groups-rings-fields

[hidden haven]

Yeah that's how you eliminate 2+4

Degree of extension should be a multiple of 3

But 2+4 says otherwise

[ivory dust]

Q(cuberoot3)(sqrt something?) = 2?

[hidden haven]

And 2+2+2 also says otherwise

[ivory dust]

I understand that but not sure how it implies

f(x) in Q cant be written as

factors of that form based on information in mod p = 3

so it must be written in the form 3+3 but im guessing thats not possible

in Q (also not sure how this is not possible)

[hidden haven]

Yeah exactly

We eliminated 3+3 using F_3

[hidden haven]

Is this the part that is not clear?

[ivory dust]

Err its more like how not being able to factor something in Fp in a certain way implies it cant be factored as such in Q

but its factorable as 3+3 in Fp

[hidden haven]

Ah

[ivory dust]

but not in Q

that too

[hidden haven]

Suppose f = gh over Q, where g and h have degree 3

We can assume then that g and h are monic

Because f is

And by Gauss we can assume integer coefficients

Reduce the whole equation mod 3

[ivory dust]

wait whats this mean?

irred in Z iff irred in Q?

[hidden haven]

f_3 = (gh)_3 = g_3 h_3 (_3 means reduced mod 3)

[hidden haven]

Yeah, a factorization of an integer polynomial over Q can be turned into a factorization over Z

[ivory dust]

ohh ok

[hidden haven]

So we can take factorizations of type m+n from Q to F_p for any p

Taking contrapositive, if f doesn't have a factorization of type m+n over F_p, it doesn't in Q

[ivory dust]

Oh wait Gauss Lemma is when g,h are primitive, and if they are monic they are primitive? so they can be factored over Z iff factored over Q??

[hidden haven]

Yep

[ivory dust]

and they must be integer polynomials over Q which they are

ok i get it

[hidden haven]

Yeah useful fact to remember

Monic integer polynomials that factor, factor into monic integer polynomials

[ivory dust]

and it doesnt have 1+5, obviously since it has no roots in Fp, it doesnt factor as 2+2+2, 2+4 either (uhh thinking as to why)

it only factors as 3+3

and hence if it doenst factor as 3+3 in Z or Q

then it cant factor as anything

2+2+2, 2+4 is cuz of degree of extension

[hidden haven]

3+3 is eliminated because if that were the case, then 2+2+2 would be a problem

[ivory dust]

3+3 we showed f(x) = (x^2-2)^3

[hidden haven]

Because 2+2+2 is the unique factorization over F_3

[hidden haven]

(x²-2)³

[ivory dust]

Oh fuck yea

mb

every polynomial has a unique factorization right

[hidden haven]

Yeah over a field

Once you factor into irreducibles

x²-2 is irreducible because no root

In F_3

[ivory dust]

so only possibility is 2+2+2 in Q?

and this is not possible as the extension field

[hidden haven]

So 2+2+2 is the unique factorization in F_3, and 3+3 is then not possible in F_3

[hidden haven]

It would be 2+4

2+2+2 doesn't necessarily pull back to the same

Like factorization type over Q gives you factorization type over F_p

Not the other way around

[ivory dust]

oh i see

[hidden haven]

2+2+2 could have come from a 2+4

Further reducing

[ivory dust]

jesus

[hidden haven]

Lol

[ivory dust]

so Q can be 2+2+2 or 2+4

[hidden haven]

Yes, and both are problematic

[ivory dust]

i.e., degree of extension field

of those minimal polnyomials over Q

are divis by 2, 4

but we need it to be divisible by 3

since Q(cuberoot2) : Q = 3

[hidden haven]

Yep

[ivory dust]

and Q < Q(cube root 2) < Q(a)

2+2+2 or 2+4 means that the minimal polynomials containing roots with those degrees over Q means Q extended over ideal generated by those polynomials with degree 2 or 4 containing a as a root is 2 or 4

[hidden haven]

Yep exactly

[ivory dust]

so Q(a):Q = 2 or 4

[hidden haven]

The degree of Q(a)/Q being 3 or 6 means that min a is degree 3 or 6

But we should deg 3 isn't possible because that would lead to a deg 3 factor over F_3

[ivory dust]

but we require deg ext Q(a)/Q = deg ext Q(cuberoot)/Q * deg ext Q(a)/Q(cuberoot)?

deg ext Q(a)/Q = 3 * deg ext Q(a)/Q(cuberoot)?

[hidden haven]

Not sure what you are doing there

[ivory dust]

its like this

[hidden haven]

Ah

[ivory dust]

degree formula

for F<B<E

[hidden haven]

Yeah using the tower

We get deg Q(a)/Q = 3 or 6

And all of this work was to show that 3 isn't possible

[ivory dust]

LMFAO

omfg

[hidden haven]

By showing that f(x) has no deg 3 factors

[ivory dust]

WHEN i saw this problem

i wanted to do it first

CUZ I THOUGHT TIS EASY

omfg

[hidden haven]

rip

[quaint tree]

I mean we know $\left[\mathbb{Q}\left(\sqrt[3]{2}\right):\mathbb{Q}\right]=3$, so the problem is just finding $\left[\mathbb{Q}\left(\sqrt{1+2\sqrt[3]{2}}\right):\mathbb{Q}\left(\sqrt[3]{2}\right)\right]$, which is either 1 or 2. That is, determine whether $\sqrt{1+2\sqrt[3]{2}}\in\mathbb{Q}\left(\sqrt[3]{2}\right)$.

[cloud walrus]

Hellinion said I'm a buffoon

[quaint tree]

Which is actually a nontrivial problem now that I think about it for longer than three seconds. Maybe there's an argument in the trace?

[hidden haven]

We got a solution by going mod 3 on the deg 6 monic that that element obviously satisfies

Mod 3 it factors as (x²-2)³, so can't have cubic factors over Q

So degree of the whole extension can't be 3

[ivory dust]

that seems kinda hard

like a+b2^(1/3) + c2^(2/3) = sqrt(1+2*2^(1/3))

i tried that

a,b,c e Q

and got some new equation pretty lengthy

but it becomes 0 = m+p2^(1/3) + q2^(2/3)

m,p,q e Q

is this impossible?

and factored it set = 0

its $0 = (4bc-1+a^2) + 2^{1/3}(2ab+2c^2-2) + 2^{2/3}(2ac+b^2)$

[cloud walrus]

Ramtin

[ivory dust]

can just rewrite as this perhaps?

$0 = m + 2^{1/3}p + 2^{2/3}q$ where $m,p,q \in Q$

[cloud walrus]

Ramtin

[ivory dust]

Not sure how to prove this is contradictory

cuz then this method would be kinda easier, or more rudimentary at least

[dusty river]

These kind of arguments are usually not easy lol

There are techniques that use trace and norm for this stuff

and you can sometimes come up with more elementary arguments too

but no standard technique

[ivory dust]

isnt like

a = 2^1/3 b + 2^2/3 c

always impossible

is that trivial?

