3.2 Separable, normal, and Galois extensions

Definition 3.6

An algebraic extension $E/F$ is separable if the minimal polynomial of every element of $E$ is separable; otherwise, it is inseparable.

Thus, an algebraic extension $E/F$ is separable if every irreducible polynomial in $F[X]$ having at least one root in $E$ is separable, and it is inseparable if

$\diamond$

$F$ is nonperfect, and in particular has characteristic $p\neq 0$ , and
$\diamond$

there is an element $\alpha$ of $E$ whose minimal polynomial is of the form $g(X^{p})$ , $g\in F[X]$ .

See 2.14 et seq. For example, the extension $\mathbb{F}_{p}(T)$ of $\mathbb{F}_{p}(T^{p})$ is inseparable extension because $T$ has minimal polynomial $X^{p}-T^{p}$ .

Definition 3.7

An extension $E/F$ is normal¹¹ 1 Bourbaki says “quasi-galoisienne”. if it is algebraic and the minimal polynomial of every element of $E$ splits in $E[X]$ .

In other words, an algebraic extension $E/F$ is normal if and only if every irreducible polynomial $f\in F[X]$ having at least one root in $E$ splits in $E[X]$ .

Let $f$ be a monic irreducible polynomial of degree $m$ in $F[X]$ , and let $E$ be an algebraic extension of $F$ . If $f$ has a root in $E$ , so that it is the minimal polynomial of an element of $E$ , then

\left.\begin{array}[c]{lcl}E/F\text{\ separable }&\implies&\text{\ }f\text{\ % has only simple roots}\\ E/F\text{\ normal }&\implies&f\text{\ splits in }E\end{array}\right\}\implies f% \text{\ has }m\text{\ distinct roots in }E.

It follows that $E/F$ is separable and normal if and only if the minimal polynomial of every element $\alpha$ of $E$ has $[F[\alpha]\colon F]$ distinct roots in $E$ .

Example 3.8

(a) The polynomial $X^{3}-2{}$ has one real root $\sqrt[3]{2}$ and two nonreal roots in $\mathbb{C}{}$ . Therefore the extension $\mathbb{Q}[\sqrt[3]{2}]/\mathbb{Q}{}$ (which is separable) is not normal.

(b) The extension $\mathbb{F}_{p}(T)/\mathbb{F}_{p}(T^{p})$ (which is normal) is not separable because the minimal polynomial of $T$ is not separable.

Theorem 3.9

For an extension $E/F$ , the following statements are equivalent:

(a)

$E$ is the splitting field of a separable polynomial $f\in F[X]$ ;
(b)

$E$ is finite over $F$ and $F=E^{\operatorname{Aut}(E/F)}$ ;
(c)

$F=E^{G}$ for some finite group $G$ of automorphisms of $E$ ;
(d)

$E$ is normal, separable, and finite over $F$ .

Proof.

(a) $\Rightarrow$ (b). Certainly, $E$ is finite over $F$ . Let $F^{\prime}=E^{\operatorname{Aut}(E/F)}\supset F$ . We have to show that $F^{\prime}=F$ . Note that $E$ is also the splitting field of $f$ regarded as a polynomial with coefficients in $F^{\prime}$ , and that $f$ is still separable when it is regarded in this way. Hence

\left|\operatorname{Aut}(E/F^{\prime})\right|\overset{\text{\ref{ft8}}}{=}[E% \colon F^{\prime}]\leq[E\colon F]\overset{\text{\ref{ft8}}}{=}\left|% \operatorname{Aut}(E/F)\right|.

According to Corollary 3.5, $\operatorname{Aut}(E/F)=\operatorname{Aut}(E/F^{\prime})$ , and so $[E\colon F^{\prime}]=[E\colon F]$ and $F^{\prime}=F$ .

(b) $\Rightarrow$ (c). Let $G=\operatorname{Aut}(E/F)$ . We are given that $F=E^{G}$ , and $G$ is finite because $E$ is finite over $F$ (apply 2.8a).

(c) $\Rightarrow$ (d). According to Theorem 3.4, $[E\colon F]\leq(G\colon 1)$ ; in particular, $E/F$ is finite. Let $\alpha\in E$ , and let $f$ be the minimal polynomial of $\alpha$ ; we have to show that $f$ splits into distinct factors in $E[X]$ . Let $\{\alpha_{1}=\alpha,\alpha_{2},...,\alpha_{m}\}$ be the orbit of $\alpha$ under the action of $G$ on $E$ (so the $\alpha_{i}$ are distinct elements of $E$ ), and let

g(X)=\prod\nolimits_{i=1}^{m}(X-\alpha_{i})=X^{m}+a_{1}X^{m-1}+\cdots+a_{m}.

