groups-rings-fields

[thorn delta]

R is an integral domain here

oh wait, nvm

we technically never showed that $R \setminus \mathfrak m$ is multiplicative so its probably looking for some explanation of that

[cloud walrus]

kxrider

[thorn delta]

... i think thats what its looking for anyway. meh, ill go with that

[tacit breach]

this is a weird one

they asked: all normal subgroups N of a normal subgroup H of G, is itself normal in G

[viscid pewter]

like, whether it's true?

[tacit breach]

and you can choose either true, false or true if N is in the center z(H)

just by the looks of it it seems like its probably no because if the first option is false so should be the third

[viscid pewter]

really? i could see 3 while not 1

[tacit breach]

but if it doesnt always happen

then its wrong

[viscid pewter]

?

[tacit breach]

you can give a counter example

is the thing

[viscid pewter]

for 3?

[tacit breach]

which makes it false so 2

but then 3 is like

weird

[viscid pewter]

if 1 is always true choose 1

else:

if 3 is always true choose 3

else: choose 2

probably

[tacit breach]

so its 3

i guess

[viscid pewter]

is it

i don't actually know

[tacit breach]

i dont either

i got asked to ask it

i dont actually take the course

[viscid pewter]

whu

why

[tacit breach]

because you see

they are now a teacher for this course, we already passed it

now they try to teach and for that they need to solve questions

and some were hard

like that one

i dont know anything about group theory even though i have a very good grade

[pallid ember]

its not true

try finding a non abelian group that gives a counter

[tacit breach]

yeah but

[viscid pewter]

uhhhh

[tacit breach]

3

why isnt 3 right

oh damn well

there are a few questions they asked me

i thought it wasnt too much but some images contain more than 1

[fading crag]

I'm having trouble writing a proof, I'm pretty sure it's really simple but I'm not sure where to start

Show that in a ring R if a + b = d and a + c = d then b = c

[chilly ocean]

to start, notice that a + b = a + c

(R, +) is a group

what's next?

[fading crag]

I'm trying to think of some property of groups that will help me here.

[chilly ocean]

well, write them all out

there aren't very many

[fading crag]

I don't think it's associativity

I suspect that it's something to do with the identity element

[chilly ocean]

can you write out what a group is

[scarlet estuary]

have you er, not covered the cancellative law of groups?

[fading crag]

I'll try, and no I haven't.

[chilly ocean]

are you doing rings before groups

[viscid pewter]

a lot of ppl do that

[scarlet estuary]

okay thats not totally unusual

[fading crag]

No, but the textbook I'm using covers both pretty sparsely

[viscid pewter]

can't fathom why tbh but

ok so

[viscid pewter]

this

define 'group'

[scarlet estuary]

"pretty sparsely" is putting it lightly

but regardless

[chilly ocean]

[next obsidian]

you have a + b = a + c

[fading crag]

A group is a finite set of elements with a binary operation that has the properties of closure, associativity, an identity element, and an inverse.

[next obsidian]

what have you been taught to do since like 1st grade

[viscid pewter]

ok so

[next obsidian]

I'm serious! I'm not trying to be condescending my point is you can just do what you think you should

[fading crag]

I mean you can't just remove a from both sides and say b = c can you?

[next obsidian]

why not?

[viscid pewter]

consider that the inverse of a is -a

[next obsidian]

they're equal

if x = y

then x - a = y - a

[fading crag]

Ok, I may have been overthinking this one. Thanks guys.

The textbook I'm using is Childs, concrete introduction to higher algebra btw

[chilly ocean]

unfortunate author name

"child's concrete introduction to higher algebra"

[latent anvil]

higher algebra

[scarlet estuary]

well

[viscid pewter]

higher algebra introduces child to concrete

[scarlet estuary]

nitpick

the given proof works for rings since their additive structure is abelian

[next obsidian]

add -a to the left

and it works

[scarlet estuary]

but in general groups just saying "add -a to both sides" is a bit too vague

yeah

you need to add it on the left

and that suffices

[tacit breach]

Turns out algebra is the hardest thing

[next obsidian]

E_infinity

[viscid pewter]

this person has never heard of inter-universal teichmuller theory

[tacit breach]

Lol what a loser

[next obsidian]

bad meem

[tacit breach]

This is how my neighbor proved abc conjecture

And riemann

[viscid pewter]

riemann more like why man amirite

[tacit breach]

Yeah why man did you not solve it

I did manage to help her

I asked people on the WhatsApps

But if this is group theory galois is the hardest course ever

I did the course last year and I don't remember this at all

[viscid pewter]

hooooboy

you're gonna have a fun time

[tacit breach]

No don't worry

I'm in useful maths

Not useless maths

[viscid pewter]

get out.

leave this server and never return.

[tacit breach]

Wait wait why do you guys talk about iutt

Did he actually prove the thing?

No like

[next obsidian]

Slimvesus understands it all tho

[tacit breach]

Did the abc take the L

[next obsidian]

cuz it doesn't use groups

[latent anvil]

do you seriously believe this? You're showing your profound ignorance of the elementary theory of heights, at the advanced undergraduate/beginning graduate level

[viscid pewter]

uhhhhhhhh

[next obsidian]

I don't know what that's a reference to

haha

[latent anvil]

it is the best meme

coming from iutt

oh sorry

[next obsidian]

lmfao

[latent anvil]

this is a direct quote from big M

[next obsidian]

is this some reddit reply to some1 saying

Oh

[viscid pewter]

ah yes i remember now

[latent anvil]

I can only say that it is a very challenging task to document the depth of my astonishment when I first read this Remark! This Remark may be described as a breath-takingly (melo?)dramatic self-declaration, on the part of SS, of their profound ignorance of the elementary theory of heights, at the advanced undergraduate/beginning graduate level.

[next obsidian]

it's even better

[latent anvil]

SS = Scholze and Stix

[next obsidian]

Wow

[tacit breach]

[latent anvil]

https://www.math.columbia.edu/~woit/wordpress/?p=10560

[next obsidian]

I thought this was going to be like

Some reddit person replying to someone on r/math saying

[latent anvil]

yeah it reads like an attack helicopter thing

[next obsidian]

"no IUTT is wrong lol"

[latent anvil]

yes lmao

[next obsidian]

yes

perfectoid space

Bruh

[latent anvil]

i watched a lecture of his for meme potential

[next obsidian]

I looked at UChicago REU 2020

[latent anvil]

but he just talked about sheaves

and i was sad

[next obsidian]

they had a special class

talking about perfectoid spaces

and I was like

[latent anvil]

bruh

[next obsidian]

O_o

perfectoid yummm

[latent anvil]

o ya i should work on my app for that

[next obsidian]

What is there to really work on for that one? The research statement?

[latent anvil]

yup

[next obsidian]

I was confused with a lot of them not asking any more questions

[latent anvil]

I mean that's really all there is to work on ever

[next obsidian]

so I was thinking like maybe I missed something

[latent anvil]

like you have your CV and transcript and an SOP

[next obsidian]

eh the ones I applied to last year like

[wind steeple]

https://www.youtube.com/watch?v=UfjaResU3tI

[next obsidian]

wanted you to answer some questions

[latent anvil]

[next obsidian]

why is he lecturing in a tank top

True

Max J moment

[wind steeple]

look at this chad

[next obsidian]

Yeh I know

are you at Uchi?

[latent anvil]

I thought they got funding this year?