[dusty river]

That is fine yeah

[ivory dust]

really

[dusty river]

Yeah because 1, cbrt(2), cbrt(2)^2 are lin ind

You should be able to do this from the field theory you know

[ivory dust]

Ohhh

thats cuz

x^3 - 2 over Q irreducible so Q/this polynomial ideal forms field, and its forms a vector space with dimension 3 (3 basis) over Q

since degree x^3-2 = 3

?

so lin independent

etc

[dusty river]

Yeah and those three form a spanning set

Another way to phrase it is that if they were linearly dependent then the linear combination of them that equals 0 will give a lower degree polynomial than the minimal polynomial

Because the highest degree in that would be 2

[ivory dust]

Thus sqrt(1+2*2^1/3) is not in Q(cuberoot2)

so the degree of the extension is not 1

so it has to be 2

it cant be more than 2 cuz

[dusty river]

oh wait I didn't see the argument

[ivory dust]

minimial polyn is at most quadratic

[dusty river]

This is a contradiction only if m, p, q are non zero

The harder thing is to prove that they can't be 0

[ivory dust]

well

for q i got

q = 2ac+b^2

if 0 = 2ac+b^2

-2ac=b^2

not possible

[dusty river]

a and c can be negative

[ivory dust]

but b cant have sqrt2

no?

b is in Q

[dusty river]

a could be 2 and c -1 for example

and b=2

[ivory dust]

oh

yikes

thats too hard then

💀

[dusty river]

Yeah lol

[ivory dust]

btw is this valid argument?

polynomial containing a as a root is degree 6

may or may not be irreducible

so thus

[Q(a):Q]<=6

[dusty river]

Yeah

[ivory dust]

And if f(x) factors in Fp in Z or Q (if its primitive in Q) it factors in similar fashion?

sorry we went over this

just tryna soldify

[dusty river]

Factoring in F_p doesn't imply factoring in Q

You use a homomorphism, Z → F_p

Which is the usual mod p homomorphism

This extends to a homomorphism Z[x] → F_p[x]

[ivory dust]

ok i follow

[dusty river]

So if you have an equation f(x) = g(x) h(x) of integer polynomials

You apply the homomorphism on both sides of the equation

The way that the homomorphism acts is that it takes every coefficient mod p

Call the homomorphism α

α(f(x)) = α(g(x)h(x)) = α(g(x))α(h(x))

So the equation remains true when you just apply the mod p thing to each coefficient

So any factoring in Z can be made into a factoring in F_p

You just have to be slightly careful because going mod p, some of the coefficients could become 0

So for example your whole polynomial could become 0 in which case it isn't much of a factoring

[ivory dust]

so just make sure everythings monic

[dusty river]

Yep

[ivory dust]

and that deals w that

[dusty river]

Yeah

[ivory dust]

it is indeed monic

since f is

so g and h must be monic

if they exist

[dusty river]

Yeah, so if the factoring was a non trivial factoring, it remains non trivial

Also there are polynomials which are irreducible over Z, but reduce modulo every prime

So the converse is very false

[ivory dust]

the non trivial factoring (x^2-2)^3 in Z3 implies if f factors in Z it will be in factors of degree 2+2+2 or degree 2+4

[dusty river]

Yeah, can you say why that would be the case

[ivory dust]

2+2+2 makes sense cuz some of those integers mod 3 may have been like diff numbers in Z (Q)

or identical?

[dusty river]

Yeah they could be different

but I am asking if you can give an argument for why no other types of factorizations are possible

[ivory dust]

like if ifs (x^2-2)(x^2-2)(x^2-2) the constant 2 terms in each could be diff in Z or the same

well 3+3 is not possible since

if it were

then there would be a root

in Zp

[dusty river]

Why

[ivory dust]

cuz everything must factor into linear or quadratic

over a field?

or something

[dusty river]

No

that's only over R

[ivory dust]

Oh

[dusty river]

or real closed fields in general

Not over Q or F_p

[ivory dust]

Well 1+5 is trivial since definetly no linear term

since no roots

[dusty river]

yep

[ivory dust]

3+3 is not possible since if it were

umm

Something abt the homomorphism?

Noo nvm

Is it cuz we know the factorization

and thats unique

[dusty river]

Yeah apply that reduction mod 3 homomorphism to the factorization equation

what do you get

(and even to apply the homomorphism you first need everything to be integer and as you said, preferrably monic)

[ivory dust]

Not sure why the homomorphism wouldnt work on two degree 3 polynomials

or would show a contradiction

[dusty river]

It would

After using Gauss lemma and all to justify that the factoring is indeed a factoring into monics in Z[x]

[ivory dust]

Polynomials reduce uniquely? isnt that true?

[dusty river]

You get a 3+3 factorization over F_3

Yeah that is true over fields

[ivory dust]

So

3+3 is impossible since we showed existence

of another factorization

[dusty river]

What if this 3+3 thing broke down further

[ivory dust]

to 2+2+2?

hm

[dusty river]

The uniqueness only holds if it is factorization into irreducibles

[ivory dust]

i see

x^2-2 is ired Z3

Hm

[dusty river]

Yep

Because no roots

[ivory dust]

3+3 case still trumps me

ahh

its not possible since

two monics

with degree 3

HM

Idk

[dusty river]

Well the two deg 3 polynomials themselves reduce to some irreducibles

What could those irreducibles be

[ivory dust]

2,1

or 3

or 1 1 1

[dusty river]

Correct, but not their degrees

[ivory dust]

but anything with 1 is nogo

[dusty river]

You can actually say what those irreducibles will be

Like as polynomials

[ivory dust]

g and h?

monic

not sure what else i can say

[dusty river]

give me a sec

[ivory dust]

their last term must multiply to 2

[hidden haven]

So x⁶-2 = (x²-2)³

Now suppose x⁶-2 = g(x) h(x)

And g(x) and h(x) themselves have some factorizations

Do you see that the product of those 2 factorizations will then give a factorization of x⁶-2 itself?

[ivory dust]

Yea

[dusty river]

ok better way to phrase it

From that factorization

[dusty river]

This one

You can list all factors of x^6 - 2

And you should see that there are no deg 3 factors

Do you see how?

[ivory dust]

If g and h have a factorization it contributes to the (x^2-2) one?

when u multiply them

[dusty river]

Too vague

Forget about what that Moldilocks guy said

useless stuff

[dusty river]

Just try this

like list all factors

[ivory dust]

(x^2-2)

[dusty river]

There are more

1 for example

[ivory dust]

(x^4-x^2+2)

[dusty river]

If you need a hint, try doing the same for natural numbers. Suppose I gave you 255, and asked you to list all of its factors. It is a hard problem unless I also tell you that 255 = 17*3*5, with which it should become very easy

[dusty river]

Right, and also the polynomial itself. Why can't there be any other factors? That is what you want to prove

[ivory dust]

well if a factor of deg3 exists

[dusty river]

Do try it with numbers first if you are struggling with the intuition

[ivory dust]

then its a factor of deg 1 * factor of 2 deg

but no deg 1 factors

[dusty river]

Right, and it can't be irreducible because

[ivory dust]

we have lower

irreducible factors

they are unique

[dusty river]

Well not just lower

We don't have irreducibles of deg 3

Like in this case yes lower

But this argument would work even if factorization over F_3 were of type 2+4

because there are no deg 3 irreducibles in that

[ivory dust]

I seeee

ok tysm 😭

[dusty river]

[signal remnant]

Concerning an exercise I'm working through ... "If r|m and s|n, find a subgroup of Z_m \oplus Z_n that is isomorphic to Z_r \oplus Z_s" (not HW, more interested in just getting a hint b/c I've been stuck today).

My current approach is to use the facts that r|m & s|n to establish |Z_r| | |Z_m| & |Z_s| | |Z_n| so that we can say |Z_m|/|Z_r| = k \in \mathbb{N} and likewise |Z_n|/|Z_s| = t \in \mathbb{N}. From here I let G = Z_k \oplus Z_t with the intention that I will first show that G is a subgroup of Z_m \oplus Z_n and then afterward show that G is isomorphic to Z_r \oplus Z_s to complete the proof.

My issue is that when I try to use subgroup tests to show that G is a subgroup of Z_m \oplus Z_n, I end up with nothing that has anything to do with G itself aside from the connection I am still struggling to put to symbols to (which is practically the original problem statement, I feel like).

