Part III · Finite fields · Lesson 9

Fields inside fields

Subfields, the Frobenius map, and why GF(q) sits inside GF(q³).

This lesson is planned. Here is what it will cover.

You’ll learn

• In characteristic $p$, $(x + y)^p = x^p + y^p$: the “freshman’s dream” is true.
• The Frobenius map $x \mapsto x^q$ fixes exactly the elements of $\mathrm{GF}(q)$ inside $\mathrm{GF}(q^3)$.
• Inside $\mathrm{GF}(q^3)$ with primitive element $\alpha$, the subfield $\mathrm{GF}(q)$ is $\{0\} \cup \langle \alpha^{q^2 + q + 1} \rangle$.
• For $q = p^e$, $\mathrm{GF}(q^3)$ is built directly as polynomials of degree below $3e$ over $\mathrm{GF}(p)$.