groups-rings-fields

[dusty river]

ah

so to split a polynomial, you split its irreducible factors

for irreducible factors its easier to find the splitting field

[ivory dust]

If it was a bunch of distinct irreducible factors

would the splitting field

be Z3[x] / (q1(x))

then that / (q2(x))

[dusty river]

almost

after you split q1

q2 might automatically split

[ivory dust]

ohh

[dusty river]

so you gotta factorize further in the new extension

and then inductively build this tower

[ivory dust]

what if its like f(x) = (x-sqrt2)(x-sqrt3) (after we factor over R) over Q

[dusty river]

thats not a rational polynomial

[ivory dust]

o true ahh ok idk where im going w this

[dusty river]

do you mean x^2-2 and x^2-3?

[ivory dust]

yes!

[dusty river]

right

then you first adjoin sqrt 2

ie adjoin x and quotient by (x^2-2)

and check if x^2-3 is irred in the new field (it is)

and then adjoin sqrt 3 to this

[ivory dust]

oh ok i see so could i write it as: Q[x]/(x^2-2) = Q(sqrt2) then (Q[x]/(x^2-2))/(x^2-3) = Q(sqrt2)(sqrt3) = Q(sqrt2,sqrt3)

where = means isomorphic im assuming

[dusty river]

which means adjoining a y and quotienting by (y^2 - 3)

[dusty river]

slightly wrong, you adjoin a different variable the second time

[ivory dust]

do you mean if (Q[x]/(x^2-2)) = F then F[y]/(y^2-3) would be proper way to write it?

[dusty river]

yes

[ivory dust]

okk ty 🙏

[dusty river]

[ivory dust]

also do uk why f(x) and f(x+a) have same splitting field in F[x], if a e F

would I use something about their bases to prove it?

[dusty river]

F[x] has an automorphism which is identity on F, and maps x to x+a

it is an automorphism because x to x-a is an inverse

and for an automorphism f of R, R/J = R/f(J)

[ivory dust]

would it be defined like phi: F[x] -> F[x] := phi(f(x)) = f(x+a)

[dusty river]

yep

do you know the substitution principle?

[ivory dust]

noo

[ivory dust]

f(J) means Image of J?

isnt it R/ker f = Im f?

[dusty river]

it says that if you want to define a homomorphism from R[x] to S, you can define what it does on R and on x

[dusty river]

yes

[dusty river]

yes

how is that relevant tho?

[ivory dust]

o i wasnt sure if it is LOL

[dusty river]

lol

one thing is we are writing equalities

[ivory dust]

Since its onto f(J) = J is that what you mean here?

[dusty river]

but all these are only isomorphisms

[dusty river]

nope

f(J)is not J if f is as I defined

x to x+a

[ivory dust]

If f(x) is irreducible over Zp then is it irreducible over Z

and hence irreducible over Q?

[dusty river]

If it is monic then yes

just has a leading coefficient not divisible by p

[ivory dust]

I need to show f(x) = x^5+2x+4 is irreducible over Q

so would a smarter approach be to show its irreducible over Zp?

for some p

[dusty river]

Yeah that would work

in general

yeah i think p=2 does it, but it is not irreducible in Z_2

you gotta work further

[ivory dust]

i think p=3 works

kinda tricky tho

[dusty river]

hmm

p=2 is somewhat easy and its a good argument

good as in you see this with many polynomials

the factorization in Z_2 tells you what the factorizations in Z look like

at least something about them

and you can get a contradiction from there

You will get 4 divides 2

[ivory dust]

but in Z2 its reducible no?

cuz its x^5

[dusty river]

yes

[dusty river]

.

you get then that in any factor of f(x) over Z, ||only the leading coefficient is odd||

[spice lance]

this is M

[dusty river]

You have to define a surjective map from M to R whose kernel is I

The intuition for defining such maps is that quotienting by I is like making all elements of I 0

[spice lance]

thank you for replying! would phi(M) --> m suffice? since it's 0 in all elements of I

[dusty river]

Yep, exactly

[spice lance]

thank you so much!!

[dusty river]

[shut halo]

What is the difference between a unital associative algebra and a unital associative ring?

[gray comet]

An algebra A over a ring R is a ring that is also an R-module "in a compatible way"

By that, I mean the multiplication in A is R-bilinear

[shut halo]

So an algebra has a "scalar multiplication" over R while a ring doesn't?

[gray comet]

Yeah

[shut halo]

Ok thanks

[gray comet]

np

[shut halo]

All these terms are so confusing😅

[gray comet]

Definitely, lmao

[chilly ocean]

A good exercise to do is on a blank sheet of paper

Try write down all the algebraic structures you can remember

And what you are allowed to do in each one

[drowsy ivy]

Could I get a hint for this? I spent a while on it yesterday but I'm having a hard time finding the right angle to tackle it.

Originally I had hoped to be able to find such a subgroup by choosing suitable members from each coset of $H$ in $G$. However, if one takes $G$ as the set ${1, -1, i, -i}$ under multiplication and $H$ as ${1, -1}$, then you can't embed $G/H$ by picking representatives from ${1, -1}$ and ${i, -i}$, so that approach isn't the way to go.

[cloud walrus]

Pan

[oblique river]

do you know the fundamental theorem of finitely generated abelian groups?

[drowsy ivy]

not by name

[oblique river]

it basically says that every finite abelian group is a direct sum of cyclic groups

that was the first thought which came to mind but I'm not sure you need that here

[drowsy ivy]

Yeah my next thought was to find a minimal generating set so that every element of $G$ could be written as a linear combintation $g = n_1 g_1 + n_2 g_2 +\dots +n_k g_k$

[cloud walrus]

Pan

[drowsy ivy]

if you could choose the generators such that $g_1,\dots,g_i$ generate $H$

[cloud walrus]

Pan

[oblique river]

that sounds like it could be a problem though

[drowsy ivy]

but I'm not sure it works

[oblique river]

if G = Z/4Z and H = Z/2Z

like G = {0,1,2,3} and H = {0,2}

[drowsy ivy]

Yeah

because I think that only works if $G=G/H + H$

[cloud walrus]

Pan

[oblique river]

yeah that sounds right

actually I think you might need to use some kind of structure theorem here

a finite abelian group is the direct sum of its p-parts

for primes p

and each of those can be written as a sum of cyclic groups

[delicate orchid]

I can't think of a way to do it without using the structure theorem

been bashing it with homomorphisms for over 10 mins now and gotten nowhere

structure theorem makes it really nice

[drowsy ivy]

Okay I will make an attempt with the structure theorem

[delicate orchid]

tbh my structure theorem proof only holds if we have some indexing set $S \subset {0, ..., n}$ $\frac{\bigoplus^n_{j=0} C_{p_j}}{\bigoplus_{i \in S} C_{p_i}}$ \cong \bigotimes_{i \in S^c}C_{p_i}$ where $S^c$ is the compliment of S in {0, ..., n}

[cloud walrus]

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

[delicate orchid]

I'll see if I can come up with a proof

[ivory dust]

Hey guys if we define r = x + (f(x)) where (f(x)) is an ideal and f(x) is irreducible over Q

where f(x) = x^5+2x+4

then K = Q/I is a field etc.

but im asked to show since r is a root of f(x) in K

f(x) = (x-r)q(x) for some q(x)

how do I find this q(x)

Im supposed to use division algorithim

but like its

x^5 + 2x + 4 = (x-r)*q(x)

and I dont understand the notation for r 🤔 do i just treat it as an arbitrary root

cuz they say r = x + I, where I is the ideal (f(x))

[chilly ocean]

hmm?

[chilly ocean]

You say in the first line we define r = x + I

[ivory dust]

yea

im copying the notation of my proff

apparently r = x + I is a root of f(x) in K

but idgi

[chilly ocean]

I guess if you take projection phi : Q[x] -> Q[x]/f then u have some f(phi(x)) =phi(f(x)) =f(x) + (f(x)) = 0 right so its a root in the bigger field

[ivory dust]

any suggestion how to find q(x)

or what that even means

😭

im thinking just let r be arbitrary?

