Fundamental Theorem of Galois Theory (finite case)

Intuition Beginner

A polynomial like $x^2-2$ has roots $\sqrt{2}$ and $-\sqrt{2}$. These two roots look interchangeable: any equation with rational coefficients that one satisfies, the other satisfies too. Swapping them is a symmetry of the number system $\mathbb{Q}(\sqrt{2})$ that leaves every rational number fixed. Galois theory studies exactly these symmetries.

Collect all such symmetries of an extension field and they form a group, the Galois group. The central discovery is that this group of symmetries knows everything about the fields sitting between the small field and the big one. Each in-between field corresponds to one block of symmetries: the symmetries that hold every element of that field still.

The bigger the in-between field, the fewer symmetries can fix all of it, so the matching runs backwards. A larger field matches a smaller block of symmetries, and the smallest field (just the rationals) matches the whole group.

Why does this matter? It converts a hard question about fields, which are infinite and continuous-feeling, into a question about a finite group, which you can write on one page. The whole tower of fields becomes a list of subgroups.

Visual Beginner

Picture two ladders standing side by side, rungs lined up. The left ladder is the tower of fields, from the small base field at the bottom to the big extension field at the top. The right ladder is the list of subgroups, from the one-element group at the bottom to the whole Galois group at the top.

The connecting lines cross, because the matching flips top and bottom. A field high on the left ladder lines up with a subgroup low on the right. Reading either ladder tells you the other one completely.

Worked example Beginner

Take the field $\mathbb{Q}(\sqrt{2},\sqrt{3})$, built by adding both $\sqrt{2}$ and $\sqrt{3}$ to the rationals. It has four basic symmetries, determined by what they do to the two square roots.

The first symmetry changes nothing. The second sends $\sqrt{2}$ to $-\sqrt{2}$ and keeps $\sqrt{3}$. The third keeps $\sqrt{2}$ and sends $\sqrt{3}$ to $-\sqrt{3}$. The fourth flips both signs at once. Each one keeps every rational number where it is.

These four symmetries form a small group with four members. Now look at the in-between fields. The field $\mathbb{Q}(\sqrt{2})$ is held still by the symmetries that keep $\sqrt{2}$ fixed: the first and the third. That pair is one of the blocks inside the group.

So $\mathbb{Q}(\sqrt{2})$ matches a two-element block, $\mathbb{Q}(\sqrt{2},\sqrt{3})$ matches the one-element block, and the rationals match all four. Larger field, smaller block, every time.

Check your understanding Beginner

Formal definition Intermediate+

Throughout, $E/F$ is a finite extension of fields ^{[Lang Algebra Ch. VI]}.

Definition (Automorphism group). $\mathrm{Aut}(E/F)$ is the group of field automorphisms $\sigma :E\to E$ that fix $F$ pointwise, i.e. $\sigma (a)=a$ for all $a\in F$, under composition.

Definition (Galois extension). A finite extension $E/F$ is Galois if it is normal and separable. Equivalently, $|\mathrm{Aut}(E/F)|=[E:F]$; equivalently, $E$ is the splitting field over $F$ of a separable polynomial; equivalently, the fixed field of $\mathrm{Aut}(E/F)$ is exactly $F$. When $E/F$ is Galois we write $\mathrm{Gal}(E/F)=\mathrm{Aut}(E/F)$.

Definition (Fixed field). Given a subgroup $H\le \mathrm{Aut}(E/F)$, the fixed field is $$ E^{H} = {, x \in E : \sigma(x) = x \text{ for all } \sigma \in H ,}. $$ This is a field with $F\subseteq E^H\subseteq E$.

Definition (The two maps). For a finite Galois extension $E/F$ with group $G=\mathrm{Gal}(E/F)$, define $$ \Phi\colon H \longmapsto E^{H}, \qquad \Psi\colon L \longmapsto \operatorname{Gal}(E/L), $$ sending a subgroup $H\le G$ to its fixed field, and an intermediate field $F\subseteq L\subseteq E$ to the subgroup of automorphisms fixing $L$ pointwise. The Galois extension $E/L$ is automatic: $E$ is a separable splitting field over $F$, hence over every intermediate $L$.

