I haven't done that in a long time, but cant you do it saying it is either prod of polynomials of degree 3, 3 or 2, 4

And then bla bla $\omega$ times roots bla bla must be 3,3

And then do direct computation

Some more fast-spoken nonsense: x^3 - sqrt(3) and x^3 + sqrt(3) are both irreducible over Q(sqrt(-3), i) = Q(sqrt(3), i) because they are irreducible over Q(sqrt(3)) by Eisenstein and if it factored in Q(sqrt(3), i) they must have a root in there as well, impossible because Q(sqrt(3), i)/Q is a degree 4 extension whilst Q[x]/(x^6 - 3) is degree 6. (@Thorgott check?)

I also don't see why Eisenstein's doesn't work directly

How?

Do i misremember it? It is basically pick a prime that divides all coefficients except the first one right

Also $p^2$ shouldnt

but this is irreducibility over Q(sqrt(-3))

not Q

Oh right wtf yeah

How did I miss that

Oh btw why I am doing this? x^3 - sqrt(3) is irreducible over Q(sqrt(-3), i) because if it factored a root would be in Q(sqrt(-3), i), which is a degree 4 extension of Q, but x^6 - 3 is a degree 6 poly

wait, isn't just $x^6-3=(x^3-\sqrt{3})(x^3+\sqrt{3})$

Q(sqrt(-3)) man

@love_sodam Let $K=\Bbb Q(\sqrt[6]{3})$ and $L=\Bbb Q(\sqrt{-3})$. If $x^6-3$ is not irreducible over $L$, then $[KL:L]<6$. Using the tower law we have $[KL:K][K:\Bbb Q]=[KL:\Bbb Q]=[KL:L][L:\Bbb Q]<12$ Now pluggin in $[K:\Bbb Q]=6$ gives $[KL:K]<2$, so $[KL:K]=1$ which implies $L \subset K$. But this is nonsense: $K$ is real and $L$ not

@BalarkaSen don't get what you do here

Forget it

Screw this

@LukasHeger this

@Krijn ?

I was mentioning that what you said makes a lot of sense, as a signal to bala

ah I see

I already gave a proof, I just couldn't be arsed to write it properly. x^3 +- sqrt(3) is irreducible over Q(sqrt(-3), i), hence x^6 - 3 is irreducible over Q(sqrt(-3))

Seems @LukasHeger make sense

@TedShifrin please flag those. I hate that. I will undelete those immediately.

@BalarkaSen yeah that works as well. Generally if $L/K$ is Galois with Galois group $\Gamma$ and $f \in L[x]$ is irreducible, then $\prod_{g \in \{\gamma f\mid \gamma \in \Gamma\}} g$ is irreducible over $K$

yeah

just fucking annoying to write, i dont give a shit

field theory is garbage

:(

I feel u undervalue the elegance of some field theory

I have learnt and forgotten it 80 times already

I don't undervalue it it just annoys the hell out of me

Ever learnt about complex multiplication?

i know how to multiply in C yes

"The theory of complex multiplication is not only the most beautiful part of mathematics but also of the whole of science."

- Hilbert

That's a reason to study it, no?

"RP^2 does not immerse in R^3"

-also Hilbert

Reason enough to not trust this man

"can't solve these 23 problems, pls fix" -Hilbert

Solve your own problems old man

"We are a university, not a bathhouse"- Hilbert

yeah Poincare didn't make a list of problems he just conjectured 1 thing

fuckin Hilbert

I have a Hilbert

I mean a question.

what is a power set map?

That's why I like Weil

Make sensible conjectures, guide a field, in the end there is a nice theory that works

And you only need a Grothendieck and Deligne to solve it

Is there algebraically closed field contained in C?

"only" Grothendieck and Deligne

which is not C

Qbar

Q

@love_sodam Yes, plenty

many embeddings of Qbar :p

what is Q bar?

"look at my fancy hat" - Hilbert

oh algebraic closure you mean

too much Hilbert in the chat man

its triggering me

new topic

@BalarkaSen Plenty of embeddings of C as well

lol

ok dude

But I want explicit example not like closure something

I heard that

"all algebraic numbers" is not an explicit example?

what does explicit mean

what does example mean

Jokes apart there's also embeddings of $\overline{\Bbb Q(x)}$ and so on

You'll need the bar right

Every algebraically closed field thats not Q bar seems more complex then Q bar

all algebraic numbers is the easiest example

it's the smallest one contained in C

Let A be the field of all algebraic numbers

@AlessandroCodenotti "Jokes apart"

wat

I heard that as an example of infinite algebraic extension, {a\in C: a is algebraic over Q}

Is this algebraiclly closed?

