groups-rings-fields

[oblique river]

https://math.stackexchange.com/questions/1649131/the-projective-space-is-not-affine-ii

[solar wyvern]

oops haha

in the ag context do "good covers" matter at all?

[oblique river]

i'm not sure what a good cover is

[solar wyvern]

the intersection of open sets in your cover is contractible ie π_1(U⋂V)= 0

(think I got that right)

[oblique river]

maybe like if you're trying to look at some cech cohomology or something

but idk that's not really what i think about

[solar wyvern]

true

[oblique river]

oh yeah there's homotopy stuff

[solar wyvern]

but it's less basic it seems haha

[oblique river]

correct

[solar wyvern]

(still interested in the cech stuff but probably gotta defer that until I understand basics)

um, so at this point it'd be tempting to say that if you can cover something with a bunch of copies of kⁿ then it's a scheme, but that definitely feels v wrong

[oblique river]

idk, i dont think about that very much

there is some gluing condition you need

like there has to be some kind of compatibility with the overlaps

also i gotta head out now sry

[solar wyvern]

np, thanks for the help!!

I'll check out that page u linked

[marble wagon]

affine schemes have a locally ringed space structure

if you can cover something with a bunch of copies of k^n in the affine scheme sense, then it's a scheme

this is stronger than just having open sets homeomorphic to k^n

because the structure sheaf must be the same

which imposes compatibility in the intersections

[bleak finch]

So I have (x, y) where x, y are members of a number ring that is not a PID.

How do I find the generator of x, y?

(I know for a fact that these x, y are principle ideals)

[fickle brook]

your question is worded confusingly

i can't parse it

can you ask it again

[somber bramble]

isn't the whole point of it not being a PID that you may not be able to find a generator

[chilly ocean]

how to approach (ii)

[marble wagon]

@bleak finch sounds tricky in general, you gotta use your proof that (x,y) is a PID to find a generator

if there's some finiteness condition (few ideals, many elements) then one of the x + ay must be a generator

[plain sequoia]

@chilly ocean invariant under conjugation

i.e. $\forall g \in G$ we have $N=gNg^{-1}$

[cloud walrus]

markus:

[timber bay]

does the primitive element theorem imply that all alg extensions are simple extensions?

or is there an alg extension that would be adjoined by like finite elements?

[solar wyvern]

um

simple extensions are either finite or transcendental

Primitive ext thm: a separable extension of finite degree is simple.
(separable means for every a∈E, its minimal polynomial f has distinct roots)

[uncut girder]

So is Q(e) isomorphic to Q(pi)

[topaz solar]

$\mathbb{Q}(e, \sqrt{d})$

[cloud walrus]

Darkrifts:

[solar wyvern]

@timber bay is the algebraic closure of Q simple?

[timber bay]

nope

thanks

also thanks flimflam

didt see that

wait i said thanks to you twice

[solar wyvern]

np I'm going over similar stuff atm anyways haha

[timber bay]

you taking an algebra course?

[uncut girder]

Answer my question

[solar wyvern]

yeah, fields/galois theory

[uncut girder]

Is Q(e) isomorphic to Q(pi)

[solar wyvern]

idk

[fickle brook]

iso to Q(s) for any transcendental s

[uncut girder]

Ok thanks

[fickle brook]

including the field of rational funcs over Q

[solar wyvern]

oh ya

[timber bay]

lol @ "answer my question"

[uncut girder]

So there's only 1 simple infinite extension of Q up to iso?

[timber bay]

yeah theyre both transcendental

so they should be isomorphic

[uncut girder]

Field of rational funcs over Q is simple because?

@fickle brook

[topaz solar]

i mean you just stick $x$ on it

[cloud walrus]

Darkrifts:

[topaz solar]

and only x

[uncut girder]

Ohh

[timber bay]

Im also taking a similar class. abstract algebra 2

[uncut girder]

Ok

[timber bay]

flimflam

[uncut girder]

Ohhhhh

Ok

[fickle brook]

what does simple mean

single generator?

[timber bay]

yeah

[uncut girder]

Yeah single extra thingy

[fickle brook]

yeah

[uncut girder]

What about Q(x)(y)

If Q(x) is rational funcs

[topaz solar]

[uncut girder]

What is Q(x)(y)

🤔

[solar wyvern]

(Q(x))(y), that is Q(x) adjoined with y
e: this is the same as Q(x,y), the rational functions in two variables

[topaz solar]

the functions of rational functions, of course

[mild laurel]

If G_1, G_2 are galois over K, then how do I show that G_1 \cap G_2 is also galois over K?

[solar wyvern]

won't Aut(G_1 ⋂ G_2/K) be exactly the automorphisms of G_1 fixing precisely K restricted to G_1 ⋂ G_2 (and thus still fixing precisely K)?

[mild laurel]

That doesn't work. You can't restrict the automorphisms to a subfield since you have to restrict both the domain and codomain. In other words there could be an automorphism of G_1 that sends something in G_1 \cap G_2 outside of G_1 \cap G_2, but restricting it wouldn't give you that automorphism.

[marble wagon]

it's separable because yes
can you see why it's normal?

[fallen bluff]

Representation theory:

Is it true that if we assume V is finite dimensional, then there exists G-submodule W such that V = U + W iff the vector subspace Z which is complement of U is a G-submodule ?

[marble wagon]

there is no canonical complement Z

you construct an appropriate one

e.g. consider complements of (1,0) in k^2

any vector (a,1) will do

[fallen bluff]

Cool ty

idk why my brain thought VS complements are unique

[marble wagon]

yeah it happens

its cuz orthogonal complements are unique

[hallow jolt]

Hi, i have a question about eucledian ring in #help-3

[timber bay]

trying to think of a proof that for y transcendental, y^2 is also transcendental

[bleak abyss]

Assume y^2 isn't transcendental

It satisfies some polynomial

[timber bay]

yeah

i thought of that

[bleak abyss]

(y^2)^n + a_1(y^2)^{n-1} + ...

That's y^{2n} + a_1y^{2n-2} + ...

Which is now a polynomial in y

[timber bay]

ohh

i see

thanks

[thorny slate]

what about the converse

:^)

it's also true but slightly trickier

[timber bay]

something with minimal polynomials i bet

check my reasoning, if F in E in K field extensions, trying to show that if y transcendental over F, then it isnt necessarily transcendental over E. E = F(y), which would be a valid simple extension that isnt algebraic over F but is algebraic over E.

[thorny slate]

yeah

[timber bay]

cool

[timber bay]

wait when is {E:F} = [E:F] = |G(E/F)|?

whats the condition?

finite normal?

anybody have tips on this/

[thorny slate]

you can try assuming the extension is not normal

and trying to get a smaller fixed field

based on the roots of a poly that doesn't split

[timber bay]

what would assuming it is not normal entail?

assuming it is not a splitting field for a polynomial

[thorny slate]

the other characterization helps more

that there is an irreducible poly in F that has a root in E but does not split completely in E

[timber bay]

that's a normal extension?