So IDK, I'd just like a hint if anyone wants to point me either forward toward the way I'm already going (ie check my work) or back to the problem statement with a hint that "There is a better way of doing this [than you have already described]..."

thnx. been stuck on this longer than i care to admit.

[delicate orchid]

have you tried showing that Z_k is a subgroup of Z_m and Z_t is a subgroup of Z_n and then showing that the direct product of subgroups is a subgroup of the direct product of the groups

that's probably how I'd proceed

[vast quiver]

How do you show C[x1,x3,x1x2,x2x3] is isomorphic to C[w,x,y,z]/(wz-xy)? My approach so far has been to define a map phi: C[w,x,y,z]->C[x1,x3,x1x2,x2x3] sending w,x,y,z to the respective generators. I've shown that (wz-xy) is in the kernel of phi, but how do I do the reverse inclusion?

More generally, is there a strategy to realize certain subalgebras of C[x1,...,xn] as quotients of C[x1,...,xm]? I'm trying to learn about affine toric varieties, and Fulton's book keeps doing this sort of thing.

[maiden ocean]

Uh is there a reason you need to do the reverse inclusion? Isnt surjectivity sufficient by the first isomorphism theorem?

Which seems clear to me because you know that both C and your generators are in the image of this map so...

[vast quiver]

uh am I brain farting

The inclusion I already have tells me that the map factoring through C[w,x,y,z]/(wz-xy) is well-defined

wouldn't the reverse inclusion tell me injectivity?

definitely surjective, I agree

I'd need that ker phi = (wz-xy) in order to get C[w,x,y,z]/(wz-xy) is isomorphic to C[x1,x3,x1x2,x2x3]. Otherwise, I only get that C[x1,x3,x1x2,x2x3] is isomorphic to a quotient of C[w,x,y,z]/(wz-xy), right?

I just get stuck as soon as I pick an element in the kernel. Let p(w,x,y,z) be in the kernel, then p(x1,x3,x1x2,x2x3)=0 in C[x1,x2,x3]. I need to show p(w,x,y,z) is in (wz-xy), or equivalently p(w,x,y,z)=0 mod (wz-xy)

I'm not sure where to go from here. I feel like I should be able to do something like, from the relation wz-xy=0 that for some indeterminate t, we have y=wt, z=tx, so that I get p(w,x,y,z)=p(w,x,wt,tx) mod (wz-xy)

Then rename things and use the fact that p(x1,x3,x1x2,x2x3)=0.

I'm not sure how to go about this though

[maiden ocean]

Oh oops im illiterate

you said that (wz - xy) is in the kernel

not that it was the kernel

[wooden ember]

completed the exercise but in doing so i never found at any point that n=6 gave a special case for the argumentation. Which subproof fails at n=6?

my hunch is b) but im not sure

yeah the next exercise shows it's b) it seems

i think i just missed that special case is all

oh it's cause i did b wrong that's why lmao

[vast quiver]

I've been thinking about this some more... I think the multiplicative monoid generated by {x1,x3,x1x2,x2x3} has a monoid presentation {w,x,y,z | wz=xy}. Then, this monoid homomorphism extends naturally to a C-algebra homomorphism.

This also generalizes a bit nicer. I don't really know anything about monoid presentations, but assuming this works out, it would be easy to then show that affine toric varieties that come from cones are cut out by binomial ideals--you just take the relations of the monoid presentation, e.g. wz=xy would go to the ideal (wz-xy).

This kinda reduces the problem, but idk in general how to find the presentation of a monoid. I guess for affine toric varieties I only need to do it for submonoids of Z^n (e.g. the one generated by {e1,e3,e1+e2,e2+e3}, which corresponds to the {x1,x3,x1x2,x2x3} example).

[wooden ember]

im not feeling too sure, but the abelian property is preserved under the 4th isomorphism theorem (correspondence theorem) right?

so if A is an abelian subgroup of G containing the normal subgroup N, is A/N abelian in G/N?

and vice versa?

(im more interested in the opposite direction)

the first question is clearly true actually but im not sure about the other direction

[mild laurel]

The other direction isn't true, the trivial subgroup is abelian in G/N, but this corresponds to the subgroup N of G, which is not always abelian

[wooden ember]

right of course

thanks

[wooden ember]

this is the moment ive been dreading

[latent night]

"find the Center of d_2n using a presentation". How does one use a presentation to do this? I am not that good with presentations, but of I understand, a presentation of the group would be {x, y} as generators, and {x^n, y^2, xyxy} would be the relations generating the kernel of the surjection F({x, y}) -> D_2n. I don't get how this tells us anything about the Center of D_2n.

Where d_2n is the dihedral group with n elements and y is a flip x is a clockwise rotation.

[wooden ember]

It's simple to show x^(n/2) is in the center if n is even. Use the relation xy=yx^-1 to show you cant have another nontrivial element in the center

was about to ask a question, realized i was being stupid

there's progress i dont realize it right after asking the question now lmao

[latent night]

Can I not say that if an element a = x^k is in the Center the relation you gave implies that ya = ya^-1 => a = a^-1, which only holds true for k = n/2?. And elements of the form (x^k)*y, k≠0 cannot be in the Center since they don't commute with elements other than x^n/2.

I don't get where a presentation is coming up. Sorry maybe I don't get what a presentation is, I've only seen the definition.

[wooden ember]

the presentation came up in the sense that you used the relation xy^=yx^-1

but yes your method works

a presentation is a set of generators and a set of relations, such that every element of the group is some product of the generators and every relation can be deduced from the generating relations

[latent night]

Yeah, the way I heard it is that F(A) is a free group, and you have a surjection from the free group to the group you care about F(A) -> G, you can take R to be a set of "relations" such that R generates the kernel of the surjection, and then the first isomorphism theorem gives that G is isomorphic to F(A)/F(R). This makes perfect sense (I think?) But I have a hard time seeing how it is useful.

[wooden ember]

idk dont have experience with free groups yet

[novel parrot]

not really sure about this one

but could i say that any element of M/N will have form of m + N in RA for some set A but N is also finite generated, so we can take the inverse of N which is also finitely generated to see that m would be generated by a finite set too

[thorny knoll]

M/N and N are both finitely generated. So as a start, you could take a look at the systems that generate each, respectively, and try to construct a set in M from those that generates M

[novel parrot]

M/N = $Ra_1 + \cdots + Ra_n$

[cloud walrus]

ActiveChapter

[novel parrot]

somewhere in the set of all sums, N must be there

so we can just inverse the generating set of N

and get a generating set just for M right?

[stone fulcrum]

N is very specifically not there

[novel parrot]

oh

okk

[stone fulcrum]

In a very lazy way of thinking M/N represents what M might look like if suddenly differences by N don't matter

[novel parrot]

oh so we can just add generating sets for M/N and N

for a set for M

yes?

[hidden haven]

That's not even lazy though

[novel parrot]

am i correct?

[stone fulcrum]

Haha fair enough I was afraid people wouldn't like how imprecise I was

[hidden haven]

This doesn't really make sense

[stone fulcrum]

Or, to put it even lazier, what structure M has left if the structure that belongs to N is removed

Sadly it isn't really the same as a "subtraction" and so combining generating sets won't do it

[hidden haven]

M/N is not a submodule of M, so you can't take elements of the former

They just aren't there in the latter

But you are very close

[novel parrot]

do we need to consider the natural homomorphism

[hidden haven]

Whenever you work with a quotient, you either implicitly or explicitly are using it

[novel parrot]

yeah

[hidden haven]

So I mean you don't need to mention it necessarily

But you can use it yes

[novel parrot]

to get M/N we make N the new identity

so just putting N back will give us M back

[hidden haven]

It seems intuitive, is true for vector spaces, but unfortunately not for modules

(M/N) x N is not necessarily isomorphic to M

M = Z, N = 2Z is a counterexample

[novel parrot]

i see

uhh

what about

taking union of generating sets

[hidden haven]

So generators on M/N, view them as cosets, then take a union of the generating cosets?

