- (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 $l \in L$.)