[small bison]

You can do the Euclidean division thingy

I don’t think you need to give an explicit q

[ivory dust]

i let q be monic degree 4 polyn

then just wrote it in terms of r

but idk if 1/r is allowed

[west violet]

if f: F--> G is an isomorphism, are orders of elements preserved?

what if F is additive and G is multiplicative?

and vice versa?

yeah

[magic owl]

the name for the operations is irrelevant

not the operations

they are quite relevant

if by “additive” you mean abelian

then an abelian group can only be isomorphic to an abelian group

again you can see this via a kernel argument

i sensed a deeper misunderstanding that might be more important hahaha

[west violet]

for example, $(\mathbb{Z}/p\mathbb{Z})^*$ is isomorphic to $(\mathbb{Z}/p-1\mathbb{Z})$, but some elements won't have the same order in both? Unless i'm misunderstanding something

[cloud walrus]

suck2015

[west violet]

p is prime btw

[gritty adder]

They do have the same order in both

[west violet]

https://proofwiki.org/wiki/Isomorphism_between_Additive_Group_Modulo_16_and_Multiplicative_Group_Modulo_17 (this is for all primes p and p-1)

ah , this is what i'm doing

[gritty adder]

They’re not even the same sets

[west violet]

so i guess the correct mapping should have elements of order k in the additive group being mapped to elements of order k in the multiplicative group

[magic owl]

yes

[west violet]

is there a method of finding that mapping or do i just brute force to find the orders ?

[magic owl]

the proof that they are isomorphic should tell you what the mapping is

it is rare to know two groups are isomorphic without first writing down the isomorphism

[gritty adder]

The idea is that you need to find an element of (Z/pZ)* of order p-1, which you can do by taking an element and “repeatedly extracting roots”

[golden pasture]

If theres a "easy" way to write down these maps crypto is fucked

[vestal snow]

If f: B --> A is integral and f^-1(J) = I for some ideal J of A and some ideal I of B, then the induced map B/I --> A/J is also integral

This is true right?

I wrote down a pretty straightforward proof but someone is saying that this statement isn't even true

[gritty sparrow]

Yes i’m pretty sure this is true. I mean, won’t the same polynomials that witness integrality of f work?

[marsh fractal]

so in my algebra textbook, they have this definition of a gcd

later they claim that you can have 2 elements which have common divisors but do not have a gcd. but wont the identity element always divide any common divisors and therefore create a gcd by this definition?

it makes sense they would exclude 1 as a common divisor but its not stated so im not sure if its because of another reason

[marsh fractal]

yes but isnt just one case required, the identity will always divide any d

[frank fiber]

if I have a double complex, then the maps in the total complex are defined by $d=d^h+d^v$

[cloud walrus]

Or x1

[frank fiber]

but why is possible to sum each $d^h$ with $d^v$ if $d^h$ and $d^v$ dont have the same codomain?

[cloud walrus]

Or x1

[magic owl]

Well what are the modules in the total complex

@frank fiber

[frank fiber]

products

the products of the modules in the diagonal

[magic owl]

Sure

So the domain of d^h is some C_{p,q}

but you also have C_{p,q} including into that product

and you can add objects in the product

[frank fiber]

thanks

[keen oasis]

Hi! I'm working on some group and subgroup exercises with focusing on the order of elements(mostly on dihedral groups). I'm wondering if exists some online tool/website which computes the order of a given element in a given group, where i can verify some of my results.

[magic sonnet]

There is https://en.wikipedia.org/wiki/GAP_(computer_algebra_system)

[keen oasis]

Thanks!

[tight otter]

I'm trying to find the integral solutions to the equation $4x^3-27=y^2$ but I'm having a hard time getting much further than showing that $y$ must be odd. I don't have too much experience with Diophantine equations so any help would be appreciated!

[cloud walrus]

panoramatopia

[thorn delta]

there are no integral solutions. First, look at the equation mod 4 to get a relation for y. Then if you look at the equation mod 2 you get a contradiction

can someone explain how the map "g" is obtained from the g_i's?

this isn't quite the regular universal property of the direct limit

[sturdy marsh]

is M the direct limit of the M_i

[thorn delta]

sure, but how do we get P?

[sturdy marsh]

you have a morphism M_i \otimes N ---> P

for all I

compose that with g_i

so you have morphisms g_i : M_i x N ---> P

for all i

[thorn delta]

oof ok yeah i see

[gritty sparrow]

I don’t see where you get a contradiction, both will just say y is odd right? In any case x=3, y=9 is a solution right?

[tight otter]

Yeah I think they both say y is odd. Because they both give you $y^2\equiv 1$

[cloud walrus]

panoramatopia

[thorn delta]

hm, yeah. I got that y should be 3 mod 4, so 4x^3 - 27 = 3 + 4n, i.e. 4x^3 - 4n = 30, but 30 is not divisible by 4. Not sure where i went wrong

[gritty sparrow]

I don’t get how you got y is 3 mod 4, i think that is where you went wrong

[thorn delta]

yep, i see now. I just saw 3^2 = 1 mod 4 and ran with it, my bad

[chilly ocean]

quick interruption

a module and a representation are the same

[steel tusk]

Sorry if this is a wrong channel, but:
Z/2Z is used for mod 2 arithmetic, but I don't understand why did we use that.
Later, I heard that multiples of phi is equidistributed on Z/Q
What does Z/Q mean?

[oblique river]

not sure what you mean by this. a representation of a group G over a field K is the same thing as a module over the ring K[G]

but not all rings look like this

[oblique river]

"mod 2 arithmetic" is by definition arithmetic in Z/2Z

and you'll have to give more context for your other question

[quaint tree]

Probably belongs more in #advanced-number-theory given that this is an elliptic curve. Anyway, this curve is the same as $16y^{2}=64x^{3}-432$ which is isomorphic to $y^{2}=x^{3}-432$ by the map $(x,y)\mapsto(4x,4y)$. By sage, this curve has rank 0 and only one integral point, $(12,36)$, so the original equation has only the solution $(3,9)$.

[cloud walrus]

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

[quaint tree]

I don't know why this isn't compiling correctly.

Here's my code if you care:

0
sage: EllipticCurve([0,-432]).integral_points()
[(12 : 36 : 1)]

[terse orchid]

for quintic formulas: the Abel proof is the full and complicated proof so how does the Galois groups relate to this?

[oblique river]

what do you mean?

[prisma ibex]

The Galois group perspective on the insolvability of quintics comes down to the fact that S_5 is not a solvable group

If you can solve a polynomial equation by taking radicals it means the Galois group of the splitting field of that polynomial is a solvable group

[kind temple]

probably an easy question, but I am trying to learn about ideals of rings. apparently, every ideal, I, of a ring R arises from a ring homomorphism.
so to prove that claim, if R is a ring and I is an ideal of R, then the map f(x) = 0 if x is in I and x is not 1, and f(x) = x otherwise should be a ring homomorphism with I as its kernel. am I missing anything here or is that sufficient?

[tough raven]

Is it a homomorphism?

[kind temple]

yes. it just seems like that was too easy for such an overarching claim, and i am unsure if i am overlooking anything

[final pasture]

how is that an homomorphism ?

What's the ring structure on the codomain, first

[kind temple]

R

[final pasture]

Ok so pick x in your ideal

f(x+x) = 1, because x + x € I, right ?

but f(x) + f(x) = 2

so this isn't an homomorphism

[kind temple]

what? if x+x is in I, then f(x+x)=0

[final pasture]

oh sry I misread

[kind temple]

its ok lol

[final pasture]

Still does not work tho

pick x in the complement of the ideal

f(x-x) = 0

f(x) + f(-x) = 2

[kind temple]

oof. okay thanks

how should i actually proceed at showing that claim? this was just a first naive attempt

[final pasture]

I don't know anything simpler than working with the quotient ring

[kind temple]

mod I ?

[final pasture]

yes

[kind temple]

hmm. that actually makes sense

[final pasture]

there's an "obvious" map from R (where R is your ring) to R/I whose kernel is precisely I

[kind temple]

should just be x maps to x + I, right?

wait, are ideals always normal subgroups of (R,+) ? its obvious if you have the claim im trying to prove, but it was not obvious until you mentioned the quotient group

[sharp sonnet]

its a subgroup right?

[final pasture]

All rings have commutative sums (atleast for my def)

[sharp sonnet]

and rings are commutative with respect to +

[kind temple]

hold on, this should be obvious tho. I is an abelian subgroup of R

[final pasture]

yes

[kind temple]

yea. thanks guys

[tough raven]

Still no?
f(x) = x for x not in the ideal

Consider f(x + i), where i in I is non-zero and x is not in I.
f(x + i) = x + i, but f(x) + f(i) = x + 0 = x.

