Normal Extensions

Normality is the load-bearing hypothesis in the Fundamental Theorem of Galois Theory: the correspondence between subgroups of $\mathrm{Gal}(K/F)$ and intermediate fields sends a subgroup $H$ to a normal subgroup exactly when the corresponding intermediate field is normal over $F$ — a beautiful dictionary between two different meanings of the word "normal" (subgroup vs. field extension) that only appears once you have this theorem in hand. This is the mechanism used to analyze solvability by radicals: a polynomial is solvable by radicals iff its splitting field sits inside a tower of normal-radical extensions, and Galois's original insight — that the symmetric group $S_5$ is not solvable, hence the general quintic has no radical formula — is a direct descendant of these ideas. In algebraic number theory, normal extensions of $\mathbb{Q}$ (equivalently, splitting fields of polynomials, equivalently Galois extensions when also separable) are exactly the number fields where the theory of decomposition and inertia groups for primes is cleanest; non-normal extensions like $\mathbb{Q}(\sqrt[3]2)$ require passing to their Galois closure (the smallest normal extension containing them — here, $\mathbb{Q}(\sqrt[3]2,\omega)$, exactly the field from Worked Example 2) before the full machinery applies. In algebraic geometry, normal field extensions underlie the notion of a Galois cover of varieties, and the splitting-field criterion is used directly to certify that a given cover is Galois. In coding theory and cryptography, finite fields $\mathbb{F}_{p^n}/\mathbb{F}_p$ are always normal (indeed Galois, with cyclic Galois group generated by the Frobenius map), a fact that underlies the algebra of Reed–Solomon codes and of many finite-field-based cryptosystems. Finally, in computer algebra systems, deciding whether a given number field is normal over $\mathbb{Q}$ — via the splitting-field criterion, by checking whether the minimal polynomial of a primitive element factors completely over the field itself — is a standard subroutine in algorithms for computing Galois groups.

:::quiz[The counterexample with $\mathbb{Q} \subset \mathbb{Q}(\sqrt2) \subset \mathbb{Q}(\sqrt[4]2)$ shows that:]

• Normality can fail at the bottom step of a tower even if the top step is normal
• Normality is not transitive up a tower: both steps normal does not force the whole tower normal ✓
• $\mathbb{Q}(\sqrt[4]2)/\mathbb{Q}(\sqrt2)$ is not normal
• $x^4-2$ is reducible over $\mathbb{Q}$

:::quiz[Why is $\mathbb{Q}(\sqrt[3]2)/\mathbb{Q}$ not normal?]

• $x^3-2$ is reducible over $\mathbb{Q}$
• $\mathbb{Q}(\sqrt[3]2)$ has infinite degree over $\mathbb{Q}$
• $x^3-2$ is irreducible over $\mathbb{Q}$, has a root in $\mathbb{Q}(\sqrt[3]2)$, but its other two (complex) roots are not in $\mathbb{Q}(\sqrt[3]2)\subset\mathbb{R}$ ✓
• $\mathbb{Q}(\sqrt[3]2)$ is not a field

Ask the tutor: Can you show me how to compute the Galois closure of a non-normal extension in general, and how its degree relates to the degree of the original extension?

Think of the minimal polynomial of $\alpha$ over $F$ as describing an "orbit" of algebraic numbers, the conjugates of $\alpha$, all of which are indistinguishable from $\alpha$ as far as $F$ can tell — they satisfy exactly the same polynomial relations over $F$. Normality says: if $K$ contains one member of the orbit, it contains the whole orbit. A field extension that is not normal is "lopsided" — it has smuggled in one root of an irreducible polynomial while leaving its algebraic siblings behind in a larger field. The reason this local condition (stated root-by-root) turns out to be equivalent to a global condition (being a splitting field) is roughly this: if $K$ is generated by $\alpha_1,\dots,\alpha_n$ over $F$, and normality holds for each $\alpha_i$, then $K$ contains all roots of $m_{\alpha_1}\cdots m_{\alpha_n}$, and in fact $K$ is generated by exactly those roots — which is precisely the splitting field condition. Conversely, splitting fields are built by adjoining all roots of a polynomial at once, so there is no way to end up with a partial, lopsided orbit.

