
## 9.15 Normal extensions

Let $P \in F[x]$ be a nonconstant polynomial over a field $F$. We say $P$ splits completely into linear factors over $F$ or splits completely over $F$ if there exist $c \in F^*$, $n \geq 1$, $\alpha _1, \ldots , \alpha _ n \in F$ such that

$P = c(x - \alpha _1) \ldots (x - \alpha _ n)$

in $F[x]$. Normal extensions are defined as follows.

Definition 9.15.1. Let $E/F$ be an algebraic field extension. We say $E$ is normal over $F$ if for all $\alpha \in E$ the minimal polynomial $P$ of $\alpha$ over $F$ splits completely into linear factors over $E$.

As in the case of separable extensions, it takes a bit of work to establish the basic properties of this notion.

Lemma 9.15.2. Let $K/E/F$ be a tower of algebraic field extensions. If $K$ is normal over $F$, then $K$ is normal over $E$.

Proof. Let $\alpha \in K$. Let $P$ be the minimal polynomial of $\alpha$ over $F$. Let $Q$ be the minimal polynomial of $\alpha$ over $E$. Then $Q$ divides $P$ in the polynomial ring $E[x]$, say $P = QR$. Hence, if $P$ splits completely over $K$, then so does $Q$. $\square$

Lemma 9.15.3. Let $F$ be a field. Let $M/F$ be an algebraic extension. Let $F \subset E_ i \subset M$, $i \in I$ be subextensions with $E_ i/F$ normal. Then $\bigcap E_ i$ is normal over $F$.

Proof. Direct from the definitions. $\square$

Lemma 9.15.4. Let $E/F$ be a normal algebraic field extension. Then the subextension $E/E_{sep}/F$ of Lemma 9.14.6 is normal.

Proof. If the characteristic is zero, then $E_{sep} = E$, and the result is clear. If the characteristic is $p > 0$, then $E_{sep}$ is the set of elements of $E$ which are separable over $F$. Then if $\alpha \in E_{sep}$ has minimal polynomial $P$ write $P = c(x - \alpha )(x - \alpha _2) \ldots (x - \alpha _ d)$ with $\alpha _2, \ldots , \alpha _ d \in E$. Since $P$ is a separable polynomial and since $\alpha _ i$ is a root of $P$, we conclude $\alpha _ i \in E_{sep}$ as desired. $\square$

Lemma 9.15.5. Let $E/F$ be an algebraic extension of fields. Let $\overline{F}$ be an algebraic closure of $F$. The following are equivalent

1. $E$ is normal over $F$, and

2. for every pair $\sigma , \sigma ' \in \mathop{Mor}\nolimits _ F(E, \overline{F})$ we have $\sigma (E) = \sigma '(E)$.

Proof. Let $\mathcal{P}$ be the set of all minimal polynomials over $F$ of all elements of $E$. Set

$T = \{ \beta \in \overline{F} \mid P(\beta ) = 0\text{ for some }P \in \mathcal{P}\}$

It is clear that if $E$ is normal over $F$, then $\sigma (E) = T$ for all $\sigma \in \mathop{Mor}\nolimits _ F(E, \overline{F})$. Thus we see that (1) implies (2).

Conversely, assume (2). Pick $\beta \in T$. We can find a corresponding $\alpha \in E$ whose minimal polynomial $P \in \mathcal{P}$ annihilates $\beta$. Because $F(\alpha ) = F[x]/(P)$ we can find an element $\sigma _0 \in \mathop{Mor}\nolimits _ F(F(\alpha ), \overline{F})$ mapping $\alpha$ to $\beta$. By Lemma 9.10.5 we can extend $\sigma _0$ to a $\sigma \in \mathop{Mor}\nolimits _ F(E, \overline{F})$. Whence we see that $\beta$ is in the common image of all embeddings $\sigma : E \to \overline{F}$. It follows that $\sigma (E) = T$ for any $\sigma$. Fix a $\sigma$. Now let $P \in \mathcal{P}$. Then we can write

$P = (x - \beta _1) \ldots (x - \beta _ n)$

for some $n$ and $\beta _ i \in \overline{F}$ by Lemma 9.10.2. Observe that $\beta _ i \in T$. Thus $\beta _ i = \sigma (\alpha _ i)$ for some $\alpha _ i \in E$. Thus $P = (x - \alpha _1) \ldots (x - \alpha _ n)$ splits completely over $E$. This finishes the proof. $\square$

Lemma 9.15.6. Let $E/F$ be an algebraic extension of fields. If $E$ is generated by $\alpha _ i \in E$, $i \in I$ over $F$ and if for each $i$ the minimal polynomial of $\alpha _ i$ over $F$ splits completely in $E$, then $E/F$ is normal.

