A Universe of Sorts

§ Normal field extensions

(1) For an extension $L/K$ , if a polynomial $p(x) \in K[x]$ and has a root $\alpha \in L$ has all its roots in $L$ . So $p(x)$ splits into linear factors $p(x) = (x - l_1)(x - l_2) \cdot (x - l_n)$ for $l_i \in L$ .
(2) [equivalent ] $L$ is the splitting field over $K$ of some set of polynomials.
(3) [equivalent ] Consider $K \subseteq L \subseteq \overline K$ . Then any automorphism of $\overline K/K$ (ie, aut that fixes $K$ pointwise) maps $L$ to $L$ [fixes $L$ as a set, NOT pointwise ].
Eq: $Q(2^{1/3})$ is not normal
Eq: $Q(2^{1/3}, \omega_3)$ is a normal extension because it's the splitting field of $x^3 - 2$ .

(1) We know that $p$ has a root in $L$ implies $p$ has all rots in $L$ .
For each $\alpha \in L$ , we take the minimal polynomial $p(\alpha)$ . Then $p$ splits over $L$ , because $L$ contains a single root of $p$ ( $\alpha$ ).
Thus, $L$ is the splitting field for the set of polynomials $\{ minpoly(\alpha) \in K[x] : \alpha \in L \}$ .

(2) says that $L$ is the splitting field for some set of polynomials.
An aut $\sigma: \overline K \to \overline K$ that fixes $K$ acts trivially on polynomials in $K[x]$ .
$L$ is the set of all roots of polynomials $\{ minpoly(\alpha) \in K[x] : \alpha \in L \}$ .
Since $\sigma$ fixes $K[x]$ , it also cannot change the set of roots of the polynomials. Thus the set $\{ minpoly(\alpha) \in K[x] : \alpha \in L \}$ remains invariant under $\sigma$ . ( $\sigma$ cannot add elements into $L$ ). It can at most permute the roots of $L$ .

(3) says that any automorphism $\sigma$ of $\overline K/K$ fixes $L$ as a set.
We wish to show that if $p$ has a root $\alpha \in L$ , $L$ has all roots of $p$ .
we claim that for any root $\beta \in L$ , there is an automorphism $\tau: \overline K/K$ such that $\tau(\alpha) = \beta$ .
Consider the tower of extensions $K \subseteq K(\alpha) \subseteq \overline K$ and $K \subseteq K(\beta) \subseteq \overline K$ . Both $K(\alpha)$ and $K(\beta)$ look like $K[x] / p$ because $p$ is the minimal polynomial for both $\alpha$ and $\beta$ .
Thus, we can write an a function $\tau: K(\alpha) \to K(\beta)$ which sends $\alpha \mapsto \beta$ .
Now, by uniqueness of field extensions, this map $\tau$ extends uniquely to a map $\overline K \to \overline K$ which sends $\alpha to \beta$ . [TODO: DUBIOUS ].
But notice that $\tau$ must fix $L$ (by (3)) and $\alpha in L$ . Thus, $\tau(\alpha) \in \tau(L)$ , or $\beta = \tau(\alpha \in \tau(L) = L$ .
Thus, for a polynomial $p$ with root $\alpha in L$ , and for any other root $\beta$ of $p$ , we have that $\beta \in L$ .

Let $L/K$ be the splitting field of $f \in K[x]$ . Let $g \in K[x]$ have a root $\alpha \in L$ .
Let $\beta \in \overline K$ be another root of $g$ . We wish to show that $\beta \in L$ to show that $L$ is normal.
There is an embedding $i: K(\alpha) \hookrightarrow \overline K$ which fixes $K$ and sends $\alpha$ to $\beta$ .
See that $i(L)$ is also a splitting field for $f$ over $K$ inside $\overline K$ .
But splitting fields are unique, so $i(L) = L$ .
Since $i(\alpha) = \beta$ , this means $\beta \in L$ as desired.

Let us have a degree 2 extension $K \subseteq L$
So we have some $p(x) = x^2 + bx + c \in K[x]$ , $L = K(\alpha)$ for $\alpha$ a root of $p$ .
We know that $\alpha + \beta = b$ for $\alpha \in L, b \in K$ . Thus $\beta = b - \alpha \in L$ .
Thus, the extension is normal since $L$ contains all the roots ( $\alpha, \beta$ ) of $p$ as soon as it contained one of them.

Consider $K \subseteq L \subseteq M$ . If $K \subseteq L$ is normal, $L \subseteq M$ is normal, then is $K \subseteq M$ normal?
Answer: NO!
Counter-example: $Q \subseteq Q(2^{1/2}) \subseteq Q(2^{1/4})$ .
Each of the two pieces are normal since they are degree two. But the full tower is not normal, because $Q(2^{1/4})/Q$ has minimial polynomial $x^4 - 2$ .
On the other hand, $Q(2^{1/4})/Q(2^{1/2})$ has a minimal polynomial $x^2 - \sqrt{2} \in Q[2^{1/2}]$ .
So, normality is not transitive!
Another way of looking at it: We want to show that $\sigma(M) \subseteq M$ where $\sigma: aut(\overline K/K)$ . Since $L/K$ is normal, and $\sigma$ is an autormophism of $L/K$ , we have $\sigma(L) \subseteq L$ [by normal ]. Since $M/L$ is normal, we must have $\sigma(M) \subseteq M$ . Therefore, we are done?
NO! The problem is that $\sigma$ is not a legal automorphism of $M/L$ , since $\sigma$ fixes $L$ as a set ( $\sigma L \subseteq L$ ), and not pointwise ( $\sigma(l) = l$ for all $l \in L$ .)