groups-rings-fields

[hidden haven]

No it's not

[static yew]

it would be easier if everybody didn't use the same 'x' in every different context

[hidden haven]

You just called them by the same name, doesn't mean that they are the same

[hidden haven]

Ye avoid

If you are working with multiple fields at once

[static yew]

right. I can grasp that concept easily, ironically, as a software engineer (recognizing the distinction between representation and identity)

the hard part is recognizing which context the author is working in

[hidden haven]

If the author doesn't make that clear then it's probably not a good author

[static yew]

maybe they think they are and I'm just a lousy reader 😄

if E, F are fields with E>=F we call E an extension (field) of F and call F a subfield of E

E is said to be finite-dimensional over F if there are elements e1,...,e_n in E such that nay e in E has the form ughhhhhhhhh this is too abstract for me to follow

[hidden haven]

Happens

Are you taking a course on this or self study?

[static yew]

self study

[hidden haven]

Then if this is too abstract you might want to do some ring theory first to get used to the abstraction

Field theory directly can be difficult

[static yew]

the book i'm reading is Stillwell's Elements of Algebra

some of this stuff is etiher super straightforward to me, or close enough so something I can grasp it easily

[hidden haven]

Never heard of it, but abstract alg is the starting point for a lot of people doing math so there are some very beginner friendly books available

[static yew]

in my head, a field is something where the algebra I learned decades ago in high school still works

so prime fields are really easy

[open sluice]

I would have said “wtf” but I did start abstract before linear so that makes sense I guess

[hidden haven]

Yeah that kind of intuition is fine

It's just that it gets hard to imagine these kinds of things with finite fields and embeddings of those into each other

Which is why doing ring theory first is better because that will give you a ton of examples to think back to every time you see a definition

[static yew]

wait so this seems to be saying that

E is finite-dimensional over F if you can "reach" any element by combining some elements of E with elements of F by a simple sum-of-products

which would be like GF(2^4) -> {1,x,x^2,x^3} paired with {0,1} from F_2

[hidden haven]

Yes, as long as you can do this using only finitely many elements from E

Which you did here

[static yew]

is F2-bar an extension field of F_2? it seems like no, then

[hidden haven]

It is, but not a finite dimensional one

[static yew]

oh right we're in finite-dimensional mode

it seems that all finite-dimensional extensions of a finite field will be finite fields

[hidden haven]

Yep

[static yew]

but e.g. C is a 2-dimensional extension of R

[hidden haven]

Yes

[white oxide]

ye

[static yew]

wooo okay I'm starting to grasp this madness

Whoops my kbd stopped working in discord again
So it looks like {1,x,x^2,x^3} is a basis

But {1,x+1,x^2,x^3} is not cuz x+1 overlaps with 1

[coral spindle]

Basis for what space?

[static yew]

GF(2^4)

Over F2

[coral spindle]

Then no, {1, x+1, x^2, x^3} is a basis I believe

[static yew]

Oh maybe it is

But {x,x+1,x^2,x^3} is a spanning set but not a basis

Oh maybe it is man this is tricky

Anyway I gotta head back to the office

L8r. Thanks for the help

[white oxide]

how does alpha^8 = 1 follow from the line with the arrow?

[long nebula]

It doesn't follow

It's just another fact

[white oxide]

oh okay

[long nebula]

You can check

[white oxide]

true

[long nebula]

It's just something you need to prove the part that follows "we see that"

[white oxide]

yea im trying to see it rn lol

[void cosmos]

if A is finitely generated R-module then A = A_t + F where F free = A/A_t. Proof: A/A_t is torsion free and finitely generated , so its free , so its projective , then we get the exact sequence A_t --> A --> A/A_t --> 0 is spit exact and we get A = F + A_t ?

is that it?

[white oxide]

oh

they all have order 2 i think

well

obv not the identity

[long nebula]

Which part are you having trouble verifying

[white oxide]

no i got it rn

thx for the help tho

[long nebula]

Np

[rocky cloak]

Is R a PID or what are your assumptions?

[void cosmos]

yes

sorry had to include that

[rocky cloak]

Then yes, that's correct.

[void cosmos]

cool af

ty

[formal ermine]

I didn't quite understand how the riemann hypothesis doesn't follow from the third weil conjecture

[chilly ocean]

So you're back from ur date already

[coral spindle]

brutal

[formal ermine]

2 ¾ hours

it went good tho

we ate and then walked around for an hour or so

we talked a lot

it was fun

[chilly ocean]

that's nice

[formal ermine]

oh gotta ping the gang @ashen heron @graceful flume

[formal ermine]

it's sad that we won't be able to see each other for the next month tho

first she's on vacation and then right when she gets back I'm going on vacation

[void cosmos]

fuck math

[chilly ocean]

you'll get through it 💀

[formal ermine]

why say it like that

now I feel cringe

[chilly ocean]

😭

[formal ermine]

anyway what's the answer to this

[void cosmos]

isnt one for finite fields

[formal ermine]

you can apply it to elliptic curves over finite fields I think

to estimate the rational points

q + 1 + error(2sqrt(q))

or something like that

[white oxide]

i get that R/M field implies that it's a division ring (and hence has the invariant dimension property), but how does that show that R has the invariant dimension property? does it have to do with the action of R/M on a free R/M module F being induced from R or something

[next obsidian]

Let N be a module, one can form N/MN

If N ≈ N’ then N/MN ≈ N’/MN’

So now assume that R^n ≈ R^m

Reduce both sides mod M

Now you know n = m because you know it over R/N

(I guess you need to note that R^n/MR^n ≈ (R/M)^n)

[white oxide]

oh neat that makes sense

thanks

[wraith cargo]

dumb question butttttt

why is R[x] flat over R?

[coral spindle]

Even better than that, it's a free module

[wraith cargo]

basis 1,x,x^2,... I assume
hm thanks!

[static yew]

it seems like:

1. you can find F4 inside F16 by taking the elements (0, 1, x^2, x^2+1)
2. which suggests that while F4 is hidden inside F16, and is therefore you can say that F4 is a subfield of F16, F16 isn't an extension of F4, because there is no basis that lets you construct F16 from just those four elements

[formal ermine]

does someone know if there's a modern tex'd version of Lectures on p-adic Differential Equations by Dwork?

[static yew]

so obviously {x,y,z} is a basis for all 3D vectors
but {x+y,y+z,x+z} also seems to be a valid basis[1]

[1] possibly required: nonzero field characteristic

[formal ermine]

hi stevie

[static yew]

i do not know anything about any p-adic diffeqs for any value of p

[formal ermine]

maybe if p = 10

lol wtf

where did 2.3.5 go

[celest furnace]

What does this say

[static yew]

you need the premium tier subscription to see 2.3.5

[static yew]

it doesn't say anything, because it's an image and cannot speak

furthermore, the text is unreadable

damn the people who I've been talking to aren't around

is Z/pZ isomorphic to Z[x] \ (x-p)

[formal ermine]

no

that's just Z

you can only get bigger by modding out a polynomial

[static yew]

damn then I misunderstood (or am misremembering) what I was told

is there a thing in Z[x] that is isomorphic to Z/pZ?

[formal ermine]

Z[x] has characteristic 0

[static yew]

oh good point

[formal ermine]

it's just 4 pages of computation lol

like boring hs algebra

[idle roost]

You may be thinking of Z[[x]]/(x-p) is isomorphic to Z_p

[static yew]

when you say Z_p you mean Z/pZ ?

