groups-rings-fields

[cloud walrus]

diligentClerk

[fossil shuttle]

@dusty river

[split escarp]

Let $G={f:f:\mathbb{R}\rightarrow \mathbb{Z}_2, \exists x\in \mathbb{R}, f(t)=0,\forall t> x}$, with addition of functions, addition modulo $2$. Let $G_A={f:f:\mathbb{R}\rightarrow \mathbb{Z}_2, f(t)=0,\forall t> 1}$. Is $G_A$ a subgroup of $G$?If so, is the index of $G_A$ in $G$ finite or infinite?

I am an Applied Math guy, please consider that if I make dumb mistakes

[hidden haven]

Thanks! It's just that I was avoiding doing things set theoretically because the relation looked scary, but phrased that way, it sounds way nicer

[cloud walrus]

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

[hidden haven]

Wait so if I'm interpreting it correctly, the forgetful functor from R-mod to Set preserves coequalizers?

That actually makes sense thinking set theoretically yeah

[fossil shuttle]

hmm i'm not sure if this is true.

[hidden haven]

Wait so when you said x ~ f(x), did you take the equivalence relation generated by all of these

[fossil shuttle]

yes

[hidden haven]

Ah yeah that's the part I'm finding daunting

Like the generated relation means I have to show there's no finite path from x to 0

[fossil shuttle]

ah. well the fact that it's filtered makes the generated relation very easy to describe. suppose that you have $M_i,M_j,M_k$ and $f :M_j \to M_i, g : M_j\to M_k$. then it's a consequence of the definition of a filtered diagram that there should be maps $f' : M_i\to M_\ell, g' : M_k\to M_\ell$ making it into a commutative square

[cloud walrus]

diligentClerk

[fossil shuttle]

right?

[hidden haven]

I'm not sure what filtered means

[fossil shuttle]

ok well maybe let me find a rigorous definition

[hidden haven]

Indexed by a directed set?

[fossil shuttle]

you can take it to be a directed set if you want but the proof goes through in just slightly more generality

obviously a directed set is filtered because by definition it satisfies axiom 1 and axiom 2 holds vacuously in a poset

[hidden haven]

Right

[fossil shuttle]

so ok yeah let's just do a directed set for simplicity

[hidden haven]

Yeah I see the square here

[split escarp]

It would be nice if someone can comment anything on this

[fossil shuttle]

right. so if you have two elements in $M_i,M_k$ which are equal because they both come from the same element in $M_j$, then we can reword this to say that they are equal because they both go to the same element in $M_\ell$

from there it's an easy induction argument to prove that $x\in M_a \sim x'\in M_b$ iff there is some $c$ with $c< a,b$ and $x,x'$ go to the same element of $M_c$

[cloud walrus]

diligentClerk

diligentClerk

[hidden haven]

Ahh nice

I see how that works now, I'll try it. Thanks!

[hidden haven]

Why the {f:f:...?

Like why 2 fs, is that a typo?

Ohh the first one is for the set builder thing

[split escarp]

Set of all f's such that

[hidden haven]

Right

Are they just set functions?

Can't be group homomorphisms because of torsion

The f

G_A is a subgroup yes, any element of G_A is in G

It has infinite index

[split escarp]

but not isomorphic I guess

[hidden haven]

Assuming that all f are set functions

It should be isomorphic

Hmm let me think

[split escarp]

then G_A is isomorphic, subgroup of infinite index?

[hidden haven]

Ah now I'm not so sure about them being isomorphic

They are if you remove that second condition on the functions in G

Like if you just took all functions

[split escarp]

But what would be the index?

[hidden haven]

Instead of eventually 0

Infinite then

Because you can get distinct cosets by adding the functions f_n which are 1 at n, 0 elsewhere, for any n>1

Like each of those f_n gives a distinct coset

[split escarp]

True, I will remove the condition

Then $G={f:f:\mathbb{R}\rightarrow \mathbb{Z}_2}$ has a proper subgroup which is isomorphic to $G$ and of infinite index(?)

[cloud walrus]

MathForEarthlings

[hidden haven]

Yes

[hidden haven]

@fossil shuttle I think this is true though, since in both cases you just quotient the codomain by f(x) ~ g(x) for each x

[fossil shuttle]

well look at $\mathbb{Z}$ and consider the two maps $\id : \mathbb{Z}\to \mathbb{Z}$ and $-1 : \mathbb{Z}\to \mathbb{Z}$. Abelian-group-theoretically the cokernel is $\mathbb{Z}/2$. But set theoretically it's $\mathbb{N}$, I think

[cloud walrus]

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

[fossil shuttle]

the argument I gave you about the equivalence relation should only work for filtered colimits, not coproducts or coequalizers in general

[hidden haven]

Ah right I see it, because you take the submodule generated by the equivalence relation, kind of

[fossil shuttle]

yeah

[hidden haven]

Ty

[cloud walrus]

MathForEarthlings

MathForEarthlings

MathForEarthlings

[sharp dirge]

i dont understand this induced k algebra homomorphism

x should be sent to x composed with phi

this is the theorem btw

i think ive understoof whats going on here but not in the example

<@&286206848099549185>

in the example above

wouldnt $x \mapsto x\varphi$

[cloud walrus]

pewdssssssss

[sharp dirge]

and (x)phi would be an element of the field?

dang this is confusing

[fierce perch]

The k-algebra morphism is determined by how it acts on the generators

[clever mountain]

I have a problem covering Sylow's theorem that I have been grappling with and
would appreciate if anyone could provide a hint.

G is a group of order $p^{k}*m$, where p does not divide m.
Say there is a natural integer r such that for any two distinct p-Sylow subgroups,
their intersection has at most $p^{r}$ elements.
The task is to show the number of p-Sylow subgroups, $n_{p}$, is congruent to 1 (mod $p^{k-r}$).

Thoughts: My idea was to let a p-Sylow subgroup, P, act on the set of all p-Sylow subgroups
by conjugation. Orbit decomposition on the set of all p-Sylow subgroups gives me then:
$n_p = 1 + \sum_i \frac {|P|}{| \text{stab}_P (H_i) |} $

Here $H_i$ is a p-Sylow subgroup. $\text{stab}_P (H_i)$ is a proper subgroup of
P, so $| \text{stab}_P (H_i) | = p^{q_i}$ for some $q_i \leq k-1$. Since $|P| = p^k$, I have

$n_p = 1 + \sum_i p^{k - q_i} $

If I can show $q_i \leq r$ then the result will follow. But I can't see how to do that.
$q_i \leq r$ would mean $\text{stab}_P (H_i)$ has at most $p^r$ elements, but
I am struggling to see how this imposition falls on the stabilizer subgroup.
Any help appreciated.

[cloud walrus]

hardisc

[sharp dirge]

in the second paragraph

it doesnt make sense?

f is a polynomial evaluted by stuff in W

phi takes values from V into W

so how would F(w ,.. wn) = something evaluted in V

$f(\varphi) \in k[V]$ this part btw $f \in K[W]$

[cloud walrus]

pewdssssssss

[ivory dust]

Let f (x) E F[x] be an irreducible polynomial of degree n, and let E / F be
a splitting field of f (x).
(i) Prove that n I [E: F].
(ii) Prove that if f (x) is separable, then n I I Gal(E/F) I.