Lol

yes

Q-bar

that's the algebraic closure of Q

Oh that's Q bar?

ye

Oh I didn't know that

Q-bar sounds like a bar

its a good exercise to show if a, b are algebraic over Q then so are a + b and ab

I don't know closure so can you prove it directly from that?

yes you can

it's actually quite tedious isn't it

I think so

which is?

ab in particular. you get the polynomial from some stupid determinant

no, that's easy

finite extension iff algebraic

^

Q(ab)/Q(a)/Q

tower law

what's the polynomial

not iff

but showing that the algebraic numbers are algebraically closed is tedious

@MikeMiller nobody cares

you dont need to know it

geometry brain

hahah

@BalarkaSen thx

@Thorgott is this what I'm thinking of?

there's a determinant somewhere

a,b is algebraic then a+b, ab is algebriac is easy but showing that collection is algebraic ally closed seems not that easy

perhaps Mike

@Thorgott that follows from transitivity of algebraicness if $E/L/K$ is a tower and $E/L$ is algebraic and $L/K$ is algebraic, then $E/K$ is algebraic

how?

yeah

what Lukas said

What are you people smoking

this is stupid

so if $E/\overline{\Bbb Q}$ is algebraic, then actually $E\subset \overline{\Bbb Q}$

Give me some of this

You can't tell me a number is algebraic and not be able to write down a polynomial

But transitivity of algebraicness is not trivial right

its tower law!!

yeah polynomials

"this is algebraic by the theory of nilpotent schemes"

finite iff algebraic

thats false balarka

it's just tower law as Balarka says

ok finite implies algebraic

yeah

Q bar is an example

you can always reduce questions about algebraicness to questions about finite extensions

yeah

direct limits

algebraic iff colimit of finite extensions

everything is a direct limit offinite extensions

COLIMIT FUCK OFF

If you can't turn this into an algorithm to find the polynomial defining x^3 + sqrt(2)x + 1 = 0 what's the point

just say union

Balarka seems aggresive tonight

I am getting triggered by (1) field theory (2) Hilbert

get this shite out of my face

Still $\Rightarrow$ CM

@BalarkaSen Just take a filtered colimit bro

@BalarkaSen let's talk about Hilbertian fields

Hilbert class field lmao

Lets talk about maximal abelian covers instead

how do you build those

and what does knot polynomials have to do with them

you can also prove it using determinants, Mike, but why would you

Please someone steer this away from Field Theory

I veto your idea balarka

@Krijn He's dielectric.

Also this ^

@AlessandroCodenotti can one use löwenheim-skolem here

@BalarkaSen Lmao you've been struggling with electric field theory all day and now it's algebraic field theory instead

when i SAY divert this AWAY from field theory I DONT MEAN SET THEORY

And now you want to move to topological quantum field theory?

haha

So confusing

DONT DO MODEL THEORY NOW

@BalarkaSen can't stop me

@MikeMiller Lmfao

I can embed the complex numbers into super duper nonstandard complex numbers of whatnot cardinality and shits

I'm laughing irl

Clearly we need to move to class field theory

Tell me something interesting that's a consequence of model theory. The only example I know is Ax-Grothendieck

what was that Tarski garbage

@user2103480 Hm not sure

existential quantifiers can be replaced by universal ones in algebraic formulas

@MikeMiller Tarski-Seidenberg

yeah that

@MikeMiller solvability of euclidean geometry

o-minimal structures

Tame Topology

In relation to what balarka said

nerdoid shit

This is the least interesting theorem I've ever read

I've been more excited by technical lemmas

LOL

@user2103480 Uninteresting unless you have an algorithm

@MikeMiller We have an algorithm

its running time is just ungodly

make it do all of Hall & Stevens or whatever

solve Euclidean geometry

@MikeMiller It's actually a natural example of a doubly exponential algorithm

We also have an algorithm to prove every true assertion coming from ZFC in finite time, believe it or not

And the complexity is optimal

It's literally just listing consequences of the axioms though kek

Lol

There's an expspace algorithm

stupidity manifest

logic brain

@user2103480 That's a semi algorithm or whatever

You can't tell me if something is false in finite time

LOL

how do people take this stuff seriously

@AlessandroCodenotti algebraically closed field with characteristic zero should be first-order axiomatizable, but not sure how to handle transcendence base