[novel parrot]

no

[hidden haven]

Then I didn't get it lol

[novel parrot]

$m = \sum r_ia_i + N$

[cloud walrus]

ActiveChapter

[novel parrot]

this is how an element of M/N will look right?

[hidden haven]

Yeah

Assuming a_i are the generators of N

[novel parrot]

so m = sum(ri ai) + n

some element of N

so we get a finite sum

i dont get why this wont work

[hidden haven]

Wait

[hidden haven]

This was stupid

What are a_i

[novel parrot]

a_i is in A, the set that will generate M/N

[stone fulcrum]

Generators of N are known? Right?

[novel parrot]

yes

[hidden haven]

So a_i are elements of M/N?

[novel parrot]

uh

no

[hidden haven]

So A is finite and generates M/N

Elements of A would then be elements of M/N. I didunt get how you got the a_i then

I actually transcribed didn't like it's pronounced

[novel parrot]

ring might not have 1?

[hidden haven]

I don't see where that comes in

[novel parrot]

hmm

maybe i dont understand what it means for a module to be generated

[hidden haven]

Everything in the module is a finite R-linear combinations of the generators

[novel parrot]

right

dont we need the ring element to have a 1

so module contains its generating set

[stone fulcrum]

Following. I don't know the answer and am interested

[novel parrot]

$M/N = {r_1a_1 + \cdots + r_na_n}$ and $N = {r_1b_k + \cdots + r_kb_k}$ so $m + N = \sum r_ia_i + N$ so for some n , $m = \sum r_ia_i + \sum r_jb_j$

[cloud walrus]

ActiveChapter

[novel parrot]

we can get all elements of M this way

so just take union of generating sets for a generating set of M/N right?

[hidden haven]

The idea is correct yes

But there is a problem with how you've written it because the last equation doesn't make sense

That should be an equation in M

[novel parrot]

it is

[hidden haven]

But a_i, as you have defined them at the beginning are not elements of M

[novel parrot]

the a_i are elements of M

[hidden haven]

They are cosets

So you have to correct the definition

Elements of M/N are cosets

Not elements of M

[novel parrot]

oh

$M/N = {r_i(a_i + N)}$

[cloud walrus]

ActiveChapter

[novel parrot]

like that?

[hidden haven]

Yeah, sums of those

[novel parrot]

oh right okay

but also

from before

it may not contain its generating set if ring doesnt have 1?

[hidden haven]

You just include it in that case

The submodule generated by A is the set A and all its finite linear combinations

Easy to check that it's a submodule

[dark lichen]

So if $\phi : G \rightarrow D$ is a homomorphism of groups with $D$ abelian, it's clear that $[G, G] \subseteq \text{Ker}(\phi)$, right (and for clarification I'm writing $[G, G]$ as the commutator subgroup)? I'm wondering in what conditions it would be the case that $\text{Ker}(\phi) \subseteq [G, G]$, whether or not $D$ is assumed an abelian group

[cloud walrus]

xdd1etteetteette

[hidden haven]

Im(phi) is an abelian subgroup iff commutator is in the kernel

[novel parrot]

im reading the part on tensor products but i just dont understand

if the R-Module N, was already an S-Module why are we defining multiplication by S

and what does "no relations" mean

[fossil shuttle]

N isn't already an S module

in the first line they say that they're not considering the case that N is an S-module because then the solution to the problem they're describing would be obvious

a relation here is some kind of equation that is imposed on elements, or a constraint, something like $\sum (s'_i, n'_i)=0$. in other words they're saying that if you have a bunch of pairs $(s_i,n_i)$ indexed by $i\in I$ for some index set $i$, and all of these are distinct (so for $i\neq j$ we have either $s_i \neq s_j$ or $n_i\neq s_j$ then $\sum r_i (s_i,n_i)=0$ for coefficients $r_i\in R$ if, and only if, each of the $r_i$ is zero.

[cloud walrus]

diligentClerk

[fossil shuttle]

sorry they said free abelian group so I should have said for coefficients $r_i\in \mathbb{Z}$, not $R$

[cloud walrus]

diligentClerk

[fossil shuttle]

you can think of elements of the free Abelian group as just being the set of all functions from $S\times N$ into $\mathbb{Z}$ which are zero on all but finitely many elements

[cloud walrus]

diligentClerk

[fossil shuttle]

there's an embedding of $S\times N$ into this Abelian group, where you send $(s,n)$ to the function $f_{s,n} : S\times N\to \mathbb{Z}$ sending $(s,n)$ to $1$ and everything else to zero. when we talk about formal sums of elements $(s,n)$ we're talking about adding them up in this larger group

[cloud walrus]

diligentClerk

[novel parrot]

uhm

when we say free abelian group of S x N

we just take finite sums of (s_i, n_i) right?

[fossil shuttle]

yeah, i'm just giving another way of thinking about it. the formal sum $\sum k_i (s_i, n_i)$ for $k_i$ an integer can be thought of as the function $S\times N\to \mathbb{Z}$ which sends $(s_i,n_i)$ to $k_i$ for each pair $(s_i,n_i)$ in the sum, and all pairs $(s,n)$ not occurring in the sum get sent to zero. then addition and subtraction of functions are done pointwise

[cloud walrus]

diligentClerk

[novel parrot]

ohh

[fossil shuttle]

note that these are "formal" sums, we don't use the additive structure on $S$ or $N$ when forming these sums

[cloud walrus]

diligentClerk

[novel parrot]

why have we defined H like that?

[fossil shuttle]

ok let me think about this. so, let's say $A$ is an $S$-module, right? What does that mean. it means that $S$ can multiply elements of $A$, so whenever you have $s\in S, a\in A$ you can multiply them to give $s\cdot a =a'$, some other element in $A$. This gives a multiplication function $S\times A\to A$.

[cloud walrus]

diligentClerk

[novel parrot]

yep

[fossil shuttle]

now part of the axioms of a module action is that it satisfies
$s\cdot (a+a') = s\cdot a + s \cdot a'$, and $(s+s')\cdot a = s\cdot a+ s'\cdot a$
That means that for this map $\mu : S\times A\to A$ to constitute a module action, it has to satisfy
$\mu(s,a+a') - \mu(s,a)- \mu(s,a')=0$
and $\mu(s+s',a)-\mu(s,a)-\mu(s',a)=0$

[cloud walrus]

diligentClerk

[novel parrot]

right

[fossil shuttle]

so like

let me write $A$ for the free Abelian group on $S\times N$, so that elements of $A$ are these formal linear combinations of pairs $(s,n)$

[cloud walrus]

diligentClerk

[fossil shuttle]

there's an obvious choice of "action" of $S$ on $A$, namely that we should define $s\cdot (s', n) = (ss', n)$ and more generally $s\cdot \sum k_i(s_i, n_i) = \sum k_i (ss_i,n_i)$

[cloud walrus]

diligentClerk

[fossil shuttle]

but there are some problems with this choice

[novel parrot]

what problems?

[fossil shuttle]

i'm trying to think of an example lol

[novel parrot]

ok

thank you so much for helping me btw

[fossil shuttle]

so let $(s,n)\in A$, and let's say we try and compute $(s_1+s_2)\cdot (s,n)$. by the definition of the action I proposed this should be $((s_1+s_2)s,n) = (s_1s+s_2s,n)$.

