groups-rings-fields

[chilly ocean]

Talking in a linear algebra class about how such and such is a Lie group

(Not saying this person is doing this)

[chilly ocean]

(And also not because they are abusing sacred maths but because many well intentioned people have ruined themselves into thinking they know more than they do)

[magic owl]

yeah fair enough

you can study a lot of stuff w/o the word smooth or Lie though

which is nice

[chilly ocean]

Yeah There is a nice book

Matrix groups

An introduction to Lie groups

One of the springer undergrad ones

@chilly ocean

Hello

[chilly ocean]

Thanks

I’m gonna search it

[lavish gale]

I am confused. It is certainly a product of 3 cycles

i should show it can be written as a product of 5 cycles?

[sharp sonnet]

how is it a product of 3-cycles?

[unique juniper]

mmmmm

[viscid pewter]

a 3-cycle is a cycle which moves 3 elements

eg. (123)

[lavish gale]

sorry i meant to say 3 transposition

so (14)(13)(12)

ya its 4 cycle

i mixed up cycles and transpositions

[chilly ocean]

have i understood this correctly about dynkin diagrams

call the left node alpha the right node beta

that there is three lines tells use that n_alpha,beta times n_beta,alpha is 3

and that the arrow points to beta tells us that n_beta,alpha is bigger than n_alpha,beta

so then n_alpha,beta must be 1 and n_beta,alpha must be 3?

or possibly both with minus signs?

oh no

they are always negative off the diagonal

[prisma ibex]

Right the Cartan matrix will be (2,-3;-1,2)

[kind temple]

exactly

[gritty hedge]

Given a group defined by a finite field GF(p), is there a simple way of finding the period of the subgroup formed by exponentiation a^i for some a in GF(p) ?

i know that the order of the subgroup is always a divisor of the order of GF(p) = p - 1

but is there a way of computing it precisely?

[chilly ocean]

May someone send some videos about Lie Algebra for me?

[chilly ocean]

Is this statement true or false ?

And how do we get to know ?

[gusty halo]

Do you know that every non-unit in a ring is contained in some maximal ideal?

[unique juniper]

If $F_1$ is the fixed field of $Aut{K/F}$ then why is $[F_1 : F] = 1$

[cloud walrus]

Yes ツ

[unique juniper]

what im trying to do is show that K:F = #Aut(K/F)

[chilly ocean]

F1 subset F?

[delicate orchid]

Or in the case of a local ring, there’s only one maximal ideal so all non-units must be contained in there

The proof follows from that

[chilly ocean]

Yes yes l know that

[chilly ocean]

I am still not able to understand how to follow after this.

a is a non unit

The a is contained in the maximal ideal

u is a unit

Then how is (u-a) is a unit ?

[gusty halo]

assume its not

then its also contained in the same maximal ideal

do you see the contradiction?

[chilly ocean]

No I still do not see. Why is there be a problem if it is in the same maximal ideal.......?

[gusty halo]

then u = a + (u-a) is in the maximal ideal

but maximal ideals must be proper ideals

[chilly ocean]

It means u (which is a unit) will get inside the ideal right ? Is that the contradiction ?

[gusty halo]

yes

[chilly ocean]

Okay

Thank you. Now I understand.

[chilly ocean]

yo how to show M is generated by kerf + img? I showed their intersection is trivial but kinda not sure about the other condition

[gritty sparrow]

For any m in M, m= (m-gf(m))+ (gf(m))

The left part is in kerf, the right part is in img

@chilly ocean

[chilly ocean]

oh ye I see ty

[gritty sparrow]

No

Problem

[chilly ocean]

what?

o

no problem ok

[gritty sparrow]

Yeah

[vestal snow]

Is S_n here just an abelian group or is it also a ring?

[gritty sparrow]

Not a ring, for example in polynomials the degree n terms are not a ring

[hot lake]

it's an abelian group

[vestal snow]

Graded rings are weird

[hot lake]

how are they weird

[vestal snow]

Multiplication behaves weirdly

[vast quiver]

There’s multiplication defined on the entire ring S, but each S_n is just an abelian group. I think a good example to think about graded rings is polynomials over Z. Then, each S_n is homogenous polynomials of degree n. Then, it’s clear that x*x=x^2 is the product of two things in S_1, but their product isn’t in S_1 (it’s in S_2)

[vestal snow]

In this exercise, is I_n supposed to be interpreted as a subgroup of S_n right?

[hot lake]

yes

[gritty sparrow]

Yes

[vestal snow]

Because there is no ring structure on I_n. Thus, it can't be an ideal

[hot lake]

hopefully, "homogeneous ideal" is defined somewhere

in between those two pictures

[vestal snow]

It is

[vestal snow]

Sum and intersection are pretty easy because we can work with just the abelian group structure. For product, let I, J be homogeneous ideals. Let ij be a generating element of IJ. Then the n^th component of i is in I and the m^th component of j is in J. Thus, i_n*j_m is in IJ. However, i_nj_m is a homogeneous element of degree n+m. Since ij expands out to terms of the form i_nj_m (each of which is homogeneous and in IJ), we conclude that each component of ij is in IJ. Thus, IJ is homogeneous

Similar argument for radical

Can someone verify that my understanding here is correct?

[gritty sparrow]

The argument for radical is a little more complicated than the argument for product and sum if i remember correctly

I remember using induction on the highest degree term of each element of the radical

[restive shard]

Is it on-topic to ask questions about matrix norms here?

[oblique river]

Yeah that’s probably fine

You might get directed to linear altebra or analysis

[barren sierra]

Why is the order of 1 in Z infinite?

Why isn't it 1?

[viscid pewter]

the identity in Z is 0

additively

Z isn't a group multiplicatively

[barren sierra]

Oh

Right duh

[proud bear]

napkin 👀

[chrome hinge]

Hi

Im trying to prove that

is a UFD

I was going to take a class f+J (with J=(xy-1) ) and factor f in Z[x,y,z] and then try to show that any to elements in f+J have the same factorization up to mult by units, i wonder if something like that would work?

Im kinda lost with this ring tbh

[gritty sparrow]

I think that approach would be a little hard to pull off, my thinking is that this would be easier for you to prove: the ring is Z[x,x^(-1)][z]. Try to show Z[x,x^(-1)] is a ufd (infact you might like to know that any localisation of a ufd is a ufd) and then say that if R is a ufd so is R[x]

@chrome hinge

[chrome hinge]

hmm

nice approach

This is a localization with respect to what multiplicative set?

[gritty sparrow]

The set {1,x,x^2,…}

Remember, setting yx=1 is like “adding an inverse to x”

[chrome hinge]

Oh, yes youre right

Im fairly new to localizations but i think i can manage this thing

Tsm!!

[gritty sparrow]

Welcome

[lavish gale]

u two guys would make friends
pfps could not hav gotten any similar

[chrome hinge]

hahahah i actually found his picture too funny and changed mine a couple of minutes ago lmao

[chilly ocean]

Is this statement true or false

And how do we decided ?

[hidden haven]

It's true because the prime subring is Z_5

[chilly ocean]

This is the definition given in the book

[chilly ocean]

Can you explain the calculation by which we get to this conclusion.....

[hidden haven]

The smallest subring of R containing 1 is known to contain 2 things

0 and 1

From here it must also contain 1+1, 1+1+1,...