Counterexamples to common slips Intermediate+

• Normal alone is not enough. In characteristic $p$, the extension ${\mathbb{F}}_p(t)/{\mathbb{F}}_p(t^p)$ is normal (it is the splitting field of $x^p-t^p=(x-t{)}^p$) but inseparable, with $|\mathrm{Aut}|=1<p=[E:F]$. The correspondence needs separability to count automorphisms correctly.

• Separable alone is not enough. The extension $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ is separable but not normal: it misses the complex cube roots. Here $|\mathrm{Aut}(E/F)|=1$ while $[E:F]=3$, and the fixed field of $\mathrm{Aut}(E/F)$ is all of $\mathbb{Q}(\sqrt[3]{2})$, larger than $F$.

• Subgroups need not be normal. In the $x^4-2$ example below, the subgroup fixing $\mathbb{Q}(\sqrt[4]{2})$ has order $2$ but is not normal in $D_4$, matching the fact that $\mathbb{Q}(\sqrt[4]{2})/\mathbb{Q}$ is not a normal extension.

Key theorem with proof Intermediate+

The engine of the correspondence is a counting theorem of Artin: a finite group of automorphisms recovers its own order as a degree.

Theorem (Artin). Let $E$ be a field and $H$ a finite group of automorphisms of $E$, with fixed field $K=E^H$. Then $E/K$ is a finite Galois extension and $$ [E : K] = |H|, \qquad \operatorname{Gal}(E/K) = H. $$

Proof. Write $n=|H|$ and $H=\{{\sigma }_1,\dots ,{\sigma }_n\}$ with ${\sigma }_1=\mathrm{id}$.

Step 1: $[E:K]\le n$. Suppose for contradiction that $E$ contains $K$-linearly independent elements ${\alpha }_1,\dots ,{\alpha }_{n+1}$. The homogeneous system ${\sum }_{j=1}^{n+1}{\sigma }_i({\alpha }_j)x_j=0$ for $i=1,\dots ,n$ has more unknowns than equations, so it has a nonzero solution in $E$. Choose one with the fewest nonzero entries, say $x_1,\dots ,x_r\ne 0$ and the rest zero; scale so $x_r=1$. The row $i=1$ (the identity) reads ${\sum }_j{\alpha }_j{x}_j=0$, and $K$-independence of the ${\alpha }_j$ forbids all $x_j\in K$, so some $x_m\notin K$. Then some $\tau \in H$ moves it: $\tau (x_m)\ne x_m$. Apply $\tau$ to every equation. Since $\tau$ permutes the rows $\{{\sigma }_i\}$, the vector $(\tau (x_j){)}_j$ also solves the system. Subtracting the two solutions kills the entry $j=r$ (both are $1$) and leaves a shorter nonzero solution, contradicting minimality. Hence $[E:K]\le n$.

Step 2: $[E:K]\ge n$. The $n$ distinct automorphisms ${\sigma }_i$ fix $K$, so they are $K$-algebra homomorphisms $E\to E$. By Dedekind's lemma distinct characters are linearly independent over $E$, so ${\mathrm{Hom}}_K(E,E)$ contains $n$ independent maps, forcing ${\dim }_{K}E\ge n$.

Together $[E:K]=n$. Every ${\sigma }_i$ fixes $K$, so $H\subseteq \mathrm{Gal}(E/K)$; and $|\mathrm{Gal}(E/K)|\le [E:K]=n=|H|$, giving $\mathrm{Gal}(E/K)=H$ and $E/K$ Galois. $\square$

Bridge. Artin's theorem builds toward the full correspondence by showing that passing to a fixed field and back returns the original group exactly, and this is exactly the round-trip identity $\Psi (\Phi (H))=H$ that makes $\Phi$ injective. The foundational reason the bijection exists is that degree counts dimension while group order counts automorphisms, and Artin welds the two counts into one equation $[E:E^H]=|H|$; the central insight is that the same number measures both ladders. The dimension bookkeeping here is dual to the degree multiplicativity of the tower law in 01.02.12, and putting these together the bridge is that a finite symmetry group of a field is rigid enough to be reconstructed from the points it leaves alone. This pattern appears again in 01.02.05, where a solvable chain of subgroups is read off the very same correspondence to decide solvability by radicals.

Exercises Intermediate+

Advanced results Master

We now assemble the full theorem and read off its structural consequences.