[thorny slate]

that's a non normal extension

[timber bay]

or not normal

okay

so if it doesnt split completely not all the roots of that poly are in E

[thorny slate]

yeah

and then you should be able to make an element in G(E/F) from there

[timber bay]

im kind of thinking out loud

[thorny slate]

that isnt in F

for example, if the poly only had 1 root in E, then that's the element you want

because the elements of G(E/F) can't move it

[timber bay]

if its not in E, its not in F.

[thorny slate]

I mean, elements of G(E/F) send roots of irreducible F polys to themselves

so it has to send that root to another one that is in E

but there's only one in E

so it's fixed

if there's more than 1 root it isn't clear to me how it works

but you should be able to do something similar

[timber bay]

oh so the fixed field would be F(that element)

[thorny slate]

it can be larger but yeah that's the idea

[timber bay]

yeah

cool

thanks for the tip

[thorny slate]

sure

I feel like you can do something like

add all the roots that are in E

that's an element that's fixed by G(E/F)

and it remains to prove it isn't in F

[timber bay]

Wouldn't G(E/F) permute the elements that are in E?

[thorny slate]

but maybe that isn't true

I can't tell

but the rough idea should work

yeah it does

so if you sum all of them

the sum isn't changed

by elements in G(E/F)

[timber bay]

so it wouldnt necessarily fix F(all those)

[thorny slate]

yeah it doesn't fix each one

[timber bay]

ah yeah

[thorny slate]

but it fixes symmetric functions in them

[timber bay]

those fixed element things

yes

symmetric functions

[thorny slate]

oh okay I see how to do it

want spoilers?

or do you wanna try

[timber bay]

let me try for a minute

you can type it out

[thorny slate]

ok

[timber bay]

ill go to another page

[thorny slate]

it's basically following the idea through

let p be irreducible in F[x], with some roots ai in E and some bj not in E

then p = Prod(x-ai) Prod(x-bj) in E[x]

what can you say about the coefficients of the polynomial in the left?

Spoilers:
||1) they are symmetric functions in the ai so they are fixed by G(E/F).||
||2) can all of them be in F? ||

[timber bay]

yeah that makes sense

can all of what be in F?

[thorny slate]

the coefficients

of the polynomial on the left

[timber bay]

arent they

since p in F[x]

[thorny slate]

p = Prod(x-ai) Prod(x-bj)

I mean the coefficients of Prod(x-ai)

[timber bay]

oh i see

whoops

they can, can't they?

i tfeel like it comes from ai in E

[thorny slate]

they can't

i'll let you figure out why

if Prod(x-ai) is in F[x] something bad happens

[timber bay]

something about it being irreducible or normal?

[thorny slate]

irreducible

the original polynomial was irreducible in F[x]

[timber bay]

oh ai arent in F for sure because they would be reducible

yeah

[thorny slate]

so some coefficient of Prod(x-ai) isn't in F

and that's what you want

[timber bay]

that makes sense

[bleak finch]

So I have (x, y) where x, y are members of a number ring that is not a PID.
How do I find the generator of x, y?
(I know for a fact that these x, y are principle ideals)

[thorny slate]

how do you know (x,y) is principal?

[bleak finch]

@thorny slate I think it is assumed in the problem statement. My professor is bad at writing out his assumptions.

[thorny slate]

if it's just assumed then it sounds hard

if there's a reason you know, then you can probably use that reason

not that I know much algebraic number theory

maybe number rings are easy for this

i mean, you could compute the whole ideal class group, and the prime ideals

and check which ideals your ideal (x,y) factors into

and so on

[vestal snow]

Is their a bijection between the sets Z/mnZ* and Z/mZ* x Z/nZ* for relatively prime n and m?

I know that by CRT, Z/mn ~ Z/mZ x Z/nZ

[quiet cave]

An isomorphism is a bijection in this case so yes

[vestal snow]

How do I prove it?

[fickle brook]

prove what

[vestal snow]

That there is a bijection from Z/mnZ* to Z/mZ* x Z/nz*

[quiet cave]

Oh I didn't see the stars

Hmm

According to what I can find, there's only an isomorphism if m and n are prime

If not, you'll have to split it into a cartesian product of prime power modulo groups before you can say you have an isomorphism

[covert vector]

they are iso if they are coprime

[vestal snow]

Can you provide a proof?

[covert vector]

ex. 8 and 15

euclid algorithm

if x,y are coprime, there are integers a,b such that ax+by=1

1 will be the generator of Z/xyZ

[quiet cave]

Woog we're talking about the multiplicative group

[covert vector]

oh woops

[quiet cave]

I don't have a proof but I think I'll trust Wikipedia on this one

[covert vector]

then I doubt it for just coprime

[vestal snow]

So is it true for coprimes or just primes?

According to Wikipedia

[quiet cave]

Just primes

Coprime is not enough

[vestal snow]

Can you send the Wikipedia link?

[quiet cave]

t!wiki multiplicative group of integers modulo n

[fossil mango]

📖 | ** https://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n **

[quiet cave]

Scroll down to Structure and then General composite numbers

[vestal snow]

Alright thanks!

[quiet cave]

🍮

Actually nah primes are too strict

If m and n are prime powers with m =/=n, your statement holds

So for example like woog's example 8 and 15 wouldn't work, but 8 and 27 would 😄

[vestal snow]

Tried with a bunch of coprimes and they all follow the same rule

[wind parrot]

There is a bijection from (Z/mnZ)* to (Z/mZ)* x (Z/nZ)* for n,m coprime but not necessarily an ismorphism

Since phi(nm) = phi(n)phi(m) whenever n and m are coprime

[vestal snow]

How do I prove that?

[thorny slate]

what

@vestal snow the iso follows from CRT

the ring version

[vestal snow]

Yeah got it

[thorny slate]

ring isomorphisms send units to units

[vestal snow]

Thanks for the help

[bleak finch]

Hint: If this ideal is principal, you can determine what the norm of a generator must be (play with divisibility). Now look for elements of that norm and show that one actually does generate the ideal.

What does this mean?

[thorny slate]

i guess the norm will be the gcd of the norms?

[bleak finch]

@thorny slate no gcd domain

[thorny slate]

what

the norms are in Z

[bleak finch]

Jacobian let me explain

I have an extension of Z that is not a PID or a euclidean domain.

This is Z[sqrt(-5)] so I don't know what the norm is.

In this non-PID, I have an ideal that is principle (because a non-PID can have SOME ideals that are principle). This principal ideal is of the form (x, y).

Because it is principle, it has a single generator.

What is the norm of this generator??