So this way you get all the elements of this form (1+...+1, and of -1-...-1)

The collection of such elements is always a subring in any ring

And in this case this collection is Z_5

And so must be the smallest subring that contains 1

[limpid edge]

why doesn’t n1 mean the product operation on the ring r between n and 1; is this just convention?

I mean n \ in Z actually might be a giveaway

but in general does ab mean b + … + b with a summands?

[chilly ocean]

Here n1 means 1 +1+1+1..... n times addition.

[limpid edge]

yes I understand but does your book denote product in a different way?

perhaps with a dot

[chilly ocean]

How is this collection Z_5 ?

Will not the collection become Z_625 ???

[hidden haven]

oh hold on

[chilly ocean]

If we are adding 1 to itself...

[hidden haven]

I was thinking of F_625

Ye you're right

It will become Z_625

[chilly ocean]

Order of 1 will be 625 right ?

[hidden haven]

Yep

[chilly ocean]

Okay okay I get it

So the conclusion is....

The given statement is false. Ringt ?

[hidden haven]

False

Ye

[chilly ocean]

😁

Ok

[weary terrace]

Hi, where can I ask about representation theory?

[scarlet estuary]

heres fine

[weary terrace]

Say we have a finite group $G$ of size $n$ with $n$ irreducible representations $\rho_i$ over $\mathbb C$. So by using the regular representation, each of the irreps is 1-dimensional.
By the character-conjugacy class correspondence, $G$ is abelian. Additionally, for each $i\in \mathbb Z, g\in G$, we have $\rho_i(g)^n = 1$ thus $\rho_i$ is a different n-th root of unity for each $i$.
Under these conditions, is it correct to say that $G$ is cyclic?

[cloud walrus]

RiesZ

[compact needle]

Every finite abelian group G has |G| irreducible representations (character-conjugacy class correspondence), so this will not imply cyclicity.

[weary terrace]

@compact needle thanks for the response!
Is it sufficient for $\rho_i$ to be faithful for all $i$ in order for $G$ to be cyclic?

[cloud walrus]

RiesZ

[compact needle]

Indeed

[sturdy marsh]

even if one of them is faithful

[compact needle]

Exactly

Not all of them can be, anyway

[weary terrace]

@sturdy marsh right of course

Do you guys have a weaker condition for G to be cyclic?

[sturdy marsh]

trivial group

[weary terrace]

lol

[compact needle]

I don't know of a nicer statement.

[sturdy marsh]

something which is weaker in principle probably exists

knowing all irreps is knowing Rep(G)

but cyclic-ness can be deduced from knowing just the group cohomology

but again all irreps in this case are characters so they might be the same information

[compact needle]

I mean there are silly statements like if the order of $\rho=|G|$ for any $\rho$, but that's just reshuffling the injectivity/faithfulness condition.

[cloud walrus]

Turgul

[weary terrace]

Thanks for the assistance, much appreciated 🙂

[sturdy marsh]

I think if you show that for all k | |G|, there are at most \phi(k) elements of order k, then G is cyclic

and this can be translated into characters

\phi is the euler totient fcn

[weary terrace]

I have additional information if you like

[compact needle]

For sure (I mean, G and G dual are isomorphic groups, so any statement about G is also true about its dual)

[weary terrace]

We also know that $\langle \chi_a \chi_i, \chi_j \rangle = 1$ for $j=i+1$ and 0 otherwise, where the $\chi$s symbolize the characters and $a$ does not divide $|G|$

[cloud walrus]

RiesZ

[compact needle]

You should be able to use that to show that $\chi_a$ has order $|G|$.

[cloud walrus]

Turgul

[weary terrace]

Well it's pretty obvious but how would it imply (if any) that G is cyclic

[compact needle]

One argument: when you know that your group is abelian (number of irr reps = |G|), then the group of characters (= representations) is isomorphic to the group itself. So if there is a character of order |G|, then G itself must have an element of order |G|

But for $\chi$ to have order $|G|$, that means that there is some $g$ such that $\chi(g)^n \neq 1$ for any $1<n<|G|$

[cloud walrus]

Turgul

[compact needle]

That g will be a generator of G

[weary terrace]

So if I'm not mistaking, it is enough to show that the set of characters contains all n-th roots of unity, which we've already done in the beginning. It follows from $\chi_i^n=1$.

[cloud walrus]

RiesZ

[weary terrace]

and the fact that G has n irreps

[compact needle]

You need to know that no smaller n than |G| works

If G is the Klein 4 group, then all characters will have max order 2

[weary terrace]

But it must be the case.. we have all n roots of unity

[compact needle]

I may be misunderstanding your assumptions

[weary terrace]

Maybe I'm missing something so let's review:
G has n irreps of dimension 1 each. G has order n and G is abelian

[compact needle]

The Klein 4 group has 4 irreps, yet each takes values only 1 and -1

[weary terrace]

So the conditions I layed here don't necessarily imply that the irreps are n-th roots of unity?

[compact needle]

Because in this case, $\chi^2=1$ for each $\chi$. You need to know that the min n satisfying $\chi^n=1$ for all $\chi$ is $n=|G|$.

[cloud walrus]

Turgul

[compact needle]

Much like if you want to show that the lcm(a,b) of 2 integers is ab, it is not enough to know that a and b both divide ab, they might divide something smaller, too

[weary terrace]

Ok.. so you agree with what I wrote in this comment?

Just verifying that I understand 😛

[compact needle]

Yes, just knowing that G is abelian of order n does not mean that the characters will hit all n-th roots of unity. It does mean that all values of a character will be n-th roots of unity, just maybe not all of them. You get all of them exactly when G is cyclic

[weary terrace]

Great! that was really helpful

[compact needle]

In general, the largest root of unity you get is the size of the largest cyclic subgroup

[delicate orchid]

ok just as a sanity check, you can always equip Hom_F(V, V) with a ring structure via standard matrix multiplication and addition right?

(where F is a field and V is a vector space)

[hidden haven]

If V is finite dimensional you can say matrix multiplication

But matrix multiplication is just linear map composition

And that notion extends to infinite dimensions

So yes if V is finite dimensional and yes* if not

[delicate orchid]

ah ok, thanks!

[hidden haven]

[delicate orchid]

I was only thinking about finite dimensions anyway but that is a very good edge case I should consider

[hidden haven]

[sinful mirage]

i'm trying to show this

what am I doing wrong/how could I proceed?

[steady axle]

Is it true even for infinite groups that for a subgroup H, $xHx^-1 \leq H$ then $x^-1Hx\leq H$?

[cloud walrus]

bert

[delicate orchid]

my intuition is telling me yes but I'll need to go actually think through a proof

if x is in N_H(H) it's definitely true as xHx^-1 = H

[compact needle]

Sadly, this is not true

At least not in general

[steady axle]

ohh

should I look into subgroups of GL_n(R) for counterexample?

[compact needle]

Counterexample: Let $G$ be the free group on generators $x$ and $y$. Let $H$ be the subgroup generated by $x^nyx^{-n}$ for $n \geq 0$. Then $xHx^{-1} \leq H$ but $x^{-1}yx$ is not in $H$.

[cloud walrus]

Turgul

[compact needle]

See https://math.stackexchange.com/questions/75613/the-set-of-all-x-such-that-xhx-1-subseteq-h-is-a-subgroup-when-h-leq-g/75620#75620 for this and another counter example in GL_2(Q)

[steady axle]

great post

[maiden ocean]

this should be in #advanced-analysis

shouldnt it

lol

[delicate orchid]

I see the word cauchy get it OUTTA HERE

[past temple]

what about cauchy's theorem

[sinful mirage]

Given $(U \psi)(x',t')=e^{if(x,t)} \psi(x,t)$, where $(x',t')=(Rx+vt+a,t+b)$, how do I compute $(U\psi) (x,t)$?