[cloud walrus]

diligentClerk

[fossil shuttle]

but on the other hand if we break up the sum $s_1+s_2$ and then multiply, we get $(s_1+s_2)(s,n) = s_1(s,n)+ s_2(s,n)= (s_1s,n)+(s_2s,n)$. These two elements aren't the same! $(s_1s+s_2s,n)\neq (s_1s,n)+(s_2s,n)$ in the group $A$. The first one is the function $S\times N\to \mathbb{Z}$ sending $(s_1s+s_2s,n)$ to $1$ and everything else to zero; the second one is the function sending $(s_1s,n)$ to 1, $(s_2s,n)$ to 1, and everything else to zero.

[cloud walrus]

diligentClerk

[fossil shuttle]

so this map $S\times A\to A$ actually fails the module-action axioms.

[cloud walrus]

diligentClerk

[novel parrot]

right

[fossil shuttle]

another problem is that we are trying to construct something which should theoretically resemble the original module $N$ in some way. and the thing $A$ we've built so far looks nothing like $N$. We're trying to build an "extension" of $N$, another Abelian group which contains $N$ but also contains some "formal products" $s\cdot n$ that don't denote elements of $N$, they're new elements. Well, we've added these new elements, but we also want to somehow preserve the original structure of $N$ as an $R$-module somehow; in particular, there should be a map $N\to A$ which is an $R$-module homomorphism. If we have such a homomorphism then we can say $N$ appears in $A$ as a submodule (if the homomorphism is injective) or at least a quotient of $N$ appears in $A$ as a submodule (perhaps forcing $N$ to admit an action by $S$ kind of introduced some relationships between elements of $N$ that weren't there before, killing them off.)
Now, there is an obvious choice of such map, i.e. the one sending $n$ in $N$ to $(1,n)$, the formal product $1\times n$; but this map is not a homomorphism because $n+n'$ goes to $(1,n+n')$, but $(1,n)+(1,n') \neq (1,n+n')$ in $A$.

[cloud walrus]

diligentClerk

[novel parrot]

ok

[fossil shuttle]

and more generally if this is an $R$-module homomorphism we would have that $r\cdot n$ gets sent to $r\cdot (1,n)$, where in the second hand we view $r$ as an element of $S$; so this would simplify to $(r,n)$. But in general $(1,rn) \neq (r,n)$.

[cloud walrus]

diligentClerk

[fossil shuttle]

Killing off the subgroup $H$ in $A$ solves all of these problems.

[cloud walrus]

diligentClerk

[fossil shuttle]

Each of the three lines in the definition of $H$ corresponds precisely to one of the three problems I just described.

[cloud walrus]

diligentClerk

[fossil shuttle]

If you quotient out $A/H$, then the new group will have the property that:

1. the action of $S$ on $A/H$ is a module action, and
2. the map $N \to A/H$ is an $R$-module homomorphism.

[cloud walrus]

diligentClerk

[novel parrot]

i see

what will elements of A/H look like?

$(s_i,n_i) + H$

[cloud walrus]

ActiveChapter

[novel parrot]

ill carry on reading

hm

how does H solve this problem

$(s_1 + s_2)(s,n) \neq (s_1 s + s_2 s ,n)$

[cloud walrus]

ActiveChapter

[novel parrot]

@fossil shuttle

[fossil shuttle]

because $(s_1+s_2)(s,n) = ((s_1+s_2)s,n) = (s_1s + s_2s,n)$, which in the quotient group is equivalent to $(s_1s,n)+(s_2s,n) = s_1(s,n)+s_2(s,n)$

[cloud walrus]

diligentClerk

[fossil shuttle]

because the first line of 10.3 forces them to be equal

[novel parrot]

hm

the first line forces it to be 0

[fossil shuttle]

as $(s_1s+s_2s,n)-(s_1s,n)-(s_2s,n)=0$ in the quotient group

[cloud walrus]

diligentClerk

[novel parrot]

hm

as cosets they will be equal?

[fossil shuttle]

as cosets, yeah.

[novel parrot]

rightt

[fossil shuttle]

i wasn't thinking in terms of cosets, sorry

more in terms of representative elements

[novel parrot]

oh its fine, i think ive understood

thank you

[fossil shuttle]

np

[vestal snow]

Since the Cech complex preserves left-exactness and right exactness (of the Cech complex) is guaranteed by the additional hypothesis, this follows immediately right?

Because then we would get an exact sequence of complexes 0 -> F* -> G* -> H* -> 0

Where F* means this complex

[wind steeple]

yes

I guess so

[vestal snow]

Thanks

[vestal snow]

Doesn't 18.2.2 imply 18.2.4?

[sly crescent]

How are Steinberg groups notated when the same diagram has multiple automorphisms of the same order?

[vestal snow]

Since we can just cover X by itself (as it is affine) and then use 18.2.2

[gritty sparrow]

We are using 18.2.4 to show 18.2.2

[vestal snow]

Oh lmao my bad

[gritty sparrow]

Np

[novel parrot]

i dont understand whats going on in theorem 8 when we take the free Z-module on S x N

our map phi that we are extending is only defined on N

[fossil shuttle]

do you understand what A is in this scenario

[novel parrot]

S x N ?

[fossil shuttle]

yes

so we just need to define a set function from S x N into L and we get an abelian group homomorphism out of it yea?

[novel parrot]

yeah

[fossil shuttle]

so what function does the proof say we should take

[novel parrot]

phi

[fossil shuttle]

no

[novel parrot]

varphi

[fossil shuttle]

no

[novel parrot]

oh

it says S x N to L

[fossil shuttle]

yes

[novel parrot]

but

wouldnt we need an initial map to extend

like the domains dont match

[fossil shuttle]

the domains don't match because you keep trying to say the map on S x N is defined to be varphi. it's not

it tells you exactly what the map is

oh

are you confused by notation lol

the phi/Phi in theorem 6 isn't the same as the phi/Phi in theorem 8

that's probably not helping

ok throw out all the notation in theorem 6 hahaha

say we have a map t: A->M

where A is a set and M is an Abelian group

then this extends uniquely to a map t' from F(A) to M

where t' is an abelian group homomorphism

you've correctly identified that A in theorem 8 is SxN

what is t?

in the context of theorem 8

(i just posted a screenshot with the answer in it)

[novel parrot]

hmm]

in theorem 6

we had a map between A -> M

and extended it to a Homomorphism between F(A) -> M

[novel parrot]

i dont know what t is

oh wait

is it N -> S (x_R) N

@fossil shuttle

[fossil shuttle]

well we agreed that A was SxN in this case. and L is M. so your answer should be a function from S x N to L.

[novel parrot]

but we wernt given such a function?

or did they extend Phi?

[fossil shuttle]

you are given such a function. it is defined as

t(s, n) = s\cdot phi(n)

[novel parrot]

ohhh

ok

lol

[fossil shuttle]

so now do you understand how we have an Abelian group homomorphism F(SxN) ->L?

this is t'

[novel parrot]

yes

but also the next part

why would H elements be sent to 0

[fossil shuttle]

so now to check that this induces a map S\otimes N->L you have to check that it sends H to zero exactly.

this is not hard. you should check that if you take any of the generators of H, and apply the function t', you get something which is zero just by applying the definition of t' and the rules for the action of S on a module L

i. e. associative, unital, bilinear

[novel parrot]

$t'(\sum a (s_i,n_i)) = \sum a s_i \varphi(n_i)$

[cloud walrus]

ActiveChapter

[novel parrot]

do you mean like expanding on this?

