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Let E be a finite separable extension of a field K and L is the smallest normal extension of K containing E. Then show that L over K is Galois.

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    "**The**" smallest normal extension of $\;E\;$ ? I think you meant something else: $\;E\;$ may have infinite normal extensions ... and I think I know what you may have meant. Please do check this. – 2017-02-11
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    It is a statement in algebra by S.Lang which is "Let E be a finite separable extension of a field k. Let K be the smallest normal extension of k containing E. Then K is finite Galois over k." – 2017-02-11
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    @sachingarg I don't think so. I think you mean corollary $\;1.6,\,\,VI\,,\,\S1\;$ , page $\;263\;$ in third revised edition. Observe that there is written the **Very important** info that $\;L\;$ is the **smallest normal extension of** $\;\color{red}K\;$ containing $\;E\;$ ...! If you mean something else please do give the exact reference. – 2017-02-11
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    I can prove L is finite and normal over K but how to prove that minimal polynomial of an arbitrary element of L in K has distinct roots @ Test 123 – 2017-02-11
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    @sachingarg You can't prove that because it makes no sense to talk of *the* smallest normal extension of $\;E\;$ ...which is $\;E\;$ itself! If you don't understand this *first*, together with my past comment, then I cannot go ahead. Perhaps someone else can, though. – 2017-02-11
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    @DonAntonio . I got it, L is the smallest normal extension of K containing E where E is separable over K. But I am still not being able to prove that L is separable over K – 2017-02-11

2 Answers 2

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If you pick a basis $ \beta_1, \beta_2, \ldots, \beta_n $ for $ E/K $, then $ L/K $ is generated as a $ K $-algebra by the $ K $-conjugates of the $ \beta_i $. Since each $ \beta_i $ is separable over $ K $ by virtue of lying in $ E $, their conjugates are all separable as well; and the result follows, as separable elements generate separable extensions.

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    Ah, nicely answered. +1\ – 2017-02-11
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Here is another approach. But first, a proposition:

Let $E$ be a finite separable extension of a field $K$. Then there exists a field $F \supset E$ such that $F/K$ is normal (actually Galois).

Proof

Write $E = K(a_1,\ldots,a_n)$. For each $i$, let $f_i$ be the minimum polynomial of $a_i$ over $K$. Then $f :=\prod_{i=1}^n f_i$ is a separable polynomial $\in K[x]$. We can construct a splitting field $F$ of $f(x)$ over $K$. Since $a_i \in F$ for all $i$, we have $E \subset F$. Finally, since $F$ is a splitting field of a separable polynomial over $K$, we get that $F/K$ is Galois.

$\square$

Now let $F$ be as in the proposition. Then $L \subset F$. We know that $F/K$ is finite, so $L/K$ is finite. Also, $L/K$ is separable since so is $F/K$. Therefore, $L/K$ is Galois.

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    Then we need to show for each $\alpha_i$, $\alpha_l$ two roots of the same $f_j$ there is $\sigma \in Aut(F/K)$ sending $\alpha_i$ to $\alpha_l$, so that $K$ is the fixed field of a finite set of automorphisms of $F$, therefore $F/K$ is Galois in the strict sense (the one describing the minimal polynomials over $K$ of elements $\beta \in F$) – 2017-08-18