Field extensions

Galois theory studies polynomial equations through the symmetries of the fields generated by their roots.

1. Algebraic extensions (Def. 1.1): finite implies algebraic (Prop. 1.2); adjoining several elements (Prop. 1.3).
2. Extending embeddings along simple extensions (Thm 1.4).
3. Splitting fields exist (Thm 1.5) and are unique up to isomorphism (Thm 1.6); their degrees (Example 1.7).
4. Normal extensions (Def. 1.8) are exactly splitting fields (Thm 1.9).

An extension $K/F$ is finite if $[K : F] < \infty$ and algebraic if every element of $K$ is algebraic over $F$. For $\alpha_1, \dots, \alpha_n\in K$, $F(\alpha_1, \dots, \alpha_n)$ is the smallest subfield of $K$ containing $F$ and the $\alpha_i$.

If $[K : F] = n$ then every $\alpha\in K$ is algebraic over $F$ of degree dividing $n$.

If $\alpha_1, \dots, \alpha_n$ are algebraic over $F$, then $F(\alpha_1, \dots, \alpha_n)$ is finite over $F$, of degree at most $\prod\deg m_{\alpha_i}$. Hence sums, products and quotients of algebraic elements are algebraic: the algebraic elements form a field.

Let $\sigma: F\to L$ be a field embedding, $\alpha$ algebraic over $F$ with minimal polynomial $m$, and write $m^\sigma$ for $m$ with $\sigma$ applied to its coefficients. The extensions of $\sigma$ to $F(\alpha)\to L$ correspond bijectively to the roots of $m^\sigma$ in $L$, via $\alpha\mapsto\beta$. So there are at most $\deg m$ of them.

A splitting field of $f\in F[x]$ is an extension $K$ in which $f$ factors into linear factors and which is generated over $F$ by the roots of $f$. Every $f$ of degree $n$ has a splitting field, of degree at most $n!$.

If $\sigma: F\to F'$ is an isomorphism, $K$ is a splitting field of $f$ over $F$ and $K'$ one of $f^\sigma$ over $F'$, then $\sigma$ extends to an isomorphism $K\to K'$, and the number of such extensions is at most $[K : F]$ (with equality when the roots of $f$ are distinct). In particular splitting fields over $F$ are unique up to isomorphism.

• $x^3 - 2$ over $\mathbb{Q}$: splitting field $\mathbb{Q}(\sqrt[3]2, \omega)$ with $\omega = e^{2\pi i/3}$; degree $6$ ($3$ from $\sqrt[3]2$, then $2$ from $\omega\notin\mathbb{R}$).
• $x^4 - 2$: $\mathbb{Q}(\sqrt[4]2, i)$, degree $8$.
• $x^p - 1$ ($p$ prime): $\mathbb{Q}(\zeta_p)$, degree $p - 1$.
• $(x^2 - 2)(x^2 - 3)$: $\mathbb{Q}(\sqrt2, \sqrt3)$, degree $4$.

An algebraic extension $K/F$ is normal if every irreducible polynomial in $F[x]$ with a root in $K$ splits completely in $K$.

A finite extension $K/F$ is normal iff $K$ is the splitting field over $F$ of some polynomial.

• Find the roots; the splitting field is generated by them (often by fewer elements, e.g. $\sqrt[3]2$ and $\omega$).
• Degree: tower law with minimal polynomials; show irreducibility (Eisenstein, rational roots) and use realness to show an element is not in a real subfield.
• Normality: is it a splitting field? Test an irreducible polynomial with a root in $K$.