@fossil shuttle

[fossil shuttle]

yeah that is the definition of t'

but like, every element in H is the sum of generators, so if you prove that the generators all go to zero it follows immediately by linearity that every element of H goes to zero

[novel parrot]

ok lets see

$t'((s_1 + s_2,n) -(s_1,n) - (s_2,n)) = (0)\varphi(n)$

[cloud walrus]

ActiveChapter

[novel parrot]

and we could check the rest

[fossil shuttle]

right. something similar for the others.

phi is R-linear and so a fortiori an Abelian group homomorphism, that's used in one of them

but they're all similar to the one you just did

[novel parrot]

right

is the kernal exactly H?

hm

i guess it would be

[fossil shuttle]

uhhhh well if phi is not injective then maybe not

like say phi was the constant zero map. then the kernel of the map t' would be everything

[novel parrot]

hm yes

ive understood that H is sent to 0

but not how this lets us factor by H for a new map

[fossil shuttle]

this is a basic technique to construct maps out of quotient groups

once you prove H gets sent to zero, in order to define the map from FA/H, you pick a representative of the coset and define where it should be sent

[novel parrot]

we could have just factored by H even if it wasnt sent to 0

[fossil shuttle]

you can still form the quotient group but a homomorphism FA/H->L will not be well-defined unless it sends H to zero.

like take a coset in FA/H. Say [p] or whatever. I build a map from FA/H to L by saying where to send p.

but if i have [p] =[p'] for distinct p, p' then the definition might depend on the choice of representative which is a problem

so you have to check that [p] =[p'] implies f(p) =f(p')

[novel parrot]

oh right

[fossil shuttle]

if [p]=[p'] then p-p' is in H

[novel parrot]

yeha

[fossil shuttle]

so if f is additive it suffices to show f kills off every element of H

[novel parrot]

right yeahp

[fossil shuttle]

as then we have f(p) - f(p') =f(p-p') =0

[novel parrot]

yes okay

idk why im struggling with modules

[fossil shuttle]

it is ok it'll become second nature eventually

this is a really cool theorem btw

this theorem constructs what, in category theoretic language, is called a universal arrow

proving the existence of and explicitly constructing universal arrows has a lot of consequences that follow for purely formal abstract nonsense reasons

[novel parrot]

its confusing :/

lol

is it ok to think of a module as just an abelian group such that a ring can act on it?

like how groups can act on sets

[mild laurel]

Yeah thats right

[fossil shuttle]

yeah, except that everything is happening in abelian group land so everything about the action has to be linear.

idk if this is helpful but you can prove that if A is an abelian group, the set of linear maps from A to itself forms a ring with multiplication given by composition and addition given by pointwise function addition

and a module action of R on A is the same as a ring homomorphism from R into this ring

[cyan marten]

I'm struggling with proving that the direct limit of modules M_i is unique up to unique isomorphism. (I don't want a categorical proof)

I assumed (M, m_i) and (N, n_i) are limits of the system (M_i, m_ij) and used the m_i to get a map from N to M and then did the opposite to get a map from M to N, but I couldn't see how they are mutually inversive.

[gritty sparrow]

What is your definition of a limit of the system?

[cyan marten]

The index set is directed and whenever we have a_i : M_i --> L with a_j o m_ij = a_i, we get a unique map a : M --> L with a o m_i = a_i.

[gritty sparrow]

Cool, now to show mutual inverse, compose the two maps to get a map from M->M and show that this map makes everything in the diagram commute (where the maps from M_i->M for both of the M’s is mi)

But notice that the identity map also makes everything commute

[cyan marten]

Okay, let f : M --> N and g : N --> M. We are granted that f, g are the unique maps that make everything commute.

I see that g f m_i = m_i, so gf does look like 1, but I'm still confused. Do we have a uniqueness restriction here too?

[gritty sparrow]

Yes so if you apply the definition of limit replacing L with M and ai with mi, you have a uniqueness restriction

Not only that, you can easily see that id:M->M makes everything commute

And you just showed that gf made everything commute, so by uniqueness gf=id

[cyan marten]

Ohhh

I kept inducing maps from N to M, and forgot to get ones from M to M

Thanks!

[gritty sparrow]

Np

Btw this proof is almost verbatim the category theory proof, (notice how we never bothered about the structure of M and N at all)

[cyan marten]

By categorical I meant mentioning the fact about adjoint functors, but yeah, it's the same if we're talking about direct limits

[signal remnant]

that makes way more sense thanks i'll try that

[maiden ocean]

What does "the coefficients are polynomials in the x_i" actually mean in this context lol

Like i can write a_i as the sum of b_1 x_1 + ... + b_m x_m?

Or i guess just as an element of K[x1, ..., xm]?

[gritty sparrow]

The coefficients of a poly are symmetric polynomials of the roots

Eg: the zero degree coefficient is just the product and the n-1 degree coefficient is the sum and whatnot

[maiden ocean]

Oh right yes by writing it as (b1 - x1)...(bn - xn) yeah

because it possess all the conjugates

Viete formulas

[gritty sparrow]

What are the bi’s? I was thinking of the same thing i think but with a variable symbol t instead of the bi’s

[maiden ocean]

Wait yes lol that should be just t

idk why i wrote it that way

[gritty sparrow]

Cool

Is this also szamuely?

[maiden ocean]

No AM

integral extensions are just AG in disguise though

[gritty sparrow]

I see, so you’re planning on doing some ag after AM?

[maiden ocean]

Kinda

I should probably do hartshorne eventually tbh

Atm my goal is to just finish up the rest of szamuely and also AM

[gritty sparrow]

Nice

[hot tinsel]

Let $R$ be a commutative ring with identity. We know that if $I+J=R$ then $IJ=I\cap J$. Is the converse true? Does $IJ=I\cap J$ imply $I+J=R$? I can see this being true in PIDs.

[cloud walrus]

King of Mykonos

[gritty sparrow]

Take one of the ideals to be the zero ideal

[maiden ocean]

yeah that is a pretty trivial counterexample

in a PID products = intersections means that (lcm(x, y)) = (xy) so if x and y are non-zero this should be true

[gritty sparrow]

That’s pretty nice

[hidden haven]

This was a hw problem in upendra's alg3 you would remember if you graded the HWs properly instead of speedrunning all in one day 😵‍💫

[gritty sparrow]

Lmao

College juniors are hardly human, I’m not gonna put that much effort

[hidden haven]

det was in your group

[gritty sparrow]

I just used his sheet as the rubric to minimize any time spent thinking

[kindred mist]

Just checking the definitions I think that, if we have an abelian group H and a group G, then H is a G-module just by the trivial action. Does that sound about right?

[carmine fossil]

Yes

[maiden ocean]

So suppose i have a field K, a subring A of K, and an algebraic closure Kbar of K

if i have a ring homomorphism f: A -> Kbar and f(x) is non-zero with inverse y in Kbar, is y necessarily in K

[mild laurel]

This isn't even true if A is a field, take A = K = Q(cube root of 2) and let f be one of the other embeddings of this field into Qbar

[maiden ocean]

yea makes sense

It was wishful thinking i found another approach though

[cyan marten]

If we have A-modules M, N such that Hom(M, X) is isomorphic to Hom(N, X) for all X (as A-modules), does this imply that M is isomorphic to N?

[quaint tree]

I'm going to be a junior this upcoming semester, but also, I agree that I'm hardly human.

[urban acorn]

have you heard of the yoneda lemma?

[cyan marten]

Yeah, but here I'm not assuming that Hom(M, --) is equivalent to Hom(N, --)

The isomorphisms need not be compatible

[urban acorn]

makes sense

my guess would be no, but I have no idea how to show it

if you reverse the direction of the arrows then it would be true, just let X be the free module with one generator

[cyan marten]

The dual Hom(X, M) = Hom(X, N) is trivial, so maybe this has to do with how Hom(M, --) is a right adjoint?