[idle roost]

p-adic integers

[formal ermine]

oh I just remembered

there's

Z[x]/(1 - x, p)

is Z/pZ

[static yew]

now you're just making things up 😄

[formal ermine]

,, \bZ_p = \varprojlim_n \bZ/p^n\bZ

[formal ermine]

or equivalently the ring of integers of the completion of Q wrt the p adic norm

or equivalently a billion other things

[static yew]

I have no idea what p-adic is and I'm afraid I'm gonna break my brain if I try to process a limit on a noncontinuous set

[formal ermine]

hm ok

[summer path]

it's an inverse limit

[static yew]

this is probably the thing I was trying to get to, although I'm having trouble wrapping my head around the 1 - x bit

[formal ermine]

do you know how the reals are constructed from the rationals by taking all cauchy sequences and modding out the zero sequence

[formal ermine]

you set x = 1

every time you see x you know "ah that's just 1"

just like C = R[x]/(x^2 + 1)

[static yew]

this sounds like I added an x but then removed it back again

[idle roost]

yes

[formal ermine]

yes

[idle roost]

that is basically what happens

[static yew]

but maybe that's what I need, to understand this

[formal ermine]

don' leave me hangin' like 'tis

[static yew]

somebody was trying to do RSA or DH stuff earlier (clue 1: "if r and p are large primes", as if the size of the prime matters. clue 2: nonsensical comparison of two different multiplicative groups)

[static yew]

buddy, I'm not even 100% convinced the reals even exist at this point

[formal ermine]

so true

so true

[static yew]

for all I know it's all Q[i]

[formal ermine]

[static yew]

or Q(i)

[formal ermine]

[static yew]

or maybe even Q[[[[[[i]]]]]]]

[formal ermine]

[summer path]

i think you dropped these [ ]

[static yew]

possibly. there were too many to carry

but anyway, the guy earlier talked about what was basically (a mod p) mod q

and I know what he meant
when he says "a mod p" he means the r such that a = qp+r, 0<=r<p

but I have recently had my eyes opened about the nature of congruence classes

and I'm trying to think of how I would write the ring of integers "mod p mod q"

the first thing I noticed is that:

• if p=q then it's just Z/pZ
• if p>q then the elements are those of Z/qZ but the operators are wonky
• if p<q then it's the same as "mod q mod p"

btw those p, q are both primes

ok so if

(a mod p)bar === a + np for all integers n
(a mod q)bar === a + mq for all integers q

then

(a mod p mod q)bar === a + np + mq for all integers n, m (with "duplicates" at multiples of lcm(p,q), for example a + qp + 0 and a + 0 + pq)

which means that "a mod p mod q" is congruent to "a mod q mod p"

[ivory trail]

veritasium actually just made a p-adics video for his (more) general audience https://www.youtube.com/watch?v=tRaq4aYPzCc

[open sluice]

I liked https://youtu.be/3gyHKCDq1YA

[ivory trail]

https://www.quantamagazine.org/how-the-towering-p-adic-numbers-work-20201019/ also quanta article

[static yew]

AHA
so Z is a field

[open sluice]

https://tenor.com/view/grass-gif-6112271

[frigid lark]

Touch field of Z

[static yew]

so now N is a field

I'm expecting the next thing to be proof that F_2 is isomorphic to C

well, shit

[static yew]

wait doesn't Cantor's diagonal argument prove the p-adics are uncountable

i'd like a refund on this headache please

[ivory trail]

yeah

[hidden haven]

There is a basis for F16 over F4: {1, x}

[frigid lark]

For part b how do I show that if K_p is infinite, it is cyclic?

[frigid lark]

Wait, is it because for each finite cyclic extension F, of degree p^n, it's Galois group is generated by sigma (restricted to F). And the Galois group of K_p will be the inverse limit of these finite Galois groups?

This feels wrong for some reason

Specifically cause part 3

Yeah, how can K_p be infinite cyclic if it's Galois group contains a subgroup of rank 2 over Z

[astral stream]

Can someone explain why the underlined part is relevant?

Thm 48.11 just says for a set of automorphism for field E, the set of fixed elements for all automorphism in that set makes a field.

[frigid lark]

I think by the prime subfield, they mean Q

[astral stream]

why should it be in it?

[frigid lark]

if sigma is a field auto of a field of char 0, then sigma(1) = 1, and sigma(-1) = -1. These additively generate Z, hence for any n in Z^+, sigma(n) = sigma(1 + ... + 1) = 1 + ... + 1 = n, and sigma(-n) = sigma(-1 - ... - 1) = -n. Hence sigma fixes Z. Finally for n in Z / {0}, sigma(1/n) = 1/n, thus form m/n in Q, sigma(m/n) = m/n

Thus sigma fixes Q

[astral stream]

Finally for n in Z / {0}, sigma(1/n) = 1/n. How is this? isn't Z/{0} just ismorphic to Z? so its the same as proving sigma(n)=n

[frigid lark]

sigma(1) = 1 => sigma(n * 1/n) = sigma(n) * sigma(1/n) = 1 => sigma(1/n) = 1/sigma(n) = 1/n

[astral stream]

oh nice, thanks

[rocky cloak]

I guess "infinite cyclic" should mean that the Galois group is the closure of a cyclic group.

[frigid lark]

Closure in the topological sense?

[coral spindle]

I'm ignorant here, but perhaps jagr means the completion in the sense of profinite groups?

[frigid lark]

As in the inverse limit of a family of cyclic groups?

[rocky cloak]

Yes, the Galois group is a profinite group and hence a topological group

[rocky cloak]

The Galois correspondence for infinite Galois extensions goes between intermediate field extensions and closed subgroups.

If you take a subgroup H of G(E/F), the closure of H is equal to G(E/E_H), where E_H is the field fixed by H.

So then cyclic extension would mean an extension of the form E/E_H where H is a cyclic group. But this does not mean that the Galois group is cyclic, just the closure of a cyclic group.

[rocky cloak]

So it's not quite the completion. For example the completion of Z is $\hat{\mathbb Z} = \prod_p \mathbb Z_p$, and Z is indeed a dense subgroup. But Z can also appear as a dense subgroup of other profinite groups. For example $\mathbb Z_p$ for a fixed p. So the ambient group matters.

[cloud walrus]

jagr2808

[formal ermine]

ah I remember this from alg nt yesterday

[minor wraith]

Does anyone have a mnemonic for memorising the field axioms?

For an exam

[formal ermine]

commutative ring and you can divide stuff

[minor wraith]

Ditto for a ring

[formal ermine]

,w define ditto

[cloud walrus]

Results provided by WolframAlpha

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

Requested by projective#0

[elder wave]

Field but you can’t divide

[formal ermine]

I'm being serious

that's how I remember what a ring, field is

[minor wraith]

Yah but like

Imagine going through a multi-choice section checking the field axioms

Isn’t it fucking exhausting

[formal ermine]

just... remember them

[minor wraith]

Trying to remember the last missing axiom

[formal ermine]

multiplication, addition

[minor wraith]

I meant like

[formal ermine]

for multiplication you have commutativity, identity, associativitiy, inverses

same for addition

and those two distribute

[minor wraith]

Wouldn’t it be much easier if you have some sort of ordering convention…

[formal ermine]

no...?

[minor wraith]

Make it a bit more algorithmic…

Unless it’s immediately obvious

Something like

MAD

[formal ermine]

[minor wraith]

CIAI

For MA

[formal ermine]

this is math not like

wtv

[minor wraith]

Well

[formal ermine]

remembering how to do stuff algorithmically is fucking stupid

[minor wraith]

Taking exams under timed conditions arguably isn’t real “math” either

[minor wraith]

Yore rite

Gotta bust out the fingers to do basic addition

[formal ermine]

ong

I had to use a calculator for 13+19 in my last exam

[hasty sinew]

The field axioms are basically just the properties of the reals you learn in middle school

[minor wraith]

Gotta get my 100 pens ready to do multiplication first

Or borrow some friend’s fingers

[hasty sinew]

Associative property, commutative property, distributive property, multiplicative identity and inverse, additive identity and inverse

[minor wraith]

Inv Dis Ass Id Com

Inverse Distributivity associativity identities commutativity

[formal ermine]

this ass is common

so true

[minor wraith]

Ah I guess it’s usually better to check associativity first

Ass Comidis

[formal ermine]

the ass commutes

[minor wraith]

I had a really racist mnemonic for another algebraic structure

So I’m never forgetting that one

[tender wharf]

this is not an eeveekawaii....

[next obsidian]

…

[next obsidian]

.pin

[warm wyvern]

dm me

[glossy crag]

If L/K is cyclic (with generator sigma) and a\in K*, why should this be a 2-cocycle? The functional equation is f(x,y)f(xy,z)=f(y,z)f(x,yz) (actually it's xf(y,z), but f(y,z)\in K, so x applied to it is the identity), so what if n=6 and you substitute x=sigma^4, y=sigma^2, z=sigma, you get a^2 on the LHS and a on the RHS.