[cloud walrus]

ProphetX

[cloud walrus]

ProphetX

[sullen zodiac]

In the following exercise, what is meant by an isomorphism $Spec A \to Spec A'$? They're topological spaces, so are they asking for an isomorphism of sheaves? Any further explanation on what they're trying to get at would be greatly appreciated!

[cloud walrus]

Kenya

[prisma ibex]

Isomorphism of locally ringed spaces (so a homeomorphism of underlying spaces that preserves the structure sheaf)

[unique juniper]

how do they know that Fpn is isomorphic to that quotient field?

[golden pasture]

show that x->\alpha(x) is both injective and surjective

alternatively

[unique juniper]

i mean $\mathbb{F}_{p^n} \cong \mathbb{F}_p/\langle f_1(x) \rangle$

[golden pasture]

For some intuition do the same proof and show like

[\frac{\mathbb Q[x]}{x^2+2}\cong\frac{\mathbb Q[x]}{x^2+8}]

[cloud walrus]

Yes ツ

ari 十年生死两茫茫，不思量，自难忘。

[unique juniper]

i think this is the part i dont understand

hm actually

is it because the splitting field for f(x) over Fp would be at most n which is F_pn over F_p

[golden pasture]

Essentially $\left[\frac{\mathbb F_p[x]}{q(x)}:\mathbb F_p\right]=\deg q$

[unique juniper]

p or n?

[golden pasture]

p

explicitly F_p[x]/q(x) is a vector space over F_p generated by 1,x,x^2,x^3,...,x^{deg q-1}

so it has cardinality p^{deg q}

i just realized

terrible idea to use p for prime and polynomial

[cloud walrus]

ari 十年生死两茫茫，不思量，自难忘。

[unique juniper]

oh yeah that makes sense

[chilly ocean]

Are you sure its false?

[unique juniper]

well

[chilly ocean]

Idk the definition of Z+ contain 0

[hidden haven]

Z^+ is the set of positive integers?

[chilly ocean]

it depends on how the author defines it

I've seen some books including it

[unique juniper]

inverses for all elements dont exist

[hidden haven]

integers arent invertible in Z^+

other than 1

you divided by g

[chilly ocean]

If you replace Z^+ with Q than that'll work

[hidden haven]

you cant do that

[chilly ocean]

Q*

[hidden haven]

like, 1/2 is not in Z+

so you cant say 2(1/2) = 1

which seems to be what you did

2 times any integer will be even for example, so cant be 1

identity is 1

under multiplication

ye

I mean theres no other integer such that g1 = 1g = g for all g

I dont get what you mean

1/g doesnt exist

ye

ye I just showed that, theres no integer n such that 2n = 1

so 2 doesnt have an inverse

[hasty elk]

ok sorry im posting this here but i find it hilarious

[paper flint]

UCB

[cursive temple]

i have no clue what is to be shown

what is n, and how are the elements ordered

this is from jacobson

[carmine fossil]

n=n_1+n_2...n_q

There are q partitions in total and ith partition has n_i elements

By that condition,you also have the first q_1 partition's have same number of elements,while the next q_2 partition's have the same number of elements and so on

Hopefully this is more clear

[cursive temple]

it is

thank you

is that which edition

[carmine fossil]

That's 3rd edition of DnF

[cursive temple]

oh lol

[steady axle]

can a linear transformation R^2->R^2 map a circle to a triangle?

[hidden haven]

no, a linear transformations maps straight lines to straight lines

nvm thats the opposite direction

oh right any such transformation has to be invertible

so you can using the above fact

[steady axle]

can there be some kinda smoothness argu

[hidden haven]

yeah that too

the triangle has no smooth parametrization

but the circle does

[steady axle]

is this easy to prove

[hidden haven]

and composing such a transformation with the parametrization of the circle gives a parametrization of the triangle

[hidden haven]

you look at the derivatives of any such parametrization at the vertices on the triangle

ah a smooth parametrization isnt enough

use local diffeomorphism

then the derivative of the transformation at the vertices has to be non zero for the inverse to be smooth

but it cant be non zero because you get that it must point along the directions of the 2 sides

but the 2 sides are linearly independent

but theres a map from R to the circle which is a local diffeomorphism

and linear transformations are also local diffeomorphisms on R^2

[steady axle]

what did you prove here exactly

[hidden haven]

so the derivative is non zero at a vertex

you can take the derivative from 2 directions separately

approaching for one side you get that the derivative has to be pointing along the side of the triangle you are approaching from

same for the other

but the derivative cant point along 2 independent sides

unless its 0

[steady axle]

yea so you proved that the derivative is not continuous right

[hidden haven]

no, that it is 0

it is a vector that points along 2 independent directions

[steady axle]

but if its 0 its discontinuous?

[hidden haven]

like if I told you that you have a vector in R^2 which is parallel to both axes

then you would get that it is 0

no not necessarily

but then its local inverse wont be differential at that point

why do you need a smoothness argument btw?

[steady axle]

i dont

just want to understand it now

[hidden haven]

I see

[steady axle]

yea

I get it now

just remembered that there can be inverse with derivative being singular but it cannot be differentiable

[hidden haven]

yeah

[steady axle]

thanks

[hidden haven]

[steady axle]

btw are you doing bachelors ?

[hidden haven]

finished it in april

[steady axle]

so what are you doing now

if i may ask

[hidden haven]

celebrating vacations will do MSc from same place next

[steady axle]

great

[hidden haven]

[steady axle]

you have lot of energy to do math

[hidden haven]

This server gives me energy, I see others studying and get motivated

[steady axle]

nice

[unique juniper]

so all of gal(K2/K1 intersect K2) will be identity when restricted

sigma may not be ?

[maiden ocean]

The identity on what?

[unique juniper]

on K1 intersected with K2

[maiden ocean]

That's true by definition

assuming K2 is galois over K1 cap K2 from context

[unique juniper]

yes

[maiden ocean]

What this is saying is that for each automorphism sigma of K1 that fix F there are |Gal(K2/K1 cap K2)| automorphisms of K2 fixing F which agree with sigma on the intersection

[unique juniper]

i dont see how that is true

[cloud walrus]

Moth In Shambles

[unique juniper]

wdym to k2

[maiden ocean]

An automorphism tau of K2 which agrees with sigma on K1 cap K2

What the statement is saying that the number of extensions of any given sigma is exactly the number of automorphisms of K2 fixing K1 cap K2

[unique juniper]

i dont understand tbh

[maiden ocean]

Try drawing a diagram to see whats going on

Which part do you not understand

[unique juniper]

the part where they know how many elements there are that are equal when restricted

[maiden ocean]

do you understand the concept of extending the restriction of sigma on the intersection to all of K2

how any such extension gives a tau in Gal(K2/F) with tau | K1 cap K2 = sigma | K1 cap K2

by definition

[unique juniper]

yes

[maiden ocean]

so you're trying to show that the # of such extensions is exactly equal to the number of automorphisms of K2 fixing the intersection

[unique juniper]

yes

[maiden ocean]

So try thinking about how you can find a relationship between these automorphisms and extensions

for a bit

hint:

Gal(K1 K2/K1) is isomorphic to Gal(K2/K1 cap K2)

and so ofc they have the same order

@unique juniper if you think about it for a bit and still struggle i can explain it fully

[unique juniper]

how do you know this?