[thorny slate]

the gcd of the norms of x and y

[bleak finch]

How can I computer the gcd in a non euclidean domain

[thorny slate]

the norms are in Z

also you should have said that your ring was Z[sqrt(-5)] before

having a concrete ring makes things much easier

[bleak finch]

but there is no euclidean algorithm

[thorny slate]

Z is an euclidean domain

[bleak finch]

Z[sqrt(-5)] is not

[thorny slate]

the norms are in Z

[bleak finch]

How does the fact that the norms are in Z let me compute the GCD of two elements in a non euclidean domain?

[thorny slate]

you compute the gcd of the norms of x and y

[bleak finch]

This tells me the norm of the gcd of (x, y) in Z[sqrt(-5)] of course. But then do I guess the elements?

[thorny slate]

then you find an element with such a norm

is what the hint says

[bleak finch]

I guess there's only a finite number of elements I have to search for.

[thorny slate]

yeah I mean the nonprincipal ideals are very easy to describe

they are like

uhhhh

whatever the factorizations of 6 are

(2, 1 +- sqrt(-5))

(3, 1 + sqrt(-5))

(3, 1 - sqrt(-5))

so it's the primes 2 and 3 that have issues

if your gcd doesn't contain 2 and 3

then you express it as products of primes in Z

and then you use those elements as your generators

so basically you know what all the prime ideals look like

and the fact that every ideal is a product of primes

[bleak finch]

Is the norm in Z[sqrt(-5)] x^2 + y^2 where e = x + yi sqrt(5)?

[thorny slate]

N(a + b sqrt(-5)) = a^2 + 5b^2

[bleak finch]

ok

I'm going to try my hand at this and see what happens.

So now I have a NEW problem.

Suppose I have the ideal (g). How do I know that it generates

(x, y)?

[thorny slate]

the ideal (x,y) is a product of prime ideals

the ideals of norm p^2 for p not 2 or 3 are (p)

I think

or something like that

so once you have the norm of g you can split it into primes

[bleak finch]

STEP 1: Norm((29, sqrt(5) ± 13)) = Norm(29) Norm(\sqrt(5) + 13)
STEP 2: Norm(29) Norm(\sqrt(5) + 13) = 174
STEP 3: Solve Pell's equation x^2 + 5y^2 = 174 has solutions (±13, ±1); (± 7, ± 5).
STEP 4: Figure out which of these solutions (e.g. the ideal (13 + sqrt(5))) generates (29, sqrt(5) ± 13)

[thorny slate]

how is it 174

also wasnt it sqrt(-5)?

I guess you meant 29^2 * 174

and yeah then you need to get the prime ideals

some of them dont split

wait what?

why are you solving that equation?

the gcd of the norms is 29

so I think you should be solving x^2 + 5y^2 = 29

[bleak finch]

Okay jacobian I got the solutions (–œ∑´®†¥¨ˆøπ“…¬˚∆˙©ƒ∂ßå¡™£¢∞§¶•ªº–––µ≤≥÷

umm ignore that

I got the solutions (±3, ±2)

How do I check if the ideal (±3 ± 2sqrt(-5)) generates the ideal (29, sqrt(-5) ± 13)?

[thorny slate]

try to get the generators from it

I mean, first you gotta pick one of them

for example (3 + 2 sqrt(-5))

and now try to get the others and try to get 29

you know 29 = (3 + 2 sqrt(-5))(3 - 2 sqrt(-5))

so you need to get 3 - 2 sqrt(-5)

and so on

just manipulation mostly

[bleak finch]

Jacobian in order to figure out whether (3 + 2 sqrt(-5)) generates (29, sqrt(-5) ± 13), I need to find a, b, c in R such that a(3 + 2 sqrt(-5)) = 29b + c(sqrt(-5) ± 13). This is Bezout's equation and I don't know how to solve it in a domain that is not euclidean.

[thorny slate]

(3 + 2 sqrt(-5))^2 = -11 + 12 sqrt(-5)
subtract 6(3 + 2 sqrt(-5)), you get -29 so you have 29

now uh

it seems like you can't get the other one?

something's wrong

oh

multiply by sqrt(-5)

to switch the parity

multiplying both of our things we get -11 sqrt(-5) - 60 and 3 sqrt(-5) - 10

and you get the idea

do some more linear algebra to get sqrt(-5) + 13

in fact you can right away add -11 sqrt(-5) - 60 to -11 + 12 sqrt(-5)

you get sqrt(-5) - 71

add some multiples of 29 and you're done

that's sqrt(-5) - 13

which is not the same as sqrt(-5) + 13 btw

they're gonna give different ideals

so be careful with writing +- so much

and you might say shit man, you're just throwing numbers around

and it just happens to work

but there is VERY little room for things here

the dimension over Z of things involved is like what, 2 at most?

torsion free Z modules are free

so you just generate two linearly independent elements

and then you can go and do linear algebra

if you notice you have a proper subspace, you multiply by stuff and you keep going

like when I noticed my parity was off

you only need to do this finitely many times

since Z is noetherian

etc etc

and yeah I only did one direction, the other one is easier

cuz you're in the field Z/29

i mean

the ring Z/29[sqrt(-5)]

you know what I mean

just one variable

whatever

[raw moth]

if you have a ring extension with 2 things, say Z[sqrt(2), sqrt(3)]

is that all elements of the form a + b sqrt(2) + c sqrt(3)

or does it also include things like sqrt(2)sqrt(3)

[thorny slate]

the latter

[raw moth]

ok cool, thanks

[oblique river]

if you have a ring extension, it had better be a ring!

so you have to be able to multiply things in it!

[thorny slate]

killingform is that you?

[raw moth]

lol

I was fairly certain it was that

but someone said the first and I doubted myself

[fringe nexus]

How would you find the degree of a field extension like Q[sqrt(2) + sqrt(3)] ?

I know for Q[sqrt(2),sqrt(3)] you can find the degree of the extension from Q to Q[sqrt(2)] and then Q[sqrt(2)] to Q[sqrt(3)]

but can i do it the same way for this?

[thorny slate]

not really

you would need to find the minimal poly of rt(2) + rt(3)

which you can do by hand

or try to express Q[rt(2)+rt(3)] in a better way

in terms of elements you know the order of

[fringe nexus]

We havent talked about finding minimal polynomials yet, is there a general way of finding it?

For sqrt(2) its simpler because we can just multiply by the conjugate

like x-sqrt(2) * x+sqrt(2)

do you just do the same process? like (rt(2)+rt(3) ) * (rt(2)-rt(3)) ... until you get a polynomial with the coefficients in your field?

