Constructing a Galois extension field with Galois group $S_n$

Given a field $F$, can you necessarily construct a field extension $E \supset F$ such that $\operatorname{Gal}(E/F) = S_n\,$?

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We will prove that there exists a finite Galois extension $K/\mathbb{Q}$ such that $S_n$ = $Gal(K/\mathbb{Q})$ for every integer $n \geq 1$.
We will follow mostly van der Waerden’s book on algebra.
You can also see his proof on Milne’s course note on Galois theory.
However, Milne refers to his book for a crucial theorem(Proposition 1 below) whose proof uses multivariate polynomials.
Instead, we will use elementary commutative algebra to prove this theorem.

Notations

We denote by |S| the number of elements of a finite set S.

Let $K$ be a field.
We denote by $K^*$ the multiplicative group of $K$.

Let $\tau$ = $(i_1, …, i_m)$ be a cycle in $S_n$.
The set {$i_1, …, i_m$} is called the support of $\tau$.
Let $\sigma \in S_n$.
Let $\sigma$ = $\tau_1\dots\tau_r$, where each $\tau_i$ is a cycle of length $m_i$ and they have mutually disjoint supports.
Then we say $\sigma$ is of type [$m_1, …, m_r$].

Definition 1
Let $F$ be a field.
Let $f(X)$ be a non-constant polynomial of degree n in $F[X]$.
Let $K/F$ be a splitting field of $f(X)$.
Suppose $f(X)$ has distinct $n$ roots in $K$.
Then $f(X)$ is called separable.
Since the splitting fields of $f(X)$ over $F$ are isomorphic to each other,
this definition does not depend on a choice of a splitting field of $f(X)$.

Definition 2
Let $F$ be a finite field.
Let $|F| = q$.
Let $K/F$ be a finite extension of $F$.
Let $\sigma$ be a map: $K \rightarrow K$ defined by $\sigma(x) = x^q$ for each $x \in K$.
$\sigma$ is an automorphism of $K/F$.
This is called the Frobenius automorphism of $K/F$.

Definition 3
Let $G$ be a permutation group on a set $X$.
Let $G’$ be a permutation group on a set $X’$.
Let $f:X \rightarrow X’$ be a bijective map.
Let $\lambda:G \rightarrow G’$ be an isomophism.
Suppose $f(gx) = \lambda(g).f(x)$ for any $g \in G$ and any $x \in X$.
Then G and G’ are said to be isomorphic as permutation groups.

Lemma 1
Let $F$ be a field.
Let $f(X)$ be a separable polynomial of degree $n$ in $F[X]$.
Let $K/F$ be a splitting field of $f(X)$.
Let $G = Gal(K/F)$.
Let $S$ be the set of roots of $f(X)$ in K.
Then $G$ acts transitively on $S$ if and only if $f(X)$ is irreducible in $F[X]$.

Proof:
If $f(X)$ is irreducible, clearly $G$ acts transitively on $S$.

Conversely, suppose $f(X)$ is not irreducible.
Let $f(X) = g(X)h(X)$, where $g$ and $h$ are non-constant polynomials in $F[X]$.
Let $T$ be the set of roots of $g(X)$ in $K$.
Since $G$ acts on $T$ and $S \neq T$, $S$ is not transitive.
QED

Lemma 2
Let $F$ be a field.
Let $f(X)$ be a separable polynomial in F[X].
Let $f(X) = f_1(X)…f_r(X)$, where $f_1(X), …, f_r(X)$ are distinct irreducible polynomials in $F[X]$.
Let $K/F$ be a splitting field of $f(X)$.
Let $G = Gal(K/F)$.
Let $S$ be the set of roots of $f(X)$ in $K$.
Let $S_i$ be the set of roots of $f_i(X)$ in $K$ for each $i$.
Then $S = \cup S_i$ is a disjoint union and each $S_i$ is a $G$-orbit.

Proof:
This follows immediately from Lemma 1.

Lemma 3
Let $F$ be a finite field.
Let $K/F$ be a finite extension of $F$.
Then $K/F$ is a Galois extension and $Gal(K/F)$ is a cyclic group generated by the Frobenius automorphism $\sigma$.

