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Intermediate fields matched one for one with subgroups, inclusion reversed: an infinite search among fields becomes a finite list of subgroups, and normal subgroups mark exactly the subextensions that are Galois.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to state the fundamental theorem, count intermediate fields by counting subgroups, compute the degree of a fixed field from the order and index of its subgroup, identify a fixed field from what its automorphisms leave alone, say which subextensions are Galois, and prove that a normal subgroup gives a normal subextension.
Two lessons ago the Galois group was a set of symmetries with no guarantee of size. Last lesson gave the conditions that make it as large as the degree. With that equality in hand, the group is no longer an approximation to the extension — it is a faithful record of it, and this lesson reads the record.
For $H \le \operatorname{Gal}(L/K)$ the fixed field is $L^{H} = \{x \in L : \sigma(x) = x \text{ for all } \sigma \in H\}$. A map between two ordered collections is inclusion-reversing when $A \subseteq B$ implies the images go the other way. Restriction sends $\sigma \in \operatorname{Gal}(L/K)$ to its action on an intermediate field, when that field is preserved.
Let $L/K$ be a finite Galois extension with group $G$, so $|G| = [L : K]$.
The theorem. The maps
$$E \longmapsto \operatorname{Gal}(L/E), \qquad H \longmapsto L^{H}$$
are mutually inverse bijections between the intermediate fields $K \subseteq E \subseteq L$ and the subgroups $H \le G$. Moreover:
What it is worth. Intermediate fields are an infinite-looking thing to search: there is no obvious list of the subfields of $\mathbb{Q}(\sqrt2, \sqrt3)$. Subgroups of a group of order $4$ can simply be enumerated. The theorem converts one problem into the other, and the answer comes with degrees attached.
For $\mathbb{Q}(\sqrt2, \sqrt3)$: the Klein four-group has five subgroups, so there are exactly five intermediate fields — $\mathbb{Q}$, three quadratic ones, and the top. The three quadratic ones are $\mathbb{Q}(\sqrt2)$, $\mathbb{Q}(\sqrt3)$ and $\mathbb{Q}(\sqrt6)$; the last is produced by the theorem rather than guessed.
Two words called normal. A subgroup closed under conjugation, and a field extension keeping whole families of roots, were named independently. Point 3 says the correspondence carries one to the other, and the proof is short: $\sigma(L^{H}) = L^{\sigma H \sigma^{-1}}$, so $H$ normal makes $\sigma(E) = E$ for every $\sigma$, which is normality of $E$.
The example where it bites. For the splitting field of $x^{3} - 2$, the group is $S_3$. Its six subgroups give six intermediate fields: $\mathbb{Q}$; $\mathbb{Q}(\omega)$ from the normal subgroup $A_3$; three copies of $\mathbb{Q}(\sqrt[3]{2}\omega^{k})$ from the three subgroups of order $2$; and the top. The three of degree $3$ are not normal over $\mathbb{Q}$, exactly matching the three non-normal subgroups — and $\mathbb{Q}(\omega)$ is Galois over $\mathbb{Q}$ with group $S_3/A_3 \cong \mathbb{Z}_2$.
It needs Galois. For $\mathbb{Q}(\sqrt[3]{2})$ the group is trivial and has one subgroup, while the degrees do not match anything. Without normality and separability there is nothing to correspond.
Galois theory replaces an infinite search among fields with a finite list of subgroups. Every question about what lies between $K$ and $L$ — how many fields, which are normal, what their degrees are — is answered by reading a subgroup lattice upside down.
Another way: picture
Draw the subgroup lattice of $G$ with the trivial subgroup at the bottom. Turn the page upside down and it becomes the lattice of intermediate fields, with $K$ at the bottom and $L$ at the top. Every line in one diagram is a line in the other, every index in one is a degree in the other, and the subgroups drawn with a special mark for normality are exactly the fields that are Galois over the base.
Another way: steps
To use the correspondence: 1. Check the extension is Galois, and compute $G$ with $|G| = [L : K]$. 2. List all the subgroups of $G$ — a finite job. 3. Each subgroup $H$ gives an intermediate field $L^{H}$ of degree $[G : H]$ over $K$. 4. To name $L^{H}$, find elements the automorphisms of $H$ leave alone, enough of them to reach the right degree. 5. Mark the normal subgroups: those fields are Galois over $K$, with group $G/H$.
Take $L$ the splitting field of $x^{3} - 2$ over $\mathbb{Q}$, of degree $6$, with $G = S_3$ acting on the three roots $\sqrt[3]{2}, \sqrt[3]{2}\omega, \sqrt[3]{2}\omega^{2}$.
The subgroups of $S_3$. The trivial one; three of order $2$, generated by the transpositions; one of order $3$, namely $A_3$; and $S_3$ itself. Six in all.
So there are exactly six intermediate fields, and the theorem hands over their degrees before any of them is identified:
| Subgroup | Order | Fixed field | Degree over $\mathbb{Q}$ | Normal? |
|---|---|---|---|---|
| $1$ | $1$ | $L$ | $6$ | yes |
| $\langle (2\,3) \rangle$ | $2$ | $\mathbb{Q}(\sqrt[3]{2})$ | $3$ | no |
| $\langle (1\,3) \rangle$ | $2$ | $\mathbb{Q}(\sqrt[3]{2}\omega)$ | $3$ | no |
| $\langle (1\,2) \rangle$ | $2$ | $\mathbb{Q}(\sqrt[3]{2}\omega^{2})$ | $3$ | no |
| $A_3$ | $3$ | $\mathbb{Q}(\omega)$ | $2$ | yes |
| $S_3$ | $6$ | $\mathbb{Q}$ | $1$ | yes |
Reading the table. The transposition fixing the first root generates a subgroup of order $2$, whose fixed field has degree $6/2 = 3$ — and $\mathbb{Q}(\sqrt[3]{2})$ is a degree-$3$ field inside it, so that is the one. The three conjugate subgroups give the three conjugate fields, none normal, matching the fact that $\mathbb{Q}(\sqrt[3]{2})$ holds one root of $x^{3}-2$ and not the others.