For this question seems very straightforward but I wanted to ask about some technical details

if f is irreducible polyn w degree n and E/F is splitting field of f

then is in E/F f splits as n linear factors correct?

hence n roots (not necessarily distinct?)

call them a1, a2,... an

so E = F(a1,a2...an)?

and by the degree formula

[E:F] =[ F(a1,a2...an):F(a1....an-1)] x [ F(a1,a2...an-1):F(a1....an-2)] x ... x [F(a1):F]

wouldnt this imply [E:F] = 2^k some k?

[gritty sparrow]

No, it doesn’t imply that, the extension F(a1)/F need not have degree 1 or 0, it could be something completely different

[ivory dust]

Oh yea true

I see why

[gritty sparrow]

Nice

[ivory dust]

Would we say

E/F = F(a1,a2....an)?

or E = F(a1,a2,...an)

[gritty sparrow]

The second one

[ivory dust]

since its splitting field of f over F

that means f splits into linear factors

over E/F

and its irred over F

so

all its roots are not in F

so can I just choose one of the roots

and adjoin it to F

say F(a)

then compare that to E

[gritty sparrow]

Look you’ve written down the correct degree formula, now just use the fact that [F(a1):F]=n (can you see why that is the case?)

[ivory dust]

Yess since f is degree n

and all roots are not in F (moreover its irreducible)

[gritty sparrow]

Correct

Now you see why n divides [E:F] right?

[ivory dust]

yess

tyty

[gritty sparrow]

Np

[ivory dust]

if f(x) = (x-a)(x-b)(x-c)(x-d) and a,b,c,d distinct e E where E is splitting field of Q

and a,b,c,d are not in Q

im asked to find Gal(E/Q)

i know each automorphism from E = Q(a,b,c,d) to E fixing Q pt. wise

has to map root to a root of f

and also

since f has 4 distinct roots

then Gal(E/Q) iso to subgroup of S4

would Gal(E/Q) be S4?

[hidden haven]

Not necessarily, there are like 6 possibilities

[hidden haven]

All you can say is this

Uh

The thing after that

[ivory dust]

well

actually the roots are sqrt2+sqrt3, sqrt2 - sqrt3, -sqrt2-sqrt3, -sqrt2+sqrt3

[hidden haven]

You should be able to figure this out by adjoining the 2 roots 1 by 1

And seeing how automorphisms extend

You have to find all extensions of the identity automorphism

[ivory dust]

f(x) = x^4 -10x^2+1

well

i was thinking phi identity obviously

but like

idg why there isnt 4! = 24 options

like say you have phi(a)

this has 4options

phi(b) then has 3

etc.

[hidden haven]

How many extensions are there to Q[√2]

[ivory dust]

one?

[hidden haven]

Nope

[ivory dust]

from E to Q(sqrt2)?

err

nvm that doenst even make sense

im reading this solution online

and I assumed it was wrong

but it maps sqrt2 to sqrt2, sqrt3 to sqrt3
sqrt2 to -sqrt2, sqrt3 to sqrt3
sqrt2 to sqrt2, sqrt3 to -sqrt3
sqrt2 to -sqrt2, sqrt3 to -sqrt3

so 4 automorphisms exist

which made like 0 sense to me

💀

[hidden haven]

No

I meant

You have identity on Q

An automorphism of Q(√2)/Q

Is exactly an extension of identity

See how many there are

Then do the same for eachof their extensions to the larger thing

[ivory dust]

would there be 2?

identity map and sqrt2 to -sqrt2?

[hidden haven]

Yep

Then for both of those you see how to extend

And then you get all the extensions

Then you see how they compose etc

To figure out grp structure

[ivory dust]

so should be 2*2 = 4

such automorphisms

since we require

roots to be perserved

of x^2-2, x^2-3

[hidden haven]

Yep

[ivory dust]

tyy 😄

[hot tinsel]

Let $p$ be a prime number. Show that the Frattini subgroup of $\Bbb Z/p^n$, $n\ge 2$, is generated by $p$.

[cloud walrus]

K零

[hot tinsel]

any help with this?

The maximal subgroups of Z are qZ for prime Z

[gritty sparrow]

qZ contains p^nZ iff q divides p^n

That should answer this

[hot tinsel]

so under canonical projection $Z\to Z/p^nZ$ the maximal subgroups of Z/p^nZ correspond to maximal subgroups of Z containing p^n Z, thus the only maximal subgroup is pZ/p^nZ

is this correct

[gritty sparrow]

Yes

[hot tinsel]

Thanks

[gritty sparrow]

No problem

[chilly ocean]

yo any ideas?

[void cosmos]

Seems straightforward just apply definitioms

Draw tower

[gritty sparrow]

I can think of a proof via zariski’s lemma

And i’m pretty sure this statement will imply zariski’s lemma

So it seems very hard

[steady axle]

I think you have to deal with transcendental degree

[chilly ocean]

most likely ye

@void cosmos lol homeboy try then

[gritty sparrow]

Have they done any commutative algebra before they gave this question? If not I’m kinda stumped for a proof that only uses stuff about fields

[dusty river]

Hmm can't you do E/K as E/L/K, where L is the algebraic part, then say L/K finitely generated, and then transcendental degree arguments for E/K finitely generated?

L/K is finitely generated because it is contained in a finitely generated algebraic and hence finite extension

The algebraic part of F/K is finitely generated because it should be what you get when you remove all the transcendental generators

[gritty sparrow]

I don’t see how you got F/K finitely generated from just that

[dusty river]

F/K is given to be finitely generated

Do you mean E/K?

[gritty sparrow]

Sorry i mean the alg closure of K on F

*in

How do you just remove transcendental generators and get the alg part?

[dusty river]

F is K({a_i} union {b_i}), where b_i are transcendental, then K({a_i}) should be the algebraic closure?

[gritty sparrow]

No?

[dusty river]

Any expression that non trivially uses the transcentals would be transcendental though

Non trivially as in not equal to 0

Assuming that the b_i's are algebraically independent

is that not true?

[gritty sparrow]

The finite generating set may have dependant bi’s

You can do trdeg then add in the alg stuff to get a generating set, but there is no guarantee that will be finite

[dusty river]

hmm I see

[gritty sparrow]

Well i think i have the argument for how this implies zariski lemma: assume this thing holds, assume there is t in F which is transcendental, then K(t) is a sub-extension, hence finitely generated. But that is a contradiction as K(t) is infinitely generated over K. Therefore F over K is algebraic. That means F over K is finite

And zariski lemma implies this is pretty straightforward

[dusty river]

I didn't get it, K(t) is finitely generated as a field extension, but not as an associative algebra

[gritty sparrow]

Yeah

[dusty river]

Aren't those the 2 things you got though?

[gritty sparrow]

I don’t ever use field extension generation

Finitely generated is always as an algebra

[dusty river]

then K(t) is a sub-extension, hence finitely generated

How is this true?