Proof:
Let $|F| = q$.
Let $n = (K : F)$.
Since $|K^*| = q^n – 1$, $x^{q^n – 1} = 1$ for each $x \in K^*$.
Hence $x^{q^n} = x$ for each $x \in K$.
Hence $\sigma^n = 1$.

Let m be an integer such that $1 \leq m < n$.
Since the polynomial $X^{q^m} – X$ has at most $q^m$ roots in $K$, $\sigma^m \neq 1$.
Hence $\sigma$ generates a subgroup $G$ of order n of $Aut(K/F)$.
Since $n = (K : F)$, $G = Aut(K/F)$.
Since $|Aut(K/F)| = n$, $K/F$ is a Galois extension.
QED

Lemma 4
Let $F$ be a finite field.
Let $f(X)$ be an irreducible polynomial of degree $n$ in $F[X]$.
Let $K/F$ be a splitting field of $f(X)$.
Let $\sigma$ be the Frobenius automorphism of $K/F$.
Then $Gal(K/F)$ is a cyclic group of order $n$ generated by $\sigma$.

Proof:
Let $\alpha$ be a root of $f(X)$ in $K$.
By Lemma 3, $F(\alpha)/F$ is a Galois extension. Hence $K = F(\alpha)/F$
By Lemma 3, $Gal(F(\alpha)/F)$ is a cyclic group of order $n$ generated by $\sigma$.
QED

Lemma 5
Let $F$ be a finite field.
Let $f(X)$ be an irreducible polynomial of degree $n$ in $F[X]$.
Let $K/F$ be a splitting field of $f(X)$.
Let $G = Gal(K/F)$.
Let $\sigma$ be the Frobenius automorphism of $K/F$.
Let $S$ be the set of roots of $f(X)$.
We regard $G$ as a permutation group on $S$.
Then $\sigma$ is an $n$-cycle.

Proof:
This follows immediately from Lemma 4.

Lemma 6
Let $F$ be a finite field.
Let $f(X)$ be a separable polynomial in F[X].
Let $f(X) = f_1(X)…f_r(X)$, where $f_1(X), …, f_r(X)$ are distinct irreducible polynomials in $F[X]$.
Let $m_i$ = deg $f_i(X)$ for each $i$.
Let $K/F$ be a splitting field of $f(X)$.
Let $G = Gal(K/F)$.
Let $\sigma$ be the Frobenius automorphism of $K/F$.
Let $S$ be the set of roots of $f(X)$.
We regard $G$ as a permutation group on $S$.
Then $\sigma$ is a permutation of type $[m_1, …, m_r]$.

Proof:
This follows immediately from Lemma 2, Lemma 3 and Lemma 5.

Lemma 7
$S_n$ is generated by $(k, n)、k = 1、．．．、n – 1$.

Proof:
Let $(a, b)$ be a transpose on {$1, …, n$}.
If $a \neq n$ and $b \neq n$, then $(a, b) = (a, n)(b, n)(a, n)$.
Since $S_n$ is generated by transposes, we are done.
QED

Lemma 8
Let $G$ be a transitive permutation group on a finite set $X$.
Let $n = |X|$.
Suppose $G$ contains a transpose and a $(n-1)$-cycle.
Then $G$ is a symmetric group on X.

Proof:
Without loss of generality, we can assume that $X$ = {$1, …, n$} and $G$ contains
a cycle $\tau$ = $(1, …, n-1)$ and transpose $(i, j)$.
Since $G$ acts transitively on $X$, there exists $\sigma \in G$ such that $\sigma(j)$ = $n$.
Let $k$ = $\sigma(i)$.
Then $\sigma(i, j)\sigma^{-1}$ = $(k, n) \in G$.
Taking conjugates of $(k, n)$ by powers of $\tau$, we get $(m, n), m = 1, …, n – 1$. Hence, by Lemma 7, $G = S_n$.
QED

Lemma 9
Let $F$ be a finite field.
Let $n \geq 1$ be an integer.
Then there exists an irreducible polynomial of degree $n$ in $F[X]$.