[delicate orchid]

you just remember shit as like

"abelian group over abelian group" for fields

"monoid over abelian group" for rings (semigroup if you're psychotic)

[ebon pine]

I am curious as to why you need a mnemonic for an algebraic structure

[minor wraith]

You don't really need to

it's more like when you're studying for an exam

and have to do like 10-20 multichoice Qs in a row

[delicate orchid]

if it works it works

[minor wraith]

it becomes much faster for me to follow the mnemonic

but yes, often the missing identity will seem obvious

[warm wyvern]

I am curious as to what the mnemonic is

[minor wraith]

I rlly prefer not to say lol

[warm wyvern]

it's that bad??

[minor wraith]

it's not even a proper sentence

[long nebula]

Don't say it here.

[minor wraith]

yeah ofc not

[long nebula]

Let's change the subject

[delicate orchid]

ok lets talk about hmmmmmmm

[hot abyss]

i wonder what you could possibly talk about in the abstract algebra channel

[delicate orchid]

a mystery for the ages

let us talk about p-groups of order p^4

[hidden haven]

Take p = 6

[delicate orchid]

considering

[tender wharf]

what about p=1

seems more fun

[delicate orchid]

considered

[long nebula]

my favorite prime is zero

[delicate orchid]

I will never understand why we exclude 0 from being a prime in Z

so.... stinky

[ebon pine]

ah yes clealy 0 = 0.p is the only factorisation of 0

[glossy crag]

Anyone?

[tender wharf]

my guy couldn't be arsed to type * instead of .

[rocky cloak]

Maybe they're Australian? Or which ever country it is that uses . For multiplication and \cdot for decimal point

[tender wharf]

[frigid lark]

I have never seen that

Used

[rocky cloak]

I feel I remember Matt Parker using it in some numberphile videos.

Wikipedia also says that in English speaking countries one uses a dot either at the bottom or halfway between bottom and top for a decimal point.

Idk

[delicate orchid]

yeah, dot in the middle for multiplication, dot at the bottom for decimals

just like the dot product (wholesome)

[rocky cloak]

I guess xy = sigma^0 or what is n?

[chilly radish]

^0

f(xy,z) would be 1, not a

[glossy crag]

Ah man, I forgot to take the actual cyclicity into account

e.g. f(xy,z) is f(1,sigma) not f(sigma^6,sigma) (i mean for the purposes of the definition of f)

[glossy crag]

If C is the algebraic closure of F_p and L/K is any subextension in it (|L| not necessarily finite), is it automatically normal and does the Galois group have some nice description?

OK, I think I've worked out the normality, is this correct:

To show L/K is normal it is enough to show L/F_p is normal. Let x\in L be arbitrary and f its minimal polynomial over F_p, then it has a splitting field K/F_p in C. We know K/F_p is cyclic with generator x\mapsto x^p (as an extension of finite fields), so all the roots of f in C are given by powers of x. But these are all contained in L, therefore f splits over L and L is normal.

[wraith cargo]

does your first line work both ways?
I know that if L/K is normal and E is some intermediate field then L/E is normal but I'm not sure the converse holds hmmmm

[glossy crag]

wdym "both ways", what I did is exactly what you're saying, not the converse. All subfields of C automatically contain F_p (it is the prime field), so if L/F_p is normal, then so is L/K.

[rocky cloak]

Take E=L

[wraith cargo]

tbh I'm not sure I'm following what you're doing
You first define C/F_p but then pull L and K out of somewhere?

[glossy crag]

C\supset L\supset K

an extension of algebraic extensions of F_p

[rocky cloak]

The Galois group should be the inverse limit of all the finite Galois groups. So in this case I guess that should be something like

$\prod_p \mathbb Z/p^{k_p}$

Where $k_p$ is the largest k for which there is an intermediate extension of degree $p^k$, and $\mathbb Z/p^\infty$ are the p-adic integers

[cloud walrus]

jagr2808

[chilly radish]

Writing Z/p^infty for the p-adics is very nonstandard

Z/p^\infty is usually the prufer group

Which is a direct limit

[rocky cloak]

Yeah, I just didn't want to have to split up my notation

[glossy crag]

OK and what if |L:K|<\infty, but K is infinite, is there a more concrete description then?

[rocky cloak]

I'm not sure the description is different, but I'm not sure if something weird happens when K is infinite

[glossy crag]

@rocky cloak you're saying the Galois group in general is the inverse limit of the finite Galois groups, so my question is what the finite Galois groups look like. If K is finite, I know it is the cyclic group of order n.

[coral spindle]

Gonna double check with yall — when people say "Abelian group of finite type," this means it's finitely generated, right?

[glossy crag]

Likely, some call finitely-generated modules "of finite type".

[rocky cloak]

I think they're still cyclic when K is infinite, but could be wrong about that

[glossy crag]

What would be a generator then

[wooden ember]

what's K in this case? An infinite algebraic extension of F_p?

[chilly radish]

Are there even nontrivial infinite subfields of F_p\bar

[wooden ember]

prufer group ought to have infinite index subgroups no?

[chilly radish]

The galois group is the profinite integers

Not the prufer group

[wooden ember]

oop my bad yeah

[wooden ember]

even then it's still the case no?

cause dont we have Zhat = prod Z_p

[chilly radish]

Probably. Z should be of infinite index no?

[wooden ember]

so take any subgroup of p adic integers

[chilly radish]

That also works yea

[lime badge]

Let's say we have Galois extensions $k \subseteq E$ and $E \subseteq F$. Its not hard to come up with examples of $k \subseteq F$ not necessarily being Galois, but I'm supposed to show that this is the case when every automorphism in $\text{Aut}_k(E)$ is a restriction of one in $\text{Aut}_k(F)$. I'm a bit lost on how to prove this though

[cloud walrus]

eternalway

[lime badge]

My dumb idea was that restriction to E of automorphisms in $\text{Aut}_k(F)$ defines a homomorphism from $\text{Aut}_k(F)$ to $\text{Aut}_k(E)$. By assumption, this is surjective and its kernel is easily seen to be $\text{Aut}_E(F)$. Then can't you use degree counting to show $\text{Aut}_k(F) = [F:k]$?

[cloud walrus]

eternalway

[wooden ember]

the argument i know shows directly that F is a splitting field of a separable polynomial

like use the fact that automorphisms are obtained from restrictions to show that your minimal polynomials split

[rocky cloak]

Should be something like for x in L, consider F_p(x) / F_p(x)\cap K and pick the generator there that you get from the Frobenius map and apply that to x. You may have to fiddle a bit to make it well defined.

[rocky cloak]

There should be a "GF(p^n)" for any super natural number n I think.

Just take the union of GF(p^m) where m ranges over every natural number that divides n

[chilly radish]

True

That seems to work

[formal ermine]

what's a super natural number

[lime badge]

Can you expand on this a bit? E is galois over k, E is the splitting field of a separable polynomial with coeffs in k. Same for E over F. How do automorphisms come into play here?

[wraith cargo]

automorphisms do permute the roots (not too hard to show)

[lime badge]

Right, that's easy to show

[wraith cargo]

(Btw the condition you listed in the problem is equivalent to the extension being normal and separability is a bit nicer since it these kinds of cases it follows so the extension if Galois)

[lime badge]

Ah, I didn't know this part

[wraith cargo]

yeah for separability you have this nice iff statement that an extension E/F is separable iff E/K and K/E are separable

and a Galois extension is defined as a normal separable extension

[lime badge]

I knew this about separability. I was referring to your comment about my problem statement being equivalent to normality

[wraith cargo]

oh yeah this is how my prof defined normality

[wraith cargo]

here's the full theorem (this is from my class' like recap notes?)

(I believe your statement is equivalent to the third point)

or it isn't wait I might have had a brain fart

[lime badge]

Yeah, I don't think it is unfortunately

[glossy crag]

Yes

[lime badge]

However, we do know that since E/k is Galois, for algebraic extension of E (like F) the automorphisms of F that fix k also map E to E

So now you know that restriction defines a homomorphism from Aut_k(F) to Aut_k(E) with kernel Aut_E(F). It's surjective by our assumption, so you can just use degree counting to prove Aut_k(F) = [F:k]

[chilly radish]

They just mean an element of \hat Z

[formal ermine]

ah

[rocky cloak]

Super natural number generalizes the unique prime factorization of natural numbers to include infinitely many primes. They correspond to principle ideals in \hat Z.

https://en.m.wikipedia.org/wiki/Supernatural_number

[chilly ocean]

UHF algebras

[void cosmos]

yoy guys

i have a very embarrassing question

but is the difference between finitely generated and free is that basis are linearly independant

while in fg u can have relatinos between the generators?

