UTM Ideals Varieties and Algorithm - Chapter 1 Section 1 Exercise 4

Problem:

Solution:

Step 1: The set $\mathbf{F} - \{0\}$ together with multiplication is a group - all axioms can be trivially verified by virtue that $\mathbf{F}$ is a finite field.

Step 2: The set $a^p$, $\forall p \in \mathbf{Z}$, $a \neq 0 \in \mathbf F$ is the cyclic subgroup generated by $a$. As such, the subgroup has an order $r \mid q-1$. Therefore $a^{q-1} = (a^r)^k = 1^k = 1$. $a^r = 1$ is because it is a cyclic subgroup.

Step 3: For $a \ne 0$, the above show that $a^{q} = a(a^{q-1}) = a$, therefore $a^q - a = 0$. It is obvious that $a^q - a = 0$ for $a = 0$ as well.

Therefore we have just proved that $a^q - a = 0$ for all $a \in \mathbf{F}$.