[thorny slate]

well you would have to show rt(2) - rt(3) is in your field

and that's as hard as the problem itself

think of it this way

you start with rt(2) + rt(3)

and you are allowed to add, multiply

and your goal is to get an element of Q

[fringe nexus]

Ok So i think the degree is four, but I'm not sure if the way i did it was valid

can i just like multiply (x - sqrt(2) - sqrt(3) * (x + sqrt(2) + sqrt(3))

and then multiply that result by its conjugate?

then you get a 4th degree polynomial x^4 - 10x^2 + 1

in Q

[thorny slate]

what kind of conjugate?

[fringe nexus]

uh so the result of that multiplication is x^2 - 2sqrt(6) + 5

then we multiply by x^2 +2sqrt(6) - 5?

to get rid of the sqrt

[thorny slate]

alright so what you have done is guess that the other root of the minpoly should be the negative

and then after multiplying you guessed that the other roots should be the conjugate ones

the result is that you have a degree 4 poly which has what you want as a root

and this is a good thing

all you need to do is either reduce it or prove it's irreducible

and you're done

[fringe nexus]

hm ok

[thorny slate]

this is also a valid technique

usually the goal is finding some polynomial with your thing as a root

and then you can try to reduce it and you're done

for big examples (the biggest you'll see should be degree 6 in extreme cases) you have to be resourceful

and play this game of guessing

[bleak finch]

Suppose I want to prove that (I) ⊆ (J, K) and (J, K) ⊆ (I). My professor said that one direction involves a simple proof by divisibility and the other side involves solving a linear diophantine equation. Which side is which?

[thorny slate]

I kinda went through it in detail

the kind of linear algebra you're doing

[bleak finch]

DEMO OMAE WA MOU SHINDEIRU.

Suppose I want to prove that (I) ⊆ (J, K) and (J, K) ⊆ (I). My professor said that one direction involves a simple proof by divisibility and the other side involves solving a linear diophantine equation. Which side is which?

[thorny slate]

you have a bad habit of generalizing your questions so they become much harder than they need to be, the context is comparing the ideals (a) and (b,c) in a quadratic number ring

anyway (b,c) ⊆ (a) is divisibility

the other one is the linear algebra argument I carried out in gruesome detail for you before

[whole hound]

@terse cradle

[terse cradle]

Yes?

[wispy leaf]

I have a problem that x^(1/x)=a, a is a constant, x1=?, x2=?
i hope someone could help me. This problem make me confused all day.

[chilly ocean]

that is not abstract algebra

Redirect to #prealg-and-algebra ?

[solar vessel]

not abstract

[chilly ocean]

@wispy leaf Please redirect your question to either #prealg-and-algebra or to a questions channel. This is not the appropriate channel for this question.

[austere vault]

Someone remind me what was the factor group

[woven delta]

It's the group of cosets of a normal subgroup

It's also called a quotient group

[austere vault]

@woven delta
Thanks now I shall prepare for the next semester

[mild laurel]

In Atiyah's commutative algebra book, he defines the submodule generated by an element x as the set of all multiplies ax, where a is in the ring. Don't we also have to consider sums such as x + x + x? Isn't there a chance the two sets could be different if the characteristic of the ring is not 0?

[fickle brook]

x + x + x = (1+1+1)x

1+1+1 is definitely an element of the ring

@mild laurel

[mild laurel]

Yeah I'm dumb

I was thinking these two elements could be different if the characteristic was 3 or something, but the module axioms require them to be the same

[fickle brook]

module axioms don't care about characteristic

[fringe nexus]

So I'm trying to show that M Maximal idea of R <-> R/M is field

For the R/M Is field -> M Maximal ideal of R

Since R/M is a field, every a + M has an inverse (a+M)^-1 S.t (a+M)(a+M)^(-1) = 1. Since we define (a+M)(b+M) = (ab + M) , we can define the projection homormophism

phi(a+M) = a

uhhh

Im actually not sure where I'm going here but im trying to use the fact that

[woven delta]

Do you know the correspondence between ideals containing an ideal I and ideals in the quotient space R/I?

[fringe nexus]

if a ideal contains 1, then its the whole ring

uhhhh

I dont think so?

Can i use the fact that a not in M implies a has an inverse, so a cannot be in the ideal

because (a+M) is in a field and the way we define multiplication of cosets

so when we have phi ( (a+M)(a+m)^(-1)) = phi(1) = 1' = phi( (a+m)) phi ( a^(-1) + m ) = a a^(-1)

[woven delta]

So you're saying that if a is not in M, then a must have an inverse?

Why?

[fringe nexus]

Because R/M is a field

and R/M is the cosets a + M

by the way we define multiplication of cosets

we have (a+M) (b+M) = (ab + M) ?

im not sure actually

oh wait yes

then we have

(a+M) (a^-1 + M ) = M

uh

ok now im super confused

[woven delta]

It should be 1+M

Not M

M is the 0 in your quotient

[fringe nexus]

Ok so I guess I'm wrong 🤔

[woven delta]

1+M is your identity

[fringe nexus]

But I was trying to think of a simpler way to prove this fact

then the one provided in my book

[woven delta]

If you know the isomorphism theorems

It should follow straight away

[fringe nexus]

uhh

can you clarify which theorems those are

Cause i might know them

[woven delta]

R/ker \phi is isomorphic to im \phi

Is the first

[fringe nexus]

yes I know that one

[woven delta]

The fourth is the most important

Because it talks about the relationship between ideals in the quotient

[somber bramble]

uh I only know three

what’s the fourth

(sry to butt in)

[fringe nexus]

phi(N) is an ideal of phi(R) if n is ideal of R?

[woven delta]

No

It is if your map is surjective

There are some obvious counterexamples if it's not surjective

So the fourth isomorphism theorem

States that if you have a ring R and an ideal I

Then there is a correspondence between ideals containing I and ideals in R/I

It's pretty easy to prove actually

If you know that the image of an ideal under a surjective map is an ideal

And that the preimage of an ideal is an ideal

Once you know that theorem

The result that @fringe nexus is trying to prove follows immediately

Or at least it does once you prove that a ring is a field iff it has exactly 2 ideals

@fringe nexus does that make sense?

[fringe nexus]

uhhhh

one sec

ok wait so I know how to prove that a ring is a field iff it has exactly 2 ideals

but Im still not sure how fourth isomorphism theorem helps here

[woven delta]

So if there is an ideal containing M

[fringe nexus]

so what it says is if you have a ideal I of a ring R, and you consider the factor ring R/I , an Ideal say A of R is an ideal iff A/I is an ideal in R/I ?

