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Let $K \leq L$ be a finite Galois extension whose Galois group is isomorphic to $S_n$. I want to show that $L$ is the splitting field of some polynomial of degree $n$ over K.

So far, I thought of picking $\alpha \in L$, letting $\alpha_1, \dots , \alpha_m$ be the orbit of $\alpha$ under the action of Gal$(L/K)$ (as in the proof of Artin's Lemma) and considering the polynomial:

$$ f(t) = \prod_{i=1}^m (t - \alpha_i)$$

I want to say, pick $\alpha \in L $ minimising $m$ ($\alpha \notin K$), and show then that $m = n$, in which case $L$ is the splitting field of $f$ over $K$. I'm not sure whether this is the right approach from here, or how to proceed. Any help or hints would be greatly appreciated. Thanks!

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    $\begingroup$ Don't you want to maximize $m$? Anyway, do you know about primitive elements? $\endgroup$ – Hagen von Eitzen May 25 '17 at 22:39
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    $\begingroup$ @HagenvonEitzen OP is asking for a polynomial of degree $n$. The primitive element theorem would guarantee a polynomial of degree $n!$ as would maximizing the degree of $f$. Was a bit confused on that point as well myself. $\endgroup$ – sharding4 May 25 '17 at 23:02
  • $\begingroup$ I don't think that would work since $L$ contains elements of a quadratic subfield ($A_n$ is a normal subgroup of $S_n$), so minimizing the degree of an irreducible polynomial of an element of $L$ will get you something of degree 2. $\endgroup$ – sharding4 May 25 '17 at 23:04
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Let $ H_i $ denote the isomorphic copy of $ S_{n-1} $ in $ S_n $ composed of elements which fix $ i $. For $ L/K $ a Galois extension with Galois group $ S_n $, consider the fixed field $ F_1 $ of $ H_1 $. Clearly $ [F_1 : K] = n $, and $ F_1/K $ is separable since it is a subextension of the separable extension $ L/K $. Pick a primitive element $ \alpha_1 $ for this extension. Then, $ \alpha_1 $ has exactly $ n $ $ K $-conjugates (counting itself), say $ \alpha_1, \alpha_2, \ldots, \alpha_n $, since its stabilizer in the Galois group is exactly $ H_1 $, which is a subgroup of index $ n $ in the Galois group. The subgroup of $ S_n $ fixing $ K(\alpha_1, \alpha_2, \ldots, \alpha_n)/K $ is exactly

$$ \bigcap_{1 \leq i \leq n} H_i = \{ \textrm{id} \} $$

By Galois correspondence, it follows that $ K(\alpha_1, \alpha_2, \ldots, \alpha_n) = L $. Therefore, $ L $ is the splitting field of the minimal polynomial of $ \alpha_1 $ over $ K $, which is a polynomial of degree $ n $.

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  • $\begingroup$ So $\alpha_j$ is a chosen primitive element for the extension $F_j / K$, right ? That is how you are getting $\alpha_1,\dots,\alpha_n$ right? $\endgroup$ – Utsav Dewan Apr 22 at 16:04
  • $\begingroup$ @reflexive Yes. $\endgroup$ – Starfall Apr 22 at 17:49

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