What is the structure of the multiplicative group \((\mathbb{F}_9)^\times\)? Construct \(\mathbb{F}_9\) as a quotient ring and find an explicit primitive generator of \((\mathbb{F}_9)^\times\).
Solution
The polynomial \[ x^2+1\in\mathbb F_3[x] \] has no root in \(\mathbb F_3\), hence is irreducible. Thus \[ \mathbb F_9\cong\mathbb F_3[i], \qquad i^2=-1=2. \] Its multiplicative group has order \(8\), and the multiplicative group of a finite field is cyclic, so \[ (\mathbb F_9)^\times\cong C_8. \]
Take \[ \alpha=1+i. \] Then \[ \alpha^2=1+2i+i^2=2i, \] and hence \[ \alpha^4=(2i)^2=4i^2=2\neq1, \qquad \alpha^8=2^2=1. \] Therefore \(\alpha\) has order \(8\), so it is a primitive generator. Hence \[ \boxed{(\mathbb F_9)^\times=\langle1+i\rangle\cong C_8.} \]