[gritty sparrow]

That is by the proposition

[dusty river]

The thing to be proved is for finitely generated as field extensions I think

[gritty sparrow]

No, i don’t think so. When finitely generated is used it is almost always as an algebra

[dusty river]

what

We used it for fields in alg4

[gritty sparrow]

No

[dusty river]

[gritty sparrow]

(For finite extensions both are the same)

[chilly ocean]

in the problem as fields I thought nnot sure if there was algebra mentioned before

[dusty river]

We proved "finite iff finitely generated algebraic"

Would be weird to not have a word for F/K being K(finite set)

[gritty sparrow]

Hmm true

Let me think about this now, hopefully it gets easier

[dusty river]

In fact it says K(finite set) is finitely generated, wouldn't that mean it is algebraic if it were as an algebra

or is it possible to somehow get quotients still

[gritty sparrow]

I have no clue how to parse that

[dusty river]

lol edited

Because the algebra generated by a finite set that contains transcentals probably won't contain inverses right

Or at least it will be a very strong hypothesis

Because then it eliminates the case of having 1 transcendental generator immediately

[gritty sparrow]

If K over k is finitely generated as an algebra, then it is algebraic: this is what zariski’s lemma says

[dusty river]

Right

I am pretty sure Ledog was studying transcendental extensions though

[dusty river]

so it seems finitely generated as a field would make sense (also given that the other interpretation is way too hard)

[chilly ocean]

ye

[dusty river]

This still doesn't work though

You might have to rewrite the generators making the transcendentals all independent

and then prove that there's a way to do this without having to add infinitely many algebraic generators

😵‍💫

[chilly ocean]

[shadow portal]

Hey Yall I’m a little lost rn. So if I have a subgroup of prime order then it’s cyclic right? And since it’s cyclic it’s abelian right? And since it’s abelian doesn’t that make it a normal subgroup? To sum it all up, is it true that subgroups of prime order are normal?

Wait nvm I think I found my answer, abelian subgroups aren’t necessarily normal right

[scarlet estuary]

correct

thats where your logic failed

[south patrol]

yeah, maybe the confusion arose because all subgroups of abelian groups are normal

a nice (ig minimal, since S3 is the smallest non-abelian group?) counterexample would be that H = {e,(12)} is a prime, non-normal subgroup of S3, as (13)H and H(13) do not coincide :)

[cursive remnant]

I feel like I might be doubting myself a bit too much here: let's say I have two (finite) groups, A and B. $A \oplus B$ is certainly isomorphic to $B \oplus A$, where $\oplus$ is the external direct product, right? In my mind, we can easily construct a bijection between the two where each element in the first just switches its components' places.

[cloud walrus]

Chris24

[south patrol]

the bijection certainly exists and you can show the map is an isomorphism directly too

[void cosmos]

@cursive remnant yes whats the problem

[cursive remnant]

No problem, specifically! Just applying my logic.

[void cosmos]

yea ur right

(a,b) --> (b,a) is a bijective homomorphism

[cursive remnant]

Well, actually, I am trying to apply that process iteratively, in a sense.

[void cosmos]

wdym

[south patrol]

Concretely, let φ: A x B -> B x A send (a,b) to (b,a) then
φ((a,b) (a',b')) = φ (aa', bb') = (bb',aa') = (b,a)(b',a') = φ(a,b)φ(a',b')

i.e. just directly compute it

[cursive remnant]

Awesome, thank you a bunch @south patrol

[south patrol]

np

now i feel weird because we always just used x for (external) direct product with groups but I suppose the difference comes in with internal direct products

[void cosmos]

yea

the notation for products sucks

like is really bad

people cant make up their mind wether to use x or this direct sum

[cursive remnant]

And @void cosmos I want to demonstrate that if I have a finite collection of groups A, B, C, etc. that $A \oplus B \oplus C . .. \oplus N$ is isomorphic to any . . . permutative product? I'm not sure how to word it.

[cloud walrus]

Chris24

[void cosmos]

yea

think of them as coorrdinates and shit

[cursive remnant]

Yeah that's what I'm going for.

[void cosmos]

and permute the elements (ordered tuples) accordingly

and tbh i dont know

if this falls

for the infinite case

ur right

[cursive remnant]

Thanks a ton!

[void cosmos]

no problem

good luck

have fun

[wooden ember]

I feel like ive taken freedoms with manipulating isomorphisms: is this argument valid? Let $P\in Syl_2(D_{2n})$, ive shown $P\cap \left<r\right>$ is a normal subgroup of $D_{2n}$. Now $\left<s,r^k\right>$ is a particular Sylow 2-subgroup of $D_{2n}$, so $P\cong \left<s,r^k\right>$. Now comes the argument im not sure about. We have hence that $P\cap \left<r\right> \cong \left<s,r^k\right>\cap \left<r\right> = \left<r^k\right>$, so that $D_{2n}/P\cap \left<r\right> \cong D_{2n}/\left<r^k\right> \cong D_{2k}$

[cloud walrus]

Little Narwhal ✓

[wooden ember]

would appreciate if someone could tell me if that final argument is valid

oh and btw k is the number such that 2n=2^a * k with k odd

[hot lake]

did you magically assume that P inter <r> is independent of P ?

[wooden ember]

yeah.... exactly why it felt fishy

[hot lake]

D2n is simple enough that we have a pretty nice picture of all its subgroups so you should just be able to brute force all the possible P

[wooden ember]

i have a tendency to treat isomorphisms like equalities too much.

[wooden ember]

all i need to show to complete the exercise i was working on is that $P\cap \left<r\right> = \left<r^k\right>$ but that's where im struggling

[cloud walrus]

Little Narwhal ✓

[hot lake]

since all Sylow subgroups are conjugate to each other you can just argue that the conjugates of <r> are <r>

[wooden ember]

oh right

[hot lake]

and same with <r^k>

[wooden ember]

right so in conjugating <s,r^k> to get any sylow 2-subgroup, the subgroup <r^k> conugates to <r^k> since it is normal in D2n and hence the intersection with <r> stays the same

[hot lake]

for any 2-Sylow P', there is g such that P' = gPg-1
then P' inter <r> = gPg-1 inter g<r>g-1 = g(P inter <r>)g-1 = g<r^k>g-1 = <r^k>

where P is the one you found

(and it might be the only one anyway)

[wooden ember]

no it's not since there are k sylow 2 subgroups (that's what the exercise asks you to show)

thanks!

[hot lake]

ah yes

[urban acorn]

TIL infinite fields never have finitely generated multiplicative groups

[kind temple]

woah kinda cool

[latent night]

If I want prove in R-mod Hom(R, M) is isomorphic to M, where M is an R-module, and R is an R-module as a a ring acting on itself, how would I do this?

My proof is like this: if we look at the underlying abelian groups of R (as a ring) and M, denoted Ar and Am, we get that Hom(Ar, Am) is isomorphic to Am. Now if we let R as a ring act on Am, we get M. This is isomorphic to R acting on Hom(Ar, Am). Where r * phi(a in Ar) = phi(r*a in Ar). Did I say anything sensible? (Probably not).

Or Hom(Ar, Am) might just be a subgroup of Am? I'm not sure.