[dusty river]

If an automorphism of F_p closure fixes all subfields setwise, then is it identity?

F_p or Q

[hot lake]

for the Fp case, like the Frobenius automorphism for example ?

[dusty river]

hmm does it fix all subfields?

[hot lake]

that depends on what you mean by "fix"

[dusty river]

setwise

as in image of subfield is itself

[hot lake]

yeah I think it does

[dusty river]

ahh I see

well I am trying to characterize all such automorphisms lol

for F_p and Q closures

and idk how to do that

[tough raven]

Maybe you classify automorphisms of their finite extensions or something?

[dusty river]

I think I see what you mean

Let me try this

So if I want to do this for Q (and there turn out to be no non id automorphisms), then it will be sufficient to show that every finite extension is contained in Q(a) where Q(a) doesn't contain any conjugates of a other than itself

because then any such automorphism of the closure restricts to identity on Q(a)

[brazen pumice]

can you come up with an automorphism of Fp-closure that doesn't fix all subfields setwise?

(hint: every finite extension of Fp is normal)

[dusty river]

oh right

So no we can't

[tough raven]

How do you pick conjugates consistently?

[dusty river]

hmm I don't get it

If you pick any subfield K, and an element x in it, then x has to map to something within F_p(x) by finiteness and hence normality of F_p(x)

and F_p(x) is in K

[brazen pumice]

correct

[dusty river]

so image of K is contained in K

surjectivity is by injectivity and the fact that everything is algebraic

ty

Now for the Q case 😵‍💫

[brazen pumice]

imma head out

[delicate orchid]

hello everyone 👋 😎 I've been messing around with finite fields and I'm kinda stuck on figuring out if it matters which irreducible polynomial you quotient by when you're doing a field extension of the form F_p/(p(x)) where p is irreducible. I've managed to show that if you use a specific irreducible polynomial of degree n you end up with a finite field of order p^n but is this the case for every irreducible polynomial?

the specific irreducibles I was using were x^n+x^(n-1)+...+x+(p-1) which I'm like 90% sure are irreducible by eisenstein

[tough raven]

Yes; ||the extension is a vector space over the original field; find a basis||.

[hidden haven]

Is there a name for elements such that adjoining one of them to a field does not add any of its conjugates?

It'd be the opposite of a primitive element in a way

Like when we adjoin 3rd root of 2 to Q, we don't get any of its conjugates, while if it's a 4th root, we also get its negative, so I want to know if the elements of the first kind have a name

[lavish gale]

any good resource on graded algebra?

[keen oasis]

I have some questions. If two rings are isomorphic then their characteristic are the same? So if i need to prove that two rings are not isomorphic one possible method is to show that their characteristic are different? If R1 is isomorphic to R2 then automatically R2 is isomorphic too with R1(i think yes)?
And i need to prove that 3Z is not isomorphic to 4Z. How can i start this? Both are infinite and without unity(so i suppose in this case the carcateristic doesn't exist). I've proved before that 2Z and 3Z are not isomorphic wiriting 4 as 2+2 and 2*2, but in this case i can't see how to approach it. Sorry if these are silly questions:)

[chilly ocean]

do you know what isomorphic means?

[keen oasis]

Yes

[chilly ocean]

what does it mean?

[keen oasis]

If f is the isomorphism then f is bijective and f(x+y) =f(x) +f(y), f(xy) =f(x) f(y)

Also f(0)=0

[unique juniper]

so f is reversable

and you can just look at the equivalence classes to see that there are not the same amount of those

[keen oasis]

Ok, and about characteristic, if for exemple 1 is not in 2Z then the caracteristic doesn't exist?

[delicate orchid]

if you just want to show that the characteristic between two isomorphic rings are the same you can do this:
let f be an isomorphism between two rings R1 and R2 where R1 has characteristic n
then f(0) = f(1*n) = n*f(1), but we know that f(0) = 0 as f is an isomorphism so n*f(1) = 0
but we know that f(1) must map to 1 in R2 as f is an isomorphism, so R2 has a characteristic of n as well

[keen oasis]

I understand, seems logic, thank you a lot! . But regard to my previous question, what happens if 1 is not in R, like 1 is not in 2Z. I'm asking because in my lecture the single mention about caracteristic is this : "the characteristic of a ring is the order of 1 in that ring"

[quaint tree]

A colloquial way to think about isomorphism is that two objects that are isomorphic are "the same up to a relabeling". Basically, if $A\cong B$ and $A$ has some property, we generally define isomorphism so that $B$ would also have that same property.

[cloud walrus]

Elliptic Curve of Rank 9001

[quaint tree]

As for how to show that $3\mathbb{Z}$ and $4\mathbb{Z}$ are not isomorphic, I can think of several ideas, but the most obvious to me is that if $3\mathbb{Z}\cong4\mathbb{Z}$ then $\mathbb{Z}/3\mathbb{Z}\cong\mathbb{Z}/4\mathbb{Z}$, which is just blatantly false.

[cloud walrus]

Elliptic Curve of Rank 9001

[keen oasis]

I find a prof for that. I calculate f(18) as f(3+3+...+3) and f (3*6) then i find a condtradiction

But, now the last question that i have is if the caracteristic of a ring is define when 1 is not in R. Or what is the term for this? Is doesn't exist, it is not defined? Like i understand the prof for equal characteristic when the rings are isomorphic, but for example if i have Z and 2Z and Z has caracteristic 0, what about 2Z?

[quaint tree]

All ideals of Z have characteristic 0.

Do you remember the definition of characteristic?

[keen oasis]

The definition that was given to me is "the characteristic of a ring is the order of 1 in that ring". This is the reason why i'm confused

[quaint tree]

The characteristic of $R$ is the least integer $n$ such that, for all $r\in R$, $nr=0$, where $nr$ means $r+r+r\dots$ $n$ times.

[keen oasis]

Nobody told what happens if 1 is not in that ring

[cloud walrus]

Elliptic Curve of Rank 9001

[quaint tree]

In rings with unity, the definitions are equivalent.

[keen oasis]

Ooo i see, now is more clear

Thanks

[solar inlet]

I'm pretty sure that's a fallacy. In general for algebraic structures $M \cong N$ such that $M,N \subseteq A$ does not imply that $A/M \cong A/N$.

[cloud walrus]

tox

[quaint tree]

Thanks for the correction. Perhaps allowing them to be isomorphic and looking at the isomorphism $\phi:3\mathbb{Z}\to4\mathbb{Z}$ and deriving a contradiction from that would be the best approach, then.

[cloud walrus]

Elliptic Curve of Rank 9001

[chilly ocean]

probably just look at image of 3 (generator) and it should follow?

[delicate orchid]

that's what I was thinking, you'd need f(3) = 4

[oblique river]

(or -4)

[delicate orchid]

or -4

[chilly ocean]

or -4

[delicate orchid]

nah

[quaint tree]

And if $\phi(3)=4$, then $\phi(9)=\phi(3)\cdot\phi(3)=16$ and $\phi(9)=\phi(3)+\phi(3)+\phi(3)=12$, which is a contradiction.

[cloud walrus]

Elliptic Curve of Rank 9001

[oblique river]

for the record, I don't think you have to phrase this like a proof by contradiction

[quaint tree]

No, but I like proofs by contradiction and it's been a while since I've made a fun one.

[oblique river]

my point is that this isn't really a proof by contradiction at all

the "contradiction" is just added on to the end and doesn't really contribute anything

what you're really arguing is that a ring isomorphism must also be a group isomorphism, and there are only two group isomorphisms from 3Z to 4Z, namely, phi(3) = 4 and phi(3) = -4. However, neither of those preserve ring multiplication.

[solar inlet]

Plus it seems like there's a lot of people here who are anti-LEM, and we wouldn't want to scare them

[oblique river]

haha

[magic owl]

not many they are just loud

also not even loud is anyone here anti lem

[chilly ocean]

yes I hate his books

[solar inlet]

pretty sure there's at least one person whose an intuitionist in #foundations

[hidden haven]

Nikita if I'm not wrong

[magic owl]

of course

[tough raven]

Still waiting for the topology references

[hidden haven]

We didn't tag him lol

[quaint tree]

Does anyone know the close-form solution to the general quintic in terms of radicals and ultraradicals/bring radicals?

The best source I can find is a 300-page-long lecture notes in German published in 1888 by Felix Klein.