(and also a basis might be infinite)

[formal ermine]

yes

[void cosmos]

cool

tysm

[south patrol]

Or needn't exist

Also, you can be free but not finitely generated e.g. infinite dimensional vector space

So the two are kinda unrelated

[stuck fiber]

So I get that the set of coset representatives is a set containing one representative from each coset, but I'm not seeing how that translates to the construction of H and G even though they feel correct

[wraith cargo]

first do you understand why cosets are disjoint?

[stuck fiber]

I think I do basically since ${x_i}$ only contains one representative for each coset then if $x_{i}\in h_{i}K,$ then $x_{i}\not\in h_{j}K$ unless $i=j.$ This would give that $x_{i}\not= x_{j}$ and therefore $x_{i}K\not= x_{j}K.$ But we have that if two cosets share any elements then they are equal and since these cosets are not equal for any index, then they must be share any elements and so are disjoint

[cloud walrus]

kenshin5334

[stuck fiber]

Sorry about that btw I had to take an important phone call I think I had a better explanation but I lost it during the call

[warm wyvern]

``if $x_{i}\in h_{i}K,$ then $x_{i}\not\in h_{j}K$''

bruh

oh that's h_j

mb

didn't notice lol

[stuck fiber]

I was about to ask what's wrong with that lol

[warm wyvern]

what you said is true

but I think irony incarnate was asking if you knew why cosets partition the group

like, {x_jK} is the set of cosets of K

anywho

the last line basically tells you that y_jx_iK add up to be the whole group

as the set of cosets should

[warm wyvern]

I assume you know why this is true

and what you're trying to understand is why each y_jx_iK is disjoint from the others?

[stuck fiber]

I'm trying to understand how we justify that $H=\bigcup_{i} x_{i}K$ and similar for G

[cloud walrus]

kenshin5334

[warm wyvern]

because the set of cosets partitions the group

and {x_iK} is a set of cosets of K

[warm wyvern]

do you understand why this is true?

[stuck fiber]

No not exactly

[warm wyvern]

it's because every element of the group belongs to some coset

namely for any element g\in G, g is also in gK

[stuck fiber]

Okay that makes sense and then since they partition the group the union of all of them makes the group

[formal ermine]

darq in abs alg channel???

[warm wyvern]

[warm wyvern]

that's not the sole reason for why they partition the group

for them to partition the group their union needs to add up to be the whole group AND they must be pairwise disjoint

[stuck fiber]

Oh no I was saying I get why H is the union of cosets now

[warm wyvern]

I hate tbh

[stuck fiber]

Thanks DarQ

[warm wyvern]

it's so neutral

[warm wyvern]

np!

we need but more upbeat

[summer path]

[tender wharf]

that looks less upbeat somehow

[coral steeple]

Is this the only way to do b)? Take G=Z/10Z. If we're just going by brute force, we can use the fact that phi(x^-1)=phi(x)^-1 to determine that 0 and 5 are fixed. If we choose something to send any of the remaining elements to, we immediately get phi(x^-1) for free, yielding 4! permutations to check

Ok I guess 4! is not that big but it seems unenlightening

[vast quiver]

I think you might figure this out after you brute force, but ||try studying where an automorphism sends 1 in particular||

[coral steeple]

Ah wait I'll brute force it for something smaller than 10 and then unspoiler your message then lol

[tender wharf]

group of automorphisms for C_n is isomorphic to ||U(n) ||

[tender wharf]

after you unspoiler it figure out why this is true

[rocky cloak]

I love that if you change _ for ^ this is still true.

The beauty of overloading notation

[lethal dune]

[frigid lark]

For question 48, are we working with W(k) or the projection of W(k) on vectors whose components are indexed by powers of p?

Or are they the same thing?

[formal ermine]

how is Z+ a group lol

[gusty thistle]

Hi im doing this exercise. I am wondering how it is clear that $\mathbb{Q} \otimes_\mathbb{Z} \mathbb{Q}$ has a (unique) $\mathbb{Q}$-module structure. I am really new to tensors lol

[elder wave]

My brother in Christ

[formal ermine]

what am I missing

what are the inverses

what is the neutral element

[elder wave]

Z

With addition

[formal ermine]

oh lmao

I thought positive integers

[cloud walrus]

__________________fold

Hi im doing this exercise. I am wondering how it is clear that $\mathbb{Q} \otimes_\mathbb{Z} \mathbb{Q}$ has a (unique) $\mathbb{Q}$-module structure. I am really new to tensors lol

[gusty thistle]

huh

what is wrong with my latex

[elder wave]

There’s a _

Ah wait nvm

[gusty thistle]

it renders in overleaf lol

[agile burrow]

I don't know if it has a unique Q-module structure, but there's certainly a natural one given by multiplication in the first factor, or q (x \otimes y) = qx \otimes y

[gusty thistle]

yeah hmm

what do i need to check to make sure it is well defined

i am having a bit of trouble understanding the subscript with the tensor product

i know it has to do with what $R$-balanced maps factor

[cloud walrus]

__________________fold

[gusty thistle]

omg lmao

but i dont have a good intuition what it means for what i can do with the actual tensors

[agile burrow]

Oh, sure. So when you tensor Q with itself over Z, the bilinearity relations only let us move around the Z factors

[gusty thistle]

yes that makes sense

oooh so it only lets us move arounf factors in Z in the tensors right

[agile burrow]

Explicitly, you could have (3/4) \otimes (5/6) = (15/4) \otimes (1/6) where I've moved the factor of 5 to the left side via balancing relations

That's right

[gusty thistle]

ah cool

this tensor stuff is rllyu cool but also sort of overwhelming it at first tho but so intersting lol

[agile burrow]

Now with this in mind, you might be able to find a simpler way to write Q \otimes_Z Q

[gusty thistle]

ok

hmm

[agile burrow]

Yeah, don't worry it took me a while to get more comfortable working with tensors

[agile burrow]

By this, I mean take a basic tensor (a/b) \otimes (c/d) and see if there's a way you can "simplify" it to a standard form

[gusty thistle]

so is it super obvious that right multiplication with elemenst of Q in the tensors gives Q otimes_Z Q a Q_module structure ?

[agile burrow]

It feels obvious to me, but maybe that's just because I've thought about this exercise before. If it isn't obvious to you, you can and should check the axioms

[gusty thistle]

ok thanks

[delicate orchid]

le extension of scalars has arrived

[gusty thistle]

id like to move the factors 1/b and 1/d out if that makes sense

but i cant do that right

because i can only move around factors in Z

[delicate orchid]

there's another way you can get rid of them though

but yes you cannot move them cause these are Z-modules

[agile burrow]

That's correct, but yeah wew is hinting at an approach to deal with these

Namely, multiplying by 1 in a clever way

[gusty thistle]

ah you can rewrite it to ab/cd (1 \otimes 1) right

by multipliying with d/d

ah yeah

[agile burrow]

That's exactly right, yeah

[gusty thistle]

and then its a Q-vectorspace with basis 1 \otimes 1

[delicate orchid]

or as I like to call it, Q

[gusty thistle]

hehe

awesome

[delicate orchid]

another fun one with a similar idea is to compute $\bZ/m\bZ \otimes_{\bZ} \bZ/n\bZ$

if the bot decides to render that is

[agile burrow]

Lol

[delicate orchid]

epic compile fail

Z/mZ (x)_Z Z/nZ

[gusty thistle]

yeah i did this one already

really awesome

[delicate orchid]

darn...

