The crown jewel -- a perfect correspondence between subfields and subgroups
Every field sitting inside a Galois extension matches exactly one subgroup of its Galois group, with the order flipped. That match turns hard questions about fields into easier ones about groups.
Click any node in the field lattice (left) or the subgroup lattice (right) to highlight its partner. Notice the top of one lattice matches the bottom of the other.
Click any node to see its correspondent. Notice the order-reversing nature of the bijection.
The Fundamental Theorem establishes a perfect, order-reversing bijection between intermediate fields of a Galois extension and subgroups of its Galois group. Bigger subgroups correspond to smaller fields, and vice versa. This single theorem is why questions about field extensions can be answered by studying groups.
Click any node in the field lattice (left) or the subgroup lattice (right) to see its correspondent highlighted. The top of one lattice maps to the bottom of the other -- this is the order-reversing property. The degree [E:K] equals the order |H| of the corresponding subgroup.
Key insight: The bijection reverses inclusion: the full group G corresponds to the base field F (smallest), while the trivial subgroup {e} corresponds to the top field E (largest). This is because a bigger subgroup fixes fewer elements.
Pick an example, then click rows (or press Reveal All) to test each basis element against each automorphism. The elements that stay fixed make up the fixed field.
Extension: Q(sqrt(2), sqrt(3)) / Q with basis {1, sqrt(2), sqrt(3), sqrt(6)}. Pick a subgroup H and test which elements it fixes.
| Automorphism | 1 | sqrt(2) | sqrt(3) | sqrt(6) |
|---|---|---|---|---|
| e | - | - | - | - |
| sigma | - | - | - | - |
Given a subgroup H of the Galois group, the fixed field E^H consists of all elements of E that every automorphism in H leaves unchanged. Select a subgroup and test each basis element against each automorphism to discover which elements are fixed.
Key insight: For Q(sqrt(2), sqrt(3))/Q, the automorphism sigma sends sqrt(2) to -sqrt(2) but fixes sqrt(3). So the fixed field of <sigma> is Q(sqrt(3)): exactly the elements built from 1 and sqrt(3) alone. The index [G : H] equals the degree of the fixed field over Q.
Pick an example and a subgroup H, then press Test Conjugations to check whether gHg^(-1) = H for every g. Normal subgroups match Galois sub-extensions.
H = <sigma> = {e, sigma}
Testing normality: is gHg-1 = H for all g in V4?
The Fundamental Theorem has a bonus: a subgroup H is normal in G if and only if the corresponding intermediate field is itself a Galois extension of the base field. Test normality by conjugating: is gHg^(-1) = H for every g in G?
Key insight: In S3, the subgroup A3 = <r> is normal (index 2 subgroups always are), so Q(omega)/Q is Galois. But <s> is not normal in S3 -- and correspondingly, Q(cbrt(2))/Q is not Galois because it is missing the complex conjugate roots omega*cbrt(2) and omega^2*cbrt(2).
Spot a mistake or have an idea?