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The general polynomial of degree $n$ has Galois group $S_{n}$

Statement

Let $F$ be a field and let $X_{1}, \dots, X_{n}$ be algebraically independent over $F$ . Write $e_{1}, \dots, e_{n}$ for the elementary symmetric polynomials in the $X_{i}$ , let

$E : = F (e_{1}, \dots, e_{n}) \subseteq L : = F (X_{1}, \dots, X_{n}),$

and let

$g (t) : = \prod_{i = 1}^{n} (t - X_{i}) = t^{n} - e_{1} t^{n - 1} + \dots + (- 1)^{n} e_{n} \in E [t] .$

Then $L$ is the splitting field of $g$ over $E$ , and

$Gal (L / E) ≅ S_{n} .$

Facts & Assumptions

Given: The field $F$ , the indeterminates $X_{1}, \dots, X_{n}$ , the rational function field $L = F (X_{1}, \dots, X_{n})$ , the subfield $E = F (e_{1}, \dots, e_{n})$ , and the polynomial $g (t)$ above.

[L1]

Every symmetric polynomial is uniquely a polynomial in the elementary symmetric polynomials (Fundamental theorem of symmetric polynomials: unique expression as a polynomial in $e_{1}, \dots, e_{n}$ ).

[L2]

For every field $A$ , the one-variable rational function field $A (t)$ is the field of fractions of $A [t]$ (For a field $F$ , $F (t) = Frac (F [t])$ is its rational function field; in particular $R (t) = Frac (R [t])$ ).

[L3]

Artin's fixed-field theorem says that if a finite automorphism group $G$ acts on a field $L$ , then $[L : L^{G}] = ∣ G ∣$ and $Aut (L / L^{G}) = G$ (Artin's fixed-field theorem: $[K : K^{G}] = ∣ G ∣$ and $Aut (K / K^{G}) = G$ ).

[L4]

A finite extension is Galois exactly when it is the splitting field of a separable polynomial (Equivalent characterizations of a finite Galois extension).

Proof

technique · direct

1.1

The polynomial $g (t)$ belongs to $E [t]$ by construction and splits in $L$ with roots $X_{1}, \dots, X_{n}$ . Since $L$ is generated over $F$ by the $X_{i}$ , it is generated over $E$ by those same roots, so $L$ is the splitting field of $g$ over $E$ .

givenalgebra

1.2

Every permutation $π \in S_{n}$ acts on $L$ by $π (X_{i}) = X_{π (i)}$ . Because each $e_{j}$ is symmetric, [L1] makes every $e_{j}$ fixed by this action, so $S_{n} \leq Aut (L / E)$ .

L1algebra

2.1

Conversely, let $h \in L^{S_{n}}$ . Iterating [L2] and clearing coefficient denominators writes $h = f / g$ with $f, g \in F [X_{1}, \dots, X_{n}]$ and $g \neq 0$ . Put $D : = \prod_{π \in S_{n}} π (g)$ . Then $D$ is a nonzero symmetric polynomial, and $h D = f \prod_{\begin{array}{c} π \in S_{n} \\ π \neq 1 \end{array}} π (g)$ is also a polynomial. Both $h$ and $D$ are fixed by $S_{n}$ , so $h D$ is symmetric as well. By [L1], both $D$ and $h D$ are polynomials in $e_{1}, \dots, e_{n}$ , hence $h = (h D) / D \in F (e_{1}, \dots, e_{n}) = E$ . The reverse inclusion is already in step 1.2, so $L^{S_{n}} = E$ . Now [L3] yields $[L : E] = ∣ S_{n} ∣ = n!$ and $Aut (L / E) = S_{n}$ .

L1L2L3algebra

3.1

Step 1.1 makes $L / E$ the splitting field of $g$ , and step 2.1 identifies its full automorphism group with $S_{n}$ . Therefore $Gal (L / E) ≅ S_{n}$ .

step 1.1step 2.1L4∎

Depends on

Used by

For n≥5 over a characteristic-zero base, the general polynomial of degree n is not solvable by radicals Corollary

Dependency tree · two levels

24 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Fields and Galois Theory, v5.10, Theorems 5.38-5.40 (standard reference, not scraped)
J. Ash, Basic Abstract Algebra, Section 6.8 (standard reference, not scraped)

The general polynomial of degree nn has Galois group SnSn​