Proof. Let $P_ i$ be the minimal polynomial of $\alpha _ i$ over $F$. Let $\alpha _ i = \alpha _{i, 1}, \alpha _{i, 2}, \ldots , \alpha _{i, d_ i}$ be the roots of $P_ i$ over $E$. Given two embeddings $\sigma , \sigma ' : E \to \overline{F}$ over $F$ we see that

$\{ \sigma (\alpha _{i, 1}), \ldots , \sigma (\alpha _{i, d_ i})\} = \{ \sigma '(\alpha _{i, 1}), \ldots , \sigma '(\alpha _{i, d_ i})\}$

because both sides are equal to the set of roots of $P_ i$ in $\overline{F}$. The elements $\alpha _{i, j}$ generate $E$ over $F$ and we find that $\sigma (E) = \sigma '(E)$. Hence $E/F$ is normal by Lemma 9.15.5. $\square$

Lemma 9.15.7. Let $L/M/K$ be a tower of algebraic extensions.

1. If $M/K$ is normal, then any automorphism $\tau$ of $L/K$ induces an automorphism $\tau |_ M : M \to M$.

2. If $L/K$ is normal, then $K$-algebra map $\sigma : M \to L$ extends to an automorphism of $L$.

Proof. Choose an algebraic closure $\overline{L}$ of $L$ (Theorem 9.10.4).

Let $\tau$ be as in (1). Then $\tau (M) = M$ as subfields of $\overline{L}$ by Lemma 9.15.5 and hence $\tau |_ M : M \to M$ is an automorphism.

Let $\sigma : M \to L$ be as in (2). By Lemma 9.10.5 we can extend $\sigma$ to a map $\tau : L \to \overline{L}$, i.e., such that

$\xymatrix{ L \ar[r]_\tau & \overline{L} \\ M \ar[u] \ar[ru]_\sigma & K \ar[l] \ar[u] }$

is commutative. By Lemma 9.15.5 we see that $\tau (L) = L$. Hence $\tau : L \to L$ is an automorphism which extends $\sigma$. $\square$

Definition 9.15.8. Let $E/F$ be an extension of fields. Then $\text{Aut}(E/F)$ or $\text{Aut}_ F(E)$ denotes the automorphism group of $E$ as an object of the category of $F$-extensions. Elements of $\text{Aut}(E/F)$ are called automorphisms of $E$ over $F$ or automorphisms of $E/F$.

Here is a characterization of normal extensions in terms of automorphisms.

Lemma 9.15.9. Let $E/F$ be a finite extension. We have

$|\text{Aut}(E/F)| \leq [E : F]_ s$

with equality if and only if $E$ is normal over $F$.

Proof. Choose an algebraic closure $\overline{F}$ of $F$. Recall that $[E : F]_ s = |\mathop{Mor}\nolimits _ F(E, \overline{F})|$. Pick an element $\sigma _0 \in \mathop{Mor}\nolimits _ F(E, \overline{F})$. Then the map

$\text{Aut}(E/F) \longrightarrow \mathop{Mor}\nolimits _ F(E, \overline{F}),\quad \tau \longmapsto \sigma _0 \circ \tau$

is injective. Thus the inequality. If equality holds, then every $\sigma \in \mathop{Mor}\nolimits _ F(E, \overline{F})$ is gotten by precomposing $\sigma _0$ by an automorphism. Hence $\sigma (E) = \sigma _0(E)$. Thus $E$ is normal over $F$ by Lemma 9.15.5.

Conversely, assume that $E/F$ is normal. Then by Lemma 9.15.5 we have $\sigma (E) = \sigma _0(E)$ for all $\sigma \in \mathop{Mor}\nolimits _ F(E, \overline{F})$. Thus we get an automorphism of $E$ over $F$ by setting $\tau = \sigma _0^{-1} \circ \sigma$. Whence the map displayed above is surjective. $\square$

Lemma 9.15.10. Let $L/K$ be an algebraic normal extension of fields. Let $E/K$ be an extension of fields. Then either there is no $K$-embedding from $L$ to $E$ or there is one $\tau : L \to E$ and every other one is of the form $\tau \circ \sigma$ where $\sigma \in \text{Aut}(L/K)$.

Proof. Given $\tau$ replace $L$ by $\tau (L) \subset E$ and apply Lemma 9.15.7. $\square$

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