[woven delta]

Then that corresponds to a nontrivial ideal in R/M

Yes

Assuming the ideal A contains I

Sorry wait

Yes, in general

But there is a special relationship between the ideals containing I and the ideals of R/I

[fringe nexus]

So our R/I in this case is a field with ideals {0} and R/I. Then, that implies A/I is either {0} or R/I Which implies A = R or A = I?

i dont think thats right but

[woven delta]

Your looking at it the wrong way around

Suppose some ideal A contains I

Then A/I is an ideal of R/I

[fringe nexus]

ok

[woven delta]

And it's nontrivial (why)

[fringe nexus]

wait

But R/I is a field implies I is a maximal ideal right

[woven delta]

Yes

[fringe nexus]

so how can it be contained

in some ideal A

[woven delta]

I'm saying

Suppose there was an ideal A

Then there would be a nontrivial ideal in R/I

Which is a contradiction

Yes

Stop deleting your correct stuff lol

[fringe nexus]

Wait i thought i was wrong

so i deleted it

ok let me try this from the start

So We have R/I is a field which implies it has only two Ideals

to show it is maximal we will use contradiction and assume that I is contained in some ideal A

then by fourth isomorphism theorem there exists a nontrivial ideal A/I in R/I

but that contradicts the first statement

thus I is maximal in R

what am i missing?

[woven delta]

That's the contradiction

I guess you don't have to phrase it as a contradiction

[fringe nexus]

Uhm

I showed it contradicted that I was not maximal right?

so that proves I is maximal

[woven delta]

You could just be like "if an ideal J contains I then J/I is an ideal in R/I. It must then be either I/I or R/I"

Which means J is either I or R

But yeah, your argument seemed fine

[fringe nexus]

for the other way round from maximal -> field

I think my original one was almost there

I just needed the fourth isomorphism theorem

so we have R/M where M is maximal ideal of R

pick some a not in M, and consider the coset (a+M), assume for contradiction a+M has no inverse

then, (R/M) (a+M) is a non trivial ideal

Its a proper ideal because it doesn't contain 1+M since a+M has no inverse

that implies that there exists an ideal in R containing M, contradicting the maximal assumption

[woven delta]

@fringe nexus this is covered by the fourth isomorphism theorem, but why can't some ideal that contains I map to 1+I under the quotient map?

[fringe nexus]

Uh

Im not sure

[woven delta]

So if the image of an ideal A that contains I is R/I

Oh sorry I mean R/I, not 1+I

Then there is some element in A that is of the form 1 + a for some a in I

Then since a is in A

We have A=R

[quiet cave]

Did I see R/I 👀

[somber bramble]

hi 🅱/J

[quiet cave]

Hi Sascha

[fringe nexus]

Why exactly is Q not a field extension of Z_2 ?

don't we just need Z_2 <= Q

[woven delta]

they have different characteristic

If Q was a field extension of Z_2, then Z_2 would imbed in Q

but Q has no element of order 2 as a group

while Z_2 does

[fringe nexus]

Do you have any examples of field extensions of Z_2 then?

Can two fields be non isomorphic even if they have the same characteristic?

finite fields

[woven delta]

so there are fields of order p^k

but the characteristic is p

[fringe nexus]

Hmm ok

[mild laurel]

Consider Z_2[x]. That's a field extension of Z_2

Yes fields can be non isomorphic if they have the same characteristic. Consider the two in the example I just gave

[fringe nexus]

Uh maybe I havent studied this enough but

is x just anything?

like when we say Q[sqrt(2)] we just mean Q/ <x^2 - 2>

but whats z_2[x] ?

[woven delta]

its the ring of polynomials over Z_2

[fringe nexus]

oh

ok

[woven delta]

@mild laurel is that a field?

[mild laurel]

nope its not im dumb

[woven delta]

a better example would be the field of rational functions maybe

[mild laurel]

Q[sqrt(2)] is isomorphic to Q[x]/<x^2-2> @fringe nexus

[fringe nexus]

yes

[fringe nexus]

For showing that every prime ideal I in a finite commutative ring R is a maximal idea, you can just prove it using R/I is integral domain -> finite integral domain is field -> I is maximal right?

[crisp totem]

||^3

[thorny slate]

yeah victoria

[crisp totem]

nani

[solar wyvern]

For a set ${F_{α}: α∈ S}$ of $k$-algebras indexed by some indexing set $S$, we should generally be able to construct some big $k$-algebra $E$ such that $E$ is a $F_α$ algebra for all $α∈S$, right?

Regardless of whether $F_α$ is trancendental or $S$ is arbitrary?

[cloud walrus]

flimflam:

[solar wyvern]

@oblique river @thorny slate

[thorny slate]

their product? their direct sum?

[solar wyvern]

idk if product works if S is uncountable

[thorny slate]

why not

[solar wyvern]

i don't know

these are all field extensions, so I still want E to be a field

[thorny slate]

you never said they were fields

🤔

[solar wyvern]

they are fields.

does this not work then?

[thorny slate]

it works

let E be the algebraic closure of k

what you want is something like E(xi) for xi whatever trascendental elements appear

in the sense that it's a free union of some trascendental bases for each Fa

that respect the relations in that Fa

[solar wyvern]

I guess that should work then

but maybe unwieldy for lots of trancendental elements

[thorny slate]

why

I mean what's the context

[solar wyvern]

wait nvm, elements of the overall closure will still be (finite) linear combinations a_1ξ_1+...+a_1ξ_n and we know how to invert those

so there's no problem

[oblique river]

there's probably some kind of axiom of choice nonsense floating around

but yes, that should be fine

[solar wyvern]

if Φ_n is the nth cyclotomic polynomial over ℚ and F/ℚ is its splitting field can you typically say anything about
[F/ℚ:ℚ]?

@oblique river

it's a simple extension of course, but that's as a ℚ-algebra, not as a vector space right?

[thorny slate]

as a vector space its Q^phi(n)

[solar wyvern]

so Φ_8 should be degree 4?

[thorny slate]

yeah

[solar wyvern]

oh darn

was thinking Φ_3 had sqrt3 and i

but it just has sqrt(-3)

so it's degree 2

rip

thanks

[oblique river]

jaco is correct

it's a theorem to prove that Phi_n is irreducible over Q

for all n

and it's degree is phi(n)

and once you have one primitive nth root of unity (i.e. one root of Phi_n) then you have all of them

and therefore the splitting field is exactly Q[x]/Phi_n(x) which has degree phi(n) over Q

[solar wyvern]

not sure if I didn't screw this up, does this work?
for $f = x^n - p \in \mathbf{Q}[x]$ we have a splitting field of degree $φ(n)$ generated by $p^{1/n}ζ_n$, namely
$$\mathbf{Q}[(p^{1/n}ζ_n)^{\phi(1)},...,(p^{1/n}ζ_n)^{\phi(n)}]$$

[cloud walrus]

flimflam:

[solar wyvern]

@oblique river

[oblique river]

that's not correct

for example, if n = 2, phi(2) = 1, but the splitting field of x^2 - p clearly has degree 2

[solar wyvern]

I don't imagine degree phi(n)+1 would amend this

[oblique river]

nope

unfortunately one data point is not enough to establish a pattern lol

[thorny slate]

forget about the p \phi

add them separately

[solar wyvern]

p^{1/n}ζ_n)^{k} should work for roots though?

