[proofplan] We prove the three parts in sequence. For (i), we observe that the elementary symmetric polynomials $e_1, \ldots, e_n$ in $x_1, \ldots, x_n$ are defined so that $f(t) = (t - x_1)(t - x_2)\cdots(t - x_n)$ in $L[t]$; since $x_1, \ldots, x_n$ generate $L$ over $F$ and all lie in $L$, this exhibits $L$ as the splitting field of $f$ over $F$. For (ii), we apply [Artin's Lemma](/theorems/1272) to the finite group $S_n$ acting as $K$-automorphisms of $L$ with fixed field $F = L^{S_n}$, which gives $L/F$ Galois with $\operatorname{Gal}(L/F) \cong S_n$. For (iii), we set $E = K(e_1, \ldots, e_n)$ and use the containments $E \subset F \subset L$ together with the [Fundamental Theorem of Galois Theory](/theorems/1274) applied to both $L/E$ and $L/F$: showing $\operatorname{Gal}(L/E) = S_n = \operatorname{Gal}(L/F)$ forces $E = F$ by the Galois correspondence. [/proofplan]

[step:Factor $f$ over $L$ and identify $L$ as the splitting field of $f$ over $F$]Recall that the elementary symmetric polynomials $e_1, \ldots, e_n \in K[x_1, \ldots, x_n]$ are defined by the identity \begin{align*} (t - x_1)(t - x_2)\cdots(t - x_n) = t^n - e_1\, t^{n-1} + e_2\, t^{n-2} - \cdots + (-1)^n e_n \end{align*} in $K[x_1, \ldots, x_n][t]$, viewing both sides as polynomials in $t$. Since $K[x_1, \ldots, x_n] \subset L = K(x_1, \ldots, x_n)$, this identity holds in $L[t]$. Therefore $f(t) = t^n - e_1\, t^{n-1} + \cdots + (-1)^n e_n$ splits completely in $L[t]$ with roots $x_1, \ldots, x_n$. Since each $e_i$ is fixed by every permutation $\sigma \in S_n$ (the elementary symmetric polynomials are symmetric by definition), we have $e_i \in L^{S_n} = F$ for each $i$. Therefore $f \in F[t]$. The roots $x_1, \ldots, x_n$ of $f$ lie in $L$, and $L = K(x_1, \ldots, x_n)$ is generated over $K$ (hence over $F \supset K$) by precisely these roots. The splitting field of $f$ over $F$ is the smallest subfield of $L$ containing $F$ and all roots of $f$. Since $L$ is generated over $F$ by the roots $x_1, \ldots, x_n$, and $L$ contains $F$, the splitting field of $f$ over $F$ is $L$ itself. This proves Part (i).[/step]

[guided]The key fact is that the elementary symmetric polynomials are *defined* by the factorisation of a generic polynomial. The polynomial \begin{align*} (t - x_1)(t - x_2)\cdots(t - x_n) \end{align*} when expanded gives $t^n - e_1 t^{n-1} + e_2 t^{n-2} - \cdots + (-1)^n e_n$ where $e_k = \sum_{1 \le i_1 < \cdots < i_k \le n} x_{i_1} \cdots x_{i_k}$. This is a polynomial identity in the ring $K[x_1, \ldots, x_n][t]$, valid for any field $K$. Viewing $x_1, \ldots, x_n$ as elements of $L = K(x_1, \ldots, x_n)$, the identity persists in $L[t]$, exhibiting $x_1, \ldots, x_n$ as roots of $f$. Why are the coefficients $e_i$ in $F = L^{S_n}$? Any $\sigma \in S_n$ acts on $L = K(x_1, \ldots, x_n)$ by permuting the generators: $\sigma(x_i) = x_{\sigma(i)}$. The expression $e_k = \sum_{1 \le i_1 < \cdots < i_k \le n} x_{i_1} \cdots x_{i_k}$ is manifestly invariant under any permutation of $x_1, \ldots, x_n$, so $\sigma(e_k) = e_k$ for all $\sigma \in S_n$. By definition of the fixed field, $e_k \in L^{S_n} = F$. Finally, why is $L$ the *splitting field* of $f$ over $F$, rather than a larger field containing the splitting field? The splitting field is the smallest extension of $F$ over which $f$ splits completely, i.e., $F(x_1, \ldots, x_n)$. But $F \supset K$, so $F(x_1, \ldots, x_n) \supset K(x_1, \ldots, x_n) = L$. Conversely, $F(x_1, \ldots, x_n) \subset L$ because $F \subset L$ and $x_1, \ldots, x_n \in L$. Therefore $F(x_1, \ldots, x_n) = L$, and $L$ is the splitting field.[/guided]