[oblique river]

have you read wikipedia? it tells you how https://en.wikipedia.org/wiki/Bring_radical

first you change variables to get it into bring-jerrard normal form

and that transformation can be done using only radicals

it doesn't write out the entire formula, but it describes how to get it

[quaint tree]

I genuinely keep forgetting that wikipedia exists.

[oblique river]

and trust me you don't want to try to write out the entire formula on one page

[quaint tree]

Is it worse than the quartic?

[oblique river]

yeah

[quaint tree]

Awesome.

[topaz leaf]

anyone here into group representation theory? (also @vocal yew / reply to my messages pls i have this server muted)

[oblique river]

If you have questions, just ask them here

or if you're looking to start a book club, you should say that (but idk how much response you would get)

[thorn delta]

if (x) is a principal ideal in a commutative ring A, isn't (x) \otimes_A A/(x) = 0? Since (rx)\otimes y = r \otimes (xy) = r\otimes 0?

Pretty sure this is wrong though since (x) \otimes_A A/(x) \cong (x)/(x)^2

[oblique river]

the issue with this is that r need not be in (x)

so you can't just leave r on the left hadn side

[thorn delta]

ah yea of course. Thanks

[oblique river]

np

[lethal cipher]

Is there some sort of notation to discuss the collection of all binary operations?

[oblique river]

meaning like

"a binary operation on a set X is a function from X x X to X"?

functions from X to Y are often denoted by Y^X

[lethal cipher]

Sure, but you consider the collection of every possible binary operation on X

[oblique river]

so I guess you could call it $X^{X \times X}$

[cloud walrus]

Buncho Bananas

[oblique river]

what else is a binary operation

but a function from X x X to X

are you working with some other definition?

[lethal cipher]

No, I am working with a trash logical statement and I am trying to use something else to make it make sense.

This statement does not make sense

What does 'make' mean?

[oblique river]

"make the following statement true"

it seems oddly worded, but maybe if you changed "a value exists" to "values exist"

it makes sense to me

[lethal cipher]

Oh, I think I understand. That was oddly worded, so thank you for clearing that up for me.

[oblique river]

I think it's saying

write a logical sentence using an existential quantifier which says "there exists values a and b such that a^2 + b^2 < 0" and then determine if this statement is true

[lethal cipher]

Yep, that's all it was.
When it says "a value exists", it made me think we had to throw in some other variable.

But I understand now.

[chilly ocean]

I'm not sure what it means by "each L-submodule of V possesses a complement"

[gritty sparrow]

For each submodule W, there is another submodule K such that V=W\oplusK

K is the complement of W in this scenario

[chilly ocean]

thank you

[chilly ocean]

I'm a little confused

Does it make sense to talk about nil potent elements in an arbitrary lie algebra?

phi(x) is an endomorphism

so here by nilpotent we mean the n-fold composition of phi(x) is the zero map

[final pasture]

idk any lie theory so maybe I'll say something dumb, but doesn't nilpotent make sense in like, any group (or even any monoid) ?

[chilly ocean]

yes you are right

it must just mean [x,[x,[x,...,x]]]]

oh no

i am wrong

[maiden ocean]

What does this concretely mean

do the big parenthesis indicate something or

am i just overthinking

this is the set of all s_i in U cap U_i for any i

right

@uncut girder can u confirm

[uncut girder]

What big parenthesis

[gritty sparrow]

I don’t really know, I haven’t heard of one yet. We can talk about ad nilpotency though ( although i assume you already knew about that)

[vestal snow]

Show that morphisms from a quasiseparated scheme are quasiseparated (the definition of qs scheme is that intersection of quasicompact opens is quasicompact)

[haughty shale]

If A is in GL(n,Q) and p is a prime such that p>n+1, then if A^p=I_n then A=I_n.

Any thoughts on how to start on this?
Ps: i deleted this from another channel since this is the more appropriate channel to ask this. Thanks!

[kind temple]

$$A^p-I_n=A^p-I_n^p=(A-I_n)(A^{p-1}+A^{p-2}+\dots+A+I_n)=0_{n\times n}$$

[cloud walrus]

coycoy

[haughty shale]

Gotcha. So the second quantity does not equate to zero?

[kind temple]

it can’t (unless x is the zero vector) since all the matrices are invertible

well wait

[haughty shale]

Nice thanks man!

[kind temple]

that’s maybe not entirely true

[haughty shale]

Oh okay

[kind temple]

i believe this is the right idea tho. just throwing some ideas around

[haughty shale]

Okay sure. I will try to work on it. Thanks!!

[gritty sparrow]

Here is a hint: ||turn A^p-I into the product over i (A-z_iI_n) were z_i are the pth roots of unity and z_1 is taken as 1. Now try to prove that for i>1, A-z_iI_n is invertible (hint: the characteristic polynomial of A has degree n<p-1 and the non identity roots of unity have a minimal polynomial over Q of degree p-1). Then we are done||

[vestal snow]

Show that morphisms from a quasiseparated scheme are quasiseparated (the definition of qs scheme is that intersection of quasicompact opens is quasicompact)

Reposting so it doesn't get lost

[chilly ocean]

question: is [a][b]=1 equal to ab ≡ 1 mod n ?

[scarlet estuary]

how are b and n related?

[chilly ocean]

few things to mention: gcd(a,n)=1 therefore there exist ab+cn=1. b and c are member of Z.

[final pasture]

no, put a = b = 2 and n = 3

[chilly ocean]

there is a typo there: ab ≡ 1 mod n

[final pasture]

where ?

oh

then yes

[chilly ocean]

mistakenly they said ab = 1 mod n

[final pasture]

I'm confused what

[final pasture]

this is correct

and where is the mistake in the screen ?

[chilly ocean]

i cannot understand why [a][b]=1 is equal to ab ≡ 1 mod n

[final pasture]

Well, that's how multiplication of equivalence classes is defined

[a][b] = [ab]

so if [a][b] = [1]

then [ab] = [1]

or, written another way, ab = 1 (mod n)

[chilly ocean]

if ab ≡ 1 mod n, then ab -1 = cn therefore it should be ab = cn + 1

[final pasture]

ab + cn = 1 <=> ab = -cn + 1 = (-c)n + 1

[chilly ocean]

so we have considered c=0 right ?

[final pasture]

wut ?

[chilly ocean]

quite confused

[final pasture]

Ok let me recap the proof, maybe that'll be clearer without the ugly notation

[chilly ocean]

yea got it

[final pasture]

oh well

ok nice then xD

[chilly ocean]

thank you.

[final pasture]

[gritty sparrow]

There is a condition that says a morphism f:X->Y is quasiseparated iff for any U,V affine opens in X that map into the same affine open of Y, U intersection V is a finite union of affine opens. With this condition the proof is easy, but for the life of me I can’t recall the proof

[vestal snow]

Oh wait

I remember proving this

Or at least something similar

a scheme is qs iff the intersection of two affine opens is a finite union of affine opens

[gritty sparrow]

Yeah this is the case when f is from X to Z i suppose. Does the proof you have generalize?

[vestal snow]

A lot of my problems when solving AG can be summarized by "I cannot recall the million different properties that quasi-normal-compact-separated-noetherian-local-connected-irreducible schemes have"

[vestal snow]

I'm afraid it is lost in the pages of my (unorganized) notebook

I'll try and find it later

[gritty sparrow]

Cool, hopefully this should work out.

[gritty sparrow]

Yeah same, I feel like there is a never ending sea of definitions in AG, I still don’t feel like I have a “handle” on the so to speak

[hidden haven]

At the pace that AG books define new things, you could almost call it a rising sea of definitions

[sturdy marsh]

LMAO

you dont need to have a perfect handle on every single adjective all the time

when you start looking at topics, you will use only a certain subset of adjectives

and it's much easier to familiarize yourself with those before you start

[scarlet estuary]

undergrad courses: learning nouns
grad courses: learning adjectives

i will not be taking questions

[oblique river]

i have a question

[vestal snow]

みつけたよ @gritty sparrow

[gritty sparrow]

Nice

Does he say anything about the more general version?

[vestal snow]

Not yet

[tough raven]

Can we learn about those at the post-doctorate level?

[safe citrus]

is it supposed to be gp(x) = {x^m | meZ}? not x

[chilly ocean]

yes

that is a typo

[safe citrus]

thanks

[vestal snow]

Can someone verify my solution to this?

