this

[obsidian sleet]

its some random shit i say ignore me

but sylow's theorem is really really nice

[viscid pewter]

not necessarily

[wise igloo]

for the set ${a \circ g : a, g \in G}$ I saw it written as $aG$ for a group $G$, can this also be written as $a \circ G$?

[cloud walrus]

gmod

[wise igloo]

and is this kind of a distribution law?

[wise igloo]

one more thing, can someone explain to me what cayley's theorem states, with the proper intuition? im lost

[lethal dune]

the operation, i.e. ∘ is implied there, when working with one group, it can be imitted because it's not ambiguous

[wise igloo]

alright I thought so

is this considered some sort of "distribution property" @lethal dune

on an entire group?

I mean it ends up being equal to the group but still

[lethal dune]

no that's kinda definition

[wise igloo]

or kinda like scalar multiple

oh

[lethal dune]

like extending the notion of a×b to a whole set

[wise igloo]

alright makes sense

also would you be able to explain what cayley's theorem is saying?

[lethal dune]

ever finite group is iso to some subgroup of Sn?

[wise igloo]

oh is that literally it

oh damn ok ty

[lethal dune]

like you were thinking of something else? I'm absolutely certain that this guys more than one theorem named after him

[wise igloo]

no it's because these videos im watching never explicitly stated it

[lethal dune]

bruh lol

[wise igloo]

and I checked wikipedia and it confused me

[lethal dune]

proof is obvious I suppose?

[wise igloo]

uh idk

this guy proves isomorphisms by doing set permutstions

[lethal dune]

yeah just enumerate the elements of the group, then see multiplication sends what element to what

[wise igloo]

but the comments sometimes say that he shouldn't be teaching groups exclusively with set permutations

[lethal dune]

and construct your permutation that way,

that's literally it

[wise igloo]

honestly sometimes doing the permutations seem time confusing

simple but time consuming

[lethal dune]

yeah

tagged wrong message

[wise igloo]

lol

[lethal dune]

yes you don't actually compute it, just state that it exists and use some properties of Sn. You'll only compute them one/twice, like when you are explicitly asked to

[wise igloo]

alright

thank you!

[past temple]

if E is a linear isomorphism of R^n that preserves orientation

how do i show this?

(the hint)

[final oasis]

what does it mean to preserve orientation, det E>0 ?

what's V and W

[past temple]

so R^n = V (+) W by assumption

[lethal dune]

maybe post the entire question and not the hint?

[past temple]

sure

this is the lemma that i want to prove

these are the first set of steps that i need to do to prove it

frankly im stuck on (a) too lol

any help would be rly appreciated

i konw that the lemma mentions homotopy, etc

but these first two steps (a) and (b)

should purely be linear algebra

im just very stupid

[lethal dune]

what is homotopy in this context?

also are you suck with why should we have such V?

i.e. there is an 1/2 dimensional invariant subspace of E

[past temple]

the continuous deformation of one map to another

[lethal dune]

ok

[lethal dune]

is that where you are stuck?

[past temple]

yeah why do we know that fact