[next obsidian]

They did but like

they try to accept like

3000000 people

so it's not enough haha

I looked at the papers written and they have such a huge range of like

sophistication of the material covered

[latent anvil]

oof

my grad student was useless this summer lmao

[next obsidian]

O O F

[latent anvil]

my project was incredibly category brained

and we had to define a natural transformation for her

[next obsidian]

l o l

wtf

[latent anvil]

she's a great low dim topologist!

it's just like

[next obsidian]

why is she on that project

[latent anvil]

very different skill set/knowledge base

yeah exactly

[next obsidian]

rip

Dman

I'm imagining getting assigned to TA for like a PDE project

and just going

¯_(ツ)_/¯

[latent anvil]

l o l

[next obsidian]

sorry guys

[fading crag]

Do they really put people into subjects they don't know like that?

[latent anvil]

Idk how exactly it happened, I think we weren't the only group she was working with

Also for this reu the grad students weren't really part of the project or doing research

They had like office hours style things

[chilly ocean]

having a big brain fart moment

How can i see that y^2-x^3-ax-b is irreducible in k[x,y]

it certainly looks like it should be irreducible since the y term is on its own but how do I know nothing funny will happen in a field of characterstic p

[next obsidian]

In general I've tried to find functions into an integral domain for which the kernel is precisely the ideal generated by that

I don't have a good guess off the bat what such a map would look like

[chilly ocean]

are you replying to me

[next obsidian]

Yes

[chilly ocean]

so i should try find a phi:k[x,y] to R some integral domain and show that the kernel is (y^2-x^3-ax-b)

[next obsidian]

To show a polynomial like that is irreducible

Well, that's what I've done

[chilly ocean]

k[x,y]/(y^2-x^3-ax-b) then has no zero divisors

[next obsidian]

but maybe there's better ways to show polynomials of > 1 variable are irreducible

Right, but like

to just show that is exactly the same as showing the polynomial is prime

<==> irreducible

[gritty adder]

I think you can view it as a polynomial in y with coefficients in k[x]

then y^2 - f is irreducible iff f is not a square or something

[next obsidian]

and then use Eisenstein or something???

oh that's a good idea too

do what Kedar said

[chilly ocean]

I thought eisenstein was only integral valued coefficients

[next obsidian]

no

it generalizes

[chilly ocean]

this is a good idea

[next obsidian]

it's just the exact same thing

hahaha

except prime is what it means in that ring

I think you need the base ring to be a UFD??

But you certainly will have that for k[x][y]

[chilly ocean]

Do you have any ideas for this? I want to find what conditions on a and b make Z(y^2-x^3-ax-b) smooth

I know this theorem

[gritty adder]

Yeah you need at least one of the partial derivatives to not vanish - think of the implicit function theorem like for differentiable maps.

In this case the Jacobian is (f_x, f_y)

[cloud walrus]

squirtlespoof

[subtle current]

You mean $\mathbb{Z}[\sqrt{-5}]$?

[cloud walrus]

evilbuggy

[golden pasture]

all these rings are orders in number fields hence 1D noetherian

so really you are asking for primes

in the case of $\mbb Z[\sqrt{-5}]$ it isnt rlly super simple cuz it isnt a PID

[cloud walrus]

a cute cat ٩(˃̶͈̀௰˂̶͈́)و

[golden pasture]

but your best hope is to basically "factor" primes in Z inside that ring

[subtle current]

$\mathbb{Z}[\sqrt{-5}] \cong \frac{\mathbb{Z}[x]}{x^2 + 5}$

[cloud walrus]

evilbuggy

[subtle current]

By correspondence theorem, the maximal ideals of $\mathbb{Z}[\sqrt{-5}]$ correspond to maximal ideals of $\mathbb{Z}[x]$ containing $x^2 + 5$

[cloud walrus]

evilbuggy

[subtle current]

The maximal ideals in $\mathbb{Z}[x]$ are of the form $(p, f(x))$ where $p$ is a prime and $f(x)$ is irreducible in $\mathbb{F}_p[x]$

[cloud walrus]

evilbuggy

[golden pasture]

iirc Z[\sqrt5] is probably a euclidean domain under the usual norm function i believe so can just characterize all primes there pretty easily

[golden pasture]

tho ig this is as good as it gets for Z[sqrt(-5)] without going too fancy into factoring ideals

[subtle current]

With some little work, we can prove that f(x) should be equal to x^2 + 5
So, the maximal ideals of Z[x] containing (x^2 + 5) are of the form (p, x^2 + 5) where p is prime such that x^2 + 5 is irreducible in F_p
The problem boils down to finding primes p such that x^2 + 5 has no roots in F_p

[subtle current]

For p = 2, clearly x^2 + 5 is reducible in F_2

Suppose p is an odd prime
x^2 + 5 has a root in F_p if and only if there is some element x such that x^2 = -5 mod p
So there are two cases

1. -1 and 5 are both squares of some numbers in F_p

-1 is a square in F_p if and only if p = 1 mod 4
From quadratic reciprocity law, 5 is square of some number modulo p if and only if p is square of some number modulo 5
So, p = 0, 1 or 4 mod 5
Therefore, if either p = 1, 5 or 9 mod 20, then x^2 + 5 is reducible in F_p

2. Both -1 and 5 are not squares of any numbers in F_p

-1 is not a square in F_p if and only if p = 3 mod 4
From quadratic reciprocity law, 5 is square of some number modulo p if and only if p is not a square of some number modulo 5
So, p = 2 or 3 mod 5
Therefore, if either p = 3 or 7 mod 20, then x^2 + 5 is reducible in F_p

So, for an odd prime p, if p mod 20 is in {11, 13, 17, 19}, then (p, x^2 + 5) is a maximal ideal in Z[x]

[golden pasture]

uhhhhhhh

wha

t

[subtle current]

Wait I might be wrong

Sorry

[subtle current]

Yeah that part is correct

[golden pasture]

jus a example of how prime ideals look like, here are the first few:

$$\left(\sqrt{-5}\right)$$
$$\left(2,\sqrt{-5}+1\right)$$
$$\left(3,\sqrt{-5}+1\right),\left(3,\sqrt{-5}+2\right),\left(7,\sqrt{-5}+3\right),\left(7,\sqrt{-5}+4\right)$$
$$\left(11\right),\left(13\right),\left(17\right),\left(19\right)$$

[cloud walrus]

a cute cat ٩(˃̶͈̀௰˂̶͈́)و

[golden pasture]

prime = maximal here except for (0)

(in fancier terms, this is a 1 dimensional noetherian ring)

[subtle current]

Okay

[golden pasture]

if you notice hard enuf, we are actually factoring x^2+5 over Z/pZ for most of the primes

so it reduces down to determining over what integer primes is 5 is a quadratic residue

tho the proper reason why this works is uh

[subtle current]

f(x) is either x^2 + 5 or a degree 1 polynomial such that f(x)(ax + b) = x^2 + 5 modulo p

I didn't see that second case is also possible

[golden pasture]

ok im dumb lol

so

$\frac{\mb F_p[x]}{x^2+5}=\mb F_{p^2}$ iff $x^2+5$ is irreducible mod $p$

[cloud walrus]

a cute cat ٩(˃̶͈̀௰˂̶͈́)و
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[golden pasture]

which means -5 is a quadratic non-residue

which is pretty simple to determine

[subtle current]

Yeah

[golden pasture]

otherwise we just factor x^2+5

[subtle current]

Okay

[golden pasture]

me cant into concrete examples lolz

and this also shows that these primes are maximal

and these also shows we found every maximal ideal cuz any maximal ideal is prime

and clearly any prime is maximal unless (0)

[subtle current]

I don't know much about noetherian rings except for the definition 😬

[golden pasture]

hahaisok 1D noetherian rings basically mean the longest ascending chain of prime ideals has length 2

and (0) is prime

so we have $(0)\subset\mathfrak p$ is the longest chain

so $\mathfrak p$ is maximal

[cloud walrus]

a cute cat ٩(˃̶͈̀௰˂̶͈́)و

a cute cat ٩(˃̶͈̀௰˂̶͈́)و

[subtle current]

Ohh so it can't be generated by more than 2 elements, right?

