Field Extension with Galois group $S_n$ is the splitting field of a polynomial of degree $n$

Question

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!

Answer 1

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 $.