[gusty thistle]

nah its a nice result

[delicate orchid]

hmm

Challenge: if A is a finite abelian group Z-module then what is Q (x)_Z A

[gusty thistle]

is Q(x) rational functions or polynomials

im asking because a someone i now use Q(x) for polynomials lol

[delicate orchid]

Q tensor A

[gusty thistle]

oh lol

[south patrol]

classic

[south patrol]

Btw doesn't this work for any A

Or am I being dumb

[gusty thistle]

ok so if an element of a has order n than q (x) a = qn/n (x) a = q/n (x) na = q/n (x) 0 = 0 :3

[south patrol]

Yeah nice

[delicate orchid]

if A is torsion free then nope

see A = Q from before

[gusty thistle]

what do u mean with doenst work ?

its always an abelian group right ?

[delicate orchid]

not equal to zero

[gusty thistle]

ah ok

[delicate orchid]

I stuck to finite groups potato just incase wheel didn't know what "torsion" meant

[south patrol]

Not what I meant lol

[delicate orchid]

then what DID you mean

you LUNATIC

oh

[south patrol]

Oh

[delicate orchid]

yeah it always kills torsion

[south patrol]

I thought you meant like

[gusty thistle]

torsion elements in abelian groups are elemesnts with finite order right ?

[delicate orchid]

yus

[gusty thistle]

cool

[south patrol]

Ye

That it kills exactly the torsion elements

[delicate orchid]

more generally in modules they're elements m such that nm = 0 and n ain't no zero-divisor

[south patrol]

But no I guess uou mean that it is 0

Ans then fair

[gusty thistle]

ah yes i remember now i did an exercise abt it i think

[delicate orchid]

trying to think how this idea continues to generalise

works for any injective Z-module instead of just Q I know that much but what happens if we change the ring

[teal vessel]

so I'm checking that the rational numbers that have denominators that are divisors of some fixed n in Z+ is a subgroup, and I've shown that it breaks down into three cases for a/b+x/y

1. yb<n, which is only possible if (y,b)=1. Since both y,b|n, then yb|n
2. yb=n, which has the same above conditions and n|n
3. yb>n, which means that (y,b)>1, but then doesn't that mean that the fraction can again be reduced until (y,b)=1 which collapses to case 1 or 2? I feel like I'm spinning my wheels at this point trying to show it to be true too symbolically.

[delicate orchid]

which is only possible if (y,b)=1
n = 16, y = 2, b = 4 is a counter example
unless you meant prime divisors

[teal vessel]

I originally was thinking about it in two cases: (y,b)=1 and (y,b)>1, sorry. Artifact from previous attempt, though that may be a better way to do it.

[delicate orchid]

the main thing to notice is that a/b+x/y = (ay+bx)/by, so you just need to show that by is another divisor of n

I don't know what you're doing with this case stuff

[teal vessel]

showing that by is another divisor of n

that's the reason it's here in the first place

[delicate orchid]

also I don't believe this result, set b = n, y = n - both clearly divide n but by = n^2 clearly doesn't

[pallid oracle]

hint: you don't need to do (ay+bx)/by. When you add fractions, the maximum number you need to place in the denominator is lcm(b, y).

[teal vessel]

suppose that y and b are copirme. It then must be the case that yb|n because y and b are products of distinct subsets of the prime factorization of n, which then means that yb is the union of those subsets, which then must itself be a subset of n's factorization.

Now suppose that they are not coprime. It then follows that the sum of the fractions a/b+x/y must be (ya+bx)/by, which is reducible by the common factor of y and b. Name this factor f, and define y=fp and b=fq. We know that factoring out f=(y,b) gives us (pa+qx)/pq with (p,q)=1, which now behaves the same as in the first scenario.

[pallid oracle]

that works too

or rather, it's another way of writing the same thing

[teal vessel]

yeah. I know that's true, but I wanted to demonstrate it to be true rather than simply take it as an assumption.

[late thicket]

How to solve this question?

[pallid oracle]

what have you tried so far?

[late thicket]

Google

I opened the bracket in matrix form

Don't know what the next step should i do

[delicate orchid]

no clue what this means, but I'd start by actually multiplying out (123)(34) to get a proper permutation

[coral shale]

Are you aware of the definition for each word in the question?

That's the 1st step

[late thicket]

No

[tender wharf]

which words do you not understand

[late thicket]

I did this only

[tender wharf]

which words do you not understand

[delicate orchid]

oh right you meant two line notation

no need for that:
(123)(34)
1 -> 2
2 -> 3
3 -> 4
4 -> 3 -> 1
so (123)(34) = (1234)

[tender wharf]

then we can just use the fact that disjoint cycles commute

I'm more concerned that you replied no to knowing the definitions

Which ones do you not know

It's imperative that you understand them otherwise you will be stuck on similar problems in the future

[coral shale]

With algebra especially - there's no point in attempting problems without knowing exactly what the question means

but that applies to all math anyways

if you don't understand the terms in the question, its time to refer to your notes on them

[solemn dew]

I asked this in book recommendations but no one answered

I think my book "Algebra from the viewpoint of Galois Theory" jumps to conclusions too fast and you really have to read something 10 times before you start to understand it

Has anyone any other book recommendations for abstract algebra?

[tender wharf]

are you trying to learn galois theory specifically

what's your preexisting knowledge on aa

[solemn dew]

Has to cover:

• basic groups

[late thicket]

Why not 3>1 ?

[solemn dew]

And this

So I think it is just basic algebra

[tender wharf]

that image is a little cropped

[solemn dew]

No I don't think Galois theory is necessary

[tender wharf]

dummit foote probably covers quite a lot of these

[delicate orchid]

that 3 is from the (123) not the (34)

[solemn dew]

thank you @tender wharf I'll look into it

[delicate orchid]

you move right to left and see where stuff goes to

[tender wharf]

no problem, also that image is cursed

[delicate orchid]

alternatively you just write them down as functions and compose them if you want to be more explicit but that's the same thing as this

[solemn dew]

(First image)

[tender wharf]

yeah wews way is the fastest

otherwise you'll have to manually plug in all the numbers

I mean it's cursed because it's so long

maybe type it up next time

[late thicket]

Why did you not write 4>3 only?

[indigo ridge]

I have a few questions with regards to the symmetric group

[delicate orchid]

because 4 first gets mapped to 3 by (34), and then that 3 gets mapped to 1 by (123)

[indigo ridge]

Im trying to find all the different subgroups for the symmetric group S_3, I know we have the trivial subgroup S_3 and the identity mapping but how will I find the others? should I use brute force?

[delicate orchid]

think about subgroups generated by a single element

i.e. the cyclic subgroups

[indigo ridge]

Yeah ok

[tender wharf]

Although S_3 is small enough to brute force

[delicate orchid]

2^6 is a lot of subsets to check

[late thicket]

So what next?

[delicate orchid]

as DerpZ said, you use the fact that disjoint cycles commute

[tender wharf]

you can check them in a smarter way no?

[delicate orchid]

probably

[tender wharf]

check all the sets of order 2,3,6

not 6

[delicate orchid]

oh you meant for S_3, yeah ok

[tender wharf]

yeah

[late thicket]

Can you give any link regarding this question

??

[tender wharf]

yes, an abstract algebra textbook

[late thicket]

Or topic name

[delicate orchid]

https://math.stackexchange.com/questions/208790/centralizer-of-a-given-element-in-s-n

[indigo ridge]

Are you asking me>

?

[indigo ridge]

why check all sets of order 2,3 and 6?

[delicate orchid]

lagrange's theorem

[tender wharf]

we ignored order 6 because thats just the whole group

[delicate orchid]

those are the only possible sizes of subgroups of a group of order 6 (not counting 1)

[indigo ridge]

gotcha

[tender wharf]

I recommend you prove this fact for yourself

Disjoint cycles commute

[delicate orchid]

there are more elements in the centraliser than just disjoint cycles though

I believe (13)(24) should also commute with (1234), for instance

[indigo ridge]

Im thinking of writing all the possiblie permutations which is 3! = 6 and then writing them in cycle notation then figuring out their order is that a reasonable way to solve my question?

[delicate orchid]

this is just considering the groups generated by elements, so sure

[tender wharf]

makes sense

[delicate orchid]

and you know that because 2, 3 are prime that all groups of that order have to be cyclic

[indigo ridge]

alright thanks, last question im trying to figure out for what n would S_n be abelian or commutative, I was thinking for all n since we can decompose every cycle

[delicate orchid]

oh no no no S_n is VERY far from abelian

[tender wharf]

n=2

[indigo ridge]

really but cant we decompose and then the transpositions are commutative?