Sorry

[dusty river]

Why are you apologizing

Hom(R, M) = M doesn't imply Hom(Ar, Am) = Am. The reason is that when you just regard these as abelian groups, you are allowed to take more general homomorphisms, so the hom set becomes bigger

While Am of course remains the same

The 2 "Hom" operators aren't the same, the first one will be in the category of modules while the second one is in the category of abelian grapes

Hint: ||Look at where the identity of R goes under such a homomorphism||

[chilly radish]

Abelian grapes

[sharp dirge]

How would be show L_d is an ideal

since it may not be closed under multiplication

[foggy merlin]

You can change the above argument, replace the polynomial
r x^e f - x^d g with just r f - g since the polynomials f,g all have the same degree already. Then the proof is essentially the same.

[near harbor]

Let $\omega$ be the solution to $x^2+x+1=0$, is the set ${a+b\omega: a, b\in\mbb{R}}$ a field

[cloud walrus]

arshsverma

[near harbor]

I'm just confused at what the inverse would be

multiplication seems to be $(a+b\omega)(c+d\omega)=(ac-bd)+(bc+ad-bd)\omega$

[cloud walrus]

arshsverma

[frank fiber]

yes

[frank fiber]

so the inverse of $a+b\omega$ is a $c+d \omega$ such that $ac-bd+(bc+ad-bd)\omega=1$

[cloud walrus]

Or x1

[hidden haven]

Do you know anything about principal ideal domains?

[foggy merlin]

wouldn't the fact that this polynomial is irreducible over R with degree two already make that set there field? I mean idk if you've had that theorem yet in your course, but it is field theory.

[carmine fossil]

Isn't it just solving 2 linear equations

bc+ad-bd=0
ac-bd=1

a and b are known constants

[near harbor]

ah, yeah, makes sense

and thanks

[tranquil parcel]

Does an isomorphism of an integral domain (in particular a ring of polynomials over a field) to itself extend to an automorphism on the corresponding field of fractions (the field of rational polynomials over a field)?

[lethal dune]

it does

[gritty sparrow]

Do you still need a solution to this one?

[chilly ocean]

Can I dm about it?

[gritty sparrow]

Sure

[patent girder]

how did i prove this?

[maiden ocean]

the noetherian condition asserts that given an arbitrary set of ideals in a ring ordered by inclusion, there is an element in the set not contained in any other ideal in the set, right?

e.g an infinitely ascending chain of ideals for which none are maximal in the set because i can just go to the next one, which is also not maximal, and so on

whereas the usual zorns lemma argument just asserts that every ideal is contained in a maximal ideal in the ring, not necessarily in a given set

[fading gulch]

@patent girder that looks like skein relations for some polynomials that are topological invariants of knots. If that's the case, most of the time this type of relations are definition of polynomials

[wooden ember]

i thought a strictly upper triangular matrix had 0s along the diagonal as well but then clearly it wouldnt be an element of the general linear group... The exercise they reference instead had 1s on the diagonal which is also strange.... So im guessing it's this last interpretation that they're referring to?

just weird because i thought usually strictly upper triangular matrices had 0s along the diagonal and the lower triangle

[hot lake]

what does exercise 17 say ?

in the context of GLn I would say it's all 1 on the diagonal, while triangular matrices would have nonzero entries on the diagonal

[patent girder]

yea, it's the conway polynomial so I'd just adjust the the C(+) - C(-) = tC(o) to get that right?

[sly crescent]

Are groups of affine Lie type things that have been considered before?

[wooden ember]

it's about upper triangular matrices with 1s on the diagonal

[prisma ibex]

What do you mean?

Like those produced by affine Dynkin diagrams?

[sly crescent]

Yes

[prisma ibex]

Yea these are incredibly well studied

Or at least the Lie algebras are

The Kac Moody Lie algebras of affine type are infinite dimensional but they have a very similar representation theory to the finite dimensional case

And there are lots of techniques for studying these

[sly crescent]

Ok

What about the analogs of the finite groups?

[prisma ibex]

Oh for the finite groups of Lie type? That’s a bit weirder of a situation

[sharp dirge]

anyone know how the part in yellow is done?

[prisma ibex]

@sly crescent there’s a paper that came out a few days ago that at least gives some idea how this might work: https://arxiv.org/pdf/2107.12727.pdf

[sharp dirge]

why do each qjgj have degree less than bi-1

[unreal portal]

It follows by definition of the division algorithm (at least in my book on groebner basis). In the previous sentence, $S(g_{i-1}, g_i)$ has multidegree less than $\beta_{i-1,i}$. In general for polynomial division, dividing $f$ by $g_1, \dots, g_n$ lets us write $f=q_1\cdot g_1+\cdots + q_n\cdot g_n + r$ where the multidegree of $f$ is greater than or equal to the multidegree of each $q_i\cdot g_i$.

[cloud walrus]

cgodfrey

[unreal portal]

It follows because $\beta_{i-1, i}>\text{multidegree}(S(g_{i-1},g_i))\geq \text{multidegree}(q_j\cdot g_j)$ for each $j$

[cloud walrus]

cgodfrey

[vast quiver]

also remember that r=0 because by definition S(g_{i-1}, g_i) = 0 mod G means that the remainder after division by G is 0, that's just the definition of congruence mod G

[viscid pewter]

heeelp

or don't bother, even if you explain this to me i will have no chance of understanding it, but one of you might find it fun lol

[dusty river]

Stuck on (a,b,c) → d. I see that d is true
iff any 2 prime ideals x and y can be separated by disjoint open sets
iff they can be separated by some disjoint X_f and X_g
iff x is in X_f, y in X_g, and fg is nilpotent (last condition is equivalent to every prime ideal containing either f or g).
Not sure how to produce such f and g though

[thorny flame]

Is this true, and how do I prove it?

[hidden haven]

Identify their underlying sets, figure out the multiplicative structure

[quaint tree]

Zariski topology?

[hidden haven]

Yeah

[ivory dust]

How to show:

any phi in Aut(GF(p^n)) is the identity map when acting on GF(p)

so phi(1) = 1 if its in Aut(GF(p^n)) since its ring/field isomorphism

and if i take any arbitrary element of GF(p)

what facts do ik about it?

like for all g in GF(p)

g^p = 1?

or p*g = 1?

[lethal dune]

g^(p^n)=g

[ivory dust]

for a g in GF(p)?

[lethal dune]

g in GF(pⁿ)

[ivory dust]

o

so

in GF(p)

every element has property:
g^p = g?

[lethal dune]

yes

[ivory dust]

ohh ok thank you

i also saw in my notes

g+g+g+....+g = 0 or 1(?)

if there are p g's

i guess just depends on the operation?

was kinda confused abt that

[lethal dune]

always 0 in GF(pⁿ)

characteristics is p so any p sum of g's is 0

[ivory dust]

ohh ok thank yous

[hidden haven]

Yeah this gives you the first thing you asked

1+...+1 kinda elements are all fixed since 1 is

[ivory dust]

so if i take an arbitrary g in GF(p)

phi(g) = phi(sum of g p times)?

er

[hidden haven]

Do you know what GF(p) looks like?

As in have you seen it expressed in other ways

[ivory dust]

Umm field w p elements

LMFAO thats all

and the binomial theorem

[hidden haven]

Do you know any fields with p elements?

[ivory dust]

Z2?

or any Zp

[hidden haven]

Yep

[ivory dust]

and like (a+b)^p = a^p + b^p

thats all ik

[hidden haven]

Are there any other such fields?