Theorem (Fundamental Theorem of Galois Theory, finite case). Let $E/F$ be a finite Galois extension with group $G=\mathrm{Gal}(E/F)$. The maps $$ \Phi\colon H \mapsto E^{H}, \qquad \Psi\colon L \mapsto \operatorname{Gal}(E/L) $$ are mutually inverse, inclusion-reversing bijections between the set of subgroups $H\le G$ and the set of intermediate fields $F\subseteq L\subseteq E$. Moreover:

1. (Degree-index dictionary) $[E:L]=|\mathrm{Gal}(E/L)|$ and $[L:F]=[G:\mathrm{Gal}(E/L)]$.
2. (Conjugacy) For $\tau \in G$, $\tau (L)=E^{\tau H{\tau }^{-1}}$ where $H=\Psi (L)$.
3. (Normality dictionary) $L/F$ is normal (equivalently Galois) if and only if $H=\mathrm{Gal}(E/L)$ is normal in $G$. When this holds, restriction gives an isomorphism$$\mathrm{Gal}(L/F)\cong G/H.$$

Theorem (Galois group of $x^4-2$). Let $E$ be the splitting field of $f(x)=x^4-2$ over $\mathbb{Q}$. Then $E=\mathbb{Q}(\sqrt[4]{2},i)$, $[E:\mathbb{Q}]=8$, and $G=\mathrm{Gal}(E/\mathbb{Q})\cong D_4$, the dihedral group of order $8$. Writing $\alpha =\sqrt[4]{2}$, the group is generated by $$ r\colon \alpha \mapsto i\alpha,\ i \mapsto i \quad (\text{order } 4), \qquad s\colon \alpha \mapsto \alpha,\ i \mapsto -i \quad (\text{order } 2), $$ with $sr{s}^{-1}=r^{-1}$. Its ten subgroups match the ten intermediate fields; the three subgroups of order $4$ are normal, corresponding to the three degree-$2$ normal subextensions $\mathbb{Q}(i)$, $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{-2})$, while the order-$2$ subgroup $⟨s⟩$ is non-normal, corresponding to the non-normal field $\mathbb{Q}(\sqrt[4]{2})$.

Theorem (Solvability by radicals). A polynomial $f\in F[x]$ over a characteristic-zero field is solvable by radicals if and only if its Galois group $\mathrm{Gal}(E/F)$ over the splitting field $E$ is a solvable group. Since $S_n$ is not solvable for $n\ge 5$ — its only proper nontrivial normal subgroup $A_n$ is simple, proved in 01.02.05 — a general polynomial of degree $\ge 5$, whose Galois group is $S_n$, cannot be solved by radicals. This is the Abel-Ruffini theorem recovered through the correspondence.

Synthesis. Putting these together, the fundamental theorem is the foundational reason that field theory and finite group theory are two readings of one object: the lattice of subgroups is the lattice of subfields turned upside down. The central insight is that Artin's count $[E:E^H]=|H|$ forces $\Phi$ and $\Psi$ to be inverse, so every field-theoretic question becomes a subgroup question and is dual to it. The normality dictionary is exactly the statement that a quotient group is a smaller Galois group, which generalises the isomorphism $\mathrm{Gal}(L/F)\cong G/H$ into the engine behind solvability: a radical tower is a chain of fields, the chain of fields is a chain of subgroups, and the chain of subgroups is a solvable series. The bridge is that the unsolvability of the quintic, an analytic-looking impossibility, becomes the plain group fact that $S_5$ has no solvable series, and this appears again in 01.02.04 where Sylow analysis supplies the subgroups whose fixed fields populate the lattice.

Full proof set Master

Proposition 1 ($\Phi$ and $\Psi$ are mutually inverse). Let $E/F$ be finite Galois with group $G$. Then $\Phi (\Psi (L))=L$ for every intermediate field $L$, and $\Psi (\Phi (H))=H$ for every subgroup $H\le G$.

Proof. Fix an intermediate field $L$. Since $E$ is the splitting field of a separable polynomial over $F$, it remains the splitting field of that same polynomial over $L$, so $E/L$ is Galois and $E^{\mathrm{Gal}(E/L)}=L$ by the fixed-field characterisation of Galois extensions. That is $\Phi (\Psi (L))=L$.