[maiden ocean]

You have an injective restriction homomorphism from Gal(K1K2/K1) into Gal(K2/F)

Or tbh too lazy to type this out I will just post the proof and you can look at it

Here I found some notes

from the internet

[unique juniper]

ok

but how can that help?

[maiden ocean]

The fundamental theorem of galois theory tells us that galois groups of galois extensions are quotients, right?

for F in L in E, Gal(L/F) is Gal(E/F)/Gal(E/L)

so thinking about F in K1 in K1K2 we have a correspondance between sigma in Gal(K1/F) and cosets of Gal(K1 K2/K1) in Gal(K1 K2/F)

Sorry I lagged out lol but

idk if you get it yet

[unique juniper]

sorry still dont

i agree with everything youve said so far

[maiden ocean]

the idea is that we have an isomorphism of Gal(K1/F) onto Gal(K1 K2/F)/Gal(K1 K2/K1) by taking the inverse of the restriction

i.e extensions

each coset is mapped to some sigma in Gal(K1/F)

thru restriction

so each sigma is mapped to a coset, and the elements of the coset are extensions of sigma

and each coset in turn has size |Gal(K1 K2/K1)| = |Gal(K2/K1 cap K2)| using the isomorphism

[unique juniper]

ah

thanks!

[maiden ocean]

np

my explanations kept getting derailed cause my wifi went out

[chilly ocean]

If we're given any group G, what does the notation G̃ corresponds to?

[magic owl]

That is def author-specific

I doubt there is a very canonical meaning

[chilly ocean]

Righty thanks

[vestal snow]

How do I prove that P^n_k is irreducible?

We know that A^n_k is

Suppose C U C' = P^n_k in P^n_k (where C and C' are closed)

[uncut girder]

profinite completion? dual group?

[vestal snow]

Then intersecting with each copy of A^n_k proves that every copy of A^n_k is contained in either C or C'

But that doesn't show that one of C or C' must equal P^n_k

[sturdy marsh]

C and C' are closed

[vestal snow]

And?

[sturdy marsh]

A^n \subset C

what is the closure of A^n

[vestal snow]

Is the closure of A^n the entire set?

[sturdy marsh]

yes

[vestal snow]

That's new info for me

Why is the closure the entire set?

[sturdy marsh]

try proving it

[vestal snow]

Can you give me a hint?

I'm not good at topology

[sturdy marsh]

assume the copy of affine space is given by X \neq 0

assume there is an open set contained in X = 0

[vestal snow]

Wait what does X = 0 mean?

[sturdy marsh]

the vanishing locus of X

[vestal snow]

so V(X)?

[sturdy marsh]

D is the nonvanishing locus right?

yeah

[vestal snow]

Wait hold on

what is X here?

[sturdy marsh]

one of the projective coordinates

[X : Y: Z: ...]

[vestal snow]

Can you do it using the gluing construction of P^n?

We haven't covered projective coordinates yet

[sturdy marsh]

okay say U_0 is one of the copies of A^n

[vestal snow]

Okay

[sturdy marsh]

assume we have some open set V in P^n

it intersects one of the U_i non-trivially

as these cover the space

[vestal snow]

Yes

We need to show that it intersects all of them non-trivially

[sturdy marsh]

well we're interested in showing that U_0 is dense

as this would imply that the closure is the entire thing

[vestal snow]

But since U_0 was an arbitrary copy, might as well show that for all U_i

[sturdy marsh]

yes, this proof will show that it will work for all the U_i

you're just relabelling

[vestal snow]

Yeah okay

So we know that V intersects some U_j

[sturdy marsh]

right

the intersection is non-empty and open in U_i

U_i is irred

so it is dense in U_i

but U_0 \cap U_i is also nonempty open

did that make sense?

[vestal snow]

Yeah

[sturdy marsh]

so they intersect

[vestal snow]

Why though

We know that U_0 and U_i intersect non-trivially

[sturdy marsh]

and the intersection is open

in U_i

[vestal snow]

and V and U_i do too

and it is open in U_i

[sturdy marsh]

and dense

both of them are dense in U_i

[vestal snow]

yes

[sturdy marsh]

U_i is irreducible

a dense set intersects any nonempty open set

[vestal snow]

why do we need this?

[sturdy marsh]

because it implies that any nonepmty open is dense

[vestal snow]

So since V intersect U_i is open and dense and U_0 intersect U_i is open, these intersect non-trivially

Is this the argument?

[sturdy marsh]

yup

[vestal snow]

got it

Thanks

Topology stuff is weird

Does the explicit construction of the underlying topological space of glued schemes ever come into play or is it enough to just think of it through the universal property?

[sturdy marsh]

This probably depends on what youre doing, but point set topology doesnt come into play a lot

a lot of it is to just make quasicoherent sheaves work

[vestal snow]

This exercise was labelled as an easy exercise in the book

And i spent almost an hour on it

[prisma ibex]

in some sense the Zariski topology is best not thought of as an honest topological space but rather a book-keeping tool, much as many other posets would be

[sturdy marsh]

Remember, topologies are just glorified semi-lattices.

If you have two semi-lattices X and Y, and a monotone function f from X to Y then an element a of X is a sufficient factor for b in Y if for any refinement of X W, refinement of Y Z and monotone function f': W -> Z that extends f, for any element w of W, w subs a => f'(w) subs b. Likewise an element a of X is a necessary factor for b in Y if for any refinement of X W, refinement of Y Z and monotone function f': W -> Z that extends f, for any element w of W, w subs a <= f'(w) subs b. An element a of X is a determining factor for b in Y if it is a necessary and sufficient factor. The map f is factorable if every element of Y has a determining factor in X. This means that there exists a function f*: Y -> X.

What it means in topology for a map F: X to Y to be continuous is that the induced map f = cl o image_F, from the closed sets of X to the closed sets of Y is a factorable map.

[vestal snow]

Topology copypasta

Hi Groupoid

[sturdy marsh]

there ya go

reacted for ng

[prisma ibex]

[chilly ocean]

can someone help me see why we do not require L to be a subalgebra of gl(V) for B and C?

[vestal snow]

@sturdy marsh I think I found a better way of doing it (it might be equivalent to the projective coordinate one). Proj k[x_0,...,x_n] = P^n_k. 0 is a homogeneous ideal. After that it's just following definitions

[surreal jacinth]

I have a question that may make sense to ask here. I'm working in analysis and have a problem where I'm looking at rational functions.
If U is a unitary matrix (whose entries are functions of x), and A is a vector whose entries are rational functions of x, and if the product UA is a vector whose entries are also rational functions of x, do the entries of U also have to be rational functions of x?

[gritty sparrow]

Look at the adjoint representation of L, that is solvable sub algebra of gl(L) then use corrolary A on the adjoint representation

[chilly ocean]

thank you that makes sense now

Is there a good way of making something into a lie algebra?