[delicate orchid]

it's abelianization is C_2 which is far smaller than S_n

since when are transpositions commutative?

(12)(23) = (123), (23)(12) = (132)

[indigo ridge]

Must be mistaken...

[delicate orchid]

disjoint cycles commute as we said before

[indigo ridge]

yes disjoint cycles is what I mean to say

[delicate orchid]

not all cycles are disjoint though

[indigo ridge]

but cant we decompose every cycle into disjoint cycles?

[tender wharf]

yes but

lets say we have S_3

[indigo ridge]

O

k

[tender wharf]

then (12) and (23) are separate distinct elements

in S_3

[indigo ridge]

Yeah

[tender wharf]

but certainly not disjoint

you might have gotten confused about that part

[indigo ridge]

Yeah that makes it alot clearer thanks

[delicate orchid]

Let $\sigma_1 = (x_1, x_2, ..., x_n)$ and $\sigma_2 = (y_1, y_2, ..., y_n)$, then [\sigma_1 \in C_{S_n}(\sigma_2) \iff \sigma_1\sigma_2\sigma_1^{-1} = (\sigma_1(y_1), ..., \sigma_1(y_n)) = \sigma_1]
this is what we need to check

[cloud walrus]

wew ladz

[delicate orchid]

and we can exploit the fact that the centraliser is a group, so we have 1, (56), (13)(24), (13)(24)(56) from before
and then obviously all powers of (1234) will commute with (1234) so we have {1, (56), (13)(24), (13)(24)(56), (1234), (1432), (1234)(56), (1432)(56)} which is isomorphic to C_2 \times C_4 and is order 8

[tender wharf]

huh this is neat

[delicate orchid]

I wrote \sigma_1, \sigma_2 as n-cycles but the same idea will still work even if they're not

[coral steeple]

Yeah I heard the notation may be a bit nonstandard. My book uses Z^+ to mean the additive group of integers

[tender wharf]

I swear I can remember what that book is but wow that must be annoying

[pallid oracle]

that is usually written (Z, +), in case you want to avoid misunderstandings

[indigo ridge]

im trying to find a homomorphism between $(Z_4, +) and (Z_5, \cdot)$ I defined a mapping $ f: Z_4 --> Z_5 to be f(x) = [0]_5. f(x+y) = [0]_5 = [0] \cdot [0] = f(x) \cdot f(y)$ is it correct?

[cloud walrus]

jayzsparrow

[indigo ridge]

Therefore f is a homomorpism

[rocky cloak]

So Z/5 does not form a group under multiplication, so probably what you actually want to do is find an isomorphism between Z/4 and the group of nonzero elements of Z/5.

If you just want a homomorphism of monoids, then what you did is correct.

[tender wharf]

theres also the bashy way of checking that U(Z/5Z) is not Z/2Z times Z/2Z

[indigo ridge]

perhaps I wrote it wrong

[tender wharf]

and then you hit it with classification of finite abelian groups

[indigo ridge]

the question in my problem set said z5

[tender wharf]

yes then what jagr said

[indigo ridge]

idk what that means lol

[rocky cloak]

Then that's not what you want. $\mathbb Z_5^\times$ refers to the elements of Z/5 relatively prime to 5. So not 0

[cloud walrus]

jagr2808

[indigo ridge]

Oh I see! thanks..

[glossy crag]

If the polynomial is written as a_nx^n+a_n-1x^n-1+...+a_0, this should be a_1/a_0 - a_{n-1}, right (basically it's s_1-s_n-1/s_n, where s_i is the i-th symmetric function of the roots of P)?

[glossy crag]

actually this assumed the polynomial is monic, one has to take a_n into account, so it's probably a_1/a_0 - a_{n-1}/a_n.

[coral steeple]

It's Artin

[vast quiver]

you need the (-1)^n to deal with the parity (just work through the example of 2 roots or 3 roots to check the sign), you don’t need to take the a_n into account for the second term because the numerator and denominator both have a factor that will cancel out, and you do need the a_n instead of the a_0 for the first term~~~

[glossy crag]

Doesn't look right to me, here's my derivation:

$\sum_j(z_j-\frac{1}{z_j})=\sum_jz_j-\sum_j\frac{1}{z_j}=s_1-\sum_j\frac{\prod_{i\neq j}z_i}{\prod_{i}z_i}=s_1-\frac{\sum_j\prod_{i\neq j}z_i}{\prod_iz_i}=s_1-\frac{s_{n-1}}{s_n}$

Then use that $f=a_n(x-z_1)\cdots(x-z_n)$ and $(x-z_1)\cdots(x-z_n)=x^n-s_1x^{n-1}+\cdots+(-1)^ns_n$, implying $-a_ns_1=a_{n-1}$ and $(-1)^{n-1}a_ns_{n-1}=a_1$ and $(-1)^na_ns_n=a_0$. Substituting this into the first identity yields $\sum_j(z_j-\frac{1}{z_j})=\frac{a_1}{a_0}-\frac{a_{n-1}}{a_n}$.

[cloud walrus]

leave_no_norm

[glossy crag]

s_i denotes the i-th symmetric function of the roots (without sign factors)

The exercise then asked to prove \phi(fg)=\phi(f)+\phi(g) and it's actually a pain in the ass to do with coefficients, but works easily with symmetric functions

[vast quiver]

I think you're right, lemme check a lil more carefully

yeah you're totally right, mb

[white oxide]

can somebody give me a hint for this question? i reduced it down to showing that the image of f = ker f, because if every a in A could be expressed as a product xy where x in Kerf and y in Imf then the image of A under f is equal to the kernel of f since Kerf is an ideal of A. i'm pretty sure that's right but man it's just on the tip of my fucking tongue

[gritty sparrow]

The claim im(f) = ker(f) is just not true, take f = id. A hint for this would be, express any a in A as a = f(a) + (a - f(a)).

also you should write in additive notation for modules, not multiplicative, since there is already a multiplication by elements of the ring

[white oxide]

oh yea good point

[white oxide]

oh right

thx

[gritty sparrow]

np

[ivory trail]

then 0 doesn't have unique factorization: 0 = 0*2

generally the prime/irreducible elements in a UFD don't correspond to the prime ideals

[coral shale]

1.5 days old

was just a dumb joke

[ivory trail]

oh my bad

if you are thinking about the prime ideals of dim n irreducible commutative rings, you might as well think about dim n integral domains. then dim 0 integral domains are obviously fields, and p \subsetneq q is a chain of primes in a dim n >= 1 UFD where p is generated by an irreducible element, then q cannot also be principal for obvious reasons

so basically this comes down to Z being dimension 1

[indigo ridge]

if the order of an element a of Zn is the lcm(a,n) then how is lcm(2,6) = 6 when 2+2+2= 6 shouldnt it be 3 not 6?

[coral shale]

lowest common multiple right

3 isnt a multiple of 2?

[indigo ridge]

yeah*

so its 6 and not 3

[coral shale]

oh let me see

where did you get that 'fact' from - just doesnt look right

[indigo ridge]

google lol.

[coral shale]

show / paste the search prompt

[ivory trail]

that entire second sentence is a disaster

[coral shale]

'By definition this is the lcm of a and n' is the error I believe

[indigo ridge]

gotcha

[ivory trail]

the other part is also wrong

[coral shale]

ab = gcd(a, b).lcm(a, b) I believe is the correct identity

right?

[ivory trail]

yeah

[coral shale]

7 upvotes smh

rare I see an upvoted wrong answer

[indigo ridge]

lol

[coral shale]

I suppose they mean ka is lcm(a, n)

Would that fix it?

[coral shale]

I missed this

[ivory trail]

wait no

it's just google mangling the thing

https://math.stackexchange.com/a/1944276

losing the division

ah i see

"this" isn't k

even though it was the object of the sentence

[indigo ridge]

I had anothe question how would i best solve the equation $x^2 + x = 0$ in mod 6 (additive)

[cloud walrus]

jayzsparrow

[coral shale]

factor

[formal ermine]

trying out things

[indigo ridge]

i am stupid

[formal ermine]

like there are only 5 possible values you can plug in lol

[coral shale]

illum is right

[formal ermine]

won't help here

[indigo ridge]

stop rubbing in my face

[formal ermine]

Z/6Z isn't an integral domain

[coral shale]

but i would factor it regardless

its faster to check for me factored

x^2+x = x(x+1), now its faster to check which x's do/don't work

[formal ermine]

[summer path]

It is easier to check factored but also there's only like one thing you can do here

[chilly ocean]

he's very mean

[ivory trail]

why is half the activity of this chat telling people that they are being cringe with emotes

it's so weird

[summer path]

Probably because it's done somewhat jokingly by people that are used to seeing each other

But also illumi doing illumi things

[ivory trail]

i think it bleeds

[coral shale]

because conveys everything we want to in a single emote without expending energy and effort into words.