Fields of cardinality p

[ivory dust]

UHHHHHHHHHHHHHH

prolly?

But wont they be iso

to Zp

[hidden haven]

Yep

[ivory dust]

so not really

Ohh so thats why

Char(GF(p)) = p

[hidden haven]

Yeah

[ivory dust]

or Char(GF(p^n)) = p?

[hidden haven]

Both

Try to see why

[ivory dust]

cuz fermats thm

i think

char is a number say n?

such that element of the ring?

[hidden haven]

Didn't get the last part

[ivory dust]

a+a+a+...+a = an = 0

[hidden haven]

0

Or you can just define it using 1

It is the least p such 1+...+1 = 0, where the sum is taken p times

A better way to define it is to use kernel of the unique map from Z to the field

[ivory dust]

So in my case since the char is p in GF(p) for any g in GF(p)
then phi(g) = phi(sum from 1 to g of (1)) = sum from 1 to g of phi(1) = sum from 1 to g of 1 = g?

[hidden haven]

Characteristic is not defined for each element separately

Or you can say that all elements would have the same characteristic

In any field, not just GF(p)

[ivory dust]

wait i dont get how this follows from anything

other than phi is automorphic

g = sum from 1 to g of (1)

is that whats unique to GF(p)?

[hidden haven]

Yeah so phi(a+b) = phi(a) + phi(b)

[hidden haven]

Yes

[ivory dust]

Ohh

im not sure why

[hidden haven]

Any ring where such a thing happens (all elements are finite sums of 1) will have a surjection onto it from Z, so it'll be a quotient of Z

And quotients of Z are fields iff quotienting is done by a prime ideal

[ivory dust]

Ohh

ok kinda confusing but i kinda get it

wow did yall know

Galois died at age 20

HE DID SO MUCH

[hidden haven]

😵‍💫

[ivory dust]

He died in a sword duel

or wounds from it*

[hidden haven]

Gun

I think

[ivory dust]

Ohh

damn

crazy

[hidden haven]

[hidden haven]

I can reduce this to the case of A having 0 nilradical, using the homeomorphism spec(A) ≅ spec(A/ℜ), but I don't see how that helps

[ivory dust]

What lvl math is that?

that looks insane

isnt Hausdorff related to topology 😩

[hidden haven]

Commutative algebra it's considered late undergrad/grad

Yeah spec A is a topological space associated to the ring A

[ivory dust]

ohh damn..

r u grad student?

[hidden haven]

Starting MSc next month

[ivory dust]

wowza

what r u specializing in or focusing on studying? in math right?

[hidden haven]

Ye 😌

[chilly ocean]

"galois died" that's what they want you to think

[ivory dust]

genius lyfe

wait for Gal(GF(4)/GF(2))

[hidden haven]

[ivory dust]

is there any theorem

to deal w this

[hidden haven]

There are many

[ivory dust]

Cuz a theorem i see splits it into two

but thats like overcomplicating it

https://puu.sh/I0Wmp/5fe16b2650.png

[hidden haven]

Big hint ||An automorphism of GF(4)/GF(2) is exactly an extension of the embedding GF(2) → GF(4)||

[ivory dust]

Uhhh idk what embedding means

[hidden haven]

The inclusion map

GF(2) has a homomorphism into GF(4), which is injective because it's domain is a field. Also it's a unique homomorphism

So we think of GF(2) as a subfield of GF(4)

[ivory dust]

Hm

Could i write GF(4) / GF(2) as E/F with f(x) e F[X]

splitting field of GF(4) / F

cuz then i have this theorem

https://puu.sh/I0WrG/a2d94da1ba.png

And apparently the answer is Z2

but if i get order(Gal(E/F)) = 2

then theres only one such group

so im done

prob wouldnt work if order was higher

[hidden haven]

Yes

[ivory dust]

and multiple groups

[hidden haven]

Yeah

I was hinting at counting extensions

Which is also done in what you did, just hidden in the proof of the theorem you cited

[ivory dust]

ohh

[hidden haven]

GF(4) is GF(2) but you adjoin a root of a quadratic irreducible

The only quadratic irreducible over GF(2) is x²+x+1

It has 2 roots in GF(4)

[ivory dust]

ooh

[hidden haven]

Because it is separable

[ivory dust]

So GF(2)(a) = GF(4) where a root of f(x) = x^2+x+1

[hidden haven]

Yeah

[ivory dust]

a is a root in GF(4)*

?

[hidden haven]

Yeah

You have to do some extra work before you talk about GF(n) though

Showing that such a field exists iff n is a power of prime, and if it does, it is unique upto isomorphism

Otherwise you can't say "the" field of n elements

[hidden haven]

In this case it's easy enough to see both existence and uniqueness

[ivory dust]

GF(2)(a) = GF(2)[x] / (f(x))
f(x) is seperable in GF(4) but irreducible in GF(2)

[hidden haven]

Separability doesn't depend on the field

[ivory dust]

if its seperable in GF(4) that means it splits into linear factors hence it has 2 roots

o

[hidden haven]

f is separable in F_2 as well

Btw GF(n) is just called F_n, easier to type the latter

[ivory dust]

o yea

finite field

[hidden haven]

And separability doesn't say anything about splitting

So you can't deduce splitting from separability

[ivory dust]

what if f(x) seperation?

oh..

(x-a)(x-b)

if a and b are in F4 doesnt it split in F4?

[hidden haven]

f(x) is separable if in all extensions of your field, f(x) has no repeated roots

[hidden haven]

How would you show that they are though

Separability is about roots being distinct in extensions

[ivory dust]

ok kinda need to review some definitions

roots can be seperable, and so can polynomials

[hidden haven]

Ye, different notions, same word

[ivory dust]

a is seperable if its transcendental

or its irred polyn is seperable

[hidden haven]

You don't define separability for transcendentals

I think

[ivory dust]

i was looking at this

[hidden haven]

oh ok

[ivory dust]

so i just need to show

f(x) which is irred in F2[X]

has no repeated roots

Wait nvm

it has no roots in F2

[hidden haven]

Ye, there's a characterisation of separability using derivates

[rustic crown]

not if you use this def

[ivory dust]

i see it too

(f, f') = 1 implies seperable

[hidden haven]

Wait what goes wrong with this def

[ivory dust]

er

[rustic crown]

(x^2+1)^100 is separable over Q

[ivory dust]

if charF = 0, then all polyn are seperable

too bad that doesnt apply here

[rustic crown]

i kinda like the other def where you say separable <=> no repeated roots

it doesn't depends on the base field then

[hidden haven]

Isn't that what is being used

[ivory dust]

Ok but no repeated roots

in

what field?

like if i use quadractic formula

[hidden haven]

Is it it has repeated roots

[rustic crown]

but each irreducible factor doesn't

[hidden haven]

It's just that the gcd criterion works only for irreducible things

ohh

[ivory dust]

btw the txt book im using is Galois theory Rotman

kinda dense

but short

[rustic crown]

that just reduces to a weird thing for irred

[hidden haven]

Damn I see

[rustic crown]

f' = 0

hanny also used this weird def

[hidden haven]

Right

[ivory dust]

https://puu.sh/I0WCl/a992548e68.png

[rustic crown]

what's the problem btw?