[golden pasture]

also that

but this doesn't imply that

i think you can have 1D noetherian rings where primes are generated with more than 2 elements?

the reason why prime ideals in this case are generated by at most 2 elements is because when you localize at every maximal element, you get a discrete valuation ring (every ideal is generated by 1 element)

(i.e. the domain is a dedekind domain)

[subtle current]

😬

I have to read this stuff yet

[golden pasture]

yee it's super cool

[subtle current]

I started reading Atiyah Macdonald's CALG

But it was very dense and I didn't get the motivation behind defining things like Nilradical, Jacobson Radical, extensions and contractions

It was as if he was just defining bunch of things

So I am planning to read Eisenbud instead

[golden pasture]

lol i just worked through all the exercises in it the definitions became like "because it has cool properties" to me

[golden pasture]

neukirch algebraic number theory introduces these in context of number theory if you're interested

[subtle current]

Okay

I will check it out

[golden pasture]

AM chap 9 and 10 is useful for neukirch if you want to see the more general picture

but at that point you already completed AM lolz

[chilly ocean]

For what number fields $K$ does every automorphism of $K$ restrict to an automorphism of $\mathbb{Q}$?

[cloud walrus]

lugita15

[golden pasture]

automorphism fixes 1

hence it fixes Z

hence it fixes Q

[subtle current]

Any automorphism of a field fixes its prime field which is Q in this case

[chilly ocean]

OK thanks guys

Now I have a more general question: if E is a finite extension of F, then does every automorphism of E restrict to an automorphism of F? If that’s not always true, under what conditions is it true?

[golden pasture]

nope

when E is F_p or Q

then it is true

[chilly ocean]

I assume you mean when F is F_p or Q. But I want to keep F an arbitrary characteristic 0 field. Is there any condition I can impose on E to make every automorphism of E restrict to an automorphism of F?

[golden pasture]

yup

for first part

for second part no

you just consider the tower

F/E/Q

oh hm

wait lemme rethink ah

i think what we really want is

every element in Gal(F^gal/Q) that permutes F also permutes E in a nontrivial way

the most simple case is when F has no non-trivial automorphism

[golden pasture]

which has a big family cuz we can consider some polynomial p(x) whose galois group is S_n then adjoin a root of p(x) to Q

not too sure if there are more nontrivial examples tho

we can run the same contruction on E/F

we can run the same contruction on E/F

suppose we have some polynomial p(x) in F[x] with galois group S_{deg p}, then adjoining a root of p(x) to F gives an extension with no nontrivial automorphsim

suppose we have some polynomial p(x) in F[x] with galois group S_{deg p}, then adjoining a root of p(x) to F gives an extension with no nontrivial automorphsim

suppose we have some polynomial p(x) in F[x] with galois group S_{deg p}, then adjoining a root of p(x) to F gives an extension with no nontrivial automorphsim

suppose we have some polynomial p(x) in F[x] with galois group S_{deg p}, then adjoining a root of p(x) to F gives an extension with no nontrivial automorphsim

suppose we have some polynomial p(x) in F[x] with galois group S_{deg p}, then adjoining a root of p(x) to F gives an extension with no nontrivial automorphsim

#chill message

discord bugging out for me

rip

[chilly ocean]

That’s exactly the case I’m interested in. Suppose that F has no nontrivial automorphisms and has infinite transcendence degree over Q, and let E be a finite extension of F. Then under what conditions do all the automorphisms of E fix F?

[latent anvil]

Hmm

If you take F = R and E = C then it's false

trying to think of other interesting examples

[cloud walrus]

squirtlespoof

[golden pasture]

it's easier to see $$\frac{\mbb Z[x]}{\left(x^2+5,2,1+x\right)}\cong\frac{\frac{\mbb Z}{2\mbb Z}[x]}{\left(x^2+5,1+x\right)}\cong\frac{\frac{\mbb Z}{2\mbb Z}[x]}{\left(x^2+1,1+x\right)}$$

[cloud walrus]

a cute cat ٩(˃̶͈̀௰˂̶͈́)و

[golden pasture]

look terra things has to be done to observe the isomorphism

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

[hybrid cove]

Hi there, would anyone know something about Calogero-Moser singular spaces and about the use/interpretation it could have in theoretical physic?

[golden pasture]

probably better to ask in physics server lol

[hybrid cove]

yep, I posted it there too :)

[gray charm]

Show that in Z_8[x], there are infinitely many square roots of 1.

What I have so far is I've found the square roots in degree 0 and 1. I know that the leading coeff must be 4, and the constant can be 3, 5, 7. My idea is try to show that for any n, there's a sqrt of 1 that is degree n, but I'm not sure how to continue, or if that's the right track. Help?

[dapper nebula]

there is a characterization of units in a polynomial ring

in general you should show that units of a polynomial ring is a unit iff the constant coefficient is a unit and the other coefficients are all nilpotent

and then from whatever construction you have of this I guess you can get square roots

[gray charm]

Oh sick, I got it; 4^2=0, 3^2=1, so for any n, consider (4x^n+4x^n-1+...+4x+3)^2

Thanks for the hint

[subtle current]

(1 + 4f(x))^2 = 1 for any f(x) in Z_8[x]

[gray charm]

Cool, thanks

[cloud walrus]

lugita15

[chilly ocean]

Never mind, I figured it out.

[rain kayak]

how to solve a tetrational equations such as this: $^n 2 = 3$?

[cloud walrus]

Deleted User 738983d

[rain kayak]

though idk if that belongs to algebra (this channel)

[rain kayak]

or in other words, how to calculate the super-logarithm, such as this: $slog _{2}{3} = n$

[cloud walrus]

Deleted User 738983d

[rain kayak]

assume slog is super-log

[wispy galleon]

How would I solve this

if x is equal to 6

[glossy yoke]

look for an element of order 3.

[wispy galleon]

@glossy yoke would 2 be one of order 3?

[glossy yoke]

yeah

[wispy galleon]

Then what would I do?

[glossy yoke]

examine the subgroup generated by 2.

[wispy galleon]

{2, 4, 6}

[glossy yoke]

yes.

[wispy galleon]

Then i'd just divide by two to get the ismoprhism of 3

z3

?

[glossy yoke]

sure.

[wispy galleon]

alright

[golden pasture]

im very sure only like 5 people in the world reallly cares about this so your best luck is reading papers on what those people have done
also idk where it should belong literally no one ever does this afaik

[vestal snow]

I had a question about terminology

I'm reading a paper on function fields and it mentions orders of the field F

Here's an example

I'm pretty sure that it means that there is a w and a point(or place) P such that ord_P(w) = p-1

can someone confirm this?

[rain kayak]

im very sure only like 5 people in the world reallly cares about this so your best luck is reading papers on what those people have done
also idk where it should belong literally no one ever does this afaik
@golden pasture ive found one, it says to use abel function of an exponential function, but i barely understand and i cannot find abel's function

[golden pasture]

abel transform*?
(may want to move to some analysis channel)

[uncut girder]

What is a split torus?

[rain kayak]

not sure, it loooks like $slog _x (z) = A _{z} [x^z] (z)$

[cloud walrus]

Deleted User 738983d

[golden pasture]

looks like a cursed ring to me

[rain kayak]

more info if you want to know. yet in munafo class system there is no such thing

only way i can solve it is "abel function of exponential function" but i cant find it anywhere, only hope is if anyone knows what is it and how to calculate it

[golden pasture]

may want to ask in an analysis channel

and give a reference for where you found it

[rain kayak]

sure

ill do it in some mins, im in class now

[rain kayak]

are here any analysis channels?