[step:Apply Artin's Lemma to $S_n \le \operatorname{Aut}_K(L)$ to show $L/F$ is Galois with group $S_n$]The symmetric group $S_n$ acts on $L = K(x_1, \ldots, x_n)$ by $K$-automorphisms via $\sigma(x_i) = x_{\sigma(i)}$ for each $\sigma \in S_n$. These are distinct automorphisms of $L$: if $\sigma \neq \tau$, then $\sigma(i) \neq \tau(i)$ for some $i$, so $\sigma(x_i) = x_{\sigma(i)} \neq x_{\tau(i)} = \tau(x_i)$ (the $x_i$ are algebraically independent over $K$, so distinct indeterminates are distinct elements of $L$). Therefore $S_n$ is a finite subgroup of $\operatorname{Aut}_K(L)$ of order $n!$, and by hypothesis $F = L^{S_n}$ is its fixed field. We apply [Artin's Lemma](/theorems/1272). Artin's Lemma states: if $G$ is a finite group of automorphisms of a field $L$, then the extension $L / L^G$ is Galois with $\operatorname{Gal}(L / L^G) = G$ and $[L : L^G] = |G|$. Here $G = S_n$ is a finite group of automorphisms of $L$, and $F = L^{S_n} = L^G$. The hypotheses of Artin's Lemma are satisfied (we need only that $G$ is a finite group of field automorphisms of $L$, which holds since $S_n$ is finite and each $\sigma \in S_n$ is an automorphism of $L$). Therefore \begin{align*} [L : F] = |S_n| = n!, \qquad \operatorname{Gal}(L/F) = S_n. \end{align*} In particular, $L/F$ is a Galois extension. This proves Part (ii).[/step]

[guided]The natural action of $S_n$ on $L = K(x_1, \ldots, x_n)$ permutes the indeterminates. Each permutation $\sigma \in S_n$ extends uniquely to a $K$-algebra automorphism of $K[x_1, \ldots, x_n]$ (by the universal property of polynomial rings), and then to a $K$-automorphism of its fraction field $L = K(x_1, \ldots, x_n)$. Why are distinct permutations distinct automorphisms? If $\sigma \neq \tau$ in $S_n$, then there exists $i$ with $\sigma(i) \neq \tau(i)$, so $\sigma(x_i) = x_{\sigma(i)} \neq x_{\tau(i)} = \tau(x_i)$. The inequality $x_{\sigma(i)} \neq x_{\tau(i)}$ holds because $x_1, \ldots, x_n$ are distinct indeterminates (algebraically independent elements) in $L$. This gives $|S_n| = n!$ distinct automorphisms. [Artin's Lemma](/theorems/1272) is now the key tool. The lemma requires a finite group $G$ of automorphisms of a field $L$. We take $G = S_n$, which is finite (of order $n!$) and acts by automorphisms of $L$. The conclusion is that $L / L^G$ is Galois with $\operatorname{Gal}(L / L^G) = G$ and $[L : L^G] = |G|$. In our situation, $L^G = L^{S_n} = F$, so we obtain $\operatorname{Gal}(L/F) = S_n$ and $[L : F] = n!$. The conclusion "$\operatorname{Gal}(L/F) = S_n$" means that every element of $S_n$ (viewed as an automorphism of $L$) belongs to $\operatorname{Gal}(L/F)$, and conversely every $F$-automorphism of $L$ arises this way. The equality $[L:F] = n!$ is the degree statement that comes along with it. Note that Artin's Lemma gives us something "for free" that would otherwise require checking: the extension $L/F$ is Galois. This is non-trivial — it means $L/F$ is both normal and separable. Normality holds because $L$ is the splitting field of $f$ over $F$ (Part (i)), and separability is guaranteed by the degree formula $[L:F] = |\operatorname{Gal}(L/F)|$ (which characterises Galois extensions among finite extensions).[/guided]