[ivory dust]

Find Gal(GF(4)/GF(2))

i was trying to use this theorem

https://puu.sh/I0WrG/a2d94da1ba.png

To show its order is 2

hence it must be Z2

and i was told irred polyn of F2 is f(x) = x^2 + x + 1

[hidden haven]

Try proving that that's the only irreducible quadratic

[ivory dust]

wait does that matter tho

cuz ive seen

[hidden haven]

(while/after you do the problem)

[ivory dust]

isomorphisms

with different

irreducibles

of same degree

and they are all equivalent

like F / (f(x))

[rustic crown]

how do you define GF(4) btw?

[hidden haven]

Wait you've seen uniqueness of F_p^n's?

[ivory dust]

F4

Galois field of degree 2^2

or finite field 4 elements

[rustic crown]

do you just say F2[x]/(x^2 + x+1) ?

[ivory dust]

yes

[rustic crown]

so you don't know the existence yet, right?

or even uniqueness?

[ivory dust]

idk how to prove its seperable

well its irreducible cuz it has no roots in F2

so that shows existence

no?

and so it must be iso to F4

[ivory dust]

uh i think you told me about it above

and i understand it

[rustic crown]

x^4 - x is separable, cuz its derivative is -1

[hidden haven]

Ah alright

[ivory dust]

but its not irreducible

[rustic crown]

you don't need irred in your definition

[ivory dust]

wait thats irrelevant

[lethal dune]

$F[x]/\langle x^2 \rangle \simeq F[x]/\langle x^2-1 \rangle$ iff $F$ is a field of characteristic 2

[hidden haven]

Well I told you about uniqueness of F_p

[rustic crown]

(f, f') = 1 if and only if f has no repeated roots

[cloud walrus]

Ryuzaki

[lethal dune]

help me solve this one

[hidden haven]

And just stated uniqueness for higher powers

Alg3 hw problem

I think

[rustic crown]

so we know x^4 - x has no repeated roots, which means non of its irreducible factors have any repeated roots, if they had any they would be roots of x^4 - x

or you can simply say x^4 - x = x(x+1)(x^2+x+1)

[lethal dune]

when you guys are done that is

[rustic crown]

notie that (x^2 - 1) = (x-1)^2

just translate by 1

[lethal dune]

in ch2 yes

need the proof of the converse

[rustic crown]

just show its false in other char

[lethal dune]

what I did was f(x^2)=p(x)(x^-1) => f(x)^2 = p(x)(x^2-1)

then f(x) is either (x^2-1)*k(x) which case it's zero so can't be isomorphic

only other case is that x^2-1 is a square

but idk if my steps are right or not

[rustic crown]

what is f here?

[lethal dune]

an isomorphism ( we assumed they are iso, so there is one such f)

the converse

[wooden ember]

What does F_4 even mean?? Z/4Z isnt a field

[rustic crown]

yea so what you're saying is f(x)^2 = 0

[wooden ember]

oh sorry discord wasnt loading i didnt see the convo

[rustic crown]

ye it means F2[x]/(x^2+x+1) a field with 4 elements

[ivory dust]

these are the 3 irreducible factors f(x) = p1(x)p2(x)p3(x) so F2/(p3(x)) somehow = F4? and by that thm

nvm ill ponder it

[golden pasture]

Suppose some degree d polynomial f(x) is irreducible mod p, what is Fp[x]/f(x)

[rustic crown]

how do you get this?

[hidden haven]

,sullycount det

[golden pasture]

that isnt even a sully-able msg

tfw

[ivory dust]

p^d?

[hidden haven]

nvm the mods hate fun 😌

Ye lo

L

[golden pasture]

the finite field with p^d yup!

[lethal dune]

assume x^2-1 is not a square, then there must be one term inside p(x) because on the we have f(x)^2.

[hidden haven]

Det's first sully?

[lethal dune]

so p(x) = k(x)^2 (x^2-1)

[rustic crown]

wait just a question, when you write (x^2 - 1) you just mean modulo that?

i was reading it multiplication 🥲

[lethal dune]

ya i'm omitting that for conv

x^2 is 0 in F1, so must be mapped to 0 in F2 too

i'm writting 0= p(x)(x^2-1) in f2

[rustic crown]

yea so we get f(x)^2 = 0 in the second ring, how does this help?

oops i called it a field xD

[lethal dune]

nice

0 in F2 means p(x)(x^2-1) for some p

as F is iso, f(x^2) = f(x)^2

i.e. p(x)(x^2-1) is a square

[rustic crown]

yep

[lethal dune]

so.

[lethal dune]

.

bcz x^2-1 is irreducible otherwise

[rustic crown]

no

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

its never irred

[lethal dune]

but we don't know that ch=2 yet

we are proving ch=2

[rustic crown]

yea, and to say not squrare => irred we need ch2

[lethal dune]

we assumed the isomorphism and we are showing ch=2

[hidden haven]

This holds in any char

[rustic crown]

but this factorization is true anywhere

[hidden haven]

[lethal dune]

oh yes

that does hold but not a square

unless ch=2

[rustic crown]

in ch 2, -1 = 1

[lethal dune]

i might have misused the term, nvm

f(x) = sqrt(p(x)(x-1)(x+1))

[rustic crown]

sqrt

[lethal dune]

p(x) must contain factor (x-1) and (x+1) unless x^2-1 is itself a square

[rustic crown]

yea

[lethal dune]

so

we can write f(x) = k(x)(x^2-1) for some k

no?

[rustic crown]

ype

[lethal dune]

but x is not zero in F1 so f(x) can't be zero in F2

[rustic crown]

yep

[lethal dune]

contradicts the iso assumption

so (x^2-1) must be square

hence ch=2

[rustic crown]

[lethal dune]

but can I even go from x^2-1 is square to ch=2?

ig ye

yes* nvm

[rustic crown]

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

if this a square then you need to have x+1 = x-1

2 = 0

[lethal dune]

yes

tx

[rustic crown]

if you wanna avoid writing all that, just say that if char is not 2, then (x+1) and (x-1) are coprime which makes the second ring iso to F x F which has no nilpotent elements (non-zero ofc)

[lethal dune]

nice

should have said it earlier,

[hidden haven]

Repostin

[nova plank]

AM?

Wait 2 months for me to catch up then I can help

[hidden haven]

nyes 😌

[unreal portal]

So it would appear that I have a misunderstanding when it comes to quotient rings, I'd appreciate some help clarifying where my thinking went wrong. Let's say I have $R=\mathbb{Q}[x,y]$ and I take the quotient with the ideal $I=\langle xy-1\rangle$: $Q=R/[xy-1]$. And then say I pick some second arbitrary ideal in $R$, in this case $J=\langle 2x^2-4y+1, 3x^3-2x^2 + y^2-1\rangle$. I've verified in Sage that if you send each generator of $J$ to $Q$ and take a groebner basis, you end up with the ideal $\langle 1\rangle=Q$. Now, my thinking was that because we get the ideal generated by 1 in $Q$, the original ideal must either contain 1 or $xy$ because of the ideal $xy-1$ that we're modding out by. But this doesn't end up happening when you try it computationally in Sage. Why is this the case?