[paper flint]

#advanced-analysis

[rain kayak]

ok thanks

[golden pasture]

lol

[unique juniper]

so

whats the recommended abstract algebra text

[golden pasture]

jacobson

(see pinned msg in #book-recommendations)

but my very biased perspective is jacobson

[chilly ocean]

jacobson is too old

[golden pasture]

wot

jacobson is like

probably one of the more recent books

lol

[obsidian sleet]

stupid question but how do i prove this

he left it as an exercise but it looks weird

idk what to do

[chilly ocean]

induction prolly

[obsidian sleet]

yea that's what i thought too but ig i'd have to split into cases

[chilly ocean]

[obsidian sleet]

one for nonnegatives

and one for negatives

[chilly ocean]

you can probably deduce the case for negatives from that for nonnegatives

[obsidian sleet]

yea yea

because i could just replace g with g^-1

and the same proof goes

[chilly ocean]

something like that, yeah

"the induction is trivial and left as an exercise to (audience)"

-your prof, and now you

that's how you prove these things

[obsidian sleet]

oh jeez am i supposed to use this...

i feel like that would be way too easy no

i feel weird using this

also this is the first time i've ever talked in these advanced channels other than the pedagogy one

[woven delta]

This is exactly what you're supposed to use

Oftentimes things are definitional so they are easy

[obsidian sleet]

i see

[cursive temple]

Im learning the basics of homology groups from fraleigh, and one of the excercises is to show that $\partial_2(\partial_3(P_1P_2P_3P_4)) = 0$. The first boundary isnt too difficult (i think), and i need to calculate $\partial_2(P_2P_3P_4-P_1P_3P_4+P_1P_2P_4-P_1P_2P_3)$. Here i immediately run into problems, as im getting $\partial_2(P_2P_3P_4) = -P_3P_4+P_2P_4-P_2P_3$, and $\partial_2(P_1P_3P_4) = P_3P_4+P_1P_4-P_1P_3$, but now the $P_3P_4$ terms wouldnt cancel as they should. I'm probably just misunderstanding how to calculate the boundary or how the indices should be treated with respect to orientation.

[cloud walrus]

Pappa

[chilly ocean]

I tried to prove these two equivalences, and for both problems and both directions I'm stock on how to show the kernel of the middle part is the subset of the image

So for (i) => I'm not sure how to show ker(bar u)=im(bar v), for (i) <= I'm not sure how to show ker(v)=im(u); for (ii) same thing

I can easily show that im(bar v) is a subset of ker(bar u) but the other direction is a bit difficult

[sturdy marsh]

M'' is the cokernel of M' ---> M

if you have something in ker(u bar)

then the composition M' ----> M ----> N is 0

so you get a map from the cokernel to N, i.e. a map from M'' to N

such that M ---> M'' ----> N is equal to the map M ---> N

which means it is in the image of v bar

[golden pasture]

in general you cant generate the whole field from repeated addition of one element unless the order of the field is prime

it is probably either referring to the α in F_q=F_p[α]

or a multiplicative generator

[south temple]

if I have a set $G$ with operation $*$ where every $a$ is guaranteed to have a right inverse $b$ and a right identity $e$, does this show that $b$ is also the left inverse of $a$?

[cloud walrus]

vov&sons

[south temple]

(the operation * is associative)

[carmine fossil]

Yes, that's correct

[leaden finch]

can someone explain to me what this means

a set S is countable if one can find a bijection between natual numbers and S

[thorn delta]

ill answer in #discrete-math

[acoustic lodge]

im reading robert ash basic algebra atm, anyone read it before or think its good?

its my second text on algebra...i just finished 1st chapter

[rustic crown]

How do I find the Galois group of $\mathbb{Q}(\sqrt{1+\sqrt{2}}, i)$ over $\mathbb{Q}$? I have checked that the extension is indeed Galois, and has degree 8. Are there nice ways of doing this or you just have to do it 'by hand'?

[cloud walrus]

det

[chilly ocean]

Let $K=\mathbb{Q}(a)$ be a number field which is not a Galois extension of $\mathbb{Q}$, and let $p$ be the minimal polynomial of $a$. Then is $Aut(K/\mathbb{Q})$ isomorphic to the group of permutations of the roots of $p$ that are in $K$, or are things more complicated?

[cloud walrus]

lugita15

[rustic crown]

things are definitely more complicated. since K is a simple extension, you precisely know that the group has order = n = number of distinct roots of p in K. In particular this is not isomorphic to Sn since n! > n.

here i'm taking n>2. But if you want this to be true, you need to have n = 1 or n = 2.

in general we probably only get that the automorphism group acts transitively and faithfully on the set of roots of p in K.

[uncut girder]

What is a rational character?

[chilly ocean]

@rustic crown Oh right, it’s not even true in the case when K is a Galois extension of Q. But is there anything different in how Aut(K/Q) acts on the roots of p in K depending on whether K is or is not a Galois extension of Q?

[uncut girder]

Are rational characters just characters have have image inside of Q, i.e. have values +1 or -1

[rustic crown]

I am currently reading Galois theory as well, so I can't say much. But since Q has char 0, any irreducible is automatically separable. So one easy thing we can say is |Aut(K/Q)| <= deg p with equality only when K is Galois.

[prisma ibex]

This is the case if \chi is assumed to be a unitary character. But in general you might have some unitary characters, in which case the statement is that these are rational iff they are a rational multiple of a rational unitary character, iff all the values of \chi are contained in Q

[uncut girder]

So the context is this

[prisma ibex]

Is k assumed to be characteristic 0 here, or completely arbitrary?

[uncut girder]

arbitrary

[prisma ibex]

okay so your representations are just assumed to be in characteristic 0

i.e. both your representation V and the character \chi_V

I think saying "defined over k" here is a little confusing when k is positive characteristic

[stark sigil]

Rational character doesn't seem to be defined up there the same as in my first google search

e.g. the identity on $\mathbb G_m$ is rational according to the usual definition but the definition given here, taken literally, implies the value field is $\bC$

[cloud walrus]

Icy001

[uncut girder]

What's the generic point of a linear algebraic group?

What's the generic point of GL(2)?

[uncut girder]

What's the generic point of affine space k^n?

[viscid pewter]

(a, b; c, d), where a, b, c, d are in R, right?

[uncut girder]

GL(2) is the group of invertible 2×2 matrices. It's a variety because its defined by the polynomial det=/=0

[viscid pewter]

ah, christ

[latent anvil]

@uncut girder, are you working with them as schemes or varieties?

I can tell you what the generic point of the scheme A^2 is

But no definition of variety I've seen has allowed for generic points

[uncut girder]

Varieties dont have generic points?

[latent anvil]

Not to my knowledge

[uncut girder]

Tell

[latent anvil]

So by definition $\mathbb{A}^1 = \mathrm{Spec}\ k[x]$

Actually let's do A^1 to start

[cloud walrus]

Shamrock -> E -> B

[latent anvil]

Assume k algebraically closed

What are the points of A^1, under this definition?