Example 1: A non-normal extension, verified carefully. Let $F = \mathbb{Q}$ and $K = \mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{R}$, where $\sqrt[3]{2}$ denotes the real cube root of $2$. We show $K/\mathbb{Q}$ is not normal, using criterion (1). The polynomial $f(x) = x^3 - 2$ is irreducible over $\mathbb{Q}$ by Eisenstein's criterion at the prime $2$. Since $f(\sqrt[3]{2}) = 0$ and $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}] = 3 = \deg f$, the polynomial $f$ is the minimal polynomial of $\sqrt[3]{2}$ over $\mathbb{Q}$. The three roots of $f$ in $\mathbb{C}$ are

Since $K = \mathbb{Q}(\sqrt[3]{2}) \subseteq \mathbb{R}$, and $\omega\sqrt[3]{2}$ has nonzero imaginary part (as $\omega \notin \mathbb{R}$ and $\sqrt[3]{2}\neq 0$), we have $\omega \sqrt[3]{2} \notin K$. Thus $f$, an irreducible polynomial in $\mathbb{Q}[x]$ with a root in $K$, does not split completely in $K$. By definition, $K/\mathbb{Q}$ is not normal. We can double check via criterion (3): the map $\sigma_0: \mathbb{Q}(\sqrt[3]{2}) \to \mathbb{C}$ sending $\sqrt[3]{2} \mapsto \omega\sqrt[3]{2}$ and fixing $\mathbb{Q}$ is a well-defined field embedding (both elements are roots of the same irreducible cubic), but its image $\mathbb{Q}(\omega\sqrt[3]{2})$ is a different subfield of $\mathbb{C}$ from $K$ — it is not even contained in $\mathbb{R}$, while $K$ is. So $\sigma_0(K) \ne K$, confirming non-normality directly.

Example 2: A normal extension of degree 6, built as a splitting field. Let $F = \mathbb{Q}$ and let $L = \mathbb{Q}(\sqrt[3]{2}, \omega)$ where again $\omega = e^{2\pi i/3}$, a primitive cube root of unity, satisfying $\omega^2+\omega+1=0$. We show $L/\mathbb{Q}$ is normal, and compute $[L:\mathbb{Q}] = 6$. Degree computation. We have $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}] = 3$ as above. The minimal polynomial of $\omega$ over $\mathbb{Q}$ is $x^2+x+1$ (irreducible over $\mathbb{Q}$ since it has no rational roots and is quadratic). Over $\mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{R}$, the polynomial $x^2+x+1$ still has no root, because its roots $\omega,\omega^2$ are non-real while $\mathbb{Q}(\sqrt[3]{2})\subseteq \mathbb R$; a degree-2 polynomial with no root in a field is irreducible over that field. Hence $[\mathbb{Q}(\sqrt[3]2,\omega):\mathbb{Q}(\sqrt[3]2)] = 2$, and by the tower law,

Normality via criterion (2). Consider $f(x) = x^3 - 2 \in \mathbb{Q}[x]$. Its roots are $\sqrt[3]2, \omega\sqrt[3]2, \omega^2\sqrt[3]2$, and

since $\omega = (\omega \sqrt[3]{2})/\sqrt[3]{2} \in \mathbb{Q}(\sqrt[3]2,\omega\sqrt[3]2)$ and conversely $\omega\sqrt[3]2 \in \mathbb{Q}(\sqrt[3]2,\omega)$. So $L$ is exactly the field generated by all roots of $f$, i.e. $L$ is a splitting field of $x^3-2$ over $\mathbb{Q}$. By $(2)\Rightarrow(1)$ of the theorem, $L/\mathbb{Q}$ is normal. Sanity check via criterion (1) directly. We check the minimal polynomial of a "generic" element, e.g. $\alpha = \sqrt[3]2 + \omega$. Rather than compute $m_\alpha$ explicitly (a degree-6 polynomial, since one can show $\mathbb{Q}(\alpha) = L$ by a primitive element argument), we can instead just re-verify the definition on generators: $\sqrt[3]2$ has minimal polynomial $x^3-2$, which splits in $L$ since $\omega, \omega^2 = \bar\omega \in L$ and thus all of $\sqrt[3]2,\omega\sqrt[3]2,\omega^2\sqrt[3]2 \in L$; and $\omega$ has minimal polynomial $x^2+x+1$, whose other root $\omega^2 = \bar\omega$ also lies in $L$ (complex conjugation restricted to $L\subset\mathbb C$ is a field automorphism fixing $\mathbb Q$, or simply note $\omega^2 = -1-\omega \in L$ directly from the equation $\omega^2+\omega+1=0$). Both minimal polynomials of the chosen generators split in $L$; combined with the argument in $(1)\Rightarrow(2)$ of the proof, this is consistent with — though does not by itself immediately prove — full normality; the splitting-field argument above is the rigorous route, and it confirms $L/\mathbb Q$ is normal of degree $6$.