advanced30 min studyLesson 157 in Mathematics

The Galois Correspondence

Translate intermediate fields into stabilizing subgroups, with degrees, reversed inclusion, and normal quotients.

On this page

What you’ll understand

  • Compute a fixed field by imposing coefficient constraints
  • Read field degrees from subgroup orders and indices
  • Recognize when an intermediate extension has a quotient Galois group

A field tower can be hard to see because its elements are infinitely numerous, even when the extension degree is small. In a finite Galois extension, a finite group records the entire intermediate-field structure. The key move is to exchange elements for the transformations that leave them fixed. Adding field elements imposes more fixing conditions, so the corresponding symmetry group becomes smaller. This reversal is the organizing principle, not a technical inconvenience to memorize.

Two operations that undo one another

Let L/K be finite Galois with group G. An intermediate field E determines the subgroup Gal(L/E), consisting of automorphisms fixing E pointwise. Conversely, a subgroup H determines its fixed field L to the H: the elements fixed by every member of H. Fixed elements form a field because automorphisms preserve sums, products, and inverses. The fundamental theorem states that these two operations are inverse bijections under the finite Galois hypotheses.

The phrase pointwise matters. An automorphism can carry a subfield onto itself while moving individual elements inside it. Such a map preserves the subfield as a set but does not belong to Gal(L/E). If E grows, fixing every element of E becomes a stronger restriction, so the subgroup shrinks. At the extremes, all of G fixes exactly K, while the identity subgroup fixes all of L. These endpoint checks quickly catch a reversed diagram.

E⟷Gal⁡(L/E),H⟷LH={a∈L:σ(a)=a ∀σ∈H}E\longleftrightarrow\operatorname{Gal}(L/E),\qquad H\longleftrightarrow L^H=\{a\in L:\sigma(a)=a\ \forall\sigma\in H\}
The two assignments form an inclusion-reversing bijection for a finite Galois extension.↔ Scroll wide equations horizontally.

Degrees are subgroup arithmetic

For H corresponding to E, the degree from L down to E is the number of automorphisms in H. The degree of E over K is the index of H in G. Their product equals the degree of L over K, matching the tower law. A subgroup of order two in a group of order twelve therefore corresponds to a field of degree six over the base, not degree two. Say which end of the tower a degree measures.

Intersections and generated fields also reverse direction. The subgroup fixing the compositum of E and F is the intersection of their fixing subgroups: a map must fix both fields. The subgroup corresponding to E intersect F is generated by the two fixing subgroups. These statements let you compute an entire field lattice from a small subgroup lattice. They apply inside the fixed ambient finite Galois extension; changing the ambient field changes the group being used.

[L:LH]=∣H∣,[LH:K]=[G:H],Gal⁡(L/EF)=Gal⁡(L/E)∩Gal⁡(L/F)[L:L^H]=|H|,\qquad [L^H:K]=[G:H],\qquad \operatorname{Gal}(L/EF)=\operatorname{Gal}(L/E)\cap\operatorname{Gal}(L/F)
Field degrees and composita have precise counterparts in the subgroup lattice.↔ Scroll wide equations horizontally.

Normal subgroups identify normal intermediate fields

Every intermediate extension L/E is Galois, but E/K need not be. The additional condition is that H = Gal(L/E) be normal in G. Conjugating H transports its fixed field, so H is normal exactly when all base-field automorphisms preserve E as a set. In that case restriction gives a surjective homomorphism from G to Gal(E/K), with kernel H, and hence a quotient group G/H.

If H is not normal, the notation G/H denotes a set of cosets but does not define the quotient group claimed by the theorem. For instance, in an S3 splitting field of an irreducible cubic, the order-two subgroup fixing one root is not normal. Its fixed cubic field is consequently nonnormal over the base. Its degree is still the subgroup index, three; the correspondence continues to work even though the quotient-group conclusion does not.

Find the fixed field of a simultaneous sign change

Take L = Q(sqrt(2), sqrt(3)) and let tau negate both square roots. Determine the fixed field of H = {identity, tau} directly, then check its degrees through the correspondence.

  1. The four elements 1, sqrt(2), sqrt(3), sqrt(6) form a rational basis of L. The degree is four because sqrt(3) is not in Q(sqrt(2)); squaring a proposed rational expression a + b sqrt(2) gives an impossible rational-square condition.

  2. Apply tau to a + b sqrt(2) + c sqrt(3) + d sqrt(6). The image is a minus b sqrt(2) minus c sqrt(3) plus d sqrt(6). Equality with the original expression forces b = c = 0 by basis independence, while a and d are unrestricted rational numbers.

  3. The fixed field is therefore Q(sqrt(6)). The subgroup has order two, giving degree [L:Q(sqrt(6))] = 2 and index two, giving [Q(sqrt(6)):Q] = 2. Since the full sign-change group is abelian, H is normal, and the intermediate quadratic extension has quotient group C2.

L⟨τ⟩={a+d6:a,d∈Q}=Q(6)L^{\langle\tau\rangle}=\{a+d\sqrt6:a,d\in\mathbb Q\}=\mathbb Q(\sqrt6)
The fixed-field condition eliminates precisely the coefficients whose basis elements change sign.↔ Scroll wide equations horizontally.
The result

A coefficient calculation and the subgroup theorem agree: fixing both radical signs simultaneously retains their product, producing exactly the quadratic field Q(sqrt(6)).

Explore the model

Interactive lab

Interactive lab · vectors

A coordinate that survives a reflection

Hold magnitude fixed and compare angles 40 and minus 40 degrees. Predict an unchanged horizontal component and a reversed vertical component. This illustrates the idea of imposing a fixing condition; the diagram does not compute a field, a subgroup lattice, or Galois automorphisms.

A coordinate that survives a reflectionPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forTo identify fixed data, apply every transformation in the chosen subgroup and solve the resulting equality conditions.

Check your understanding

A finite Galois extension L/K has group of order 24. If H has order 6, what is the degree of its fixed field E over K?

Try it yourself

In L = Q(sqrt(2), sqrt(3)), let sigma negate sqrt(2) while fixing sqrt(3). Find L raised to the subgroup generated by sigma, and determine whether that field is normal over Q.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Write a general element in the four-element basis and compare coefficients after applying sigma.

Compare with the solution

Sigma sends a + b sqrt(2) + c sqrt(3) + d sqrt(6) to a minus b sqrt(2) plus c sqrt(3) minus d sqrt(6). Fixedness requires b = d = 0, leaving exactly Q(sqrt(3)). Its fixing subgroup has order two and is normal because the full group C2 times C2 is abelian. Therefore Q(sqrt(3))/Q is Galois, with quotient group C2, as its splitting polynomial x squared minus three also confirms.

Take it with you

  • Larger intermediate fields correspond to smaller fixing subgroups
  • Subgroup order measures the upper degree and subgroup index measures the lower degree
  • An intermediate field is Galois over the base exactly when its fixing subgroup is normal

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