@user2103480 Sure, it's even a model complete theory with QE, but I still don't see how to get embeddings out of LS

@BalarkaSen Meh it's their game just like we have ours

@MikeMiller if it's false, its converse will show up as well

negation

since that is true

independence intensifies

If we know it's true, it will be proven in finite time. But we cannot in finite time decide whether it will be proven, generally

@MikeMiller agree but it is my individual right to call some games stupid

for example Dungeons & Dragons

Call of Cthulhu pen and paper RPG

@user2103480 that's useless right if you don't know if is independent from ZFC

exactly

@BalarkaSen like banach-mazur games

looking at the general topology guy in this round

Markov winning strategies

@TedShifrin $\frac\pi4$. I didn't see another answer, but I could have missed it.

@AlessandroCodenotti ah we can just take ultraproducts instead of löwenheim-skolem (which is about the same, but ensures the embedding of C into the field)

something something cardinality then shows that it's not actually C, hopefully

@BalarkaSen well, I guess I won't be inviting you to our weekly get together, when we can have them again.

haha

@user2103480 Uncountable algebraically closed field of fixed characteristic are classified by cardinality

@robjohn This was for geocalc to see an application of integrals being independent of a parameter, but I don’t think he thought about it even one second.

And I did flag that question, robjohn. Don't know if any mod dealt with it.

@AlessandroCodenotti I read it backwards lol I thought it was about fields that contain C, sorry bout that

@TedShifrin Yeah, but I don't know if he would have seen why it was independent of $n$, though the proof is simple, just not necessarily obvious.

I see what should be $n$ but not why it's independent of $n$

@BalarkaSen Substitute $x\mapsto\frac1x$, then add the two.

Oh, insane

Nice

the part that is hard is that often one goes down a rat hole looking at complex analytic approaches.

I did for a few minutes after I saw the question.

I assigned it to my class for differentiation under the integral.

@TedShifrin Ouch! another rat hole. I was thinking of looking at that, but tried $x\mapsto\frac1x$ first, thank heavens.

Nah, it works fine.

@TedShifrin really? I'll have to look at that.

So I recollect.

@TedShifrin Okay, you use the same substitution to show that the derivative is $0$.

I think I used $\pi/2-x$

Need to think again. It's been zillions of years.

@TedShifrin Ah, yes... you are using $\tan(\theta)$. I had substituted $\tan(\theta)=x$

Then $x\mapsto\frac1x$ is the same as $\theta\mapsto\frac\pi2-\theta$

Right. The log suggests it strongly.

$$\int_0^{\pi/2}\frac{\mathrm{d}\theta}{1+\tan^n(\theta)} =\int_0^{\pi/2}\frac{\tan^n(\theta)\,\mathrm{d}\theta}{1+\tan^n(\theta)}$$ upon substituting $\theta\mapsto\frac\pi2-\theta$. Average the two.

Question: If $T$ is an operator on a Hilbert space $\mathcal{H}$ with the property that, for any bounded sequence $(x_n)$ in $\mathcal{H}$ the sequence $(Tx_n)$ has a convergent subsequence, does this imply that $\text{Im}(T)$ is complete? I feel like the answer is "yes, clearly," but I'm in a headspace of doubting myself at the moment.

@Rithaniel compact you mean

$$\int_0^\infty\frac{\mathrm{d}x}{\left(1+x^2\right)\left(1+x^n\right)}=\int_0^\infty\frac{x^n\,\mathrm{d}x}{\left(1+x^2\right)\left(1+x^n\right)}$$ upon substituting $x\mapsto\frac1x$. Average the two.

Yeah, the overall goal is to show that $T$ is compact, but this completeness thing is a pit-stop I'm making in my proof.

Then you're done pretty quickly

The idea is "complete $\to$ orthonormal basis $\to$ finite dimensional $\to$ compact"

Is this too much of a convoluted path, maybe?

For compact operators you consider the closure of the image

So you consider the closure of the image of the unit closed ball, in which every bounded sequence is the image of a bounded sequence in the domain, yada yada

@user2103480 So, my attempted proof failed because I falsely thought that complete implied separable. However, your proof also doesn't make any sense to me. I'm trying to go from "all sequences in the image of $T$ have convergent subsequences" to "if $x_n$ is a weakly convergent sequence, then $Tx_n$ is a strongly convergent sequence."