I want to show that the sum of two lie subalgebras need not be a lie subalgebra. I don't know an example of this, but I see that what goes wrong is - say S and T are our subalgebras - we have an element [s,t] not in S+T

So i am trying to make a lie algebra that goes wrong like this

Let V have a basis {s1 , s2, t1, t2, e1, e2}

define [s1,s2]=s1 , [t1,t2]=t1

[si,ti]=ei

Now S and T are sub spaces but their sum is not a subspace

But i have not checked that this is a lie algebra

is there an easy way to do this, or perhaps say something like "extend this to be a lie algebra"

[gritty sparrow]

Well i don’t know of a way of extending a lie algebra without defining the multiplication on all basis elements, but i do know of a condition that you can check once you’ve defined a multiplication on basis elements, to see if that actually is a lie algebra multiplication on all elements

It is given in the first chapter of jacobson’s lie algebra book, and it might be there in humphrey’s book as well but i don’t really remember too well

Also, as a more direct hint for your question, you can get an example of this sort of thing without trying to generate a lie algebra with some generators and multiplication. Some very natural examples exist

[unique juniper]

so uhh

[chilly ocean]

yes

[unique juniper]

$[\overline{\mathbb{F}}_p(x,y) : \overline{\mathbb{F}}_p(x^p , y^p)] = p^2$

[cloud walrus]

Yes ツ

[unique juniper]

so im trying to show that its dimension is p^2

but its not what im getting

i want to say its {1,x,...x^p-1} x {1,y,..,y^p-1}

[hidden haven]

So that gives you a spanning set of the extension as a vector space

degree <= p^2

And for the other inequality you can either try to prove that that same set is linearly independent

or you can use the fact that degree is multiplicative

and you extend this by adjoining x and y one by one, and prove that each step is degree p

[unique juniper]

ooh

yea

[vestal snow]

Show that affine schemes are quasi-separated (intersection of quasi-compact open sets is quasicompact open). Let U, V be quasicompact open sets of Spec A. Then U = U D(f_i) and V = D(g_j) (both indexed over finite sets as they are quasicompact). Then their intersection is U (D(f_i) intersect D(g_j)) = U D(f_ig_j). However, D(f_i g_j) is isomorphic to Spec (A_(f_ig_j))

Thus, U intersect V can be written as a finite union of affine schemes

This means that it is quasicompact

Is this proof correct?

[compact needle]

Looks good

[void knot]

I know that Cayley theorem states that every group is isomorphic to a subgroup of some group of symmetries. What are the applications of this theorem?

[vestal snow]

It unified the earlier theory of symmetric groups with modern group theory

Other than that, it is pretty useless

[void knot]

Ohh I see

[chilly ocean]

Is this statement true or false ?

And how do we decided

[magic owl]

You want to exhibit (x) as the kernel of a surjective ring homomorphism Z[x]->Z

Basically until you build up intuition any time you want to know whether R/I = S you should try to actually construct such a surjective map R->S with kernel exactly I.

[chilly ocean]

Okay I get the idea of constructing a onto homomorphism with kernel = (x)

[magic owl]

So, let's try to build one and call it f

do we know anything about f already?

[chilly ocean]

Wait

[chilly ocean]

This thing happens in case of groups right ?
Does this also happens in case of rings ?

[magic owl]

Yes rings have a similar first isomorphism theorem

[chilly ocean]

It was the 1st isomorphism theorem in case of group I guess

[magic owl]

https://en.wikipedia.org/wiki/Isomorphism_theorems#Theorem_A_(rings) @chilly ocean

that should take you straight to the correct part of the wiki page

but its theorem A for rings if not

[chilly ocean]

Ok

I will see the theorem and then come back

[magic owl]

feel free to ping me im just doing some other work

[magic owl]

Yes

[chilly ocean]

Ok I saw it. Now we can proceed I guess

[magic owl]

Do you know what the generators of Z[x] are (or what that means?)

We can still solve this without that but its much shorter

[chilly ocean]

I do not clearly rebember the actual definition

[magic owl]

then lets table that

Do you know what a generic element of Z[x] looks like?

[chilly ocean]

Yes I know that

[magic owl]

generic isnt a technical term i just mean like, what are the elements of Z[x] as a set

[chilly ocean]

Polynomial with integer cofficients

[magic owl]

Sure, so explicitly like $a_0 + a_1z + ... + a_nz^n$ right?

[cloud walrus]

MaxJ (Cali Surfer Arc)

[chilly ocean]

Yes yes

[magic owl]

Okay so we want to define $f(a_0 + a_1z + ... + a_nz^n)$

[cloud walrus]

MaxJ (Cali Surfer Arc)

[magic owl]

We know f (if it exists) is a ring homomorphism, so let's simplify a little

so we know

$f(a_0 + a_1z + ... + a_nz^n)=f(a_0)+f(a_1z)+...+f(a_nz^n)=a_0f(1)+a_1f(z)+...+a_nf(z^n)=a_0f(1)+a_1f(z)+...+a_nf(z)^n$

[cloud walrus]

MaxJ (Cali Surfer Arc)

[magic owl]

take a second to make sure that makes sense to you

[chilly ocean]

The line after the last equal to sign we get because it is a ring homomorphism ?

[magic owl]

Every line comes from that

First I used the additive distribution of f

then I used the multiplicative one twice

[chilly ocean]

Wait

Are not "z" only suppose to be place holders ?

[magic owl]

oh shoot

sorry

I'm reading a complex analysis thing right now so I have z on the brain

replace z with 'x'

[chilly ocean]

No no

I understand that

[magic owl]

anyway x is an element in Z[x] not really a placeholder

[chilly ocean]

But the "x" are just place holders

[magic owl]

You should think of x as a formal symbol

but it is still an element of the ring

and follows all the ring-rules

[chilly ocean]

x = (0,1,0,0,0,0..................)

[magic owl]

?

I'm not sure what that means

x is just a symbol

[chilly ocean]

Yes

[magic owl]

but its also an element of the ring

[chilly ocean]

But this is what it means

[magic owl]

i have no idea what the right hand side of that means

or what you're trying to say, to be honest. Sorry

[chilly ocean]

Infinite sequence of numbers.

[magic owl]

Yeah that doesn't mean anything to me

x doesn't mean anything

its just x

[chilly ocean]

No no wait

[magic owl]

If you have a polynomial $a_0+...+a_nx^n$ you can evaluate it at some number like $7$ by replacing $x$ with $7$ but this is really just a ring homomorphism with $x\mapsto 7$

[cloud walrus]

MaxJ (Cali Surfer Arc)

[sharp sonnet]

you can construct a polynomial ring as the sequences that are eventually zero, then x is the sequence (0, 1, 0, 0, ...)

that's what they mean max

(i think)

[chilly ocean]

Yes yes

That what I meant

[magic owl]

Oh wow

I've never seen that construction

I think thats honestly terrible lol

but all right

[sharp sonnet]

i think it's standard in ug algebra lol

[magic owl]

I never saw it

Anyway

[chilly ocean]

And I am asking from that perspective what that thing means

[sharp sonnet]

its often skipped because there is no reason to be that formal, but i see it in ug books a lot

[magic owl]

Okay so

(0,1,000...) is an element of your ring Z[x]

and it follows all the ring rules

[chilly ocean]

Yes

[magic owl]

so we can call it x and manipulate it like i did

im not sure I understand the confusion, sorry

[chilly ocean]

Wait

[magic owl]

Its not even more formal

its just bad

[chilly ocean]

f(z)^n = f(1)^n right ?

[magic owl]

no

lets use x to be consistent

that was my fault

but i dont wanna confuse

x and 1 are different elements of the ring, and there is no way to relate them only using addition and multiplication

so a ring homomorphism doesn't have to send them to the same thing

[chilly ocean]

Ok

What I tried to write

And what it looked like

Are two different things

[magic owl]

?