[step:Show $K(e_1, \ldots, e_n) = F$ using the Galois correspondence]Let $E := K(e_1, \ldots, e_n)$. We establish the chain of inclusions \begin{align*} K \subset E \subset F \subset L. \end{align*} The inclusion $E \subset F$ holds because each $e_i$ is symmetric in $x_1, \ldots, x_n$ and hence lies in $F = L^{S_n}$, so $E = K(e_1, \ldots, e_n) \subset F$. We now show $L/E$ is Galois. The polynomial $f(t) = t^n - e_1\, t^{n-1} + \cdots + (-1)^n e_n$ has coefficients in $E$ (since $e_1, \ldots, e_n \in E$). From Part (i), $f$ splits in $L[t]$ as $(t - x_1)\cdots(t - x_n)$, and the roots $x_1, \ldots, x_n$ are distinct. (Distinctness: $L = K(x_1, \ldots, x_n)$ is a rational function field in $n$ indeterminates, so $x_1, \ldots, x_n$ are pairwise distinct elements of $L$.) Since $L = K(x_1, \ldots, x_n) = E(x_1, \ldots, x_n)$ and $x_1, \ldots, x_n$ are all roots of $f \in E[t]$, the field $L$ is the splitting field of the separable polynomial $f$ over $E$. Therefore $L/E$ is a Galois extension. Since $F \subset L$ and $E \subset F$, the inclusion $E \subset F$ gives $\operatorname{Gal}(L/F) \le \operatorname{Gal}(L/E)$: every $F$-automorphism of $L$ fixes $F$ pointwise, hence fixes $E \subset F$ pointwise, so it is also an $E$-automorphism. From Part (ii), $\operatorname{Gal}(L/F) \cong S_n$. Therefore $S_n \le \operatorname{Gal}(L/E)$ as a subgroup. On the other hand, every $\sigma \in \operatorname{Gal}(L/E)$ is an automorphism of $L$ fixing $E$ pointwise. Since $f \in E[t]$, each such $\sigma$ permutes the roots of $f$, i.e., $\sigma$ permutes $\{x_1, \ldots, x_n\}$. Because $L = K(x_1, \ldots, x_n)$ is the field of rational functions in the $x_i$, an automorphism $\sigma$ that permutes the generators $x_1, \ldots, x_n$ is completely determined by the permutation it induces on $\{x_1, \ldots, x_n\}$. Therefore the map \begin{align*} \operatorname{Gal}(L/E) &\to S_n \\ \sigma &\mapsto \text{the permutation } i \mapsto j \text{ where } \sigma(x_i) = x_j \end{align*} is an injective group homomorphism, so $\operatorname{Gal}(L/E)$ is isomorphic to a subgroup of $S_n$. In particular, $\operatorname{Gal}(L/E) \le S_n$. Combining both containments: \begin{align*} S_n \le \operatorname{Gal}(L/E) \le S_n, \end{align*} which forces $\operatorname{Gal}(L/E) = S_n = \operatorname{Gal}(L/F)$. Now $L/E$ is Galois (verified above), and the [Fundamental Theorem of Galois Theory](/theorems/1274) gives a bijection between intermediate fields of $L/E$ and subgroups of $\operatorname{Gal}(L/E)$, sending an intermediate field $M$ to $\operatorname{Gal}(L/M)$ and a subgroup $H$ to $L^H$. Under this correspondence: \begin{align*} F &\mapsto \operatorname{Gal}(L/F) = S_n, \\ E &\mapsto \operatorname{Gal}(L/E) = S_n. \end{align*} Since $E$ and $F$ are both intermediate fields of $L/E$ (we have $E \subset F \subset L$ and $E \subset E \subset L$), and both correspond to the same subgroup $S_n$, the injectivity of the Galois correspondence forces $E = F$. Therefore $F = K(e_1, \ldots, e_n)$, which proves Part (iii).[/step]

[guided]This is the critical step where we promote the containment $E \subset F$ to equality. The strategy is to use the [Fundamental Theorem of Galois Theory](/theorems/1274) applied to the extension $L/E$. We first verify that $L/E$ is Galois. The polynomial $f$ lies in $E[t]$ (its coefficients are $e_1, \ldots, e_n \in E$). It splits in $L$ with roots $x_1, \ldots, x_n$, and the roots are distinct because they are algebraically independent indeterminates — in particular, they are pairwise unequal as elements of $L$. Since $L = E(x_1, \ldots, x_n)$ is generated over $E$ by the roots of $f$, and $f$ is separable (having distinct roots), $L$ is the splitting field of a separable polynomial over $E$, hence $L/E$ is Galois.[/guided]