Am back

Hoi

ohi

Want to work a bit on Galois?

Let's DO IT

Aight

Okay so earlier we mentioned that a finite extension is finitely generated

By taking the basis elements

Yup

Now, how the frickity frop is a finitely generated extension not finite?

I mean look at $\Bbb C(z)/\Bbb C$ :)

That's an infinite dimensional vector space

Oh... Okay I guess yeah

I'll see if I can prove that finite field extensions are algebraic

Yup, try it

Oh I think I'm starting to see it by analogy with $\mathbb{C}$

Okay so

Let $L$ be a finite extension of $k$

We know that a linear polynomial in $k[x]$ has a root

Or wait no this isn't necessarily the right way to go about it actually

The intuition I have is that you should have some kind of similar thing to $\mathbb{C}$ where you can't really distinguish between $i$ and $-i$

That's true

Like if you have some finite dimensional vector space over $k$, we know that you can pick a bunch of bases for it, and somehow you ought not be able to distinguish when writing $L$ as an extension. This ambiguity might express itself as a $k$-polynomial with these as roots

Sort of right intuition

Like if you can find some field automorphism of $L$ which fixes $k$ pointwise, then you can just pair up stuff. Now just gotta find out how to make this work precisely

The tricky part is what distinguishes finite from infinite

Okay so maybe I have it?

OK?

Choose a basis $b_1, \ldots, b_k$ for $L$ such that $b_1 \in k$

Then consider the map $\phi: b_i \to -b_i$ for $i > 1$

Extending linearly

This definitely fixes $k$ pointwise, now I want to show that this is a field automorphism.

Let's pretend I did that already. Then given $z\in L$, we should have $(x-z)(x-\phi(z)) = x^2 - x(z + \phi(z)) - z\phi(z)$

Or meh

I guess we need that $z\phi(z) \in k$ even then

Oh wait I see how we can use finite maybe

It's not entirely clear to me what that particular automorphism will give you

If you define $k-1$ automorphisms, one for each basis that takes you out of $k$

WAIT A SECOND

Okay okay this is gonna be funny

So let $z\in L$

We know that $1, z, z^2, z^3, \ldots , z^k$ are linearly dependent

Bingo

Oh my god that is beautiful

Right?

Okay so, $\mathbb{C}$ is also a separable extension. What a nice field!

It indeed is

@Daminark just changing the sign of the basis elements won't define a field automorphism in general

Right, I wasn't paying attention due to the h-principle thing earlier. You want to take conjugates of roots to elsewhere. Not all conjugates are related by a sign.

Eg, think about the splitting field of $x^3 - 2$ over $\Bbb Q$ (that means the smallest extension of $\Bbb Q$ over which $x^3 - 2$ splits as a product of linear polynomials)

That is of course $\Bbb Q(\sqrt[3]{2}, \zeta)$. The conjugates of $\sqrt[3]{2}$ are related by multiplication by $\zeta, \zeta^2$, etc

Ah, I see