[chilly ocean]

I understood this much

We can continue after that

[magic owl]

$a_0f(1)+a_1f(z)+...+a_nf(z)^n$ okay so this is what we ended up with

[cloud walrus]

MaxJ (Cali Surfer Arc)

[magic owl]

sorry for the z's

if you look closely

the only "f" things left

are $f(1)$ and $f(x)$

[cloud walrus]

MaxJ (Cali Surfer Arc)

[magic owl]

so it suffices to define f(1) and f(x), because the ring homomorphism properties make everything else

[chilly ocean]

You use z

[magic owl]

does that make sense?

[chilly ocean]

It is better

[magic owl]

I mean we are talking about Z[x]

so we should use x

[chilly ocean]

Ok understand

[magic owl]

All right

let's remember our goal

what should the kernel of f be?

[chilly ocean]

f(1)=0 and f(x)=0

{1,x}

[magic owl]

Is that gonna end up being surjective?

[magic owl]

Hint: compute this

with your idea

[chilly ocean]

Ok

It will not be surjective

[magic owl]

Okay, so what did we say we wanted the kernel to be

what Ideal?

[chilly ocean]

Kernel of f should be (x)

[magic owl]

Okay, so what should f(x) be?

(you already guessed this one)

Maybe the better question for me to ask is, is 1 an element of (x)

[chilly ocean]

f(x)=0

[magic owl]

Yes

[chilly ocean]

No I think. Because there is no inverse of "x"

[magic owl]

Yep!

(in fact, if I is an ideal containing 1 it must be the whole ring)

[chilly ocean]

Yes yes l rebember

[magic owl]

Okay now compute this with what we know (leaving f(1) as it is)

[chilly ocean]

Every polynomial is going to a_0f(1)

[magic owl]

perfect

Do you have a guess for what f(1) should be?

[chilly ocean]

f(1)=1

[magic owl]

Yep

Okay so now I guess what is left is to convince yourself than f as we defined it is a homomorphism

and has kernel exactly (x)

and also that its surjective but thats not hard

I think you can do this yourself but feel free to ping if you have questions

[chilly ocean]

Surjective is done I think

[chilly ocean]

Okay I will think for some time

[chilly ocean]

This one will also work in the same way I guess

[chilly ocean]

I got this one btw. Thank you for your help

[magic owl]

No problem!

If you want a cool observation

If we require that ring maps be unital, i.e. f(1)=1 then a map Z[x]->S is determined completely by where we send x

if we drop this requirement, its determined completely by where we send 1 and x

[chilly ocean]

This observation will also hold for Z[[x]] right ?

Yeah ok I think it will.

Thank you

[latent anvil]

It's not clear to me that it's true for Z[[x]]

[devout crow]

I've read that the `infinite general linear group' over some field is $\text{GL}(\mathbb{F})$ , the direct limit of $\text{GL}n(\mathbb{F})$. this can be thought of as the group of vector space isomorphisms of $\bigoplus{n \in \mathbb{N}} \mathbb{F}$. with this in mind, is there a name for the group of automorphisms of $\bigoplus_{x \in \mathbb{R}} \mathbb{F}$, i.e. the uncountable direct sum of $\mathbb{F}$?

[cloud walrus]

∧res

[gritty sparrow]

For any vector space V, Gl(V) is the notation for the group of automorphisms of V

[devout crow]

hmm, is there anything that's easier to google?

[gritty sparrow]

I’m not really sure tbh

[loud root]

can someone briefly explain this reasoning to me please

[sharp sonnet]

if it were not irreducible, you could factor it as a product of two polynomials of degree 1, but then this would give you rational roots

[loud root]

thanks#

[vestal snow]

Let X be a scheme which is glued together from reduced schemes X_i. Then X is reduced. Let p be in X. Then p is in X_i for some i. Since each X_i is open, the stalk at p in X is the same as the same as the stalk at p in X_i. Since the latter is reduced (reducedness is stalk local), so is the former. This implies that X is reduced.

How do I make this argument rigorous?

Or rather, is this level of rigor okay when starting out with scheme theory or should I justify some stuff?

Is there any benefit to writing this out in detail?

[cloud walrus]

Yes ツ

[unique juniper]

$x^3 + ax^2 +bx + c = (x - \alpha)(x - \beta)(x - \gamma)$

[cloud walrus]

Yes ツ

[unique juniper]

why will adjoining 1 root and discriminant give a field that contains the roots?

[compact needle]

This looks good. There really isn't much more to say, given the definition of reduced for a scheme is stalk-local. If the question where that X is covered by affine X_i=Spec(A_i), then the A_i being reduced implies X reduced, you would want to convince yourself that A_i being reduced is the same as all of its localisations being reduced, which is less tautological.

[compact needle]

If you know some Galois theory, you can break the problem into 2 cases: when adjoining a single root already gives you the splitting field, and when it doesn't. When $F(\alpha)$ is already a splitting field, then it contains all three roots $\alpha,\beta,\gamma$, hence it contains the square root of D already (since this is just $(\alpha-\beta)(\alpha-\gamma)(\beta-\gamma)$.

[cloud walrus]

Turgul

[compact needle]

When $F(\alpha)$ is not the splitting field, then the splitting field is $F(\alpha,\beta)$ and is degree 6 over $F$, which will again contain $\sqrt{D}$. So all we need to know is that $\sqrt{D}$ is not contained in $F(\alpha)$, because $F(\alpha,\sqrt{D})$ is contained in the splitting field, and is the same dimension over F, so must be the whole splitting field.

[cloud walrus]

Turgul

[vestal snow]

Thanks!

[compact needle]

But the Galois group of $F(\alpha,\beta)/F(\alpha)$ must exchange $\beta$ and $\gamma$, which sends $\sqrt{D}$ to $-\sqrt{D}$, hence $\sqrt{D}$ is not contained in $F(\alpha)$.

[cloud walrus]

Turgul

[unique juniper]

thanks

!

got it

[vestal snow]

Show that a scheme is integral iff it is irreducible and reduced. Suppose $X$ is irreducible and reduced. Since irreducibility and reducedness are inherited by open subschemes, it suffices to check that $O_X(X)$ is an integral domain. Suppose $fg = 0$. I claim that $C_f \cup C_g = X$ ($C_f$ refers to the points at which $f$ vanishes (the image of f in the local ring $k(p)$ is zero)). $C_f$ and $C_g$ are closed (we know this). Let p be in X and let $U_p$ be an affine neighborhood of p with $U_p \cong Spec A$ for some ring A. Let $f'$ and $g'$ be the restriction of f and g to $U_p$. Under the isomorphism of schemes, $f'$ and $g'$ correspond to elements of $a_f$ and $a_g$ of A and p corresponds to a prime ideal $P$ of A. We see that $0=(fg)_p=(f'g')_p=f'_pg'_p$. The right side corresponds to $\frac{a_fa_g}{1}$ in $A_P$. This would imply that $a_fa_g=0$. In particular, $a_fa_g$ is contained $P$. WLOG, let $a_f$ be contained in P. This would imply that the $\frac{a_f}{1}+PA_P=0$. This corresponds to the image of f' in k(p) being 0. That is, the image of f in k(p) is 0. Thus, p is in $C_f$.

Using irreducibility of X, we conclude that (WLOG) $X = C_f$. This means that f disappears at every point. Since the scheme is reduced, this would imply that f is 0

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

Is this proof correct?