[uncut girder]

(x-a)

[latent anvil]

You're missing one

[uncut girder]

(0)

[latent anvil]

Right

So $\xi = (0)}$ is called the generic point because it has the property that $\overline{{\xi}} = \mathbb{A}^1$

[cloud walrus]

Shamrock -> E -> B

[latent anvil]

In A^2 we also have a generic point (0)

But also any irreducible curve has a corresponding generic point

A point on the curve whose closure is the curve itself

and generally since point = primes and primes = irreducible closed subsets and closure of point = set of all primes containing p we get generic points of all affine subvarieties

More generally any scheme has a unique generic point for each irreducible subset

[uncut girder]

Wdym prime = irreducible closed subset

[latent anvil]

in an affine variety

prime ideals of the coordinate ring are the same as irreducible closed subsets

unless I'm fucking something up lol

[uncut girder]

So the generic point of k^2 is not really a point of k^2

[latent anvil]

It is if k^2 is a scheme

but yeah lol idk what they mean if they're sticking to varieties

[uncut girder]

What's the difference between scheme and variety

[latent anvil]

Quite a bit

schemes are locally ringed spaces

Which locally look like Spec A

Varieties are just like open subsets of closed subsets of P^n for some n where P^n is projective space over a(n algebraically closed) field

People who are more schemepilled than me will define a variety as a certain kind of scheme

[uncut girder]

Bad

[uncut girder]

Open subsets of closed subsets?

[sturdy marsh]

in terms of schemes, a variety is an integral separated scheme of finite type over a field

some authors drop the irreducible condition

some might ask for the field to be algebraically closed

[latent anvil]

lol

[latent anvil]

Yeah so like, I would call A^2 \ {0} a variety

Wait no lol

So if you just take a closed subset of A^n

We can make A^n an open subset of P^n

lmfao nope this one is backwards too

okay take 3

Yes okay consider the curve y - x^2 in A^2

This is an open subset of the curve zy - x^2 in P^2

which is itself a closed subset of P^2

there we go

So this is an example of an open subset of a closed subset of P^n

Which we might want to call a variety

[uncut girder]

Okay w.e

Okay let's say you have a group and a vectorspace that this group acts on. So V is a representation. Now if V is over an algebraically closed field then V = k^n is affine space and so has a generic point. And then what does it mean to say the generic point is a single orbit? Does it mean, that under the group action, the generic point stays fixed?

Idek if my question makes sense...

[vestal snow]

How do I do this?

I know it has something to do with row and column reduction

But I suck with matrices

the form in prop 2.11 is where it's a diagonal matrix with d_i dividing d_{i+1}

[rustic crown]

you can compute the gcd of all n*n minors to find the product d1...dn

[vestal snow]

We haven't really done that yet

I thought the procedure was: Do row ops, multiply all matrices that you obtained from doing row ops

Do the same for column ops

[rustic crown]

yep

[vestal snow]

After that, you use those matrices to somehow get the desired basis

[rustic crown]

so to C2 --> C2 + 2C1 and C3 --> C3 + 3C1

[vestal snow]

Wait

[rustic crown]

this gives,

-6  0 0
-15 6 9

[vestal snow]

Those aren't row ops

We can't multiply by non-units

[rustic crown]

we are allowed to do column operations

[vestal snow]

Yeah, but even in column ops we're not allowed to multiply by non-units

And 2 is a non-unit

[rustic crown]

yea, but we are allowed to add any multiple of one row to any other row

[vestal snow]

Oh okay

[rustic crown]

I'm not allowed to do C1 --> 2C1

but the other 2 operations are alright, since you can reverse what you just did

[vestal snow]

I did the procedure

Here's what I got

R2-->R2 - 3R1; R1--> R1+2R2, R1 --> R2 and R2 --> R1

What are the corresponding matrices for these?

[rustic crown]

so you got,

3  0  0
0 12 18

[vestal snow]

I got [1 -3] for the first row op
[0 1]

idk if that's correct

[vestal snow]

Yeah

[rustic crown]

you should get,

[1  0]
[-3 1]

for the operation R2 --> R2 - 3*R1

doing this row operation on a matrix is same as multiplying by that 2*2 matrix from the left

[vestal snow]

yep

Wait left or right?

no it's left

you're right

[rustic crown]

[ 1 0][ -6 12 18]    =   [-6 12 18]
[-3 1][-15 36 54]        [ 3  0  0]

[rustic crown]

lol this made me laugh

[vestal snow]

Okay so I now have the matrices for all 3 rows

[vestal snow]

lol

Do i just multiply them now?

to get one of the bases?

[rustic crown]

yea but make sure you don't multiply them the wrong order

[vestal snow]

[0 1]
[1 0]

this appears in the left

Since we did it last

[rustic crown]

yea

but multiplying matrices is kinda a pain, so its generally nicer to just augment an identity matrix and do the operation on the matrix along with that identity

[vestal snow]

$\begin{pmatrix}-3&1\ -5&2\end{pmatrix}$

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

I got this

[vestal snow]

Symbolab

so is my basis for Z_2 (-3,-5) and (1,2)?

or is it (-3,1) and (-5,2)?

[rustic crown]

i need to check, but probably this

[vestal snow]

Okay

Let's try the column

Is the matrix this?

[rustic crown]

lol i was doing by hand 😛

[vestal snow]

Hmm what do i do to get rid of the 18?

Because I can't multiply a column by 2

actually wait

[rustic crown]

yea but you can do C2 --> C2 - C3; C3 --> C3 + 3*C2; C2 --> -C2

[vestal snow]

I don't think you need the last step

[rustic crown]

yea, but +6, looks cuter

[vestal snow]

let me figure out the matrices

[1 0 0]
[0 1 0]
[0 -1 1]

For the first transformation?

pls respond I sent you my matrices

[rustic crown]

yea

sorry, i was working out the calculations

[vestal snow]

haha it was a joke

$\begin{pmatrix}1&0&0\ 0&-1&3\ 0&1&-2\end{pmatrix}$

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

This is what i got

[rustic crown]

yea

to see the basis you need to probably invert this

[vestal snow]

$\begin{pmatrix}1&0&0\ 0&2&3\ 0&1&1\end{pmatrix}$

[cloud walrus]

Have a Banana, Bitch

[rustic crown]

so say P = be that earlier 2x2 matrix and Q be the inverse of that 3x3

[vestal snow]

So in the inverted matrix

[rustic crown]

i have a bad feeling that i got it backwards lol

[vestal snow]

is (1,0,0); (0,2,1); (0,3,1) my basis?

Or is it the row vectors

[rustic crown]

i think we need to invert the 2x2 and not this

so your basis should be {(1, 0, 0), (0, -1, 1), (0, 3, -2)} for Z^3 and {(-2, -5), (1, 3)} for Z^2

[vestal snow]

=\begin{pmatrix}-2&1\ -5&3\end{pmatrix}

[cloud walrus]

Have a Banana, Bitch
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[rustic crown]

let's just quickly check

[vestal snow]

So the basis in the rows for both of them?

[rustic crown]

columns for both

[rustic crown]

we just wrote M = P^-1 * A * Q

so the answer is either that or
{(1, 0, 0), (0, 2, 1), (0, 3, 1)} for Z^3 and {(-3, -5), (1, 2)} for Z^2

[vestal snow]

It's neither

[rustic crown]

yea my calculations says this is the one

[vestal snow]

Is this what I should be doing?

Because then I get this

Which is different from the original

[rustic crown]

the first one should be
-2 1
-5 3

[vestal snow]

But we're taking the inverse of that right?

[rustic crown]

could you send me the latex for this?

[vestal snow]

$\begin{pmatrix}-3&1\ ::::-5&2\end{pmatrix}\begin{pmatrix}3&0&0\ :::0&6&0\end{pmatrix}\begin{pmatrix}1&0&0\ :::0&2&3\ :::0&1&1\end{pmatrix}$

[cloud walrus]

Have a Banana, Bitch

[rustic crown]

$\begin{pmatrix}-6&12&18\-15&36&54\end{pmatrix} = \begin{pmatrix}-2&1\-5&3\end{pmatrix}\begin{pmatrix}3&0&0\0&6&0\end{pmatrix}\begin{pmatrix}1&0&0\0&-1&3\ 0&1&-2\end{pmatrix}^{-1}$

[cloud walrus]

det

[rustic crown]

this is what we got from those elementary operations

[vestal snow]

Yes

So are the bases just the rows of the two square matrices there?