[thorny slate]

p^(1/n), z^phi^1, ..., z^phi^whatever

[oblique river]

those are the roots, yes

[thorny slate]

aka add p^(1/n) and then add the roots of unity

then you have all n roots

[solar wyvern]

degree of the splitting field over Q should be n?

[thorny slate]

adding p^(1/n) is degree n

[oblique river]

noooooo :(

[thorny slate]

then adding the other stuff is degree phi(n)

[oblique river]

(unless p = 2 and n is divisible by 8)

[solar wyvern]

ooooh

[thorny slate]

is it degree phi(n)? maybe not always

but you get the idea

[oblique river]

the splitting field has degree n*phi(n) except in the case p = 2, n = 0 mod 8

[thorny slate]

cool

[oblique river]

in that last case, the degree is n*phi(n) / 2

[solar wyvern]

what happens there

[thorny slate]

is this hard to prove?

[oblique river]

zeta_8 involves a sqrt(2) already

[solar wyvern]

ah

[oblique river]

so if you wanted to add, say, 8th root of 2 to that

you only need to add 4th root of sqrt(2)

i.e. it's a degree 8/2 extension of Q(zeta_8)

I don't think it's that hard to prove. You basically need to establish that the intersection of Q(zeta_n) and Q(p^(1/n)) is just Q except in the special case

[solar wyvern]

it's like Q[sqrt2, sqrt-2, i]?

for the zeta_8 case?

[oblique river]

that is Q(zeta_8), yes

[solar wyvern]

whew

just wanted to check I didn't mess that up

the rest....i will attack after dinner

thanks @oblique river @thorny slate

[oblique river]

jaco's idea is right though. you have to realize that the splitting field is exactly Q(zeta_n, p^(1/n))

so you just need to check how zeta_n and p^(1/n) interact

[solar wyvern]

realizing that now...

is it much different if you replace Q with F_p

[oblique river]

OH WAIT OH NO

[solar wyvern]

[oblique river]

sorry, the p = 2 and n = 0 mod 8 case is not teh only special case

it is a special case

if p = 2, then the answer depends on if n is divisible by 8 or not

but for other primes weird things can also happen

[solar wyvern]

oh

oh no

[oblique river]

for example, take p = 5 and x^10 - 5

Q(zeta_10) does contain sqrt(5)

so in that case the degree would be phi(10)*10/2

[solar wyvern]

if your power is also prime it's fairly uncomplicated though, i.e., if you're looking at
x^ℓ - p?

[oblique river]

alright so what's going to happen is that for x^n - p, you will get degree n*phi(n), unless n is even and Q(zeta_n) contains sqrt(p)

correct

if it's just x^l - p then you're going to be l(l-1) always

[solar wyvern]

also can ℓ=p^n

[oblique river]

in the case that n is even and Q(zeta_n) contains sqrt(p), then you will get degree n*phi(n)/2

yes if l = p you're still fine (edit: or if l = p^n as long as p is not 2, see the p = 2 n = 8 example)

but those are teh only two possible cases

it gets much worse when you want to talk about x^n - a

where a is not necessarily prime

but for x^n - p you only have those two cases

[solar wyvern]

yeah...

[oblique river]

maybe jsut focus on x^n - 2

that's still an interesting example and you get some cool stuff (like I said if n = 0 mod 8)

[solar wyvern]

so if there are any degree issues will degree always be either n*phi(n) or n*phi(n)/2?

[oblique river]

as long as p is prime, yes

as I said, there are 2 cases:

if n is even and sqrt(p) is contained in zeta(n), you get the /2 case

otherwise, you get n*phi(n)

[solar wyvern]

okay, I think I have a fair idea what sorta badness can happen...just not totally convinced that's the only thing that could go wrong

[oblique river]

(and furthermore, if you want to know when sqrt(p) is contained in Q(zeta_n):

if p = 1 mod 4, then it is iff p divides n.
if p = 3 mod 4, it is iff 4p divides n)

[solar wyvern]

O:

[oblique river]

here's the idea of how badness can happen:

if you have two fields K and L and you want to know [KL : Q]

where KL is the compositum

[solar wyvern]

ok

[oblique river]

you can use [KL : Q] = [K : Q]*[L : Q]/[K cap L : Q]

so basically it's like the product but you have to take into account their intersection

now here K = Q(zeta_n) and L = Q(p^(1/n))

we have to ask: What is K cap L?

well it's certainly a subfield of both K and L. All the subfields of K are galois over Q because Gal(K/Q) is abelian and thus all subgroups are normal

all of the subfields of L are of the form Q(p^(1/m)) for some m dividing n (you have to prove that)

but the only time Q(p^(1/m)) is galois over Q is if m = 1 (in which case it's just Q) or 2 (in which case it's Q(sqrt(p)))

so therefore the only time that K cap L couldn't be Q is if m = 2 and sqrt(p) is contained in Q(zeta_n)

(and thus n is even since m divided n)

and in that bad case, K cap L would be quadratic over Q and thus that's why you get the division by 2

[solar wyvern]

subfields of K aren't necessarily simple extensions of Q right?

[oblique river]

what?

every finite extension of Q is simple

but that's not relevant here anyway

[thorny slate]

simple <=> finitely many intermediate fields

keep this in mind

[solar wyvern]

yah, prim el thm

[oblique river]

the primitive element theorem really isnt explicit at all

like usually the primitive element of your extension will look kinda garbage

like for Q(zeta_n, p^(1/n))

the primitive element might be something like zeta_n + p^(1/n) or something

but like why would you ever consider that element lol

[solar wyvern]

you wouldn't?

[oblique river]

correct haha

[solar wyvern]

I thought if you had like Q(a) and Q(b) with coprime degree in Q you might try like Q(a+rb)

[thorny slate]

even if the degree isnt coprime that works

[oblique river]

yeah, in what what is true is that Q(a, b) will be generated by Q(a + cb) for all but finitely many rational numbers c

but finding c is hard/impossible

[solar wyvern]

oh, that's what you meant

[oblique river]

in my experience, the primitive element theorem is only useful in doing very general things

[solar wyvern]

PET isn't constructive?