$A_3$ is normal of index $2$, so $\mathbb{Q}(\omega)$ is Galois over $\mathbb{Q}$ with group $S_3/A_3$ of order $2$ — and indeed $\mathbb{Q}(\omega) = \mathbb{Q}(\sqrt{-3})$ is a splitting field.
What was gained. Every intermediate field of a degree-$6$ extension has been found, with its degree and its normality, by listing the subgroups of a group with six elements. Searching for subfields directly would have had no obvious place to stop.
Getting the direction wrong. A bigger subgroup fixes a smaller field. The correspondence reverses inclusion, always.
Swapping degrees and orders. $[L : L^{H}] = |H|$, and $[L^{H} : K] = [G : H]$. Mixing them up gives answers that do not multiply to $|G|$, which is the check.
Expecting every intermediate field to be Galois over $K$. Only the normal subgroups give those.
Applying it to a non-Galois extension. Nothing corresponds. $\mathbb{Q}(\sqrt[3]{2})$ has a trivial group and a genuine structure the group cannot see.
Forgetting the two ends. The trivial subgroup corresponds to $L$ and the whole group to $K$; both count as intermediate fields in the usual bookkeeping.
Assuming distinct subgroups can fix the same field. They cannot — the correspondence is a bijection, and that injectivity is what the proof of the normal-subgroup statement uses.
$\operatorname{Gal}(\mathbb{Q}(\sqrt2,\sqrt3)/\mathbb{Q})$ is the Klein four-group, which has exactly three subgroups of order $2$.
Three subgroups.
So there are exactly three intermediate fields of degree $2$ over $\mathbb{Q}$ — no more and no fewer.
Three quadratic subfields.
$\mathbb{Q}(\sqrt2)$ and $\mathbb{Q}(\sqrt3)$ are two; the third is fixed by the automorphism flipping both roots, which leaves $\sqrt2\sqrt3 = \sqrt6$ alone. So it is $\mathbb{Q}(\sqrt6)$.
The theorem found it.
$\mathbb{Q}(\zeta_5)$ over $\mathbb{Q}$ has degree $4$ and Galois group $(\mathbb{Z}/5\mathbb{Z})^{\times}$, cyclic of order $4$.
A cyclic group of order four.
A cyclic group of order $4$ has three subgroups, of orders $1, 2, 4$.
Three subgroups.
So there is exactly one field strictly between, of degree $2$: it is $\mathbb{Q}(\sqrt5)$, fixed by the subgroup of order $2$. A cyclic group of prime order would have given none at all.
One intermediate field.
The group is cyclic of order $9$, and a cyclic group has exactly one subgroup for each divisor of its order.
Subgroups match divisors.
The divisors of $9$ are $1, 3, 9$, so there are three subgroups.
Three subgroups.
So three intermediate fields counting both ends: the base, one of degree $3$ over it, and the top. All three subgroups are normal because the group is abelian, so every one of those fields is Galois over the base.
$L = \mathbb{Q}(\sqrt2, \sqrt3)$ is Galois over $\mathbb{Q}$ with the Klein four-group as its Galois group. The order of each subgroup is given. Fill in the degree of $L$ over the field that subgroup fixes, and the degree of that fixed field over $\mathbb{Q}$.
| Order of the subgroup | Degree of $L$ over the fixed field | Degree of the fixed field over $\mathbb{Q}$ | |
|---|---|---|---|
| The trivial subgroup | 1 | ||
| The first subgroup of order two | 2 | ||
| The second subgroup of order two | 2 | ||
| The third subgroup of order two | 2 | ||
| The whole group | 4 |
The Galois group is a cyclic group of order $4$. How many fields lie between the base and the top, counting both ends?
Answer:
Is this right: every subgroup of the Galois group is the fixer of some intermediate field?
Select every statement that the fundamental theorem makes about a finite Galois extension $L/K$ with group $G$.
This task has no paper form; do it on a device.
In $\mathbb{Q}(\sqrt2, \sqrt3)$ over $\mathbb{Q}$, match each subgroup of the Galois group to the field it fixes.
| $\mathbb{Q}(\sqrt2, \sqrt3)$ | $\mathbb{Q}(\sqrt2)$ | $\mathbb{Q}(\sqrt3)$ | $\mathbb{Q}(\sqrt6)$ | $\mathbb{Q}$ | |
|---|---|---|---|---|---|
| The trivial subgroup | |||||
| The subgroup flipping only $\sqrt3$ | |||||
| The subgroup flipping only $\sqrt2$ | |||||
| The subgroup flipping both |
Build the proof that the intermediate field of a normal subgroup is itself normal over the base.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Is this right: the correspondence works for any finite extension?
You can list the intermediate fields of a Galois extension by listing the subgroups of its group, with the right degrees. Say in your own words why a bigger subgroup fixes a smaller field. Next: the cyclotomic extensions, where the group is abelian and everything is visible at once.
9. Your turn: how many intermediate fields has a Galois extension of degree $9$ with cyclic group?, step 3