[cloud walrus]

cgodfrey

[rustic crown]

wait for a few months until i read about grobner basis 😓

[unreal portal]

The groebner basis isn't relevant to the question, I think, it's just a way to compute a 'better' representation of the ideal

[wooden ember]

how would one go about labelling vertices/edges/faces of a dodecahedron an icosahedron to make the isomorphism they have with a5 clearer?

it's easy to do with the cube and octahedron for example: label opposite vertices in the cube and opposite faces in the octahedron to get 4 each time, clearing up the relationship with A4 a little

but opposite faces/vertices in the dodecahedron/icosahedron respectively come in 6

(and they correspond more clearly to the 6 sylow 5 subgroups of A5 than to some 5 element set that A5 acts on)

[rustic crown]

oh okie, if i understand it right, then what we're saying is (J + I)/I = R/I, that is so we need to show that the ideal J + I contains 1 or xy, that may not necessarily be the case for just J as sage says.

[unreal portal]

but why wouldn't that be the case? I don't see how J + I contains neither 1 nor xy, and yet (J + I)/I is the ideal generated by 1

[rustic crown]

J + I has to contain 1 or xy if your arguement/calculation about the image of the ideal J to Q is (1) is correct

i don't know how to use sage very well, so maybe try once with J + I.

what i was saying is that J may not contain 1 or xy but the bigger ideal J + I does

[unreal portal]

okay, I see

I'll play around with it

[rustic crown]

notice that dodecahedron has 30 edges that's a multiple of 5!!!!!!!!!! we need a nice way to group 6 edges

and if you stare at a dodecahedron for long enough

there is only one natural choice!

(i did that in like 2019 or something , very fun)

make a dodecahedron with origami or something and have fun

else i'll spoil you

||so the argument is basically you can achieve any even permutation of these colors||

||if you're seeing these messages, and can't find the pattern, look at the yellow color, i think it its in a very nice position to see the pattern||

||to complete the argument, we need to identify all the rotations of this dodecahedron, there are 5 ways to choose the color of edge at the very front! (yellow in the image) once you choose a color, notice there are 6 edges of that color and you can bring any of them to the front, also you can rotate by 180 degrees. In total there are 2 * 6 ways to rotate if a particular color is at the front. its not hard to see that all these 12 are different. This will give you a total of 60 stuff in S5, which must to be A5! and this would complete the proof||

[crimson falcon]

So, not sure if this is the proper channel. Basically i want to understand what would be the pre-requisites to tackle a curse with this topics and in particular if i need any projective geometry for it :

Rings and commutative fields, . Notios of sub-rings ring morphisms, the notion of ideal and quotient rings. Divisibility in the frame of principal rings and factorials, ring extensions, galois' theory and is applications to algebraic equations.

[prisma ibex]

Just some linear algebra and intro to proofs

And “mathematical maturity”

[crimson falcon]

So i have a small shimmer of hope, thanks 😄

(or not, we will have to test the mathematical maturity part..)

[prisma ibex]

It’s probably fine

[rustic crown]

wait, if you're doing galois theory, then knowing group theory is a prereq i would say. it would help a lot in understanding the constructions for rings as well.

[urban ice]

I think this fits in this channel best. Is the dimension of $\bQ(\pi)$ (when considered as a $\bQ$-vector space) countable or uncountable?

[cloud walrus]

Syst3ms

[urban ice]

On second thought, it is countable

{1,π,π²,π³,...} is a basis

(i wasn't sure, i had the set of all sequences in mind, but the dimension of that is only uncountable because linear combinations have to be finite, so the countable apparently-basis doesn't work)

(but in this case, the result of a series involving πⁿ isn't included in ℚ(π))

[runic hemlock]

This is not a basis, though there exists a countable basis

[urban ice]

huh? why isn't it?

[runic hemlock]

how do you express $1/\pi$?

[cloud walrus]

Phorphyrion

[urban ice]

ahh, fair

oh yeah, you just have to enumerate ℤ

[runic hemlock]

you need to include all possible polynomial denominators

[urban ice]

well that is annoying

countable but not easily enumerable, i suppose?

anyway, i'm further along in this course, and i've stumbled upon this which confuses me

(where <f> is the ideal generated by f)

i get about everything here, except why they call alpha a generator

i don't see how its powers generate 𝔽_p[x]/<f>

(if that's the definition used here)

and if it's not, then i'd want to know why the term is used for those two things

[runic hemlock]

there are countably many polynomials with rational coeffs

[urban ice]

i already have trouble enumerating the polynomials with integer coeffs

(which not-so-coincidentally is precisely an enumeration of ω^ω)

[runic hemlock]

Do you agree that there are countably many linear polynomials in integer coeffs?

of the form ax+b

[urban ice]

yeah

[runic hemlock]

The same is true for any degree, there are countably many degree 17 polynomials

[urban ice]

yeah, yeah

i don't doubt that fact in the slightest dwai

[runic hemlock]

So its a countable union of countable sets, so countable

[urban ice]

that isn't what i was talking about

of course it's countable

constructing a way to enumerate it perhaps eludes the mind is all

anyway, that is really not what i want to focus on

[urban ice]

but rather this

[maiden ocean]

maybe this is a silly question but if f: A -> B is a ring homomorphism is f*(V(q)) = V(q^c) for all prime q in B not sufficient for f * to be closed?

Every closed set of Spec(B) is the intersection of some such V(q), no?

i ask because it seems like this should be sufficient to say that going up implies closed

but it is not

@upper pivot do you remember enough AM to understand this

I feel like i am going insane

[upper pivot]

I wish i was at my dorms

all my typed up sols are on my computer at dorms kek

let me see if i can solve this

[maiden ocean]

its not a problem im just confused

i feel like going up implies that f*(V(q)) = V(q^c) right

but then shouldnt this imply closed

[upper pivot]

is this an

image of intersection is not intersection of image

meme

[maiden ocean]

Oh

Yes

it is

F

[upper pivot]

rip

[maiden ocean]

Sad!

but i think that makes sense

[gritty sparrow]

This property will imply that the spec map is closed, the proof is a little tricky though

[vestal snow]

Let X be a ringed space and F be a finite rank locally free sheaf on O_X. Does this mean that there exists an n and a cover U_i such that F|_U_i is isomorphic to O_X|U_i ^ n or does this mean that there is a cover U_i and natural numbers n_i such that F|_U_i is isomorphic to O_X|U_i ^ n_i?

[gritty sparrow]

I think the second definition is the right one

[vestal snow]

To show a, I need to show that X(C,w_C) + h^0(C,w_C (x) w_C^*)

How do I do that?

[vestal snow]

By X(C,w_C) I mean the Euler characteristic of w_C

where w_C is the invertible sheaf of Serre duality

[gritty sparrow]

Here is a hint: the dual of O(C )is itself. So use serre duality on h^1(C,O(C))

[maiden ocean]

late reply but u need B to be noetherian i think

or Spec(B) to be noetherian as a space at least

[gritty sparrow]

No Atiyah does it for noetherian, but it is true in general i’m pretty sure

[maiden ocean]

monkaS is it really

I have found the stacks lemma

[gritty sparrow]

Lmao stacks

[maiden ocean]

nGroupoid stackspilled me ok its not my fault

do u now how much i had to read it for szamuely bc of the nonsense AG in that book

algebraic geometry be like

[vestal snow]

Can you tell me the solution?