Proof:
Let $|F| = q$.
Let $K/F$ be a splitting field of the polynomial $X^{q^n} – X$ in $F[X]$.
Let $S$ be the set of roots of $X^{q^n} – X$ in $K$.
It’s easy to see that $S$ is a subfield of $K$ containing $F$.
Hence $S = K$.
Since $X^{q^n} – X$ is separable, $|S| = q^n$.
Hence $(K : F) = n$.
Since $K^*$ is a cyclic group, $K^*$ has a generator $\alpha$.
Let $f(X)$ be the minimal polynomial of $\alpha$ over $F$.
Since $K = F(\alpha)$, the degree of $f(X)$ is $n$.
QED

Lemma 10
Let $f(X) \in \mathbb{Z}[X]$ be a monic polynomial.
Let $p$ be a prime number.
Suppose $f(X)$ mod $p$ is separable in $\mathbb{Z}/p\mathbb{Z}[X]$.
Then $f(X)$ is separable in $\mathbb{Q}$.

Proof:
Suppose $f(X)$ is not separable in $\mathbb{Q}$.
Since $\mathbb{Q}$ is perfect, there exists a monic irreducible $g(X) \in \mathbb{Z}[X]$ such that $f(X)$ is divisible by $g(X)^2$. Then $f(X)$ (mod $p$) is divisible by $g(X)^2$ (mod $p$). This is a contradiction.
QED

Proposition 1
Let $A$ be an integrally closed domain and let $P$ be a prime ideal of $A$.
Let $K$ be the field of fractions of A.
Let $\tilde{K}$ be the field of fractions of $A/P$.
Let $f(X) ∈ A[X]$ be a monic polynomial without multiple roots.
Let $\tilde{f}(X) \in (A/P)[X]$ be the reduction of $f(X)$ mod $P$.
Suppose $\tilde{f}(X)$ is also wihout multiple roots.
Let $L$ be the splitting field of $f(X)$ over $K$.
Let $G$ be the Galois group of $L/K$.
Let S be the set of roots of $f(X)$ in $L$.
We regard $G$ as a permutation group on $S$.
Let $\tilde{L}$ be the splitting field of $\tilde{f}(X)$ over $\tilde{K}$.
Let $\tilde{G}$ be the Galois group of $\tilde{L}/\tilde{K}$.
Let $\tilde{S}$ be the set of roots of $\tilde{f}(X)$ in $\tilde{L}$.
We regard $\tilde{G}$ as a permutation group on $\tilde{S}$.

Then there exists a subgroup $H$ of $G$ such that $H$ and $\tilde{G}$ are isomorphic as permutation groups.

Proof:

Corollary
Let $f(X) \in \mathbb{Z}[X]$ be a monic polynomial of degree $m$.
Let p be a prime number.
Suppose $f(X)$ mod $p$ is separable in $\mathbb{Z}/p\mathbb{Z}[X]$.
Suppose $f \equiv f_1…f_r$ (mod $p$), where each $f_i$ is monic and irreducible of degree $m_i$ in $\mathbb{Z}/p\mathbb{Z}[X]$.
Let $K/\mathbb{Q}$ be a splitting field of $f(X)$.
Let $M$ be the set of roots of $f(X)$.
$G = Gal(K/\mathbb{Q})$ can be regarded as a permutation group on $M$.
Then $G$ contains an element of type [$m_1, …, m_r$].

Proof:
By Lemma 10, $f(X)$ is separable in $\mathbb{Q}[X]$.
Let $F_p$ = $\mathbb{Z}/p\mathbb{Z}[X]$.
Let $\tilde{f}(X) \in F_p[X]$ be the reduction of $f(X)$ mod $p$.
Let $\tilde{K}/F_p$ be a splitting field of $\tilde{f}(X)$.
Let $\tilde{G}$ be the Galois group of $\tilde{K}/F_p$.
Let $\tau$ be the Frobenius automorphism of $\tilde{K}/F_p$.
Let $\tilde{M}$ be the set of roots of $\tilde{f}(X)$.
We regard $\tilde{G}$ as a permutation group on $\tilde{M}$.
By Lemma 6, $\tau$ is a permutation of type $[m_1, …, m_r]$.
Hence the assertion follows by Proposition 1.
QED

Theorem
There exists a finite Galois extension $K/\mathbb{Q}$ such that $S_n$ = $Gal(K/\mathbb{Q})$ for every integer $n \geq 1$.

