The Galois Correspondence

By now you have two very different objects attached to a field extension E/F. On one side sit the intermediate fields — every field K squeezed between the base and the top, F \subseteq K \subseteq E. On the other side sits the Galois group G = \mathrm{Gal}(E/F), the symmetries of E that hold F fixed, together with its subgroups. The two lists look unrelated: one is made of fields, the other of groups.

The Fundamental Theorem of Galois Theory says they are the same list in disguise. For a finite Galois extension the intermediate fields and the subgroups are in perfect one-to-one correspondence — a complete dictionary translating every field question into a group question and back. This is the summit of the module. Once you own it, a problem about which numbers you can build inside E becomes a problem about the subgroups of a finite group, which you can simply enumerate. Geometry, in effect, becomes bookkeeping.

The idea is the last gift of Évariste Galois, who died in a duel in 1832 at the age of twenty. The night before, expecting to die, he scrawled a letter to a friend sketching exactly this bridge between fields and groups. It went unread for over a decade. What he saw that night is the theorem on this page.

The two maps

The dictionary runs in both directions, and each direction is something you already know how to compute.

From a field to a group. Given an intermediate field K, look at all the automorphisms of E that fix every element of K (not just F). They form a subgroup of G — the Galois group of the top extension E/K:

K \;\longmapsto\; \mathrm{Gal}(E/K) \;\leq\; G.

From a group to a field. Given a subgroup H \leq G, look at all the elements of E that every automorphism in H leaves alone. They form a field sitting between F and E, the fixed field of H:

H \;\longmapsto\; E^{H} = \{\, x \in E : \sigma(x) = x \text{ for all } \sigma \in H \,\}.

The theorem's punch line is that these two maps are inverse to each other: start with a field, take its group, take that group's fixed field, and you land back on the field you started with — and likewise starting from a subgroup. Nothing is lost in translation.

Bigger field, smaller group — the mirror

There is one twist that makes the whole picture click, and it is worth saying slowly. The correspondence is inclusion-reversing. The larger a field K is, the harder it is for an automorphism to fix all of it, so the fewer automorphisms survive — i.e. the smaller the group \mathrm{Gal}(E/K). Growing the field shrinks the group; shrinking the field grows the group. The two lattices are reflections of one another, one hanging upside-down beneath the other.

The two extreme ends make this concrete. The smallest intermediate field is the base F itself, and everything in G fixes F by definition, so F corresponds to the whole group G. The largest intermediate field is E, and only the identity automorphism fixes all of E, so E corresponds to the trivial subgroup \{e\}. Bottom of one lattice meets top of the other.

Let E/F be a finite Galois extension with G = \mathrm{Gal}(E/F). Then:

The showcase: \mathbb{Q}(\sqrt 2, \sqrt 3)

Nothing sells the theorem like the smallest example that has room to breathe. Take E = \mathbb{Q}(\sqrt 2, \sqrt 3) over F = \mathbb{Q}. This is a degree-4 extension with basis \{1, \sqrt 2, \sqrt 3, \sqrt 6\}, and its Galois group is the Klein four-group V_4 = \{\,\mathrm{id},\, \sigma,\, \tau,\, \sigma\tau\,\}, where

\sigma:\ \sqrt 2 \mapsto -\sqrt 2,\ \sqrt 3 \mapsto \sqrt 3, \qquad \tau:\ \sqrt 2 \mapsto \sqrt 2,\ \sqrt 3 \mapsto -\sqrt 3.

Each automorphism independently chooses a sign for \sqrt 2 and a sign for \sqrt 3 — 2 \times 2 = 4 choices, and every one of them is its own inverse, so the group really is V_4. Now V_4 has exactly three subgroups of order 2, and each fixes exactly one of the three "middle" fields:

And the ends close the story: the trivial subgroup \{\mathrm{id}\} fixes all of E, while all of V_4 fixes only \mathbb{Q}. Five subgroups, five intermediate fields, matched exactly — and because V_4 is abelian, every subgroup is normal, so every one of these intermediate fields is itself a normal extension of \mathbb{Q}. Here is the whole correspondence, drawn as the two mirrored lattices it really is.

Read across at any height and you have a matched pair. At the middle level, \mathbb{Q}(\sqrt 2) (degree 2 over \mathbb{Q}) pairs with the order-2 subgroup \{\mathrm{id}, \tau\}, and indeed [E : \mathbb{Q}(\sqrt 2)] = 2 = |\{\mathrm{id},\tau\}|, exactly as the theorem promises. The whole of Galois theory is contained, in miniature, in this one square-shaped picture.

The single most common error is to run the dictionary the wrong way and expect a bigger subgroup to match a bigger field. It is the opposite. The correspondence reverses inclusions. Ask yourself what a subgroup has to do: its job is to fix things. A large subgroup contains many automorphisms, and to be fixed by all of them an element must be very special — so only a small field survives. A small subgroup demands almost nothing, so a large field is left untouched. Whole group G \leftrightarrow tiny field F; trivial group \{e\} \leftrightarrow whole field E. Draw the group lattice upside-down and everything lines up.

A second, subtler trap: it is tempting to conclude that every intermediate field is a normal extension of the base. It is not! In our V_4 example it happens to be true only because V_4 is abelian, so all its subgroups are normal. For a Galois group like S_3 (which arises for \mathbb{Q}(\sqrt[3]{2}, \omega)), the order-2 subgroups are not normal, and the matching intermediate fields — such as \mathbb{Q}(\sqrt[3]{2}) — are not normal extensions of \mathbb{Q}. Normality of the field is exactly normality of the subgroup, no more and no less.

Here is the deep reason the whole scheme works. An F-automorphism of E must send each root of a polynomial over F to another root of the same polynomial — it can only shuffle roots among themselves, because it preserves every algebraic relation with coefficients in F. So the Galois group is a group of permutations of the roots that respects all their arithmetic. An element of E lands in the base field F precisely when every such shuffle leaves it alone — "fixed by the whole group" is the algebraic fingerprint of "already a rational number." That is why fixed fields recover fields, and why the symmetry group knows everything about the tower.