Field Automorphisms

Central Question

Which movements of algebraic numbers preserve every field operation and fix the base field?

Why This Matters

The symmetries of a field extension preserve algebraic relationships, replacing the search for formulas with the study of allowable transformations.

Mathematical Notes

An automorphism cannot ignore algebraic relations

An $F$-automorphism of $L$ is a bijective field homomorphism $\sigma:L\to L$ with $\sigma(a)=a$ for every $a\in F$. If $m\in F[x]$ and $m(\alpha)=0$, then

Thus a generator must move to another root of its minimal polynomial that belongs to $L$. Once its image is fixed, the images of all expressions in that generator are forced.

For $L=\mathbb Q(\sqrt2)$, the only possibilities are

Both preserve the defining relation and are genuine automorphisms. Rational numbers cannot move: any field automorphism fixes $1$, hence integers and their quotients.

Two independent sign choices

In $L=\mathbb Q(\sqrt2,\sqrt3)$, define $\sigma_{\varepsilon,\delta}$ by

On a general element,

The unique four-term representation makes these maps well-defined; the square relations show they preserve multiplication. Applying each map twice is the identity, proving bijectivity. These four choices exhaust the possibilities.

The crucial limitation

For $K=\mathbb Q(\sqrt[3]2)\subset\mathbb R$, an automorphism must send the real cube root to a root of $x^3-2$ inside $K$. Only the real root is available, so

There are nevertheless three $\mathbb Q$-embeddings of $K$ into $\mathbb C$, one for each conjugate root. The two nonreal embeddings leave the original field; they are not automorphisms of it.

For a finite extension, the number of base-fixing embeddings into an algebraic closure is at most its degree, with equality exactly for separable extensions. Normality ensures these embeddings land back in the field. Together they explain why a finite Galois extension has exactly as many automorphisms as its degree.

Learning Prompt

Introduce automorphisms of field extensions. Start with $\mathbb Q(\sqrt2)$. Make me discover why an automorphism fixing $\mathbb Q$ can send $\sqrt2$ to $-\sqrt2$, but cannot arbitrarily move rational numbers. Then do $\mathbb Q(\sqrt2,\sqrt3)$. I want to see automorphisms as symmetries of algebraic relationships.

Ideas

Examples

Questions

Connections

• When Different Constructions Give the Same Field
• Tower Extensions
• From Automorphisms to Groups
• Week overview

One Thing That Surprised Me

Further Rabbit Holes

Sources

Ian Stewart, Galois Theory, fifth edition, CRC Press, §§8.1–8.6, 11.1–11.2. Publisher and edition details.