The Stacks project

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 $M/E_ i/F$, $i \in I$ 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 $\sigma , \sigma ' \in \mathop{\mathrm{Mor}}\nolimits _ F(E, \overline{F})$ 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 $\sigma \in \mathop{\mathrm{Mor}}\nolimits _ F(E, \overline{F})$. 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 $\sigma _0 \in \mathop{\mathrm{Mor}}\nolimits _ F(F(\alpha ), \overline{F})$ mapping $\alpha $ to $\beta $. By Lemma 9.10.5 we can extend $\sigma _0$ to a $\sigma \in \mathop{\mathrm{Mor}}\nolimits _ F(E, \overline{F})$. 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 any $K$-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 $[E : F]_ s = |\mathop{\mathrm{Mor}}\nolimits _ F(E, \overline{F})|$. Pick an element $\sigma _0 \in \mathop{\mathrm{Mor}}\nolimits _ F(E, \overline{F})$. Then the map

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

is injective. Thus the inequality. If equality holds, then every $\sigma \in \mathop{\mathrm{Mor}}\nolimits _ F(E, \overline{F})$ 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 $\sigma \in \mathop{\mathrm{Mor}}\nolimits _ F(E, \overline{F})$. 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. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 09HL. Beware of the difference between the letter 'O' and the digit '0'.