Existence of Finite Fields

For every prime power q = p^n, there exists a field with q elements.
Existence of Finite Fields

Theorem (Existence).
Let q=pnq=p^n be a prime power. There exists a with qq elements, denoted Fq\mathbb{F}_q.

One construction: choose an irreducible polynomial g(t)Fp[t]g(t)\in \mathbb{F}_p[t] of degree nn and set

FqFp[t]/(g(t)). \mathbb{F}_q \cong \mathbb{F}_p[t]/(g(t)).

Uniqueness up to isomorphism is treated in .

Examples.

  1. q=4q=4: take g(t)=t2+t+1F2[t]g(t)=t^2+t+1\in\mathbb{F}_2[t] (irreducible), so F4F2[t]/(t2+t+1)\mathbb{F}_4\cong \mathbb{F}_2[t]/(t^2+t+1).
  2. q=9q=9: t2+1t^2+1 is irreducible in F3[t]\mathbb{F}_3[t], giving F9F3[t]/(t2+1)\mathbb{F}_9\cong \mathbb{F}_3[t]/(t^2+1).
  3. q=pq=p: FpZ/pZ\mathbb{F}_p\cong \mathbb{Z}/p\mathbb{Z} is the of characteristic pp.

Related. .