Fix a subgroup $H\le G$ and set $K=E^H$. Artin's theorem applies to the finite automorphism group $H$ acting on $E$: it gives $\mathrm{Gal}(E/E^H)=H$, that is $\Psi (\Phi (H))=H$. Both composites are the identity, so $\Phi$ and $\Psi$ are mutually inverse bijections. Inclusion-reversal was shown in Exercise 3 for $\Phi$, and the inverse of an inclusion-reversing bijection reverses inclusions as well. $\square$

Proposition 2 (Normality dictionary and quotient isomorphism). With notation as above, $L=E^H$ gives a normal extension $L/F$ iff $H⊴G$, and then $\mathrm{Gal}(L/F)\cong G/H$.

Proof. By Exercise 6, $\tau (L)=E^{\tau H{\tau }^{-1}}$, and $L/F$ is normal iff $\tau (L)=L$ for all $\tau \in G$ iff $\tau H{\tau }^{-1}=H$ for all $\tau$ iff $H⊴G$.

Assume $H⊴G$. Each $\tau \in G$ satisfies $\tau (L)=L$, so restriction $\rho :G\to \mathrm{Gal}(L/F)$, $\tau \mapsto \tau {|}_L$, is a well-defined group homomorphism. Its kernel is $\{\tau :\tau {|}_L={\mathrm{id}}_L\}=\mathrm{Gal}(E/L)=H$. For surjectivity: $L/F$ is Galois, so $|\mathrm{Gal}(L/F)|=[L:F]=[G:H]$ by the degree-index dictionary, while the image of $\rho$ has order $|G|/|\ker \rho |=|G|/|H|=[G:H]$. The image therefore exhausts the target. By the first isomorphism theorem, $\mathrm{Gal}(L/F)\cong G/H$. $\square$

Proposition 3 ($\mathrm{Gal}(\mathbb{Q}(\sqrt[4]{2},i)/\mathbb{Q})\cong D_4$). The splitting field $E$ of $x^4-2$ over $\mathbb{Q}$ has degree $8$ and Galois group $D_4$, with the subgroup-subfield lattice described above.

Proof. The roots of $x^4-2$ are $\pm \alpha ,\pm i\alpha$ with $\alpha =\sqrt[4]{2}$, so $E=\mathbb{Q}(\alpha ,i)$. Now $x^4-2$ is irreducible over $\mathbb{Q}$ by Eisenstein at $2$, so $[\mathbb{Q}(\alpha ):\mathbb{Q}]=4$; and $i\notin \mathbb{Q}(\alpha )\subseteq \mathbb{R}$, so $[E:\mathbb{Q}(\alpha )]=2$. By the tower law $[E:\mathbb{Q}]=8$, and as a splitting field of a separable polynomial $E/\mathbb{Q}$ is Galois with $|G|=8$.

An automorphism is determined by its action on the generators $\alpha$ and $i$. It must send $\alpha$ to a root of $x^4-2$, one of $\{\alpha ,i\alpha ,-\alpha ,-i\alpha \}$, and $i$ to a root of $x^2+1$, one of $\{i,-i\}$, giving at most $4\times 2=8$ candidates; all eight occur since $|G|=8$. Let $r$ be $\alpha \mapsto i\alpha ,i\mapsto i$ and $s$ be $\alpha \mapsto \alpha ,i\mapsto -i$. Then $r^4=\mathrm{id}$, $s^2=\mathrm{id}$, and computing $srs$ on $\alpha$: $s(\alpha )=\alpha$, $r(\alpha )=i\alpha$, $s(i\alpha )=-i\alpha =r^{-1}(\alpha )$, while both fix or invert $i$ consistently, so $srs=r^{-1}$. These are the defining relations of $D_4$, and the eight elements $r^k,r^{k}s$ exhaust $G$, so $G\cong D_4$.