Proof(van der Waerden):
By Lemma 9, we can find the following irreducible polynomials.

Let $f_1$ be a monic irreducible polynomial of degree $n$ in $\mathbb{Z}/2\mathbb{Z}[X]$.

Let $g_0$ be a monic polynomial of degree 1 in $\mathbb{Z}/3\mathbb{Z}[X]$.
Let $g_1$ be a monic irreducible polynomial of degree $n – 1$ in $\mathbb{Z}/3\mathbb{Z}[X]$.
Let $f_2 = g_0g_1$.
If $n – 1 = 1$, we choose $g_1$ such that $g_0 \ne g_1$.
Hence $f_2$ is separable.

Let $h_0$ be a monic irreducible polynomial of degree 2 in $\mathbb{Z}/5\mathbb{Z}[X]$.
If $n – 2$ is odd, Let $h_1$ be a monic irreducible polynomial of degree $n – 2$ in $\mathbb{Z}/5\mathbb{Z}[X]$.
Let $f_3 = h_0h_1$.
Since $h_0 \neq h_1$, $f_3$ is separable.

If $n – 2$ is even, $n – 2 = 1 + a$ for some odd integer $a$.
Let $h_1$ and $h_2$ be monic irreducible polynomials of degree $1$ and $a$ respectively in $\mathbb{Z}/5\mathbb{Z}[X]$.
Let $f_3 = h_0h_1h_2$.
If a = 1, we choose $h_2$ such that $h_1 \ne h_2$.
Hence $f_3$ is separable.

Let $f = -15f_1 + 10f_2 + 6f_3$.
Since each of $f_1, f_2, f_3$ is a monic of degree $n$, f is a monic of degree $n$.

Then,

$f \equiv f_1$ (mod 2)

$f \equiv f_2$ (mod 3)

$f \equiv f_3$ (mod 5)

Since $f \equiv f_1$ (mod 2), $f$ is irreducible.
Let $K/\mathbb{Q}$ be the splitting field of $f$.
Let $G = Gal(K/\mathbb{Q})$.
Let $M$ be the set of roots in $K$.
We regard $G$ as a permutation group on $M$.

Since $f$ is irreducible, $G$ acts transitively on $M$.

Since $f \equiv f_2$ (mod 3), $G$ contains a $(n-1)$-cycle by Corollary of Proposition 1.

Similarly, since $f \equiv f_3$ (mod 5), $G$ contains a permutation $\tau$ of type [$2, a$] or [$2, 1, a$],
where $a$ is odd.
Then $\tau^a$ is a transpose.
Hence $G$ contains a transpose.

Hence, $G$ is a symmetric group on $M$ by Lemma 8.
QED

There is no general solution to your question.
It depends on the base field $F$.
I will show that your problem is solved affirmatively when $F$ is a field of rational functions over any field.

Let $k$ be a field.
Let $K$ = $k(X_1, …, X_n)$ be the field of rational functions over k.
Each element of $S_n$ acts on $K$ as an $k$-automorphism.
Hence $S_n$ can be regarded as a subgroup of Aut($K/k$).
Let $F$ be the fixed field by $S_n$.
By Artin’s theorem, $K/F$ is a Galois extension and its Galois group is $S_n$.
Let $s_1, …, s_n$ be the elementary symmetric functions of $X_1, …, X_n$.
Then $F = k(s_1, …, s_n)$.
It can be easily proved that $s_1, …, s_n$ are algebraically independent over $k$. Hence $F$ can be regarded as the field of rational functions of $n$ variables.

Therefore we get the following proposition.

Proposition
Let $k$ be a field.
Let $F = k(x_1, …, x_n)$ be the field of rational functoins of $n$ variables over $k$.
There exists a Galois extension $E$ of $F$ such that $S_n = Gal(E/F)$.