[rustic crown]

yea so v1 = (1, 0, 0); v2 = (0, -1, 1) and v3 = (0, 3, -2)
u1 = (-2, -5) and u2 = (1, 3) we get,
Mv1 = PAQ^-1 v1 = P A e1 = P 3e1 = 3u1
Mv2 = PAQ^-1 v2 = P A e2 = P 6e2 = 6u2
Mv3 = 0

[vestal snow]

Ah okay

I think that does it

Thanks so much

[rustic crown]

uwu

yw

[vestal snow]

Is every finitely generated abelian group finitely presentable?

[rustic crown]

yea, that's because Z is noetherian

[vestal snow]

Wait

Yeah okay

How do I do the second part of this problem?

[vestal snow]

I can't use this without proof because we haven't covered this yet

Is there an easy proof of every finitely generated abelian group being finitely presentable?

[sturdy marsh]

Z is noetherian

[woven delta]

Can't you just take the map from Z^n to your group G given by the generators

[vestal snow]

We haven't proved that Z noetherian implies all fin. gen. Z-modules have finite presentations

[woven delta]

Then the kernel of that map is an ideal which is finitely generated because Z is noetherian

[sturdy marsh]

it's a one-liner

[vestal snow]

Oh huh

I'll try it out

[sturdy marsh]

it's the thing liquid said

[woven delta]

I can expand on this a bit if you want. The map is given by mapping the basis elements of Z^n (the coordinate vectors) to the generators of G

[vestal snow]

No, I think I got it

Thanks

[leaden finch]

can someone help me with 1

[chilly ocean]

Assume that they're not unique

And arrive at a contradiction

[leaden finch]

what does it mean by unique?

[chilly ocean]

That there are not two or more additive and multiplicative inverses

That there's only one

[paper flint]

Contradiction is redundant. For an element a, suppose b and c were both its inverses. Demonstrate b=c using definition of inverse and some algebraic manipulations.

[stark sigil]

I’m late to the party but I prefer the view that varieties have generic points, you just aren’t aware of them if you study them as varieties

And that a good substitute for a genetic point in the case you are working over Q is a point with transcendental coordinates

Well actually even over C, you can just take a point with coordinates in C(t)

[rigid cave]

Hello! I am struggling to understand this section, specifically the part where the author mentions that the splitting field for f(x) over F(s_1 ... s_n) is F(x_1 ... x_n). I am confused because there is a theorem that says that if f(x) is a polynomial over a field F with roots a_1 ... a_n then F(a_1 .. a_n) is the splitting field for f(x). This makes me think that the splitting field for the polynomial in the image above should be F(s_1 .. s_n, x_1 .. x_n). In the wikipedia article about Galois theory they state the same thing expect that they use Q (the rationals) instead of F. Can someone explain why this is true?

[hot lake]

you can show fairly easily that F(s1,...,sn,x1,...,xn) = F(x1,...,xn)

[rigid cave]

How would I do that? If F(s_1,...,s_n, x_1,...,x_n) = F(x_1,...,x_n), wouldn't it imply that F(s_1,..,s_n) = F?

[rustic crown]

no, the latter won't happen unless s_i in F... which doesn't happen since x1, ..., x_n are indeterminates

the splitting field of f(x) is the smallest extension E of F(s1, ..., s_n) over which f(x) factors completely. So E must contain all roots of f(x). Can you continue from here?

[rigid cave]

Well if E must be the smallest extension of F(s_1,...,s_n) and also contain the roots of f(x) then I would simply adjoin the roots of f(x) to F(s_1,...,s_n) to obtain F(s_1,...s_n,x_1,..x_n). What am I missing here?

[rustic crown]

that much is correct. But now can you see why F(s_1, ..., s_n, x_1, ..., x_n) = F(x_1, ..., x_n)?

[oblique river]

I think the point is just that each s_i is contained in F(x_1, ..., x_n).

There's nothing you need to prove here

[rigid cave]

Ohh okay! So this happens because of how s_i is defined, right? Since s_i is some combination of x_i then it surely must be in F(s_1,..s_n), right?

[rustic crown]

i think you want the last thing to be F(x_1, ..., x_n)...

[rigid cave]

Yes, I was little too fast there! Thank you guys, you really helped me!

[oblique river]

np 👍

[leaden finch]

can someone help me with this one

[chilly ocean]

#proofs-and-logic

[viscid pewter]

i asked this once before but one more time shouldn't hurt too much

how exactly do you use coxeter groups/reflection groups to show things about polyhedra and polytopes and so forth? like what would an example of a nice property be that's proved by group theory?

also, same question but group actions

[vestal snow]

Show that one can think of the grassmannian of dimension $n-1$ of $k^n$ as $\mathbb{P}^{n-1}$ (defined here to be the set of one dimensional subspaces, or lines, of $k^n-1$.

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

i have the intuitive idea of identifying each such subspace as the line "perpendicular" to the space

But I don't know how to state that for an arbitrary field $k$

[cloud walrus]

Have a Banana, Bitch

[sturdy marsh]

what is your definition of projective space

oh wait nvm

yes, you roughly have the right idea

you can construct a "pairing"

[latent anvil]

the inner product V × V -> k determines orthogonality when V = R^n and k = R

but there's a natural bilinear map that always exists, with no extra structure

It goes V × V^* -> k

So the "orthogonal complement" should really be thought of as a subspace of V^*, namely the annihilator

That's my intuition at least

and for V = k^n there's an iso V ≈ V^*

Since we have a canonical basis

[vestal snow]

Hmm so what should my map be?

Note: We haven't studied dual spaces yet

Neither have we done bilinear maps

So I don't think we can use those

Maybe I can do something like this

Fix a basis $w_1,\cdots,w_{n-1}$ for $W$. Let $w_i=\sum_{j=1}^nk_{ij}e_j$

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

We need to find a vector $v=\sum_1^nc_ke_k$ such that $\sum_{j=1}^nk_{ij}c_j=0$ for all $i\leq n-1$

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

So we have n unknowns (c_k) and we have n-1 equations

Can't we say something about the solution set of this?

Like there is a non-zero solution to it

and all solutions are on the same line?

Were you talking about something like this @sturdy marsh ?

I guess the we still have to show that the solution set is invariant under base change of W

[latent anvil]

Ah sorry for the confusion

[latent anvil]

This sounds correct to me

[vestal snow]

I think I almost got it.

I just need to prove the following

Given $n$ unknowns and $n-1$ homogeneous equations, the solution set exactly equal to a line

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

Say W is an $n-1$ dimensional subspace of $k^n$. Given a basis $S=w1,\cdots, w{n-1}$ define the matrix $M(S)$ as the matrix having $w_1$ for the first row (in standard basis representation of $w_1$), $w2$ as second,..., $w{n-1}$ as the $n-1^{th}$. Then the solution set of M(S)x=0 will be the same as the solution set of M(S')x=0 for some other basis S' of W

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

I feel like both of these are basic facts from linear algebra that I have forgotten

[rustic crown]

i found the answer to this. The automorphism group must act on the roots faithfully and transitively which means it can be identified with a transitive subgroup of S4 of order 8, so it must be a sylow-2 subgroup, and since all sylow-2 subgroups are conjugates, and D8 also acts faithfully and transitively on 4 vertices, we get Aut group is D8.

[unique juniper]

I dont see how that theorem lets them rearrange the last part. That map doesnt "Fix" anything?

