Galois Groups of Polynomials

Galois's original object was the group of permutations of the roots of a polynomial. The Galois correspondence recovers that viewpoint concretely: label the $n$ roots of a separable polynomial, and each automorphism of the splitting field permutes them, embedding the Galois group as a subgroup of the symmetric group$S_n$. Which subgroup it is can be read from two computable invariants — the discriminant and, for quartics, a resolvent cubic — and settled over $\mathbb{Q}$ by reducing modulo primes.

The Galois group as a permutation group

Let $f(x)\in F[x]$ be separable of degree $n$ with distinct roots ${\alpha }_1,\dots ,{\alpha }_n$ in its splitting field $K$. An automorphism $\sigma \in \mathrm{Gal}(K/F)$ sends each root to another root, and since $K=F({\alpha }_1,\dots ,{\alpha }_n)$, the whole of $\sigma$ is determined by the permutation it induces. Fixing an ordering of the roots gives an injective homomorphism $\mathrm{Gal}(K/F)\hookrightarrow S_n$.

Irreducibility corresponds to a group-theoretic property of the action.

If $f$ is irreducible, any two roots ${\alpha }_i,{\alpha }_j$ have the same minimal polynomial $f$, so an isomorphism $F({\alpha }_i)\cong F({\alpha }_j)$ fixing $F$ extends to an automorphism of $K$ carrying ${\alpha }_i$ to ${\alpha }_j$. Conversely a factorization of $f$ splits the roots into blocks no automorphism can cross.¹

Because the Galois group embeds in $S_n$, its order divides $n!$, and transitivity forces $n$ to divide the order — a first constraint on which groups can occur. The extreme case is when nothing constrains the roots at all: over the field of rational functions in the coefficients, the group is the full symmetric group.

Over $\mathbb{Q}$ this is the typical case: most polynomials of degree $n$ have Galois group $S_n$.²

The discriminant and the alternating group

The subgroup $A_n$ of even permutations is detected by a single element of the splitting field.

The number $D$ vanishes exactly when two roots coincide, so $D\ne 0$ is separability. The quantity $\sqrt{D}$ is the same object used to define the sign homomorphism: a permutation fixes it precisely when it is even. This gives the discriminant test.

Since $\sqrt{D}$ always lies in the splitting field, $F(\sqrt{D})$ is the fixed field of $\mathrm{Gal}(f)\cap A_n$ — the unique quadratic subextension detecting parity.

Cubics

A cubic $f(x)=x^3+a{x}^2+bx+c$ becomes the depressed form $y^3+py+q$ after $x=y-a/3$, with the same splitting field and discriminant. Its discriminant is

The classification is complete once irreducibility and the square-class of $D$ are known.

• Reducible. A linear-times-quadratic factorization gives group $1$ or $\mathbb{Z}/2$; three linear factors give the trivial group.
• Irreducible, $D$ a square. The splitting field has degree $3$; the group is $A_3\cong \mathbb{Z}/3$, and adjoining a single root already splits $f$.
• Irreducible, $D$ not a square. The splitting field has degree $6$; the group is $S_3$, and the splitting field is $F(\theta ,\sqrt{D})$ for any root $\theta$.

Cardano's explicit root formulas are the computational face of this solvability — see solvability by radicals.

Quartics and the resolvent cubic

A quartic $f(x)=x^4+a{x}^3+b{x}^2+cx+d$ depresses to $y^4+p{y}^2+qy+r$. Its Galois group is one of the five transitive subgroups of $S_4$, since transitivity is forced by irreducibility.

To distinguish them, form the three elements ${\theta }_1=({\alpha }_1+{\alpha }_2)({\alpha }_3+{\alpha }_4)$, ${\theta }_2=({\alpha }_1+{\alpha }_3)({\alpha }_2+{\alpha }_4)$, ${\theta }_3=({\alpha }_1+{\alpha }_4)({\alpha }_2+{\alpha }_3)$. The symmetric group only permutes these three among themselves, so their elementary symmetric functions lie in $F$: they are the roots of the resolvent cubic.

Reading the group off $h$ combines its factorization with the square-class of $D$.

The last case is resolved by one more factorization: when exactly one ${\theta }_i$ lies in $F$, the group is $D_8$ or $C_4$, and $F(\sqrt{D})$ is the fixed field of $\mathrm{Gal}(f)\cap A_4$. Since $D_8\cap A_4=V$ is transitive on the roots while $C_4\cap A_4$ is not, the group is $D_8$ exactly when the quartic stays irreducible over $F(\sqrt{D})$.³

Worked example: $x^4-2$ over $\mathbb{Q}$

Computing Galois groups by reduction modulo $p$

For an integer polynomial, reducing modulo a prime turns a Galois-group computation into polynomial factorization over a finite field. Take $f\in \mathbb{Z}[x]$ separable, with nonzero discriminant $D$. For any prime $p\nmid D$, the reduction $\bar{f}\in {\mathbb{F}}_p[x]$ is separable, and its Galois group over ${\mathbb{F}}_p$ is cyclic, generated by the Frobenius. The cycle type of that generator is dictated by the factorization of $\bar{f}$.

The Frobenius acts transitively on the roots of each irreducible factor ${\bar{f}}_i$, so it contributes a cycle of length $\mathrm{deg}{\bar{f}}_i$; assembling the factors gives the full cycle type. Different primes populate the Galois group with different cycle types, and once enough cycle types are collected, the group is often forced.⁴

For example, $f(x)=x^5-x-1$ has discriminant $2869=19\cdot 151$, so any prime other than $19,151$ is usable.

• Mod $2$: $x^5-x-1\equiv (x^2+x+1)(x^3+x+1)$, giving a $(2,3)$ cycle. Its cube is a transposition.
• Mod $3$: $x^5-x-1$ is irreducible, giving a $5$-cycle. This also proves $f$ irreducible over $\mathbb{Z}$.

Since a transitive subgroup of $S_5$ containing a transposition and a $5$-cycle is all of $S_5$, the Galois group of $x^5-x-1$ over $\mathbb{Q}$ is $S_5$. The same method builds polynomials with $S_n$ as Galois group for every $n$, by prescribing factorization types modulo $2$, $3$, $5$ that force a transposition and an $n$-cycle.

The fundamental theorem of algebra, group-theoretically

The permutation viewpoint yields a proof that $\mathbb{C}$ is algebraically closed from two analytic facts: every odd-degree real polynomial has a real root (intermediate value theorem), and every complex number has a complex square root. Given $f\in \mathbb{R}[x]$, let $K$ be the splitting field of $f$ over $\mathbb{R}$ adjoined with $i$, a Galois extension with group $G$. A Sylow $2$-subgroup $P$ has odd index, so its fixed field has odd degree over $\mathbb{R}$; the odd-degree fact forces that fixed field to be $\mathbb{R}$, so $G=P$ is a $2$-group. Then $\mathrm{Gal}(K/\mathbb{C})$ is also a $2$-group, and a $2$-group with a nontrivial part would have a subgroup of index $2$, producing a quadratic extension of $\mathbb{C}$ — impossible. Hence $K=\mathbb{C}$.⁵ This argument of Artin reduces a theorem of analysis to the structure of $p$-groups, using analysis only for the two starting facts.