My First Genuine Galois Group

Central Question

How can we discover a polynomial’s symmetry group by constructing its splitting field?

Why This Matters

A complete hand-calculable example brings roots, extension degrees, automorphisms and group structure into a single investigation.

Mathematical Notes

Start with the polynomial

Take $f(x)=x^3-2$. Eisenstein at 2 proves irreducibility over $\mathbb Q$. Let $\alpha=\sqrt[3]2$ and $\omega=e^{2\pi i/3}$. The splitting field is

Its degree comes from the real cubic field followed by the nonreal quadratic adjunction. Since the extension is normal and characteristic zero is separable, its Galois group has order six.

Construct the automorphisms rather than guess the name

Define

The second map is complex conjugation restricted to $L$. For the first, use $[L:\mathbb Q(\omega)]=3$: $x^3-2$ remains irreducible over $\mathbb Q(\omega)$, and replacing its generator by the root $\omega\alpha$ defines an isomorphism. Its image is all of $L$, since $\omega$ and $\omega\alpha$ recover $\alpha$.

The maps $r^k$ and $r^ks$, $k=0,1,2$, are six distinct automorphisms. Every automorphism has one of three choices for the image of $\alpha$ and one of two for the image of $\omega$, so there are no others.

Read the relations on generators

For example, $srs(\alpha)=\omega^2\alpha$, whereas $r(\alpha)=\omega\alpha$. Thus $r$ and $s$ do not commute.

On the roots $(\alpha,\alpha\omega,\alpha\omega^2)$, $r$ is a three-cycle and $s$ swaps the last two roots. These generate every permutation of three objects:

The semidirect product records that conjugation reverses the direction of the three-cycle; it is not $C_3\times C_2$.

A second way to identify an irreducible cubic's group

For a monic separable polynomial with roots $\alpha_i$, its discriminant is

The unsquared product changes sign under odd root permutations. Over a characteristic-zero field $F$, $\Delta$ is a square in $F$ exactly when the Galois group lies in $A_n$. For an irreducible cubic, transitivity leaves only $A_3\cong C_3$ and $S_3$ as possibilities. Thus a square discriminant gives $C_3$; a nonsquare gives $S_3$.

For $x^3+ax+b$, $\Delta=-4a^3-27b^2$. Here $\Delta=-108$, a nonsquare in $\mathbb Q$, confirming $S_3$. This criterion can identify the group without explicitly finding every radical root.

The first fixed fields

The subgroup $\langle r\rangle$ fixes $\omega$, so its fixed field is $\mathbb Q(\omega)$, of degree two over $\mathbb Q$. The subgroup $\langle s\rangle$ fixes $\alpha$, so its fixed field is $\mathbb Q(\alpha)$, of degree three. The correspondence in the next session proves that these contain every element fixed by the respective subgroup.

Learning Prompt

Take a polynomial whose Galois group is interesting but still calculable by hand. Starting from the polynomial, make me find its splitting field, field extensions, automorphisms, and finally its Galois group. Don't reveal the group at the beginning. Let me discover it.

Ideas

Examples

Questions

Connections

• Splitting Fields
• From Automorphisms to Groups
• Fundamental Theorem of Galois Theory
• Week overview

One Thing That Surprised Me

Further Rabbit Holes

Sources

Ian Stewart, Galois Theory, fifth edition, CRC Press, §§9.1, 11.1, 13.1, 22.2–22.3. Publisher and edition details.