[scarlet estuary]

uh

not really sure what your confusion is

do you not see that (1 5 4) and (2 7 6 3) are disjoint?

[unique juniper]

no

[scarlet estuary]

they dont share any elements.

alternatively, to use the definition they gave in 1.3.4

clearly for each number from 1 to 7

they are in at most one permutation

[unique juniper]

$X = (1 5 4)(2 7 6 3)$

[cloud walrus]

Yes

[scarlet estuary]

for example, the cycle (2 7 6 3) maps 1 |-> 1

yes, the point is that sigma can be written as a product of two permutations

[unique juniper]

The definition says that for every 1 ... n

[scarlet estuary]

right.

and here n = 7

and the definition of disjoint holds true

if we let $\alpha = (1\ 5\ 4)$ and $\beta = (2\ 7\ 6\ 3)$

[cloud walrus]

Namington

[unique juniper]

ye

[scarlet estuary]

then $\beta(1) = 1, \alpha(2) = 2, \alpha(3) = 3, \beta(4) = 4, \beta(5) = 5, \alpha(6) = 6, \alpha(7) = 7$

[cloud walrus]

Namington

[scarlet estuary]

so the definition of disjoint is satisfied

an easier way to "think about" disjoint though is just saying "the cycles share no elements"

[unique juniper]

i see

I misundertstood the definition

ty

[chilly ocean]

[unique juniper]

Are they supposed to suggest a binary operation ?

[carmine fossil]

The binary operation is composition

[unique juniper]

Is that just assumed ?

[carmine fossil]

Assuming that section is about S_n, yes

[unique juniper]

ok

thanks

this is long -_-

[tight otter]

scratching my head over this one a bit. i tried finding some sort of relationship or pattern within the coefficients as I expanded the polynomial but I couldn't find anything that seemed easily identifiable. Am I missing out on some theorem that applies here?

[gray charm]

x-c is a factor of p(x) iff c is a root of p(x)

Consider also ||Fermat's little theorem||

[tight otter]

ohhhhhh

i think i get it

What about when x=p=0? My memory is telling me that (p-1)! = -1 mod p but i can't remember if that's an actual theorem or not

[carmine fossil]

That's Wilson's theorem

[rustic crown]

don't put that p=0 lol

[tight otter]

it's supposed to be an equiv sign

but i'm too lazy to use latex

[carmine fossil]

Also use the fact that a polynomial of degree n can have atmost n roots in F[x]

[tight otter]

thank you

[rustic crown]

oh okie

[carmine fossil]

Actually, This is a nice way of proving Wilson's theorem

[rustic crown]

yea, but you also have the straight way directly by pairing up a with a^-1 and only elements that left unpaired are the ones that satisfy a^2=1. So this reasoning works in any (finite) abelian group...

that is product of all elements = product of all order 2 elements

[carmine fossil]

That works too,but this is good too

[rustic crown]

true

[carmine fossil]

Are you familiar with McKay's proof of Cauchy Theorem?

[rustic crown]

i don't know by the name, but are you talking about the one that uses coloring necklace by elements of the group satisfying that product of the beads = 1?

[carmine fossil]

This can be used to prove fermat's little theorem

[rustic crown]

yea, this argument was too cute

[carmine fossil]

Take |G| such that p doesn't divide |G|

[rustic crown]

This is generally said using class equation or by using Burnside's lemma

[carmine fossil]

There is exactly one equivalence class of size 1,i.e., |G|^(p-1)=1 mod p

[rustic crown]

the nice thing is that this proves a seeming stronger statement, that number of elements of order p is (p-1) mod p

or termed differently, number of subgroups of order p is 1 mod p.

this looks a lot like sylow's third theorem, and indeed iirc its true that number of subgroups of order p^k such that p^k | n, is 1 mod p.

[carmine fossil]

Yes

[narrow ibex]

If G is an infinite group, does it necessarily have a subgroup isomorphic to the integers? I'm thinking this will fail if every element has finite order, but I'm not sure how to provide a specific example.

[rustic crown]

(Z/2Z) x (Z/2Z) x ....

or another way to say this is,
set functions from N to Z/2Z under point-wise addition.

[narrow ibex]

got it, thanks

[thorn delta]

Det’s example works because it is not finitely generated. An infinite finitely generated abelian group will have a subgroup iso to Z though @narrow ibex

[narrow ibex]

thanks for clarifying, I was actually getting hung up on finitely generated abelian groups specifically

so the infinite product of Z/2Z w/ itself didn't even come to mind until det mentioned it

[thorn delta]

I think this is true for infinite finitely generated groups in general, not just abelian ones. Consider the cyclic subgroup generated by each generator. At least one of them has to generate a subgroup of infinite order

[glossy yoke]

there exist finitely generated infinite groups where each element has finite order. i don't know how to construct it, but it's called the Burnside problem: https://en.wikipedia.org/wiki/Burnside_problem

even if generators have finite order, you can have an infinite group. like, consider the free product of Z/2Z with itself. Though a and b might be order 2, ab has infinite order.

[thorn delta]

ah, right okay

[junior furnace]

I brought this up a week or two ago and no one had anything to say, I hope it's ok if I ask again:

Suppose you have a ring of non-commutative polynomials of {i, j, k}, an orthonormal basis for R3: for example, D(i, j, k) = i^2 + j^2 + k^2 or C(i, j, k) = ijk + jki + kij - ikj - jik - kji.

Then define a substitution map as a square matrix A, so a transformation that maps i, j, and k to some three arbitrary vectors; is there some way to find all the eigenvectors of A? By which I mean, the polynomials P for which P(Ai, Aj, Ak) = nP(i, j, k) for some scalar n. For example, C(Ai, Aj, Ak) = det(A) C(i, j, k) for any A.

And all the ordinary eigenvectors of A will still be eigenvectors of A ofc

Does this make no sense the way I phrased it lol

I swear this is a valid concept, for example take $A = \begin{bmatrix}
1 & 0 & 0\
0 & 0 & 1\
0 & 1 & 0
\end{bmatrix}$

Then, Ai = i, Aj = k, Ak = j

So D(Ai, Aj, Ak) = i^2 + k^2 + j^2 = i^2 + j^2 + k^2 = D(i, j, k)

And C(Ai, Aj, Ak) = ikj + kji + jik - ijk - kij - jki = -1 * C(i, j, k)

[cloud walrus]

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

[vestal snow]

I had a question about chain complexes and cochain complexes

Given a chain complex, can't we just flip it around to get a co-chain complex?

For example, $\cdots \longleftarrow A_i \longleftarrow A_{i+1} \longleftarrow A_{i+2}\longleftarrow \cdots $can be made into $\cdots \longrightarrow A_{i+2} \longrightarrow A_{i+1} \longrightarrow A_i \longrightarrow \cdots$

So why even bother studying both of these when they are essentially the same?

[scarlet estuary]

by "flip it around" do you mean

apply Hom(X, -)?

because you cant naively just "flip the arrows" due to dimensionality issues

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

I mean like what I did in the example

What do you mean by dimensionality issues?

[scarlet estuary]

those are the same complex

both of those are chain complexes since their differentials (arrows) are decreasing dimension

[vestal snow]

Does the indexing matter?