The coefficients $a_{j}$ are symmetric polynomials in the $\alpha_{i}$ , and each $\sigma\in G$ permutes the $\alpha_{i}$ , and so $\sigma a_{j}=a_{j}$ for all $j$ . Thus $g(X)\in F[X]$ . As it is monic and $g(\alpha)=0$ , it is divisible by $f$ (see the definition of minimal polynomial, p. 1.11). Let $\alpha_{i}=\sigma\alpha$ ; on applying $\sigma$ to the equation $f(\alpha)=0$ we find that $f(\alpha_{i})=0$ . Therefore every $\alpha_{i}$ is a root of $f$ , and so $g$ divides $f$ . Hence $f=g$ , and we conclude that $f(X)$ splits into distinct factors in $E$ .

(d) $\Rightarrow$ (a). Because $E$ has finite degree over $F$ , it is generated over $F$ by a finite number of elements, say, $E=F[\alpha_{1},...,\alpha_{m}]$ , $\alpha_{i}\in E$ , $\alpha_{i}$ algebraic over $F$ . Let $f_{i}$ be the minimal polynomial of $\alpha_{i}$ over $F$ , and let $f$ be the product of the distinct $f_{i}$ . Because $E$ is normal over $F$ , each $f_{i}$ splits in $E$ , and so $E$ is the splitting field of $f.$ Because $E$ is separable over $F$ , each $f_{i}$ is separable, and so $f$ is separable. _□

Definition 3.10

An extension $E/F$ of fields is Galois if it satisfies the equivalent conditions of (3.9). When $E/F$ is Galois, $\operatorname{Aut}(E/F)$ is called the Galois group of $E$ over $F$ , and it is denoted by $\operatorname{Gal}(E/F)$ .

Remark 3.11

(a) Let $E$ be Galois over $F$ with Galois group $G$ , and let $\alpha\in E$ . The elements $\alpha_{1}$ , $\alpha_{2},...,\alpha_{m}$ of the orbit of $\alpha$ under $G$ are called the conjugates of $\alpha$ . In the course of proving the theorem we showed that the minimal polynomial of $\alpha$ is $\prod(X-\alpha_{i})$ , i.e., the conjugates of $\alpha$ are exactly the roots of its minimal polynomial in $E$ .

(b) Let $G$ be a finite group of automorphisms of a field $E$ , and let $F=E^{G}$ . By definition, $E$ is Galois over $F$ . Moreover, $\operatorname{Gal}(E/F)=G$ (apply 3.5) and $[E\colon F]=|\operatorname{Gal}(E/F)|$ (apply 3.2).

Corollary 3.12

Every finite separable extension $E$ of $F$ is contained in a Galois extension.

Proof.

Let $E=F[\alpha_{1},...,\alpha_{m}]$ , and let $f_{i}$ be the minimal polynomial of $\alpha_{i}$ over $F$ . The product of the distinct $f_{i}$ is a separable polynomial in $F[X]$ whose splitting field is a Galois extension of $F$ containing $E$ . _□

Corollary 3.13

Let $E\supset M\supset F$ ; if $E$ is Galois over $F$ , then it is Galois over $M.$

Proof.

We know $E$ is the splitting field of some separable $f\in F[X]$ ; it is also the splitting field of $f$ regarded as an element of $M[X].$ _□

Remark 3.14

An element $\alpha$ of an algebraic extension of $F$ is said to be separable over $F$ if its minimal polynomial over $F$ is separable. The proof of Corollary 3.12 shows that every finite extension generated by separable elements is separable. Therefore, the elements of an algebraic extension $E$ of $F$ that are separable over $F$ form a subfield $E_{\text{sep}}$ of $E$ that is separable over $F$ . When $E$ is finite over $F$ , we let $[E\colon F]_{\text{sep}}=[E_{\text{sep}}\colon F]$ and call it the separable degree of $E$ over $F$ .

An algebraic extension $E$ is purely inseparable over $F$ if the only elements of $E$ separable over $F$ are the elements of $F$ . If $E$ is a finite extension of $F$ , then $E$ is purely inseparable over $E_{\mathrm{sep}}$ . See Jacobson 1964, Chap. I, Section 10, for more on this topic.

Definition 3.15

An extension $E$ of $F$ is cyclic (resp. abelian, resp. solvable, etc. $)$ if it is Galois with cyclic (resp. abelian, resp. solvable, etc.) Galois group.