[urban acorn]

Okay, I think I have an idea

Hom(N, A) = Hom(M, A) by letting X = A, but then Hom(Hom(N, A), A) = Hom(Hom(M, A), A)

and you should be able to do a double dual thing

to deduce that N = M

is there a hole here? I don't usually work with modules over arbitrary rings

so I don't have the intuition to know for sure that this makes sense

[dusty river]

Double dual is not isomorphic to the module

even for infinite dim vector spaces

[maiden ocean]

I have seen an MO post about this i think in general it can be quite complicated

like even in just vector spaces over F_2 this can be undecidable

[dusty river]

Undecidable?

[dusty river]

Like this statement?

undecidable as in independent of ZFC?

[maiden ocean]

Yea

https://mathoverflow.net/questions/101431/are-there-two-non-isomorphic-modules-such-that-all-the-hom-sets-are-isomorphic

[urban acorn]

the cited part of that mathoverflow link

And if you do not assume the isomorphisms between Hom-sets to be natural, then for example over a field the question boils down to whether it is possible for two non-isomorphic vector spaces to have isomorphic duals. Over the field with two elements this is simply a question about the cardinality of power sets, which might very well be independant of ZFC

[cyan marten]

Does it really "boil down," though? Hom modules being isomorphic is more restrictive

[dusty river]

Damn

Everything is k^n for some n so like hom(k^n, k^m) should split as a product of hom(k^n, k) I guess?

or product of hom(k, k^m) I think

because k^n is coproduct of n copies of k

Then you can probably do something

[cyan marten]

I think dimension of the dual of k^a has dim 2^a

[urban acorn]

exactly

so this is just "can nonisomorphic A,B have P(A) isomorphic to P(B)"

and I'm pretty sure I heard it's independent of ZFC

[dusty river]

Yeah it is

[maiden ocean]

mhm

[dusty river]

Martin's axiom is about that

[maiden ocean]

Wacky

[dusty river]

and it is independent of ZFC + not GCH as well

[urban acorn]

damn

i like how when the isomorphisms are natural this is just yoneda's but when they're not it's independent of ZFC

(or at the very least, it's not provable within ZFC because a special case is independent of ZFC)

[maiden ocean]

General results on things like this are often hhhhhh

[daring ibex]

iiiiiiiiii

[maiden ocean]

Moldi and I were sort of talking about a similar result earlier to day where you can have two groups with the same order isomorphic subgroup lattices and isomorphic quotient group lattices that are still not isomorphic

[urban acorn]

are these groups finite?

[maiden ocean]

a lot of the time just having nice relationships between all hom sets or all subgroups or all subrings is not usfficient

It doesnt even work for finite groups intel

https://math.stackexchange.com/questions/14588/does-the-order-lattice-of-subgroups-and-lattice-of-factor-groups-uniquely-det here is a counterexample

p group moment

[dusty river]

Actually we haven't proved the non provability, at least I don't see if we did

we want to show all hom sets are the same → isomorphic

even if we assume that duals are isomorphic → isomorphic is independent

I don't see how the original statement becomes independent

[urban acorn]

it doesn't become independent

[dusty river]

Yeah I mean non provable

[urban acorn]

the negation might be provable

[dusty river]

It has stronger hypotheses

The original one

[urban acorn]

hmm, I think you're right

this question leads to a weird rabbithole

[cyan marten]

True, we have only dealt with vector spaces so far, which are more rigid

[dusty river]

Is this like a HW problem, or are you just thinking about it randomly?

That would be a clue as to whether it's doable or not

[cyan marten]

Just thinking randomly lol

[dusty river]

ok I see

[cyan marten]

I think the vector space case reduces to something like the existence of two cardinals a, b such that ac = bc for all large c

[dusty river]

That is the case for any 2 a and b

Because cardinal multiplication is just taking maximum

[cyan marten]

Ohh right

Maybe it's 2^(a + c) = 2^(b + c) then

But again a + c is the maximum

[dusty river]

Finding the counterexample is not possible

In the case of F_2

[cyan marten]

Okay, what is dim(k^m, k^n)?

It's the sum of dim(k^m, k) n times I think..?

[dusty river]

|k^n|^|m|

Also you have to be very careful with |k^m| btw

[cyan marten]

I see

[dusty river]

because |F_2^m| is not the same as |powerset of m| for infinite m

So much notational annoyance

[cyan marten]

Yeah

[dusty river]

or wait my interpretation of the notation might be wrong here lol

[novel parrot]

this seems a bit vague? we could have multiple maximal submodules?

[dusty river]

Yeah you could, what's the issue with it then?

[novel parrot]

oh nothing

i was just making sure there could be multiple max submodules

[dusty river]

ah ok

[cyan marten]

If M and N have this property, then so do M (x) A/m and N (x) A/m for all maximal m

I'm not sure how this reduces it to the case of vector spaces, if it does at all

[chilly ocean]

how do I show every field automorphism of field K extends to automorphism of algebraic closure of K?

[hidden haven]

Do you know that it extends to K(1 algebraic element)

Like viewing them as homomorphisms into K closure

K → K closure extends to a homomorphism K(1 elt) → K closure

[chilly ocean]

hmm not sure tbh

okay I guess take splitting field of that element and map to some root of that polynomial?

[hidden haven]

Use the 1st isomorphism theorem. ||Map K[x] into the closure such that the kernel is the ideal generated by min(x)||

Yeah pretty much that

So now you use zorn's lemma. Partially order all homomorphisms from extensions of K into K closure by what restricts to what

Rest should be clear

[chilly ocean]

ok I see ty

[split escarp]

Is it necessary to use Zorn's lemma for the problem?

[hidden haven]

Some form of choice is required yes

It is consistent with ZF that Q closure have a trivial galois group

If I recall correctly

[golden pasture]

holupwha

thats wild

[hidden haven]

[golden pasture]

how does that work tho

[hidden haven]

Lol actually I'm wrong

[golden pasture]

i feel like you can argue complex conjugation exists as an automorphism

[hidden haven]

Because conjugation

[golden pasture]

oof

[hidden haven]

Yeah

Then it's not Q

But in general for fields

I'm trying to find the exact statement lol

For C, only 2 automorphisms is consistent

Oh I got it

Algebraic closure are not unique upto isomorphism without choice

Without choice, it is consistent that there exists some closure of Q with trivial galois group

@golden pasture

[golden pasture]

wait but cant you have some form of complex conjugation

tho ig embedding in C is hard to define

[hidden haven]

Yeah exactly

Because you can't extend the embedding of Q

This is mentioned in Milne

On pg91

But nothing more than this statement lol

[golden pasture]

ahh

huh weird

lol

ill stick to my choice world for now

[split escarp]

Is it consistent with ZF that Q closure intersection reals have a trivial galois group?

[hidden haven]

To intersect Q closure with R you need to embed it into C

[golden pasture]

you need to choose an embedding with complex in that case

by which you have conjugation

[hidden haven]

But then it's intersected with R

So it's not galois

[golden pasture]

oh

right

then form the galois closure and get back Qbar

heh

[split escarp]

My bad

[golden pasture]

i feel like this isj ust adding sqrt(-1)

[split escarp]

We will not have i in Q closure intersection R though it is a root of x^2+1=0

[hidden haven]

Usually it would be but without choice I don't even wanna think about it

[hidden haven]

That's not the issue for galois extensions

It's not a normal extension then

[golden pasture]

brein hurty

need choice

[hidden haven]

Because it contains cbrt(2) but not its conjugate ω cbrt(2)

[sly crescent]

ω?

