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Let $E$ be finite separable extension of a field $k$. Let $K$ be the smallest normal extension of $k$ containing $E$. Does $K$ be separable?

Actually this is from the statements of Lang's the algebra, Corollary 1.6 at p.263., to show that $K$ is Galois extension. Lang states it obvious since $K$ is finite composite of finite number of conjugates of elements in $E$.

I showed that $K$ is such components, using Lang's remark on separable extension. But still have no idea to see $K$ is separable. By Lang's notation, $K = (\sigma_{1}E)(\sigma_{2}E)\cdots(\sigma_{n}E)$ where $n = [E:k]$, $\{ \sigma_{i} \}_{i=1}^{n}$ is embeddings $E \to k^{a}$, a algebraic closure of $k$ over $k$. Could you give me some hint for the reason why each $(\sigma_{i}E)$ is separable over $k$?

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  • $\begingroup$ In fact, $K$ is obtained by adjoining to $E$ the roots of all minimal polynomials over $k$ of elements from $E$. Can you see now why is it separable? $\endgroup$ – user26857 Sep 26 '15 at 9:02
  • $\begingroup$ @user26857 What if some irreducible polynomial of an element $\alpha \in E$ is $(x-\alpha)(x-\beta)^n$? $\endgroup$ – user124697 Sep 26 '15 at 9:11
  • $\begingroup$ It seems you forgot that $k\subset E$ is separable. $\endgroup$ – user26857 Sep 26 '15 at 9:11
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    $\begingroup$ Yes, an element is separable iff its minimal polynomial has no multiple roots. $\endgroup$ – user26857 Sep 26 '15 at 9:20
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    $\begingroup$ See also here, Definition 8.5.2. $\endgroup$ – user26857 Sep 26 '15 at 9:21
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Definition. Let $E \supset k$ be an algebraic extension. Fix an algebraic closure $k^a$ of $k$ containing $E$.

  • The separable degree $[E:k]_s$ is the cardinality of the set of embeddings of $E$ into $k^a$ over $k$, that is, $$ [E:k]_s = \#\{ \sigma \colon E \to k^a \mid \sigma \text{ is a field homomorphism and } \sigma|_k = \operatorname{id} \}. $$
  • $\alpha \in E$ is separable over $k$ if $[k(\alpha):k]_s = [k(\alpha):k]$.
  • A polynomial $f \in k[X]$ is separable if it has no multiple roots in $k^a$.

Theorem. Let $E \supset k$ be an algebraic extension. Then, the following are equivalent:

  1. $[F:k]_s = [F:k]$ for every subfield $F$ of $E$ such that $F \supset k$ and $F$ is finitely generated over $k$.

  2. Every element $\alpha$ of $E$ is separable over $k$.

  3. For every $\alpha \in E$, the irreducible polynomial $\operatorname{Irr}(\alpha,k,X)$ is separable.

  4. There exists a set of generators $\{ \alpha_i \}_{i \in I}$ of $E$ over $k$ such that each $\alpha_i$ is separable over $k$.

Definition. If an algebraic extension $E/k$ satisfies any of the above equivalent conditions, then we say that $E/k$ is a separable extension.


Now, suppose that $E/k$ is a finite separable extension. Let $\sigma \colon E \to k^a$ be an embedding of $E$ over $k$. To show that $\sigma(E)$ is a separable extension of $k$, we will show equivalently that the irreducible polynomial of each element of $\sigma E$ is separable.

So, let $\alpha \in E$ and let $f(X) = \operatorname{Irr}(\alpha,k,X)$ be its irreducible polynomial. Since $E$ is separable over $k$, $f(X)$ is separable. Since $\sigma$ is an embedding over $k$, $\sigma(f) = f$. Since an embedding maps a root of $f$ to a root of $\sigma(f)$, $\sigma(\alpha)$ is a root of $\sigma(f) = f$. Since $f(X)$ is irreducible over $k$, $f(X)$ is the irreducible polynomial of $\sigma(\alpha)$ as well.

Hence, the irreducible polynomial of $\sigma(\alpha)$ is separable. Since $\sigma(\alpha)$ is an arbitrary element of $\sigma(E)$, every element of $\sigma(E)$ is separable over $k$ and hence, $\sigma(E)$ is separable over $k$.

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