[gritty sparrow]

Yeah this looks correct, but since C_f is just the points x where the germ of f at x is in the maximal ideal, the union of C_f C_g being X is a lot simpler right?

In the sense we just have to say (f_x)(g_x)=0 hence one of them has to be in the maximal ideal

[tough raven]

I don't think it will determine the result of applying the map to infinite power series, although I don't have a counterexample.

[vestal snow]

Can someone verify my proof?

Since Spec(A) is a subset of X and the closure of n is X, we know that for every point p in Spec(A), all open sets containing p must contain n. In particular, Spec(A) contains n. However, we know that the closure of the [(0)] ideal in the spectrum of an integral domain is Spec(A). Thus, n is contained in the closure (in Spec A) of [(0)]. This implies that all open sets containing n in Spec A must contain [(0)]. Since we can "zoom in" on Spec A instead of all of X when computing the stalk, this proves that the stalk of X at n is the same as the stalk of Spec A at [(0)] which is the fraction field.

[vestal snow]

"If s vanishes on V, it also vanishes on U"

Why is this true?

the scheme is assumed to be integral

[compact needle]

Because the vanishing set of s should be a closed subset of U, but it vanishes on V and V is open in U, hence dense (via irreduciblity)

[compact needle]

This looks okay to me

[vestal snow]

Thanks

[unique juniper]

$[\mathbb{R}(\theta) : \mathbb{R}] = finite$

[cloud walrus]

Yes ツ

[unique juniper]

why will the minimal polynomial for theta have odd degree?

[hot lake]

what's theta

[unique juniper]

just anything

anything to make this a simple finite extension of R

[hot lake]

how many simple finite extensions of R do you know ?

[unique juniper]

hmmm

none

lol

[hot lake]

btw, the intermediate value theorem easily shows that any odd degree polynomial with real coefficients has a real root

[unique juniper]

yeah

im trying to understand why there are no finite extensions of R that have odd degree

except 1

[hot lake]

well how do you make field extensions

[unique juniper]

quotient by an irreducible element ?

[hot lake]

yes, quotient of R[x] by an irreducible polynomial

so if you want to show that R has no nontrivial finite extension with odd degree

you have to show that R[x] doesn't have any irreducible polynomial of degree 3,5,7,...

[unique juniper]

ohhhh

okkeeeee

got it lmao

[hot lake]

I'm glad you got the answer to why will the minimal polynomial for theta have odd degree?

[unique juniper]

😅

[hot lake]

do you know an irreducible polynomial of even degree

[unique juniper]

in R ?

[hot lake]

yeah in R[x]

[unique juniper]

yeah

[hot lake]

so then, you know a nontrivial finite simple field extension of R ?

[unique juniper]

yes

no

hmmmm

is this a trick question

[hot lake]

I don't think it's a trick question

[unique juniper]

C is the algebraic closure of R right?

so

no ?

[hot lake]

yeah

?

wdym, no

[unique juniper]

hmmm

C/R is finite and seperable

so simple

hmm

C = R(theta)

[hot lake]

well usually C is introduced as R[i], which makes it a simple extension

[unique juniper]

yeah

i thought you were asking me a trick question!!

[hot lake]

no I was just wondering why you said you didn't know of any simple finite extension of R

[unique juniper]

oh

lol

yes

i thougt

you were asking if there was any intermediate simple fields

or atleast thats what i thought

[hot lake]

ah

[frank fiber]

if $a/b \cong 0$ then $a \cong b$?

[cloud walrus]

Or x1

[frank fiber]

with a and b groups

[tough raven]

b is a subgroup of a?

[frank fiber]

yes, a normal subgroup

[tough raven]

If a/b \cong 0 then b = a

[final pasture]

lol the notation is so confusing

[native orbit]

what happens with stuff like ℝ[x]/(x-α)²

that is no longer isomorphic to an extension

i was told that this is now in the territory of algebraic geometry

[hidden haven]

Idk what kind of answer you expect lol

but it is a 2 dimensional vector space over R and an R algebra ig

[vestal snow]

Here's an different argument that I came up with: Let $\nu$ be the generic point of the scheme. Notice that it suffices to show that the map from $O_X(U) \rightarrow O_{X,\nu}$ is injective. Assume that the stalk of $s\in O_X(U)$ is $0$. For each $p\in U$, there is an affine scheme $U_p\cong Spec A$ for some $A$ an integral domain. For each $q\in U\cap U_p$, let $B$ be a basis element of $U_p$ corresponding to a basis element of $D(f)$ of Spec A. Then the map $O_X(B) \rightarrow O_{X,\nu}$ corresponds to the map $A_f \rightarrow Frac(A)$. Since the latter is injective, so is the former. Since the stalk of $s\in O_X(U)$ is $0$, we have that the stalk of the restriction of s on B is 0. However, this must mean that the restriction of s on B is 0. Gluing over $q \in U\cap U_p$, these glue to 0 in $O_X(U\cap U_p)$. Since $p$ was arbitrarily chosen from $U$, these then glue to give us that $s$ is 0.

[cloud walrus]

Have a Banana, Bitch

[compact needle]

This looks reasonable to me

[vestal snow]

Thanks

[vestal snow]

Also, i realized that my proof of the stalk being the same as the fraction field was unnecessarily long. Since generic points are unique and closure in open sets = closure in ambient set intersected with the open set (provided the set we are taking closure of is contained in the open set), n would have to be the generic point of every open affine subscheme. However, each spec A an open subset of X would need to have A be an integral domain. Thus, n would literally be the point [(0)] in each affine open. So the stalk at it would be the localization which is the frac field

[compact needle]

Nice

[tough raven]

I think it should be isomorphic to the dual numbers?

[native orbit]

wat

[tough raven]

Which are intuitively the real numbers plus infinitesimals "to first order" in that there is a non-zero element (x - alpha) whose square is 0, mimicing a first-order infinitesimal whose square is second-order and hence 0.

[tough raven]

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

[native orbit]

I see... so what about when this α is a multiple root of a larger polynomial

i.e. what is the behaviour of R[x]/P(x) where P(x) has a multiple root

[tough raven]

😳 IDK

[native orbit]

the Chinese remainder theorem

direct sum of extensions

I think

[tough raven]

🤔

That's fine if there are products of powers of different irreducibles

But for /(power of irreducible)

🤔

Seems a bit weird TBH

[native orbit]

well it was the source of my original question

[tough raven]

Maybe you could think of it as (dual numbers)/(irreducible - epsilon)
i.e. adding a 1st-order infinitesimal epsilon, and a "almost root" of the irreducible, alpha, so that p(alpha) isn't 0, but is epsilon?

Not sure if those are isomorphic though

[hidden haven]

Isn't alpha in R?

[tough raven]

No it was generalised from there

[tough raven]

.

[tough raven]

.

[hidden haven]

I see

[tough raven]

Is this right?

I think, more formally, it would state that
( R[x]/(x^2) )[y]/(P(y) - x) isomorphic R[y]/(P(y)^2)
at least for an irreducible P

[vestal snow]

Can someone explain the proof to me?

[vestal snow]

I figured it out

[sinful mirage]

can someone help me see why this remark is true?

[vestal snow]

Seems like an analysis question?

Though I'm not sure

Maybe try asking on the analysis channel too

[sinful mirage]

sure,might ask there,thanks

thought its algebra cause it's rep theory

[steady axle]

any hints for part b ?

