Field extensions, splitting fields, finite fields

In one sentence

The degree of an extension is a dimension, the tower law multiplies degrees, and a splitting field is one that contains all the roots — which $\mathbb{Q}(\sqrt[3]{2})$ conspicuously does not.

Why the exam asks it

Because degree computations are concrete and self-checking, and because "contains a root" and "contains all the roots" are easy to conflate. $\mathbb{Q}(\sqrt[3]{2})$ is the standard test of whether that distinction has landed.

The idea

For $K \subseteq L$, the degree $[L:K]$ is the dimension of $L$ as a $K$-vector space.

Tower law: $[M:K] = [M:L]\,[L:K]$.

This is the workhorse. Degrees multiply, so intermediate fields have degrees dividing the total — which immediately rules out, say, a degree-$2$ subfield of a degree-$3$ extension.

$\alpha$ is algebraic over $K$ if it satisfies some polynomial. Its minimal polynomial $m_\alpha$ is the monic generator of the ideal of polynomials killing it; it is irreducible, and

$$K(\alpha) \cong K[x]/(m_\alpha), \qquad [K(\alpha):K] = \deg m_\alpha .$$

So computing a degree means finding a minimal polynomial and proving it irreducible — which is why the previous chapter's tests matter here.

$[L:K]$ finite $\Rightarrow$ $L$ algebraic over $K$. The converse fails: $\overline{\mathbb{Q}}$ is algebraic over $\mathbb{Q}$ and infinite-dimensional.

Worked degree. $[\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}] = 4$, by the tower $\mathbb{Q}\subset\mathbb{Q}(\sqrt2)\subset\mathbb{Q}(\sqrt2,\sqrt3)$ with degrees $2$ and $2$. And $\sqrt2+\sqrt3$ generates the whole thing: squaring gives $5+2\sqrt6$, so it satisfies $x^4-10x^2+1$, which is irreducible over $\mathbb{Q}$.

Splitting fields and normality

The splitting field of $f$ over $K$ is the smallest extension in which $f$ factors completely. It exists and is unique up to isomorphism.

$L/K$ is normal if every irreducible polynomial over $K$ with one root in $L$ splits in $L$ — equivalently, $L$ is the splitting field of some family.

$x^3-2$ is irreducible over $\mathbb{Q}$, so $[\mathbb{Q}(\sqrt[3]2):\mathbb{Q}] = 3$. But $\mathbb{Q}(\sqrt[3]2)$ lies inside $\mathbb{R}$, and the other two roots $\omega\sqrt[3]2$ and $\omega^2\sqrt[3]2$ are not real. One root in, two out: not normal. The splitting field is $\mathbb{Q}(\sqrt[3]2,\omega)$, of degree $6$.

Contrast $\mathbb{Q}(\sqrt2)$: the other root $-\sqrt2$ is already there. Every degree-$2$ extension is normal, since the second root is determined by the first and the coefficients.

Finite fields

• Every finite field has $p^n$ elements for a prime $p$, and one exists for each such size.
• $\mathbb{F}_{p^n}$ is unique up to isomorphism — it is the splitting field of $x^{p^n}-x$.
• $\mathbb{F}_{p^n} \subseteq \mathbb{F}_{p^m} \iff n \mid m$.
• $\mathbb{F}_q^{\times}$ is cyclic of order $q-1$.
• The Frobenius $x \mapsto x^p$ is a field automorphism fixing $\mathbb{F}_p$ pointwise.

The containment rule is the trap: $\mathbb{F}_{p^4}$ does not contain $\mathbb{F}_{p^3}$ despite being larger, because $3 \nmid 4$. The subfield lattice is the divisor lattice of the exponent.

Ruler and compass

A constructible number lies in a tower of quadratic extensions, so its degree over $\mathbb{Q}$ is a power of $2$. Hence:

• Doubling the cube needs $\sqrt[3]2$, of degree $3$. Impossible.
• Trisecting a general angle needs a root of $8x^3-6x-1$, degree $3$. Impossible.
• Squaring the circle needs $\sqrt\pi$, and $\pi$ is transcendental. Impossible.
• Regular $n$-gon: constructible $\iff$ $n$ is a power of $2$ times distinct Fermat primes.

Where intuition breaks

"$L$ contains a root of $f$, so $f$ splits in $L$." $\mathbb{Q}(\sqrt[3]2)$ contains one root of $x^3-2$ and not the others. This is the topic's central distinction.

"$\mathbb{F}_{p^n}\subseteq\mathbb{F}_{p^m}$ when $n \le m$." Divisibility, not size. $\mathbb{F}_{16}$ does not contain $\mathbb{F}_8$.

"Algebraic implies finite degree." $\overline{\mathbb{Q}}/\mathbb{Q}$ is algebraic of infinite degree. Finite $\Rightarrow$ algebraic runs one way only.

"There are several fields with $p^n$ elements." Exactly one up to isomorphism. Different constructions — different irreducible polynomials of the same degree — give isomorphic fields.

"$\mathbb{F}_{p^n}$ is $\mathbb{Z}/p^n$." $\mathbb{Z}/p^n$ is not a field for $n>1$; it has zero divisors. $\mathbb{F}_{p^n}$ is $\mathbb{F}_p[x]/(f)$ for an irreducible $f$ of degree $n$.

"$[\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}] = 2$ because both generators have degree 2." Degrees multiply, giving $4$. The tower law is exactly what stops them merging.

"The additive group of a finite field is cyclic." The multiplicative group is. The additive group is elementary abelian, $(\mathbb{Z}/p)^n$.

The exam's angle

1. Find the minimal polynomial and prove it irreducible. The degree falls out.
2. Use the tower law both ways — to compute a degree, and to rule out intermediate fields whose degree does not divide.
3. For normality, count roots inside. All of them or it is not normal, and a degree-$2$ extension always is.
4. For finite fields, work with exponents. Containment, subfield count and Galois group all read off divisibility of $n$.
5. For constructibility, compute the degree and check whether it is a power of $2$.

The night before

• $[L:K] = \dim_K L$; tower law $[M:K] = [M:L][L:K]$; intermediate degrees divide.
• $K(\alpha)\cong K[x]/(m_\alpha)$, degree $= \deg m_\alpha$, $m_\alpha$ irreducible.
• Finite $\Rightarrow$ algebraic; converse false ($\overline{\mathbb{Q}}$).
• Splitting field exists and is unique. Normal $=$ splitting field of a family.
• $\mathbb{Q}(\sqrt[3]2)$: degree $3$, not normal — one real root, two complex ones missing. Splitting field $\mathbb{Q}(\sqrt[3]2,\omega)$ has degree $6$. Degree $2$ is always normal.
• $[\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}] = 4$; $\sqrt2+\sqrt3$ has minimal polynomial $x^4-10x^2+1$.
• Finite fields: order $p^n$, one per size, $\mathbb{F}_{p^n}\subseteq\mathbb{F}_{p^m}\iff n\mid m$, $\mathbb{F}_q^\times$ cyclic, Frobenius $x\mapsto x^p$.
• Constructible $\Rightarrow$ degree a power of $2$. Doubling the cube, trisection and squaring the circle all fail.