I don't see it

[gritty sparrow]

How does szamuely introduce ag in the book? Does he actually start from the definitionsor is it all informal?

[maiden ocean]

He literally goes from defining what a scheme is in the start of chapter 5 to doing etale pi_1, specialization maps, tame pi_1, abelianization of pi_1, etc

in a chapter

It is utterly insane

He does this by piling like 20 definitions into the first section and then skipping over all the details in every proof

[gritty sparrow]

Well h^1(C,O_C)=genus=h^0(C,wc) (the second thing uses serre duality)

[gritty sparrow]

Holy shit, how does he expect anyone to understand lmao

[maiden ocean]

devil book

[vestal snow]

How did you get this equality?

I can see why 1 and 3 would be equal

but how did you get the equality of the genus to either one

[gritty sparrow]

h^0(C,O_C)-h^1(C,O_C)=euler char and h^0 term is 1

Therefore h^1=1-eulerchar=genus

[vestal snow]

Wait why is the h^0 term 1?

Isn't H^0(C,O_C) just the global sections of O_C?

Am I missing something here?

[gritty sparrow]

Yes, but this will be isomorphic to the underlying field k

[vestal snow]

Why?

[gritty sparrow]

This is just some fact, i forgot the exact proof

[vestal snow]

Do you recall what the theorem is?

[gritty sparrow]

Should be true for any proper connected scheme over an algebraically closed field

[vestal snow]

I don't think we're in an algebraically closed field

[gritty sparrow]

Hmm, then I’ll have to think about this some more, sorry about that

[vestal snow]

No prob.

This is exercise 18.5.A in Vakil

I'll post it again just so other people can see

[gritty sparrow]

Oh I think that the h^0(C,O_C)=1 should work for geometrically integral curves, they usually have the exact same sort of properties that curves over an algebraically closed field do

[vestal snow]

Do you know where I could find a proof?

[gritty sparrow]

I’m searching stacks rn

It will almost surely be there

[vestal snow]

Vakil doesn't prove it at least

So I would be interested to see if there is some other proof

[gritty sparrow]

https://stacks.math.columbia.edu/tag/0366

Found it

[vestal snow]

Do we know that the curve is geometrically connected and proper?

[gritty sparrow]

Yes it will be

Geometrically integral will give geometrically reduced and projective will give us properness

[vestal snow]

Cool thanks

For b), I need to somehow show that h^0(C, w_C (x) w_C^*) = 1

Do you know why this might be true?

[gritty sparrow]

Oh instead of geo reduced i mean geometrically irreducible which implies connected. Anyway, wc tensored with its dual is just O_C

[vestal snow]

Why is w_c (x) w_c^* = O_C?

Is this another theorem?

Wouldn't that mean that the inverse of an invertible sheaf is its dual?

[gritty sparrow]

Yes

[vestal snow]

Is this true in full generality?

[gritty sparrow]

Yep, true for any invertible sheaf

[vestal snow]

Do you know where I can find a proof of this?

[gritty sparrow]

Is this not in vakil?

[vestal snow]

Oh nvm

It is

It was a while back

Thanks for the help

[gritty sparrow]

Np

[fossil shuttle]

this is basic... we all learned this in 8th grade...

(Nb: I do not have any idea what the etale fundamental group is)

[maiden ocean]

it is like pi_1

but for schemes

[prisma ibex]

algebraic geometers really be like "a single point can have a lot of nontrivial covering spaces"

[urban ice]

Posting it again, but why is α called a generator here?

[hidden haven]

The quotient field is the smallest field containing α (and 𝔽_p which all characteristic p fields must contain anyway)

[urban ice]

ahh, in that way

[wooden ember]

I have a way to group them i think but it certainly doesnt feel natural

[hidden haven]

30 is a multiple of 5!!!!!!!!!

[urban ice]

I don't understand this, isn't the splitting field unique?

[hidden haven]

Yeah, that's what they are saying

Wait how are you defining the splitting field

It's just any extension of f such that f splits and it is generated by the roots of f over the base field

[urban ice]

(internet is having trouble)

And right after that they go ahead and use "the"

Is the splitting field unique, or just unique up to isomorphism? In the latter case, what would be an example of two isomorphic splitting fields?

[hidden haven]

Upto isomorphism

You can always rename elements

[urban ice]

an example would really help here

[hidden haven]

A splitting field of x²+x+1 over F_2 is F_2[x]/(x²+x+1), and another one is {0,1,a,b} where you define the operations mimicking the ones in the other one

Just renaming the new elements of the first one to some arbitrary elements a and b

[rustic crown]

why so

[rustic crown]

[wooden ember]

image is loading...

shit connection sorry

it doesnt feel natural because 2 of the groups are of one type while the 3 others are different

[rustic crown]

yea sad

[wooden ember]

like i dont have a single canonical grouping that "tesselates" the dodecahedron

[rustic crown]

but there is a better way to group ❤️

[wooden ember]

so there's a way with a single pattern?

[rustic crown]

yup

[wooden ember]

hmm alright ill keep trying

gotta admit i liked my grouping

[rustic crown]

if you have a miniature dodecahedron lying around, then play with it

[wooden ember]

im playing with my rough drawing lmao

[urban ice]

Ah I see, abstract symbols strike again

[wooden ember]

so it's a bit hard

[rustic crown]

what did you do to Senko!!

[urban ice]

greatness, is what i did

[rustic crown]

[rustic crown]

yep, very hard

i made one with origami and played with it for a few hours

[hidden haven]

[wooden ember]

when the original grouping i sent uploads lemme know what you think

[gritty sparrow]

I wonder if this is the shit euclid was doing all day

[urban ice]

is it true that for a polynomial P irreducible over F, F[x]/P is a splitting field of P?

[rustic crown]

lol okei

[gritty sparrow]

Also, hullo det

[rustic crown]

henno

[gritty sparrow]

Your uwu language is getting worse

[rustic crown]

[hidden haven]

F[x]/(P) is F but with one root of P adding in

You have to repeat this process with the new factorization of P's image in the quotient to get the splitting field

[rustic crown]

dwon't wowry i cwan mwake it a lwrott worse

[urban ice]

right, x+(P) is a root

[gritty sparrow]

Lwrott is i think the worst thing i’ve ever read

[rustic crown]

[urban ice]

so then, is the fact that 𝔽₂[x]/(x²+x+1) a splitting field sort of a fluke?

[gritty sparrow]

No, quadratics will always split

[rustic crown]

is there a way to type this directly from keyboard? 𝔽₂

[urban ice]

oh right i'm stupid

take one root out and you're left with a linear factor

[hidden haven]

If you use GBoard, you can import dictionaries, and there are latex dictionaries available on github

[gritty sparrow]

Wrong again, king

[hidden haven]

On phone that is

[rustic crown]

typing on phone is too hard

[wooden ember]

eyy just tried and checked with your spoiler: it's correct

the perspective i drew my dodecahedron in helped cause two of the groups look like a sort of vortex from that perspective

so it jumps out easier

ill try send

[rustic crown]

❤️