It seems much simpler than the hints

let pi be the morphism. Then pi^-1(Spec k)= Spec A where A is a finite k-algebra

However, that must mean that A is a field

Then X = pi^-1(Spec k) = Spec A is just a single point

And the rest follows trivially

This feels to easy

And also suggests something stronger than what the hints suggest

[brazen pumice]

Look at the example. Is your conclusion true (for that example)?

[vestal snow]

Hmm so it seems that A is not necessarily a field

Though I don't see where exactly my proof goes wrong

[brazen pumice]

is the ring in the example a finite k-algebra?

[vestal snow]

Ah I see

Thanks

[safe citrus]

I finished 1, but i'm stuck on 2 now

does anyone know how to show n>0?

[unique juniper]

what would happen if n = 0

[safe citrus]

uh

nothing?

[hot lake]

how did you define n

[safe citrus]

it's the number of times the binary operation occurs?

[unique juniper]

then only the identity would be in the group, and since its finite, there exists such a n, otherwise it would be infinite

[unique juniper]

yes

well

[safe citrus]

did i do smthing wrong lol

[hot lake]

I don't see how that's a rigorous definition of anything

[unique juniper]

i think they mean that n means the amount of times that x has been combined with itself

[hot lake]

where ? when ?

[unique juniper]

the order is the smallest integer such that x^n = e

[hot lake]

well yeah but you need to show that such a number exists first

(smallest positive integer such that x^n = e)

what if there is no positive integer such that x^n = e

[safe citrus]

it's subgroup of G tho

[unique juniper]

but we are assuming its finite, if the order was infinite then the cyclic subgroup would be infinite -> gp is infinite

[hot lake]

that's what the exercise wants you to prove properly

[unique juniper]

thats what i was trying to say :D

[unique juniper]

i think they r on part 2

[safe citrus]

uh

[hot lake]

we are not assuming the order exists and is finite, we are assuming gp(x) is a finite set

[safe citrus]

oh yeah i am

i dont know if i even did 1 right anymore now

[unique juniper]

?

[unique juniper]

yeah but the order always exists tho?

[hot lake]

?????

arguably not if it the order is infinite

[unique juniper]

it could be infinite or a finite positive integer

[hot lake]

besides they haven't defined the order of a group element yet

that's the point of question 2

[unique juniper]

okay

[hot lake]

student : I have trouble doing this exercise that properly defines this notion
you : hey it's easy you have to use the notion

[unique juniper]

lol

😅

[safe citrus]

sry im still confused

uhh

[unique juniper]

suppose that there didnt exist such an n>0

[safe citrus]

ok

[safe citrus]

OH

i finally get what you mean

lmfao i am slow ty 🙏

[vestal snow]

To use the affine communication lemma, I need to prove that if pi^-1(Spec A) is quasicompact, then pi^-1(Spec A_f) is quasicompact as well

Any ideas?

[gritty sparrow]

Yeah, write pi^-1. (Spec(A)) as the finite union of affine opens, spec(B_i) then in the ring maps phi:A->B_i it is easy to see that that the if a prime doesn’t contain phi(f), phi^-1(p) doesn’t contain f, and all p st phi^-1(p) is in D(f) come this way. Hence pi^-1(A_f) intersection with spec(Bi) will be the affine open spec(Bi_phi(f)). So we can cover pi^-1(A_f) by finitely many affine opens

[haughty shale]

Thank you!!

[gritty sparrow]

Np

[vestal snow]

I don't understand

What is phi?

[gritty sparrow]

The ring map corresponding to the map we get from spec(Bi)->spec(A) (which is the restriction of pi to spec(Bi))

[gritty sparrow]

Does this make sense now?

[vestal snow]

Yeah

I think so

[gritty sparrow]

Nice

[reef agate]

Hi guys quick question: If I have a GF(2^4) field generated through Q(y)=y^4+y+1. And a composed field GF((2^4)^2) generated by P(x)=x^2 + x + w^14 while Q(w) = 0. What form does a primitve element of the composed field have and how can I construct a basis for the composed field?

[chilly ocean]

Are Hamel bases on an infinite dimensional vector space uncountable?

ℝ[x] and {1, x, x^2, ...}

or are you asking for the number of hamel bases

[chilly ocean]

no, just take any set of countably many linearly independent vectors, their span is infinite dimensional vector space with countable hamel basis

[brazen pumice]

what do elements of a quadratic extension L/K look like?

[chilly ocean]

Nah, the basis itself.

If I tried to construct a Hamel basis for an infinite dimensional vector space, would the basis be an uncountable set?

homie I just told you and look at tterra example

i just gave an example of a countable one, as did ledog

[chilly ocean]

What's R[x]?

polynomials in x with real coefficients

[chilly ocean]

But, would that basis be able to express every polynomial as a finite linear combination?

[tough raven]

There is a vector space over any field with dimension any cardinality.

[tough raven]

Yes.

[solar inlet]

Part of the definition of polynomials is being finite, so inherently they must be composed of a finite linear combination of monomials.

[reef agate]

They are pairs of of elements of the base field right?

[chilly ocean]

Of course! smart.

[chilly ocean]

Would the same construction be possible for every infinite dimensional vector space?

[chilly ocean]

Can every infinite dimensional vector space have a countable Hamel basis?

[tough raven]

No.

[tough raven]

By this, there is a vector space with any cardinality as basis.
And all bases have the same cardinality.

[chilly ocean]

Cool, thanks!

[solar inlet]

Hmmm, I wonder can we talk about maximal vector spaces then? Like how the surreals are a proper class that behaves like a field and includes an embedding of every cardinality field, is there a class that includes an embedding of every cardinality dimension?

[brazen pumice]

how does that produce an element of L?

[reef agate]

Yeah well thats my question because I have this polynomial to generate the elements but I don't know how.

[brazen pumice]

so let's say L = K[x]/(x^2+ax+b) where a,b in K

then what does an element of L look like?

[solar inlet]

In general, I think if you have K[x]/(p(x)) you can think of it as K[c] where c is some constant such that c is not an element of K and p(c) = 0.

for example R[x]/(x^2+1) is R[i] because i is not an element of R and (i^2+1) = 0

[reef agate]

Ok thank you guys this is very helpful. I will try to figure out tomorrow at work how this applies to my problem.

Do you have any literature resources a paper or something that explains this concept in a practical way like you two just did?

[chilly ocean]

I found the book written by authors Dummit&Foote really nice, it has a lot of examples and problems.

[reef agate]

Add this book also covers extension fields and irreducible polynomials like we disscussed now?

[chilly ocean]

Yes.

There's a rings chapter and a fields one soon after

[reef agate]

Ok thanks I will try it out!

[reef agate]

Can you continue on your example? I think it will help me understand better ^^

[chilly ocean]

Basically elements of that ring are polynomials of degree 1 less than the thing you take quotient of.

So like when you have a ring K[x]/(x^2 +ax +b) the elements are f(x) + (x^2+ax+b), but they simplify to some cx + d

[tough raven]

You can't ... oh, proper classes. I have no idea then.

[tough raven]

Possibly if you construct the free vector space on the class of all sets?
I'm not sure that you can construct a (proper class) free vector space on a proper class, though.

Actually, I think you can.
Let V := { f | f is finite and f is a function and the range of f is a subset of the base field F }.

Given f, g; let dom(f) = A and dom(g) = B then define h as a function on A U B which is the sum of f and g.

Similarly for scaling.

[tough raven]

Actually restrict that f never takes the value 0 to get a canonical representative.

[tough raven]

h is then defined on A U B \ { x in A U B : f(x) + g(x) = 0 }.

The zero function would have to be viewed as a function on the empty set (i.e. the "empty" function).

[solar inlet]

This seems like it would still be a regular set of cardinality 2^|F| though?

Oh wait I see what you mean now

One possible basis would include every f such that dom(f) is {x} and f(x) = 1. This wouldn't necessarily encompass the whole basis, but at least part of it. And if you were to let F be the surreal numbers then there is a basis vector corresponding to every surreal number, and since the number of surreals is greater than every possible cardinal number, the dimension of V would be greater than every possible cardinal number. Thus all vector spaces over any ordered field can be embedded into V.

I wonder if that construction is strictly necessary though? Perhaps the surreals would directly satisfy such a vector space. If F is an ordered field, then F is a subfield of No. Now suppose you wanted a vector space of dimension n on F, where n is a cardinal. You could find a larger field by finding a set of n linearly independent numbers, B, in No\F and considering F[B] as a vector field over F.