[scarlet estuary]

well they do if youre reusing indices

lmao

if your i is supposed to be a different i then its not a big deal

[vestal snow]

Like say that the complex was indexed by the integers. Can't we fix the issue by just taking every index and turning it into its negative

[scarlet estuary]

but if you use "i+1" twice in the same sentence im gonna assume theyre the same i

[vestal snow]

One sec I'll make it more precise

[scarlet estuary]

but yes, (the differential in) a chain complex decreases the degree whereas a cochain complex increases the degree

this matters more than "the direction of the arrows"

[vestal snow]

Let $\cdots \longleftarrow A_i \longleftarrow A_{i+1} \longleftarrow A_{i+2}\longleftarrow \cdots$ be a chain complex indexed by the integers. Then define $B_i=A_{-i}$ (do the same for the maps, but idk how to write stuff over the arrows. Then, $\cdots \longrightarrow B_i \longrightarrow B_{i+1} \longrightarrow B_{i+2}\longrightarrow \cdots$ is a co-chain complex

[scarlet estuary]

okay sure

[vestal snow]

Yes, so is there a point to not just study one instead of both?

[cloud walrus]

Have a Banana, Bitch

[scarlet estuary]

well cohomology gives you different invariants

than homology does

the formal interpretation here is that cochain complexes are (categorical) duals of chain complexes

[vestal snow]

Does that mean there is an isofunctorism (is that a word?) between the categories?

[scarlet estuary]

wikipedia can probably exposit this better than i https://en.wikipedia.org/wiki/Cohomology#Singular_cohomology

right basically @vestal snow

[vestal snow]

Gotcha

[scarlet estuary]

duality is a "functor" from a space to its dual space

and its "inverse" is called a "pullback"

(this is a slightly theory-of-modules-biased viewpoint but its good enough)

[vestal snow]

So any category theoretic info we get about chains will also be true for cochains

[scarlet estuary]

making the appropriate modifications of course

yes

[vestal snow]

However, when we think of chains and co-chains as some constructions of, say, topological spaces

We are fixing a point of reference, so to speak

So we can't say that they are the "same" from a fixed point of reference

Thus, the need to study them separately?

Is this the intuition?

[chilly ocean]

Can someone help me with this from atyiah macdonald, I've found this solution but I'm a bit confused.

We are trying to show that if $f$ is a unit then $a_0$ is a unit and $a_i$ for $i\neq 0$ are nilpotent

[cloud walrus]

lime_soup

[scarlet estuary]

@vestal snow sure, and oftentimes when we talk about chains, intuitively our constructions are becoming "more degenerate"

like going from a space into its boundary

whereas cochains go the other direction

for an analogy, its kinda like differnetiation vs integration

[chilly ocean]

I don't understand the comment about since A/p is an integral domain, f must have degree zero

[scarlet estuary]

differentiation is easier to compute but often less nice/powerful

the invariants cohomology gives us are typically very powerful relative to homological ones

[vestal snow]

One more question

Say we define define a way to construct a chain from any topological space. Now, we reverse the arrows from the category of topological spaces, and create a cochain from this new category in the same way as we did originally.

Will the chains from the original encode the same (topological) properties as the cochains from the new (up to translating these properties from the topological spaces category to the cotopological space category) ?

[vestal snow]

When do two polynomials over an integral domain multiply to give you a unit?

What can you say about their degrees?

[chilly ocean]

Sorry yes this is exactly where my confusion is, I believe the following two things to be true but they contradict each other

1. if A is an integral domain, then deg(-):A[x] to Z is an additive function. This would give us the result.
2. Consider F_p[x]. then x^{p-1}*x=x, but the degree function is not additive then

[vestal snow]

Ah I see you're making the mistake I once did

Do not think of polynomials as functions

$x^p\neq x$ as polynomials

[cloud walrus]

Have a Banana, Bitch

[chilly ocean]

okay so we don't apply the characterstic rule to the formal variable

[vestal snow]

What is the characteristic rule again?

Sorry I'm not familiar with that term

[chilly ocean]

oh maybe i am using the wrong term

[vestal snow]

Do you mean fermat's little theorem?

[chilly ocean]

just this thing when you have a field of characteristic p that a^p=a

[latent anvil]

This is not true

This is only true for the field Fp

[vestal snow]

Yeah, that only works for elements of F_p

[latent anvil]

Not for like, F_p^2 or Fp(t)

(everyone I know who's learned field/galois theory has made mistakes like this in positive characteristic though, you're in good company)

[oblique river]

just hopping on board here to say that this doesn't matter anyway, what you wrote above that this doesn't apply to the formal variable is correct

x^p and x are not the same polynomials over F_p

[vestal snow]

The way I like to think of polynomials is formal expressions (until you learn category theory, then you should think of it as an initial object in a special category)

[oblique river]

one has degree 1 and the other has degree p

[chilly ocean]

yeah i understand the exercise now

wait so

what is the story with field of prime charaterstic

i thought charaterstic being p was that a^p=a

[latent anvil]

Nope

[oblique river]

(not trying to be dismissive when I say "it doesn't matter" -- this is a very good thing to think about and the explanations shamrock and banana are giving are correct -- just wanted to point out to you that there are two separate questions being discussed)

[latent anvil]

Characteristic p mean pa = 0

for all a

[oblique river]

in fact, F_p is the only field which satisfies a^p = a for all a

[vestal snow]

Consider something like F_4

[chilly ocean]

so a field of characterstic p, is not torsion free as a Z module

[vestal snow]

And you will see that a^2 does not equal a for all a in F_4

[latent anvil]

Yup

I mean for example, Fp = Z/pZ

[vestal snow]

If it did, that would violate the number of roots theorem over fields

[latent anvil]

This has p torsion

[chilly ocean]

god i really regret pretending char p didn't exist earlier in my mathematical journey

[latent anvil]

Haha I know what you mean

I still don't understand (in)separable extensions

someday I will learn number theory and see how it all fits today

[vestal snow]

If you want a more algebraic treatment of inseparable extensions, consider checking out Stichtenoth's book

[latent anvil]

I mean I've seen the algebraic treatment lol

[chilly ocean]

what do i actually have to know about fields of characterstic p

[latent anvil]

It didn't stick

I think seeing it in an arithmetic or geometric context would help more

[latent anvil]

What do you want to do with them?

Very little if you want to do pdes

Even less if you're interested in architecture

[chilly ocean]

read atyiah macdonald

[vestal snow]

Isn't there an analog of PDE for p-adic analysis?

[latent anvil]

¯\_(ツ)_/¯

you got me I guess

[vestal snow]

Not a whole lot about them specifically, but keep in mind that finite fields exist

so don't treat polynomials as functions

[latent anvil]

I think it's worthwhile to know two things about fields of positive characteristic

(1) existence and uniqueness of finite fields

(2) if k = Fp(t) and K = k[x]/(x^p - t) then K/k is inseparable

[chilly ocean]

I also have this bad thing of thinking characterstic p means F_p

[latent anvil]

Yeah haha also very common when learning this stuff

Characteristic p actually means Fp, Fp^2, or Fp(t)

[vestal snow]

If it helps, every char p field contains F_p

[chilly ocean]

i remember doing some alg geom problem

so it was an algebraically closed field

show that such and such a zero set was irreducible

then it asked about the case what about char 3

and then i started trying to show that the polynomial had some root in F_3

[vestal snow]

Unrelated, but every alg. closed field is infinite (proof is a nice exercise to try out)

[chilly ocean]

knowing that i was working over an algebraically clsoed field

yeah

[chilly ocean]

this is 'funny' part of me mistake

i knew this to be true

and i knew that F_3 was certainly not infinite

but it didn't click that what i was doing was wrong

[vestal snow]

Happens to everyone I think

Which is why I've made a point to work out one example and one "counterexample" to every theorem I see

[latent anvil]

This is what I was trying to get at with the two ideas above

The first tells you every possible example and counterexample to anything about finite fields

The second is what you should think of whenever you see the word "separable"

[chilly ocean]

Thank you kings

[obsidian sleet]

hello

i need help understanding again

is this free or do i wait

i assume free

i will post my pain

i am new

😄

i don't understand how the yellow part works

or how this test works