Proof involving field extensions

[daon]

I have to prove that all quadratic extensions are normal. I think my problem here is the wording.

In my mind this is what it is asking: If f(x) is an irreducible quadratic polynomial over a field F, and f(x) splits in E, then the extension E/F is normal.

Would that be right?

So for example \(\displaystyle f(x)=x^2+1\) is irreducible over \(\displaystyle \mathbb{R}\), but since f(x) splits in \(\displaystyle \mathbb{C}\), \(\displaystyle \mathbb{C}\)/\(\displaystyle \mathbb{R}\) is normal.

 

[daon]

Okay, I am given that TFAE, which makes any extension E/F normal:

1) E is a splitting field of some polynomial f(x) in F[x]
2) For every algebraic extension K/E and ever F-homomorphism \(\displaystyle \phi:E \rightarrow K, \,\, \phi(E)=E\)
3) For every algebraic extension K/E and ever F-homomorphism \(\displaystyle \phi:E \rightarrow K, \,\, \phi(E) \le E\) (\(\displaystyle \le\) meaning subfield)
4) Every irreducible polynomial p(x) in F[x] which has a root in E, splits in E.

(2 and 3 seem redundant to me, but they're there.)

So I was thinking about letting \(\displaystyle p(x)=ax^2+bx+c \in F[x]\) be irreducible and showing #4... that if \(\displaystyle p(x)\) has a root in E, then both roots of \(\displaystyle p(x)\) must be in E.

If \(\displaystyle p(x)\) has a multiple root, then its trivial, so assuming p(x) separable (having no multiple roots). So if \(\displaystyle E/F\) is an extension with \(\displaystyle \alpha \in E\) a root of p(x) then \(\displaystyle p(x)=(x-\alpha)(\beta x + \gamma) \in E[x]\).

Then I must have that \(\displaystyle \beta\) and \(\displaystyle \gamma\) both be in E and therefore the second root is \(\displaystyle \frac{-\gamma}{\beta} \in E\). That seems too easy and I think I'm missing something.