[chilly ocean]

by endomorphisms in L we just mean elements in L?

(because L is a subalgebra of gl(V) )

[oblique river]

that's hwo I interperet it

@chilly ocean

[chilly ocean]

thanks

[hasty sorrel]

Is this related to "Lie Algebras" or is that something else?

[scarlet estuary]

"this" meaning abstract algebra?

a lie algebra is a certain type of nonassociative algebra (vector space with an extra operation)

so its an object one would study in abstract algebra

but in practice lie algebras are mostly used by analysts

and physicists

well, thats misleading, theyre studied plenty in abstract algebra too

but my point is that theyre not solely confined to #groups-rings-fields even though they are an algebraic structure

[junior moth]

By "this" they probably meant the question posted above

[scarlet estuary]

then still no

[gritty sparrow]

The above question is with reference to the lie algebra on gl(V)

[scarlet estuary]

wait wtf

i missed lime soups question

and was referring to the one by doodle/tox above it

sorry

im causing needless confusion

[chilly ocean]

unforgivable

[chilly ocean]

the nullstellensatz reverses inclusions

this should change smallest into biggest?

as in

im trying to show that I(Y_1)+I(Y_2)=\sqrt{I(Y_1\cap Y_2}

opps radical on the wrong side

but I'd like to argue that \sqrt{ I(Y_1)+I(Y_2) } is the smallest radical ideal containing I(Y_1) and I(Y_2)

Y_1\cap Y_2 is the biggest algebraic set contained in Y_1 and Y_2

so I(Y_1\cap Y_2) should be the smallest radical ideal which contains I(Y1) and I(Y2)

[chilly ocean]

yes never mind it works

[quaint tree]

A tough problem that I can't figure out: Let $\pi$ and $e$ be the familiar constants, i.e. $\pi\approx3.14\dots$ and $e\approx2.71\dots$. Show that $\mathbb{Q}(\pi)(e)/\mathbb{Q}(\pi)$ is a transcendental field extension, or show that it is an algebraic field extension.

[cloud walrus]

Elliptic Curve of Rank 9001

[quaint tree]

(Edited because the book that I'm using right now was published in the '70s and uses notation and spellings of words that I haven't seen anywhere else)

[sturdy marsh]

Is this not an open problem

[quaint tree]

Well maybe that's why I was having such a hard time of it.

[sturdy marsh]

This was a problem in a book?

wtf

[quaint tree]

No, I read the chapter in a book on transcendental field extensions and made my own problem for fun

There was a proof for the transcendence of $\mathbb{Q}(e)/\mathbb{Q}$ and for the transcendence of $\mathbb{Q}(\pi)/\mathbb{Q}$ and I thought it wouldn't be that much harder.

[vestal snow]

Try this next: Prove that any finite group can be realized as a Galois group of some extension over Q or give a counterexample.

[cloud walrus]

Elliptic Curve of Rank 9001

[quaint tree]

I think I can do any finite abelian group.

[vestal snow]

I'd rather do the Tits group

[quaint tree]

The more general case sounds a bit harder.

[sturdy marsh]

It’s an open problem

[quaint tree]

Well that would explain why it sounds harder.

[chilly ocean]

[vestal snow]

Can someone verify my solution for b)?

I think it's just writing out the definitions, but there are so many of them that I'm not sure I applied them correctly

Suppose pi: X --> Y is integral and of finite type. Let Spec B be an open affine subset of Y. We get that pi^-1(Spec B) = Spec A where the induced ring morphism pi^# from B to A is integral. However, since pi is of finite type, A must be a finitely-generated B algebra through pi^#. Now use that an element is integral over B iff it is contained in a subalgebra of A which is a finite B-module on the generators of A.

[sly crescent]

Do all Chevalley groups contain the corresponding Weyl groups as subgroups?

[stark obsidian]

let $C(\mathbb{R)}$ commutative ring with unit of $f: \mathbb{R} \rightarrow \mathbb{R}$, and $M_a={ f | f(a)=0, \quad a\in \mathbb{R} }$. how can i prove that $M_a$ its the only maximal ideal?(i supose that )

[cloud walrus]

Polux12

[rustic crown]

don't think this is true, it would be true if you look at C(K) where K is a compact space, but not true in general.

we just need to find a proper ideal that isn't contained in any of those M_a, and then the maximal ideal containing this would be different from any M_a

[cloud walrus]

det

det

[gritty sparrow]

Yeah this seems correct

[gritty sparrow]

Also, you should maybe state that the fi’s are positive or something, otherwise they won’t necessarily be zero outside of (-n,n).

[rustic crown]

ahh oopsie

[vestal snow]

Is the map on the underlying topological space just the identity and the map O_X(U) --> O_X(U) given by a |--> a^p?

This map clearly satisfies the appropriate commutative diagram and induces the map a |--> a^p at stalks as well (if a is a non-unit, then so is a^p so it is local). Thus, it is a scheme morphism

[final pasture]

Hey

Is there an easy proof of the fact that $\mathbb{K}[X]/(\pi^m)$ is a local algebra (unique maximal ideal) if $\pi$ is an irreducible polynomial ?\
My proof goes like this:\
I show that for some ring $R$ and a two-sided ideal $I$, $f: A \to B, J \to J/I$ is bijective, where $A$ is the set of all ideals of $R$ containing $I$ and $B$ the set of all ideals of $R/I$.\
Then I say that an ideal containing $\pi^m$ is of the form $P\mathbb{K}[X]$, since $\mathbb{K}[X]$ is a PID.\
Since $P\mathbb{K}[X]$ contains $\pi^m$, $P \mid \pi^m$, and because $\mathbb{K}[X]$ is factorial, $P$ must be of the form $\pi^n$ with $n < m$ (up to multiplicative constant), and then I get the chain of ideals
$$(\pi^m)/(\pi^m) \subset (\pi^{m-1})/(\pi^m) \subset \cdots \subset (\pi)/(\pi^m)$$
and I'm done

[cloud walrus]

Shika

[final pasture]

the book I'm reading vaguely says (not really says, but that seems to be what the author meant) that we could do something easier "in the fashion of the classification of the ideals of Z/p^nZ, with p prime", but I don't really know how to do it in the case of Z/p^nZ either (in a simpler way than this, I mean)

oh wait

I guess we can just use Bezout 🤔

Yeah bezout should work.\
Quotient of PID is a PID, so for any ideal $I$ of $\mathbb{K}[X]/(\pi^m)$, there is some $P \in \mathbb{K}[X]$ such that $\bar{P}(R/I) = I$.\
Now, because Bezout, we get that for any $Q, U \in \mathbb{K}[X]$, $gcd(P, \pi^m) \mid PQ + U\pi^m$ and that there is $Q, U$ such that $gcd(P, \pi^m) = PQ + U\pi^m$, and that shows $\bar{P}(R/I) = gcd(P, \pi^m)(R/I)$

[cloud walrus]

Shika

[final pasture]

And then since $gcd(P, \pi^m)$ must be a divisor of $\pi^m$, we're done

[cloud walrus]

Shika

[final pasture]

does that look correct ? 🤔

(\bar{P} being the equivalence class of P)

[vestal snow]

What does fancy K mean here?

[final pasture]

K is a field

[vestal snow]

I think I proved this in December 2019

Hold on. I'll try and find it in my old notebooks

It might take a while

[gritty sparrow]

Yes this is correct

[vestal snow]

Is this from Dummit and Foote?

I vaguely recall proving that the image of (pi) is the unique maximal ideal

[gritty sparrow]

I think you can argue like this: In K[x] the maximal ideals are exactly (f(x)) for irreducible f. f^n is contained in (f) and if f^n is contained in g for some other irreducible g, g divides f

Does this seem correct?

[vestal snow]

Yeah that's what I think too

Any maximal ideal containing pi^m will also contain pi which forces it the ideal to be (pi)

[gritty sparrow]

Isn’t that the same argument? (Since g is a prime element)

[vestal snow]

Yeah it is.

I think this generalizes to PIDs

[final pasture]

you're sure we don't need the ring to be factorial ?

🤔

[vestal snow]

PIDs are factorial, right?