The three index-two (order-four) subgroups $⟨r⟩$, $⟨r^2,s⟩$, $⟨r^2,rs⟩$ are normal; their fixed fields are the degree-two extensions $\mathbb{Q}(i)$, $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{-2})$ respectively (each fixed by the listed generators by direct substitution). The order-two subgroup $⟨s⟩$ fixes $\alpha$, so its fixed field is $\mathbb{Q}(\alpha )=\mathbb{Q}(\sqrt[4]{2})$; it is not normal in $D_4$ (conjugating $s$ by $r$ gives $rs{r}^{-1}=r^{2}s\ne s$), matching the failure of $\mathbb{Q}(\sqrt[4]{2})/\mathbb{Q}$ to be normal. $\square$

Connections Master

• Algebraic field extension, degree, splitting field 01.02.12. That unit stops at the door of the correspondence: it builds normality, separability, splitting fields, and the tower law, observing only informally that $|\mathrm{Gal}(E/F)|=[E:F]$ for separable splitting fields. The present unit turns that single equality into Artin's counting theorem and the full bijection, so 01.02.12 supplies the fields and degrees while this unit supplies the groups that index them. The separable-splitting-field characterisation of "Galois" used throughout is exactly the existence-and-uniqueness machinery proved there.

• Solvable, nilpotent groups, Jordan-Hölder theorem 01.02.05. The solvability-by-radicals criterion reads a radical tower as a subnormal series with abelian quotients, which is the definition of a solvable group transported across the correspondence. The non-solvability of $S_n$ for $n\ge 5$, resting on the simplicity of $A_n$ proved in that unit, is what blocks a general radical formula for the quintic. The Jordan-Hölder uniqueness of composition factors is the group-side shadow of the well-definedness of the lattice of normal subextensions.

• Sylow theorems 01.02.04. Sylow theory is the standard tool for locating subgroups of a Galois group of given prime-power order, and each such subgroup pins down a specific intermediate field through $\Phi$. In the $x^4-2$ example the $2$-group $D_4$ is itself a Sylow $2$-subgroup picture, and its subgroup lattice — produced by Sylow and normal-subgroup analysis — is read off directly as the lattice of subfields of $\mathbb{Q}(\sqrt[4]{2},i)$.

Historical & philosophical context Master

Galois 1832, writing the night before a fatal duel, introduced the group attached to an equation and asserted that the solvability of the equation by radicals is governed by a property of this group ^{[Galois 1832]}. His "groupe de l'équation" permutes the roots while respecting all rational relations among them, and the chain of subgroups he constructs is precisely the subnormal series whose abelian quotients signal radical solvability. The manuscript was dense and was understood only after Liouville published it in 1846; the modern bijective formulation between subgroups and subfields is a later distillation.

The abstract field-theoretic version — fixed fields, the correspondence as a lattice anti-isomorphism, and Artin's fixed-field counting theorem as its engine — is due to Emil Artin, whose Notre Dame lectures recast the whole subject as linear algebra over the base field ^{[Artin 1942]}. Artin's reorganisation replaced Galois's resolvent-polynomial computations with the clean statement that a finite group of automorphisms satisfies $[E:E^H]=|H|$, after which the entire correspondence is a formal consequence. This is the form codified in Lang's treatment ^{[Lang Algebra Ch. VI]} and the version a contemporary algebra course transmits. Philosophically, the theorem is the prototype of a duality: a hard, infinite-feeling object (the continuum of subfields) is shown to be the mirror of a finite, combinatorial one (the subgroup lattice), a move that recurs throughout modern mathematics from Galois descent to the Grothendieck correspondence between covering spaces and fundamental groups.

Bibliography Master

@article{Galois1832,
  author = {Galois, \'Evariste},
  title = {M\'emoire sur les conditions de r\'esolubilit\'e des \'equations par radicaux},
  journal = {Journal de Math\'ematiques Pures et Appliqu\'ees},
  year = {1846},
  note = {Written 1832, published posthumously by Liouville},
}

@book{Artin1942,
  author = {Artin, Emil},
  title = {Galois Theory},
  series = {Notre Dame Mathematical Lectures 2},
  publisher = {University of Notre Dame},
  year = {1942},
}

@book{Lang2002,
  author = {Lang, Serge},
  title = {Algebra},
  edition = {3rd revised},
  publisher = {Springer},
  year = {2002},
  series = {Graduate Texts in Mathematics 211},
}

@book{DummitFoote2004,
  author = {Dummit, David S. and Foote, Richard M.},
  title = {Abstract Algebra},
  edition = {3rd},
  publisher = {Wiley},
  year = {2004},
}

@book{Stewart2015,
  author = {Stewart, Ian},
  title = {Galois Theory},
  edition = {4th},
  publisher = {CRC Press},
  year = {2015},
}