# 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 $\operatorname{Gal}(E/F) = S_n\,$?

• Depends a little on $F$ - don't try it with finite $F$ - but pick a polynomial of degree $n$, check to see whether its group is $S_n$, if not, pick a different polynomial of degree $n$, try again, iterate until successful. There are lots of $S_n$-polynomials, chances are you'll hit one pretty quick. – Gerry Myerson Jul 2 '12 at 12:20
• Hmm.. Is there any general way to construct? From arbitrary $F$ ? – minwoo Jul 2 '12 at 12:26
• It's not possible for arbitrary $F$ and $n$. For example, it's impossible if $F$ is finite or $\mathbb{R}$, and $n$ is at least $3$. – Chris Eagle Jul 2 '12 at 12:30
• Even where it's possible, I'm not sure if there's any general way (other than the trial-and-error method). Note that if $f$ of degree $n$ works over $F$, then it doesn't work over any extension of $F$ that has one or more roots of $f$, so your construction is going to have to know a lot about $F$ in order to work. – Gerry Myerson Jul 2 '12 at 12:52
• If $k$ is any field, and $x_1,x_2,\ldots,x_n$ are variables (=algebraically independent transcendental elements over $k$), then the permutations of the variables realize $S_n$ as a group of automorphisms of $E=k(x_1,x_2,\ldots,x_n)$. Let $F$ be the fixed field (you should know what it looks like?)... – Jyrki Lahtonen Jul 2 '12 at 13:01

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.

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 $S_n = Gal(E/F)$.

• So this gives an $S_n$ extension of $E$, but not of the given field $F$. – Gerry Myerson Jul 3 '12 at 2:01
• @GerryMyerson Yes, of course. There's no general solution. However, my answer solves the question when the base field is a rational function fields of n variables over an arbitrary field. – Makoto Kato Jul 3 '12 at 4:45