[split escarp]

cuberoot of unity

[hidden haven]

Cube root of 1, not equal to 1

[sly crescent]

Ah

[golden pasture]

e^{2πi/3}

[sly crescent]

Why does it have its own symbol?

[hidden haven]

Comes up a lot

Like i

[golden pasture]

cuz why not

[split escarp]

Z[w] is Eisenstein integers

[golden pasture]

a good lattice in fact

hits the hermite constant

[split escarp]

Eisenstein integers Z[w] is a Unique Factorization Domain and there is a proof of x^3+y^3=/=z^3 for natural x,y,z using Eisenstein integers

[sharp dirge]

does this mean (r1, .. rn) multiplying by k instead of R ?

[hidden haven]

Yeah, all elements of R are polynomial expressions of the r_i with coefficients in k (ie smallest subring of R containing k and the r_i's is R itself)

[sharp dirge]

oh cool

[dusty river]

This is on direct limits of modules, the mu_ij are the maps of the directed system, M_i → M_j. I can't figure out the second part.

Do I just explicitly show that the generators of the submodule that we quotient by in the construction can't generate x_i if none of its images are zero 😵‍💫 I am hoping there is something nice that can be done using the universal property

I think I got it. We look at the kernels in M_i, of all the maps going out of it in the diagram. Assume none of them contain x, then their union doesn't, and its a submodule because for any 2 kernels, you can find an upperbound kernel using the directedness

Then M_i/this union of kernels should make a cone below the diagram, by everything in M_i mapping by the quotient, and the things that remain undecided after this, all going to 0. I just need to show that this is well defined, and then this cone will factor through the universal one

[dusty river]

I am not sure how to make this cone precise. Even if M_i and M_j are not comparable, there are still some constraints enforced on M_j if they have lower and upper bounds

When it is a linear directed diagram then I think my solution works

[sly crescent]

Is it possible for a group to be a proper subgroup of itself of infinite, but not finite, index?

[split escarp]

by definition a proper subgroup of G should be a proper subset of G, so a group cannot be proper subgroup of itself

[magic owl]

I assume they mean an isomorphic copy of itself

I want to say the answer is yes and that you can demonstrate it with some free group construction

but I am not sure

[split escarp]

$\mathbb{Z,+}$ is isomorphic to $\mathbb{2Z,+}$

[magic owl]

But Z/2Z has finite index

[split escarp]

Need to think about it then

[magic owl]

Yes okay I think you can do this by taking a free group on 3 generators and finding a subgroup that is free on three generators but only uses two of the original genereators

[hidden haven]

Z^|N| should contain its copies with infinite index

Because |N| = |N| x |N|

[magic owl]

ah yeah similar idea i guess

[hidden haven]

Also possible with free group on 2 generators, by using an embedding of the one on infinitely many generators then taking a 2 generator free subgroup of that

wack

[sly crescent]

But doesn’t it also contain copies with finite index?

[hidden haven]

Oh only

[magic owl]

oh

[hidden haven]

Thats not possible

[magic owl]

that won't be possible

[hidden haven]

Because itself

[split escarp]

Does $F_2^{\mathbb{N}}$ work?

[hidden haven]

But if you exclude that then maybe

[cloud walrus]

MathForEarthlings

[magic owl]

i think that changes this problem from a cute exercise to genuinely hard hahaha

[sly crescent]

Nope. Index 2.

[hidden haven]

Have to use torsion creatively 😵‍💫

Assuming it's true

[fair obsidian]

I'm not sure it'll work, but maybe the direct sum of Z/nZ for naturals n

[split escarp]

Does $F_2^{\mathbb{R}}$ work?

[cloud walrus]

MathForEarthlings

[hidden haven]

Does that have proper copies?😵‍💫

[hidden haven]

No

Infinite powers of any group won't

Of any finite group

Sorry

[fair obsidian]

Say $a_n$ is the generator of $\bZ/n\bZ$ in the direct sum, then I think the map defined by $a_n \mapsto 2a_{2n}$ should have an image of infinite index.

[cloud walrus]

The_Vman

[hidden haven]

For that to be an isomorphism you only do this for odd n I suppose?

[split escarp]

But why?

[hidden haven]

But wait I don't think that works

[fair obsidian]

Yeah whoops

[cloud walrus]

woof ✓

[hidden haven]

I mean isomorphic ofc

And whoops

I'm viewing k as an ordinal here

So should have written 1+k

Ahh

All I'm saying is

[split escarp]

I think it is about index

[hidden haven]

You can take an extra copy

[split escarp]

Not about cardinality

[hidden haven]

In the product and that doesn't change the group

Yeah so if G is finite

[cloud walrus]

woof ✓

[split escarp]

Got it

[hidden haven]

And both sides of the inclusion are isomorphic

[sly crescent]

Wait

[fair obsidian]

Actually I may have confused myself. Does the map I suggested not give a proper isomorphic copy with infinite index?

[sharp dirge]

what does composition of polynomials mean?

or how does it work

[hidden haven]

I don't see how it's injective

For even n that goes wrong

[sly crescent]

Why not just take an infinite group not containing itself and raise that to an infinite power?

[hidden haven]

And if you only do that for odd n, it's surjective

[hidden haven]

Yeah seems like the play, but won't work with Z

Because you can multiply one of its factors by 2

[sly crescent]

Only because Z contains itself.

[sharp dirge]

anyone know whats up here

[prisma ibex]

It’s a composition of functions in the usual sense

[sharp dirge]

but why in n variables?

i wanna say it should be m variables

[prisma ibex]

The resulting polynomial is in n variables yes

Since you’re creating a polynomial expression in polynomials in n variables

[sharp dirge]

well

this is what im thinking

[prisma ibex]

Just try an example and you’ll see why it’s n variables

[sharp dirge]

$F \circ \varphi ((a_1, \cdots , a_n)) = F \circ (\varphi_1((a_1, \cdots , a_n)) , \cdots, \varphi_m((a_1, \cdots , a_n))) = F(w_1, \cdots , w_m)$

[cloud walrus]

pewdssssssss

[sharp dirge]

that didnt work

but we have m inputs for F

[prisma ibex]

Sure, but the corresponding composition has n variables

[chilly ocean]

woof can I ask in dm about one thing cuz this has been taken for like an hour and its probably some 1 line shit I cant come up with

[prisma ibex]

This is reflected in what you’ve written

[sharp dirge]

what do you mean?

[hidden haven]

sure

[prisma ibex]

F circ varphi is a function in a_1,…,a_n

[sharp dirge]

so like expanding the F

[prisma ibex]

Again this follows from what you’ve written

Yes

[sharp dirge]

wed get varphis

i see

whoops

my bad

[fair obsidian]

What about Q/Z? Or the dyadic rationals mod Z.

[sly crescent]

That would work.

[fossil shuttle]

hey this might be overkill but you might consider a very different approach, which is a total recharacterization of the colimit. What you've proved in the first part of the question is that, on a set theoretic level, the map from the disjoint union $\coprod M_i\to \operatorname{colim}(M_\bullet)$ is surjective, ergo set theoretically we can write the colimit as $\coprod M_i/\sim$ for the obvious equivalence relation, i.e. $x\sim y$ if both go to the same point in the colimit.

I claim that this equivalence relation can be described as follows: it is the weakest equivalence relation on $\coprod M_i$ such that if $x\in M_i$ and $f : M_i\to M_j$ is a morphism in the diagram, then $x\sim f(x)$. Just try proving concretely that this has the universal property of the colimit on a purely set theoretic level. then check that you can define a natural notion of addition on this thing, that all the maps involved are $R$-linear, etc. once you've checked that this really is the colimit, the answer to the second part of this question is trivial, as it follows by construction