[oblique river]

like sometimes it's convenient to be like "I wanna prove things for all fields over Q, so I can just prove them for things of the form Q(a)"

but if you're working with an explicit field Q(a,b), finding a primitive element usually isn't going to help you do anything at all

I dont think so, no

update: it can be made explicit if you know the minimal polynomials of a and b

http://pi.math.cornell.edu/~kbrown/6310/primitive.pdf

[solar wyvern]

[bleak abyss]

Lol, so in the section that thankfully wasn't mine for Galois theory, the professor after the midterm gave a problem set with just one question

He gave a quadratic polynomial and said okay, the Galois group is S_4, no need to prove that. Find a primitive element for every intermediate subfield, and then the minimal polynomial of it

[solar wyvern]

how does a quadratic have galois group S_4

[bleak abyss]

Quartic

Whoops

[solar wyvern]

@oblique river can one say anything about the splitting field of x^n - a over F_p? hazarding p not since just figuring out n=2 not completely trivial

but I have a long list of splitting fields to compute for psets so I'll try anything tbh

[oblique river]

so every extension of F_p is cyclic

that's nice

the extension will certainly contain the nth roots of unity over F_p, and you can figure out the degree of that extension just using some modular arithmetic

other than that I don't see anything super nice off the top of my ehad

[wind steeple]

Wait, every finite extension of F_p is primitive bc it has necessary a finite number of sub-extension ?

[bleak finch]

So how do I show that the ideal p O_k ramifies in a quadratic field extension where p is a rational prime

[bleak finch]

Bump

[thorny slate]

yeah @wind steeple

[wind steeple]

that's nice

[solar wyvern]

@thorny slate so I feel like this should be super basic but is there a clean way to show that every automorphism of an ordered field is trivial?

@oblique river

[oblique river]

that's just not true though?

Q(sqrt(2)) is ordered

because it's a subfield of R

@solar wyvern

[solar wyvern]

uhoh

😦

oh yeah, you should still have an involution right @oblique river

[oblique river]

what do you mean

there are subfields of R with galois group Z/3Z

which is not an involution

do you want your automorphism to preserve the order?

cuz that's not what happens on Q(sqrt(2))

[solar wyvern]

original question was for R, but I thought this might be a general prop for ordered fields

[oblique river]

oh

yeah no it's just a theorem that Aut(R) is trivial

and it doesn't erally have anything to do with the order

well maybe you could use the order to help you

[solar wyvern]

wait what's an automorphism on Q(sqrt2) which doesn't preserve order

does a + bsqrt2 to a - bsqrt2 work

[oblique river]

yeah actually using the order is best

lol how does that preserve order?

[solar wyvern]

idk

[oblique river]

sqrt(2) > 0 but -sqrt(2) < 0

[solar wyvern]

v confuse atm

[oblique river]

what

i just showed that that map doesn't preserve the order

with a simple example

[solar wyvern]

it's an automorphism tho right?

[oblique river]

what?

yes

[solar wyvern]

😰

[oblique river]

the map from Q(sqrt(2)) to itself that takes sqrt(2) to -sqrt(2) is the only nontrivial automorphism

i thought you knew some galois theory, maybe i just forgot haha

my b

[solar wyvern]

[oblique river]

wait you definitely do cuz you were asking about galois groups

[solar wyvern]

sorry just freaking out since assignment due soon

[oblique river]

you should know what Gal(Q(sqrt(2))/Q) is...

in any case if you want to prove something about Aut(R)

idk why you wanna go to the generality of ordered fields

cuz Aut(R) is trivial, period. not even just "any automorphism that preserves the order is trivial"

now, part of the proof will require you to consider the order

but there's no need to try to do some crazy generalization when the problem is only asking about the reals.

[thorny slate]

@solar wyvern maybe you meant Gal(R/Q) = {id}

that's where the order comes in

but yeah buncho explained it well

dont think about orders

[plain sequoia]

Ask in #calculus 😃

[scarlet coyote]

Okay

[cloud walrus]

mathDE:

[chilly ocean]

a

[simple valley]

okay

$\sigma = {0, 1, 2}$. A model of this signature is a triple of functions: $f_0 = 1 \to X$, $f_1 : X \to X$, $f_2 : X \times X \to X$, with more familiar names $f_0 = e$, $f_1 = {}^{-1}$, $f_2 = \cdot$

[cloud walrus]

mniip:

[simple valley]

then you start stating the axioms:

$\forall x_1, \forall x_2, \forall x_3, f_2(f_2(x_1, x_2), x_3) = f_2(x_1, f_2(x_2, x_3))$

$\forall x_1, f_2(f_0(), x_1) = x_1$

$\forall x_1, f_2(x_1, f_0()) = x_1$

$\forall x_1, f_2(f_1(x_1), x_1) = f_0()$

$\forall x_1, f_2(x_1, f_1(x_1)) = f_0()$

[cloud walrus]

mniip:

[simple valley]

the thing to note here is that you have a finite amount of universal quantifiers then an equality using the operations and the quantified elements

[chilly ocean]

right

[simple valley]

the reason this is nice is because if you're considering B a subset of A such that you can restrict f_k to B

then the axioms are automatically satisfied

[lethal tiger]

🤓

[chilly ocean]

i see

[simple valley]

I don't have the rigor to prove that right now

but you can "see" it

the universal quantifeirs quantify over a subset of A

and all expressions stay within B

and equality of expressions in B is the same as in A

because it is a subset

ok so thus you have subobjects

[chilly ocean]

yes

[simple valley]

you also always have the terminal algebra

a singleton set where all f_k are trivial

oh yeah I didn't explain the fields

if you try the same with fields,

you will find that the trick fails

the axioms of fields, as usually stated, don't allow us to do this "subset" trick

[chilly ocean]

because there's no terminal object?

[simple valley]

nah

[chilly ocean]

then

[simple valley]

ok so suppose you have inversion in your signature

your axiom looks something like $\forall x \in F, x \ne 0 \implies x \cdot x^{-1} = 1$

[cloud walrus]

mniip:

[simple valley]

where $\cdot$, ${}^{-1}$, $1$ are the supposed operations

[cloud walrus]

mniip:

[simple valley]

so to begin with, in this formulation we can't say that $0^{-1}$ is not defined. All we can say is that we don't care it to be a specific value

[cloud walrus]

mniip:

[simple valley]

that messes with subobjects because you can't restrict inversion

you can say $0^{-1} = 0$ and then subobjects will be fine

[cloud walrus]

mniip:

[simple valley]

but then quotient objects will fail

quotient objects can't ``handle'' the implication from $x \ne 0$

[cloud walrus]

mniip:

[simple valley]

anyway back to terminal algebra?