[final pasture]

wait yes

you're right

I'm so bad at ring theory, I should really try learning it properly some day

[vestal snow]

Might I suggest Aluffi chapter 0 for it

[final pasture]

I don't exactly get your proof tho, are you arguing about this in K[X] or in the quotient K[X]/(pi^m) ? 🤔

wait 🤔

[vestal snow]

My bad

I should probably be more explicit

I'm using that ideals in the quotient are images of the ideals in the original ring which contain the thing you're quotienting by

In fact, these two sets are in a natural bijection

[final pasture]

oh right, so one of the thing I've proved above right ?

[final pasture]

here

that's precisely what I prove

[vestal snow]

Oh lmao

You're basically developing the theory yourself

[final pasture]

Well I don't really know ring theory, except basic defns, so I don't know "the theory" lol, I do what I can with what I know

[vestal snow]

It's pretty cool ngl

It doesn't behave very nicely sometimes, but modules behave VERY nicely and you need ring theory to study those

[gritty sparrow]

Btw what book are you reading from shika?

[final pasture]

it's a book in french, about "advanced" (not that much) linalg

(https://www.amazon.fr/Algèbre-linéaire-endomorphismes-Roger-Mansuy/dp/2311002856, this one, in case anyone's interested )

[gritty sparrow]

And it teaches ring theory? Damn that is interesting, is this for the jordan decomposition or whatever?

[final pasture]

no, it doesn't teach ring theory, that's the point, it remains fairly light on the abstract algebra prerequisites in most of the book, except at the end of each chapter where there is a dedicated section for "comments and developments" where the author (intentionally) forget about the prerequisites and just give interesting facts, rephrase some stuff by using some more theory (quotients instead of just doing everything by hand, for example) etc.

[vestal snow]

Do you know french?

[final pasture]

I'm french, so yeah lol

[vestal snow]

Nice. That means you can read EGA pretty easily (if you decide to)

[final pasture]

and yes, jordan and froebenius stuff, diagonalization, trigonalization, cayley hamilton, and all the theory around that + some stuff about the topology of Mn(R) and Mn(C)

(oh and markov chain stuff but idk probability so I won't read that part )

[gritty sparrow]

Oh no, don’t do this to him, he’s too young to take the AG pill

[vestal snow]

Abandon sanity, become Goomer

[final pasture]

Saketh is right, I'm still too young and too innocent

Maybe in a year or two

oh woops, I can't write

[gritty sparrow]

Goomer? Never heard that one before, but its nice to know math has wojacks too

[final pasture]

(nobody saw that typo)

[gritty sparrow]

How come?

[final pasture]

wait

reee

no

reeeeeeeeeeeee

I did a typo

while writing about my typo

[gritty sparrow]

Now I look like a dumbass, thanks

[final pasture]

(well not really a typo, more like a brainfart, but w/e )

[final pasture]

no problem !

[gritty sparrow]

Anyway, do you have any interest in learning about ag at all?

[final pasture]

yes, AG looks really interesting

I don't even know what a good starting point would be tho, any recommendation ? (I'm not planning on starting in a near future, there's some stuff I want to learn before, but I'll keep the recommendations in mind )

[vestal snow]

you definitely need a good handle on algebra before you start

Or maybe

Instead of using ring theory to understand schemes, you draw schemes to understand ring theory

[gritty sparrow]

I think the standard route is atiyah mcdonald for commutative algebra, then move on to hartshorne /vakil for ag

[final pasture]

hartshorne

[gritty sparrow]

Already scared of it? How come?

[final pasture]

I've heard so many things about that book lol, now I see it as a sacred book that can only be read by the chosen ones

[vestal snow]

Tbh, I don't think you need to read A-M for AG

Only a few topics from it

Because the stuff on dedekind domains and stuff is primarily for NT

[gritty sparrow]

Dimension theory will get really hard without a good commutative algebra background, i really don’t know how you’ll manage without it

[gritty sparrow]

Good or bad things?

[final pasture]

"the book is super hard" kind of things

[final pasture]

(btw is that correct ? )

[gritty sparrow]

Yeah some of the exercises are pretty brutal, but it isn’t impossible (especially if you’re willing to skip stuff and come back later).

Although i’m yet to finish it fully, so I can’t really say wether it gets much worse even later on

[final pasture]

I see

[vestal snow]

Vakil is much gentler

[final pasture]

http://math.stanford.edu/~vakil/216blog/FOAGapr2915public.pdf
this ?

[gritty sparrow]

Oh yeah, I was about to ask you what you thought about vakil. Do you think it’s better than hartshorne?

[vestal snow]

I only read ch 1 and 4 from Hartshorne, but I think Vakil is better

I also don't like exercises being piled up at the end of each section so I really liked Vakil for not doing that

[gritty sparrow]

I see

[final pasture]

Also I've heard great things about that book from a friend:
http://userpage.fu-berlin.de/aconstant/Alg2/Bib/Shafarevich.pdf
do you have any opinions on it ?

(that's actually 3 books, only the first one is included in the link, but the toc of the others are there too )

[gritty sparrow]

I haven’t really heard of it before. Seems pretty interesting, but it looks like it has a lot of focus on the concrete algebraic varieties rather than the generalities (that probably isn’t a bad thing depending on your pov)

[final pasture]

I don't know what's an algebraic variety and what's supposed to be the more general stuff you're speaking about, but okay

[quaint ivy]

a prof in my department loves this book lol, Book 1 is supposed to be more concrete indeed. According to a friend it's quite well written though the "old Russian book" style really shows

[sour plume]

If a Lie subalgebra $\mathfrak{p}$ of a real, semisimple Lie algebra $\mathfrak{g}$ complexifies to a Lie subalgebra $\mathfrak{p}\mathbb{C} \subset \mathfrak{g}\mathbb{C}$ which contains a Cartan subalgebra of $\mathfrak{g}_\mathbb{C}$, did then $\mathfrak{p}$ necessarily contain a Cartan subalgebra of $\mathfrak{g}$?

[cloud walrus]

Lartomato

[chilly ocean]

wdym excluding rotations and mirroring?

https://www.whitman.edu/Documents/Academics/Mathematics/Huisinga.pdf anyways, this might be of use. (I found it pretty helpful, but not sure what exactly you're looking for.)

[chilly ocean]

[carmine fossil]

if a and b are conjugate they should have the same cycle type

In other words a^i should be a 9-cycld

So everything except integers of form 3k

When you write a permutation as a product of disjoint cycles note the number of elements in each cycle

That sequence is cycle type

For example,cycle type of (1 2 3 4 5) is 5

for (1 2) (3 4) it's 2,2

11 is the general statement

Hint: follow the disjoint cycle containing 1

If i is not relatively prime,That cycle has less than m elements

Otherwise,it will have m elements

This works because of a modular arithmetic property

[quaint tree]

I just realized that $\sqrt[3]{1}\in\mathbb{Q}(\sqrt{-3})$ and I'm not sure how I feel about that. The math checks out; I proved it like 5 different ways, but it just genuinely weirds me out that a cube root of an integer is contained within a degree 2 field extension of $\mathbb{Q}$ even though I proved it a bunch of times.

[cloud walrus]

Elliptic Curve of Rank 9001

[quaint tree]

(I should maybe clarify that I'm talking about the complex roots, by the way)

[sturdy marsh]

this is almost always the case

a primitive nth root of unity lives in an extension of order phi(n)

where phi is the euler totient function

x^n - 1 is not irreducible

in the case n=3, x^3 - 1 = (x-1)(x^2 + x + 1)

the primitive cube roots of 1 are roots of x^2 + x + 1

which is why they live in a degree 2 extension

[quaint tree]

Ah yes, that helps my intuition a lot. Thanks.

[vestal snow]

Does E follow directly from D or am I applying it incorrectly?

Here's the definition of a closed embedding

[brazen pumice]

I mean, you still need to talk about localizations

[vestal snow]

Using D, can't we just say that since we can cover Spec B with, well Spec B such that the inverse image is isomorphic to Spec A and the induced map B --> A is surjective, we're done

[vestal snow]

A property is affine local if given an affine open cover U such that each set of U has the property, then every affine open has the property

[brazen pumice]

so using your cover, to prove that Spec A -> Spec B is a closed embedding, you need to show that Spec A -> Spec B is a closed embedding?

