the smallest normal extension of separable extension is separable.

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 $k$?

• 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? – user26857 Sep 26 '15 at 9:02
• @user26857 What if some irreducible polynomial of an element $\alpha \in E$ is $(x-\alpha)(x-\beta)^n$? – user124697 Sep 26 '15 at 9:11
• It seems you forgot that $k\subset E$ is separable. – user26857 Sep 26 '15 at 9:11
• Yes, an element is separable iff its minimal polynomial has no multiple roots. – user26857 Sep 26 '15 at 9:20
• See also here, Definition 8.5.2. – user26857 Sep 26 '15 at 9:21

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 $$k$$.