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

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!

• Don't you want to maximize $m$? Anyway, do you know about primitive elements? – Hagen von Eitzen May 25 '17 at 22:39
• @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. – sharding4 May 25 '17 at 23:02
• 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. – sharding4 May 25 '17 at 23:04

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 $n$.
• 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? – Utsav Dewan Apr 22 at 16:04