[vestal snow]

Yes

Well

No

We need to show that the inverse image of Spec B is isomorphic to Spec C for some ring C and the induced morphism B --> C is surjective

In this case, C is just A and the induced map is what we started with

[brazen pumice]

but your definition says "every" affine open subset

[vestal snow]

But we can use D to require just proving it on an affine open cover

[brazen pumice]

D is talking about the property "is a closed embedding"

if you want to use D, you need to prove that every element in your cover has the property "is a closed embedding"

i.e. you need to prove that Spec A -> Spec B is a closed embedding

[brazen pumice]

you're confusing it with the property "(the ring map) is surjective"

[vestal snow]

What is the statement of D in precise terms?

[brazen pumice]

if I have a map X -> Y and if {Yi} cover Y and if π^-1(Yi) -> Yi is a closed embedding for each i, then X -> Y is a closed embedding

[vestal snow]

I misinterpreted it

Thanks

[brazen pumice]

np

[torpid harbor]

Hello! How would one prove that any irreducible element of a Bezout Domain is a prime element ?

[torpid harbor]

(I'm asking here cuz I saw this and I thought it was related to field extensions, but if not I'm sorry, where should I post my question? I'm a bit new here)

[sturdy marsh]

is this true?

oh wait it might be

hmm

[torpid harbor]

this book asked me to prove it

[sturdy marsh]

ye it's fine

fix any irreducible element x

assume there is an ideal J containing (x)

pick an element in J which is not in (x)

say y

then (x) + (y) = (z)

[torpid harbor]

yes

and z should be in J right

[sturdy marsh]

z is in J, sure

[torpid harbor]

wait oooo can't we do smth like

[sturdy marsh]

z divides x

but z also divides y

x does not divide y

so z cant be x

or a unit times x

[torpid harbor]

wait but x is irreducible right so in order for it to be in the ideal, every other element has to be a multiple of x right?

every other element in the ideal

[sturdy marsh]

which ideal

[torpid harbor]

J

[sturdy marsh]

no

J contains (x)

(x) subset J

[toxic mortar]

How do i do this?

im confused

[torpid harbor]

like if x is irreducible and x is in J, J has to be equal to xR right ? where R is the bezout domain we're working on

cuz otherwise

x would be a multiple of a number

oo wait okay I'm dumb

[sturdy marsh]

I mean after we're done with the proof that's true

[torpid harbor]

J can be equal to R times a unit right

[sturdy marsh]

J can be R

[torpid harbor]

ofc

oh

but I mean like

this has to be a process we can't do on non Bezout domains right? how do we use the fact that R is a bezout domain ?

I think that's what I don't understand

[sturdy marsh]

.

[torpid harbor]

oh

[sturdy marsh]

that's where we use that it's a bezout domain

again, here's the argument from the start

assume there's an ideal J which properly contains (x)

assume J is not R

so there is a non-unit y in J \setminus (x)

(x) + (y) = (z) as we're working with a bezout domain

(x) + (y) is not the entire ring as it is contained in J

so z is not a unit

but z divides x and z is not equal to x (as z divides y)

did that make sense @torpid harbor

[torpid harbor]

right I think I see

by (x) you mean xR right

like the ideal formed by x?

[torpid harbor]

but isn't this assumption equivalent to x not being irreducible ?

I think you proved the other direction innit

but I kinda see the idea, I think I could do the same for the direction I want (after all it's clearly an iff right)

I still have to formalize it in my head tho

[sturdy marsh]

no, this is equivalent to saying that (x) is not maximal

[torpid harbor]

but ty lots @sturdy marsh! I think I have a clearer picture

[torpid harbor]

ohh

mmm

[vestal snow]

If B --> A is surjective, then Spec A --> Spec B is injective. Does this work the other way around?

Since Ring isn't an abelian category, I'm not sure

[hidden haven]

The inclusion Z → Q should be a counterexample

[vestal snow]

Any idea on how to do this?

Let Spec C be affine open and let Spec D be it's preimage

I know that the induced map Spec D --> Spec C is injective

But how do I show that C --> D is surjective?

Given B --> A is surjective

[prisma ibex]

Take G=Sp_4 and its two standard maximal parabolics with Levi decompositions P_1=M_1N_1 and P_2=M_2N_1. Then M_1=GL_2 and N_1 is a length 2 unipotent algebraic group 0≤N'_1≤N_1 with M_1 acting by N_1/N'_1=V_2 the standard representation and N'_1=det, and M_2=SL_2xGL_1 and N_2 is a length 1 unipotent algebraic group with M_2 acting by N_2=Ad=Sym^2(V_2)\otimes det^{-1}.

This persists to dual groups, after switching the long and short roots. Now what do these decompositions look like for G=Sp_6 and its dual group? Now one has three standard maximal parabolics with Levis GL_3, SL_2xGL_2, and Sp_4xGL_1, and intermediate parabolics with Levis GL_2xGL_1, GL_2xGL_1, and SL_2xGL_1xGL_1.

[vestal snow]

Is surjectivity here using that we can check epimorphismness of a sheaf morphism by checking if it induces surjective stalk morphisms?

[prisma ibex]

yea morphisms of sheaves are surjective iff surjective on stalks

to see surjective -> surjective on stalks use skyscraper sheaves

[vestal snow]

Thanks

Can someone help me prove that an open immersion is locally of finite type

I have proved that it suffices to show this when the target is affine

Let pi: U --> Spec B be an open immersion

Then U can be written as a union of distinguished open sets of Spec B

Each of these induce the map B --> B_f for some f

and this is clearly a finitely gen B algebra

How do I continue?

[vestal snow]

This would immediately follow if locally of finite type was affine local on the source, but it seems to good to be true

[gritty sparrow]

Open immersions aren’t necessarily of finite type i think

[vestal snow]

Vakil disagrees

[gritty sparrow]

Locally of finite type

[vestal snow]

Ah my bad

Sorry

[gritty sparrow]

Which is what you showed basically

[vestal snow]

Did I?

I still need affine localness on the source to finish this

Which I don't think holds in general

[vestal snow]

Let B --> C be a morphism such that B --> C --> C_f_i realizes C_f_i as a finitely generated B-algebra and (f_1,...,f_n)=1, then is it true that C is a finitely generated B-algebra?

[gritty sparrow]

Sorry, I lost signal for the internet, but aren’t we done by what you’ve done because for any u in U, we can cover u by some D(f) inside U and the map B->B_f makes B_f fg as a B module

@vestal snow

[vestal snow]

Not exactly sure what you're saying

I proved that U can be covered by affine sets, each having corresponding ring a finitely generated B algebra

[gritty sparrow]

I mean for each u in U we’ve shown that f is of finite type at u, hence f is locally of finite type

[vestal snow]

I think you're misunderstanding what locally of finite type means

[gritty sparrow]

I used an alternate equivalent definition

[vestal snow]

I see

We haven't covered that

yet

[vestal snow]

Do you know if this is true?

This would solve it

[gritty sparrow]

Yes it is true

[vestal snow]

Do you know how to prove it?

[gritty sparrow]

I remember it was a lot of index tracking, but other than that i don’t really recall much of the proof

[vestal snow]

I'm not sure that this is true

I thought I proved it, but (f_1,...,f_n) = 1 does not imply the coefficients of the linear equation of the f_i which gives 1 need to be in the image of B

Do you know any book where this is mentioned?

[gritty sparrow]

I don’t really recall, you can try matsumura. In any case, i’m pretty sure putting in the coefficients along with the fi and the c_i,j which are in C whose image in C_fi generate C_fi will give you the generating set

[gritty sparrow]

Yes this works

[vestal snow]

Can you provide a proof?

[south storm]

As far as I am aware is $GL_n(\bZ)$ finitely generated, is the same true for $M_n(\bZ)$ viewed as a monoid?

[cloud walrus]

27182818284tropy

[south storm]

Here GL_n(Z) denoting M_n(Z) with the condition that the determinant is +-1

[brazen pumice]

hint: let n=1

[tough raven]

Is M1(Z) not Z

[final pasture]

(yes that's the joke)

wait

[tough raven]

Is that not finitely generated?

[hidden haven]

Is it a joke tho?

[final pasture]

yes

[hidden haven]

Z isn't I think

[final pasture]

not it's not

[tough raven]

👀

[hidden haven]

As a multiplicative monoid

[tough raven]

Oh multiplicatively

Right

[final pasture]

I read that as a joke

and then

I noticed when Raghuram said something

[hidden haven]

Probably because of my

[final pasture]

that it was actually not dumb

lol

yes

because of your

[hidden haven]