[chilly ocean]

mid emote tho

[summer path]

I remember watching blue people avatar and only being able to think of

[chilly ocean]

idk whats worse

avatar or that emote

💀

[summer path]

💀

scrapes the same villain "it'll be even funnier the second time!"

[summer path]

Review definition of order of an element

[past temple]

if M is a free module with k basis elements, and N is a free submodule of M, does N have at most k basis elements?

[formal ermine]

I would say so, yes

dim M = dim N + dim M/N

wait are free submodules injective?

[chilly radish]

Not necessarily

[formal ermine]

oh oop

[delicate orchid]

this is not the case when the modules are not over an integral domain right

[formal ermine]

then ignore what I said

[south patrol]

Are you tacitly assuming all bases of M have the same size

[chilly radish]

Z is Z-free but not injective

[glossy crag]

#groups-rings-fields is probably the least dedicated to its named purpose of the advanced channels, just observing from what little I've seen of the others. It's basically #discussion or #chill with some basic algebra mixed in.

[elder wave]

the submodule being assumed free saves this doesn't it

can't you like

[delicate orchid]

not sure

[elder wave]

use that R^m doesn't inject into R^n

[chilly radish]

I think if you assume the ring has IBN then what they said it's true

[indigo ridge]

I have to work on a project for my abstarct algebera class basically just discussing a topic in algerba or a theorem, I really want to wow my Professor so I can geta perfect mark any suggestions?

[chilly radish]

This isn't true if your ring is not nice enough

[delicate orchid]

the bar resolution

[elder wave]

Non commutative rings don't exist shin

[glossy crag]

What topics are you familiar with?

[elder wave]

stop being silly

[chilly radish]

Timo go touch grass

[indigo ridge]

you guys are the only people here unironically who are not discussing AA rn

[chilly radish]

I'll go tell my linear algebra class matrices aren't real

[past temple]

wait, the bases of a free module can have different sizes?

[indigo ridge]

Basic group and ring theory

[chilly radish]

For some rings, yes

[glossy crag]

I realise this, it was just a comment on the general atmosphere in this channel.

[indigo ridge]

but im willing to learn alot of i have to

[past temple]

what about for commutative unital rings

[formal ermine]

how "advanced" is your class?

[elder wave]

so all rings

[chilly radish]

Commutative rings have invariant basis number

[chilly ocean]

If both are finitely generated, then yes

If not then no

[formal ermine]

ah

[ivory trail]

what do you mean "you guys"

[chilly radish]

So all bases have the same size

[glossy crag]

How expansive should your project be? If it's a presentation in a class, you could do Witt's proof of the little Wedderburn theorem. This is a very good blend of basic algebra, involving the class equation for finite groups, divisibility, and cyclotomic polynomials. For a source see Herstein's "Topics in Algebra", "Proofs from the BOOK", and this video by Borcherds https://youtu.be/O_XtW3iaogc.

[elder wave]

"k basis elements"

[formal ermine]

you could look into localization/tensor products, if you really want to go far

[indigo ridge]

it's a second course in Abstarct Algebra but it doesn't go very far in my opinion

just basic group and ring theory

[past temple]

sure yeah im taking k to be a finite number

[chilly radish]

What did you do in the first course

[indigo ridge]

Ok interesting

[indigo ridge]

the same thing!

its crazy

[chilly ocean]

This basically reduces to showing that there is no injection from A^m to A^n if m > n

[indigo ridge]

applications would also be a +

[past temple]

right

[elder wave]

Good suggestion

wonder who else thought of that

[chilly ocean]

bruh

;-;

[south patrol]

Doss IBN hold for comm ring with infinite bases

[glossy crag]

If you want applications, you could do something about finite fields and algebraic codes. For more ideas see Lidl's "Applied Abstract Algebra".

[south patrol]

Oh yeah it does

[chilly radish]

If you replace number with cardinal the. It should

[past temple]

what is IBN?

[glossy crag]

Again, this depends heavily on how expansive the project should be.

[south patrol]

I guess assuming an appropriate form of choice

Just the nice proof I know that comm rings have IBN uses exterior powers

[chilly radish]

I fail to see why this necessarily holds.

[south patrol]

And that wouldn't work for infinite cardinals

[chilly ocean]

The proof is very annoying

[indigo ridge]

Sure of course, the professor hasnt talked much of it, but its going to be a presentation so I think somehwere around 30 mins should be good

[chilly ocean]

I think you can do it with Cayley Hamilton tho

[indigo ridge]

a 30 minutes presentation*

[chilly radish]

The proof I know is just quotient by a maximal ideal and then use the fact it holds for vector spaces

[south patrol]

But yeah the proof using exterior algebra is hot

[chilly radish]

Which works for cardinals too

[indigo ridge]

Will do

[chilly radish]

Oh I remember seeing this in commie alg

You're right

[south patrol]

Yeah that's the proof I'd use for infinite ones

[indigo ridge]

is it on archive?

[elder wave]

it's one of those

[glossy crag]

So it's just a presentation then, nothing more? Then Wedderburn's theorem is of perfect length, I'd say. It's a beautiful proof that draws on multiple ideas from a first course in algebra.

[elder wave]

seems obvious but very annoying to prove

things

[glossy crag]

libgen.rs is your friend.

[indigo ridge]

perfect ill have a look at it, thanks

[south patrol]

The nice proof I know using exterior powers;
Kth exterior power of A^k surjects to to A, whilst the k+1th is zero

So you can find the basis number using exterior algebra

[glossy crag]

See the sources above.

[elder wave]

is there no easier proof of the original statement though

[south patrol]

Very nice

[indigo ridge]

already am!

[south patrol]

And it works for vector spaces

So it provides a nice proof that dimension of vector spaces is well defined

At least for finite dim ones lol

[chilly radish]

I think, once you know rank is well defined, if N has a basis of m>k elements then they have to be R-linearly dependent, but proving rank is well-defined even over commutative rings requires a bit of work

[glossy crag]

Do we have any Lie algebra bigwigs in the channel?

[chilly radish]

Like in particular by this I mean showing any set of k generators is a basis

For a free module of rank k

Which requires the Cayley Hamilton thing

[elder wave]

so it's just the same thing

pretty much

i guess there's no way around it

[chilly radish]

I think so, yea

[elder wave]

Modules are a silly thing

[formal ermine]

modules are just rings with extra steps

[south patrol]

Yrah

[south patrol]

No

Jk hm

A ring is just a monoid

[chilly radish]

There's no sense in which this is a true statement

[elder wave]

you have addition and multiplication

seems true to me

[formal ermine]

you're just not based enough

[chilly ocean]

Just don't care about modules and focus on rings 🤝

[chilly radish]

You could maybe say this about (R,R) bimodules

[coral steeple]

Is $|\operatorname{Aut}(\mathbb{Z}/n\mathbb{Z})|=\phi(n)$ true?

[cloud walrus]

person2709505

[formal ermine]

yes, $\on{Aut} C_n = C_n^\times$

[coral steeple]

Nice

[white oxide]

Is it true that $\mathbb{Q}(\sqrt{2}, \sqrt[3]{2}) = \mathbb{Q}(\sqrt{2} + \sqrt[3]{2})$? I followed the construction as in the proof of the primitive element theorem, and chose $1 \in \mathbb{Q}$ to be the given $a$ as stated in the proof. (The zeros of irr$(\sqrt[3]{2}, \mathbb{Q})$ taken as $\beta_j$ and the zeros of irr$(\sqrt{2}, \mathbb{Q})$ taken as $\gamma_i$. If not, what did I do wrong?