[magic owl]

use part a to represent an element in gH as g(g^-1h^-1gh)

and then compute the nth power

[steady axle]

ya that works.
why didnt I think of that. it was not very difficult.

[chilly ocean]

I've been struggling to see why the reduced expression must involve all sigma_alpha?

intuitively something like, if it is missing sigma_beta then beta will not get sent to -beta, but thats not quite right

[vestal snow]

Show that locally Noetherian schemes are quasiseparated

Can someone help me out with this?

[vestal snow]

I probably need to use this somewhere, but the fact that noetherian spectrum does not imply noetherian ring seems to be causing problems

[latent anvil]

What do you mean by locally noetherian ring? A_p is noetherian for all p or Spec A is a locally noetherian scheme?

[vestal snow]

My bad

That was a typo

[latent anvil]

ah okay

[vestal snow]

I have shown that any affine open in the scheme can be written as a finite union of noeth. spaces

[latent anvil]

Spaces? Not schemes?

How did you do this?

[vestal snow]

Let U be an open affine

[latent anvil]

Yup

[vestal snow]

There is a cover U_i where U_i = Spec A_i for A_i noeth.

[latent anvil]

Yup

[vestal snow]

Intersect U with U_i

[latent anvil]

Yee

[vestal snow]

We can cover this intersection by open sets that are distinguished in both U_i and U

[latent anvil]

Yup

[vestal snow]

These must be noetherian because A_i is noetherian so (A_i)_f must also be noeth.

[latent anvil]

Yes

[vestal snow]

Union these covers over i to get a cover of U by distinguished open sets which are noeth.

[latent anvil]

Right

[vestal snow]

Use compactness of U to get a finite subcover

[latent anvil]

Yes, so U is actually noetherian

I'm confused by why you said you could only cover by noetherian spaces

[vestal snow]

This gives us that U can be written as a finite union of noetherian things

[latent anvil]

Use this

[vestal snow]

Where though?

[latent anvil]

You have U covered by distinguished opens

If U = Spec B call these opens D(f1),...,D(fn)

Since they cover U you have (f1,....,fn) = (1)

[vestal snow]

Then (f_1,...,f_n) = B

[latent anvil]

Yeah

And B_fi are all noetherian

[vestal snow]

Why?

[latent anvil]

You chose them to be simultaneous distinguished opens with a noetherian ring

So B_fi ≈ C_g for some noetherian ring C, the functions on one of your global cover elements

[latent anvil]

You already said they're noetherian

[vestal snow]

Ahh I see

Hold on let me process this

[vestal snow]

I totally forgot that these were schemes

Thanks for the help

[ivory dust]

hey guys if x^8-x = x(x-1)(x^3+x+1)(x^3+x^2+1) over Z2[x]

How do i construct a field of 8 elements by using roots of this polynomial and Z2

Cuz there are 2 irreducible factors that dont split over Z2

So which do I choose to provide the root?

x^3+x+1 or x^3+x^2+1 or does it not matter?

Im supposed to use Kroneckers theorem which allows me to assume there a larger field that contains these factors, Z2 and the root of the factors

field extension*

[vast quiver]

so this method of talking about finite fields is a little bit different than what I think you're trying. You seem to be familiar with how we can start with a field, like Z2, and then get a field extension by doing Z2[x]/(f(x)) where f(x) is an irreducible polynomial. Then, we get an extension with the same degree as the polynomial

instead, over here we're working in the algebraic closure of Z2, maybe call it F. The algebraic closure of Z2 is actually infinite. Then, we are looking for subfields of Z2 with cardinality 8.

in the closure, x^8-x has 8 roots (you can check that the polynomial has no double roots because it is separable)

these 8 roots actually form a subfield of the closure

you don't need to do Z2[x]/(x^3+x+1) (though that would get you an isomorphic finite field of order 8), because we are starting with an algebraic closure, then going down (rather than starting with Z2 and going up)

the thing you need to prove is that these 8 roots actually do make a subfield of the algebraic closure

from factoring, you already see that it has 0 and 1 as roots, which is a good first step

then, you need to show it is closed under addition, multiplication, and respective inverses

[ivory dust]

damn so i have to find 6 more roots from those other 2 degree3 polynomials?

How does one go about that

[vast quiver]

you don't have to do it super concretely, you just need to show abstractly that it works

it might be easier to generalize?

[ivory dust]

LIke the question is asking me to construct a field with 8 elements by adjoining suitable roots from that polynomial to Z2

[vast quiver]

oh

well if that's all it wants then I think your original strategy is fine

take an irreducible polynomial of degree 3 over Z2

[ivory dust]

Z2[x]/(x^3+x+1) is isomorphic to field w 8 elements but isnt Z2[x]/(x^3+x^2+1) ?

whats the diff b/w these

[vast quiver]

they're actually isomorphic

[ivory dust]

Ohhh

fuck

ok thank you

😂

[vast quiver]

yeah no problem lol

[vestal snow]

I am trying to prove the part in b) which says that if A is Noetherian then U is quasicompact. I proved that U is locally Noetherian using the fact that U is covered by things of the form Spec((S_f)_0) where S is graded ring and that S must be finitely generated as an A algebra. There was an earlier exercise where we proved that locally noetherian implies quasiseparated. I don't know how to go from here

[prime gale]

I don't know much about this stuff and don't know what book you are using. But if X is union of a finite number of quasicompact things, then it is quasicompact, right? U is covered by a finite number of your Specs, which are each quasicompact, right?

[compact needle]

You should be able to leverage S being noetherian (finitely generated over noetherian implies noetherian) to show that Proj(S) is a noetherian space, hence every open (in particular, U) is quasicompact

[warm holly]

To show that two applications $\phi, \phi' \in \mathcal{L}(V)$ have the same eigenvectors, is it sufficient to show that ker$(\phi-\lambda.Id)=$ker$(\phi'-\lambda'.Id)$? For $\lambda$ some eigenvalue of $\phi$ and $\lambda'$ of $\phi'$

[cloud walrus]

𝔻аniil

[kind temple]

what are your thoughts on this?

[warm holly]

I think that it is true but I would like to ensure myself, and if not to have a counter example...

What I wrote above must hold for all eigenvalues lambda,lambda' i guess

[kind temple]

yes. it’s saying that every eigen vector of $\phi$ is an eigen vector of $\phi’$ and vice versa. further, you can show that $\lambda\phi’(x)=\lambda’\phi(x)$ for any eigen vector of $\phi$ or $\phi’$

[cloud walrus]

coycoy

[warm holly]

alright, thank you

[chilly ocean]

Is this statement true or false ?

I tried to think of a isomorphism but couldn't

[kind temple]

what does the /(3) mean ?

mod three?

[chilly ocean]

Ideal generated by 3

3*7=1 so the ideal is the whole ring

Z_10[x] is a polynomial ring.

3*7 is still 1 there

[kind temple]

im not really inclined in this area, but what is significant about 3*7?

[chilly ocean]

ideals are closed for multiplication by members of the ring

3 is in the ideal and 7 is in the ring

and 3*7=1

so since 1 is in the ideal

every other ring element is in it too

[kind temple]

oh so you can kind of generate everything else?

[chilly ocean]

(3)=Z_10

(3)=Z_10[X]

yea

[chilly ocean]

O ok yes

[chilly ocean]

This means the statement is false right ?

[final pasture]

yes

[chilly ocean]

Because there is only one element in Z_10[x]/(3)

yea

And in Z_3[x] there are more than one element