[chilly ocean]

yes

but wait a sec

let me process this

[simple valley]

in an algebraic theory, a singleton set (with trivial operations) is always a model

if * is the single inhabitant of the singleton set, then in any axiom, any universally quantified variable will be *, and any expression using those variables and operations will be *

and thus all equalities will be satisfied

for any algebra <A, f> there's a homomorphism to it, just send every element of A to *

h(f_k(...)) = *

g_k(h(...), ..., h(...)) = g_k(*, ..., *) = *

commutes, hence homomorphism

the homomorphism is unique because, well, any homomorphism is also a function, and there's only one function to the singleton set (it is terminal in Set)

the category of models of an algebraic theory has products

if you have models <A, f> and <B, g> then you can make a model on AxB

$fg_k((a_1, b_1), \dots, (a_j, b_j)) = (f_k(a_1, \dots, a_j), g_k(b_1, \dots, b_j))$

[cloud walrus]

mniip:

[simple valley]

axioms are again satisfied because of algebraicness

ok

these notes are really scattered in my scratchbook

the defining equation for fg_k is this commutative diagram

[cloud walrus]

mniip:

[simple valley]

you get "projections": homomorphisms into A and B

to see that they're homomorphisms ponder on this diagram

[cloud walrus]

mniip:

[simple valley]

and the last bit of magic is that given C with homomorphism C -> A and C -> B, you can take the function C -> AxB generated by the universla property of product in Set, and it will be a homomorphism

therefore the category of models of an algebraic theory has products

next we define a congruence on $A$ as a subobject of $A \times A$ such that when interpreted as a set it is an equivalence relation on $A$

[cloud walrus]

mniip:

[simple valley]

we can denote such a congruence by $\sim$ with a notation $x \sim y \iff (x, y) \in {\sim} \subseteq A \times A$

[cloud walrus]

mniip:

[chilly ocean]

hm

so when we have equiv relations on sets

we can define quotient sets right

[simple valley]

yes

[chilly ocean]

so is there something like quotient cats

[simple valley]

yes? but that's beside the point here

[chilly ocean]

just wondering lel

[simple valley]

what we're about to do is construct quotient objects in the category of models

consider the set of equivalence classes $A/{\sim}$: it can be upgraded to an algebra: $[f]_k([x_1], \dots, [x_j]) = [f_k(x_1, \dots, x_j)]$

[cloud walrus]

mniip:

[simple valley]

it somehow turns out to be a function?

[chilly ocean]

[solar vessel]

[simple valley]

ah yes

trivial

suppose $x_1 \sim y_1, \dots, x_j \sim y_j$, then $(x_1, x_1) \in {\sim}, \dots, (x_j, y_j) \in {\sim}$, then because ${\sim}$ is an algebra, a subalgebra of $A \times A$, $ff_k((x_1, y_1), \dots, (x_j, y_j)) = (f_k(x_1, \dots, x_j), f_k(y_1, \dots, y_j)) \in {\sim}$, ergo $f_k(x_1, \dots, x_j) \sim f_k(y_1, \dots, y_j)$

[cloud walrus]

mniip:

[simple valley]

thus the equation up there for [f]_k is well defined

also [] as a function A -> A/~ is a homomorphism of algebras, from <A, f> to <A/~, [f]>

you also have images in this category

set-theoretical Im h, for some homomorphism h, can be shown to be a subobject of the codomain of the homomorphism

we need to prove that $g_k$ can be restricted to $\operatorname{Im}h$: given $y_1, \dots, y_j \in \operatorname{Im} h$, there exist $x_1, \dots, x_j$ such that $y_1 = h(x_1), \dots, y_j = h(x_j)$, thus there exists $x = f_k(x_1, \dots, x_j)$ such that $y = g_k(y_1, \dots, y_j) = h(x)$

because h is a homomorphism

[cloud walrus]

mniip:

[simple valley]

thus $g_k(y_1, \dots, y_j) \in \operatorname{Im}h$

[cloud walrus]

mniip:

[void dragon]

Honest question from lurker, is A^\sigma(k) in the diagram up there denotes product of A's?

[simple valley]

yes

the diagrams are in Set

so your regular old cartesian product

[void dragon]

Ok just clarifying since my notational memory is a bit bad

[simple valley]

what's left is to show that uhhh

pullback of h along itself gives a congruence

that commutes with h

fuck that I'm gonna go play minecraft

oxide isn't listening anyway

I should condense all of this into a seminar in the future...

[void dragon]

Lol

You should make it up in a pdf

Or maybe Dropbox paper. The latex support is not bad

[simple valley]

well I have started a series on abstract algebra themed category theory

for our reading group on abstract algebra

it's up on youtube too

[mild laurel]

there's a algebra reading group?

[simple valley]

yes

do you want in?

[void dragon]

Didn't even know you have reading groups here

[mild laurel]

yeah

what algebra are you guys doing

[void dragon]

Hmm. I'm interested. What's the YouTube channel?

[simple valley]

@mild laurel intro level abs alg currently

we're following Judson: abs alg: thry and apps

@void dragon I think it's the only one currently

wish we did more

[void dragon]

Can't really commit into something like reading group but would be interested in category theory applied to algebra

[simple valley]

oof

there isn't much content yet

because we need to cover the algebra first

[mild laurel]

yeah I'm actually studying commutative algebra out of atiyah rn so I could help out for now

[solar wyvern]

$$\begin{tikzcd}
& & & & E \arrow[r, dotted] & \overline{k} \
& F_1 \arrow[rr, dashed] \arrow[rrru] & & F_1F_2 \arrow[ru, "m"] & & \
& & F_1 \cap F_2 \arrow[lu, dashed] \arrow[rd, dashed] & & & \
... \arrow[r, "1"] & k \arrow[uu] \arrow[rr] \arrow[ru, "n", dashed] & & F_2 \arrow[uu, dashed] \arrow[ruuu] & &
\end{tikzcd}$$

[cloud walrus]

flimflam:

[solar wyvern]

o hopefully this is less dumb and more comprehensible but what happens when n,m \neq 1 here (degree of the field extension):
(k is a prime field, kbar its algebraic closure, so all the extensions should be separable I think)

@oblique river @thorny slate if any of these aren't galois what sort of degree arguments do we have even

each extension besides E to kbar assumed finite

[thorny slate]

are you making the diagram complicated on purpose?

[solar wyvern]

no, not especially

[thorny slate]

what's the question?

[solar wyvern]

how do you find [F_1F_2:F_1 ⋂F_2]

or when rather

[thorny slate]

could be anything

[solar wyvern]

what nice things happen when n=m=1

[thorny slate]

m doesn't do anything

if n = 1 then [F_1 F_2: k] is the product of both

[solar wyvern]

if m=1, F_1 is a splitting field for f_1, F_2 is a splitting field for f_2, F_1F_2 is the minimal field over which both split, right?

[thorny slate]

why if m = 1?

I don't understand why m and E exist

[solar wyvern]

E could be a finite extension where those two (but maybe other) polynomials split in?

which you may know or be given?

[thorny slate]

E could be anything

and it doesn't affect any other part

[solar wyvern]

or it could be a simple extension

but not the minimal one containing whatever you want?

I mean given two k-extensions you always get the compositum and intersection? Doesn't seem that outlandish to wonder how they interact with other extensions.

[thorny slate]

they dont in this case

because you expressly say that E has nothing to do with anything

looks intentional to me

but idk