[cloud walrus]

okeyokay (analysis is cool)

[white oxide]

(I'm trying to verify it now and it's not looking so good lol)

[coral steeple]

Before I embark on a pointless exercise, how much machinery does it take to prove this?

[frigid lark]

It's a pretty chill proof

[open sluice]

birb drink is bloo

[frigid lark]

Yeah

[coral steeple]

Thanks, brb (hopefully)

[frigid lark]

Okey, I believe the proof for the primitive element theorem just states there exists some distinct a, b in Q, such that Q(asqrt(2) + 2^(1/3)) = Q(bsqrt(2) + 2^(1/3))

Not explicitly what they are

[glossy crag]

This is kind of stupid, but why in K[G] is the set of elements on which G acts trivially a submodule?

If gx=x for all g, then why should gyx=yx for all g and y in K[G]?

[delicate orchid]

yx is another element of K[G]

so g acts trivially on it

[glossy crag]

Huh?

[delicate orchid]

oh ok nvm I misread

[frigid lark]

And in that case Q(asqrt(2), 2^1/3) = Q(sqrt(2), 2^1/3)

[delicate orchid]

you're asking about the kernel of a rep not the trivial rep

[glossy crag]

Forget reps, just treat K[G] as a ring

[delicate orchid]

easier to think about it with reps

[glossy crag]

In it identify the subset of elements such that gx=x, why is that a left ideal?

[glossy crag]

Reps and modules over K[G] are absolutely equivalent concepts, so it should be equally easy to formulate a proof in either setting

[delicate orchid]

yes thank you I'm well aware

and I don't believe it is a left ideal - is your group acting on the right?

[glossy crag]

I'm not saying you aren't, I'm simply looking for a ring-theoretic explanation

[glossy crag]

Actually I don't even know, I simply assumed K[G] is a left module over itself

[glossy crag]

And my first thought was this set would obv be a right ideal

Let me check

[delicate orchid]

which I agree with

yxg = yx
(x+x')g = xg+x'g = x+x'

we can also potentially view this as a stabiliser subgroup

[white oxide]

oh i thought it was there exists a d in F such that c + a + db and F(c) = F(a, b) or is that equivalent to what you just stated

i don't think it is right

[glossy crag]

It's a left module

[delicate orchid]

group must act on the right then

which is standard

[frigid lark]

Oh shit, sorry Q(asqrt(2) + 2^1/3) = Q(bsqrt(2) + 2^1/3)

[agile burrow]

Does it suffice to check that it's preserved by left-multiplication by basis elements g

[delicate orchid]

it would, by linearity

[delicate orchid]

yeah that's the left regular representation, which corresponds to taking K[G] as a module over itself

which is obviously a left-right module

[agile burrow]

And these are elements invariant under left-multiplication by elements of G

[glossy crag]

As a left module over itself, yeah. So how is {gx=x} then a left ideal?

[delicate orchid]

right yeah, it's just this lol

[glossy crag]

Ah, wait a minute, it's that easy lol

[chilly radish]

Ocean man

[delicate orchid]

the map r |-> gr-r is an algebra homomorphism is it not?

so just take the kernel

[chilly radish]

I think your confusion is in the fact that the action of g on K[G] is K-linear

So you can pull elements of K out

It then suffices to check then gx is in M whenever x is in M, but gx=x so this is obvious

[glossy crag]

Yep, I was scratching my head over literally nothing

[delicate orchid]

I have absolutely no idea what just happened but I'm glad you get it now

[glossy crag]

How is that an algebra hom tho, it's not multiplicative

[glossy crag]

My thanks to you and the others

[chilly radish]

It happens to all of us

[delicate orchid]

f(zr) = (1-g)zr = zf(r)?

am I on the vvizard vveed or something 2nite

ah wait yeah it's just a module hom

[glossy crag]

Huh? Aren't you talking about the map $\varphi(x)=gx-x$? Then $\varphi(xy)=gxy-xy$ and $\varphi(x)\varphi(y)=(gx-x)(gy-y)$/

[delicate orchid]

which is all we actually needed, kernel is a module

[cloud walrus]

leave_no_norm

[delicate orchid]

and this kernel is exactly the fixed points

got there in the end chat

[chilly ocean]

Looks correct

[frigid lark]

The d can come from (the edited version) of what I said, specifically d = a. Since bsqrt(2) + 2^(1/3) is in Q(asqrt(2) + 2^(1/3)) = K, we have that (b-a)sqrt(2) in K (b neq a), hence sqrt(2) and 2^(1/3) in K. The proof your text gives is a lot more constructive than the one I have seen.

[hollow mica]

For an element x of a commutative ring X, and a subring Y, we say that x is integral over Y if x is the root of some monic polynomial with coefficients in Y. Moreover, we say that the integral closure of Y is the set of all elements in X that are integral over Y (this of course contains Y and is itself a subring). With this in mind, the definition of integrally closed domain above doesn’t make sense to me

Ohhh shit wait

is it just saying like no fraction (i.e. not in A) isn’t integral over A?

Yeah ok nvm I just can’t read

[frigid lark]

Is this argument correct for well definition of $\sum_{i = 0}^\infty(Vy)^i$. Since the n'th ghost component of $(Vy), (Vy)^{(n)} = 0^{p^n} + py_0^{p^{n-1}} + ... + p^ny_{n-1} = 0$ as char k = p, and addition and multiplication is done componentise with respect to ghost components, the ghost components of $\sum_{i = 0}^n(Vy)^i$ is 0 for all integers n, hence $\sum_{i = 0}^\infty(Vy)^i$ is just the element in $W(k)$ defined by having ghost components all be 0.

[cloud walrus]

parrottea

[frigid lark]

Wait, that sounds wrong, cause then $\sum_{i=0}^\infty (Vy)^i = 0$

[cloud walrus]

parrottea

[chilly ocean]

Hey guys, would like some confirmation on this. Suppose $k$ is a finite field with characteristic $ \neq 2$. Then every quadratic extension of $k$ is equal to the quadratic extension $K := k(\sqrt a)$ where $a$ is a primitive root of $k$. This is simply because the squares of $k$ are precisely even powers of $a$ and the nonsquares odd powers; then every quadratic extension is obtained by taking a square root of an odd power of $a$, which is contained in $K$, so by degree reasons they are all equal.

[cloud walrus]

janohel

[coral spindle]

Seems right to me

[teal vessel]

question: given that the dihedral group is generated by a rotation r and reflection s, is the subset of reflections in a dihedral group the set of all elements that include s as a factor (for lack of a better term)?

[rocky cloak]

Yes, every element of the dihedral group is off the form r^n or sr^n

[coral spindle]

The subset of reflections is the only nontrivial coset of the subgroup of rotations.

[teal vessel]

it's what I thought, but just making sure.

[teal vessel]

give an explicit example of a group G and an infinite subset H of G that is closed under the group operation, but is not a subgroup of G
N, Z+, both subsets of the group (Z,+)

sometimes it feels like they throw a gimme question in the middle of harder questions just to give you a break.

[south patrol]

Nice

In fact this is basically the only example lol

In that if G is infinite, H an subset closed under operation and x is in H, then x^-1 is in H unless x has infinite order

[teal vessel]

well, any group isomorphic to Z would technically have such subsets, but yeah, up to isomorphism this is the one (or the only family of them)

[south patrol]

Corresponding to x having infinite order so that <x> iso to Z whilst {1, x, x^2,...} is iso to N

I mean that every example contains what you just gave

For a suitable element

[teal vessel]

I could technically have it be any set {x in Z | x > n; m|x} for some arbitrary n and m in N

[south patrol]

Good point yeah

Though I like the N example

Because then it is a submonoid but not a subgroup [I.e. it contains 0 and is closed under addition]

Which is more interesting imo

[teal vessel]

I like the N example because I spent a month going through axiomatic set theory before this stuff and I love the VN construction of the numbers

and to annoy the mathematicians who say N starts with 1

[south patrol]

Lol

Oop

n+1 = n u {n} lol

[teal vessel]

if I interpret that last statement correctly as n+=n U {n} then yeah

[south patrol]

Ye

[teal vessel]

gives a beautiful definition of natural numbers: a natural number is a set that belongs to every inductive set.

[chilly ocean]

beautiful

[frigid lark]

nvm, I found an article that's helping me through these problems